Inspiring Mathematics: Lessons from the Navajo Nation Math Circles 9781470453879, 9781470454500

Citation preview

Inspiring Mathematics Lessons from the Navajo Nation Math Circles

Mathematical Circles Library

Inspiring Mathematics Lessons from the Navajo Nation Math Circles Dave Auckly, Bob Klein, Amanda Serenevy, and Tatiana Shubin, Editors

Berkeley, California

Providence, Rhode Island

Advisory Board for the MSRI/Mathematical Circles Library Titu Andreescu David Auckly H´el`ene Barcelo Zuming Feng Tony Gardiner Nikolaj N. Konstantinov Andy Liu Alexander Shen

Tatiana Shubin (Chair) Zvezdelina Stankova James Tanton Ravi Vakil Diana White Ivan Yashchenko Paul Zeitz Joshua Zucker

Scientific Editor: David Scott This volume is published with the generous support of the Simons Foundation and Tom Leighton and Bonnie Berger Leighton. 2010 Mathematics Subject Classification. Primary 00-XX, 97-XX.

For additional information and updates on this book, visit www.ams.org/bookpages/mcl-24

Library of Congress Cataloging-in-Publication Data Library of Congress Cataloging-in-Publication Data Names: Auckly, Dave, editor. | Mathematical Sciences Research Institute (Berkeley, Calif.) Title: Inspiring mathematics : lessons from the Navajo Nation math circles / Dave Auckly [and three others]. Other titles: Lessons from the Navajo Nation math circles Description: Providence, Rhode Island : American Mathematical Society, [2019] | Series: MSRI mathematical circles library ; 24 | “MSRI, Mathematical Sciences Research Institute, Berkeley, California.” | Includes bibliographical references. Identifiers: LCCN 2019034509 | ISBN 9781470453879 (paperback) | ISBN 9781470454500 (ebook) Subjects: LCSH: Mathematics–Study and teaching (Elementary)–West (U.S.) | Mathematics– Study and teaching (Middle school)–West (U.S.) | Mathematics–Study and teaching–West (U.S.) — Navajo children–Education. | Navajo youth–Education. | Mathematics–Social aspects– West (U.S.) | Mathematical recreations. | AMS: General. | Mathematics education. Classification: LCC QA13.5.W47 I57 2019 — DDC 372.7–dc23 LC record available at https://lccn.loc.gov/2019034509

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to [email protected]. c 2019 by Mathematical Sciences Research Institute. All rights reserved.  Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines 

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ Visit the MSRI home page at htpp://www.msri.org/ 10 9 8 7 6 5 4 3 2 1

24 23 22 21 20 19

Contents Foreword Henry Fowler

vii

Introduction

xi

The Scripts

1

Grid Power Tatiana Shubin

5

What is the sum?—A 5-card Magic Trick Gabriella Pinter

25

Decanting Problems (A.K.A Euclidean Algorithm) Tatiana Shubin and Elgin Johnston

31

Bean Bag Tossing Amanda Serenevy

47

Toilet Paper Math Phil Yasskin and Dave Auckly

61

Golomb Rulers Elgin Johnston

75

The Cookie Monster Problem Gabriella Pinter

89

Math Blocks Dave Auckly

103

Counting Diagonals Tatiana Shubin

123

Liar’s Bingo Bob Klein

137

Parity and Other Invariants Bob Klein and Tatiana Shubin

149 v

vi

CONTENTS

Pancake Problem Rebecca Bycofski, Bob Klein, and Sierra Knavel

171

The H − L Protein Folding Model Amanda Serenevy

183

SOMA and Friends Dave Auckly and Stan Isaacs

199

How To Fold A Tie Into Sevenths James Tanton

219

Boomerang Fractions Amanda Serenevy

235

From Hats to Codes Joe Buhler and Bob Klein

243

From a Magic Card Trick to Hall’s Theorem M. Kawski and H. A. Kierstead

263

Foreword Dr. Henry Fowler In 1972 I was sent to a Bureau of Indian Education (BIA) boarding school in Kaibito, Arizona. In order to receive a formal American education, I had to leave my homestead in Tuba City, Arizona, at the age of four years old. This was the only option for my family because we lived in a rural area where our life was centered around farming and livestock. My formal education at the BIA school was far different than the informal Navajo education I had received at home. The education I experienced at the BIA school was foreign to me and far removed from the familiarity and practicality of my education at home. The BIA school provided basic academic classes in grammar, arithmetic, penmanship, and spelling. In addition, I attended a supplemental class that included correct table manners. My formal education journey consisted of many challenges that included learning the English language. In grade school I was labeled as a special education student, and I was taught to leave my tradition and culture behind in order to acquire the formal American education. The concept of being bi-cultural was not an option. Today, I am using my experiences to create change in the education system, particularly in mathematics. I believe the Navajo people can acquire an excellent education that embraces the Navajo culture. I am dedicated to improving the living condition for my Navajo people and have made it my mission through the quality of re-envisioned math education for native learners. My intent is to inspire young minds through the powerful lens of mathematics. Several years ago I was introduced to Math Circles by my sister Tatiana Shubin, a mathematician from San Jose State University. Tatiana and I bonded immediately. Her perspective was similar to mine, which was to promote math literacy as a means of empowering youth. I could not say “no” to Tatiana’s offer to help enlighten my Navajo people by shining a new light using mathematics. I care about my people — each day is a huge struggle for them. They live a difficult life surrounded by domestic abuse, alcoholism, drug abuse, unemployment, and poverty. To confront this social ill, we can use mathematics and strengthen our math education for grades

vii

viii

FOREWORD

K–12. Math Circles has provided me the necessary resources to re-energize the spirit of mathematics in the Navajo land. Math Circles’ focus on the art of teaching math from an open-ended perspective closely relates to my Navajo cultural teaching. Before, I was challenged to study and investigate curriculum as it related to my native population and to create new approaches that could make a difference in the lives of young Navajo people. Even though I have long been motivated to teach math, throughout the years of my teaching career, I began to have mixed feelings about teaching math. My enthusiasm about teaching math had begun to lessen. Each year in my math classes, I observed students who were quiet and unmotivated to learn mathematics. My teaching was unattractive to them and they found my questions meaningless. My daily challenge was to teach math to students who lacked knowledge of basic math facts, were unmotivated, had high absenteeism and tardiness, were unprepared for class, lacked parental support, lacked current math books, had no access to technology, were in large classes, and were disruptive. The sum of these challenges weighed heavily on me, and my passion for teaching began to stall. Fortunately, Math Circles afforded me a new platform for thinking critically about my teaching experience. The application of Math Circle math lessons and other inquiry methods has lead me to discover there are alternative methods available to improve my situation. As a direct result with Math Circles and my interaction with mathematicians, I have made inquiries and gained clear insights about teaching math to Navajo students. This has set the stage for invigorated research about and development of new instructional strategies to energize both my students to learn math and me to teach it. Math Circles helped open the opportunity for me to address the dismal outlook of the Navajo high school performance in mathematics. As a direct result of the Math Circles, I am more aware of my surroundings and how they impact teaching delivery and reception. I bring an enlivened critical thinking mindset to my intellectual endeavors, and I feel empowered as an educator to lead efforts to change the math education on the Navajo Nation. I am encouraged to broaden the perspective of my immediate horizon and I am challenged to actively pursue my interest in improving math education for Navajo students. Spirit of Math My mind is embedded with the spirit of math My mind is filled with numerations I use parts of my body to measure and numerate I weave using the sacred spirit of math I use math in my weaving Walk in harmony with the spirit of math

FOREWORD

ix

Walk in beauty with the spirit of math Walk in all directions with the spirit of math My children...may the spirit of math glow in your Life. –Dr. Henry H Fowler and Grandma Sally Fowler

Introduction I grew up knowing it’s wrong to have more than you need. It means you’re not taking care of your people. Navajo Proverb. The Navajo Nation, or more appropriately, the Din´e Bikeyah (“Din´e” means “people” and “Bikeyah” means “land of”), is home to more than 200,000 Din´e. Bounded by four sacred mountains in Arizona, Utah, Colorado, and New Mexico, the Din´e Bikeyah is a land of rich mesas, plateaus, wildlife, and arroyos, all under an immense, deep blue sky. As defined by the U.S. Federal government, the “Navajo Reservation” consists of land whose boundaries are in Arizona, Utah, and New Mexico, and encompass a land area larger than 10 U.S. States, including West Virginia. The Din´e have an elected President, Vice President, Council of Elders (legislature), and Supreme Court, as well as regional Chapters with voting blocks of “chapter house members.” Those who live in Din´e Bikeyah live in a dry climate, yet they know rain (including a monsoon season), snow, heat and cold. They know the vastness of desert prairies stretching to the horizon, but also the pine-forested highlands around the Chuska Mountains. They attend rodeos and thrill at the roping and riding skills of Din´e rodeo men and women, and they enjoy camping out by Wheatfields Lake, enjoying a good picnic following a refreshing summertime swim. The kids carry smartphones and listen to Pandora while Snapchatting or Facebooking, knowing which of the few cellular service providers have cell towers in their region. Basketball and running are popular sports to watch and participate in, and until recently, the Navajo Nation had very active chess clubs in the Tuba City and Kayenta regions. Navajo families live and work all over the United States, as nurses, engineers, ranchers, entertainers, poets, medicine men, counselors, and professors. Din´e might attend a Christian church service in the morning and a Din´e dance/sing in the afternoon of the same day. In brief, the Din´e live with one foot securely planted in modernity — defying attempts to “romanticize the native experience” as that of “simple lives” lived “in harmony with nature.” The other foot, for many Din´e, is rooted in the extensive culture, language, and tradition handed down from ancestors through successive generations. Those traditions are also connected to a long history of struggle against colonial intrusion and genocide, and against cultural and economic isolation.

xi

xii

INTRODUCTION

The challenges for the Din´e are real and include cultivating an economic base, providing social services including health and education, managing the scars of extractive industries (and the U.S. Government) whose hunger for coal and uranium was fed at the expense of safety standards and adherence to regulations that might have left the land and water safe and intact. Knowledge of the richly descriptive language, “Din´e,” has skipped Generation X, though it is making a return in the Millennials and the Gen Zs thanks to the advocacy of elders and the strength of teachers who find ways to engage students in “Din´e Studies.” The opportunities for the Din´e are equally real. During World War II, a group of Navajo men created a battlefield code using the Din´e language and then trained other Din´e men to use it in the field. It was the only code never broken by the Japanese or Germans (nor later by the Koreans or Vietnamese – the code was used through the early part of the Vietnam war). Today, the Code Talkers are revered for their bravery as U.S. Patriots and Din´e Warriors. Moreover, they are revered for their ingenuity as creators, educators, and cryptographers. Navajo science, expressed in distinctly Navajo cultural and linguistic terms, has led to agricultural advances, engineering feats (including large-scale construction projects), and astronomical techniques (to name but a few areas, and with distinctly Western terminology). Contributions to the arts, and especially weaving (rugs, blankets, shawls, clothing) express tradition, spirituality, and style of the Din´e broadly, and the artists individually. A blanket might combine the pragmatism of being woven in a double-layer to insulate against a cold, hard ground, with the idealism of the Grey Hills pattern. Indeed, weavers learned the art (and science) of weaving from one of the original Din´e deities, Spider Woman. Knowledge of the arts, the sciences, lifeways, and spirituality, is profound among the Din´e and Western eyes gloss over its Din´e expression at their own peril. Many of the national challenges of the U.S. are shared by the Din´e and they have a distinct advantage in contributing to the resolution of these challenges if both their traditional and Modern/Western knowledge are developed and valued. This book is a glimpse into one effort to share the love and value of mathematics and the beauty and power of Din´e culture, tradition and knowledge.

Baa H´ ozh´ o For the Din´e, the term “Baa H´ ozh´o” speaks to “balance and harmony” as a way of living, and this phrase is seen everywhere in Din´e Bikeyah. For the mathematician too, balance and harmony are the most fundamental and beautiful elements of good mathematics. An equation’s most fundamental task is to communicate that two expressions (right and left) are in balance; if you add or subtract something, you have to do it to both sides. The importance of symmetry (“sameness”) pervades geometry, algebra, and analysis

INTRODUCTION

xiii

not only because of its utility, but for its beauty and communication of harmony. Even today, the word “Algebraist” may be found on signs over rural storefronts in Arabic-speaking countries, where it refers to a bone-setter, one able to “restore the balance” of a broken bone back into harmony. Yet there is in the United States right now a deeply asymmetric pattern to be found in the resources and educational outcomes in mathematics and science between rural and non-rural students, and especially in Indigenous and non-Indigenous students. 1 For all of the talk of “leaving no children behind,” U.S. Federal and State policies, and especially unfair funding formulas, have left entire peoples behind. In 2015, Grade 8 Mathematics scores (as measured on the National Assessment of Educational Progress (NAEP)) showed “American Indian/Alaskan Native” having only 20% of students score “at or above proficient,” compared to 43% of “White” students at or above proficient. This is comparable to “Hispanic” students (19%) and higher than “Black” students (13%), but still significantly below “Asian/Pacific Islander” (59%), “Asian” (61%), and “Two or More Races” (36%).[1] Further evidence of this gap can be found in access to and participation in Advanced Placement courses, one indicator of a student’s likelihood of pursuing a college education. The 10th Annual AP Report to the Nation (College Board, 2014) reports that though “American Indian/Alaskan Native” students represented 1% of the graduating class of 2013 in the United States, yet constituted only 0.6% of AP test takers and only 0.5% of successful AP test takers. This indicates a significant gap in performance, and mirrors an equally distressing gap in student access to AP courses in their High Schools to begin with. Unlike the NAEP results, “American Indian/Alaskan Native” students have no “near neighbors,” demographic categories showing anywhere near the lack of access to AP courses experienced by Indigenous students. This persistent asymmetry (or gap) worsens generation after generation as it entrenches a stereotype of Indigenous people “not being good at math” or “not being college material.” Students failing to see any evidence of Indigenous American mathematicians, engineers, scientists, college professionals, etc., has the effect of reifying the perception that “Navajos can’t be mathematicians” (or similar). It represents a growing “poverty of perceptions” that is fed by stereotypes even as it is starved by a failure to provide resources needed to prepare the next generation of Indigenous role models 1

This section uses the term “Indigenous” rather than “Native American” for many reasons, including that the word “American” applies to two continents worth of peoples despite being claimed by citizens of the United States, and the word “Native” is highly contested. More to the point, we use “Indigenous” because we were told by a group of close collaborators from the Din´e and Hopi people, that this was the word that they preferred. No label is adequate, and labels are inherently exclusionary. We recognize the inescapability of this and recognize that some readers will prefer other terms to describe the inhabitants of a place prior to the arrival of Western colonial and imperial people (equally contested terms).

xiv

INTRODUCTION

and heroes. Parents and community members are deeply committed to their children and their education, but often must prioritize finding food for the table and overcoming a median family income of $27,389, [2] or about half of the median household income of Arizona or the U.S. generally. Despite these figures, these measures of poverty, the promise of a generation of Din´e scholars and artists persists. One of the editors (Klein) has grown to understand during the time of his participation in this project, that the relationships between economic poverty and happiness, between opportunity and persistence, are far more complicated than his experience would have led him to believe. Indeed, this perception of cultural capital in the consumerist U.S. may be one of the most overt narratives serving to marginalize Indigenous students as it instills the belief that the key to participation in “modern life” requires a smart phone and Amazon Prime. This trend is not new, and the contrast between rampant and inexhaustible consumerism and peaceful balance can be seen in Godfrey Reggio’s cathartic 1982 film Koyaanisqatsi (from the Hopi word meaning “life out of balance”). Mathematics shares in this tension in as much as its power makes possible the science, engineering, and technology that shape the human-built world even as mathematics celebrates beauty, harmony, and balance. The recent book (and film) Hidden Figures is a good example of how marginalization and exclusion presents a significant challenge in the sciences, including mathematics. Moreover, it demonstrates the extent to which we must actively work toward inclusion, both to acknowledge the underlying morality of inclusion, but also the valuable contributions of voice and talent that are lost when not all people have equal access to mathematics. Mathematics itself contains all of the “seeds of inclusion” in as much as it is built on an ethos of questioning assumptions, embracing logical reasoning, and eschewing authority in favor of peer review — in short, it embodies a democratic disposition that, when fully realized, points toward inclusion. Emerging stories (such as Hidden Figures) that challenge the white male stereotypes of Western mathematicians suggest that it is far past time to question the cultural axiomatics of mathematics as we redefine participation and access. This book arises from an effort by mathematicians to use “Math Circles” as a way to share beautiful and joyful mathematics, and to therefore rewrite the cultural axioms of participation in mathematics, to challenge the notion that someone’s personal statistics should not define their mathematical opportunities.

MATH CIRCLES

xv

Math Circles Math Circles are almost 100 years old, having started in Bulgaria and finding great visibility in the former Soviet Union. They grew out of concern by mathematicians that “school mathematics,” being shaped by curricular state mandates, lacked necessary elements to prepare and engage the next generation of mathematicians and scientists. Weekly meetings of mathematicians with students were (usually) extracurricular interventions wherein mathematicians would facilitate investigation of problems that intrigued and delighted them. Over the course of a century, a robust set of problems and sessions (collections of thematically linked problems that might span 1–3 hours) has developed, fostering more and more Math Circles. Often but not exclusively, Math Circles helped to prepare mathematics competitors, and some Circles continue to serve this function. It was only in the 1990s that Math Circles for students came to the United States, with groups like Boston’s “The Math Circle,” and two circles in the San Francisco Bay area in California serving as pioneers in shaping a U.S. math circle “movement” leading to more than 200 student math circles and an umbrella organization, the National Association of Math Circles https://mathcircles.org/). One coeditor of this volume, Tatiana Shubin, was among this group of pioneers. In all such Math Circles, a group of students join with mathematicians and other mathematics professionals on some regular basis to explore joyful, beautiful mathematics. Math circles celebrate the problem over the exercise, the difference being that an exercise is something you know how to complete when you start whereas a problem requires the solver to try many paths or strategies without knowing ahead of time which might be fruitful. Math Circle leaders recognize that every discipline of study, and every endeavor in life is marked by problems (some good, some bad), making Math Circles an ideal training ground for better living. Students (usually aged 11–18, though some groups include students as young as 3 years old) are the primary focus of Math Circles and adults are sometimes allowed to attend as quiet observers. In 2006, teachers of elementary and secondary mathematics started collaborating in Math Teachers’ Circles (hereafter MTCs) as a result of the efforts of Mary Fay-Zenk who had been attending a Math Circle as an observer and declared that “teachers are people too!” There are now more than 100 MTCs in the United States, with the idea spreading to other countries. Like student Math Circles (hereafter MCs), MTCs meet regularly and generally are facilitated by mathematics professionals who share interesting investigations. MTCs now have an umbrella organization, the Math Teachers’ Circle Network (https://mathteacherscircle.org/ ). This book collects a set of “scripts” from some of the most experienced and engaging math circle leaders from around the United States into one collection with each script written with a level of detail and support to allow teachers in K-12 schools to use and adapt the sessions to use for after-school math clubs or Math Circles.

xvi

INTRODUCTION

I have been to the end of the earth, I have been to the end of the waters, I have been to the end of the sky, I have been to the end of the mountains, I have found none that are not my friends. Navajo Proverb.

The Navajo Nation Math Circle Project Din´e Bikeyah’s Math Circles and Math Teachers’ Circles continue to evolve, but they all involve an awareness that you have to respect and value someone’s total identity if you are to change his or her mathematical identity. The novel idea behind the Navajo nation Math Circle Project (NNMCP) is to engage with Din´e Hatalii (Medicine Men/Women), elders, and other Din´e educators, to integrate Din´e culture sessions in parallel with Math Circle sessions. This acknowledges the co-development of identity as a key to overcoming stereotypes that might otherwise block participation in mathematics or other STEM subjects. It recognizes that mathematics is but another side rail in Manuelito’s Ladder.2 The history of the NNMCP describes one such hand steadying that ladder. Tatiana Shubin lived in Kazakhstan before coming to the US. The network of Math Circles in Russia and Kazakhstan spurred her interest in mathematics. When she moved to the U.S., she started building Math Circles. Her San Jose Math Circle was formed as part of the same initiative that formed the Berkeley Math Circle and the Bay Area Math Olympiad in 1998. While traveling the Southwest U.S., Shubin discovered and fell in love with the Din´e and the Din´e Bikeyah — the culture, the environment, and the people reminded her of Kazakhstan. In January of 2011, Shubin approached Dave Auckly with an idea for a sabbatical to be spent engaging Navajo students in great mathematics. Auckly suggested that if she was going to spend half a year in the area, they ought to use that time to create a lasting program for teachers and students in the Navajo Nation. The program would include mathematical visitors who would lead teacher workshops, and special Math Circles at local schools. To make this work, it would take contacts in the community and money. At the time, Auckly was the Associate Director of the Mathematical Sciences Research Institute. One of the first outreach activities he organized was the initial Circle on the Road workshop in Tempe, AZ (2010). One of the explicit goals of this Circle on the Road program was to create mathematics outreach programs for under-served populations. Matthias Kawski was one of the co-organizers of this workshop, and he made a serious effort to advertise the associated Julia Robinson Math Festival across the Southwest. Kawski has continued to run mathematics festivals in Tempe, AZ every year since, and also started a successful Math Circle program there. The White 2

Hastinn Ch’il Haajiin, aka Chief Manuelito, one of the leaders of the Din´e through the Long Walk, said in 1893, “My grandchild, education is the ladder. Tell our people to take it.”

THE NAVAJO NATION MATH CIRCLE PROJECT

xvii

family in Ganado, AZ heard about the math festival and reached out to Kawski about attending. Jennifer White was exactly the connection that Shubin and Auckly sought, and she was generous in making introductions in the Din´e Community. In particular, she introduced Shubin to Henry Fowler, a professor in the mathematics department at Din´e College who became one of the founding directors of the NNMCP. With community contacts and funds from groups such as the National Science Foundation, the National Security Agency, and others, the NNMCP was born in the Fall of 2012. During the Fall of 2012 Shubin led school-based Math Circles at five schools. She met with Fowler’s pre-service mathematics teachers class each week. Other stars aligned to help the continuation of the program. In 2009 the University of Utah received an NSF Noyce Fellowship program grant to develop a section of the Math for America teacher preparation/enhancement program to be directed by Hugo Rossi. From the onset, Rossi wished to include Navajo Teachers in the program, and had begun to establish contacts. Because of this collaboration with Rossi and his colleagues, the emerging Navajo Nation Math Circle Project was able to offer professional development workshops for Navajo teachers, in addition to offering a summer camp for students. One of the first mathematician visitors to the program was Amanda Serenevy. Serenevy formed the Riverbend Community Math Center in South Bend, IN in the Fall of 2006, and her duties there include running Math Circles. Serenevy first came to the Navajo Nation to do work on the project in the Spring of 2013, leading several sessions on Mathematical Origami and Rational Tangles at Ganado, St. Michaels, and Many Farms High Schools. She also led professional development sessions for teachers at Din´e College. Since that time, she has continued to visit regularly. Shubin met Bob Klein, a professor at Ohio University, at an American Institute of Mathematics (AIM) How to Run a Math Teachers’ Circle workshop in Washington, D.C., in Summer 2012. Then, at a Circle on the Road in Puerto Rico in 2013, they met again, and this time with Fowler. Upon hearing about the NNMCP program, Klein, a native of Albuquerque instantly felt a desire to contribute to the project. In 2014, while at the Joint Math Meetings, Shubin asked Klein to join the NNMCP as a co-director and to contribute to the summer session beginning in July 2014. Since that time, Klein has returned many times. All of the present and past people involved with the Navajo Nation Math Circles are working to sustain and grow math circle programs in Indigenous Communities. Beginning in 2017, Klein, Shubin, Serenevy, and Fowler began a new effort to expand this work to other Indigenous Nations, called the Alliance of Indigenous Math Circles (https://aimathcircles.org). At the same time, Auckly, Fowler and Jayadev Athreya launched another project with a similar but slightly different name—Indigenous Math Circles—to sustain the original NNMC, and create a mirror program in Washington State. Information about this program can be found at https://navajomathcircles.org/.

xviii

INTRODUCTION

Since its founding, the NNMCP has had the active participation of more than 50 mathematicians from around the United States and internationally, who have been inspired by the Din´e and Din´e Bikeyah to share in joyful, Baa H´ozh´o mathematics. This book is a collection of sessions celebrating those who were inspired by the experience as well as those who continue to inspire students to see the beauty of great mathematics.

References [1] Nation’s Report Card. https://www.nationsreportcard.gov/reading_math_2015/# mathematics/acl?grade=8 [2] Navajo Nation community Profile: 2016. http://nptao.arizona.edu/sites/nptao/ files/navajo_nation_2016_community_profile.pdf

The Scripts

Who is this book for? Briefly, this book is for the kid within all of us who can’t help but be fascinated by patterns. It’s for everyone who found themselves bored for a moment only to find joy and wonderment in the floor tiles under their feet, wondering why they are usually square and if they could be any kind of shape. For in the pages that follow, the authors share that playful fascination via “simple questions” that often lead to deep insight into the “why” and “how” behind beautiful patterns. Every script that follows explores deep mathematical themes and connects mathematical ideas, many of which don’t seem connected initially. Each script celebrates the “problem” over the “exercise”, and student curiosity and effort over teacher didactics and lecture (indeed, the best-led scripts are ones in which the teacher celebrates their role as a student to the problem!). The scripts that follow don’t fit neatly into the siloed school curriculum (“geometry” over here, “algebra” over there, “counting” back there. . .) but instead adhere to a model of first posing a question and following that question into requisite “math content” as opposed to starting with a “topic” and looking for a good question. All of the sessions in this volume have been used with students and teachers of the Navajo nation. As a reader, you are now part of an expanding circle of sharing that began with a mathematician sharing a given problem or session with Din´e students (or teachers), and often thereafter refined and shared again. Now, it is being shared with you, as a gift from the mathematicians who initiated the session and the Din´e students and teachers who subsequently helped to refine and reshape the problems and sessions. This book is for teachers, home-schooling parents, and others who want some support for facilitating good Math Circle sessions. While doing sessions in Math Teachers’ Circles or other teachers workshops, the editors often hear one of two comments from teachers: “My students could never do this math” and “This is great, but I could never facilitate this math.” This volume is written to address both of these questions in several ways. First, the scripts vary greatly in terms of requisite mathematics, topics, modalities, and age ranges—there should be something in here for everyone. Second, detailed “teachers’ notes” help to guide the facilitator toward a great session and around pedagogical potholes. That said, each editor has led sessions that didn’t go as planned. The key to a great Math Circle is treating the Circle as a place for everyone to engage, fail, succeed, notice, wonder, and question — challenge only becomes frustration when the challenge is no longer fun. Three of the editors have watched the fourth smiling while looking at the participants, only to have the smile turn to a sudden realization that she had just made an error. Her reaction to that error, though, is always one of a small laugh followed by the necessary correction or discussion about the error itself as a means of further inquiry. Klein has led the Liar’s Bingo script more times than he can recall and every time is different. One of the amazing

THE SCRIPTS

3

things about the scripts that follow is that they unfold in different ways every time. The editors encourage you to try these scripts multiple times and observe how different participants, ages, etc., change your experience of facilitating them. Finally, this book provides a means for facilitators to see the genius in their kids and to cause serious reconsideration of “my kid can’t.” How should this book be used? This book is not like a novel. Nor is it like a textbook. The novel generally is read linearly, from start to finish, and the textbook is an attempt to codify knowledge by organizing it in a specific order to structure learning along some kind of progression. Were it not for the limitations of binding pages in order, we might publish each script as a bright, tempting piece of candy, all thrown into the jar that we think of as a book. The reader is therefore highly encouraged to give in to temptation and reach into the jar. Whether you pick a script at random or just browse the scripts for some word or idea that tickles your mathematical fancy, you are bound to find something good. Once you have settled on a script, read with a pencil and some paper. Shubin, in particular, recommends grid paper (page 5). Get lost in the problems as they arise and avoid the temptation to seek out someone else’s solutions at first (though hints, encouragements, and solutions are included in the teachers’ notes). One of the editors (Klein) frequently spends days pondering this or that question arising in a script, highlighting one important caveat: scripts are not intended to be timed to sessions of a specific length. Good problems are impossible to time because it is impossible to specify a priori how long, say, Bob will take to find insight, relative to Amanda for a given problem. So give yourself lots of preparation time so that you can enjoy wondering. But do read through the teachers’ notes. They represent each author’s best advice for what running the session might look like and for how to make that most meaningful. Scripts should feel more like one of those old “Choose Your Own Adventure” (“if you decide to slay the dragon, turn to page 8; otherwise, turn to page 41”) books than a prescription of activity. The Teacher’s Notes will provide enough background to support the facilitator in the many midstream changes of direction that can occur in a Math Circle session. So don’t be afraid to use just part of a script or to change some aspects of it to suit your students. Ask for help! Don’t be afraid to collaborate with other teachers, the authors or editors of this volume, or students, as you prepare. You will find greater fun by doing so, and you will be sharing the joy of great problem solving. Then, once you facilitate the session(s), share again. Post to a blog about your experience, write in a journal, or write to one of the editors with

4

THE SCRIPTS

your experience. The editors would love to hear about your experiences with the sessions. Don’t expect this to be school curriculum! The greatest source of disappointment is poorly-tuned expectations. Try to step out of the role of teacher and feel what it feels like to facilitate. Encourage your students to step out of the role of students to become mathematicians. The impacts of this are long lasting. A colleague of ours, Duane Bollenbacher, once remarked that “classrooms shouldn’t be places where young people come to watch old people work.” Math Circles should be places where old people come to be young again and to rediscover the joy of doing and sharing great mathematics. In this way, you know that you will be sharing in Baa H´ozh´o. Ah´ehee’ (Thanks!) The Editors.

Grid Power Contributed by Tatiana Shubin Short Description: In this session, a simple sheet of grid paper is fertile ground for generating questions that lead to combinatorial reasoning, finite arithmetic series, algebraic identities, and the Pythagorean Theorem. Materials: Participants will need lots of grid paper and pencils. You may choose to use the included handout or you may simply write the questions where participants can see them. Mathematics Beneath & Beyond: This material is a treasure trove of mathematical ideas. Among them: We see combinatorics, or at least combinatorial reasoning (e.g., Solution 2 of Problem 1). Along the way we find sums of a finite arithmetic series, and of the squares and cubes of integers. We discover some useful tools such as proving algebraic identities by means of representing quantities involved as geometric objects. In passing we find out a fact that the slopes of perpendicular lines are negative reciprocals of one another. And there is this astonishing but undeniable phenomenon that an inquiry which started by simple counting led us to geometry — we’ve proven the Pythagorean Theorem, felt the importance and meaning of area as a measure (solving Problem 1), and had a chance to use properties of parallel lines and similar triangles (using our approach to Problem 3). Finally there is number theory — while solving Problem 1, we run across two peculiar sets of integers — one consists of integers which can be represented as sums of two perfect squares, and another contains only integers lacking such representation. Studying these sets can become a worthwhile endeavor which would bring students quite deep into the fascinating field of elementary number theory.

5

6

TATIANA SHUBIN

Grid Power Student Handout Warm-up Problems: (A) Is it possible to place dots into cells of an 8 × 8 square (no more than one dot per cell) so that the number of dots in every column is the same, while no two rows have the same number of dots? (B) Is it possible to cover all squares of a chessboard with 32 dominoes (each domino covering exactly two squares) in such a way that no two dominoes cover a 2 × 2 square? (See Figure 1.)

Dominoes

Chessboard

Figure 1. Chessboard and Dominoes

GRID POWER

7

Problems: (1) (a) How many squares are there in a 7 × 7 square? (b) How many squares are there in an n × n square? (2) (a) How many rectangles whose sides lie on grid lines are there in a 7 × 7 square? (b) How many rectangles whose sides lie on grid lines are there in an n × n square? (3) Connect the vertices of a square in a cyclic way to the midpoints of the opposite sides. (See Figure 2.) These four lines form an innermost quadrilateral inside the original square. What type of quadrilateral is it? What part of the area of the original square does it have? What happens if you replace the midpoint with a point that is a fraction r along each side?

e

Figure 2. What is the innermost quadrilateral?

8

TATIANA SHUBIN

Grid Power Teacher Guide Warm-up Problem Solutions: (A) Since there are 8 rows and each row can contain no less than 0 and no more than 8 dots, the number of dots in each of 8 rows must be chosen from among the following numbers: 0, 1, 2, 3, 4, 5, 6, 7, and 8. So we can omit exactly one of them. Which one? The sum of all of them is 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 Since each column is to have the same number of dots, the sum of our chosen eight numbers (i.e., 36 minus the omitted number) must be divisible by 8, and only 4 does the job: 36 − 4 = 32 = 8 · 4 Figure 3 shows a possible way of placing 0, 1, 2, 3, 5, 6, 7, and 8 dots so that each column has exactly 4 dots:

Figure 3. Placing the dots so that each column has 4 dots.

GRID POWER

9

(B) We will try to avoid placing dominoes in a position covering a 2 × 2 square. First, let us label all squares of the chessboard with numbers from 1 to 64 as shown in Figure 4.

Figure 4. Labeled Chessboard Since all squares of the chessboard must be covered, a domino has to cover the bottom left corner, square 57. This can be done placing a domino vertically so that it covers squares 49 and 57, or horizontally, covering squares 57 and 58. Both situations can be analyzed in a similar manner; we will look at the former case, as shown in Figure 5.

Figure 5. The first domino.

10

TATIANA SHUBIN

We must also cover square 58, and the only way to do it and avoid covering a 2 × 2 square would be by placing a domino on squares 58 and 59. Now we need to cover square 50 avoiding covering a 2 × 2 square, which can only be done by placing a domino on squares 50 and 42. (See Figure 6.)

Figure 6. The first three dominoes.

Continuing this process, we have no choice but to place the dominoes as shown in Figure 7.

Figure 7. The forced pattern of dominoes.

GRID POWER

11

Now we have to put a domino on squares 8 and 16, thus creating a 2 × 2 square. Therefore, it is impossible to cover the chessboard and avoid covering a 2 × 2 square with two dominoes. Problem Solutions: (1) (a) There are 140 regular squares and 196 tilted squares for a total of 336 grid squares. See the Presentation Suggestions section below for a detailed solution. (n − 1)n2 (n + 1) n(n + 1)(2n + 1) regular squares and (b) There are 6 12 n(n + 1)2 (n + 2) grid squares. Again, tilted squares for a total of 12 see Presentation Suggestions for a detailed solution. (2) (a) Solution 1: Let’s label cells in a 7×7 square as shown in Figure 8.

Figure 8. Labeled 7 × 7 grid.

Now count the number of rectangles that have cell 1 in the bottom right corner; add the number of rectangles that have cell 2 in the bottom right corner; add the number of rectangles that have cell 3 in the bottom right corner; continue this process until we add the last number — that of the rectangles that have cell 49 in the bottom right corner. We get the following sum:

12

TATIANA SHUBIN

+ + + = + + + = =

(1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 4 + 6 + 8 + 10 + 12 + 14) (3 + 6 + 9 + 12 + 15 + 18 + 21) + (4 + 8 + 12 + 16 + 20 + 24 + 28) (5 + 10 + 15 + 20 + 25 + 30 + 35) + (6 + 12 + 18 + 24 + 30 + 36 + 42) (7 + 14 + 21 + 28 + 35 + 42 + 49) (1 + 2 + 3 + 4 + 5 + 6 + 7) + 2(1 + 2 + 3 + 4 + 5 + 6 + 7) 3(1 + 2 + 3 + 4 + 5 + 6 + 7) + 4(1 + 2 + 3 + 4 + 5 + 6 + 7) 5(1 + 2 + 3 + 4 + 5 + 6 + 7) + 6(1 + 2 + 3 + 4 + 5 + 6 + 7) 7(1 + 2 + 3 + 4 + 5 + 6 + 7) (1 + 2 + 3 + 4 + 5 + 6 + 7)(1 + 2 + 3 + 4 + 5 + 6 + 7)   7·8 2 2 = 784 (1 + 2 + 3 + 4 + 5 + 6 + 7) = 2 Note: It might be helpful to start by looking at smaller cases — a 2 × 2 square, then a 3 × 3 square. In these two cases counting rectangles using the described method should present very little difficulty and will make the above computation clear. Let’s look at the 2 × 2 case as shown in Figure 9.

1 2 3 4 Cell # X 1 2

All rectangles with cell # X in the # of these bottom right corner rectangles 1 2

3

2

4

4 Figure 9. Counting rectangles.

Using this table we obtain the total number of all rectangles in a 2 × 2 square as follows: (1 + 2) + (2 + 4) = (1 + 2) + 2(1 + 2) = (1 + 2)(1 + 2) = (1 + 2)2 Solution 2: Every rectangle is defined by two vertical and two horizontal sides. Thus, to specify a rectangle, we need to choose 2 vertical lines out of 8 vertical lines in the 7 × 7 square — this can be done ( 82 ) ways — then we need to choose 2 horizontal lines,

GRID POWER

13

which again can be done ( 82 ) ways. Hence the total number of  2 2 = 784 rectangles is ( 82 ) · ( 82 ) = ( 82 ) = 8·7 2 (b) As in part (a), to specify a rectangle we need to choose 2 vertical sides out of (n + 1) vertical grid lines in an n × n square, then choose 2 horizontal sides out of (n + 1) horizontal grid lines. This can be done 

   2   (n + 1)n 2 (n + 1)2 n2 n+1 n+1 n+1 = = = 2 2 2 2 4 ways. Therefore, the number of such rectangles in an n × n square is (n + 1)2 n2 . 4 Note: Counting these rectangles by a different method we can express their number as 13 + 23 + 33 + . . . + n3

(See [1]). Thus we have proven that   (n + 1)n 2 3 3 3 3 = (1 + 2 + 3 + . . . + n)2 . 1 + 2 + 3 + ... + n = 2 This last identity can be also illustrated as shown in figure 10.

14

TATIANA SHUBIN

Figure 10. Rearranging the sum of cubes.

Figure 11 illustrates the case when n = 5.

Figure 11. The case when n = 5.

GRID POWER

15

These illustrations would also be helpful if you want to prove the identity rigorously and use induction on n, and notice that according to the illustration it might be a good idea to consider two cases, when n is odd, and when n is even. (3) Let us draw a tilted square as shown below. A grid line through a vertex crosses the opposite side at its midpoint. Thus Figure 12 represents the construction required in the problem.

Figure 12. A tilted square with area 5.

We can see that the quadrilateral inside is a square of area 1 while the tilted square has the area of 22 + 12 = 5. Hence its area is 1/5 of the area of the original figure. Now consider a few other tilted squares as shown in Figure 13.

16

TATIANA SHUBIN

Figure 13. Some other tilted squares.

In the tilted square with a = 3 and b = 1, we have: r = 13 ; the area of the inner square is (3 − 1)2 = 4, while the area of the tilted square is 32 + 12 = 10. Thus the ratio of the area is 4/10. In the tilted square with a = 5 and b = 2, we have: r = 25 ; the area of the inner square is (5 − 2)2 = 9, and the area of the tilted square is 52 + 22 = 29. Hence the ratio of the areas is 9/29. In general, using the same idea of representing the given square by a tilted one we get the following result: if r = m n , where m < n and m and n are positive integers, the area of the inner square is (n − m)2 , the area of the outer square is m2 + n2 , and so the ratio of the areas is (n − m)2 /n2 (1 − r)2 (n − m)2 = = . n2 + m2 (n2 + m2 )/n2 1 + r2 The formula remains true even when r is irrational. Presentation Suggestions The main goal of the warm-up problems is to familiarize students with the grid and start “respecting the grid” so that they are truly mindful of the grid lines, grid points and cells. You can easily skip these problems if you don’t have sufficient time. It would normally take a 75–90 minute session just to do Problem 1. This problem is rich with ideas and can lead to deep and intriguing investigations in a variety of subjects such as combinatorics and number theory. Don’t rush — the time spent on truly understanding this problem is time well spent. Start by posing Problem 1. Let students work on it for 5–10 minutes; make sure that they draw a 7 × 7 square and start counting. After some students get an answer, have them come to the board and write their answers for all to see. There might be many different answers, such as 49, or 50, or even “infinite.” Hopefully, some students will have found the number to be

GRID POWER

17

140. Discuss with them why some (or even all) of these answers are ‘correct’ — for example, if we only count the grid cells (i.e., 1 × 1 squares) then we clearly have exactly 49 squares; if we totally disregard grid lines then there are, indeed, infinitely many squares which can be drawn in a 7 × 7 square. Then have a student who found 140 squares explain what he or she did to find that number. Prompt them to list the number of squares according to their sizes, thus creating the following table on the board: Square size # of these squares 1×1 49 2×2 36 3×3 25 4×4 16 5×5 9 6×6 4 7×7 1 Total 140 Students may notice something interesting: the number of squares of each size is a square number! But why? Let students think about it for a few minutes or until someone explains the phenomenon. If nobody comes up with a good explanation, show them the reason using a particular size — say, 3 × 3 squares. Suppose that you have a model of a 3 × 3 square with a colored dot at some place, for example, at the center of the upper left cell. Place this square at the upper left corner of the 7 × 7 grid. What are all other possible positions for a 3 × 3 square in the grid? To get all these positions systematically, start moving your square horizontally as far to the right as possible; make a mental note of the position of the colored dot after each move. When it is no longer possible to continue this process of sliding the square horizontally, return it to its leftmost position. Now slide it just one row down, and start moving it again to the right. Repeat these steps as long as possible, keeping in mind where the colored dot resides for each position of the 3 × 3 square. We get the array of dots shown in Figure 14, and so there are twenty-five 3 × 3 squares in the grid.

18

TATIANA SHUBIN

Figure 14. Marking locations of the upper left cell of every 3 × 3 square. Now we have a good explanation for the total number of 140 squares. But have we actually counted all squares? Are there any squares defined, in some way, by the grid that we have not accounted for as yet? Ask students in what way did the grid ’define’ all the squares we have counted so far. The answer is that their edges were on grid lines. For convenience, we will call these “regular squares.” But the grid has something else besides lines, i.e. grid points (the points of intersection of grid lines). Of course, the vertices of every regular square are grid points, but there are squares whose vertices are grid points but whose edges do not lie on grid lines. We will call these “tilted squares.” Challenge the students to draw a number of different tilted squares; give them time to discover a fair amount of different orientations and sizes. Figure 15 shows just three possible examples.

Figure 15. Three tilted squares. So now we will try to find the number of all tilted squares in a 7 × 7 grid. Is there any good systematic way to count them?

GRID POWER

19

Students might come up with several suggestions, such as (a) the side length; (b) the slope of one side; (c) the size. At this point, it will be a very natural place to discuss what is it that we call the size of a square, or more generally, the size of a flat figure. You should wait until someone says that it is the area (if need be, lead them to this conclusion by asking, “What do we mean by the size of a piece of straight line? A piece of a curve?” To be sure, it is the length. If we know that one piece of a curve has the length of 2 units, and another has the length of 3 units, it tells us that the latter piece is of a bigger size). Thus we can systematize our counting in a way similar to that used for counting regular squares — by listing possible areas (sizes) of tilted squares, then finding out what is the total number of squares of each area. This seems to be a good plan. Are there any problems with this approach? Let students think for a while; then ask what is the area of the smallest tilted square? It might help if they find the areas of all three tilted squares pictured above. Let students work trying to find the answer for 10–15 minutes. If nobody comes up with a general method, give them a hint: without saying anything, just outline the smallest regular square containing the tilted one, as shown in Figure 16.

Figure 16. A titled square inscribed in a regular square.

They should notice that the circumscribed regular square has an area of 9, and to get the area of the tilted square we need to subtract the total area of four triangles that lie outside of the tilted square. Each of these four triangles is exactly half of a 2 by 1 rectangle; if we pair them up, we see that the total area is that of two such rectangles. Therefore, the area of the tilted square is 32 − 2 · (2 · 1) = 9 − 4 = 5. Similarly, the area of the smallest tilted square is 22 − 2 · (1 · 1) = 4 − 2 = 2,

20

TATIANA SHUBIN

and the area of the third tilted square shown above is 52 − 2 · (3 · 2) = 25 − 12 = 13. Now we can generalize the above observation as follows. Let us consider any tilted square. Start with its lowest vertex and go to the next one (counterclockwise). Since it is tilted, we should go a cells to the right and b cells up. To get to the next vertex, we must go a cells up and b cells to the left (check that students see the necessity of this — otherwise, the edges will not be of the same size and perpendicular to each other (why?)). Continuing in this manner, we get the picture shown in Figure 17.

Figure 17. A generalized tilted square inscribed in a regular square.

Thus the area of the tilted square is (a + b)2 − 2 · (a · b) = a2 + 2ab + b2 − 2ab = a2 + b2 . Note: If we denote the length of an edge of the tilted square by c, then obviously its area is c2 . So from the above computation, we see that c2 = a2 + b2 . So we just proved Pythagorean Theorem! (in the special case when the length of each side of a right triangle is an integer) We conclude that a positive integer is the area of a tilted square if and only if it can be written in the form a2 + b2 , where a and b are positive integers. Ask students to check which of the consecutive integers starting with 1 can be written in this form. Give them a few minutes for the task; they should get the following possible areas.

GRID POWER

21

2 = 12 + 12 5 = 12 + 22 8 = 22 + 22 10 = 12 + 32 13 = 22 + 32 17 = 12 + 42 Let us give regular and tilted squares a collective name of ‘grid squares.’ We can state that a positive integer is the area of a grid square if and only if it can be written in the form a2 +b2 , where a and b are integers, a is positive, and b is either positive (for a tilted square) or 0 (for a regular square). We summarize our observations as follows: • The following numbers represent areas of grid squares: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, . . . • The following numbers can’t be areas of grid squares: 3, 6, 7, 11, 12, 14, 15, . . . A profound question is, “what characterizes each of these two groups of numbers?” You should challenge your students to investigate this question. (Even though it leads deep into number theory, students could extend the lists and then try to make conjectures of their own. Using computers might help to get large lists thus providing them with more data for conjectures.) Returning to the task of counting the tilted squares in a 7 × 7 grid, we know that every tilted square can be inscribed in a regular square, so we should look at regular squares and see how many tilted squares can be inscribed in each of them. Each 2 × 2 regular square contains only 1 tilted square. Each 3 × 3 square contains two different tilted squares, since 3 = 1 + 2 = 2 + 1 (Figure 18 illustrates this).

Figure 18. Two tilted squares inscribed in a regular 3 × 3 square.

22

TATIANA SHUBIN

Likewise, each 4 × 4 square contains three different tilted squares, since 4 = 1 + 3 = 2 + 2 = 3 + 1. (Have students draw the corresponding pictures.) Now we can add two additional columns to our first table: Grid square size

# of grid # of tilted squares squares per grid square

1×1 2×2 3×3 4×4 5×5 6×6 7×7

49 36 25 16 9 4 1 140

Total:

0 1 2 3 4 5 6

Total # of tilted squares inscribed in regular squares of this size 49 · 0 = 0 36 · 1 = 36 25 · 2 = 50 16 · 3 = 48 9 · 4 = 36 4 · 5 = 20 1·6=6 196

Total:

Hence the total number of all grid squares in a 7×7 square is 140+196 = 336. If students are ready to do some algebraic computations we are now in a position to generalize our previous work. The total number of regular squares in an n × n square is n(n + 1)(2n + 1) n2 + (n − 1)2 + (n − 2)2 + · · · + 22 + 12 = 6 The total number of tilted squares in an n × n square is (n − 1)2 · 1 + (n − 2)2 · 2 + (n − 3)2 · 3 + · · · + (n − (n − 1))2 · (n − 1) = (n2 − 2n · 1 + 12 ) · 1 + (n2 − 2n · 2 + 22 ) · 2 + (n2 − 2n · 3 + 32 ) · 3 + · · · + (n2 − 2n · (n − 1) + (n − 1)2 ) · (n − 1) = n2 (1 + 2 + 3 + · · · (n − 1)) − 2n(12 + 22 + 32 + · · · + (n − 1)2 ) + (13 + 23 + 33 + · · · + (n − 1)3 ) (n − 1)n2 (n + 1) n2 (n − 1)n 2n(n − 1)n(2n − 1) (n − 1)2 n2 − + = 2 6 4 12 Finally, the total number of all grid squares in an n × n grid is

=

n(n + 1)2 (n + 2) n(n + 1)(2n + 1) (n − 1)n2 (n + 1) + = 6 12 12 Comparing the last two formulas, we see that the number of all grid squares in an n × n grid is exactly the same as the number of tilted squares in an (n + 1) × (n + 1) grid. But why? Can we find a ‘natural’ one-to-one correspondence between these two sets? Pose this question to your students and let them brood over it.

GRID POWER

23

An interesting and surprising use of tilted squares can be seen in solving Problem 3. Having drawn the given square as a tilted one, a solution to the problem requires no calculation. Every statement we made in the presented solution can be justified using some elementary geometry. If your students are interested and ready you might ask them to identify and prove every statement in the official solution which needs justification.

References [1] Arthur T. Benjamin, Jennifer J. Quinn, and Calyssa Wurtz. “Summing Cubes by Counting Rectangles” https://www.math.hmc.edu/~benjamin/papers/rectangles. pdf [2] Sergey Dorichenko, A Moscow math circle, MSRI Mathematical Circles Library, vol. 8, Mathematical Sciences Research Institute, Berkeley, CA; American Mathematical Society, Providence, RI, 2012. Week-by-week problem sets; Translated from the 2011 Russian original and with a foreword and an addendum by Tatiana Shubin; Translation edited by David Scott and Silvio Levy. MR2883119

What is the sum?—A 5-card Magic Trick Contributed by Gabriella Pinter Short Description: This is a simple card trick that can inspire young students to make up their own tricks. Along the way they find different patterns, and can pose and investigate their own questions. Materials: Chalk board with colored chalk, playing cards that can be taped together, or cards from template (1 set for groups of 3–4 students). Have some blank pieces of paper (preferably card stock or construction paper) and colored pens around so students could make their own set of cards for their trick. Mathematics Beneath & Beyond: Starting with a base 5-card trick, the session leads to investigating triangular numbers as well as linear functions defined on finite sets and their graphs. The key idea is how a certain quantity (in this case the sum of the numbers face up) changes if an action is performed (a card is turned from one side to the other). This relates the problem to many others, among them those where a quantity does not change, i.e., problems involving the idea of invariance.

25

26

GABRIELLA PINTER

What is the sum?—A 5-card Magic Trick Student Handout Prepare five cards numbered 1 through 5 in red on one side and 6 through 10 in black on the other side. Make sure that the 6 is on the reverse side of the 1, the 7 is on the reverse of the 2, and so on. Pass the cards to a volunteer, who shuffles them and throws them up in the air without you looking. Ask the volunteer how many black cards she sees, and amaze your audience by announcing the sum of the numbers showing on the five cards. Repeat the trick a few times, then give groups of 3 to 4 students a set of cards to explore the trick. (1) Figure out how the trick works. (2) Make and perform a new trick by adding a sixth card with 6 in red on one side and 11 in black on the other. What if a seventh card with 7 in red and 12 in black is also added? How does the trick change? (3) Can you make the trick work if the red number on the 6th card is 9, and the black one is 14? (4) Make up your own trick, and be ready to perform it.

WHAT IS THE SUM?—A 5-CARD MAGIC TRICK

27

What is the sum?—A 5-card Magic Trick Teacher Guide Solutions: When the trick is performed, a frequent misunderstanding is that students think that the ‘magician’ tells them the sum of the black cards only. It is important to stress that the sum announced is the sum of the numbers facing up on all five cards. (1) If the trick is performed a few times, then students realize that the sums announced are numbers in the set {15, 20, 25, 30, 35, 40}. They see that if there is one black card the ‘magician’ will say that the sum is 20, with two black cards 25, and so on, but may not be very clear on why this happens. If students are struggling with their explanations, a good question to ask is the following: “How does the sum change if one card is turned from red to black?”. This usually helps, and a table can be put on the board with the sums corresponding to the number of black cards. no. of black cards sum of numbers 0 1 + 2 + 3 + 4 + 5 = 15 1 20 2 25 3 30 4 35 5 40 It is easy to see that when no black numbers are face up, the sum must be the sum of all the red numbers: 1 + 2 + 3 + 4 + 5 = 15. Now the key idea is that no matter which card is turned from red to black, the black number is always 5 more than the red one, so the sum will increase by 5. This will happen each time a card is turned from red to black, since the black number is always 5 more than the corresponding red one by the way the cards are constructed. A trick question can also be asked: “What is the sum if you see six black cards?” Students will probably say that it is 45 (even though at this point there are only five cards, so there cannot be six black cards face up). This transitions well to Problem 2. (2) Even if a sixth card with 6 in red and 11 on the black side is added, the sum with six black cards face up is not 45. This is surprising to young students, but they quickly realize that there is a new ‘base case’, that is, when all cards are red face up, the sum is 1+2+3+4+5+6 = 21, while in the 5-card trick it was 15. Thus, the resulting sums in this 6-card trick are in the set {21, 26, 31, 36, 41, 46, 51}. Similarly, if seven cards are used with red numbers 1 through 7, then the relevant sums are in the set {28, 33, 38, 43, 48, 53, 58, 63}. At this point a small detour to triangle

28

GABRIELLA PINTER

numbers (that is, numbers that are in the form 1+2+3+· · ·+n, n ∈ N) can be added depending on interest, motivated by the need to calculate the base case for the n−card trick. Students can also be asked to work out a relationship between the number of black cards showing and the sum of the numbers on the cards. This leads to a linear function F : X → R, F (b) = 15 + 5b for the original 5-card trick, and S : Y → R, S(b) = 21 + 5b for the 6-card trick, where X = {0, 1, 2, 3, 4, 5} and Y = {0, 1, 2, 3, 4, 5, 6}. These are natural examples for functions defined on a finite set, and it is instructive to draw their graphs that consist of isolated points. (3) The point in this problem is to help students understand that it is not essential for this trick to have consecutive integers on the red cards. First, students may say that the trick will not work, because there is already a card with the number 9 on it. However, a more careful examination reveals that the base case (no black cards facing up) is now 1 + 2 + 3 + 4 + 5 + 9 = 24, and since we still have the property that each black number is 5 more than the corresponding red one, the trick will work as before with sums in the set {24, 29, 34, 39, 44, 49, 54}. (4) In the course of trying this activity in several classrooms lots of new tricks were found. The most typical one that students make is a trick with five cards with red numbers 1 through 5 and each black number 4 more than the corresponding red one. Another common idea was to use more cards. If students do and understand Problem 3, then more varied tricks may be expected. Figures 19–21 show a few examples:

Figure 19. Trick 1: The base case shows the connection between the sum of consecutive odd integers and perfect squares. Each black number is 3 more than the corresponding red one.

Figure 20. Trick 2: The red numbers do not follow an obvious pattern here, emphasizing that what makes the trick work is the relationship between the corresponding red and black numbers.

WHAT IS THE SUM?—A 5-CARD MAGIC TRICK

29

Figure 21. Trick 3: The red numbers are larger than the black ones.

Further Exploration: In one (middle school) classroom a student made the following trick: the five cards were {1, 2, 3, 4, 5} in red with corresponding black sides {3, 6, 9, 12, 15}, respectively. Thus, instead of adding the same number to each red number (5 in the original trick), the student multiplied each red number by 3. The trick does not work as before, since, for example, both {1, 2, 3, 4, 15} and {1, 2, 3, 12, 5} would involve four red and one black number, and the two sums, 25 and 23, are not the same. Thus, the information that one black number is face up does not let the ‘magician’ determine the sum of the numbers on the five cards. Faced with this dilemma two ideas emerged. Students felt that there may be a way to ‘save’ this example, and they started thinking about multiplying the numbers instead of adding them. Returning to the key question of what is happening when a red card is turned to black is very helpful here. By the way the cards are made up, we can see that whenever a red card is turned to its black side, the red number gets multiplied by 3. If all cards are red, the base product is 1·2·3·4·5 = 120. Each time a red number is turned to black the product grows by a factor of 3, so when all cards are black, their product is 120 · 35 = 29160. Thus, the product of the numbers is in the set {120, 120 · 3 = 360, 120 · 32 = 1080, 120 · 33 = 3240, 120 · 34 = 9720, 120 · 35 = 29160}. Of course, this makes a much more cumbersome trick both for the ‘magician’ and for the audience. When this problem was presented in a professional development seminar, teachers remarked that even though the sum cannot be determined uniquely, it comes from a relatively small set of numbers, and they sought to figure out the probability of guessing correctly. This can be an interesting exercise involving basic combinatorics and probability. We will assume that each card has the same chance of landing on its red or black side. For this problem, the sum when all cards are red face up or black face up is uniquely determined, 15 and 45, respectively, so in this case no guessing is needed. If one card is black face up, then depending on which card it is, we could get five different sums {17, 19, 21, 23, 25}. Thus the magician has a 20% chance of guessing the sum correctly if she chooses one of these numbers. The case of two black cards face up is more interesting. There are 10 ways to choose the two black-side-up cards out of the five cards. However, some of these choices will result in the same sum. Namely, if black numbers 3 and 12 or 6

30

GABRIELLA PINTER

and 9 are up, then the sum in both cases is 25. Similarly, 3 and 15, as well as 6 and 12 result in the sum 27, while 6 and 15 as well as 9 and 12 give 29. The other possible sums, 21, 23, 31 and 33 occur in just one case each. Thus, the magician has a higher chance, 1/5, of guessing correctly, if she guesses one of 25, 27 or 29. Guessing 21, 23, 31 or 33 has only a 1/10 chance to be correct. The case of three cards black side up can be done in a similar way. One can also realize, that three black cards face up mean that there are two red ones face up, so the previous analysis can be carried out for the red cards with similar results. That is, guessing 31, 33 or 35 has a 1/5 chance of being correct, while guessing 27, 29, 37 or 39 has a 1/10 chance of success. The case of four black cards face up is again easier, since there are only five possible sums, each occurring just once. Thus, we can see that if the magician thinks a bit, then her chance of being correct is always at least 1/5. Of course, many new questions can stem from this discussion. For example, could the black numbers in a 5-card trick be chosen in such a way that guessing has less of a chance to succeed than in the previous case? Suggestions: • Students enjoy throwing up the cards in the air, and are usually instantly motivated to figure out the trick. After a few demonstrations in front of the whole class, give each group of 3 to 4 students a set of cards, and go around ‘guessing’ their sums. If time allows, let them perform the trick to each other once they start catching on to how it works. • Always plan to have enough time for the students to come up with their own tricks. If card stock or construction paper is available, they can make a nice set of cards to take home, so they could show the trick to their family. Problem posing is an important aspect of learning mathematics, and students will almost always come up with something unexpected. Many times these new problems can start fruitful discussions, as in the example above.

Decanting Problems (A.K.A Euclidean Algorithm) Contributed by Tatiana Shubin and Elgin Johnston Short Description. Decanting problems provide a hands-on introduction to the Euclidean Algorithm and Diophantine Equations. Materials: You may wish to provide students with cups (representing “jugs” of different sizes), sharpies for labeling the cups, and cubes or beans to represent pints of water. Students will also need paper and pencils. Mathematics Beneath & Beyond: This set of problems provides an introduction to at least two deep mathematical ideas — the Euclidean Algorithm and Diophantine Equations. The Euclidean Algorithm is used to find the Greatest Common Divisor (GCD)3 of two numbers. A more familiar way which uses representing numbers as products of prime powers isn’t practical for large numbers while Euclidean algorithm remains surprisingly efficient. It is based on a very simple idea of dividing with remainders. Besides being a practical way of finding the GCD of large numbers, the Euclidean Algorithm plays very important role in establishing many numbertheoretical results. In particular, it can be used to show that the GCD of two numbers can be written as their linear combination (this concept is mentioned in the solutions to problems 1 and 4). A Diophantine Equation is an equation for which we seek only integer solutions. Probably the most famous one is an equation xn + y n = z n . For nearly four hundred years this equation had defied many generations of first-class mathematicians; only a few years ago it was proved that such an equation does not have solutions in positive integers if n is larger than 2. Simplest Diophantine Equations are the ones known as linear equations — they are of the form ax + by = c, where a, b, and c are known and one needs to find all suitable integer values of x and y. Most problems in this set are some variations of this general idea. Despite the relative ease with 3 School textbooks often use the term Greatest Common Factor (GCF) in place of GCD. Mathematicians usually use the term GCD instead.

31

32

TATIANA SHUBIN AND ELGIN JOHNSTON

which Diophantine Equations can be stated, there are many deep and even unsolved problems in this area of mathematics.

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

33

Decanting Problems (A.K.A Euclidean Algorithm) Student Handout Warm-up problem: After the Dynamic Jug Duo finished playing a song, Joe asked if the audience had any requests. “Yeah!” came a shout from the audience. “Using just that 5-pint jug and that there 12-pint jug, measure me one pint of water!” Is this possible with just two jugs? How? The problem above is known as a Decanting Problem. It is a liquid measuring problem that begins with two unmarked decanters with given capacities a and b. Usually a and b are integers (in the problem above, a = 5 and b = 12, and we assume that we have an unlimited supply of water that we can use to fill jugs). The problem is to measure a certain amount c of water (in the warm-up problem c = 1) using three actions: (1) Fill up an empty jug (or decanter). (2) Dump out a full jug. (3) Pour from one jug to the other until either the receiving jug is full or the poured jug is empty. Problems: (1) Using the 5- and 12-pint jugs can you get exactly 16 pints? Nine pints? Two pints? Are there any quantities between 1 and 17 pints that you cannot get? How do you know? (2) Suppose we have jugs of size 17 pints and 7 pints. How can you get any volume from 1 to 17 + 7 = 24 pints? Hint: First, come up with a repetitive process that leads to any volume from 1 to 7 − 1 = 6 pints. (3) A farmhand was sent to a nearby pond to fetch 8 gallons of water. He was given two pails — one 11 gallons and the other 6 gallons. How can he measure the requested amount of water? (4) Suppose you have jugs of size 15 pints and 36 pints. What volumes between 1 and 15 + 36 = 51 pints can you get? Which can’t you get and why? (5) Suppose you have jugs of size 15 pints and 25 pints. What volumes between 1 and 15 + 25 = 40 pints can you get? Which can’t you get and why? (6) Now suppose we have two jugs, one of p pints and another of q pints with 0 < p < q, and suppose that p and q are relatively prime (i.e., have only 1 as a common divisor, like the numbers 12 and 35). Is it possible to measure exactly 1 pint? How? Formulate a good description of the repetitive process you used. (7) Can you use your observations in problems 4 and 5 to state a general problem similar to problem 6? (8) Two friends have an eight-quart jug of water and wish to share it evenly. They also have two empty jars, one holding five quarts, the other holding three quarts. How can they measure out exactly four

34

TATIANA SHUBIN AND ELGIN JOHNSTON

quarts of water? Note: in this problem there is no other source of available water; we must work only with the original 8 quarts of water. (9) There are three jugs each having an integer number of pints of water. A move consists of doubling the amount of water in a jug by pouring water from one other jug provided the latter has sufficient water in it. Assuming that each jug is large enough to hold all the water in the three jugs, prove that one can empty one of the jugs. (This problem was in the 1971 All-Union Math Olympiad (USSR).) (10) Chicken McNuggets can be purchased in quantities of 6, 9, and 20 pieces. You can buy exactly 15 pieces by purchasing a 6 and a 9, but you can’t buy exactly 10 McNuggets. What is the largest number of McNuggets that can NOT be purchased?

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

35

Decanting Problems Teacher Guide Solutions: For convenience, we will use notation x, y to mean that the larger jug contains x pints and the smaller jug, y pints, and we will use descriptions similar to the one below: If we have the large 12-pint jug and the small 5-pint jug, the chain 0, 0 → 12, 0 → 7, 5 → 7, 0 has the following meaning: we start with two empty jugs; fill up the large jug; pour from the large jug into the small jug (filling it); then dump out the liquid from the small jug. In the chain, every pair of numbers is a state and every arrow is an action. We will count them from left to right. In the chain above, the first state is 0,0; the second state is 12,0, the third state is 7,5, and the fourth state is 7,0. Warm-up problem: 0, 0 → 12, 0 → 7, 5 → 7, 0 → 2, 5 → 2, 0 → 0, 2 → 12, 2 → 9, 5 → 9, 0 → 4, 5 → 4, 0 → 0, 4 → 12, 4 → 11, 5 → 11, 0 → 6, 5 → 6, 0 → 1, 5 In the 19th state we see that there is one pint in the large container, so we have produced 1 pint as asked. This 19 state process is a solution to our problem. It is also worth noting that if we had asked for 4 pints instead of 1 pint, we would have been done in the 11th state, 4,5. (1) Let’s use the chain in the Warm-Up Problem to make the following table. The second column contains three numbers — the first two are the number of pints in each container and the last one is their sum. The third column contains any numbers appearing for the first time.

36

TATIANA SHUBIN AND ELGIN JOHNSTON State 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Quantities represented at this state 0,0,0 12, 0, 12 7, 5, 12 7, 0, 7 2, 5, 7 2, 0, 2 0, 2, 2 12, 2, 14 9, 5, 14 9, 0, 9 4, 5, 9 4, 0, 4 0, 4, 4 12, 4, 16 11, 5, 16 11, 0, 0 6, 5, 11 6, 0, 6 1, 5, 6

New quantities at this state 0 12 5, 7 2

14 9 4

16 11 6 1

Thus the 19 state chain produces all volumes in pints from 1 to 17 except for 3, 8, 10, 13, 15, 17. Continuing the chain from the 19th state, we get: 1, 5 → 1, 0 → 0, 1 → 12, 1 → 8, 5 → 8, 0 → 3, 5 → 3, 0 Hence we now get 13 (22nd state), 8 (23rd state), and 3 (25th state). To get the remaining three numbers — 10, 15, and 17 — we can use the following two independent chains: 0, 0 → 0, 5 → 5, 0 → 5, 5 → 10, 0 → 10, 5, and 0, 0 → 0, 5 → 12, 5. Therefore, we can obtain all quantities from 1 through 17 pints. Note: We also observe that going from state 1 to state 6, we filled up the large jug once, and dumped out the small one two times, ending up with 2 pints. This corresponds to the equation 1 · 12 − 2 · 5 = 2, that can be rewritten as 1 · 12 = 5 · 2 + 2, which, in turn, corresponds to the fact that 2 is the remainder when 1 · 12 is divided by 5. Similarly, going from state 1 to state 12, we get 2 · 12 − 4 · 5 = 4; i.e., 4 is the remainder when 2 · 12 is divided by 5. Likewise, the chain from state 1 to state 20 corresponds to the fact that 1 is the remainder when 3 · 12 is divided by 5, and from state 1 to state 26 it corresponds to the fact that 3 is the remainder when 4 · 12 is divided by 5. There is one more interesting observation: Looking at the first three chains, we notice that we filled 12-pint jug 3 times and dumped out the 5-pint jug 7 times, and the resulting amount of water was 1 pint. This can be written as 12 · 3 + 5 · (−7) = 1. This demonstrates that 1, the Greatest Common Divisor (GCD) of 12 and 5, can be written as a linear combination of 12 and 5.

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

37

(A linear combination of two numbers, a and b, is any expression of the form ax + by, where x and y are integers.) (2) Use a similar method: 0, 0 → 17, 0 → 10, 7 → 10, 0 → 3, 7 → 3, 0 → 0, 3 → 17, 3 → 13, 7 → 13, 0 → 6, 7 → 6, 0 → 0, 6 → 17, 6 → 16, 7 → 16, 0 → 9, 7 → 9, 0 → 2, 7 → 2, 0 → 0, 2 → 17, 2 → 12, 7 → 12, 0 → 5, 7 → 5, 0 → 0, 5 → 17, 5 → 15, 7 → 15, 0 → 8, 7 → 8, 0 → 1, 7 → 1, 0 → 0, 1 → 17, 1 → 11, 7 → 11, 0 → 4, 7 → 4, 0 Those states (i.e. pairs of numbers) in which a new number (quantity) including the sum of the amounts in the two jugs appears for the first time are written in boldface. Let us list all the new non-zero quantities we obtain in each row, reading from left to right and listing each number only once: 1st row: 17, 10, 7, 3 2nd row: 20, 13, 6 3rd row: 23, 16, 9, 2 4th row: 19, 12, 5 5th row: 22, 15, 8, 1 6th row: 18, 11, 4 The only missing numbers are 14, 21, and 24. We get them as follows: 0, 0 → 0, 7 → 7, 0 → 7, 7 → 14, 0 → 14, 7; and 0, 0 → 0, 7 → 17, 7. Note: As in problem 1, we can write arithmetic equations illustrating the number of times container is filled (positive) and dumped (negative). In what follows we are writing the equation for the new amount that ends each row but always starting with 0,0 in the first row. Row 1: 1 · 17 − 2 · 7 = 3, or 1 · 17 = 2 · 7 + 3, or (1 · 17) ÷ 7 = 2 r 3 Rows 1–2: 2 · 17 − 4 · 7 = 6, or 2 · 17 = 4 · 7 + 6, or (2 · 17) ÷ 7 = 4 r 6 Rows 1–3: 3 · 17 − 7 · 7 = 2, or 3 · 17 = 7 · 7 + 2, or (3 · 17) ÷ 7 = 7 r 2 Rows 1–4: 4 · 17 − 9 · 7 = 5, or 4 · 17 = 9 · 7 + 5, or (4 · 17) ÷ 7 = 9 r 5 Rows 1–5: 5 · 17 − 12 · 7 = 1, or 5 · 17 = 12 · 7 + 1, or (5 · 17) ÷ 7 = 12 r 1 Rows 1–6: 6 · 17 − 14 · 7 = 4, or 6 · 17 = 14 · 7 + 4, or (6 · 17) ÷ 7 = 14 r 4 (3) This can be solved in exactly the same manner: 0, 0 → 11, 0 → 5, 6 → 5, 0 → 0, 5 → 11, 5 → 10, 6 → 10, 0 → 4, 6 → 4, 0 → 0, 4 → 11, 4 → 9, 6 → 9, 0 → 3, 6 → 3, 0 → 0, 3 → 11, 3 → 8, 6 → 8, 0

38

TATIANA SHUBIN AND ELGIN JOHNSTON

Note: This corresponds to the equation 4 · 11 − 6 · 6 = 8. The equation can be rewritten in the form 11 · 4 + 6 · (−6) = 8, which demonstrates a solution x = 4, y = −6 for the Diophantine equation 11x + 6y = 8. Intermission. If you study the chain solutions in problems 1, 2, and 3 you might make the following curious observations: (i) The big container is the only one we fill from the unlimited water source — we never fill the small jug from the source. (ii) Water from the big container is only poured into the small container — it is never emptied by pouring the water onto the ground. (iii) We only put water into the small container from the big container and the only way we take water out of the small container is to dump it on the ground. Can you change things? Can these problems be solved by only filling the small container from source, only taking water out of the big container by emptying all contents on the ground, and only putting water in the big container from the small container? Think about it! (4) Suppose we filled up a 36-pint jug x times and we dumped out a 15-pint jug y times. The total amount of water we have now is 36x − 15y = 3(12x − 5y), a multiple of 3. Thus the only quantities of water we can hope to obtain are 3, 6, 9, . . ., 51 pints. All these amounts are indeed possible: 0, 0 → 36, 0 → 21, 15 → 21, 0 → 6, 15 → 6, 0 → 0, 6 → 36, 6 → 27, 15 → 27, 0 → 12, 15 → 12, 0 → 0, 12 → 36, 12 → 33, 15 → 33, 0 → 18, 15 → 18, 0 → 3, 15 → 3, 0 → (*) 0, 3 → 36, 3 → 24, 15 → 24, 0 → 9, 15 → 9, 0 Listing all new quantities by the row (as we did in problem 2), we get: 1st row: 36, 21, 15, 6 2nd row: 42, 27, 12 3rd row: 48, 33, 18, 3 4th row: 39, 24, 9 The only multiples of 3 missing in these four rows are 30, 45, and 51; and these numbers can be obtained as follows: 0, 0 → 0, 15 → 15, 15 → 30, 15; and 0, 0 → 0, 15 → 36, 15 Note: Writing an equation corresponding to the first long chain from state 1 to state 20, we get: 36 · 3 + 15 · (−7) = 3 We observe that GCD(36, 15) = 3. Thus, the GCD of two numbers can be written again as their linear combination.

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

39

Actually, we already solved this problem when we solved Problem 1. Note that 36 = 3 · 12 and 15 = 3 · 5 and in Problem 1 we had containers holding 12 pints and 5 pints. Now go back to the chain (*) shown above. Notice that for the first three rows of the chain, if you divide every number in the chain by 3 you will have the chain produced in the warm-up solution and used to solve Problem 1! (5) Similarly to the previous problem, getting a certain amount c of water in this case is equivalent to solving an equation c = 25x − 15y = 5(5x − 3y), where x and y are integers. But then c is a multiple of 5, and hence the only possible quantities would be 5, 10, 15, . . ., 40. Each of these numbers can actually be obtained – use a process similar to the ones above. (6) This problem generalizes problems 1, 2, and 3. We start by filling up the larger jug, pouring from it into the smaller jug and emptying the smaller jug whenever it is full, until we have no water in the larger jug and some water left in the smaller jug. Then we repeat this process again and again. As a result, we obtain in the smaller jug the following sequence of volumes: r1 , which is the remainder when 1 · q is divided by p (q pint jug filled and emptied the 1st time); r2 , which is the remainder when 2 · q is divided by p (after q pint jug is filled and emptied the 2nd time); r3 , which is the remainder when 3 · q is divided by p (after q pint jug is filled and emptied the 3rd time); .. . rp−1 , which is the remainder when (p − 1) · q is divided by p (q pint jug emptied the (p − 1)st time); The amazing thing is that when p and q are relatively prime all of these remainders are different. Indeed, if two of them, say, ri and rj are the same, with j < i, then we must have i · q − m · p = j · q − n · p, and hence (i − j) · q = (m − n) · p, and so (i − j) · q is a multiple of p. But p and q are relatively prime, so p must divide (i − j). This is clearly impossible because the absolute value of (i − j) is smaller than p. Now, since we have exactly (p−1) different remainders and each one is bigger than 0 and less than p, these remainders must be 1,2,3,. . . (p− 1), in some order. In particular, this method gives us all volumes from 1 pint to (p − 1) pints. (7) Problem: If p and q are any integers such that 0 < p < q, and we have two jugs, one of p pints and one of q pints, is it possible to measure exactly d pints, where d is the GCD of p and q? How? Formulate a good description of the repetitive process that could be used.

40

TATIANA SHUBIN AND ELGIN JOHNSTON

Solution: The answer is yes. Moreover, d is the smallest amount that can be measured, and all other amounts that could be measured are multiples of d: d, 2d, 3d, . . . , kd, where kd = p + q. Similarly to problem 6, we start by filling up the larger jug, pouring from it into the smaller jug and emptying the smaller jug a number of times, until we have no water in the larger jug and some water left in the smaller jug. Then we repeat this process again and again. As a result, we obtain in the smaller jug the following sequence of volumes: r1 , which is the remainder when 1 · q is divided by p; r2 , which is the remainder when 2 · q is divided by p; r3 , which is the remainder when 3 · q is divided by p; .. . rt−1 , which is the remainder when (t − 1) · q is divided by p, where p = t · d. Each of these remainders is a multiple of d (for example, r1 = 1 · q − np for some integer n, and since both q and p are multiples of d, so is r1 , too). Exactly as in the previous problem, it can be shown that all these remainders are different and therefore they must be the numbers d, 2d, 3d, . . . (t − 1)d, in some order. (8) Let’s use the notation x, y, z to express the fact that the 8-quart jug contains x quarts of water, 5-quart jug contains y quarts of water, and 3-quart jug contains z quarts. The following chain represents a solution: 8, 0, 0 → 5, 0, 3 → 5, 3, 0 → 2, 3, 3 → 2, 5, 1 → 7, 0, 1 → 7, 1, 0 → 4, 1, 3 → 4, 4, 0

(9) The notation (a, b, c) means that there are a pints in container 1 (C1 ), b pints in container 2 (C2 ), and c pints in container 3 (C3 ). Let m(a, b, c) denote the minimum of a, b, c. (For example m(3, 6, 3) = 3 and m(0, 4, 19) = 0.) We show that if m(a, b, c) > 0, then there is a pouring scheme that produces a pint distribution (a , b , c ) with m(a , b , c ) < m(a, b, c). Because m(a, b, c) is always a nonnegative integer, it can only decrease a finite number of times, that is, we will reach a state (a , b , c ) with m(a , b , c ) = 0. In that case one of the containers is empty. Assume that 0 < a ≤ b ≤ c. Divide a into b and let the quotient be q and the remainder r, with 0 ≤ r < a. In other words, b = qa + r. We will show that by strategic pouring we can end with r < a pints in C2 . We start by writing q in base 2 (that is, as a sum of unique powers of 2). Suppose q = 2k1 + 2k2 + · · · + 2kn where 0 ≤ k1 < k2 < · · · < kn ,

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

41

so b = qa + r = (2k1 + 2k2 + · · · + 2kn )a + r. The integers 0 ≤ k1 < k2 < . . . < kn may not be consecutive, that is, there may be gaps between some of the pairs ki , ki+1 . Let j1 < j2 < · · · < jm be the missing integers, between 0 and kn . Thus the sets {k1 , k2 , . . . , kn } and {j1 , j2 , . . . , jm } are disjoint and {k1 , k2 , . . . , kn } ∪ {j1 , j2 , . . . , jm } = {0, 1, 2, . . . , kn }. Because jm < kn we have (2j1 + 2j2 + · · · + 2jm )a ≤ (1 + 21 + 22 + · · · + 2jm )a = (2jm +1 − 1)a < 2kn a ≤ b ≤ c, so C3 has at least 2j1 a + 2j2 a + · · · + 2jm a pints. We are now ready to pour! We do this in a series of kn + 1 steps. The first step is Step 0. Step 0. C1 has a = 20 a pints. If k1 = 0, then we pour 20 a pints from C2 into C1 . If k1 = 0, then j1 = 0 and we pour 20 a pints from C3 to C1 . C1 now has 21 a pints. .. . Step 1. If ki = 1 for some i, then pour 21 a pints from C2 into C1 . Otherwise ji = 1 for some i and we can pour 21 a pints from C3 into C1 . C1 now has 22 a pints. .. . Step p. C1 now has 2p a pints. If ki = p for some i, then pour 2p a pints from C2 into C1 . Otherwise ji = p for some i and we can pour 2p a pints from C3 into C1 . C1 now has 2p+1 a pints. .. . Step kn . C1 now has 2kn a pints so we can pour 2kn a pints from C2 into C1 . After this final step there are r < a pints in C2 , that is, we have fewer than a pints in one of the containers. Thus we have reduced m(a, b, c) as desired. (10) Solution 1: The answer is 43. Indeed, 43 pieces cannot be purchased: to purchase 43 pieces we could only buy 0, 1, or 2 packages of 20. If we buy 0, then 43 must be represented as a linear combination of 6 and 9, which is impossible since 3 is a common divisor of 6 and 9, but 3 does not divide 43. If we buy one 20-piece package, then 23 must be represented as a linear combination of 6 and 9, which is again impossible by the same reason. And if we buy two 20-piece packages then we’re left with 3 pieces which cannot be bought. On the other hand,

42

TATIANA SHUBIN AND ELGIN JOHNSTON

44 45 46 47 48 49

= = = = = =

4 · 6 + 0 · 9 + 1 · 20 3 · 6 + 3 · 9 + 0 · 20 1 · 6 + 0 · 9 + 2 · 20 0 · 6 + 3 · 9 + 1 · 20 5 · 6 + 2 · 9 + 0 · 20 0 · 6 + 1 · 9 + 2 · 20

Every larger number can be written as one of these six numbers plus a multiple of 6, and therefore it can be bought. Solution 2: First note that if we limit ourselves to purchasing only packages of 6 and 9 pieces, then we cannot purchase 3 pieces, but we can purchase any other multiple of three pieces (e.g., 6, 9, 12, 15, 18,. . .). Next note that if we purchase 3 boxes of 20 pieces then we can replace these 60 pieces with 10 boxes of six pieces. Thus we can purchase any achievable total with 0, 1, or 2 twenty piece boxes. Taking no boxes of twenty gives us the multiples of 3 greater than or equal 6 discussed above. Taking one box of 20 and allowing any number of 6 and 9 piece boxes gives us any of the totals 20, 26, 29, 32, 35, . . .. In particular we can purchase any amount C for C ≥ 26 and C congruent to 2 modulo 3. Taking 2 boxes of 20 and allowing any number of 6 and 9 piece boxes gives us any of the totals 40, 46, 49, 52, 55,. . ., in particular we can purchase any number of pieces C with C ≥ 46 and C congruent to 1 modulo 3. Combining these three cases we see we have captured all numbers greater than or equal to 44. Let us now check that 43 pieces cannot be purchased: indeed, 43 leaves a remainder of 1 when divided by 3 so we would need two boxes of 20, but then could not purchase the needed 3 pieces. Presentation Suggestions: The comments below draw on our experience at various Math Circles for students not unlike yours. (1) It is helpful to start with the given warm-up problem — the problem is engaging and allows experimentation. It might be helpful to give them some physical objects to play with. For example, a couple of plastic cups of different sizes could represent the jugs, and dry beans, little wooden cubes, etc., to represent pints of water. Let students actually solve the problem — don’t rush to help. There is no such thing as ‘too much time’ when it comes to understanding simple ideas deeply.

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

43

Figure 22. An illustrated solution to a decanting problem with containers holding 12 and 5. (2) After students have solved the warm-up problem, let some of them explain their solution to the rest of the class. At this point it might be useful to introduce a convenient notation, e.g., notation used in the solutions section. You might ask them to illustrate a solution as shown in Figure 22 using a suitable picture or just do it yourself so that it is clear that the numbers under the jugs are all that is needed — thus the proposed notation. (3) Having discussed the warm-up problem and introduced a convenient notation, let them work through as many problems as they want. For most students, problems 1 through 5 followed by problem 8 might be a reasonable plan. (4) An important part of the session is making a connection between actions depicted by our notation and numerical equations leading to algebraic equations. At this point you might spend some time talking

44

(5) (6)

(7)

(8)

TATIANA SHUBIN AND ELGIN JOHNSTON

about Diophantine equations (see Mathematics Beneath & Beyond and More Mathematics for a Curious Reader sections). In particular, challenge your students to find out when a linear Diophantine equation ax + by = c has a solution. Problems 6 and 7 might be attractive for students who like algebra and slightly deeper thinking. Problem 10 is ideal for experimentation — students can spend some considerable time trying to tackle it, and even if they don’t solve it completely they might gain some insight which you can build on at a subsequent session. If they don’t get to work on the problem at the current session encourage them to investigate it at home. Problem 9 is difficult — it asks for a general proof independent of particular initial distribution of water. But again, just trying to understand it and experimenting with a number of initial distributions (of their own making) will be interesting enough and might lead to a considerable intellectual growth. Last but not least: Do not expect to go through all the problems and ideas mentioned above in one session. Students need time to experiment, to think, to absorb ideas. We would suggest spending from 1–3 (possibly more) 90 minute sessions allowing students to investigate the problems and ideas, to ask their own questions, to pose and study their own related problems.

More Mathematics for a Curious Reader To find the GCD of 234 and 425, we perform the following calculations: 425 = 234 · 1 + 191

(we divide 425 by 234; the remainder is 191); 234 = 191 · 1 + 43 (now divide 234 by 191; the remainder is 43); 191 = 43 · 4 + 19 (now divide 191 by 43; the remainder is 19); 43 = 19 · 2 + 5 (divide 43 by 19; the remainder is 5); 19 = 5 · 3 + 4 (divide 19 by 5; the remainder is 4); 5=4·1+1 (divide 5 by 4; the remainder is 1); 4=1·4+0 (divide 4 by 1; the remainder is 0). The GCD of 234 and 425 is the last non-zero reminder, i.e., 1. In general, we first divide the larger number by the smaller one; then divide the smaller one by the first remainder; then divide the first remainder by the second remainder; and so on, until we finally get the remainder of

DECANTING PROBLEMS (A.K.A EUCLIDEAN ALGORITHM)

45

0. The previous remainder — the last non-zero one — is the GCD we’re seeking. Here is one more example: find the GCD of 600 and 189: 600 189 33 24 9 6

= = = = = =

189 · 3 + 33 33 · 5 + 24 24 · 1 + 9 9·2+6 6·1+3 3·2+0

Since the last non-zero remainder is 3, therefore it’s the GCD of 600 and 189. The process of repeatedly filling one container with another illustrates the idea of a generator for a finite cyclic group. Finite cyclic groups can be modeled using modular (or “clock”) arithmetic on a finite set of integers of the form Zn = {0, 1, 2, ..., n − 1}. An operation on this set can be defined through addition modulo n. This means that if k and j are two elements in {0, 1, 2, ..., n − 1}, then the sum k + j is the remainder when k + j is divided by n. For example, using the set Z5 = {0, 1, 2, 3, 4} and addition modulo 5, we have 3 + 4 = 2 (the remainder when 3 + 4 = 7 is divided by 5.) Under this kind of operation, (Zn , +) is an Abelian (i.e., commutative) group. This means that: (i) The set Zn is closed under the operation + (addition modulo n) (ii) The operation + is associative: (a + b) + c = a + (b + c) (iii) The operation + is commutative: a + b = b + a (iv) There is an identity element. This element is 0: note that a + 0 = a (v) For each a in Zn there is an element a with a+a = 0 (note a = n−a.) In the warm-up problem, as we repeatedly pour from the 12 pint jug into the 5 pint jug, we see that the first time we empty the 12 pint jug we end with 2 pints in the 5 pint jug. Notice that 12 = 2, modulo 5. When we empty the 12 pint jug for the second time, we have 2 + 2 = 4 pints in the 5 pint jug. Each cycle of emptying the 12 pint jug into the 5 pint jug “adds 2” to the pints in the small jug: 2 → 2 + 2 = 4 → 4 + 2 = 1 (modulo 5) → 1 + 2 = 3 → 3 + 2 = 0 (modulo 5). Note that the numbers produced, 2, 4, 1, 3, 0 are all of the numbers in Z5 = {0, 1, 2, 3, 4}. We say that 12 = 2 (modulo 5) is a generator for the cyclic group Z5 . Conversely, if we repeatedly pour the 5 pint container into the 12 pint container, emptying the 12 pint container each time it is full, then we produce the following volumes: 5 → 5 + 5 = 10 → 10 + 5 = 15 = 3 (modulo 12) → 3 + 5 = 8 → 1 → 6 → 11 → 4 → 9 → 2 → 7 → 0,

46

TATIANA SHUBIN AND ELGIN JOHNSTON

where in each step above we have added 5 (pints) and taken the remainder on division by 12. Notice that this process produces all 12 of the elements of Z12 . Thus 5 is a generator for the cyclic group Z12 . If p and q are relatively prime, the repeated sums, p, p + p, p + p + p, . . . when divided by q and the remainders are saved, produce all of the elements of Zq . Conversely the repeated additions q, q + q, q + q + q, . . . on division by p produce all of the elements of Zp . In particular, given containers of m and n pints, repeated pouring of the m pint container into the n pint container produces all of the volumes 0, 1, 2, . . . , n − 1 if and only if m and n are relatively prime. Likewise, m is a generator of Zn if and only if m and n are relatively prime. However, if m and n are NOT relatively prime then the numbers produced by repeatedly adding m form what is called a subgroup of Zn . This subgroup consists of all of the multiples of d = GCD(m, n) that are in Zn .

Bean Bag Tossing Contributed by Amanda Serenevy Short Description: Students will explore modular arithmetic — a type of arithmetic that results from arranging numbers in circles instead of on a number line. Students make and prove conjectures about patterns relating to factors and remainders. Materials: A bean bag or ball. An outdoor space, sidewalk chalk, postersized chart paper, markers, handouts, and pencils. Alternatively, number disks (such as a roll of shelf liner rubber) can be used indoors, or dry erase boards or paper can be used to draw the number rings. Preparation: Prepare handouts and materials, if necessary. Mathematics Beneath & Beyond: Students will gain a deeper understanding of factors, multiples, and the Fundamental Theorem of Arithmetic. This session is related to other sessions, such as Decanting Problems (page 31), involving greatest common divisors (or GCD, which is sometimes also called GCF, or greatest common factor) and the Euclidean Algorithm. Modular arithmetic is an important topic in number theory and is also useful in computer programming. This topic can also be related to Nim games. Navajo Nation Math Circle Connection: This was one of the topics explored at the Tsehootsooi Intermediate Learning Center Math Circle during the 2015–2016 academic year. Devon Lynch, one of the students in that group, said that this was one of his favorite activities. A year later, he told us that he still enjoyed “tricking people” by asking them whether there are any other reasonable answers to 2 + 1 besides 3. He then teaches them modular arithmetic and shows them that when working modulo 3, 2 + 1 is equivalent to 0.

47

48

AMANDA SERENEVY

Bean Bag Tossing Student Handout During this session, you may use the chart below to record your results. Detailed instructions are on the next page.

Modulus 1

2

3

4

5

6

7

8

9

10

1 2

Tossing Number

3 4 5 6 7 8 9 10 Conjectures: What patterns did you notice in the chart? Record some of your conjectures (guesses) about these patterns.

BEAN BAG TOSSING

49

We will explore a type of arithmetic that results from arranging numbers in circles instead of on a number line. On some of our circles, 4 + 5 is not equal to 9! Modular arithmetic is useful in situations where some quantity cycles. Time on a clock is one example of this. You use modular arithmetic when you try to figure out what time is 5 hours after 9 AM. In computer programming the “mod” operation is used for programs that need to assign values cyclically. Modular arithmetic is one of the first examples of an algebraic system that undergraduate math majors encounter in abstract algebra. Algebraists study underlying patterns to determine when two systems that look different on the surface really have the same structure. Some systems that algebraists study include arithmetic of real numbers, modular arithmetic, arithmetic of string braids, composition of motions, and operations based on symmetries of physical objects. Emmy Noether was a mathematician who made important contributions to abstract algebra and modern physics. Instead of thinking in terms of long and complicated computations, Emmy looked for simple underlying patterns. This enabled her to create new and more powerful ways of thinking about difficult problems in mathematics and physics. Bean Bag Tossing • Make a circle of numbers from 0 up to some number between 1 and 10. Have one person stand on each of the numbers in the circle. • How many numbers are in your circle? (Remember to count the zero.) This number is called the “modulus” of the circle. The plural of “modulus” is “moduli”. • Choose a number between 1 and 10 as the “tossing number”. • Find the space in the chart corresponding to the modulus and the tossing number you chose. Put a large circle in that box so you remember where you are working. • Give the bean bag to the person standing on the 0. That person counts around the circle (in the direction the numbers go) up to the tossing number, and then tosses the bean bag to that person. Do not count the person who is doing the tossing. • The person who catches the bean bag uses the same tossing number to count, and throws the bean bag to the new person. This pattern continues around the circle until the bean bag returns to the person standing on the zero. • Did everyone in the circle catch the bean bag? If the answer is “yes” then write a Y in the chart where your circle is. If you did not land on all the numbers, then write an N. • Work together to find out whether each square in the chart should have a Y or an N. It is a good idea for more than one team to check the

50

AMANDA SERENEVY

answer in each box. Put a check mark or a star in the corner of a box if your team confirms the answer that another team already found. • As the chart develops, look for patterns that are developing. Do you notice any patterns that you think will continue? If so, record your “conjecture” (your guess) about patterns that might be true. You might , then .” phrase your conjecture as “If Post those conjectures where other people can see them. • From time to time, read a conjecture to see whether you can disprove it by showing a counter-example (an example which does not fit the pattern). You might also find more evidence to support the conjecture. • You might be able to explain why it is that a conjecture MUST be true. This kind of argument is called a “proof” if it explains WHY the pattern happens.

BEAN BAG TOSSING

51

Modular Arithmetic Modular Counting Use a circle with modulus 5 (0 through 4). (1) Start on the 0. If you take 5 steps, where do you land? (2) Start on the 0. If you take 6 steps, where do you land? (3) Start on the 0. If you take 9 steps, where do you land? (4) Start on the 0. If you take 13 steps, where do you land? (5) Start on the 0. If you take 15 steps, where will you land? (6) Start on the 0. If you take 100 steps, where will you land? Why? (7) Start on the 0. If you take 147 steps, where will you land? Why? (8) Start on the 0. If you take 284 steps, where will you land? Why? Now use a circle with a modulus of 7. (9) Start on the 0. If you take 15 steps, where will you land? (10) Start on the 0. If you take 482 steps, where will you land? Why? (11) Suppose you use a circle with modulus m and you take k steps starting from zero. What procedure can you use to figure out where you will land? Arithmetic on a Number Line (12) How could you use a number line (13) How could you use a number line (14) How could you use a number line (15) How could you use a number line (16) How could you use a number line Addition Use a circle with modulus 5. (17) 2 + 2 ≡5 (18) 3 + 4 ≡5 Subtraction Use a circle with modulus 7. (19) 6 − 4 ≡7 (20) 3 − 6 ≡7 Multiplication Use a circle with modulus 10. (21) 2 × 4 ≡10 (22) 5 × 3 ≡10

to to to to to

find find find find find

the the the the the

answer answer answer answer answer

to to to to to

6 + 2? 6 − 2? 2 − 6? 6 × 2? 6 ÷ 2?

52

AMANDA SERENEVY

Division Use a circle with modulus 8. (23) 6 ÷ 2 ≡8 (24) 5 ÷ 3 ≡8 (25) Modular division is not as straightforward as the other arithmetic operations. Sometimes there is no answer to a modular division problem. Find an example of a modular division problem with no solution.

BEAN BAG TOSSING

53

Bean Bag Tossing Teacher Guide With a large group, it is ideal to use a large room or outdoor area for this exploration, and use number disks made out of non-slip shelf liner paper (if working inside), or sidewalk chalk (if working outside). Make a circle using the numbers 0 through any number up to 9. Count the number of numbers in the circle (including the 0). This number is called the “modulus” of the circle. For example, if the circle has the numbers 0 through 5, then the modulus would be 6. Gather the students around the circle and ask for volunteers to stand on each number in the circle. Ask the person standing on the 0 to choose a “tossing number” between 1 and 10. Ask all of the students to take out their charts and draw a large circle in the square corresponding to the chosen modulus and tossing number. The person standing on 0 should count in the direction the numbers go until they reach the tossing number. The person who catches the bean bag repeats this process using the same tossing number. Continue this pattern until the bean bag returns to the person on 0. Ask the students “Did everyone in the circle catch the bean bag?” If the answer is “yes”, the students should write a Y in the circled square. Otherwise, they should write an N. Work out several other examples using various circles and tossing numbers until the students catch on to the process. It can help to show the students how to model the process without tossing the bean bag by letting the numbers on the ground stand for the people and marking the positions that would catch the bean bag. After the students seem to understand the process, ask them to scatter around the area to find more answers, recording them in their charts and returning periodically to fill in a giant chart shared by the class. Tell the students that the group will work together to complete and confirm the whole chart. Because some students may make mistakes, it is helpful to ask students to place a check mark or star in the corner of any completed box that they have confirmed. Alert the students briefly that they should look for patterns as they work and try to formulate “conjectures” (or guesses) about them. One way to formulate their conjectures is to use an “If . . ., then . . .” structure. Correct answers for the chart, and some possible theorems and proofs are described below.

54

AMANDA SERENEVY

Modulus and Tossing Number Solutions

Tossing Number

Modulus 1

2

3

4

5

6

7

8

9

10

1

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

2

Y

N

Y

N

Y

N

Y

N

Y

N

3

Y

Y

N

Y

Y

N

Y

Y

N

Y

4

Y

N

Y

N

Y

N

Y

N

Y

N

5

Y

Y

Y

Y

N

Y

Y

Y

Y

N

6

Y

N

N

N

Y

N

Y

N

N

N

7

Y

Y

Y

Y

Y

Y

N

Y

Y

Y

8

Y

N

Y

N

Y

N

Y

N

Y

N

9

Y

Y

N

Y

Y

N

Y

Y

N

Y

10

Y

N

Y

N

N

N

Y

N

Y

N

Conjectures, Theorems, and Proofs Conjectures are the mathematical equivalent of a hypothesis, or guess. In this case the students’ conjectures will be guesses about patterns that they see in the chart, which they think will continue for numbers larger than 10. There are many patterns they may notice. Some of their conjectures will likely be correct, while others may turn out to be incorrect. A conjecture can be turned into a theorem if the students give a reason why they know that the pattern will continue. The reason needs to be logically complete and convincing to others to be a proof. Some potential conjectures and the overarching theorem for this activity are given below. Refining conjectures sometimes needs to occur during a separate session depending on how quickly the group is moving. Students should first draft their conjectures on their own papers, and then post them where everyone can see them. Gather the students together for a moment, and ask them

BEAN BAG TOSSING

55

to read each conjecture like a lawyer. They should nitpick holes in the way it is written. Is it written using good grammar and spelling? Is it clear and complete? Can they come up with a counter-example — an example that shows that it is wrong? On the other hand, the students may agree with a conjecture as it is written and may think that the writing cannot be improved. Here is one example of a conjecture that students might find. “If the tossing number is 2 and the modulus is even, then not everyone in the circle will catch the bean bag.” That conjecture is clear and complete. When moving on to the proof, some students might say something like “This is true because 2 is a factor of any even number.” That is true, and it even gets to the heart of the matter, but it does not really explain how the bean bag is moving around the circle and why. Push them to write proofs that really explain what is going on and which refer to the numbers in the circles and the tossing process. Here are a few of the smaller conjectures/theorems that students might find. The overarching theorem that determines the answer to this question in general is farther down because students usually do not find that one until later. Theorem: If the modulus is 1, then everyone in the circle will catch the bean bag. Proof: The only number in the circle is a zero, and that person starts holding the bean bag and then tosses it to him or herself, so clearly everyone in the circle will catch the bean bag. Theorem: If the tossing number is 1, then everyone will catch the bean bag. Proof: When you use a tossing number of 1, the bean bag starts at the 0 and is tossed to the next number, which is 1 (unless the modulus is only 1). As the tossing pattern continues, the bean bag never skips anyone in the circle until it returns to 0. Thus, everyone will catch the bean bag. Theorem: If the tossing number is 2 and the modulus is even, then the bean bag will not be caught by all of the people. On the other hand, if the tossing number is 2 and the modulus is odd, then everyone will catch the bean bag. Proof: When tossing by 2s the first time around, the bean bag will land on all of the even numbers and skip the odd numbers. If the modulus is even, then the number right before the 0 will be an odd number, and so the bean bag will not land on it the first time around. Instead, it will land on 0 again and the pattern will start over. Thus, if the modulus is even, the bean bag will never land on any of the odd numbers. On the other hand, if the modulus is odd, then the number right before the 0 will be an even number and the bean bag will land on that number the first time around.

56

AMANDA SERENEVY

The bean bag will then skip over the 0 and land on the 1. During the second time around the circle, the bean bag will land on all of the odd numbers and then return to 0. So all of the people in the circle will catch the bean bag. Theorem: If the modulus is 2 and the tossing number is even, then the bean bag will not land on all of the numbers. On the other hand, if the modulus is 2 and the tossing number is odd, then all of the people will catch the bean bag. Proof: When the modulus is 2 there is only a 0 and a 1 in the circle. Every time the bean bag is tossed 2 starting from 0, it returns to 0. Any even tossing number can be thought of as adding 2 a certain number of times (because any even number e is equal to 2 × n for some whole number n, which can be interpreted as making 2 steps n times). Since each 2-step takes the bean bag back to 0 without landing on 1, it is clear that tossing the entire even number also takes the bean bag back to 0 without landing on 1. On the other hand, any odd number can be thought of as adding 2 a certain number of times followed by adding 1. After tossing all of the 2s the bean bag will be at 0, but adding one more step takes it to 1. Thus, the bean bag will land on both 0 and 1, and so all of the people will catch the bean bag. Theorem: If the modulus is a multiple of the tossing number, and the tossing number is not 1, then the bean bag will not land on all of the numbers. Proof: If the modulus is a multiple of the tossing number (which is not 1), then on the first time around the bean bag will skip all of the numbers which are not multiples of the tossing number and land back at 0 again. This is true because if we applied the tossing number starting at 0 on the number line, we would land on the number representing the modulus (because it is a multiple of the tossing number). If we coil the number line around the circle, the number representing the modulus would land in the same position as the 0. So it is not possible for the bean bag to land on all of the numbers. Students may come up with other conjectures and/or proofs for specific cases. The Tossing Number Theorem The big theorem that determines the answer for every number of people and every tossing number is shown below. When I explore this topic with middle school students, they usually discover this conjecture after making less general conjectures, but we may or may not go through the complete proof depending on their stamina and the amount of time we have.

BEAN BAG TOSSING

57

Two numbers are relatively prime if they have no common factors other than 1. This is a key concept in the main theorem of this investigation. Tossing Number Theorem: When the modulus and the tossing number are relatively prime, all of the people will catch the bean bag. Otherwise, some people will not have a turn to catch it. Note A more technical way to state this theorem is to say that the group of integers modulo m is generated by a number t if and only if m and t are relatively prime. Proof: Suppose that the modulus and the tossing number are not relatively prime. That means that they have some factor f , which is not equal to 1, in common. Because f is a factor of the modulus which is greater than 1, if f was the tossing number, then on the first time around the bean bag would land on all of the multiples of f , and then back at 0. The bean bag would not land on any of the other numbers. Since the tossing number is a multiple of f , the bean bag must land on one of the numbers which is a multiple of f or 0 itself (since we make the tossing number by stringing together some number of f -tosses). No matter how many times we repeat the tossing number, the bean bag can never leave the multiples of f , and so it will never land on the other numbers. Now we need to prove that if the modulus m and the tossing number t are relatively prime, then everyone will catch the bean bag. One way to think about this is to picture the number line being wrapped around the circle over and over again. When the number line is wrapped around the number circle of modulus m, all of the multiples of m coincide with the 0 in the circle, and these are the only numbers which coincide with 0. Claim: the smallest number which is a multiple of both m and t is their product, mt. This is true because of the Fundamental Theorem of Arithmetic, which tells us that each number has a unique prime factorization. This theorem means that each number is built from its prime factors. If m and t have no prime factors in common, then the only way to build a number which can be divided by all of the prime factors in m and by all of the factors in t is to multiply m by t. Because mt is the smallest number which is a multiple of both m and t, if the bean bag starts at 0 on the number line and is tossed t positions each time, it will not arrive back at 0 until we have made m jumps, which is the number of numbers in the circle. Now we need to show that the bean bag could not have landed at the same number more than once during those m tosses. If it had landed on the same number during those m jumps, then there would be a way to travel a distance equal to a multiple of m by using fewer than m jumps of size t. Since there is no multiple of m and t that is smaller than mt, this is not possible. Thus, everyone in the circle will catch the bean bag.

58

AMANDA SERENEVY

0 7

1

6

2 5

3 4

Figure 23. Example with modulus 8 and tossing number 3. Examples Here are some specific examples that illustrate the Tossing Number Theorem. First, let the modulus be m = 8 and the tossing number be t = 3. These two numbers are relatively prime, and so the theorem tells us to expect that all of the people will catch the bean bag. Moreover, because 8 × 3 = 24 is the smallest number which is a multiple of both 8 and 3, we know that the bean bag will move 24 steps around the circle, or in other words it will make 3 revolutions around the circle, before it returns to 0. (See Figure 23.) Now, let us suppose that the modulus is still m = 8, but the tossing number is t = 6. These numbers are not relatively prime, and so we know that some people will not be able to catch the bean bag. Because the common factor of these two numbers is 2, the bean bag can never land in any of the odd-numbered positions and the tossing situation will be essentially the same as if the modulus were m = 4 and the tossing number were t = 3 (which are the numbers we obtain after dividing by the greatest common divisor of the two numbers). (See Figure 24.)

BEAN BAG TOSSING

59

0 7

1

6

2 5

3 4

Figure 24. Example with modulus 8 and tossing number 6.

Modular Arithmetic Solutions Use a circle with modulus 5 (0 through 4). (1) 0 (2) 1 (3) 4 (4) 3 (5) 0 (6) 0. Because 100 is a multiple of 5, and every time we go 5 steps we return to 0. (7) 2. After 145 steps we will be back at 0 because 145 is a multiple of 5. Two more steps will take us to 2. (8) 4. After 280 steps we will return to 0 because 280 is a multiple of 5. Four more steps will take us to 4. (9) 1 (10) 6. If we went 490 steps, we would land back at 0 because 490 is a multiple of 7. That is 8 steps too many, so if we go backwards around the circle 7 steps and then one more, we would land at 6. We could also divide 482 by 7 and then find the remainder. (11) You can divide k by m and find the remainder. The remainder tells you which number you will land on. This works because every m steps bring you back to 0, so any multiple of m takes you to 0 also. The remainder tells us the steps that are left over out of k steps after we go a multiple of m steps. (12) You could start at 0, go 6 in the positive direction, and then go 2 more (landing at 8). Alternatively, start on the 6 and then go 2 in the positive direction on the number line. (13) You could start at 6 and then go 2 on the number line in the negative direction (landing at 4). Alternatively, you could start at the 2 (the

60

(14)

(15)

(16)

(17) (18) (19) (20) (21) (22) (23) (24) (25)

AMANDA SERENEVY

second number) and count the number of steps that you need to walk in the positive direction to reach the 6 (the first number). You could start at 2 and then go 6 on the number line in the negative direction to land at -4. Alternatively, you could start at the 6 (the second number) and count the number of steps that you need to walk in the positive direction to reach the 2 (the first number). You actually need to walk backwards and so you need negative four of these steps. You could interpret this as starting at the 0 and going 6 two times in the positive direction (to land at 12). Alternatively, you could start at 0 and go 2 six times in the positive direction. You could interpret this as starting at the 6 and going by jumps of 2 in the negative direction until you land at 0. Since you need to make 3 jumps, this is the answer. Alternatively, you could start at the 0 and make jumps of 2 in the positive direction until you land at the 6, keeping track of the number of jumps needed. You will still make 3 jumps. 4 2 2 4 8 5 3 7 A modular division problem will have no solution if it is impossible to reach 0 by repeated subtraction of the second number starting from the first number. If the second number is a factor of the modulus and the first one is not, this will happen. For example, 5 ÷ 4 ≡8 has no solution because if you start at 5 and make jumps of 4 in the negative direction, then after one jump you will land at 1 and after two jumps you will land back at 5 again. This pattern will continue forever and it is not possible to land at 0.

Toilet Paper Math Contributed by Phil Yasskin and Dave Auckly This is a collection of problem which use toilet paper as the manipulative. The first block of problems is suitable for middle school students. A later block of problems is suitable for primary school students. We recommend having several rolls of toilet paper on hand to properly get the feel for these problems. Phil Yasskin was inspired by toilet paper prices and combined it with other practical observations on the subject. Dave Auckly heard Phil’s ideas and decided Phil was on a roll. . . (Single ply toilet paper works slightly better for this activity as students won’t be tempted to count ply.)

Toilet Paper Math Student Handout Toilet Paper is a wonderful tool. Consider the following three ways people use toilet paper: The Halves Method: Fold a strip in half, fold the result in half and repeat some number of times. See Figure 25. The End Method: Fold some paper from one end and fold that over repeatedly. See Figure 26. The Chaotic Method: Crumple it up.

Figure 25. The Halves Method

Figure 26. The End Method Further variants are possible. One can arrange the “packet” and then rip it off of the roll, or can rip a strip off and then arrange it. One can have the paper come off the top of the roll, or off the bottom. We will concentrate on the three main headings above without worrying about when the paper is separated from the roll. 61

62

PHIL YASSKIN AND DAVE AUCKLY

I know this is getting rather personal, but: How many of you use the chaotic method? How many use the halves method? How many use the end method? We can make a bar graph on the board (with toilet paper of course!).

Toilet Paper for Middle School Students (1) Most toilet paper comes perforated into sheets. I went to the store to buy toilet paper and saw that the package said “Now 33 13 % more sheets on each roll for the same price.” I bought a pack and took it home where I found my old package and compared the “truth in packaging” labels. I found the new package told the truth. The old package said there were 900 sheets on each roll. How many sheets are on each roll in the new package? Then I looked closer and saw that both packages said the total length of paper on each roll was 10,800 cm. How could both rolls have the same length of paper but one roll have 33 13 % more sheets? What percent shorter is each sheet of paper on the new roll compared with the old roll? (2) I found the sheets of paper on the roll were exceedingly short. So I decided to fold from the end (using the end method ) so each layer would be longer than 1 sheet but shorter than 2 sheets. I folded it 1, 2, 3 times, and miraculously the final edge of the folded paper was exactly at a perforation. Now for the big questions. How long was each layer? How many sheets did I rip off in total? After the first fold, what fraction of the next sheet was covered? I’ll demonstrate if it helps. You may also play with some toilet paper on your own. (3) Going a bit further with the previous questions: (a) How do the answers to the three questions (length of layer, number of sheets, first overlap) change if I fold it 4 times and see that the final edge of the folded paper is exactly at a perforation? 5 times? 6 times? n times? (b) Which rational numbers between 1 and 2 can you produce for the length of a layer by appropriately choosing the number of sheets you start with and the number of times you fold? Let k be the number of sheets and n be the number of folds. (c) Is there more than one way to produce the same rational number by making different choices of k and n? (4) Each group of 3 or 4 people should have a full roll of paper and a hallway long enough to unroll it completely or at least half way. Unroll the full roll of paper and measure its length. Now fold it in half repeatedly. How long is the roll after each fold? How many layers

TOILET PAPER MATH

63

are in the stack after each fold? How thick is the stack after each fold? How many folds can you make before you cannot fold it any more? Why? (If you are using double-ply toilet paper, you can assume you already made 1 fold.) The world record for the number of folds is 13. Watch the YouTube video: https://www.youtube.com/watch?v=ZQ0QWn7Z-IQ

Toilet Paper for All Put a roll of toilet paper on the ground so it can roll. Consider two cases: (1) First, orient the paper so the loose end comes off the top. See the left of Figure 27. (2) Second, orient the paper so the loose end comes off the bottom. See the right part of Figure 27. In each case, what will happen to the roll when you pull on the end? (Don’t pull. Think about it first. Why? Will it do what you say?) Your answer should involve the laws of physics as well as math. Now try pulling the loose end to see if you were right.

Figure 27. The roll

See More Mathematics for a Curious Reader at the end of this section for a more detailed analysis of how the roll rolls.

Toilet Paper for Primary School Students (1) Tape a strip of toilet paper horizontally across the front of the room. Start counting the squares from the left end. Label the numbers you see. (a) Rip off a separate 3 squares. Use them to generate the skip count by 3 (count by 3’s) sequence. (b) Rip off 2 squares and rip off 3 squares. Use these to compute 2 + 3 and 3 + 2. Can you see that 3 + 2 = 2 + 3? Repeat with 3 squares and 5 squares. (2) Tape a strip horizontally across the front of the room. Start counting the squares in each direction from the middle. Label the numbers you see positive on the right and negative on the left.

64

(3)

(4)

(5)

(6) (7)

PHIL YASSKIN AND DAVE AUCKLY

(a) Demonstrate 5 − 2, 6 − 3, and 7 − 4. What do you notice? What would you have if you worked out (5 + 51) − (2 + 51)? How about (5 + DOG) − (2 + DOG)? (b) Demonstrate 2 − 5, 3 − 6, and 4 − 7. What do you notice? What would you have if you worked out (2 + 51) − (5 + 51)? How about (2 + DOG) − (5 + DOG)? If you fold the paper (using the end method) a fixed amount from the end repeatedly: 1 time, or 2 times, or 3 times, or CAT times, how many layers of paper will you have? Make a table (chart) of the numbers of layers as a function of the number of folds. Make a graph of this same data. Describe the shape of the graph. If you fold the paper in half repeatedly (the halves method): 1 time, or 2 times, or 3 times, or CAT times, how many layers of paper will you have? Make a chart (table) of the numbers of layers as a function of the number of folds. Make a graph of the data and describe the graph. We wish to fold packets of paper (using the end method) so each layer is 1 1/2 sheets long. (a) If we want to have 2 layers, how many sheets should we rip off? (b) If we want to have 4 layers, how many sheets should we rip off? (c) Construct three more packets that are 1 1/2 sheets long. For each packet, how many sheets did you rip off and how many layers were there? Make a graph of this data. Describe the shape of the graph. What does this have to do with equivalent fractions? Why is it more convenient to write the mixed number 1 1/2 as the improper fraction 3/2? Demonstrate 3 × 2 in six different ways using toilet paper. Hint: Skip count, make a rectangle or make layers. Take 8 squares, then repeatedly fold (with the end method) a length of 3 squares. How many times can you do so? How many layers will it have? Will this use up all the squares? If not, how many squares will be left? What is this? Generalize to different numbers of starting squares and different layer lengths.

Toilet Paper Teacher Guide Toilet paper is a fun manipulative that has many mathematical possibilities. After running a toilet paper math session in one fourth grade classroom, one of the authors noticed that the students decided to skip recess to continue to play with the problems! We encourage you to see what other problems/classroom applications you can devise for this marvelous tool. Our first application was to make a bar graph. This may be done with magnets on a magnetic whiteboard, or with tape on almost any surface. The result of the poll expressed as a toilet paper bar graph might look like Figure 28.

TOILET PAPER MATH

65

Figure 28. The Toilet Paper Bar Graph

Toilet Paper for Middle School Students Solutions (1) This question is a fairly typical story problem with toilet paper rolled in. When asked how there are 33 13 % more sheets with the same total length, students will immediately answer “They made each sheet shorter!” But when you follow that with “What percent shorter?” they will incorrectly answer “33 13 % shorter.” In fact, the correct answer is 25% shorter.” Here is the reasoning: The original roll had 900 sheets. So the new roll has 900 + 33 13 % × 900 = 1200 sheets. Both rolls have length 10,800 cm. So the length of each sheet of the original roll is 10800 900 = 12 cm while the length of each sheet of the new roll is 10800 = 9 cm. The difference is 12 − 9 = 3 cm. So the percent change is 1200 3 12 = 25%. If the students can handle a little algebra, redo it without numbers. The original roll had n1 sheets. So the new roll has n2 = 43 n1 sheets. Both rolls have length L. So the length of each sheet of the original roll is L1 = nL1 while the length of each sheet of the new roll is L2 = nL2 = 34 nL1 . The difference is L1 −L2 = nL1 − 34 nL1 = 14 nL1 . So the percent change is

1 L 4 n1 L n1

=

1 4

= 25%. Equivalently, since L = L1 n1 = L2 n2 ,

the number of sheets and the length of each are inversely proportional. So when the number of sheets is multiplied by 43 , the length of each sheet is multiplied by 34 . (2) This question is more typical of a math circle question. It is a bit different from a typical story problem and invites exploration, generalization and modification. Perhaps the first thing to notice is that the number of layers is one more than the number of folds, e.g. 0 folds has 1 layer, 1 fold has 2 layers, etc. Since it is folded 3 times there are 4 layers. Since each layer is longer than 1 sheet but less than 2

66

PHIL YASSKIN AND DAVE AUCKLY

sheets, the total ripped off must be between 4 and 8; therefore 5, 6 or 7. These correspond to layer lengths of 5/4, 6/4 = 3/2 and 7/4. After 1 fold, there are 2 layers with a total lengths of 5/2 = 2 12 , 6/2 = 3 and 7/2 = 3 12 . The fraction of the next sheet covered is the remainders: 12 , 0 and 12 , respectively. (3) (a) With 4 folds, there are 5 layers. So the possibilities for (rip off length, layer length) are (6, 6/5), (7, 7/5), (8, 8/5), and (9, 9/5). For 5 folds, with 6 layers, we have (7, 7/6), (8, 8/6), (9, 9/6), (10, 10/6), and (11, 11/6). For 6 folds, we have (8, 8/7), (9, 9/7), (10, 10/7), (11, 11/7), (12, 12/7), and (13, 13/7). In general, with n folds, there are n + 1 layers, so one must rip off between n + 2 and 2n + 1 sheets. It follows that the possible layer lengths are k/(n + 1) with n + 2 ≤ k ≤ 2n + 1. The analysis that we have done is not that sophisticated mathematically, but there are a number of items that are not typical in a middle school math problem: there is no unique answer, the numbers that you are given do not line up well with the numbers that you need to use (number of folds, number of layers). (b) It is possible to produce any rational number between 1 and 2. Indeed write the rational number as k/(n+1) and start by ripping off k sheets, and then fold n times. (c) It is possible to make the same rational number in more than one way. For example one gets a layer length of 3/2 = 6/4 = 9/6 by either ripping off 6 and folding 3 times or ripping of 9 and folding 5 times. Notice that in the first case it lands exactly on a perforation at 2 layers, and then again at 4 layers. Similarly, in the second case it lands exactly on a perforation at 2 layers, 4 layers, and then again at 6 layers. One can also get a layer length of 3/2 by ripping off 3 sheets and folding once to produce 2 layers. In this case it lines up with the perforation only at the last step (which is also the first step!). (4) Assuming each layer of toilet paper is 0.005 cm thick, and the roll is 8000 cm long, the following table encodes the length, the number of layers and the thickness of the stack after each fold using a fairly idealized mathematical model of the situation. Number of Folds

0

1

2

3

..

7

8

..

13

CAT

Number of Layers

1

2

4

8

..

27

28

..

213

2CAT

Length of Layer

8000

4000

2000

1000

..

62.5

31.3

..

.977

8000 · 2−CAT

Thickness of Packet

0.005

0.01

0.02

0.04

..

0.64

1.28

..

41.0

0.005 · 2CAT

We will also comment here that we find it useful to jump from specific numbers in order, to other numbers that are not in order, and

TOILET PAPER MATH

67

then to words as we prepare students to start working with algebra. Hence our “CAT” variable. In practice it is very difficult to fold the packet once it gets as thick as it is wide. This is an example of an applied mathematics problem. In applied math the objectives are different than the objectives in pure math. One is interested in a model that can describe/predict what will happen and maybe design a widget that works better. One may prove some formula based on assumptions used in the model, but sometimes a model will describe a situation more accurately, but one will not be able to rigorously prove the formula or conclusions based on the model. Of course it is pleasing to see an accurate model that can be analyzed rigorously. The first repeated folding model completely ignores the end effects. It captures the qualitative features of the process, but will not produce numbers that agree with experiment. The toilet paper will bend around the ends. Modeling each bend as a semicircle will be more accurate. At the start of the new Millennium, then high school student Britney Gallivan created what is now the generally accepted more accurate model for repeated paper folding, based on the semi-circular ends [1]. One of her equations relates the thickness t, length L, and number of folds n as: π L = t(2n + 4)(2n − 1). 6 Perhaps your students would wish to re-derive this formula. There are many other ways to use toilet paper with middle school and even high school students. Toilet Paper For All Solutions: This is also an example of an applied math problem. It is important to have participants guess what will happen before running the experiment. The guess should come with a reason. This is at the very heart of the scientific method. One develops a hypothesis, perhaps based on a theory, performs an experiment and iterates the process. Here we imagine that the pull on the end is slowly increased until the roll starts to move In each case, there are three behaviors we imagine participants will guess: the roll will roll in the direction of the pull, the toilet paper will unroll but the roll will stay where it is, and the roll will roll in the opposite direction of the pull. In the first case, where the loose end comes off the top (as in Figure 25), the paper rolls in the direction of the pull, since this is both the direction we pull and the way the paper wants to turn along the floor. In the second case, our experiments indicate the behavior depends on the angle the free end of toilet paper makes with the ground. For small angles the roll slides in the direction of the pull, and for larger angles the toilet paper rolls in the opposite direction of the pull. A mathematical analysis of this is given at the end of the solutions section.

68

PHIL YASSKIN AND DAVE AUCKLY

Toilet Paper for Primary School Students Solutions: Toilet paper turns into a flexible number line. In addition to the usual number line explorations, one is able to rip off pieces, fold and rearrange. We explore some of the possibilities in the first couple of questions below. This material may not seem like a math circle to folks in the math circle community. The point to remember is that the material in a math circle does not have to be new. It should just be new to the participants. A six-year old may not have considered equivalent differences before. (1) These two problems deal with the positive part of a number line. (a) See Figure 29 for the toilet paper number line!

Figure 29. The toilet paper number line (b) See Figure 30 for toilet paper addition.

Figure 30. Add and discover 3 + 2 = 2 + 3 (2) These two problems deal with the full number line, positive and negative. We can call these “equivalent difference” problems. (a) See Figure 31 for toilet paper subtraction. We have the concrete example 5 − 2 = (5 + 1) − (2 + 1) = (5 + 2) − (2 + 2). It is possible to put a roll at the end and see that you can add as many sheets to the first number in a difference as you like as long as you subtract the same number of sheets from the second number in a difference and you will still line up in the same place. There is an analogy between the notion of equivalent differences and the notion of equivalent fractions where we would have 5·3 5 = . 2 2·3 After using some small numbers we jump up to larger numbers that serve as place holders. After students see the pattern with large numbers, we use fun words. This prepares students for algebra later in life. (b) The first part had an answer in the positive part of the number line. The second ends in the negative part. Mathematically, negative numbers are defined as equivalence classes of non-negative integers representing equivalent differences in exactly the same way as non-zero rational numbers are defined as equivalence classes of non-zero integers representing equivalent fractions. While everyone who has gone far enough in school has heard of equivalent

TOILET PAPER MATH

69

fractions, this is not the case for equivalent differences. We argue that primary school teachers should have their students explore and discover this idea in the context of subtraction and differences in order to make it more natural in the context of division and fractions later.

Figure 31. Subtract and discover 5 − 2 = 6 − 3 = 7 − 4 (3) Here is the (end) fold table. Number of Folds

1 2 3 .. .. 17

CAT

Number of Layers 2 3 4 .. .. 18 CAT + 1 It is important to get students to make a jump to a higher number to see the pattern. This prepares the way for algebra. Using a cute word, or better yet stick figure is even closer to algebra. The graph appears in Figure 32. It takes the shape of a line. This is why we call such relationships linear relationships.

Figure 32. The (End) Fold Graph (4) Here is the half fold table. Primary school students will not know the term exponential, but they can understand that the numbers get big fast. There will be some six year-olds who are able to double up to 32. Number of Folds

1 2 3 .. ..

17

CAT

Number of Layers 2 4 8 .. .. 217 2CAT The graph of the data is shown in Figure 33.

70

PHIL YASSKIN AND DAVE AUCKLY

Figure 33. The (Half) Fold Graph (5) Fourth graders in one class really loved this problem. Even though it seems fairly standard to folks who have spent time exploring fractions, there is a fair amount to discover, especially for students who have not studied fractions. In particular, students should try to convert between mixed fractions and improper fractions, see decimals, compare different fractions and equivalent ways of writing fractions. (a) We need 3 sheets. Then each layer is 3/2 = 1 12 = 1.5 sheets long. (b) We need 6 sheets. Then each layer will be 6/4 = 3/2 = 1 12 = 1.5 sheets long. (c) Use the fact that 9/6 = 12/8 = 15/10 = · · · (9 squares and 6 layers, 12 squares and 8 layers, 15 squares and 10 layers). Of course this is a model of equivalent fractions. A graph of the numerator (number of squares) vs the denominator (number of layers) for the length of a layer being 3/2 is displayed in Figure 34.

Figure 34. Numerators vs Denominators for 3/2 (6) Skip count in two ways, make a grid in two ways, fold from the end in two ways. Play and see Figure 35. (7) One fold gives two layers. These two layers take up six squares leaving a remainder of two squares. This is division with remainders, of course. See Figure 36.

TOILET PAPER MATH

71

Figure 35. Two Times Three

Figure 36. Eight Divided by Three Rows of squares are useful to explain place value in exploding dots. Or analyze some combinatorial games such as frogs and toads. Toilet paper is cheap and perhaps a bit funny, so it can enliven a math lesson/exploration.

More Mathematics for a Curious Reader Here we describe our model of this system. Consider the limiting case when the end of the toilet paper runs along the floor, so the angle between the floor and the end is 0. It is displayed on the left of Figure 37. When you start pulling, at first nothing will happen. There will be a force that you apply to the end represented by the green arrow, the force due to gravity acting at the center of mass represented by the blue arrow, the reaction force of the floor acting at the point where the roll meets the floor, and the friction force of the floor (both of these last represented by red arrows. For small pulls, the forces will all balance and the roll will not move. When the pull on the end gets larger than the maximal static friction force, the roll will start to slide in the direction of the pull.

Figure 37. Horizontal Pull and Ideal Slanted Pull

72

PHIL YASSKIN AND DAVE AUCKLY

Now consider the diagram for the ideal slanted pull on the right side of Figure 37. In this case we resolve the force along the end of the paper into a horizontal component and a vertical component. The (green) vertical component of the pull plus the (red) vertical reaction force will exactly balance the vertical component of gravity. However, the pull is off to the side compared to the other two forces and this would introduce a torque in the clockwise direction (that is not canceled out by the torque due to friction) inducing the roll to roll without slipping away from the direction of the pull. To be more quantitative, use the following notations: Interpretation Variable P Magnitude of the force (pull) on the end of the toilet paper Mass of the toilet paper m Magnitude of acceleration due to gravity g r Radius of the toilet paper Coefficient of static friction μ Angle between the toilet paper and the floor θP Magnitude of the friction force on the toilet paper F R Magnitude of the reaction force of the floor on the toilet paper The vertical component of the pull is P sin θP upward and the force of gravity is mg downward. Since the toilet paper goes neither up nor down the reaction force must have magnitude R = mg − P sin θP . The standard model of the magnitude of friction is that it is a multiple (the coefficient of static friction) times the magnitude of the reaction force. Thus F = μR = μmg − μP sin θP . Assuming the toilet paper does not roll, it will slide if the horizontal component of the pull is larger than this and not move if it is smaller than this. The horizontal component of pull is P cos θP . Thus the critical amount of pull which will induce the toilet paper to slide Ps is determined by Ps cos θP = μmg − μPs sin θP . This implies that

μmg . cos θP + μ sin θP The counterclockwise torque from the friction is no more that rF = rμR = rμmg −rμP sin θP . The toilet paper will roll forward if the clockwise torque from the pull (rP ) exceeds this. Call the critical point where the toilet paper will start to roll Pr . It is determined by Ps =

rPr = rμmg − rμPr sin θP .

TOILET PAPER MATH

73

One solves to obtain

μmg . 1 + μ sin θP As 1 ≥ cos θP one has Pr ≤ Ps and concludes that the toilet paper will always start to roll first when the angle θP is positive. This all sounds very convincing. The only problem is that this is not what happens when one pulls the end of the toilet paper at a moderate angle (even 15◦ off of the floor.) As is often the case in applied mathematics, we have to refine our model. To do so, we introduce the notion of stability. A system is said to be in equilibrium if it will not change. The equilibrium is stable if the system will remain near the equilibrium even after a small change of state. It is unstable otherwise. A zero mode is an unstable equilibrium that is nearly stable. These are all illustrated for a ball on various surfaces in Figure 38. The point is that the only equilibria one observes in day to day life are stable equilibria. Life will conspire to disturb the unstable ones. A perfectly round roll would represent a zero mode. If we really had a perfectly round roll, if anything bumped it, then it would just roll off. This is not what we observe with a roll of toilet paper. These rolls will sit still, thus we conclude that they are in stable equilibria. Pr =

Figure 38. Stability We conjecture that a toilet paper roll has subtle “flat spots” and it rests in one. Thus, when we pull up on the end at a slight positive angle the reaction force from the table will act on the front of the roll as in Figure 39. To model this, we do not need to carefully model the geometry of the “flat spot.” It suffices to assume that the angle between vertical and the point where the reaction force and friction force acts is θc displaced in the direction the toilet paper would roll. At first nothing will happen. Then the horizontal component of the pull will over come the static friction, but the vertical components will remain balanced, and the total torque will remain zero, so the roll will slide in the direction of the pull. When the angle from the floor passes a critical level, the offset from the flat spot will not be sufficient to balance out the torque and the roll will roll away from the direction of the pull. We can now revise our prior analysis. The horizontal components and critical pull for a slide will not change. The counterclockwise torque generated by the friction will only see

74

PHIL YASSKIN AND DAVE AUCKLY

Figure 39. Correct Slanted Pull the component of the friction force tangent to the circle and is thus rF cos θc . The reaction force will now also induce a counterclockwise component of torque corresponding to the tangential component of the reaction force: rR sin θc . Thus the critical amount of pull for the toilet paper to start rolling is determined by rPr = rF cos θc + rR sin θc = rRμ cos θc + rR sin θc = r(μ cos θc + sin θc )(mg − Pr sin θP ). Solving gives mg(μ cos θc + sin θc ) . 1 + sin θP (μ cos θc + sin θc ) Notice that this agrees with the earlier expression in case θc = 0. The toilet paper will start move by sliding if Ps < Pr and start to move by rolling if Ps > Pr . Call the angle off the floor θP where these quantities the transition angle θt . We find it by solving for θP in Ps = Pr : μ cos θt = . μ cos θc + sin θc This is equal to one at θc = 0, but it is also a decreasing function around θc = 0. We conclude that Pr =

The toilet paper will first move by sliding if the pull angle satisfies θP < θt The toilet paper will first move by rolling if the pull angle satisfies θP > θt . This agrees with observation. There are similar questions one may analyze. If the end of the toilet paper comes off the top as in Figure 25 and one pulls the end some length, how far will the toilet paper roll? Instead of slowly increasing the pull force, imagine suddenly applying a large force. What will happen? This may even be analyzed when the toilet paper is on a spindle. Will it unroll, or rip off?

References [1] Pomona Historical Society Folding Paper in Half 12 Times: The story of an impossible challenge solved at the Historical Society office. 2005. http://web.archive.org/web/ 20170119085235/http://pomonahistorical.org/12times.htm

Golomb Rulers Contributed by Elgin Johnston Short Description: This activity requires very little mathematical background. It provides students with wonderful opportunities to experiment with pencil and paper and to collaborate in teams, to explore and discuss new problems with other students. The activity leads students to work with triangular numbers and to an example of “proof by contradiction.” The activity also exposes students to easy-to-understand but unsolved problems on which mathematicians are actively working. Materials: Chalk board with colored chalk or whatever is appropriate for your classroom. Some rulers for students to use as a manipulative can be helpful. Also prepare several strips of paper that students can use to make their own rulers. Mathematics Beneath & Beyond: The Golomb ruler is named for mathematician Solomon Golomb, though the idea was first investigated by S. Sidon (1932) and Wallace C. Babcock (1953).4 This topic is very rich and provides connections to various mathematical ideas and fields such as graph theory, combinatorial designs, and construction of affine and projective planes. Golomb rulers have applications in information theory, error correcting codes, and the design of phased arrays for radio telescopes. References/Authorship: This activity is inspired by and based on the article “Golomb Rulers” by Roger C. Alperin and Vladimir Drobot that appeared in the February 2011 issue of Mathematics Magazine.

4 Thus, the Golomb ruler is an excellent example of “Stigler’s Law of Eponymy”, which states that no scientific discovery is named after its original discoverer.

75

76

ELGIN JOHNSTON

Golomb Rulers Student Handout Warm-up problem: Markus Metrik, the Chief Executive Officer of the Four Corners Ruler Company wants to cut production costs for the 12 rulers that his company manufactures. He tells the Board of Directors “A standard 12 ruler is a useful tool, but in a sense it is also inefficient. For example, to measure a length of 6 you do not need a 6 mark on the ruler. If this mark were removed you could still measure 6 by measuring from the 2 to the 8 mark. And any other whole number of inches from 0 to 12 can be measured without the 6 mark.” One example of a ruler with fewer marks is shown in Figure 40.

Figure 40. A ruler with seven marks (including the ends of the ruler.) Can every length from 0” to 12” inches be measured with this ruler? Markus challenges his Research Department to construct the most efficient 12 ruler. What is the minimal number of marks on the ruler, and where should they be, if we want to measure every integral distance from 0 to 12 ? By the way, we will count the two ends (0 and 12 ) as two of the marks. (Thus a ruler with only a 5 mark will be counted as having three marks: 0 , 5 , and 12 .) (1) For the ruler that you designed in the Warm-up, how do you know that it is the most efficient? Can you give an argument to convince someone that with fewer marks you cannot measure all distances from 1 to 12 ? (2) For the ruler you have designed, are there any distances that can be measured in two or more ways? How many such distances are there? (3) Suppose you want a 12 ruler that measures any integer distance only once, even if there are some distances it cannot measure. What is the greatest number of different distances that can be measured under these conditions and what might such a ruler look like?

GOLOMB RULERS

77

Definitions and Notation: The largest distance that can be measured by a ruler is called the length of the ruler. The number of marks on the ruler (including the two ends) is called the size of the ruler. We will use the notation {a1 , a2 , . . . , am } with 0 = a1 < a2 < . . . < am = L to describe a ruler of length L and size m (i.e., with m marks). Thus, we denote the ruler in Figure 40, which is not a Golomb ruler, by {0, 2, 4, 5, 8, 10, 12}; the length of this ruler is 12 and its size is 7. A ruler is called a Golomb ruler if each distance that it measures can be measured in only one way using two of the marks on the ruler. For example, {0, 5, 12} is a Golomb ruler of length 12 and size 3. Notice that for a Golomb ruler of length L there might be integer distances less than L that cannot be measured, but any distance that can be measured can be measured in exactly one way. (4) A Golomb ruler of length L is called maximal if it has the largest possible number of marks for that length. What is the largest possible number of marks on a Golomb ruler of length 12 ? Is there an example of such a ruler? Answer the same question for rulers of length 5 , 6 , 8 , 11 , and list all possible maximal Golomb rulers for these lengths. (5) A Golomb ruler of length L is called perfect if it can be used to measure each of the distances 1 , 2 , 3 , . . . , L . How many perfect Golomb rulers can you find? Can you find all perfect Golomb rulers? Why or why not? (6) Is there a perfect Golomb ruler of infinite length? Such a ruler has one end (the 0 mark) and extends infinitely in one direction. Is there a way to put infinitely many marks at some of the integer points so each positive integer distance can be measured in exactly one way?

Definition. Now suppose that we wish to make a Golomb ruler with m marks. There are many such rulers but we are interested in the ruler of minimal length. We denote the length of this ruler by G(m).

(7) Calculate G(2), G(3), G(4), and G(5). (8) For a positive integer m > 5 what can you say about G(m)?

78

ELGIN JOHNSTON

Golomb Rulers Teacher Guide Warm-up Problem Solution: This is a good activity for teams of 3 or 4 students. The optimal ruler has 6 marks (two of which are the ends.) Some groups will likely think they are done with a 7 or 8 mark ruler. Encourage them to keep working, and if they are adamant that a 7 or 8 mark ruler is the best they can do tell them that they can do better. Give students 12 to 15 minutes on this problem. When they find a 6 mark solution, ask them how they know that 6 is the best and that no efficient 5 mark ruler exists. Challenge them to find other 6 mark solutions. Have students present their optimal ruler(s). Also have students make any observations about the rulers and ask if there were any strategies that they employed in constructing the rules. A couple of important points are: • If an efficient ruler is reflected about the center point, then we get another efficient ruler. For example the ruler with marks {0, 1, 2, 3, 8, 12} when reflected becomes the ruler with marks {12 − 12, 12 − 8, 12 − 3, 12 − 2, 12 − 1, 12 − 0} = {0, 4, 8, 10, 11, 12}. • A good strategy is to put on marks to measure the longer lengths first. Starting with the marks at the ends, 0 and 12 , we know that we must be able to measure 11 . Thus we must have a mark at 1 or 11 . These are symmetric situations so we can assume we have a mark at 1 . To measure 10 we need a mark at 2 , 10 , or 11 . For each of these cases there are only two more marks to place to make a 6 mark efficient ruler. It is easy to run through the possibilities. There are 14 six mark rulers that measure every integral distance from 0 to 12 . Seven of these mark sets are given here, and the other seven are obtained by reflection: {0, 1, 2, 3, 8, 12}, {0, 1, 2, 6, 9, 12}, {0, 1, 3, 5, 11, 12}, {0, 1, 3, 7, 11, 12}, {0, 1, 4, 5, 10, 12}, {0, 1, 4, 7, 10, 12}, {0, 1, 7, 8, 10, 12}.

Problem Solutions: (1) The key to answering this problem is to ask: “Suppose we had only five marks. How many distances could we measure?” To get students thinking along these lines have them consider a ruler of length 3 and ask: ”How many marks are needed to measure the distances one, two, and three inches?” Students should have no trouble answering three (with mark sets {0, 1, 3} or {0, 2, 3}.) Then ask: “How do you know you cannot do it with two marks?” This line of questioning usually gets students to think about how many distances can be measured with a set of marks.

GOLOMB RULERS

79

If we have a five mark ruler, say, for example {0, 2, 7, 11, 12}, then using the 0 mark as one mark, we can measure 4 distances. If use the 2 mark, there are at most 3 more (new) distances that can be measured. With 7 as one of the marks, we have at most 2 more new distances and with 11 as a mark, at most one more. Thus, with five marks we can measure at most 4+3+2+1 = 10 distances. So with just five marks on the ruler, it is impossible to measure the twelve integral lengths from 1 to 12 . Ask a couple of follow up questions about the maximum number of distances that can be measured with a four mark ruler (3 + 2 + 1 = 6) or a six mark ruler (15). Students may recognize these numbers as triangular numbers (See the Mathematics Beneath & Beyond section for more information.) Familiarity with triangular numbers can be useful with later problems in this activity so you are encouraged to spend a little time discussing these. For more advanced students you might bring in the combinatorial idea: The maximum possible number of different distances that can be measured with an m mark ruler is equal to the number of ways to select a pair of marks from the set of m marks:   m m(m − 1) = 1 + 2 + 3 + · · · + (m − 1). C = = m 2 2 2

(2) From the work in Problem 1, we know that with six marks we can measure between pairs of marks in 5 + 4 + 3 + 2 + 1 = 15 ways. Because we know our ruler measures all distances from 1 to 12 , three of the 15 distances must be duplicates. For example, for the ruler with marks {0, 1, 2, 3, 8, 12} the distance 1 can be measured in 3 ways (for two duplicates) and the distance 2 can be measured in two ways (for one more duplicate). Each of the distances from 3 to 12 can be measured in exactly one way for this ruler. (3) For applications in information and coding theory, and in the design of radio arrays, the duplication we observed in Problem 2 should be avoided if possible. As seen in Problem 2, if we have 6 marks on our 12 ruler then there must be duplication in the distances that can be measured between two marks. However, if we have only 5 marks, then we will be able to measure at most 4 + 3 + 2 + 1 = 10 distances, so it may be possible that all of these distances can be different. It is easy to construct such examples in this case. For example take the 6 mark ruler {0, 1, 4, 5, 10, 12} and erase the 5 mark to get {0, 1, 4, 10, 12}. This set measures the ten distances 1 , 2 , 3 , 4 , 6 , 8 , 9 , 10 , 11 , 12 and each of these distances is measured exactly once.

80

ELGIN JOHNSTON

Suppose that a Golomb ruler of length L has m marks. We first count the number of distances that are measured. Starting from the first mark (e.g. the left end) we can measure to any of the other m − 1 marks. If we start at the second mark, then we can measure from it to any of the m − 2 marks to its right, for m − 2 new distances. (Note, we do not consider the distance from the second to the first mark because that distance was already measured in the first part of the count.) From the third mark from the right we can measure m − 3 new distances. Continuing this process we conclude that with an m mark Golomb ruler we can measure m(m − 1) (m − 1) + (m − 2) + · · · + 3 + 2 + 1 = 2 different distances. However, because the ruler has length L, all distances measured must be from the set {1, 2, 3, . . . , L − 1, L}, that is, no distance greater than L can be measured. Thus the total number of distances measured cannot exceed L, which implies that m(m − 1) ≤ L. 2

(4) If a Golomb ruler of length L has m marks, then it must be the case that m(m − 1) ≤ L. (m − 1) + (m − 2) + · · · + 3 + 2 + 1 = 2 Because 3 + 2 + 1 = 6 > 5, a Golomb ruler of length 5 can have at most 3 marks. The three mark Golomb rulers of length 5 are {0, 1, 5}, {0, 2, 5}, and their reflections, for a total of four such rulers. The Golomb rulers of length 6 have four marks and are {0, 1, 4, 6} and its reflection {0, 2, 5, 6} for a total of two such Golomb rulers. A Golomb ruler of length 8 can have at most four marks and rulers of length 11 or 12 can have at most five marks. The maximal Golomb rulers of length 8 are {0, 1, 3, 8}, {0, 1, 5, 8}, {0, 1, 6, 8}, {0, 2, 3, 8}, and their reflections for a total of eight. The maximal Golomb rulers of length 11 are {0, 1, 4, 9, 11}, {0, 3, 4, 9, 11} and their reflections for a total of four such rulers. We saw an example of a 5 mark Golomb ruler of length 12 in the discussion of Problem 3. (5) If a perfect Golomb ruler has length L and m marks, it must measure each of the L distances 1, 2, 3, . . . , L exactly once. Because an m mark perfect Golomb ruler must measure m(m − 1) (m − 1) + (m − 2) + · · · + 3 + 2 + 1 = 2

GOLOMB RULERS

81

distances uniquely, it must be the case that L = m(m−1) is a triangular 2 number. Thus any perfect Golomb ruler must have length equal to a triangular number, that is L ∈ {1, 3, 6, 10, 15, 21, 28, . . . }. It is easy to produce perfect Golomb rulers of length 1, 3, or 6, but there are no finite length perfect Golomb rulers of length greater than 6. A perfect Golomb ruler of length 10 has marks at 0 and 10 and, to measure a length of 9 , must have a mark at 1 or 9 (but not both.) We may assume the mark is at 1 since the choice of 9 is a reflection of this choice. To measure 8 we need mark pairs (0, 8) or (1, 9) or (2, 10), so must add a mark at 2 , 8 , or 9 . The choices 2 or 9 each result in two ways to measure one inch so they must be discarded. With a mark at 8 our ruler now has marks at 0, 1, 8, 10. It is easy to check that there is no way to place the fifth point to make a Golomb ruler. This proves that there is no perfect Golomb ruler of length 10. A similar argument can be used to show that there is no perfect Golomb ruler of length 15. The general argument is similar. Consider the construction of a perfect Golomb ruler of length L > 15. As in the previous argument we can start with marks at 0, 1, L, allowing us to measure lengths of 1 , (L − 1) and L . To measure (L − 2) we need one of the mark pairs (0, L − 2), (1, L − 1), or (2, L). Only the addition of a mark at L − 2 avoids a duplication. With mark set {0, 1, L − 2, L} we can measure distances of 1 , 2 , (L−3) , (L−2) , (L−1) , L . To measure (L − 4) we need one of the mark pairs (0, L − 4), (1, L − 3), (2, L − 2), (3, L − 1), or (4, L). Only the addition of a mark at 4 does not result in duplication. We now have the mark set {0, 1, 4, L − 2, L} and can measure distances of 1 , 2 , 3 , 4 , (L − 6) , (L − 4) , (L − 3) , (L − 2) , (L − 1) , L . (These marks are all distinct if L − 6 > 4, that is, if L > 10 but are not distinct if L = 10.) Now we add a mark to measure (L−5) . We need one of the mark pairs (0, L−5), (1, L−4), (2, L−3), (3, L − 2), (4, L − 1), or (5, L). A quick check shows that adding any of the listed points not already in the mark set leads to a duplication. This shows that there are not perfect Golomb rulers of length 10 or longer. (6) Rather surprisingly there is an infinite length perfect Golomb ruler. Our construction of such a ruler demonstrates how to find as many marks as we want on such a ruler, though it does not give us a complete description of the set of all marks. Our construction is in a series of steps. Step 1. Start with marks at 0 ,1, and 3 so our mark set at this point is {0, 1, 3}. Note with this set we can measure each of the distances 1 , 2 , 3 .

82

ELGIN JOHNSTON

Step 2. We note that the smallest distance that we cannot measure is 4 , so we add marks 3 · 3 = 9 and 3 · 3 + 4 = 13. Our set of marks with these marks we can measure the is5now {0, 1, 3, 9, 13} and  , 3 , 4 , 6 , 8 , 10 , 12 , 13 , and none , 2 = 10 distances 1 2 of these distances can be measured in more than one way. Step 3. After Step 2, the smallest distance that we cannot measure is 5 and the largest mark we have made is 13. We next add marks 3 · 13 = 39 and 3 · 13 + 5 = 44. We now have the mark set {0, 1, 3, 9, 13, 39, 44}. With this mark set we can now measure a distance of 5 and it is not hard to check that the set measures  7 2 = 21 different distances with no repetitions. .. . Step k + 1. that after Step k ≥ 2 our (2k + 1)-mark set mea  Suppose different distances (with no repetitions.) Let d be sures 2k+1 2 the smallest distance not measured by this set and let mk be the largest mark in this set. Note that d < mk because mk > 6 and there is no perfect Golomb ruler of length greater than 6. Add the two marks 3mk and 3mk + d to obtain the new (2k + 3)-mark set {0, 1, 3, . . . , mk , 3mk , 3mk + d}. With this new mark set we can measure distance d, so the smallest distance not measured is now greater than d. In addition, as we show below, any distance measured by this mark set can only be measured in one way. We know that there are no repeated distances measured using only the mark set produced in Step k. If a, 0 ≤ a ≤ mk is a mark from the Step k set {0, 1, . . . , a, . . . , mk }, then 3mk − a ≥ 2mk

and (3mk + d) − a > 2mk ,

so neither of these distances is measured by the Step k mark set (which can only measures distances ≤ mk ). Thus distances measured by one of the new points and one of the points in the Step k set are new. Furthermore, no two such measurements are the same. If two of these measurements were equal then there would be two marks a, b from the Step k set with 3mk − b = (3mk + d) − a

which leads to

a − b = d.

However this is impossible because distance d was the smallest distance not measured by two marks in the Step k set. This completes the construction and establishes the existence of an infinite perfect Golomb ruler. The mark set for this ruler can be described as S = (Step 1 set) ∪ (Step 2 set) ∪ · · · (Step k set) ∪ · · · .

GOLOMB RULERS

83

To see that this mark set does result in a perfect Golomb ruler, note that (a) Every distance can be measured. In particular if d > 0 is an integer, then distance d is measured using marks from the Step d mark set. (b) Every distance is measured in only one way. Pick any three or four marks w < x, y < z and consider the distances x − w and z − y. Find k greater than each of w, x, y, z. Then x, y, z, w are in the Step k mark set. Because this mark set does not repeat any distances, x − y = z − y. Thus, the marks in S measure no distances in more than one way. (7) The (very hard) problem of determining the value of G(m) for large values of m is a problem on which some mathematicians are working. There are no known efficient methods or algorithms for determining G(m) for these large values. Many of the recent values have been found through “distributed computing,” that is, by having many computers involved in the investigations, each computer working on some small part of the larger problem. The value of G(2) is the length of the shortest Golomb ruler with two marks. Thus G(2) = 1 and the only example of such a ruler is the one with mark set {0, 1}. Next, we have G(3) = 3. The ruler with mark set {0, 1, 3} is a perfect Golomb ruler of length 3, and there is no three mark Golomb ruler of length 2. Similarly, G(4) = 6 and we have seen the four mark perfect Golomb ruler of length 6 and having mark set {0, 1, 4, 6}. With four marks we have 6 pairs to measure distances. Thus there is no four mark Golomb ruler of length less than 6. We have seen that a five mark ruler measures 10 distances and that there is no perfect Golomb ruler of length 10. Therefore G(5) ≥ 11. There is a five mark Golomb ruler of length 11: its mark set is {0, 1, 4, 9, 11}. different distances, so (8) An m mark Golomb ruler must measure (m−1)m 2 (m−1)m . If m ≥ 6, then (m−1)m ≥ 15. must have length equal to at least 2 2 Because there are no perfect Golomb rulers of such lengths, it must be the case that m2 − m + 2 (m − 1)m +1= . G(m) ≥ 2 2 Further Explorations: This section contains problems that the author has not investigated. Math Circlers might have fun exploring these questions.

84

ELGIN JOHNSTON

An Efficient Circular Protractor. Figure 41 shows a circular protractor that can measure any angle that has measure a multiple of 30◦ . What angle markings could be used for a most efficient such protractor?

Figure 41. An inefficient circular protractor. One or Two Measurements Allowed. If a 12 ruler has marks at 1 and at 9 , then you can measure 1 and can measure 4 by first measuring 1 and then measuring 3 . If you are allowed one or two measurements to measure a distance, what markings could be used for a most efficient such ruler? Welcome to the Fold. Suppose your 12 ruler is a strip of foldable paper. As seen in Figure 42 by folding at the 9 mark you can generate a length of 6 . Figure 43 shows that if you fold to the 9 mark you can produce a length of 1.5 ! What questions can you ask (and answer) about folding rulers?

Figure 42. Fold at the 9 mark to measure 6 .

Figure 43. Fold to the 9 mark to measure 1.5 .

GOLOMB RULERS

85

More Mathematics for a Curious Reader: • The Wikipedia page on Golomb Rulers [2] provides a table of known values and examples of G(m) for large values of m. See also the Wolfram Math World article [3] and the DataGenetics blog entry on this topic [4]. • With n + 1 marks on a ruler, we can measure at most n(n + 1) (1) Tn = n + (n − 1) + · · · + 2 + 1 = 2 th different distances. The number Tn is the n triangular number. These numbers are so named because they manifest in sums of objects in a triangular array. For example Figure 128 illustrates that T6 = 1 + 2 + 3 + 4 + 5 + 6 = 21.

Figure 44. The 6th triangular number T6 . The closed form for Tn can be derived by writing the sum of Tn in two ways, then adding the numbers in pairs, Tn

=

1

+

2

+

3

+

...

+

(n − 1)

+

n

+Tn

=

n

+

(n − 1)

+

(n − 2)

+

...

+

2

+

1

=

(n + 1)

+

(n + 1)

+

(n + 1)

+

...

+

(n + 1)

+

(n + 1)

=

n(n + 1)

2Tn

n(n + 1) . 2 The triangular numbers show up (sometimes surprisingly) in many problems. As an example, consider the handshake problem: In a room with n people, each person shakes hands with every other person. How many handshakes take place? Number the people 1, 2, 3, . . . , n. Person 1 shakes hands with each of the other for (n − 1) handshakes. Person 2 shakes hands with persons 3, 4, . . . , n (having already shaken with person 1) for a total of n − 2 more handshakes. Person 3 shakes with 4, 5, . . . , n for n − 3 handshakes. Finally, person (n − 1) shakes with person n. This gives a total of (n − 1) + (n − 2) + · · · + 1 = Tn−1 handshakes. Triangular numbers are just one example of polygonal numbers, that is numbers that correspond to the number of dots in certain polygonal configurations. There are square numbers (which are well known), Dividing by 2 we find Tn =

86

ELGIN JOHNSTON

pentagonal numbers, hexagonal numbers, etc. You can have your students figure out the first few such numbers and perhaps come up with formulas for some of these numbers. Or students might research these numbers on their own, starting with the Wolfram Math World article [5]. • In the discussion of the solution to Problem 7 we noted that the problem of determining G(m) for large m is a very hard problem. In fact, mathematicians have a way of classifying some problems by the length of time it takes to solve them. There is a conjecture that the problem of determining G(m) is NP-hard (NP stands for Non-Polynomial deterministic.) This classification means that as m grows, the time needed to determine G(m) grows much much faster. Presentation Suggestions: Prepare enough handouts for all participants. Consider distributing the handout at the end of the session, and instead prepare a PowerPoint (or other) presentation so you can present the questions one at a time to pace the session and make sure all students are working on the same problem. You might also consider producing extra copies of the Teacher Guide for distribution to interested parents or teachers. • Start by telling the story in the Warm-up Problem, then have students break into teams to come up with the most efficient 12” ruler. Give students 12 to 15 minutes to work on this and check in with each team a couple times during the discussions. It is not unusual for students to claim that an 8 or 7 mark ruler is the most efficient. Have them keep working, perhaps announcing the “best result so far” to keep teams trying to do better. After most of the groups have found an efficient 12 inch ruler, have groups present their solutions. Likely there will be different solutions. This can lead to a discussion of how many such rulers there are. In addition students will likely notice that a reflection an efficient ruler gives another efficient ruler. • The problem of proving that there cannot be an efficient 5 mark 12 ruler is harder. Read the solution to the warm-up problem for possible questions to ask students to get them thinking about how many distances can be measured with 5 marks. • One of the most important and difficult things that mathematicians do is ask questions. Give each group (or individual) 5 minutes to come up with a question about rulers. Write the questions on the board. It may well be the case that students ask some of the questions that you intend to pursue during the activity. (This is why I prefer to not hand out the questions until the end of the session. Students will feel more ownership if they are working on questions that they have formulated. And of course you can always ask questions from the handout if needed.)

GOLOMB RULERS

87

• When students try to construct Golomb rulers or perfect Golomb rulers they generally start by placing marks to measure distance of 1, then distance of 2, etc. This leads to many possibilities and it can be hard to keep track of progress towards the best selection of marks. Suggest to students that they find marks for the longer distances first, and ask them why this might be an easier way to start. • Question 6 (the infinite Golomb ruler) is not difficult but can be challenging for students to understand. Give students 10 to 15 minutes to think about this problem and try constructing some initial marks for such a ruler. Have teams present their ideas and have other students ask questions and possibly weigh in on whether or not the construction will work. If some team comes up with a working construction, great! Otherwise it can be very educational to work through the presented solution with the class as a whole. Take the students through the first three steps of the construction, then ask them to do the next two steps. After these discussions they might be ready to describe the general k th step on their own. Once the students seem to understand the construction, ask them how they would convince someone that the marks produced really do give an infinite ruler that measures each positive integer distance in exactly one way.

References [1] Alperin, Robert and Drobot, Vladimir. “Golomb Rulers”, Mathematics Magazine, Vol. 84, February 2011, pp. 48-55. [2] “Golomb Ruler” Wikipedia page. https://en.wikipedia.org/wiki/Golomb_ruler [3] “Golomb Ruler” Wolfram Math World article. http://mathworld.wolfram.com/ GolombRuler.html [4] “An Inefficient Ruler” Data Genetics blog entry. http://datagenetics.com/blog/ february22013/index.html [5] “Polygonal Number” Wolfram Math World article. http://mathworld.wolfram.com/ PolygonalNumber.html

The Cookie Monster Problem Contributed by Gabriella Pinter Short Description: This session is a wonderful example of a problem with a low entry point and deep mathematics. Students who have just learned to count can play with it, yet there is sufficient depth to engage graduate students. The problem also provides a great example of a compelling story for the intended audience. In the session you let participants experiment and practice unstructured problem solving. Students can make up and test conjectures, and can develop different strategies for the problem. By asking more and more questions, students can see how a simple problem can evolve and lead to open-ended explorations. Materials: • Chalk board with colored chalk, or white board with colored pens, or similar to display computations to participants; • Writing supplies (including two colors) for participants; • Participant hand-out. For younger audiences it is helpful to have 15 jars or cups labeled with the numbers 1–15 and 120 cardboard cookies or other counters. For the youngest audiences, each group of three students could use a set of six jars labeled 1–6, and 21 cardboard cookies. Mathematics Beneath & Beyond: This set of problems can be used in a general problem solving circle, and illustrates many problem solving strategies: experimentation, trying with smaller numbers, formulating conjectures, proofs, generalization, and asking new questions. The problems can be done at different grade levels, since even elementary school students can experiment with different steps that empty jars. They can ask their own questions too, that may lead the discussion in different directions. It is very informative for students to realize that such a simply stated, understandable problem can lead to an unsolved question in mathematics. Binary representation of non-negative integers comes up naturally in the original problem, but it is certainly not a ‘prerequisite’ for students to know. c 2019 by the author 89

90

GABRIELLA PINTER

They can be introduced to this representation through the problem. A nice way to do this with very young students is illustrated in Figure 45.

Figure 45. Binary Stars Here we start with thirteen stars. They are bagged by twos in green bags, then green bags are put by twos in yellow bags, and yellow bags are put by twos in blue bags. Once we no longer have a pair of bags to make a larger bag, the procedure stops, and we tally what we end up with. In this case, 1 blue bag, 1 yellow bag, 0 green bags, and 1 star. Considering how many stars are in each type of bag, this shows that 13 = 1·8+1·4+0·2+1·1. The original cookie jar problem [3] has been extended in many different ways, and has many connections to research problems. In particular, high school students and their mentors in MIT’s Primes and summer research programs published some interesting extensions to what happens when the number of cookies in the jars are Fibonacci numbers [4]. Indeed, some ideas for the storyline are adapted from this paper. Another paper investigates the properties of a set of m jars that can be emptied in less than m steps [8]. In yet another paper, the author considers sets of jars that can be emptied in two or three steps, and describes connections of the problem to Wythoff’s game [7].

THE COOKIE MONSTER PROBLEM

91

The Cookie Monster Problem Student Handout Cookie Monster’s Mommy is very organized when it comes to cookies. She labeled 15 jars on a shelf with numbers 1–15, and placed one cookie in the first jar, two cookies in the second, three in the third, and so on, finishing with fifteen delicious cookies in jar number 15. Cookie Monster — who loves cookies — can get some cookies from the jars every hour, as long as he obeys his Mommy’s rule: he can choose some jars on the shelf, and he has to take the same number of cookies from each of his chosen jars. For example, the first hour he could choose the jars numbered 3, 4, and 8, and take two cookies from each, but if he is hungrier, he could take three from each (but not more, since jar number 3 has only three cookies). Cookie Monster would like to eat all the cookies. Can you help him figure out how to choose the jars and the number of cookies to take from them each hour? Explore different ways of emptying all the jars. How many hours do the cookies last? (1) Cookie Monster would like to eat the cookies as fast as he can, but he needs to obey his Mommy’s rule. How should he choose the jars, and how many should he take from his chosen jars each hour? (2) Cookie Monster’s Mommy is very disappointed that the cookies did not last even for half a day, so next time, she doubles the number of jars and for good measure adds one. Now there are 31 jars with 1, 2, . . . , 31 cookies in them. How long will it take for Cookie Monster to eat all the cookies, if he tries to eat them as fast as he can? How long would 7 jars filled with 1, 2, . . . , 7 cookies last? (3) What is the shortest time it takes to empty all the jars according to Cookie Monster’s Mommy’s rule, if there are n jars with 1, 2, . . . , n cookies in them? (4) Cookie Monster’s Mommy would like to have her cookies last as long as they can. She would like to fill just three jars so they last for at least three hours (that is, they cannot be emptied in less than three hours according to the rules, even if Cookie Monster is really, really hungry and clever). Can you help her? What if she would like to fill four jars to last for at least 4 hours? (5) Can you fill n jars in some way so that they are guaranteed to last for at least n hours? (6) How fast can Cookie Monster empty jars that contain {1, 5, 33, 36} or {2, 6, 7, 13}, or {33, 34, 36, 40, 48}, or {1, 2, 4, 11, 16, 17} cookies? What kinds of strategies could you use for emptying jars in these cases? (7) What is the shortest time required to empty k jars containing different numbers of cookies? (8) Is there a strategy for emptying jars that is optimal no matter how many jars we have with cookies in them?

92

GABRIELLA PINTER

(9) What if ‘negative cookies’ can also be taken — that is, cookies can be added to certain jars, but the same number to each of them. How fast can jars with {1, 2, 4, 8, 16} be emptied in this case? (10) Cookie Monster’s Mommy proposes a game. She fills two jars with 9 and 10 cookies, respectively. She and Cookie Monster take turns removing cookies from the jars still keeping to the rule. (In this case, they either take any number of cookies from one jar, or take the same number of cookies from both jars.) The one who cannot take any more cookies loses the game. If you were Cookie Monster would you like to start or go second in this game? (11) Cookie Monster’s Mommy now has so many jars that they do not fit on any shelf. So she arranges them in a rectangular array on a table as in the left panel of Figure 46. However, now she needs new rules! Let us say that a “step” consists of blocking consecutive jars from the left and/or from the right in any number of rows, and then removing the same number of cookies from all ‘unblocked’ jars. For example, in the right panel of Figure 46 jars with an X are blocked, so cookies will not be taken from them. In each step, different jars can be blocked from the left and/or from the right, and then the same number of cookies are removed from all jars that are not blocked.

Figure 46. Cookie Jars on a Table How many steps does it take according to these rules to empty all the jars?

THE COOKIE MONSTER PROBLEM

93

The Cookie Monster Problem Teacher Guide Solutions: (1) Try the problem with a smaller number of jars to get ideas on how to empty the jars. Cookie Monster can empty all fifteen jars in four hours/steps. He can take 8 cookies from jars 8–15, then 4 cookies from jars 4–7 and jars 12–15. Next, take 2 cookies from jars 2–3, 6–7, 10–11 and 14–15. As a last step, take the one remaining cookie from jars 1, 3, 5, 7, 9, 11, 13 and 15. Note that these steps can be taken in any order, as long as you remove the stated number of cookies from the jars indicated. This is an important point, so explore it with the students. For example, Cookie Monster might want to eat a smaller number of cookies first, so he could end up with a big feast. A good notation is useful when we try to record our steps. At first, one could keep track of all the cookies in all the jars, so a representation might look like this: Results after Step 1 Step 2 Step 3 Step 4

1 1 1 1 0

2 2 2 0 0

3 3 3 1 0

4 4 0 0 0

5 5 1 1 0

6 6 2 0 0

7 7 3 1 0

8 0 0 0 0

Jars 9 1 1 1 0

10 2 2 0 0

11 3 3 1 0

12 4 0 0 0

13 5 1 1 0

14 6 2 0 0

15 7 3 1 0

At this point, students should realize that it does not matter whether there are one or more jars with the same number of cookies in them. For example, after the second step, there are four sets of jars with 1–3 cookies in them. However, four jars with say, 3 cookies can be dealt with exactly the same way (take 2 from each in Step 3 and 1 in Step 4). Thus it does not really matter how many such jars we have. Consequently, the notation can be made more economical, listing only jars with different numbers of cookies in them. One possible notation would be the following: {1, 2, 3, 4, 5, 6, 7, 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 }8 → {1, 2, 3, 4 , 5 , 6 , 7 }4 → {1, 2 , 3 }2 → { 1 }1 → {} . Note that the jars from which we take the number of cookies indicated in the subscript are boxed. This notation illustrates the key idea developed above that even though we are left with two sets of jars with 1, 2, . . . , 7 cookies in them after the first step, these two sets can be treated the same way, so it is sufficient to consider them just once. That is, one only needs to keep track of jars with different numbers of cookies in them. Figure 47 below is another representation of the steps taken to empty the jars. In Step 1 we remove the cookies represented by the orange square; in Step 2 we remove the cookies represented by the two blue squares; in Step 3, the cookies represented by the four

94

GABRIELLA PINTER

green squares get removed; finally, in Step 4 we remove the cookies represented by the eight yellow squares.

Figure 47. A geometric representation of the jars and the steps taken to empty them. It is important to establish that all cookies cannot be removed in a shorter amount of time. (If referring to time is cumbersome, you can say that selecting a set of jars, and removing cookies from them constitutes a ‘step’. Here we are trying to minimize the number of steps needed to remove all the cookies from all the jars. This terminology is often used in the rest of these notes.) Thus, we need to show that three steps cannot remove all cookies from all jars. One way to see that at least four steps are needed would be by realizing that in any scheme, there has to be a step when one cookie is taken from a set of jars, otherwise jar number 1 could not be emptied. Let us call this set of jars S1 . Now consider jar number 2: if this jar is in the set S1 , then another step of taking 1 will be needed, and then three steps clearly will not be enough — just think of emptying the jar with say, 4 cookies, and consider what happens in the jar that started with 7 cookies. If jar number 2 is not in S1 , then either there is a step in which two cookies are removed from this jar (and possibly some others), or two other steps of ‘one cookie removed’ are used to empty this jar. In either case, it is again clear that three steps cannot empty all cookies from all fifteen jars. Thus, at least four steps are necessary, and the scheme above does remove all cookies from the fifteen jars in four steps, so this is the minimal number that we sought. Another way to show that at least four steps are necessary to empty fifteen jars would be the following: When considering how optimal or efficient the steps above are, we can observe that by taking 8 cookies from jars 8–15, one jar became empty, and the remaining number of jars with different numbers of cookies got halved, since we have two sets

THE COOKIE MONSTER PROBLEM

95

of jars with 1–7 cookies in them. That is, we have (15 − 1)/2 = 7 jars with different numbers of cookies. In general, if you take any three jars with different numbers of cookies in them, then there is no step that will make all three jars contain the same number of cookies. So a single step cannot reduce more than two jars containing different numbers of cookies to one, so ‘halving’ is the best option. Thus after one step in Problem 1, the number of jars with different numbers of cookies is at least 7. After two steps, we have at least 3 jars, after three steps at least 1 non-empty jar. So again, at least four steps are necessary, and we have seen that four steps are indeed enough. This is a more general argument than the previous one, and can be helpful in Problems 2 and 3. Another proof can be found in (3) below. (2) Take sixteen cookies from jars 16–31. This leaves two sets of jars with one to fifteen cookies in them, which, according to Problem 1, can be emptied in four steps. Note again that the number of steps is not affected by having two of each type of jar. Thus the 31 jars can be emptied in 5 hours (or 5 steps), so Cookie Monster’s Mommy will not be very happy. The proof above can be extended to this case to show that four steps are not enough — again consider jars number 1 and 2, and this leaves at most two steps to deal with everything else. You can also use the ‘halving’ argument to see that four steps are not enough to empty all jars. Similarly, seven jars can be emptied in three steps. Note that after removing 8 cookies from jars 8–15 in Problem 1, we had two sets of jars with one to seven cookies in them, so we can use the scheme in Problem 1 for emptying these jars. (3) A conjecture for the general case can be developed by making a chart with the number of jars, n, and the minimum number of steps needed to empty the jars, M (n): n

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

M (n) 1 2 2 3 3 3 3 4 4 n M (n)

4

4

4

16 17 18 19 20 21 22 23 24 · · · 5

5

5

5

5

5

5

5

5

···

4

4

4

31 32 33 5

6

6

Based on this data, we develop the following proposition and then prove it. Proposition 1. If 2k ≤ n < 2k+1 , then M (n) = k + 1. That is, k + 1 is the minimum number of steps needed to empty n jars with 1 through n cookies.

96

GABRIELLA PINTER

Proof. First, we will show that k + 1 steps are sufficient to remove all cookies from all jars. Consider k+1 steps that remove 1, 2, 22 , 23 , . . . , 2k cookies from sets of jars S0 , S1 , S2 , S3 , . . . , Sk , respectively. (We will determine these sets below.) Consider an arbitrary integer p, 1 ≤ p ≤ n. It can be expressed as p = ak · 2k + ak−1 · 2k−1 + · · · + a2 · 22 + a1 . . . 21 + a0 · 1, where ai is 0 or 1 (this is the representation of p in binary). Now let jar p be included in the set Si if and only if ai = 1 in the binary representation of p. Since p is included in all sets Si for which ai = 1, removing 1, 2, 22 , 23 , . . . , 2k cookies from sets of jars S0 , S1 , S2 , . . . , Sk will empty jar p. Moreover, this is true for all jars p, 1 ≤ p ≤ n, so all jars will be empty in k + 1 steps. It might be helpful if the reader tried it out by applying this argument in some particular case, say, when n = 15. Construct all sets S0 , S1 , S2 , S3 , and take a particular value of p, for example, p = 11. Next, we show that k + 1 steps are needed to empty all the jars 1–n, that is, it cannot be done in k steps. This argument is based on the following observation: In Problem 1, the fifteen jars could be emptied in 4 steps, because the set of steps {8, 4, 2, 1} were such that every number from 1 to 15 could be expressed as a sum of one or more of these steps. For example, 6 = 4 + 2, 11 = 8 + 2 + 1, etc. So if we have steps {m1 , m2 , m3 , m4 }, the maximum number of jars that can possibly be emptied by using these steps will be the determined by how many numbers can be expressed as a sum of one or more of these mi . Now, each mi is either included or not included in a sum, so the number of such sums is 24 − 1, since at least one mi has to be in a sum (it cannot happen that none of them are included). The general case goes exactly like this. With steps {m1 , m2 , . . . , mk }, at most 2k − 1 numbers can be expressed as a sum of one or more of these steps. However, n ≥ 2k , so k steps, no matter how they are chosen, cannot be sufficient to empty all jars with 1, 2, . . . , n cookies in them. Thus we showed that to empty n jars where 2k ≤ n < 2k+1 , with 1, 2, . . . , n cookies, k + 1 steps are necessary and sufficient.  (4) For 3 jars, either of the sets of cookies {1, 2, 4} or {1, 5, 25} would work. For 4 jars, {1, 2, 4, 8} or {1, 5, 25, 125} would work. (5) For general n, either of {1, 2, . . . , 2n−1 } or {1, 10, . . . , 10n−1 } would work. In general, put the jars in increasing order by the number of cookies in them. If the number of cookies in the ith jar is greater than the total number of cookies in jars 1 through i − 1 for all i, then there is no way to empty all the jars faster than one by one. To see this, consider the jar with the most cookies. Since this number is larger than the sum of the cookies in all the other jars, this means that there has to be a step that touches only this jar (even after all other jars are empty there must still be cookies in this jar). Without adversely influencing the removal scheme, this step can take all cookies from this maximal

THE COOKIE MONSTER PROBLEM

97

jar. (It simply means that the maximal jar will not be included in any other step, even if before it may have been.) However, now we have one less jar with the same property, so the previous argument can be repeated for the jar that contains the second most cookies. Going on like this, the jars are emptied one by one. (6) There are several strategies that might work to empty the jars. Here we summarize a few, and show that these strategies can help to answer the specific question in Problem 6. Greedy Monster Algorithm (GMA):. In every step, take the most cookies possible. For example, in the original problem with cookies 1– 15 in jars 1–15, students can figure out what step will result in the most cookies eaten. A table can be drawn, and a pattern can be established. If k cookies are taken from each jar, where 1 ≤ k ≤ 15, then there are 15 − k + 1 jars that k cookies can be taken from. Thus, we want to maximize the product P (k) = k(15 − k + 1), where k is an integer between 1 and 15. This can be done by trying each possible k, or by considering the quadratic function P (k), or by using the inequality between the arithmetic and geometric mean. Students realize that taking 8 cookies from jars 8 through 15 will result in the most cookies taken in this step. In the next step, it is 4 cookies from all the jars that have at least 4 cookies in them: jars 4–7 and 12–15, and so on. Trying this method on our problem leads to: {1, 5, 33 , 36 }33 → {1, 5 , 3 }3 → {1, 2 }2 → { 1 }1 → {}

4 steps

{2, 6 , 7 , 13 }6 → {2, 1, 7 }7 → { 2 , 1}2 → { 1 }1 → {}

4 steps

{ 33 , 34 , 36 , 40 , 48 }33 → {1, 3, 7, 15 }15 → {1, 3, 7 }7 → {1, 3 }3 → { 1 }1 → {}

5 steps

{1, 2, 4, 11 , 16 , 17 }11 → {1, 2, 4 , 5 , 6 }4 → {1, 2 }2 → { 1 }1 → {}

4 steps

Note that the case of a tie can be investigated by students. Binary Algorithm (BA):. We have seen that powers of 2 were significant in our previous discussions, and it can give rise to the following algorithm. The idea is that for k as large as possible, take 2k cookies from all jars that contain at least that many cookies. This algorithm amounts to writing the cookie numbers in binary, and then going from left to right (or in any order), and ‘zero-down’ the 1s at the different place values. To use this on our problem write {1, 5, 33, 36} = {1, 101, 100001, 100100} in binary, so there are 1s at the place values 1, 4 and 32. Thus, this algorithm will result in 3 steps: {1, 5, 33 , 36 }32 → {1, 4 , 5 }4 → { 1 }1 → {}

3 steps

98

GABRIELLA PINTER

To better visualize this situation, the binary representation of the number of cookies in the jars can be written in a column: 1 101 100001 100100. Using the Binary Algorithm we get: {2, 6, 7, 13} = {10, 110, 111, 1101}, 4 steps {33, 34, 36, 40, 48} = {100001, 100010, 100100, 101000, 110000}, 6 steps {1, 2, 4, 11, 16, 17} = {1, 10, 100, 1011, 10000, 10001}, 5 steps Least Jars Algorithm (LJA):. Since the goal is to empty all jars, we could try to reduce the number of non-empty jars with different number of cookies in them as much as possible in each step. For example, the following steps all reduce the number of jars from state {1, 2, 3}: { 1 , 2, 3}1 → {2, 3},

{ 1 , 2 , 3}1 → {1, 3},

{1, 2 , 3}2 → {1, 3},

{1, 2 , 3 }2 → {1},

{ 1 , 2 , 3 }1 → {1, 2}, {1, 2, 3 }3 → {1, 2}.

The Least Jars Algorithm would choose the step {1, 2 , 3 }2 → {1} since this results in the least number of jars. With this method: {1, 5, 33 , 36 }31 → {1, 2, 5 }5 → {1, 2 }2 → { 1 }1 → {}

4 steps

{2, 6, 7 , 13 }7 → {2, 6 }6 → { 2 }2 → {}

3 steps

{ 33 , 34 , 36 , 40 , 48 }33 → {1, 3, 7, 15 }15 → {1, 3, 7 }7 → {1, 3 }3 → { 1 }1 → {} {1, 2, 4, 11, 16 , 17

16

5 steps

→ {1, 2, 4, 11 }11 → {1, 2, 4 }4 → {1, 2 }2 → { 1 }1 → {}

5 steps

Sometimes there are several steps that would result in the least number of jars possible. In general, we can specify that if two steps in the Least Jars Algorithm lead to the same number of non-empty jars with different number of cookies in them, then we choose the step that minimizes the number of cookies in the jar that has the most cookies in it. For example, {3, 4, 5, 6 , 7 }2 → {3, 4, 5} leaves three jars, the same number as the step {3, 4 , 5 , 6 , 7 }4 → {1, 2, 3}. We compare {3, 4, 5} and {1, 2, 3}, and since 5 > 3, we choose the step that results in {1, 2, 3}. Before comparison, take out common factors of cookies in

THE COOKIE MONSTER PROBLEM

99

the jars, since {4, 8, 12} behaves the same in terms of these steps as {1, 2, 3}. If two steps would leave the same minimum in the largest jars, then we go to the next largest jars, compare them, and choose the step that results in a smaller number of cookies in this jar, and so on. (7) Note that the second part of the argument in Problem 3 answers this question. No matter what steps are used, k steps can empty at most 2k − 1 jars. (8) This is a very difficult question, generally believed to be an open problem (no one knows how to solve it — yet!). Note that none of the strategies in Problem 6 are optimal in all cases; we also include the set {2, 4, 5, 11} for which the optimal strategy is none of the GMA, BA, or LJA. Bold in the table indicates the corresponding strategy is optimal for the particular case. GMA BA LJA Optimal {1, 5, 33, 36}

4

3

4

3

{2, 6, 7, 13}

4

4

3

3

{33, 34, 36, 40, 48}

5

6

5

5

{1, 2, 4, 11, 16, 17}

4

5

5

4

{2, 4, 5, 11}

4

4

4

3

(9) Under normal rules, five jars with {1, 2, 4, 8, 16} cookies could not be emptied in less than 5 steps by Problem 5. However, if cookies can be added to certain jars, then we could start by adding 1 cookie each to the jars that contained 1, 4 and 8 cookies, resulting in {2, 5, 9, 16}. (Note that at this point we would have two jars with 2 cookies, but as before, these two jars can be dealt with the same way, so we list it just once.) Now the key is that 2 + 5 + 9 = 16, so three additional steps of 2, 5 and 9 will empty the jars, making a total of 4 steps. Many new questions could be asked about what happens if one of the steps can be an addition of cookies like in this case. There are some important differences too, since the order of steps may now matter. (10) This game is known as Wythoff’s Game [6]. The game can be represented on a grid, indicating the number of cookies in Jar 1 on the horizontal, the number of cookies in Jar 2 on the vertical axis (see Figure 48). Cells in green are so-called P-positions — if a player leaves the jars in a P-position after her turn, she can win the game no matter what the other player does. Students can work out P-positions starting with (0, 0). It turns out that these positions are ( kτ , kτ 2 ) and

100

GABRIELLA PINTER

Figure 48. Grid representing Wythoff’s Game √ ( kτ 2 , kτ ), where k is a non-negative integer, τ = (1 + 5)/2, the golden ratio, and x denotes the integer part of x. More details can be found at the Cut the Knot website [5]. Thus with (9,10) as the beginning configuration, Cookie Monster would like to start, and take 3 cookies from Jar 1, leaving position (6, 10). (11) One way to represent this is to use matrices. If your students have been introduced to the basic arithmetic of matrices, you can use a matrix to describe the number of cookies in the different jars in this problem. We could empty all the jars using the following decomposition of the original matrix: ⎡

⎤

⎡

⎢5 ⎢ ⎢ ⎢2 ⎢ ⎢ ⎢ ⎢5 ⎢ ⎢ ⎣ 2

5 3 3⎥ ⎢1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢1 2 5 2⎥ ⎢ ⎥=2·⎢ ⎥ ⎢ ⎢1 3 3 2⎥ ⎥ ⎢ ⎥ ⎢ ⎦ ⎣ 5 5 3 1

⎤

⎡

1 1 1⎥ ⎢1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0 1 1 1⎥ ⎢ ⎥+1·⎢ ⎥ ⎢ ⎢1 1 1 1⎥ ⎥ ⎢ ⎥ ⎢ ⎦ ⎣ 1 1 1 0

⎤

⎡

1 1 1⎥ ⎢1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0 0 1 0⎥ ⎢ ⎥+2·⎢ ⎥ ⎢ ⎢1 1 1 0⎥ ⎥ ⎢ ⎥ ⎢ ⎦ ⎣ 1 1 1 0

⎤ 1 0 0⎥ ⎥ ⎥ 0 1 0⎥ ⎥ ⎥. ⎥ 0 0 0⎥ ⎥ ⎥ ⎦ 1 1 0

Zeros in the matrices indicate blocked jars, while 1s denote jars that cookies are taken from. Thus it would take 3 steps. More Mathematics for a Curious Reader: The idea in problem 11 was to model what happens when multileaf collimators are used in radiation therapy. The matrix stands for an area of the body that gets radiation treatment, but not every point needs the same amount. The numbers in the original matrix stand for the amount of radiation that needs to reach each place/cell. The machine sends equal amount of radiation through each point when it is turned on. To achieve the goal,

THE COOKIE MONSTER PROBLEM

101

the operator can push in blockers/leaves from the left or from the right in each row, otherwise each cell would get the same amount of radiation. If the leaves are from the top and bottom then a rotated version of the matrix can be used. The problem is to figure out how to block cells, and what dose to send in each ‘step’ so that the smallest number of steps are used, that is, the smallest number of rearrangements of the leaves are needed, and each cell gets its required total amount of radiation. This is a very real applied problem that can be thought of as a two-dimensional generalization of the original cookie jar problem. Instead of minimizing the number of ‘steps’, sometimes it is desirable to minimize the total ‘beam-on’ time, that is the total amount of radiation used. This means that if in step i you send ri amount of radiation (remove ri cookies from non-blocked jars), then you want to minimize the sum of ri until all cells receive the prescribed amount of radiation (all jars are empty). The following example of a column vector shows that the two minimization problems are not the same: ⎡

⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎡ ⎤

⎢33⎥ ⎢ 0⎥ ⎢ 0⎥ ⎢1⎥ ⎢0⎥ ⎢0⎥ ⎢0⎥ ⎢1⎥ ⎢ 1⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢34⎥ ⎢ 0⎥ ⎢ 1⎥ ⎢1⎥ ⎢0⎥ ⎢1⎥ ⎢1⎥ ⎢1⎥ ⎢ 0⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢36⎥ = 33 ⎢1⎥ + 1 ⎢1⎥ + 2 ⎢1⎥ + 4 ⎢0⎥ + 8 ⎢0⎥ = 8 ⎢0⎥ + 11 ⎢1⎥ + 15 ⎢0⎥ + 25 ⎢1⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢1⎥ ⎢40⎥ ⎢ 1⎥ ⎢ 0⎥ ⎢1⎥ ⎢0⎥ ⎢1⎥ ⎢1⎥ ⎢0⎥ ⎢ 1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 1 48 1 0 1 1 1 1 1

The first decomposition corresponds to a minimum ‘beam-on’ time (48), and the second to a minimum ‘step’ (4) scenario. Practical decomposition algorithms and theoretical considerations can be found in the PhD thesis The Segmentation Problem in Radiation Therapy by Celine Engelbeen [9]. Presentation Suggestions: • It is useful to prepare a set of ‘jars’ (cups or bowls) and cookies (wooden blocks, counters or pebbles), so that students can experiment with the problem. • After students spend some time with the warm-up problem, let them explain their solutions to the rest of the class. At this point it might be useful to introduce notation for indicating how many cookies are taken, and from which jars, so that a sequence of steps can be described more easily. It is important to discuss why the prescribed steps can be carried out in any order, and why we only need to keep track of jars that have different numbers of cookies in them. (That is, it does not matter if we have multiple copies of jars with the same number of cookies.)

102

GABRIELLA PINTER

• Proving that a certain number of steps is minimal is not easy, and students may struggle with their explanations. It is again helpful to start with a small number of jars, and build from there. If students know about logarithms, their answer can be formulated as M (n) = log2 n + 1, where m stands for the integer part of m. • Problems 7 and 8 generalize the original question by considering arbitrary numbers of cookies in the jars. As noted above, many other generalizations are possible. Students could consider questions when the number of cookies in the jars form a general arithmetic or geometric sequence. At this point students may come up with their own questions, too. • Problem 6 asks for different strategies to empty jars. Students may come up with different ideas than the ones presented, but these three strategies are still worth discussing. • Problem 8 is very difficult, and is generally considered to be an unsolved problem. While different upper and lower bounds can be worked out for the optimal number of steps emptying the jars, the exact number for the general question is still unknown. The Cookie Monster problem was among the finalists in the K-12 Unsolved Problems in Mathematics collection [2]. Dr Gordon Hamilton, who is spearheading the movement of including at least one unsolved problem in each K-12 classroom has some informative videos and descriptions of the cookie jar problem and others at his website [1].

References [1] http://mathpickle.com/ [2] http://mathpickle.com/unsolved-k-12/ [3] Paul Vaderlind, Richard K. Guy, and Loren C. Larson, The inquisitive problem solver, MAA Problem Books Series, Mathematical Association of America, Washington, DC, 2002. MR1917371 [4] Leigh Marie Braswell and Tanya Khovanova, Cookie Monster devours Naccis, College Math. J. 45 (2014), no. 2, 129–135, DOI 10.4169/college.math.j.45.2.129. MR3183035 [5] “Wythoff’s Game” Cut-the-Knot entry. http://www.cut-the-knot.org/pythagoras/ withoff.shtml [6] “Wythoff’s Game” Wikipedia page. https://en.wikipedia.org/wiki/Wythoff%27s_ game [7] M. Belzner. Emptying sets: The cookie monster problem. 2013. http://export.arxiv. org/pdf/1304.7508 [8] L.M. Braswell. Angles of the cookie monster problem. 2013. http://math.mit.edu/ research/highschool/rsi/documents/2013Braswell.pdf [9] C Engelbeen. The segmentation problem in radiation therapy. 2010. PhD Thesis, University Libre de Bruxelles

Math Blocks Contributed by Dave Auckly Short Description: Everyone loves blocks. Blocks are a wonderful way to see mathematical patterns. This session explores many patterns that may be built from blocks. The patterns include sequences of numbers, addition and multiplication tables, and others. Materials: A collection of 9/16 inch wooden cubes or 1 inch wooden cubes are required. A dark color spray paint helps. Wooden cubes may be purchased at many craft stores, and on the internet. Participants may also wish to use graph paper and scrap paper. Glue is also helpful for some problems. Mathematics Beneath & Beyond: The problems in this session all fit under the category of “proof without words.” There are many sources for such proofs. The books Proofs without Words: Exercises in Visual Thinking (Classroom Resource Materials) (v. 1, 2, 3) by Roger Nelsen is one source of similar materials. Proofs Without Words and Beyond — Proofs Without Words 2.0 by Doyle, et al. is another. Mathematics Magazine and the College Mathematics Journal each regularly include proofs without words. A proof without words is similar to a bijective proof, which is the gold standard of proof in combinatorics. In a bijective proof, one represents one expression as a formula that counts something, and another expression as a formula that counts something else. If one can describe a one-to-one and onto (bijective) correspondence between the two sets, the two expressions must be equal. To see an example, let ( nk ) represent the number of ways to choose k items from a set of n items. For example, ( 52 ) = 10 because the following are all ways of choosing 2 elements from 5: {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}. n ). Indeed, given any way to choose k elements We claim that ( nk ) = ( n−k from n, one may just consider the set of elements that were not chosen. In the example, we can associate the set {3, 4, 5} to the set {1, 2}, et cetera. There are many remarkable bijective proofs. One could even wildly suggest that any true mathematical fact would have a bijective proof. Similarly, one could go wild and imagine that The Book6 has proofs without words for 6

Mathematician Paul Erd¨ os frequently referred to “The Book”, an imaginary book containing the most beautiful proof for every conceivable theorem in mathematics. His 103

104

DAVE AUCKLY

every true mathematical statement. Such speculation is indeed wild, but it is fun to try to find such proofs. In particular, this session leads students to discover the following formulas: n(n + 1) 2 n(n + 1)(2n + 1) = 6   n(n + 1) 2 = 2

1 + 2 + ...+ n = 12 + 22 + . . . + n2 13 + 23 + . . . + n3

References/Authorship: The idea of drawing geometric figures to represent mathematical patterns has been around for several thousand years. Joshua Zucker commented that it is fun to build the addition table and multiplication table out of Legos and that inspired Dave Auckly to develop this math circle session.

highest compliment of another mathematician’s work was to say “That’s straight from The Book.”

MATH BLOCKS

105

Math Blocks Student Handout In this session we will play with blocks to explore interesting numerical patterns. There are four families of shape numbers which we will investigate. The sequence for each family is illustrated in Figure 49. • Squares (The number of little squares in each member of a sequence of squares that grows steadily larger. We use the following notation: S1 = 1, S2 = 4, S3 = 9, . . . , Sn ) • Triangles (T1 = 1, T2 = 3, T3 = 6, . . . , Tn ) • Cubes (C1 = 1, C2 = 8, . . . , Cn ) • Tetrahedra (T et1 = 1, T et2 = 4, T et3 = 10, . . . , T etn )

Figure 49. Squares, Triangles, Cubes, and Tetrahedra. In addition to these families of shapes, we will also explore two kinds of models of addition and multiplication tables (in addition to the usual addition and multiplication table). • Big City Model. In the “big city” model of an addition or multiplication table, we first make an ordinary table and then stack the number of blocks corresponding to either the sum or product on top of each square in the table. For example, for the square corresponding to 4 + 5 we would put a tower of 9 cubes. This model is depicted in Figure 50. In this figure, the addends increase from front to back and from right to left. • Farmland Model. We will only use the “farmland” model to represent the multiplication table. Figure 51 shows this model of a multiplication table for 1×1 through 4×4. Notice that each table entry is represented by a rectangle with area equal to the corresponding product.

106

DAVE AUCKLY

Figure 50. Big city model of the addition table.

Figure 51. “Farmland” model for a multiplication table.

MATH BLOCKS

107

As you look for interesting numerical patterns in these figures and models, you may wish to draw on grid paper, or build models with the small wooden cubes. Problems: (1) How could you compute the square of 7 (S7 ; this number is denoted by 72 )? What is S17 ? See Figure 49. (2) What can you say about the difference between the number of little squares in S2 and S1 ? What about the difference between S3 and S2 ? What about between S16 and S15 ? (3) What can you say about the difference between the number of little squares in triangle T2 and T1 ? What about the difference in T3 and T2 ? What about T16 and T15 ? See Figure 49. (4) Can you see a way to compute T4 ? Can you see a second way? What about T15 and Tn ? (5) What is a good way to compute/see the sum 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 using the blocks? (6) How many little cubes would we need to build a “big city” model of the addition table from 0 + 0 all the way up to 49 + 49 including both versions of each sum (for example 13 + 23 and 23 + 13 would both appear)? (7) How does the value of the sum change when you move up or down a column in the addition table? What about left or right across a row? (8) How does the value of the product change when you move up or down a column in the multiplication table? What about left or right across a row? (9) What can you say about the sum of the numbers in the third row of the multiplication table from 3 × 1 to 3 × 7? (10) What can you say about the sum of the numbers in a multiplication table from 0 × 0 all the way up to 29 × 29 including both versions of each sum (for example 13 × 23 and 23 × 13 would both appear)? (11) Can you compute the number of small cubes in cube C12 ? How? (The number of little cubes in C5 is called the cube of 5. It is denoted by 53 .) See Figure 49. (12) Can you compute/see the difference in the number of small cubes in C15 and C14 ? How? (13) Can you see the difference between the number of small cubes in T et9 and T et8 ? See Figure 49. (14) What is a good way to compute/see the sum 12 + 22 + 32 + 42 + 52 + 62 + 72 ? Hint: Can you represent this kind of sum as a pyramid? Try gluing cubes together to make three identically-sized pyramids. Now try to assemble these to get as close as you can to making a rectangular solid.

108

DAVE AUCKLY

(15) Some interesting patterns appear if you look at diagonals or L-shaped regions in the multiplication table (see Figure 52). What can you say about the sum of the numbers in the multiplication table along the diagonal from 1 × 4 to 4 × 1? What about the analogous diagonals? Can you see the pattern? 0

1

2

3

4

5

6

7

8

9

10 11 12

1

1

2

3

4

5

6

7

8

9

10 11 12

2

2

4

6

8 10 12 14 16 18 20 22 24

3

3

6

9

12 15 18 21 24 27 30 33 36

4

4

8

12 16 20 24 28 32 36 40 44 48

5

5

10 15 20 25 30 35 40 45 50 55 60

6

6

12 18 24 30 36 42 48 54 60 66 72

7

7

14 21 28 35 42 49 56 63 70 77 84

8

8

16 24 32 40 48 56 64 72 80 88 96

9

9

18 27 36 45 54 63 72 81 90 99 108

10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144

0

1

2

3

4

5

6

7

8

9

10 11 12

1

1

2

3

2

2

4

6

4

5

6

7

8

9

10 11 12

8

10 12 14 16 18 20 22 24

3

3

6

9 12 15 18 21 24 27 30 33 36

4

4

8

12 16 20 24 28 32 36 40 44 48

5

5

10 15 20 25 30 35 40 45 50 55 60

6

6

12 18 24 30 36 42 48 54 60 66 72

7

7

14 21 28 35 42 49 56 63 70 77 84

8

8

16 24 32 40 48 56 64 72 80 88 96

9

9

18 27 36 45 54 63 72 81 90 99 108

10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144

Figure 52. Multiplication Ls and Diagonals. (16) What can you say about the sum of the numbers in the multiplication table along the L-shaped regions shown in Figure 52? (17) What is a good way to compute/see the sum 13 + 23 + 33 + 43 + 53 + 63 + 73 ?

MATH BLOCKS

109

Math Blocks Teacher Guide Solutions: (1) A square of side 7 would have seven rows of seven squares so 7 × 7 = 72 = 49 small squares. Similarly a square of side 17 would have 17 rows of 17 for 17 × 17 small squares. (2) S2 − S1 = 4 − 1 = 3. One also has S3 − S2 = 9 − 4 = 5. In fact, the picture shows this in a nice way. When we take the small square S2 in the upper left (in case of the Figure 53) away from the larger square, in this case S3 we are left with two 1 by something rectangles plus one extra block. Here the something is the side of the smaller rectangle.

Figure 53. Difference of squares version 1. Thus, we get S3 − S2 = 2 × 2 + 1 = 5. If we do this with squares S16 and S15 we get S16 −S15 = 2×15+1 = 31. Quick, what is 1 + 3 + 5 + 7 + 9 + 11 + 13 + . . . + (2 × 15 + 1)? Working algebraically we see it is 1 + (S2 − S1 ) + (S3 − S2 ) + . . . + (S15 − S14 ) + (S16 − S15 ) = S16 = 16 × 16 = 256. Alternatively, just look at Figure 54.

Figure 54. Sum of odds. Notice that this is 1 + 3 + 5 + . . . + (2 × 6 + 1) = 7 × 7 = 49. Figure 55 shows that the difference of two squares of any size can be related to a rectangle. In a formula one has a2 − b2 = (a + b)(a − b).

110

DAVE AUCKLY

Figure 55. Difference of squares version 2. Thus (reading the last formula from right to left) we have: 93 × 87 = (90 + 3)(90 − 3) = 8100 − 9 = 8091. (3) T2 − T1 = 2, T3 − T2 = 3, and T16 − T15 = 16. (4) Look at the number in the top row, then the second row, et cetera. We have: T15 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15. Now look at the Figure 56 below. It shows two copies of T3 fit together to make a (3 + 1) × 3 rectangle. Thus 2T3 = (3 + 1) × 3, and so T3 = 12 (3 + 1) × 3. Likewise, T15 = 12 (15 + 1) × 15 = 120.

Figure 56. Two copies of a triangular number.

MATH BLOCKS

111

(5) The two answers produced in problem 4 must be the same. Thus, 1 + 2 + 3 + . . . + 13 + 14 = T14 1 (14 + 1) × 14 = 2 = 105. (6) The addition table from 0 + 0 up to 2 + 2 is displayed in Figure 57. The top two layers form a tetrahedral configuration. If it is removed, it may be placed in the corner that remains to complete a 3 × 3 × 2 brick. The analogous thing will happen with the addition table from 0 + 0 up to 49 + 49. Namely, we move the top 49 layers to fill in the corner to get a 50 × 50 × 49 brick. Thus there are 122,500 blocks in the “big city” model of the addition table from 0 + 0 up to 49 + 49. Alternately, two copies of the addition table may be put together to produce a 50 × 50 × (49 + 49) “brick.” The sum will be one half the number of blocks in this brick.

Figure 57. Addition Table. (7) The elements in the third row are 2 + 0, 2 + 1, 2 + 2, . . .. Each step we take to the right adds one to the result. The same occurs in each row. Each step down a column the number will get one larger. Since we start the addition table at 0 + 0, the first row is the 0+ row, the second row is the 1+ row, et cetera. (8) We start the multiplication table at 1 × 1. Thus the third row will be 3 × 1, 3 × 2, 3 × 3, . . .. Thus it gets 3 units larger each step we take to the right in the third row. It gets 13 units larger each step we take to the right in the 13th row. Similarly, it will get 3 units larger each step we take down the third column. (9) The third row from 3 × 1 to 3 × 7 will just be 3T7 : 3 × 1 + 3 × 2 + . . . + 3 × 7 = 3 × (1 + 2 + . . . + 7) = 3T7

112

DAVE AUCKLY

(10) If we lay out the multiplication table as in Figure 51 we see that it fills up a square of side 1 + 2 + · · · + 29. This sum is just the 29th triangular number so it equals T29 = 12 (29 + 1) × 29. Thus, the number 2  of little squares in this flat model would be 12 (29 + 1) × 29 = 4352 = 189, 225. (11) There are 12 layers of squares of side 12 in the cube of side 12. Each such square contains 12 × 12 small cubes, so there are 12 × 12 × 12 small cubes in C12 . This is why it is commonly called the 12 × 12 × 12 cube. (12) Look at Figure 58. It shows a decomposition of a C3 into a C2 , three 1 × 2 × 2 bricks, three 1 × 1 × 2 bricks and a final C1 . The same thing will happen with C15 . It may be decomposed into a C14 three 1 × 14 × 14 bricks, three 1 × 1 × 14 bricks and a final C1 . Thus, 153 − 143 = 3 · 142 + 3 · 14 + 1.

Figure 58. Difference of adjacent cubes. (13) It is a triangular pattern of cubes. Thus T et9 − T et8 = T9 . (14) It is natural to put three pyramids together. The left over bit will be a triangular number. It can be cut in half and moved to complete a brick as in Figure 59. Using the idea shown in this figure, we see that 12 + 22 + 32 + 42 + 52 + 62 + 72 = 13 7(7 + 1)(7 + 12 ) = 140. (15) The sum 1 × 4 + 2 × 3 + 3 × 2 + 4 × 1 is the fourth tetrahedral number T et4 . See Figure 60. (16) The terms in the third L are 1 × 3, 2 × 3, 3 × 3, 3 × 2 and 3 × 1. Notice that 1 × 3 + 2 × 3 + 3 × 3 = T3 × 3, and 3 × 2 + 3 × 1 = 3T2 . Thus the number of little cubes over the third L in the multiplication table

MATH BLOCKS

113

Proof without words: Sum of squares

12 + 22 + . . . + n2 = 13n(n + 1)(n + 12)

n n

n

n+

1 2

n+1

n

—MAN-KEUNG SIU University of Hong Kong

Figure 59. Three pyramids forming a rectangular block. c Mathematical Association of America, 1984. All rights ( reserved.) is 3(T3 + T2 ) = 3 · 32 = 33 . Similarly, the sum of the cubes over the fourth L is 43 . (17) By the answer to (10) it would just be the number of cubes in a model of the multiplication table up to 7 × 7. Now the answer to (16) makes this  3

3

3

3

3

3

3

1 +2 +3 +4 +5 +6 +7 =

1 (7 + 1) × 7 2

2 = 282 = 784.

Presentation Suggestions Part 1: To begin have a counting race. Pick two participants and tell them that they are going to race to count some number of cubes. Then put a disorganized heap in front of one and an array in front of the other as shown in Figure 61. Of course the student with the array should win. Ask why and have everyone understand that it is easier to count items in an array.

114

DAVE AUCKLY

Figure 60. Decomposing T et4 .

Figure 61. It is much harder to count the blocks when they are in a jumbled heap. After the race ask participants to make a model of (2 + 3)2 out of the blocks so that they can see 22 and 32 . You are hoping someone will create a model that looks like Figure 62. Ask if (2 + 3)2 is equal to the sum of 22 and 32 . Write (2 + 3)2 = 22 + 32 + 2 × (2 × 3) where all can see it. Draw a cartoon model of (21 + 33)2 (such as the one shown in Figure 63) and discuss it. Demonstrate the following multiplication trick: ask someone to tell you a two-digit number that has a 5 in the ones place, e.g., 75. Tell them 75 × 75 = 5625. Repeat with a couple more numbers. Then describe the following estimate: 75 is between 70 and 80, so it is not unreasonable to guess that 75 × 75 is close to 70 × 80 = 5600. The trick is to multiply the tens place with one more than the number in the tens place (so 7 × 8 in our example) then append 25 hence 5625.

MATH BLOCKS

115

Figure 62. A model of (2 + 3)2 .

Figure 63. A cartoon sketch for (21 + 33)2 . Ask for a pair of numbers that add to 35. Ask what “thirty five” means. (Answer: three tens plus five.) Then partially cut a square piece of paper into a (30 + 5)2 configuration as in Figure 64. Label the edges of the specific rectangles (perhaps draw the same figure on the board.) Rip away the small square representing 5×5 = 25 then give the rest of the paper to a participant and ask them to rearrange the paper recalling that it is easier to compute the area of a rectangle. You should get a configuration like the one shown in Figure 64. Ask if the same trick would work for 24 × 28. Ask if a similar trick would work for 45 × 75. Ask if a similar trick would work for 62 × 68. Ask them to justify any trick that works. (The last trick will work using the same geometric reasoning as the 75 × 75 trick.) An algebraic derivation is as follows:

116

DAVE AUCKLY

Figure 64. Rearranging (30 + 5)2 − 52 to make a rectangle.

(10n + k)(10n + (10 − k)) = 100n2 + 10nk + 10n(10 − k) + k(10 − k) = 100n2 + 10nk + 100n − 10nk + k(10 − k) = 100n(n + 1) + k(10 − k). Part 1 could be optional, but it does lead to a nice conclusion at the end of the session. Part 2: Ask the participants to make a model of 1 + 2 + 3 + 4 where they can see each of the numbers. Have some of the students make the model out of all white blocks and some make it out of all black blocks. Remind the students that it is easier to compute the number of items in an array. The students might then be motivated to combine a white model with a black model as in Figure 56. Instead of combining two different models together, some participants may rearrange the cubes in one model to make an array. This can be done in many different ways; make sure that they notice a systematic way which leads to discovering a formula. After they get the first sum ask for 1 + 2 + · · · + 6 + 7. See if they can figure it out without making a model. Then ask for 1 + 2 + · · · + 16 + 17, and 1 + 2 + · · · + 1000 and 1 + 2 + · · · + n. (It is very beneficial to generalize up from small numbers to medium numbers where the arithmetic is still easy, then up to even larger numbers without doing the arithmetic and finally to general place holders. Doing so before studying algebra will make the ideas easier. Doing so during the study of algebra will help people remember that letters represent numbers. As a placeholder they can use anything they want — “Last Number”, or “LN” for short, or just “n”, or even “Cat” if they like.)

MATH BLOCKS

117

Ask them to tell you the answer for each of the last three without simplifying any of the arithmetic. Thus the answers to two of the last three problems will be: n × (n + 1) 17 × (17 + 1) and . 2 2 After students figure out the formula for the sum of the first so many positive integers (such sums are called triangular numbers, denoted by Tn ) they could move on to find the sum of the first so many squares. We slightly prefer asking them to build a model of the addition table. Give them a grid as at the top of Figure 50 and ask them to build stacks up representing the values of the sums. Their model will look like the image at the bottom of Figure 50. Once the students build a model for the addition table ask them to explain why some of the stacks of blocks are the same height, e.g. 3 + 0 and 1 + 2. (They simply have the same number of little cubes.) Ask why the model is symmetric along the diagonal. Ask them for the sum of the numbers in the addition table 0 + 0 up to 3 + 3. Then ask for the sum of the numbers in the addition table 0 + 0 up to 13 + 13, and then 0 + 0 up to n + n, i.e. the total number of the little cubes in their model. (This brings us to Problem (6) of the handout. See the Solutions section above.)

118

DAVE AUCKLY

Figure 65. Building the sum of the sums in an addition table from 0 + 0 up to 4 + 4. The white blocks form one copy of this sum and the black blocks form another copy of this sum. Thus the sum of the sums in this range is 12 ×5×5×8 = 100. After they compute the sum of the sums in an addition table, ask the students to make a model of the multiplication table 1 × 1 up to 3 × 3. Two interesting models they can make are the “big city” model which shows each product as a stack of cubes, or the checkerboard “farmland” model, displayed in Figure 51, which shows each product as the area of a rectangle. The “big city” model can be (optionally) extended in a nice way to show products involving negative factors. The finished Two Towers model, which the students will help to build, is shown in Figure 66. It helps to build a large base of black “bedrock” before the session, with weakly attached black cubes that the students will dig out to represent negative products. Figure 68 shows the part of the base that should be glued securely. The left picture shows the model right side up. Other black cubes should be loosely attached to this base to fill in the rest of the rectangular solid as viewed from above. The picture on the right shows the inside of the bedrock base. The inside of the base can be left hollow as shown to save cubes. Studying the towers shown in Figure 66 is a wonderful way to demonstrate that the product of a pair of negative numbers is positive. Take the

MATH BLOCKS

119

time to build and explain it. In this model, the white cubes represent positive numbers and the top of the black platform represents 0. Taking a to represent the first factor and b to represent the second factor, this model uses height to show the products for all pairs a × b with a and b running from −3 to 3. The pairs (a, b) in this range are displayed in Figure 67 in the same orientation as the products in the model.

Figure 66. Two Towers.

120

DAVE AUCKLY

Figure 67. The multiplication table base for the Two Towers.

MATH BLOCKS

121

Figure 68. The Bedrock Base. The base consists of all the black blocks in the model, representing the non-positive values in the multiplication table. The model may be hollow to save blocks, but all of the black “bedrock” should be glued together securely into one unit. The left image shows the base in the upright position. The right image shows how the base can be left hollow to save cubes. The white tower on the right is just the “big city” model for products of two positive factors. The tallest part of the white tower on the right shows that 3 × 3 is 9. Build the white tower on the right with the students. When this is done, use your fingers to walk down the right front face. The stack of 6 cubes shows 3 × 2 is 6, and then we find that 3 × 1 is 3. Finally, we reach the top of the black platform where 3 × 0 is 0. To continue going down by 3s we need to excavate some black blocks from the base, showing that 3 × −1 = −3, 3 × −2 = −6 and 3 × −3 = −9. We can repeat this process to fill in the rest of the quadrant with a > 0 and b < 0 (seen as the quadrant closest to the front in the picture). Similarly, it is easy to use the pattern going down the back right face to excavate the blocks in the quadrant with a < 0 and b > 0 at the back of the model. Now the students must figure out how to finish the model in the quadrant with two negative factors. Starting with 3 × −3 = −9, which is the lowest corner of the model at the front of the photo, use your fingers to walk up and to the left. As you do this, the second factor will remain −3, and the first factor will decrease from 3. As we climb, we see that 2 × −3 = −6, that 1 × −3 = −3, and that 0 × −3 = 0. If the pattern continues climbing in this way, we see that we must use white cubes to represent the products of two negative factors. In particular, we see that −1 × −3 = 3, that −2 × −3 = 6,

122

DAVE AUCKLY

and −3 × −3 = 9. Ask the students to figure out how to complete the rest of the final quadrant. Ask the students to compute the sum of the entries in the multiplication table 1 × 1 up to 3 × 3. The “farmland” model will make it clear that the answer is (1 + 2 + 3)2 . Ask the same thing for the multiplication table 1 × 1 up to 10 × 10. This time the answer will be

10(10 + 1) 2 = 552 = 3025. (1 + 2 + · · · + 10)2 = 2 This brings us back to where the session began. It may happen that some participants sum the entries of the multiplication table by putting several (indeed, four) together. This would be done by using a second table to complete the “triangles” along the rows of the first multiplication table to get a triangular prism. Two more multiplication tables may be combined into a second prism. The two prisms then fit into a “brick”. More Mathematics for a Curious Reader: One may wonder if there is a closed formula for 1p + 2p + . . . + np in general. The answer is yes. Here is a brief exposition that uses series manipulations: The Bernoulli numbers are defined as the coefficients of the exponential generating function below: ∞

 Bk z k z = ez − 1 k! k=0

−1 z as a sum of exponentials, and then expand the z ez − 1 −1 ∞ N   eN z − 1 zn j p . Expand exponentials into power series to obtain N + n! z n=1 j=1 z and z into power series. Multiply the resulting series to get a series for e −1 eN z − 1 z . Compare this with the prior formula to obtain: z ez − 1  N −1 n    n + 1 Bk n N n−k+1 . j = n+1 k Expand

eN z

j=1

k=0

Comments/Questions? Contact: Dave Auckly [email protected], Math Department, Kansas State University

Counting Diagonals Contributed by Tatiana Shubin Short Description: Even though the main goal of this session is to introduce students to combinatorial reasoning, and especially to counting the number of possible ways to choose several objects from a given collection, there are many interesting ideas and observations worth noting along the way. In particular, students enjoy discovering facts about Platonic solids, such as their names and duality. Moreover, this session gives us a chance to prove something interesting, i.e., the total number of Platonic solids, and to point out that the existence of a model doesn’t prove the existence of a mathematical object. In what follows we will discuss these and other interesting and important side trips. Materials: Bringing models of many solids and giving students time to play with them is a good ice breaker at the beginning of the session and also helps them to solve the problems later in the session. Bring several models of each Platonic solid as well as other polyhedra, including a buckyball. Material doesn’t matter; you can have a mixture of models made of Zome Tools (http://www.zometool.com/), Polydrons (http:// www.polydron.co.uk/), paper, wood, gumdrops and toothpicks, et cetera. Mathematics Beneath & Beyond: Combinatorics is an important part of modern mathematics; moreover, it has very useful applications. It is instrumental in computing probability, and it is used on a daily basis by computer scientists. It is a vast and rapidly developing field with an ever growing array of special techniques for solving intricate problems, yet its core is simply the idea of counting. This session introduces one of the basic combinatorial ideas — that of counting the number of ways to choose a certain number of objects from a given total number of such objects. To achieve this goal, we start by getting familiar with Platonic solids. We prove that there are exactly 5 of them, and learn some of their properties, such as duality. We count their geometric components — vertices, edges, and faces, and make use of Euler’s Formula: V − E + F = 2. To count diagonals, we identify them with pairs of vertices, thus coming to the central question — how many ways can these pairs be chosen? 123

124

TATIANA SHUBIN

Counting Diagonals Student Handout For convenience, we will use some standard geometry terms; a short reminder is below. A polygon is a 2D figure with straight sides. Each side of a polygon is called an edge; the point where two edges meet is called a vertex (plural vertices). A polyhedron (plural polyhedra) is a 3D figure (i.e., solid) with polygonal sides. Each side of a polyhedron is called a face; a line segment where two faces meet is called an edge; a point where edges meet is called a vertex. A figure (2D or 3D) is convex if the whole line segment connecting any two points of the figure lies inside the figure. A Platonic solid (a.k.a. regular solid) is a convex polyhedron whose faces are congruent (identical in shape and size), convex, regular (all angles and all sides equal) polygons with the same number of faces meeting at each vertex. There are five different Platonic solids: tetrahedron, cube, octahedron, dodecahedron, and icosahedron. Problems: (1) Why are there only five Platonic solids? (2) Place a dot in the center of each face of a cube and connect those dots which lie on adjacent faces. These line segments are the edges of a solid. What is the solid? What happens if you repeat the process starting with an octahedron? a dodecahedron? an icosahedron? a tetrahedron? (3) What is left if you place a dot at the midpoint of every edge of a triangle, connect two dots on the adjacent edges with a line segment and then chop off every corner along these lines? What if you start with a regular tetrahedron instead of a triangle and do the same thing? What happens if you start with a regular icosahedron, trisect each edge, and then use those marks to chop off the vertices? (4) How many diagonals does a convex polygon have? (3-gon(a triangle), 4-gon, 5-gon, 6-gon, 11-gon, 100-gon, n-gon) (5) How many vertices, edges, and faces does each Platonic solid have? What about the buckyball? (6) How many spatial diagonals are there in each Platonic solid? In a buckyball? (A spatial diagonal is a line segment connecting two vertices of a solid which isn’t an edge and which does not lie on a face of the solid.) (7) There are 10 towns in a small country, and every pair of them is connected by a road. How many roads are there? What if there are 73 towns? (8) Two teams, Monster Slayer and Born for Water, have decided to hold a Math Wrangle every Saturday for the entire year.7 Monster Slayer is a 7

In the 2015 Navajo Math Circle Summer Camp, our students were divided into two teams to compete in Math Wrangles. They chose as their team names, Monster Slayer and

COUNTING DIAGONALS

125

team of 12, and they want to appoint two co-captains every Saturday. Can they do this without repeating any pair of co-captains on any Saturday? (9) On the street map in Figure 69, Devon is located in the lower left hand corner and he wants to get to his hooghan in the upper right hand corner.8 Devon must travel along the roads. If he only travels either north (up) or east (to the right) at each intersection, how many different ways can he travel from where he is to his home?

Figure 69. Devon must get to his hooghan. Additional problems for those who like 3D geometry: (10) Color some facial diagonals of a cube red and the rest of them blue in such a way that you form two congruent tetrahedra, one with red edges and another with blue ones. What solid is formed by the intersection of these two tetrahedra? (11) On a dodecahedron, color one diagonal on each face red so that these red diagonals form a cube. Repeat the process with different colors until you get five different cubes. What solid is formed by the intersection of all these cubes? (12) Suppose that every edge of a regular dodecahedron is of length 1. Find the diameter of a sphere circumscribed about the dodecahedron.

Born for Water, the “Hero Twins” and sons of Changing Woman in the Navajo Creation Story. 8 A hooghan (sometimes spelled “hogan”) is a traditional Navajo dwelling.

126

TATIANA SHUBIN

Counting Diagonals Teacher Guide Presentation Suggestions: Start the session by placing a number of Platonic and other solids on a table. This should include all Platonic solids (PSs) (several of each type) and some other solids including a buckyball. Ask students to split the models into two groups. If the time is limited, tell them right away that they should split them not by material but by shape. On the other hand, if you have enough time, let them do what they want. If they start by splitting the models by material it offers an opportunity to talk about what is mathematically important (material is not in this case). When they are done, ask students to explain their reasons. Repeat until splitting into PSs and other solids has been achieved. From here, go into the question of why there are only five Platonic solids as described in the handout and the solutions below. Solutions: (1) When discussing the number of Platonic solids, there are two important questions to address: • Why isn’t it possible to have more than five different PSs? And • How do we know that these five possible PSs actually exist? The impossibility of more than five PSs can be proven in a number of ways. For example, imagine that we have unlimited supply of regular polygons of various shapes. If we want to use only equilateral triangles for the faces, what is the minimal number of faces that could meet at a vertex? The answer is three (resulting in a tetrahedron). If four triangles meet at a vertex we get an octahedron, and if we have five triangles at each vertex then we obtain an icosahedron. When we try to use six triangles, we see that it is impossible — each angle of an equilateral triangle is 60◦ , so putting six of them together results in a flat surface. Putting more than six triangles at a vertex is not going to work, either. (Why not?) Thus the only PSs whose faces are equilateral triangles are the tetrahedron, octahedron, and icosahedron. For the same reason, the only PS with square faces is the cube, and the only one made of regular pentagons is the dodecahedron. If we try to use regular hexagons we are out of luck since putting just three of them at a vertex results in a flat surface. Using regular polygons with more sides will not be possible. Have students justify these cases if time allows. Thus we’ve proven that we can’t have more than five Platonic solids. But do these five actually exist? Models don’t really prove it since no physical artifact can ever be made so precisely that each side of a model is exactly regular — it is always just a more-or-less close approximation.

COUNTING DIAGONALS

127

It is easy to believe that the ideal tetrahedron and cube actually exist. The existence of an icosahedron is harder to prove. Fortunately, we can use the construction shown in Figure 70.

Figure 70. Constructing the icosahedron. As can be seen in the figure, all we need are three mutually perpendicular rectangles; we connect their vertices with line segments as shown. It is not hard to calculate the length of these line segments given the length of the sides of the rectangles. It turns out that all these line segments are of the same length provided that we started with golden rectangles, i.e., rectangles whose side lengths are in the √ golden ratio 1 : 1+2 5 . Thus all faces of the constructed solid are equilateral triangles, making it an icosahedron. At this point, the idea of duality arises naturally. In addition, the use of duality helps to prove existence of the rest of PSs. Once we know that an icosahedron exists it implies that a dodecahedron exists, as well. Likewise, the existence of the cube implies the existence of its dual, an octahedron. Duality is also helpful for counting faces and vertices — the number of vertices of a solid is the same as the number of faces of its dual. (2) When we place a dot in the center of each face of a cube, and connect two points if they lie on adjacent faces of a cube, we form an octahedron. If we instead start with an octahedron and repeat the process, we obtain a cube. Doing the same with a dodecahedron, the resulting solid is an icosahedron, and vice versa. However, if we do this with a tetrahedron, the resulting solid is again a tetrahedron.

128

TATIANA SHUBIN

We say that a cube and an octahedron are dual to each other; likewise, a dodecahedron and an icosahedron are dual; a tetrahedron is self-dual. (3) If we draw an equilateral triangle, mark midpoints on each side, and chop off all three corners (each time along a line connecting two midpoints), we obtain a new equilateral triangle (see Figure 71). Now we do a similar thing with a tetrahedron. You might want to discuss how a plane in 3D space is similar to a line in the plane (2D space). For example, a line in the plane divides it into two parts, and if you were a 2-dimensional creature you wouldn’t be able to cross from one part into the other (since you cannot “hop” over the line). Likewise, 3-dimensional space is split into two parts by a plane. So while we can cut off a corner of a triangle with a line, we need to use a plane to cut off a vertex of a tetrahedron. Now ask the students to figure out what is left if we cut off each of its vertices by a plane through the midpoints of edges coming out of this vertex. This is not an easy task; allow ample time to think, and encourage students to make drawings and/or look at a model. Most likely, some students will be tempted to use the 2D example and will say that the remaining part is another tetrahedron. If this happens, challenge the students by asking how many faces the remaining solid has. In this case, four triangular faces are inherited from the original faces of the tetrahedron (see Figure 71).

Figure 71. Four triangular faces from the tetrahedron. But cutting off each of the four vertices of the tetrahedron adds four new triangular faces as shown in Figure 72. Thus the new solid has eight triangular faces.

COUNTING DIAGONALS

129

Figure 72. Creating the octahedron by cutting. Now it is easy to see that it is an octahedron. Ask what happens when the corners of an icosahedron are chopped by marking each edge with points trisecting it; then cutting off a vertex along the plane through five marks closest to the vertex. Ask students to identify a solid obtained this way. Ask students what it resembles (a soccer ball). Give its mathematical name — a buckyball — and share with them that it is a real molecule (but don’t yet reveal the C60 formula). (4) Discuss the question, “what is a diagonal?” (A line segment connecting two non-adjacent vertices, i.e., two vertices that are not connected by an edge.) Draw the table shown below on the board without filling in the numbers of diagonals for each polygon, and ask students to copy it. You might want to first discuss a triangle and square with the whole group, then let them work on the larger polygons. (Regular) polygon # of diagonals triangle 0 square 2 pentagon 5 hexagon 9 After some (most?) students fill in the table, add two more rows to the table: the 11-sided polygon (or 11-gon for short) and 100-gon. Ask them what they would do to count the diagonals in these polygons. Obviously, we need a better approach than drawing and counting! Look at a pentagon again and choose a vertex. How many diagonals arise? Draw all diagonals in this systematic way, going through all vertices in a chosen direction, e.g., counterclockwise. Can we do the same with a hexagon? A polygon with 11 or 100 sides? Can we improve the process? Discuss and write on the board:

130

TATIANA SHUBIN

The Main Idea To produce a diagonal or an edge, one needs to choose two vertices. Thus, to count all the diagonals and edges is the same as to count all possible pairs of vertices. Now we can count diagonals using the main idea. Some students might do the following: Consider the pentagon again: we can choose the first vertex five different ways; after we have chosen one vertex, the second vertex can be chosen four different ways (why?). Thus the total is 5 · 4 = 20, an absurd number. What is wrong with this count? To see what happened, label the vertices as shown in Figure 73 and list all pairs systematically:

Figure 73. Pairs of vertices in the pentagon. AB AC AD AE BC BD BE BA CD CE CA CB DE DA DB DC EA EB EC ED Note: Do only the first line AB . . . AE on the board; let students complete the table (urge them to be as systematic as possible). What do you notice? Each edge and each diagonal is listed exactly twice. So the correct way to count all the edges and diagonals would 5·4 . be 2 Since five of these are edges rather than diagonals, the total number 5·4 − 5. (Check that the result is 5, as expected.) of diagonals is 2 6·5 − 6 = 9. Now try it for a hexagon: 2 Be sure to discuss each part of the expression: • What does the 6 in the numerator of the fraction represent? The number of ways to choose a vertex. • What does the 5 represent? The number of ways to choose another vertex after the first one has been chosen.

COUNTING DIAGONALS

131

• What does 6·5 represent? The total number of all pairs of vertices where each pair is listed twice. 6·5 represent? The total number of all edges and • What does 2 diagonals. 6·5 − 6 represent? The number of all diagonals. • What does 2 Ask students to apply the same method to find the number of diagonals in an 11-sided polygon (a hendecagon!). The method above 11 · 10 would give us − 11 = 44. For a 100-sided polygon (a hectagon!), 2 100 · 99 − 100 = 4850. In general, for an n-sided polygon we would get 2 n(n − 1) − n. (sometimes called an n-gon), the number of diagonals is 2 Note: Some students might come up with different ideas, such as: n(n − 3) is an expression for counting only diagonals, since we (a) 2 can choose the first vertex in n different ways, and the second, non-adjacent vertex in (n − 3) ways. For example, in the case of a pentagon, we have the following list of pairs: AC AD; BD BE; CE CA; DA DB; EB EC, where each diagonal appears exactly twice. (b) (n−1)+(n−2)+. . .+1 is an expression for counting all edges and diagonals starting with one vertex and only adding missing pairs, e.g., for a pentagon, it corresponds to the following list: AB AC AD AE BC BD BE CD CE AE These observations prove the following algebraic identities: n(n − 1) n(n − 3) = − n; and (a) 2 2 n(n − 1) . Note that this is a triangular (b) (n−1)+(n−2)+. . .+1 = 2 number Tn−1 . Triangular numbers are discussed in some other sections of this book — in particular in “Golomb Rulers” (page 75), “Math Blocks” (page 103), and “What is the sum? — A 5-card Magic Trick” (page 25). Thus, the completed table will be as follows: (Regular) polygon # of diagonals triangle 0 square 2 pentagon 5 hexagon 9 11-gon 44 100-gon 4850 n(n−1) n-gon −n 2

132

TATIANA SHUBIN

(5) It is important in all mathematics to have a clear, shared vocabulary, but especially important when discussing the geometric components of a polyhedron. Though we have already used these names earlier in the session, this is a good place to repeat the discussion. Flat surfaces bounding a polyhedron are called faces. A line segment where two faces meet is called an edge. A point where three or more edges meet is called a vertex. Draw the table shown below on the board and ask the students to help you fill in the numbers. Finding V (the number of vertices), E (the number of edges), and F (the number of faces) in a tetrahedron and a cube is straightforward. Using duality, we get the V and F of an octahedron from the F and V for the cube. The E will be the same for both polyhedra. Checking this is easy if you hold the octahedron with one vertex on top and one at the bottom — there are 4 edges coming down, 4 going up, and 4 around the middle, for a total of 12. Ask students to count the faces of a dodecahedron (thus obtaining the number of vertices of an icosahedron, at the same time). You might ask someone to explain how they get the number (e.g., when a dodecahedron is placed on a table there is 1 face at the bottom, 5 faces coming up from this bottom face, 1 face on top, and 5 faces coming down from the top one, for a total of 12). In a similar manner, ask them to count the number of faces in an icosahedron (which, by duality, is the same as the number of vertices in a dodecahedron). What is the number of edges in a dodecahedron and an icosahedron? Let students look at the table once again and challenge them to notice any relationship between V, E, and F based on the first three rows for a tetrahedron, a cube, and an octahedron. With some little nudging, they will notice the famous Euler’s Formula: V − E + F = 2. Tell them that it holds for any convex polyhedron. Note: An alternate approach is to use a combinatorial method to find the number of edges. We illustrate it by counting the number of edges of a dodecahedron. Every face of a dodecahedron has 5 edges, and we know that there are 12 faces. If we simply multiply 5 times 12, we would count every edge twice since it belongs to 2 faces. Thus the number of edges of a dodecahedron is 5 · 12 = 30. 2

COUNTING DIAGONALS

133

Of course, we can now find E for the dodecahedron and the icosahedron. Let students do it, either using Euler’s Formula, or the counting arguments used above, or both. We are now ready to turn to the buckyball. Remember that it is obtained by chopping off the vertices of an icosahedron with planes going through points trisecting its edges. What do we see if we look at a face of the icosahedron after this slicing off the vertices? Wait to hear students’ answers. The correct answer is a regular hexagon, as shown in Figure 74:

Figure 74. Slicing the vertices of an icosahedron. Thus, every face of the icosahedron gives rise to a hexagonal face of a buckyball. Will we have any other faces there? When we chop off a vertex, do we create a face? Wait to hear students’ answers. The correct answer is a regular pentagon, since five congruent triangles meet at a vertex, and our cutting plane intersects each of these faces along a straight line, thus producing five edges. Therefore, every vertex of the icosahedron gives rise to a pentagonal face of a buckyball. So, for a buckyball, we have: F = (20 hexagonal faces) + (12 pentagonal faces) = 32. Make sure to wait for the students to come up with their own count! How many vertices does a buckyball have? Remind students to think of the way vertices of the buckyball are created: we trisect each edge of the icosahedron. Thus V = (# of edges of an icosahedron) · 2 = 30 · 2 = 60. Recall that a buckyball has the shape of a real molecule! One atom of carbon is situated at each vertex of the buckyball. Thus the molecule is C60 . How do we find the number of edges? Answer: Use Euler’s formula: 60 − E + 32 = 2, so E = 90. The completed table will therefore be:

134

TATIANA SHUBIN

SOLID

V vertices E edges F faces 4

6

4

8

12

6

6

12

8

20

30

12

12

30

20

60

90

32

tetrahedron

cube

octahedron

dodecahedron

icosahedron

buckyball (6) We now turn to counting diagonals in the 3D case. Notice that there is a little twist when counting polyhedron diagonals — in a polyhedron, some diagonals lie on its faces (and are called facial diagonals), while the others do not (those are called spatial (or non-facial, or inner ) diagonals). Our goal is to count the number of spatial diagonals of a buckyball. Let students start by attempting to determine the number of spatial diagonals in a tetrahedron, a cube, and an octahedron. At some point, remind them of the Main Idea (counting the number of all the diagonals and edges of a buckyball is the same as counting all possible pairs of the vertices) and ask them how it might help. With time, they will conclude that: (# of spatial diagonals) = (# of pairs of vertices) − (# of edges) − (# of facial diagonals). They also note that (# of facial diagonals) = (# of faces) × (# of diagonals per face). Using this equation and the two tables (both tables should remain on the board) we arrive at the following results (make sure that students do it on their own): 4·3 −6−4·0=0 For a tetrahedron: D = 2

COUNTING DIAGONALS

135

8·7 − 12 − 6 · 2 = 4 2 6·5 − 12 − 8 · 0 = 3 For an octahedron: D = 2 20 · 19 − 30 − 12 · 5 = 100 For an dodecahedron: D = 2 12 · 11 For an icosahedron: D = − 30 − 20 · 0 = 36 2 60 · 59 − 90 − 20 · 9 − 12 · 5 = 1440 For a buckyball: D = 2 n(n − 1) for counting (7) Now that we have repeatedly used an expression 2 the number of ways to choose 2 vertices from the set of n vertices, we might notice that the same expression can be used for finding the number of ways to choose any 2 objects from a collection of n objects, regardless of the nature of these objects. This observation makes solving the rest of the problems much simpler — let students do it! Finding the number of roads connecting a certain number of towns is equivalent to finding the number of pairs of towns, i.e., the number of ways to choose 2 towns out of the total number of towns. If there 10 · 9 = 45. If there are 73 towns, are 10 towns, the number of roads is 2 73 · 72 = 2628. the number of roads is 2 12 · 11 = 66, so Monster (8) The number of possible pairs of co-captains is 2 Slayers will be able to choose their co-captains without repeating the same pair for the entire year. (9) Let’s look at one possible way, shown in Figure 75, that Devon can go from where he is to his hooghan. Along the path he passes 10 road intersections (counting the intersection where he is, but not counting the intersection where his hooghan is located). If we mark each of these 10 intersections either U or R according to whether Devon goes up or to the right at this intersection, then this path can be described as URRRRRRURR. Clearly, every possible path can be described by a sequence of 10 letters R and U, of which exactly 2 must be U. In fact, all we need to do to specify a particular path is to choose 2 places out of 10 and to put a U in each of these two places. Thus the total 10 · 9 = 45. number of ways Devon can travel is 2 For a cube: D =

136

TATIANA SHUBIN

Figure 75. One path for Devon to get to his hooghan. (10) A regular octahedron. (11) The name of the solid obtained is a Rhombic Triacontahedron (see Figure 76). It has 30 faces, 60 edges, and 32 vertices. All its faces are congruent rhombuses. It’s interesting to note that the longer diagonals of the faces form a regular icosahedron, and the short diagonals of the faces form a regular dodecahedron.

Figure 76. A Rhombic Triacontahedron (12) Consider any one of the five inscribed cubes (as in problem 11). It can be seen that a spatial diagonal of the cube is a diameter of the circumscribed sphere. The length of an edge of the cube is that of a diagonal of a regular pentagon with√the edge of length 1, and therefore is equal to the golden ratio φ = 1+2 5 (this can be derived by considering two pairs of similar triangles formed by edges and diagonals of the pentagon). Using Pythagorean Theorem twice, one finds that the √ diameter of the sphere is φ 3.

Liar’s Bingo Contributed by Bob Klein Short Description: Participants examine a handful of paper strips covered with red and black numbers. After identifying any patterns they see, the facilitator performs a magic trick (plays Liar’s Bingo), and the hunt begins for the “secret” to the magic trick. This session involves recognizing pattern and searching for underlying structure, number theory, numeration, and potentially binary arithmetic. Materials: Liar’s Bingo cards. These can be downloaded at www.ams.org/ bookpages/mcl-24, printed in color, and cut into strips. The facilitator should do this in advance because the strips are printed in a convenient order that reveals too much to the participants if they cut up the strips. Giving a complete set of strips to each group of 3–4 students might be optimal. Also, a classroom board or a chart paper are useful, and a document camera might be helpful. Mathematics Beneath & Beyond: In base-ten, our place value positions proceed according to powers of ten from right to left. Binary numeration is similar, except that it records the numbers using only two digits, 0 and 1, with place-value positions according to powers of two — ones, twos, fours, etc. The session’s connections to binary and base-8 representation are evidence of the game’s connections to computer engineering — a connection worth making to the participants (a search of online videos with the terms “binary computers” turns up some helpful background materials). References/Authorship: This session was shared with me by Mr. Steve Phelps, Madeira High School. He learned it from Chuck Sonenshine (http://www.keynotemath.com/).

137

138

BOB KLEIN

Liar’s Bingo Teacher Guide: Presentation The human brain is an amazing pattern-processing machine. Many games depend on this and the best music, art, and stories use disruption of patterns to add interest — think of syncopation in music (perhaps a switch to a minor key in an upbeat tune), a bright red balloon held by a little girl in an otherwise black-and-white photograph, or the twist of a plot when we learn that Luke Skywalker has a sister. Sometimes, as in the game of Liar’s Bingo, order seems to arise magically from something we first assume to be random or chaotic. In this case, we use the game of Liar’s Bingo to engage participants’ desire to find patterns, and supercharge that desire by demonstrating a magic trick that captivates attention by disrupting participants’ expectations that the Liar’s Bingo cards are without pattern. I have led this activity more than twenty times with audiences ranging from fifth-grade participants to retirees, with backgrounds ranging from not yet ready for fraction arithmetic to PhDs in mathematical research. As my colleague Steve Phelps points out, “Patterns are the great leveler,” and Liar’s Bingo is suitable for that fifth grader and the PhD in mathematics to work side-by-side discovering the patterns at relatively the same pace. It exemplifies what Jo Boaler calls “low-floor, high-ceiling” problems: everyone can get started (low-floor) and everyone can find challenge (high-ceiling). Don’t be surprised if every time you lead this session, that a participant points out a new pattern in the cards that you have never seen before. Of the 20+ times I’ve led this, only once have I not been presented with a pattern previously unknown to me. This makes it fun and fresh for the facilitator each time. (1) Objective 1: Get participants to engage with the cards and to begin noticing patterns. Don’t mention the “magic trick” at this stage. Launch the activity by asking participants if they would like to learn a new game called “Liar’s Bingo.” I usually then tell them that I’m going to hand them a handful of strips of paper and that they should lay them out on the table in front of them. There are two prompts that work well for launching this. Either (1) ask them to put the cards at their table in order, or (2) ask them to write down everything that they observe about the cards in front of them, no matter how straightforward. In the case of (1), participants will inevitably ask what you mean by “in order” and you should instruct them, “I’m not going to say. Whatever ‘in order’ means to you.” In the case of (2), participants often start by looking for something deep, but here everything is important so find a way to encourage them to note that, for instance, there are six “cells” of numbers, or that some of the numbers are red and some black. The objective of this (10–15 minutes) is to get the class to note as many observations or patterns as they can find. (2) Objective 2: Collect these patterns for the whole class. Have groups share out, perhaps noting the patterns on the classroom board or chart

LIAR’S BINGO

139

paper for all to see. In some cases, a document camera may be helpful as participants have to use cards and pointing to evidence the pattern they see. (3) Objective 3: Introduce the game. I usually ask participants to “keep these patterns/observations in mind” and that “they may be useful and we’re going to revisit them.” Ask participants if they would like to learn to play Liar’s Bingo. In order to play Liar’s Bingo, you must first teach them to play Truth-teller’s Bingo. Show the class a card that has a mixture of red and black numbers, such as the one below (a document camera may be helpful here if it is available). To play Truth-teller’s bingo we read the colors only (not the numbers) from left to right. In the case of the card below, we would say, “Black, Red, Black, Black, Red, Red.”

63 3 33 27 21 22 Pick another card and have participants play Truth-teller’s Bingo one more time. When you are certain that they understand, congratulate them on being great players of Truth-teller’s Bingo and assure them that they are ready to learn Liar’s Bingo. The game of Liar’s Bingo is just like that of Truth-teller’s Bingo except that the player lies about just one color. For instance, in the card above, the player might decide ahead of time that they are going to lie about the third cell from the left (the 33) and they would therefore read “Black, Red, Red, Black, Red, Red.” Pick another card, pick a cell, and ask them, “if we were to lie about this cell [indicating one of the cells], how would we read this card?” When you are convinced that they understand, ask someone across the room to play Liar’s Bingo to convince you that you understood. Have their table make sure that he/she is playing correctly. After they read the colors, ask that participant’s table, “did he/she do it correctly?” If they say yes, then respond with, “so he lied about 27?” (or whatever the corresponding number is). Since you could not see the card and since the participant only read the colors, they will be astounded. Let the room understand what just happened. Sometimes when the participant is reading the colors I will pretend to be busy, not looking at them, by erasing the board with my back to them, adding to the effect. Invite more participants to play against you and guess the number each time. I usually ask them how they think I know. “Do I have ESPN?” (this usually gets a laugh). “Did I memorize all of the cards?”

140

BOB KLEIN

(4) Objective 4: participants should uncover the trick. In some instances, the class may need overnight to continue working on this. After 5–8 correct guesses during the class, tell them that they are going to have to use those patterns they figured out (in Objective 1) to understand the trick. Give them some time to talk in their groups and circulate, offering to play with anyone. The following prompts help participants to focus on keys to unlocking the trick: • What’s the smallest number you see? The largest? What’s the smallest digit you see? The largest? • What would be a useful sequence of colors to ask me about? (Going for all black (0) or all red (77) here as a starting point). Have participants write these down. • If you were to lie to me about zero, what would you say to me? About one? Two? • participants usually develop the notation BRBBRR (using letters to represent the colors) but if they don’t, you might encourage its use. Timing is hard to predict and every group of participants is different. As a few participants indicate that they “get” the trick, encourage their peers to test them by having those participants play and guess the number. Use those participants to help others begin to understand the trick (not to tell them how to do it but to help them discover it). You may have the feeling that the participants will never get it, but in the many times I have led it, almost all of the participants understand this by the end of 1–1.5 hours. Avoid the urge to say too much — they have far more fun when left to discover it with your nudging than if you tell them how to do it. (5) Objective 5 (Optional but important): Connect the understanding of how to do the trick with the patterns identified in Objective 1. There is underlying structure that makes the trick work and this structure is coded into the cards. I have had participants who claim to understand the trick go home and make their own set of cards to convince me. I have also given participants something like the card below, with only the first cell completed and asked, “Is this enough? Can you tell me what all the other cells should be based on just that cell?” (Without referring to the pre-made cards).

23 Alternatively, you might give them a card with only the third or other cell shown and ask the same question. A participant who can

LIAR’S BINGO

141

complete the card above could be challenged to do the same with a card that begins with a black 23 (instead of the red 23 above). As explained in the Understanding the Trick section below, a Liar’s Bingo card is strictly determined by the choice of colors for each cell. If the cell containing the red 23 is lied about then in order to obtain 23 one needs to hear the sequence BRBBRR. Therefore, the correct colors on the strip are RRBBRR. Using the analysis similar to that represented in the Mathematics Beneath & Beyond we see that the strip which starts with a red 23 must be

23 43 73 67 61 62 A strip which starts with a black 23 will only work if we assume that Red stands for 0 while Black stands for 1. Depending on the list of patterns identified, some important connections should be probed: • why we need two colors. What if we had three colors? • how many cards constitute a complete set? • why there are no digits higher than 7 and no numbers higher than 77? • why we have six cells. (How) could we use this same system to represent three digit numbers? Which three digit numbers could we (not) represent in that case? These connections (to the underlying structure and patterns of the existing cards) and modifications of the existing system (more colors, more cells, etc.) are investigations in their own right that give participants ways to keep thinking about the session. In classroom use, they could be extensions for participants who are working at a faster pace and want further challenge in class. They could also be used in a computer programming context to challenge participants to code the algorithms necessary to (1) create a complete set of cards and/or (2) play the game. Steve Phelps has programmed GeoGebra, for instance, to supply a random Liar’s Bingo card each time a button is clicked.

142

BOB KLEIN

Liar’s Bingo Teacher Guide: Explanations Liar’s Bingo is, at its heart, about finding joy in discovering patterns and using them to unlock a mystery. The facilitator’s role then is to provide just enough nudging to allow participants the time and space to experience the joy in discovering. There is a fine line between productive struggle and frustration leading to resignation. Facilitators should celebrate insights widely to keep participants motivated and believing that they will figure it out. Understanding some of the patterns that participants may notice and struggles they may encounter will help facilitators to guide participants in this way, to keep things oriented toward fun and productive struggle. The notes below may be helpful to that end and are given in list format for easy reference. Understanding the Trick. There are several ways to understand the trick. Perhaps the simplest involves dividing each color sequence in half (the first three colors and the last three colors). Understand that the first three colors determine the tens digit of the number being guessed and the last three colors determine the ones (unit) digit of the number being guessed. When a participant gives you the sequence, BRRRBR, the first three colors (BRR) determine the tens digit, and the last three colors (RBR) determine the ones digit. To determine the tens digit and ones digit, we use binary numbers. Recall that for base-ten numeration, we have names for the first ten whole numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and then we use place value to represent any other whole number (other number sets aren’t discussed here). Hence, the number 3140 represents 3 thousands, 1 hundred, 4 tens, and 0 ones. In binary, we use only the digits 0 and 1. For place values, instead of ones, tens, hundreds, we have ones, twos, fours, eights, and so on, or 20 = 1, 21 = 2, 22 = 4, 23 = 8, etc. Hence, the number 11010 would represent 1 sixteen, 1 eight, 0 fours, 1 two, and 0 ones. We know this in base ten as the number 16 + 8 + 2 = 26. If we wanted to count in base-2 (binary) up to 26, it would look like: 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010. In Liar’s Bingo, we are using Red to indicate a ‘1’ and Black to indicate a ‘0.’ Hence writing the first seven numbers in binary (0, 1, 10, 11, 100, 101, 110, 111) can be done using these colors: B, R, RB, RR, RBB, RBR, RRB, RRR. It takes only three cells to represent the numbers 0 to 7. This is a key insight for why there are no 8’s or 9’s on the cards and why the Liar’s Bingo cards can represent the set of numbers present on the cards. So the sequence RBR represents 4 + 0 + 1 = 5. If these are the first three colors of the six-color sequence, they represent 5 × 10 = 50. If they are the last three colors of the six-color sequence, then they represent 5. Hence, the sequence RBRRBR represents the number 55.

LIAR’S BINGO

143

Suppose a participant tells you the sequence BRRRRR. The first three colors are BRR indicating 0 fours, 1 two, and 1 one, or 2 + 1 = 3. But since these first three colors represent the tens-digit of the number we are guessing, we know that the “magic” number is thirty-something. The last three colors in the sequence are RRR indicating 1 four, 1 two, and 1 one, or 4 + 2 + 1 = 7. Because these represent the ones digit of the “magic number”, we now know that the number the participant lied about is 37. If a participant says BBBBBB we know there are 0 tens and 0 ones (represented by BBB and BBB respectively) so the number lied about is 0. If they give the sequence RRRRRR then they are lying about the number 77 (RRR indicates 7, in this case 7 tens and 7 ones). In fact, a Liar’s Bingo card is strictly determined by the choice of colors for each cell. Let’s consider an example. Suppose that the cells are BRRBBB. If the first (from the left) color is lied about then the leader will hear the sequence RRRBBB which encodes the number (4 + 2 + 1) × 10 + (0 + 0 + 0) = 70. Thus, the first number on the card must be 70. In fact, we obtain the following table: Color lied about (counting left to right) 1 2 3 4 5 6

Sequence obtained

Number encoded in the cell

RRRBBB BBRBBB BRBBBB BRRRBB BRRBRB BRRBBR

(4 + 2 + 1) × 10 + (0 + 0 + 0) = 70 (0 + 0 + 1) × 10 + (0 + 0 + 0) = 10 (0 + 2 + 0) × 10 + (0 + 0 + 0) = 20 (0 + 2 + 1) × 10 + (4 + 0 + 0) = 34 (0 + 2 + 1) × 10 + (0 + 2 + 0) = 32 (0 + 2 + 1) × 10 + (0 + 0 + 1) = 31

Thus, the card must be

70 10 20 34 32 31

Notes by Objective (1) Objective 1: Get participants to engage with the cards and to begin noticing patterns. • The Launch Prompt. It is difficult to know in advance whether the first prompt (“order the cards”), or the second (“what do you notice?”) will work best. Having participants order the cards (whatever that might mean to them) is often a more focused task and can work better for participants who might not have had much experience with “what do you notice?” questions before. That said, if in some alternate reality I were asked to lead Liar’s

144

BOB KLEIN

Bingo at a conference for poets, I might use the second prompt since the idea of ordering might be more intimidating than that of creative observation. The choice should be motivated by the facilitator’s knowledge of the participants and their background. I often start with the first (order these) prompt and then use the second (what do you notice) prompt as I walk around the room from group to group. Groups will almost always order the cards numerically and in increasing order according to the first cell. Some striking patterns can emerge from this (see below). Note, though, that this is not the only interpretation of order. I have had groups order by getting all cards with a zero and lining them up so that the zeros form a diagonal (zero in the first cell in the first card, in the second cell on the second card, and so on); they then do similarly for cards with a ‘1’ and so on. Other groups have ordered them according to the number of red cells or black cells. In each case, interesting patterns emerge. • A sample of patterns: – There are six cells. – Some numbers are black and some are red. – The smallest digit is 0 and the largest is 7. There are no 8’s or 9’s on any card. – The largest number is 77. – Black numbers decrease left-to-right and red numbers increase left-to-right. – The first three cells (left-to-right) have equal ones digits and the last three cells have the same tens digit (in some cases an implied zero for the tens digit). Each of the patterns the class generates can be empirically tested as a conjecture. When a group offers a pattern such as “Black numbers decrease left-to-right,” tell the participants that we now have a conjecture and we can test that conjecture by looking at the cards for a counterexample. Someone may ask whether or not you have handed out “all of the cards,” that is, a complete set. Assure them that you have. In such a case, conjectures can be verified by the whole class determining if a counterexample exists. Each of the patterns the class generates may be connected later to the structure underlying the trick. This session demonstrates well that knowing how is not always knowing why. In this case, participants might know how to do the trick, but not understand the underlying structure that explains why the trick works and where the patterns above come from. (2) Objective 2: Collect these patterns for the whole class. Ask each group to contribute one pattern or observation they noticed regarding the cars. Keep polling the groups until you have them

LIAR’S BINGO

145

all (or until you deem it important to move on). The facilitator, having walked around the room during the observation time should be aware of what the groups noticed and can help motivate them to contribute if necessary. Keep each group engaged by asking them to look at their cards to verify or to find a counterexample each time a group makes an observation. For instance, “Look at your cards. Does anyone have an example where the black numbers do NOT decrease reading left to right?” Use vocabulary like “conjecture” and “counterexample” as they are key to reasoning mathematically. (3) Objective 3: Introduce the game. If you were careful not to mention “magic trick” before then participants should be astounded when you guess the number. At some point the participant might ask about common features of games that aren’t actually present here, like, “how do you win?” or “do you take turns?” I usually tell them simply, “it’s not that kind of game.” It is important to move past the “magical wonder” and shock they have at your ability to correctly guess the number to focus them on wondering how the trick is done. Push them to ignore the cards after a while and to start to think about the sequence of colors and what makes sense to ask (like all black or all red). One misconception that arises each time I do this is that participants will look at the all black strip 40-20-10-4-2-1 and wonder where zero is. Why does BBBBBB represent zero when there isn’t a zero on the all black strip. Remind them that when we say, “BBBBBB,” that we are lying about one of the colors. That is, the person who says, “BBBBBB” has to be looking at a card with one red (and lying about it). Participants will also want to know if they can lie about the colors of more than one cell at a time. The answer is, “no.” We are trying to grow great mathematicians after all, not great liars. (4) Objective 4: Participants should uncover the trick. This objective is the hardest to determine how long it takes. Some participants discover the trick more quickly than others. Have these participants’ fellow group members play the game with them to be sure that they understand. Then use them to help guide the others toward discovery (this is different from simply teaching the others how to do it). Sometimes this portion of the session takes 20 minutes and sometimes it takes longer. In some cases, it has taken 30 minutes to have most understand how to do the trick. As mentioned above, celebrate people’s observations in the service of continuing their motivation to keep discovering the trick. Emphasize using notes and writing down sequences of colors and the values they represent. This is a good time to make a table of sequences and values (BBBBBB is 0, BBBBBR is 1, and so on). (5) Objective 5 (Optional but important): Connect the understanding of how to do the trick with the patterns identified in Objective 1.

146

BOB KLEIN

In the 1–1.5 hour sessions I have used in the past, we don’t always get to work on these, but it is important to emphasize that the patterns identified by the class can be explained by the underlying structures of the cards, including properties of binary numbers and the use of 3 + 3 cells to represent the tens + ones digits of the “magic number.” Some notes below explaining the patterns above should get you started. It’s okay not to know in advance all of the patterns and why they might not work. I frequently am confronted with a pattern someone discovers in this activity that I can’t explain at the outset. It’s good to mention new patterns when they arise and to admit when you can’t explain it. But it is not okay to evade the issue — encourage the group as a whole to consider those moments important open problems that the group should work on. Be sure to follow up in subsequent sessions to see if anyone made progress on those open problems. Sample patterns from above: • There are six cells. We only need three cells to represent the digits 0–7 but we have two decimal places so we need three cells for the ones and three cells for the tens. If we wanted to expand this to three-digit numbers (using only 0–7 as our digits) then we would need 3 + 3 + 3 = 9 cells to represent the number. Hence BRRBBRRRB would represent the number 316. It might be fun to encourage students to manufacture a set of 9-celled Liar’s Bingo cards. Students often ask about using 4 + 4 = 8 cells (four cells for the tens digit and four cells for the ones digit) so that we could then represent 8s and 9s. What would happen if we did that? Some of the numbers wouldn’t have unique representations. For instance, looking only at the four cells representing the ones place, what would RRBR represent? In binary, we know that this would be 1 eight, 1 four, 0 twos, and 1 one, or 8 + 4 + 1 = 13. Can we have a 13 in the ones place? Consider two cards:

B B B B R R B R and

B B B R B B R R Both of these sequences represent the number 13 even though they are different sequences. What would the numbers on the cards have to be? Would it be possible for the students to lie about these cards? The answer is left as a gift to you, dear reader, to figure out. • Some numbers are black and some are red. These represent 0 and 1 respectively. But what if we could use three colors? Could we then use base-three representation by having Black, Red, and Yellow represent 0, 1, and 2 of each place value? How would we have to construct the cards in such a case? Which numbers could

LIAR’S BINGO

147

we then represent? Again, the answers to these questions are a gift to the reader and beyond the scope of the Liar’s Bingo script offered here. • The smallest digit is 0 and the largest is 7. There are no 8’s or 9’s on any card. As mentioned above, three places can only represent 0–7 in binary. Representing 8 or 9 in binary (1000 and 1001) requires a fourth place. • The largest number is 77. Per the line above, if the maximum tens or ones digit is 7, then the maximum number must be 77. • Black numbers decrease left-to-right and red numbers increase left-to-right. Reading left to right, suppose one of the cells has a black number in it. In the example on page 139 and reprinted below, the first black number is 63. It is 63 because lying about the cell means saying RRBBRR which corresponds to 40 + 20 + 2 + 1 = 60 + 3. Moving to the right means decreasing in place-value from 40, to 20, to our next black number (in the case of this strip it is in the 10 place-value or third square from the left) that is 33. Lying about this cell means changing the 10 place-value from black to red, adding only 10 to the eventual sum rather than the 40 that we added by changing the first cell to red. We are only turning on the 10s place as opposed to the 40s place before. Hence our eventual magic number is worth 40 − 10 = 30 less than when we lied about the first cell. So moving left to right, changing black to red means adding smaller and smaller amounts to the sum indicating the “magic number” so it makes sense that the black numbers decrease left-to-right. The case for reds increasing left-to-right is similar.

63 3 33 27 21 22 • The first three cells (left-to-right) have equal ones digits and the last three cells have the same tens digit (in some cases an implied zero for the tens digit). Because the first three cells represent the tens place only, the ones digit must be the same. Similar reasoning justifies why the tens digit of the last three cells are equal.

Parity and Other Invariants Contributed by Bob Klein and Tatiana Shubin Short description: The session presents an introduction to a very powerful technique of problem solving, i.e., finding and using invariants. We investigate invariants of several types: parity (evenness/oddness), a remainder upon division by a given number, a recurring pattern in a sequence, and also invariants of a geometric nature such as area and perimeter. Materials: Students will certainly need lots of scratch paper, pencils and/or pens. In addition, you might need some other materials depending on your choice of problems: • plastic caps for Warm-up problem • some tokens (e.g., dry beans, or pennies, etc.) for Warm-up game • ’hats’ of two colors (they might be made by cutting paper into several strips and stapling them into bands as in the picture below; cardboard stock works very nicely) for problem 4 • grid paper is useful for problems 6, 7, 8, 14. Mathematics Beneath & Beyond: Invariants are powerful tools in mathematics and beyond. They play an important role in computer science, physics, and other sciences. In higher mathematics, many important investigations are concerned with finding the full set of invariants of some objects under a given set of transformations. Some examples of invariants from the school curriculum are: • Cutting and rearranging parts of a flat figure does not change the area. (The Dissection Principle). • The area of a triangle does not change if one of its vertices moves along a line parallel to its base. • If a point P is not on a circle, and a line through P intersects the circle at points X and Y , then the product P X · P Y is an invariant (i.e, such product will have the same value for any line through P ). • For any convex polyhedron, V − E + F = 2, where V is the number of vertices, E is number of edges, and F is the number of faces. (Euler’s Formula). After introducing the definition of invariant in one of our math circles, one of our more thoughtful and engaging students wrinkled her brow, raised 149

150

BOB KLEIN AND TATIANA SHUBIN

Figure 77. Teachers at a Math Teachers’ Circle session practicing the solution to problem 4. her hand, and said, “so an invariant is like Taylor Swift always being unlucky in love?” Exactly.

Parity and Other Invariants Student Handout (A) There are 17 upside-down cups on a table. During each turn you must make exactly 4 flips (one cup can be flipped four times; or one cup can be flipped three times and another cup flipped once; or two cups can be flipped twice each; or one cup can be flipped twice while other two cups flipped once each; or each of the four cups can be flipped once). The goal is to put all 17 cups upright after several rounds. Can you do it? (B) A number of participants sit around a table. Each starts with an even number of tokens in front of them — some may have 20, some 14, some just 2, etc. The initial distribution is random, but every person must have an even number of tokens. There is also a reserve supply of the tokens in a bowl in the middle of the table. Each participant divides their tokens in two equal piles, and shifts one of their piles to the person on their right. They do the shift simultaneously. Everyone counts their tokens and each of the participants who has an odd number of tokens takes one additional token from the bowl in the middle of the table. Now they repeat the previous step — divide their tokens in two equal piles and shift one of the piles to the

PARITY AND OTHER INVARIANTS

151

right. Repeat this process until something interesting happens. What is it? Why does it happen? (C) Each cell of a square contains a digit as shown in Figure 78.

1

8

4

6

3

9

5

7

2

Figure 78. A grid of digits. You can start at any cell of the square and walk into any neighboring cell (they must share a wall like 6 and 3 above — 6 and 7 are not neighbors above). You may not visit any cell more than once. Ashly walked through the cells as shown in Figure 79 and wrote down all digits she encountered along the way, thus getting the number 84937561.

1

8

4

6

3

9

5

7

2

Figure 79. Ashly’s path through the cells. Draw a path that results in the largest possible number. Explain how you know that the number you’ve got is, indeed, the largest. (D) Is it possible to cut a 13 × 5 rectangle into pieces and then rearrange those pieces so that they form an 8 × 8 square?

152

BOB KLEIN AND TATIANA SHUBIN

Problems: (1) There are 5 red marbles and 6 green marbles in a jar. Devon plays a strange game. He removes two marbles at a time, with the following rules: (a) If the marbles are both green, he puts one green marble back. (b) If there is one marble of each color, he puts one red marble back. (c) If the marbles are both red, he puts one green marble (from his infinite supply of different marbles) back in the jar. If Devon starts playing this game, will it ever end? If it does end, how will it end? (2) Numbers 1, 2, 3, . . ., 2999, 3000, 3001 are written on a board. You are allowed to replace any two of these numbers by one number, which is either the sum or the difference of these two numbers. You perform this operation 3000 times. How many numbers are left on the board? Can that number be 6000? (3) A grasshopper jumps along a line. His first jump is 1 cm in length, his second jump is 2 cm in length, and so on. Each jump can take him to the left or to the right. Is it possible that after 3002 jumps the grasshopper returns to the point at which he started? (4) A witch caught 100 dwarves and she decided to give them a test. If they pass, they all go free. Otherwise, they all become the witch’s captives for life. The test is as follows: the dwarves stand in a line, all facing to the right. The witch puts either a red or a blue hat on each dwarf. Each dwarf can only see the colors of the hats in front of him; he can’t see the color of his own hat or the hats of the dwarves behind him. Then, in any order they want, each dwarf guesses the color of the hat on his head, and says either “Red” or “Blue.” No other words are allowed and no other signals like coughing, throwing pieces of paper, touching each other, or anything of that sort is permitted. To pass the test, no more than one of them may guess incorrectly. That is, they can only make one mistake if they are to go free. If the dwarves can agree on their strategy beforehand, can they be sure that they will all go free? (5) Of 101 coins, 50 are counterfeit, and, although some can be heavier and some lighter, each differs from the genuine coins in weight by 1 gram. Jaime has a balance scale that shows the difference in weights between the objects placed in each pan. He chooses one coin, and wants to find out using the scale only once whether the coin is counterfeit or not. Can he come up with a method guaranteed to work every time? (6) You start with an 8 × 8 chessboard. For some reason, your chessboard is missing two squares at opposite corners. You have 31 dominoes, and you want to tile what’s left of the chessboard with these dominoes. (A domino is a 2-by-1 tile and so it covers exactly two squares of the chessboard.) Can you do it? (7) Is it possible to tile a 10 × 10 grid with twenty-five 1-by-4 tiles?

PARITY AND OTHER INVARIANTS

153

(8) A rectangular floor is covered by 2 × 2 and 1 × 4 tiles. One tile got smashed, but we have one more tile of the other kind available. Can we retile the floor perfectly? (9) A dragon has 100 heads. A knight can cut off either 15, 17, 20, or 5 heads, with one blow of his sword. In each of these cases, 24, 2, 14, or 17 (respectively) heads grow on its shoulders. (Cutting 15 heads off regrows 24 heads in its place, cutting 17 heads regrows 2 heads . . .) If all heads are cut off, the dragon dies. Can the dragon ever die? (10) Prince Ivan is on his way to free his younger brother imprisoned by their enemies when he is confronted by an evil dragon. Ivan has two magic swords. With each blow of the first sword Ivan can cut off 21 heads of the dragon. With each blow of the second sword he can cut off 13 heads but the dragon will immediately grow 594 new heads. In order to continue his journey, Ivan must cut off all the dragon’s heads. Could Ivan succeed if originally the dragon had 100 heads? (11) (a) A mad veterinarian has invented an animal transmogrifying machine. If you put in two cats or two dogs, then one dog comes out of the machine. If you put in one cat and one dog, then one cat comes out. The veterinarian’s goal is to end up with only one cat and no other animals. If he starts with 3 cats and a dog, can he achieve his goal? What if he starts with 13 cats and 10 dogs? Could you describe all starting collections which allow the veterinarian to achieve his goal? (b) The veterinarian’s old machine breaks. Now he has cats, dogs, and mice. The new transmogrifying machine can take in any two different animals and then out comes the third animal. Can the veterinarian achieve his goal of ending up with exactly one cat and no other animals if he starts again with 3 cats and 1 dog? With 4 of each animal? (c) Suppose now that you can use the machine going in any direction, forward or backward at your wish. The forward machine is described in (b). The backward machine receives one animal and produces two new animals, one of each of the two other kinds — for example, if you put in a dog then you receive a cat and a mouse. Can you get exactly one cat and no other animals if you start with 3 cats and 1 dog? With 4 animals of each kind? (12) On a planet far away, the only inhabitants are chameleons. They come in three colors: green, yellow, and red. If two chameleons of the same color meet nothing changes. But if two chameleons of different colors meet, they both change to the third color. If there were 4 green, 5 yellow, an 5 red chameleons to begin with, is it possible that all chameleons will eventually be of the same color? What if there were 13 green, 15 yellow, and 17 red chameleons to begin with? Try some other initial numbers; what do you observe?

154

BOB KLEIN AND TATIANA SHUBIN

(13) (a) Numbers 1,2,3, . . ., 100 are written on the board. Each minute, we erase any two of them (say, x and y), and write their sum, x + y. What happens in the long run? (b) What if in the previous problem x + y is replaced by xy? (c) What if x + y is part (a) is replaced with xy + x + y? (14) An infection spreads among the squares of a 10×10 grid in the following way: if a square has two or more infected neighbors, then it becomes infected itself (a neighbor being a square which shares an edge). If initially there were 9 infected squares, is it possible that all 100 squares will eventually become infected?

Parity and Other Invariants Teacher Guide Warm Up Problem Solutions: (A) After n turns, the total number of flips is 4n, an even number. It takes an odd number of flips to put one upside-down cup upright. Hence the total number of flips required to put all 17 cups upright needs to be also odd (as the sum of an odd number of odd numbers). Thus, it is impossible to make all cups upright. (B) The number of tokens will eventually equalize — everyone will end up with the same number. A rough explanation of this astonishing result is that if we look at the largest number anyone has at each round, that number can only remain the same or become smaller. Likewise, the smallest number of tokens someone has at a given round will either remain the same or get bigger. The process will continue until the biggest number will stop getting smaller and the smallest number will stop getting bigger — and this will happen when all numbers are the same. (C) The largest number is 573,618,492. Here’s an explanation that it is indeed the largest. First, color a 3×3 square like a checkerboard as shown in Figure 80

Figure 80. Coloring a 3 × 3 square like a checkerboard. If you’re at a white square (W) you can only move to a black one (B), and if you’re at a black square you can only go to a white one. Thus, if you start at a black square you can’t get a 9-digit number — if you did, your path would look like BWBWBWBWB which is impossible since there are only four black squares. Every 9-digit number is

PARITY AND OTHER INVARIANTS

155

bigger than any 8-digit number, so we must start with a white square, and we need to pick the largest number on a white square, which is 5. From there we can either go to 6 or to 7 so to make a larger number we would have to choose 7. From there, we have a choice between 2 or 3, so choosing 3 will make the number bigger. And from 3 we can only go to 6 — otherwise we will not be able to go through all the remaining squares. We do not have any choices for the rest of the path: 6 → 1 → 8 → 4 → 9 → 2. Therefore, we see that the largest obtainable number is 573,618,492. (D) The area of a 13 × 5 rectangle is 65, while the area of an 8 × 8 square is 64. Since cutting and rearranging pieces do not change the area, therefore the task is impossible. Problem Solutions: (1) Consider what happens at a certain step: If Devon removes two green marbles and puts one green marble back, then the number of red marbles in the jar remains the same. If Devon removes one marble of each color and puts back a red marble, then again, the number of red marbles in the jar stays the same. If Devon removes two red marbles and replaces them with a green one, then the number of red marbles in the jar goes down by 2. Hence, in any case the parity of the number of red marbles does not change. Since the initial number of the red marbles was odd, it means that the last marble remaining in the jar cannot be green — if it is, then the number of red marbles is 0, which is even, and this is impossible. Hence the last marble remaining in the jar is red. Question: What if Devon starts with 6 red and 6 green marbles? (2) Notice that the solution here is virtually identical to the previous problem’s solution. At each step, the total number of the numbers on the board goes down by one. Hence after 3000 steps exactly one number remains on the board. The sum and the difference of two numbers of the same parity is even; the sum and the difference of two numbers of different parity is odd. Thus, if at a certain step we choose two even numbers they will be replaced by another even number; after this step, the total number of odd numbers on the board does not change. If we choose two odd numbers, they will be replaced with an even number; hence the number of odd numbers will go down by two. If we choose an even and an odd number, they will be replaced by another odd number; hence the total number of odd numbers remains the same. Thus, at each step the total number of odd numbers written on the board either remains unchanged or decreases by 2, making its parity an invariant (a feature of the problem that remains unchanging). Initially, there were 1501 odd numbers, which is odd. Hence the last number which remains on the board after 3000 steps must also be odd, and therefore

156

BOB KLEIN AND TATIANA SHUBIN

it cannot be 6000. Question: What if we started with 1, 2, . . ., 3000 instead? (3) Every odd-numbered jump moves the grasshopper an odd number of centimeters, and every even-numbered jump moves him an even number of centimeters. If he takes 3002 jumps, he has taken 1501 evennumbered jumps: they were numbered 2, 4, 6, . . ., 3002. The total displacement caused by these jumps must be even. The total number of odd-numbered jumps is 1501 which is odd; hence the displacement caused by these jumps is odd. Thus, the total displacement after all 3002 jumps has to be odd (as the sum of an even and an odd number), and hence cannot be 0. So the grasshopper cannot come back to his starting place. Question: Obviously, he can come back after 3 jumps — he can take the first two jumps to the right, then the 3rd one to the left. Can you find some other number of jumps which allow him to come back to the starting point? Can you describe all such numbers? (4) The dwarves agree that they would guess starting with the last dwarf in the line and then going forward. The last dwarf counts the total number of red hats in front of him; if the number is even, he says “Red”; if the number is odd, he says “Blue.” This dwarf might make a mistake (with the probability 12 ), but all the other dwarves know their color precisely. Hence all the dwarves will go free. Here is an example. Suppose, there are only 7 dwarves, and their hats are RRBRBBB (they all stand facing right, so the last dwarf is the leftmost, and his hat is red). The last dwarf sees 2 red hats, so he says “Red.” The next dwarf knows that the number of red hats in front of him plus his own is even, but he sees only 1 red hat. Therefore, his hat is red. So he says “Red.” The next dwarf now knows that the number of red hats seen by the last dwarf went 1 down, so now the number of red hats in front of him plus his own is odd. He sees 1 red hat in front of him, so his own hat is blue, and he says “Blue.” The next dwarf in the line knows that the number of red hats in front of him plus his own is odd, and he sees 0 red hats; hence his own hat must be red, and he says “Red.” The next dwarf knows now that the number of red hats in front of him plus his hat is even. He sees 0 red hats, so his hat must be blue, and he says “Blue.” Each of the front two dwarves argues in the same manner, and says “Blue.” Thus, in this case 7 dwarves made no mistakes at all, and they will all go free. (5) First, run a few experiments. Suppose we have just 5 coins, 2 of which are counterfeit. There are three possibilities: (A) one bad coin is too heavy and one is too light; (B) both bad coins are too heavy; (C) both bad coins are too light. Let us assume that the weight of a genuine coin is w grams. We will consider all three possible cases, and in each case we’ll make a table describing all possible distributions of coins between the left and right pans.

PARITY AND OTHER INVARIANTS

157

Case A. We need to analyze two situations: (1) when the chosen coin is genuine, and (2) when the chosen point is counterfeit. (1) The chosen coin is genuine. Then the remaining four coins contain 2 good ones, 1 heavier, and 1 lighter coin. coins on left pan coins on right pan weight difference 2 good 1 heavier, 1 lighter 2w − ((w + 1) +(w − 1)) = 0 1 good, 1 heavier 1 good, 1 lighter (w + (w + 1)) −(w + (w − 1)) = 2 1 good, 1 lighter 1 good, 1 heavier (w + (w − 1)) −(w + (w + 1)) = −2 1 heavier, 1 lighter 2 good ((w + 1) + (w − 1)) −2w = 0 (2) The chosen coin is counterfeit. Then there are two possible cases: (i) The chosen coin is lighter; and (ii) the chosen coin is heavier. (i) In this case among the remaining coins 3 are good, and 1 is heavier. coins on left pan coins on right pan weight difference 2 good 1 good, 1 heavier 2w − (w + (w + 1)) = −1 1 good, 1 heavier 2 good (w + (w + 1)) − 2w = 1 (ii) In this case the remaining coins include 3 good ones and a lighter one. coins on left pan coins on right pan weight difference 2 good 1 good, 1 lighter 2w − (w + (w − 1)) = 1 1 good, 1 lighter 2 good (w + (w − 1)) − 2w = −1 Case B. (1) The chosen coin is genuine. Then the remaining coins include 2 good and 2 heavier ones. coins on left pan coins on right pan weight difference 2 good 2 heavier 2w − 2(w + 1) = −2 1 good, 1 heavier 1 good, 1 heavier 0 2 heavier 2 good 2(w + 1) − 2w = 2 (2) The chosen coin is counterfeit. Then 3 good and 1 heavier coin remain. coins on left pan coins on right pan weight difference 2 good 1 good, 1 heavier 2w − (w + (w + 1)) = −1 1 good, 1 heavier 2 good (w + (w + 1)) − 2w = 1 Case C. (1) The chosen coin is genuine; 2 good and 2 lighter coins remain. coins on left pan coins on right pan weight difference 2 good 2 lighter 2w − 2(w − 1) = 2 1 good, 1 lighter 1 good, 1 lighter 0 2 lighter 2 good 2(w − 1) − 2w = −2

158

BOB KLEIN AND TATIANA SHUBIN

(2) The chosen coin is counterfeit; 3 good and 1 lighter coin remain. coins on left pan coins on right pan weight difference 2 good 1 good, 1 lighter 2w − (w + (w − 1)) = 1 1 good, 1 lighter 2 good (w + (w − 1)) − 2w = −1 We see an interesting phenomenon: in all cases, if the chosen coin is genuine, the weight difference is even, and if the chosen coin is counterfeit, the weight difference is odd. In fact, the opposite is also true — if the weight difference is even, it means that the chosen coin was genuine; if the weight difference is odd, then the chosen coin must have been counterfeit. This is encouraging, but would the same result hold for the original problem with 101 coins of which 50 are counterfeit? As before, we assume that the weight of a genuine coin is w grams, so that the weight of a counterfeit coin is either w + 1 or w − 1. Also, let us suppose that there are exactly H heavier coins of weight w + 1 each. Then there are 50 − H lighter coins of weight w − 1 each. Jaime can choose either a genuine or a counterfeit coin, so let us analyze these two cases separately. Case 1. If Jaime chose a genuine coin, then he has a remaining pile of 100 coins, 50 of which are counterfeit. Let him put any 50 coins on the left pan and the other 50 coins on the right one. The following table summarizes the distribution of coins on the left and the right pans. Left pan Right pan Genuine coins num- x 50 − x ber Heavier coins number y H −y Lighter coins number 50 − x − y (50 − H) − (50 − x − y) = −H + x + y We can now calculate the difference between the weights of the coins on the left pan and those on the right pan: Left pan weight Right pan weight Weight difference xw (50 − x)w (2x − 50)w = 2(x − 25)w y(w + 1) (H − y)(w + 1) (2y − H)(w + 1) = 2y(w + 1) − H(w + 1) (50 − x − y)(w − 1) (x + y − H)(w − 1) (50−2x−2y +H)(w −1) = 2(25 − x − y)(w − 1) + H(w − 1) So the total weight difference is 2(x − 25)w + 2y(w + 1) + 2(25 − x − y)(w − 1) − H(w + 1) + H(w − 1) = 2(x − 25)w + 2y(w + 1) + 2(25 − x − y)(w − 1) − 2H which is an even number.

PARITY AND OTHER INVARIANTS

159

Case 2. If Jaime chose a heavy counterfeit coin, then he has a remaining pile of 100 coins, 49 of which are counterfeit. As before, let him put any 50 coins on the left pan and the other 50 coins on the right one. The following table summarizes the distribution of coins on the left and the right pans in this case. Left pan Right pan Genuine coins num- x 51 − x ber Heavier coins number y H −1−y Lighter coins number 50 − x − y (50 − H) − (50 − x − y) = −H + x + y Let’s calculate the difference between the total weight of the coins on the left and right pans: (xw + y(w + 1) + (50 − x − y)(w − 1)) − ((51 − x)w + (H − 1 − y)(w + 1) +(−H + x + y)(w − 1)) = (x − (51 − x))w + (y − (H − 1 − y))(w + 1) + ((50 − x − y) −(−H + x + y))(w − 1) = (2x − 51)w + (2y − H + 1)(w + 1) + (50 − 2x − 2y + H)(w − 1) = 2(x − 25)w + 2y(w + 1) + 2(25 − x − y)(w − 1) − 2H + 1 And this number is odd. Case 3. If Jaime chose a light coin, then again he has a remaining pile of 100 coins, 49 of which are counterfeit. He puts 50 of the coins on each pan. In this case, we have the following distribution. Genuine coins number Heavier coins number Lighter coins number

Left pan x y 50 − x − y

Right pan 51 − x H −y (49 − H) − (50 − x − y) = 1 − H + x + y

This time, the difference between the total weights is (xw + y(w + 1) + (50 − x − y)(w − 1)) − ((51 − x)w + (H − y)(w + 1) + (1 − H + x + y)(w − 1)) = (x − (51 − x))w + (y − (H − y))(w + 1) + ((50 − x − y) − (1 − H + x + y))(w − 1) = (2x − 51)w + (2y − H)(w + 1) + (49 + H − 2x − 2y)(w − 1) = 2(x − 25)w + 2y(w + 1) − 2(x + y)(w − 1) + 48w − 2H − 49 This is an odd number again. Hence, we can conclude that if the difference of the weights is even, then Jaime chose a genuine coin, and if the difference is odd, then he chose a counterfeit one.

160

BOB KLEIN AND TATIANA SHUBIN

Question: Can you solve the problem with different number of coins — say, 235 coins with 117 counterfeit? Can you make a general conjecture? (6) A chessboard contains 32 white and 32 black squares. The missing corners are both the same color. Suppose they are white (the case when they are black can be analyzed similarly). So now the board contains 30 white and 32 black squares. Each domino covers one white and one black square. To tile the 62 squares of the chessboard with two missing corners we would need 31 dominoes. But they would cover 31 squares of each color, and we have only 30 white squares. Hence such tiling is impossible. Question: Is it possible to tile a chessboard if we remove one white and white black square randomly chosen? (7) To solve this problem, let us paint the grid using four colors as shown in Figure 81.

Figure 81. The painted grid. No matter where a 1 × 4 tile is placed, it will cover one square of each of four colors. Hence if 25 tiles are placed on the board, they would cover 25 squares of each color. But we have only 24 squares of color 4. Hence such a tiling is impossible. (8) Let us represent the floor by an m × n grid as illustrated in Figure 82. We can see that a 2 × 2 tile always covers exactly 1 black square, while a 1 × 4 tile either covers 2 black squares or none. Therefore, if one 2 × 2 tile gets smashed, regardless of a rearrangement of the remaining tiles, we would need to cover 1 additional black

PARITY AND OTHER INVARIANTS

161

Figure 82. The tiled floor. square. But one additional 1 × 4 tile will either cover 2 black squares or none. Hence, this replacement is impossible. Likewise, if one 1 × 4 tile gets broken, then we would either need to cover 2 additional black squares or none, which a 2 × 2 tile is incapable of doing. Therefore, retiling the floor is impossible. (9) There are four types of blows. The results of blows of each type are listed below. Type 1: −15 + 24 = 9, so it creates 9 additional heads. Type 2: −17 + 2 = −15, so it takes off 15 heads. Type 3: −20 + 14 = −6, so it takes off 6 heads. Type 4: −5 + 17 = 12, so it creates 12 additional heads. Thus, after x blows of type 1, y blows of type 2, z blows of type 3, and w blows of type 4, the dragon has 100 + 9x − 15y − 6z + 12w heads. We want this number to be 0, so we need to solve the following equation in integers: 100 + 9x − 15y − 6z + 12w = 0 Moving all terms but 100 to the right side, we get: 100 = −9x + 15y + 6z − 12w = 3(−3x + 5y + 2z − 4w) For any integers x, y, z, and w, the right-hand side of the equation is divisible by 3, while 100 is not. Thus, it is impossible to kill the dragon. (10) Suppose that Ivan used first sword n times and second sword m times (in some order, which does not matter). Then the number of the remaining heads is 100 − 21n + (−13 + 594)m = 100 − 21n + 581m

162

BOB KLEIN AND TATIANA SHUBIN

We want this number to be 0, thus we need to find to integers n and m which solve the following equation: 100 − 21n + 581m = 0 Moving two last terms to the right and factoring out 7 we have: 100 = 7(3n − 83m) The right-hand side of the equation is a multiple of 7, while the left-hand side isn’t. Hence, the equation has no solution; Ivan can’t kill the dragon. (11) (a) Record the number of cats and dogs the veterinarian has as an ordered pair, first number being the number of cats, and the second the number of dogs. For example, the pair (7,3) means that he has 7 cats and 3 dogs. If he puts 2 dogs in the machine, he will have 7 cats and 2 dogs, and we will record is as follows: (7, 3) → (7, 2) If he starts with 3 cats and 1 dog, he can put in his machine 2 cats, then 2 dogs, then 1 cat and 1 dog, and he will achieve his goal: (3, 1) → (1, 2) → (1, 1) → (1, 0). If he starts with 13 cats and 10 dogs, he will be able to achieve his goal, as well — look at the following sequence: (13, 10) → (11, 11) → (9, 12) → (7, 13) → (5, 14) → (3, 15) → (1, 16)

→ (1, 15) → (1, 14) → (1, 13) → (1, 12) → (1, 11) → (1, 10) → (1, 9)

→ (1, 8) → (1, 7) → (1, 6) → (1, 5) → (1, 4) → (1, 3) → (1, 2)

→ (1, 1) → (1, 0). Notice that at every step the total number of animals decreases by 1, hence to go from the total of 23 animals to 1 animal requires 22 steps. But what is done at each step may vary. For example, our first step could be (13, 10) → (13, 9) instead of (13, 10) → (11, 11). Also, the actual action represented a step is not determined by our notation. Putting 2 dogs in the machine results in reducing the number of dogs by one while keeping the number of cats unchanged, exactly like putting a dog and a cat into the machine. We can see that this problem is identical to problem 6 — just replace cats with red marbles and dogs with green marbles. Recalling the solution of that problem, we see that the veterinarian would be able to achieve his goal if the starting collection of his animals contains an odd number of cats.

PARITY AND OTHER INVARIANTS

163

(b) If we start with 3 cats and 1 dog, we don’t have any choices. At the first step, we must put 1 cat and 1 dog in the machine, thus getting now 2 cats, 0 dogs, and 1 mouse. All the subsequent steps are recorded in the table below. Cats Dogs Mice 3 1 0 2 0 1 1 1 0 0 0 1 So the veterinarian cannot achieve his goal. The second case isn’t as straightforward since there are many choices. But we can notice that at every step the difference between the numbers of different pairs of animals either remains the same or changes by 2. Indeed, suppose we record the numbers of three types of animals as an ordered triple (x, y, z). Then one step can be recorded as (x, y, z) → (x − 1, y − 1, z + 1). The differences are listed below: x − y and (x − 1) − (y − 1) = x − y y − z and (y − 1) − (z + 1) = y − z − 2 x − z and (x − 1) − (z + 1) = x − z − 2 Hence if the difference in the beginning was even it must remain even. The difference between the number of cats and dogs was 4−4 = 0, an even number. Hence it can’t become 1−0 = 1, which is odd. Therefore, in this case the veterinarian can’t achieve his goal again. (c) Let (x, y, z) denote the situation when we have x cats, y dogs, and z mice. With the machine which we have now — the one that can take two different animals and convert them into one animal of the third kind or the other way around, start with one animal and transform it into two animals of the other two kinds — the triple (x, y, z) can only be transformed into one of the following 6 triples: • (x − 1, y − 1, z + 1) (one cat and one dog have transformed into a mouse) • (x − 1, y + 1, z − 1) (one cat and one mouse transformed into a dog) • (x + 1, y − 1, z − 1) (one dog and one mouse transformed into a cat) • (x − 1, y + 1, z + 1) (one cat transformed into one dog and one mouse) • (x + 1, y − 1, z + 1) (one dog transformed into one cat and one mouse) • (x + 1, y + 1, z − 1) (one mouse transformed into one cat and one dog).

164

BOB KLEIN AND TATIANA SHUBIN

Consider the sum of cats and dogs. Before the transformation, it was x + y; after the transformation it becomes either x + y − 2 (in the first case), or x + y + 2 (in the last case), or it remains x + y (in all other cases). Thus we see that the parity of the sum never changes. If we start with 3 cats and 1 dog, the sum would always remain even, hence it is impossible to end up with one cat and no other animals. Similarly, if we start with 4 animals of each kind we will never reach the situation with one cat only. (On the other hand, if we start with 3 cats, 0 dogs, and 0 mice we can reach the desired outcome: (3, 0, 0) → (2, 1, 1) → (1, 0, 2) → (0, 1, 1) → (1, 0, 0). (12) This problem is similar to 11(b), but not the same. Here one step can be described as (x, y, z) → (x − 1, y − 1, z + 2). Now, comparing the differences we get the following results: x − y and (x − 1) − (y − 1) = x − y y − z and (y − 1) − (z + 2) = y − z − 3 x − z and (x − 1) − (z + 2) = x − z − 3 So the differences either remain the same or differ by 3. Hence if we divide the difference by 3, then the remainder before and after taking a step will be the same. For example, consider the following step: (17, 15, 13) → (16, 14, 15). The differences before the step were 2, 2, 4; after the step, 2, -1, 1. Now, dividing with the remainder we have: 2 ÷ 3 = 0, remainder 2, and 2 ÷ 3 = 0, remainder 2; 2 ÷ 3 = 0, remainder 2, and −1 ÷ 3 = −1, remainder 2; 4 ÷ 3 = 1, remainder 1, and 1 ÷ 3 = 0, remainder 1; If we start with the triple (4, 5, 5), the difference 5 − 5 = 0 has the remainder of 0 when we divide by 3, and if all chameleons turn green at the end, then we must have the triple (14, 0, 0), where again 0 − 0 = 0, which has the same remainder as before. Thus, it is plausible that this triple is achievable. Indeed, we can see it in the following table. Green Yellow Red 4 5 5 6 4 4 8 3 3 10 2 2 12 1 1 14 0 0 Notice that we cannot get all yellow or all red chameleons (can you see why?).

PARITY AND OTHER INVARIANTS

165

In the second case, all differences have non-zero remainders, and hence we can never obtain a triple with two 0 entries. Hence all the chameleons can never become of the same color. (13) We notice that at each step the total number of numbers written on the board decreases by 1, so at the end there will be only one number left. The question now is what is that last number? (a) In this case, the sum of all the numbers currently on the board is an invariant. Hence the last number on the board must be = 5050. 1 + 2 + 3 + . . . + 100 = 100·101 2 (b) This time, the invariant is the product of all numbers. Hence the last number left on the board must be 1 · 2 · 3 . . . · 100 = 100! (c) We can start by considering examples with smaller numbers. Numbers on the board Two chosen numbers New number 1, 2, 3 1 and 3 1·3+1+3=7 2, 7 2 and 7 2 · 7 + 2 + 7 = 23 23 Try again: Numbers on the board Two chosen numbers New number 1, 2, 3, 4, 5 2 and 5 17 1, 3, 4, 17 1 and 17 35 3, 4, 35 4 and 35 179 3, 179 3 and 179 719 719 The key observation is that xy + x + y = (x + 1)(y + 1) − 1. This can be seen algebraically or by considering Figure 83. The letters outside the rectangle denote the length, and the expressions inside the rectangles indicate their areas. The area of the large rectangle is the product of its side lengths, (x + 1) and (y + 1), and at the same time it is the sum of the areas of the smaller four rectangles. x y

1

xy x

1 y

1

Figure 83. xy + x + y = (x + 1)(y + 1) − 1 Hence we have (x + 1)(y + 1) = xy + x + y + 1, and so xy + x + y = (x + 1)(y + 1) − 1. Now, suppose we start with any three numbers x, y, z. Then we will have:

166

BOB KLEIN AND TATIANA SHUBIN Numbers on the board x, y, z (x + 1)(y + 1) − 1, z

Two chosen numbers

New number

x, y (x + 1)(y + 1) − 1, z

(x + 1)(y + 1) − 1 [((x + 1)(y + 1) − 1) + 1] [z + 1] − 1 = (x+1)(y +1)(z +1)−1

(x+1)(y+1)(z+1)−1

Continuing in the same way, we see that if we start with the numbers 1, 2, 3, . . . , 100 written on the board, then the last remaining number will be (1 + 1)(2 + 1)(3 + 1) · . . . · (101) − 1 = 2 · 3 · 4 · . . . · 101 − 1 = 101! − 1 (14) Look at the perimeter of the shape(s) that results from joining all of the infected squares. For example, in Figure 84 the perimeter of the infected area is 18.

Figure 84. The perimeter of the infected area (shown in black) is 18. It is easy to see that if a square which neighbors 2 infected squares gets infected, the perimeter does not change; if the square which neighbors 3 infected squares gets infected, then the perimeter decreases by 2; and if a square which neighbors 4 infected squares gets infected, then the perimeter decreases by 4. Figure 85 illustrates this statement. Therefore, the perimeter either remains the same or it decreases. But the perimeter of the union of 9 infected squares is at most 9 × 4 = 36, while the perimeter of the entire 10 × 10 square is 40. Hence it is impossible for all 100 squares to become infected. Presentation Suggestions: Warm-up problems and games can be used to introduce the ideas of parity (even/odd) and invariants (features or values of the problem that don’t change) in a gentle yet captivating way. Game 1 introduces parity as a strong tool for analyzing the situation. Students enjoy playing Game 2, and they are pleasantly surprised with the outcome — an equal distribution of the tokens. It takes some effort to see why such an outcome is inevitable, and it’s where the idea of a monovariant enters. Puzzle 3 brings in an idea of coloring; it will be useful in further problems.

PARITY AND OTHER INVARIANTS

167

Figure 85. The change in the perimeter depending on the current infection status of the neighborhood. Problem 4 is very easy yet it introduces a totally different invariant — the area of a flat figure. We recommend using at most two warm-up activities; though in most cases, one is enough. Then it is time to move on to the more challenging problems. For young and/or inexperienced students, it might help to ask them what is the sum of two even numbers? two odd numbers? an even and an odd number? Then write on the board and leave there the following equations: even + even = even odd + odd = even even + odd = odd odd + even = odd Some students might need an explanation so be sure to provide it. After this short discussion, you might move to problem 1. Let students play with it in small groups long enough so that they form some opinion.

168

BOB KLEIN AND TATIANA SHUBIN

Then discuss the problem with the whole class. Push students to justify their answer. Make sure that the idea of parity sinks in. Sometimes a student will have the right idea yet struggle to communicate it. In such cases you might have another student put the argument in their words for the whole class to hear. Next, let students do problem 2. Usually they realize quickly that it’s exactly like problem 1. Students with weaker backgrounds might need to add to the equation above similar ones with subtraction instead of addition. Problem 3 is very similar to 1 and 2; it makes sense to simply suggest it as homework. Next, do problem 4. It works well if you break students into groups of size 5–6 and tell them that they should imagine that their group has been caught by a wicked witch and they would be put to the test described in the problem. Motivate them by telling them that they have only 15 minutes to find a strategy that will liberate them. You might give them three hints at your discretion: (a) it is possible to find such strategy; (b) the strategy works equally well no matter how many people are involved; (c) it has to do with evenness/oddness. After 15 minutes, if any group claims that they’ve found a strategy, let them show it. Put them in a line, and place at random “hats” of two different colors on their heads. You can use real hats, paper hats, sheets of cardstock of two colors (kids would simply hold them above their heads), or just sheets of paper with color marks or letters indicating color. If they pass the test, they should explain their strategy. If no group finds a strategy, tell them the answer: the last person in a line counts the number of red hats she sees, and if the number is even, she says “Red”, but if the number of red hats is odd, she says “Blue.” Then tell the students that this last person is the one who might be wrong about her color because what she says is not a guess but a signal that allows every other person to know their colors precisely. Let students figure out how it works. In this case, the answer is not the solution. Telling them the method doesn’t justify why it works (the real magic!). Now it’s time to tell the students that the problems they’ve been dealing with were solved using a powerful tool of invariants. An invariant is something that remains unchanged throughout some process. Invariants can come in many different forms, and we have used a couple so far. It might be good to write on the board something like: Invariants: • a pattern (such as . . .BWBW. . .) • parity (evenness/oddness) Of course, students know that evenness/oddness is simply a property of division by 2. But there are other integers besides 2. Let them now do problem 9. In this problem, we use divisibility by 3, so when they have

PARITY AND OTHER INVARIANTS

169

solved and discussed the problem, you could add one item to your list of invariants: • divisibility (and remainders) Problem 10 is very similar; suggest it as homework. Most likely, this will be all you can do in one 90-minute session (but it’s also likely that you would actually do much less — it all depends on your particular circle; the main thing is not to rush). You can use other problems for a follow-up session (or sessions). In fact, the following three sequences make for nice sessions. Sequence I: (1) Problem 11 (parts (a) and (b) only — part (c) might take more time, so it might be better to tell students that it’s their new homework). Hopefully, the students would recognize that part (a) is exactly like problem 1 from the previous session. Part (b) is a little more complicated, since here it’s the difference between the numbers of animals that retains parity. (2) Problem 12 — it is easy to see the similarity with part (b) of the previous problem, but in this case it’s divisibility by 3 (and remainders when dividing by 3) that matters. (3) Problem 5. It is an interesting problem that requires time and effort (and some algebraic dexterity) to solve as stated. It is often useful to suggest to students that they start with a simpler version with just 5 coins of which 2 are counterfeit. Ask them to analyze all possible situation and to note their observations. This analysis might take the rest of the 90-minute session; in this case, tell the students to finish the problem at home. Sequence II: (1) Problem 6 — after it’s solved, there is a number of questions to investigate (for example, what happens if we remove any two squares of different color; any two black and any two white squares). (2) Problem 7 — here a smartly-chosen coloring with four colors instead of two will help. (3) Problem 8 — here again a Black and White coloring is helpful, but not done in a checkerboard pattern. Sequence III: (1) Problem 13 — other types of invariants. (2) Warm-up game (B). The game will lead to the idea of a monovariant, which would be useful in solving the next problem: (3) Problem 14 — a delightful problem where kids can do some experimentation.

Pancake Problem Contributed by Rebecca Bycofski, Bob Klein, and Sierra Knavel Short Description: The “pancake problem” uses a story of a sloppy shortorder cook to motivate an investigation that, on the surface, seems like sorting a disordered stack of pancakes of various sizes but, on a deeper level, is an investigation into an open problem related to something called “prefix sorting.” Prefix sorting is studied for its relationship to list-sorting algorithms and evolutionary genetic mutations, and is the topic Bill Gates (of Microsoft fame) chose for his only published academic paper. Materials: Pencils, paper, and at least four small disks of different sizes per group to resemble pancakes. In some instances, we have used actual pancakes (this can be messy and the manipulatives sometimes disappear), while in others we have used a circle cutter to make cardboard or foam board cut-outs. In other instances, groups have preferred not to use any physical objects to represent the pancakes. Objectives. Students will: • understand that mathematics still contains many unsolved problems that they can work on; • look at variations of a problem to better understand the original problem; • begin with a simple problem that becomes increasingly difficult with further exploration. As such, they will see that it is questioning rather than answering that motivates great and engaging mathematics. References/Authorship: The Pancake Problem originates from Jacob Goodman (under the name Harry Dweighter), though it appears in an academic paper from Bill Gates & Christos Papadimitriou, and has further been analyzed by many including John Conway and Ivars Peterson. Jacob Goodman, using the name “Harry Dweighter” (harried waiter), sent in his pancake problem to American Mathematical Monthly to investigate his real-life problem of folding towels into piles and rearranging those piles. 171

172

REBECCA BYCOFSKI, BOB KLEIN, AND SIERRA KNAVEL

Papadimitriou & Gates further investigated prefix sorting by offering the burnt pancake problem, studying how many flips are required to reorder a stack of burnt pancakes such that the burnt side faces down. See the references for further suggested resources.

PANCAKE PROBLEM

173

Pancake Problem Student Handout A waiter at a restaurant is tasked with serving a stack of n pancakes. The chef is very sloppy. She makes pancakes of all different sizes and puts them on a plate in no specific order. To ensure that he gets the best possible tips from his customers, the waiter wants to make the stack look better by putting the pancakes in order by size, with the largest pancake on the bottom of the stack. He has the pancakes in one hand and a spatula in the other. The only thing he can do to change the order of the pancakes is to insert the spatula somewhere in the stack and flip all of the pancakes above it. What is the maximum number of flips required to put a stack of n pancakes in the desired order? Let’s call this the pancake number for n pancakes and denote it Pn (1) If the waiter is only carrying one pancake, how many flips are required to put the stack of one pancake in desired order? In other words, what is P1 ? (2) (a) If the waiter carries only two pancakes, what are the possible arrangements of the two pancakes? (b) Can you find the arrangement of pancakes such that no flips are required? (c) In the other case, how many flips are required so that the pancakes are stacked in order (with the smaller pancake on top)? (d) What is the maximum number of flips needed to put a stack of 2 pancakes in order? In other words, what is P2 ? (3) (a) How many different ways could a stack of three pancakes be arranged? What technique was used to figure this out? (b) If the waiter carries the three pancakes stacked in any of the previous ways listed, what is the maximum number of flips required to reorder the pancakes? In other words, what is P3 ? (4) Continue on to find P4 and P5 . (a) How many arrangements of pancakes there are for a 4-pancake stack? For a 5-pancake stack? (b) What are the worst case scenarios for 4-pancake and 5-pancake stacks? What are P4 and P5 ? (5) (a) Can you come up with an algorithm for sorting pancakes? In other words, if you are given an arbitrary stack of n pancakes, what steps would you take in order to sort them correctly? (Hint: how can you reduce a large stack of pancakes to a case that has been solved before?) (b) (Do not read this problem until you have solved the previous one. One algorithm for sorting pancakes is the “largest-pancake-to-thebottom” method (henceforth “LPttB”). In this method, the goal is to keep moving the largest pancake not already where it is supposed to be, to the top, then flipping part of the stack to move it to the bottom. For instance, if we have four pancakes (1 being

174

REBECCA BYCOFSKI, BOB KLEIN, AND SIERRA KNAVEL

the smallest, and 4 the biggest), and if they are arranged top-tobottom: 3-2-4-1, then our first goal is to move 4 to the bottom. To do this, we insert the spatula between pancakes 4 and 1 and flip, resulting in the stack: 4-2-3-1. Now we put the spatula below the 1 and flip, giving a stack of 1-3-2-4. Now pancake 4 is where it should be so our next task is to move pancake 3 where it goes (immediately above 4). Insert the spatula between pancakes 3 and 2 and flip, giving a stack of 3-1-2-4. To move pancake 3 into position, insert the spatula between pancakes 2 and 4 and flip, giving us 2-1-3-4. Keep going until the pancakes are in there appropriate positions. If we use the LPttB method as described here, with n pancakes, what is the maximum number of flips required? The same waiter is serving pancakes on another day, but the chef is having a very bad day. All of the pancakes are coming out burnt on one side! Today, the waiter still wants to put them in order of size with the largest on bottom, but he also wants to make sure that the burnt side isn’t showing. What is the maximum number of flips required to put a stack of n burnt pancakes in the desired order with the burnt side down? Let’s call this the burnt pancake number for n pancakes and denote it Bn (6) What is the worst possible configuration for one burnt pancake? What is B1 ? (7) What is the worst possible configuration for two burnt pancakes? What is B2 ? (8) What is the worst possible configuration for three burnt pancakes? What is B3 ? (9) Do you notice a pattern in these worst possible configurations? (10) Can you come up with an algorithm for sorting burnt pancakes in the scenario where the pancakes are correctly sorted by size but all the burnt sides are up? (11) How many steps would your algorithm take on n pancakes in the case where all of the pancakes start with burnt sides up? (12) How does this problem compare to the regular “pancake problem”? Is it easier or more difficult? Why?

PANCAKE PROBLEM

175

Pancake Problem Teacher Guide It is incredibly important to give students a chance to present their solutions to various problems using their own representations and notations. Do your best not to offer too much help early on with representations. Have students present their solution at the board but encourage them to explain their representations. Often students will write something down on their paper but consider it “scratch work,” leaving it out of presentations. Yet often that “ scratch work” contains ways of thinking about and representing the problem that is very valuable to share. Solutions: (1) If the waiter is only carrying one pancake, then the stack is already in the proper order and no flips are required. P1 = 0. (2) (a) With two pancakes, there are two possible starting cases as shown in Figure 86. (b) Case 1: The arrangement of pancakes is such that the smaller pancake is on top. Therefore, the pancakes are already in correct order and no flips are necessary. (c) Case 2: The larger pancake is on top. This requires one flip to move the smaller pancake on top, leaving the larger pancake on the bottom. (d) Case 1 required 0 flips to order the pancakes, while Case 2 required 1 flip. Therefore, the maximum number of flips required for two pancakes is P2 = 1.

Figure 86. The 2 possible starting arrangements of 2 pancakes. (3) (a) A way of arranging items is called a permutation. There are 3 ways to choose one pancake to be on the bottom of the stack. For each of these three options, there are 2 remaining pancakes that might be in the middle. Finally, there is only one option for a pancake to be on top. The formula for the number of permutations possible in this case is 3! or 3 × 2 × 1, which equals 6. The six possible arrangements for three pancakes are shown in Figure 87.

176

REBECCA BYCOFSKI, BOB KLEIN, AND SIERRA KNAVEL

Figure 87. The 6 possible starting arrangements of 3 pancakes. (b) To find P3 , we need to find the “worst case scenario” from these six possibilities (see Figure 88).

PANCAKE PROBLEM

177

Figure 88. Number of flips needed to sort each possible stack of 3 pancakes. So, the largest number of flips required is P3 = 3. (4) (a) When the waiter has four pancakes, there are 4! or 4 × 3 × 2 × 1 possible arrangements. This equals 24 possibilities. When the waiter has 5, there are 120 possibilities.

178

REBECCA BYCOFSKI, BOB KLEIN, AND SIERRA KNAVEL

(b) It is impractical to determine the number of flips required for every single possible arrangement, so it is useful to try to figure out which arrangement would be the “worst case scenario.” It turns out that there are three worst case scenarios in the case of a 4-pancake stack.9 All three are shown in Figure 89. For 5-pancake stacks, there are 20 different worst case scenarios. One of them is shown in Figure 89: 5 2

3

4

4

1

2

3

1

1

4

3

2

3

2

1

4

Figure 89. Worst Case Scenarios for 4-pancake and 5pancake stacks. So P4 = 4, and P5 = 5. (5) (a) One example of an algorithm is the “largest-pancake-to-the-bottom” method. In this method, the goal is to put the pancakes in order from the bottom up. In each step, you move the largest pancake not already where it is supposed to be to the top, then flip part of the stack to move that pancake into position. If we start with a stack of n pancakes that are out of order, we can get the largest one on the bottom by inserting the spatula below the largest pancake, flipping so that the largest pancake is now on top, and then flipping the entire stack. This reduces the problem from an ordering problem with n pancakes to an ordering problem with the n − 1 pancakes that are on top. This procedure could then be repeated for the second largest pancake. Insert the spatula below the second largest pancake, flip so that it is on top, and then insert the spatula just above the largest pancake on the bottom so that the second largest pancake ends up next to it. Repeat with the third largest pancake, and so on, until the stack is in order. (b) With this method, suppose the pancakes are in their worst possible position, requiring us at every stage to move the biggest pancake into position — that is, none of the pancakes fall into position inadvertently. Consider four pancakes as an example and let’s 9

When the number of pancakes is small, this can be done by listing all possibilities. As the number of pancakes increases, this strategy is actually helpful in seeing how results from the previous case (n − 1 pancakes) can be used to generate the worst-case stack for n pancakes. For instance, once the largest pancake is on the bottom, you are left with the n − 1 pancake scenario.

PANCAKE PROBLEM

179

think of each stage as getting the largest pancake not already in it’s correct position into position. To move pancake 4 to the bottom requires at most two flips: move pancake 4 to the top, then flip the stack to move it to the bottom. Moving pancake 3 to just above pancake 4 (we assume it isn’t already there so that we get a maximum number of flips) requires at most two flips to move pancake 3 to the top, then another flip to put it just above pancake 4. Now once we get down to the last two (smallest) pancakes, we will need at most one flip since the worst case scenario is that they sit at the top of the rest of the ordered stack with the smallest one under the next smallest. Hence, with four pancakes, we had 4 − 2 “stages” consisting of two flips each stage (one to move the pancake to the top, then one to move it into position) followed by one final flip to order the top two pancakes. Thus, with n pancakes, we would have n − 2 stages with 2 flips per stage, plus one more flip to order the last 2 pancakes. While we have a method for ordering the pancakes and we can tell how long it would take in the worst case scenario, there might be other methods that require a smaller number of flips. Thus, we have only shown that Pn ≤ 2(n − 2) + 1 = 2n − 3 flips. The solution presented here is explained as 2(n − 2) + 1; students may be able to interpret the same approach as “twice the number of pancakes less three moves” or Pn ≤ 2n − 3. For example, for five pancakes, if you use this algorithm, the maximum number of flips needed would be 7, so we know that P5 ≤ 2(5) − 3, or in other words that P5 ≤ 7. In fact, analyzing all of the 5-pancake stacks individually, it turns out that P5 is only 5. (6) For the “burnt pancake problem” with one pancake, there are two options: either the burnt side is on top or the burnt side is on the bottom. The case where the burnt side is on the bottom is trivial. The worst-case scenario occurs when the burnt side is on top. In this case one flip is required, so B1 = 1. (7) To find B2 , we first need to know how many different possible configurations there are for two burnt pancakes. The 8 possible configurations are shown below. The smallest pancake is denoted by the number 1, and the largest is denoted by the number 2. The lines show which side of each pancake is burnt. Depending on your students, you may choose to represent the burnt pancakes as trapezoids instead of using this notation. 1 1 1 2 2 2 2 1 2 2 2 1 1 1 1 2 (i.) (ii.) (iii.) (iv.) (v.) (vi.) (vii.) (viii.) In this case, there are eight possible configurations versus only two configurations in the regular “pancake problem.” The worst possible configuration is case when the larger pancake is on the bottom and

180

REBECCA BYCOFSKI, BOB KLEIN, AND SIERRA KNAVEL

both burnt sides are facing up (case (iv.) above). One example of a sequence of flips to solve it is shown below. 1 2 2 1 1 2 1 1 2 2 The first flip moves 2 to the top and 1 to the bottom, with both burnt sides down. Flipping the top (2) pancake moves the burnt side up on 2 but leaves the pancake order otherwise unchanged. Flipping the entire stack yet again moves 1 to the top and 2 to the bottom, with the burnt side of 2 where it should be, pointing down. All that remains is to flip the top (1) pancake so that the burnt side is down. This totals four moves. Some experimenting with all eight possible configurations of burnt pancakes shows that configuration (iv.) above is indeed the worst case scenario, and that the fewest number of flips required is four, so the burnt pancake number for two pancakes is B2 = 4. (8) For three burnt pancakes, the worst possible configuration is when the largest pancake is on bottom, the medium pancake is in the center, and the smallest pancake is on top, with the burnt side of each facing up. One sequence that orders the burnt pancakes properly is shown below. 1 2 3

3 3 1 2 2 1 1 2 2 2 1 1 2 2 1 1 3 3 3 3 3 So B3 = 7. (9) There is a pattern between the worst-case scenarios for one, two, and three burnt pancakes. For each of these, the worst-case scenario occurs when the pancakes are in the correct configuration (largest on bottom), but with the burnt side of each pancake facing up. (10) One example of an algorithm for sorting a stack of pancakes that is ordered correctly by size but which has all burnt sides up is as follows: First, flip the entire stack so that the largest pancake is on top with the burnt side facing down. Then, flip just the top pancake so that the burnt side is facing up. Next, flip the entire stack again so that the largest pancake is on the bottom with the burnt side facing down. Then, insert the spatula above the largest pancake to flip all of the other pancakes. Now, the second largest pancake will be on top with the burnt side facing down. Flip just this pancake so that the burnt side is facing up. Then, insert the spatula above the largest pancake again to flip the pancakes so that the second largest is put in the correct position with the burnt side facing down. Repeat this process, each time bringing the next largest pancake to the top, flipping it so that the burnt side is up, and then flipping the top portion of the stack in order to put the top pancake in its correct location. Once the smallest pancake is on top, all that will be required is to flip this pancake over so that the burnt side is facing down.

PANCAKE PROBLEM

181

(11) The “largest-pancake-to-the-bottom” algorithm will take three steps for each pancake except the smallest, which will only take one step. For n pancakes, this algorithm shows that Bn ≤ 3(n − 1) + 1 steps, or Bn ≤ 3n − 2 steps. (12) There are several nuances which make the “burnt pancake problem” different from the regular “pancake problem.” For example, there are many more possible configurations for this problem. In the “pancake problem,” there is only one possible configuration for one pancake, while in the “burnt pancake problem” there are two. For two pancakes, two configurations turn into eight, and for three pancakes, six configurations turn into forty-eight. This makes the “burnt pancake problem” much more difficult than the regular “pancake problem.” Presentation: This activity can fit into a 90-minute session, but it can last much longer if time permits. It is easy to get caught up in the activity and spend too much time on certain problems, so it is important for the leader of the activity to push participants to move on to the other problems when necessary. For a 90-minute session: Focus on the intro for about 5 minutes. Let students discuss what the “maximum number of flips required to put a stack in order” means. For example, you could have an infinite number of flips if you simply flipped the top pancake continuously. However, this flipping is not required to put the pancakes in the correct order. Then, move on to questions 1–4. Let students work on these problems for about 20 minutes. For the first 5 minutes, talk about problems 1 and 2 together. Have students answer the problems aloud and discuss them as a class. Then, split them into smaller groups to work on problems 3 and 4. Make a chart on the board listing the class’ results for P1 , P2 , P3 , etc. When one group comes up with an answer, have other groups check it and confirm. Often, it helps to have the groups show the class the configuration of the pancakes that was the “worst case scenario” that they found. After the class reaches a consensus on P3 and P4 and some groups have moved on to P5 or maybe even P6 , have them move on to problem 5. Have them spend about 10 minutes coming up with an algorithm for sorting pancakes and trying to figure out the maximum number of flips using this algorithm. Next, move on to the “burnt pancake problem.” Explain the problem and have students try to find B1 , B2 , B3 , etc. Let them work on problems 6–8 for about 15 minutes and then have them spend about 20 minutes on problems 9–12 and try to come up with an algorithm for sorting pancakes that are piled correctly by size but the burnt sides are all up. Spend the rest of the time (5–10 minutes) closing the activity by talking about what is known about the “pancake problem” and “burnt pancake problem” so far and showing students the known pancake numbers:

182

REBECCA BYCOFSKI, BOB KLEIN, AND SIERRA KNAVEL

P1 0 P11 13 P2 1 P12 14 P3 3 P13 15 P4 4 P14 16 P5 5 P15 17 P6 6 P16 18 P7 8 P17 19 P8 9 P18 20 P9 10 P19 22 P10 11 P20 ??? To split this into two smaller sessions of 45 minutes each, introduce the problem and work on the “pancake problem” for the first session, and focus on the “burnt pancake problem” and the conclusion during the second session. Hidden mathematical agenda: This problem was first posed in 1975 by mathematician Jacob Goodman under the pen name “Harry Dweighter.” Bill Gates’s only academic paper is on the topic of pancake flipping. He co-authored the paper with Christos 5n + 5 flips required Papadimitriou and the paper proved an upper limit of 3 to reorder n pancakes [1]. The “pancake problem” is an open problem, which means it has not yet been solved. The Pancake Number Pn is currently unknown for more than 19 pancakes.

References [1] William H. Gates and Christos H. Papadimitriou, Bounds for sorting by prefix reversal, Discrete Math. 27 (1979), no. 1, 47–57, DOI 10.1016/0012-365X(79)90068-2. MR534952 [2] Singh, S. (2013). Flipping pancakes with mathematics — Simon Singh. Retrieved July 27, 2016, from https://www.theguardian.com/science/blog/2013/nov/14/ flipping-pancakes-mathematics-jacob-goodman [3] Haymaker, Katie. (Winter/Spring 2016). “Flipping pancakes.” MTCircular Winter/Spring 2016. 16-18. Available at http://scimath.unl.edu/nebraskamath/ newsletters/extras/2016-02/MTCircularWS2016_pancakes.pdf [4] D. J. Kleitman, Edvard Kramer, J. H. Conway, Stroughton Bell, and Harry Dweighter, Problems and Solutions: Elementary Problems: E2564-E2569, Amer. Math. Monthly 82 (1975), no. 10, 1009–1010, DOI 10.2307/2318260. MR1537911 [5] A078941. The Online Encyclopedia of Integer Sequences of Neil Sloane. https://oeis. org/A078941 includes table of burnt pancake numbers, (known up to 12?)

The H − L Protein Folding Model Contributed by Amanda Serenevy Short Description: This session is based on a simple model for protein folding that biochemists created to explore some of the features of real proteins without worrying about all of the complexities. This model represents protein folding as an embedding of a chain in a square grid. An energy is associated with each folded protein, and students explore questions about when the minimal energy corresponds to a unique embedding. This topic allows students to learn about the role that toy models play in mathematical modeling, and exposes them to mathematical thinking in a context that is unlike traditional school mathematics. Materials: Handouts, grid paper, and pencils are essential, and it is also useful to have a board and writing implements for sharing ideas. It can be helpful to make a physical model by cutting red and blue circles, white rectangles, and yellow bars from card stock to represent the components. Brass fasteners and sticky tack can then be used to hold the chains together and represent the hydrogen bonds. Some students find it helpful to have a physical chain to manipulate. History: Biochemists Ken Dill and Hue Sun Chan invented the hydrophobic-hydrophilic model in the early 1990s [1]. Since that time, several research groups have used the model to explore questions about the complexity and decidability of protein folding. In the early 2000s, Oswin Aichholzer, David Bremner, Erik Demaine, Henk Meijer, Vera Sacrist´an, and Michael Soss used this model to explore the stability of protein foldings and considered unique optimal foldings [2]. This session is based on the questions and ideas presented in their papers. Dr. Anna Swan, professor of biomedical engineering at Boston University, assisted with the creation and initial pilot-testing of this activity. During the 2017 Navajo Prep Math Camp, the following students assisted by contributing solutions: Kami Atcitty, Irvilinda Bahe, Amber Cly, Leyonah Endischee, Sky Harper, Sierra Knavel, Kamia Leano, Devon Lynch, Kayleigh Paddock, Aidan Pioche-Lee, Anastasia Todacheeny, Mandi Wheeler.

183

184

AMANDA SERENEVY

The H − L Protein Folding Model Student Handout In the early 1990s, biochemists Ken Dill and Hue Sun Chan invented the H − L10 protein folding model to explore some of the features of real proteins without worrying about all of the complexities [1]. This model and variations of it have been studied since then by many biochemists, mathematicians, engineers, and other scientists interested in protein folding. In the early 2000s, Oswin Aichholzer, David Bremner, Erik Demaine, Henk Meijer, Vera Sacrist´an, and Michael Soss explored many of the questions presented here [2]. Amino acids can be divided into two types: • H stands for hydrophobic amino acids – they Hate water • L stands for hydrophilic amino acids – they Love water

Amino Acid Types Alanine Asparagine Aspartic acid Glutamine Glutamic acid Histidine Isoleucine Leucine

hydrophobic (H) hydrophilic (L) hydrophilic (L) hydrophilic (L) hydrophilic (L) hydrophilic (L) hydrophobic (H) hydrophobic (H)

Lysine Methionine Phenylalanine Serine Threonine Tryptophan Tyrosine Valine

hydrophilic (L) hydrophobic (H) hydrophobic (H) hydrophilic (L) hydrophilic (L) hydrophobic (H) hydrophobic (H) hydrophobic (H)

In the H − L model, a chain of amino acids is placed onto a square grid. This means that at each amino acid, the chain may continue straight ahead or may make a 90◦ bend to the left or right. If two H amino acids are next to each other in the grid, but are not already connected by a link in the original chain, a hydrogen bond forms between the two H’s. No diagonal bonds are allowed. The total number of hydrogen bonds is called the bond score (B) of the configuration. Amino acid chains fold themselves so that their bond scores are as large as possible, because these situations correspond to lowenergy configurations for the protein and the water around it. The energy is lower in these configurations because when a chain has more hydrogen bonds, the hydrophobic amino acids are tucked inside the protein and are not exposed to the surrounding water molecules. When water molecules are disrupted by hydrophobic amino acids, they must orient themselves to create a water cage in order to be stable, which uses energy. (1) Figure 90 shows an example of a folded protein chain. Using dashed lines, draw in the hydrogen bonds and determine the bond score (B) for the configuration. (Remember that diagonal bonds are not allowed and you may not draw hydrogen bonds between H’s that are connected in the chain.) The H −L Protein Folding Model is usually called the H −P Protein Folding Model in publications, but we use H to stand for amino acids that Hate water and L to stand for amino acids that Love water. 10

THE H − L PROTEIN FOLDING MODEL

185

H

H

H

H

H

H

H

H

L

H

H

L

Figure 90. Example folded protein chain. What is the bond score?

(2) Translate the following chains of amino acids into strings of H’s and L’s: (a) Valine – Lysine – Tryptophan – Asparagine – Isoleucine – Histidine – Lysine – Phenylalanine (b) Leucine – Alanine – Aspartic acid – Glutamine – Tryptophan – Valine – Threonine – Tyrosine (c) Serine – Valine – Histidine – Lysine – Alanine – Aspartic Acid – Glutamic Acid – Tyrosine – Glutamine – Threonine – Leucine – Asparagine (d) Phenylalanine – Glutamic acid – Methionine – Glutamine – Alanine – Asparagine – Lysine – Valine – Threonine – Tryptophan – Aspartic acid – Glutamic acid – Phenylalanine – Isoleucine (3) Make or draw a model of one of the chains in the previous section. Try to find a configuration that has as large a bond score as possible. Is there more than one way to get the maximal score? (4) Here are some other questions to consider. While working, make a list of new questions, observations, and conjectures that occur to you. (a) Make up your own strings of length 8, 9, 10, 11, or 12 and decide how many configurations they have which maximize B. Explore this question for other strings of the same length. (b) What is the largest bond score B that is possible for chains consisting of h H’s? Determine the maximal bond score for H-chains of many lengths, beginning with length 1 and going up from there. (c) Consider a chain configuration picture and its mirror image. How do their corresponding H − L strings relate? Consider an H − L string and a string written in the reverse order. How do their chain configuration pictures relate? What does this suggest about how people exploring chains of a given length should count configurations?

186

AMANDA SERENEVY

(d) How would you describe a particular configuration to someone without drawing the picture? Can you invent notation that would describe the picture using words or symbols? Does every possible configuration correspond to a description in your system? Is there more than one description in your system for any configuration? (e) Find the optimal folding for the chain L − H − L − L − H − L − L − H − L − L − H − L. We could write that chain as (LHL)4 . Can you say anything about repeating pattern chains of the form (LHL)4n (meaning that the string L − H − L gets repeated 4n times)? (f) Make up your own repeating pattern chain along the lines of the previous question. Can you say anything about the maximal bond scores for your chain? (g) Choose a chain with a unique (up to symmetry) maximal bond score configuration, and call its specific bond score B. How many other chains of the same length would have bond score B if they were placed in that configuration? Is B the maximal bond score for all of those other chains or do they have better configurations? If B is the maximal score, is this configuration unique (up to symmetry) for the other chains, or do they have other configurations that also have bond score B?

THE H − L PROTEIN FOLDING MODEL

187

The H − L Protein Folding Model Teacher Guide If you choose to make physical models for the students to manipulate, here is one way to do it. Cut 1-inch diameter red and blue circles from card stock, and use an exacto knife to make a small slit in the center of each one. Cut 1/2 inch by 4 inch rectangles from white card stock, and make a slit near each end of each piece. Finally, cut 1/4 inch by 4 inch rectangles from yellow card stock. Students will need brass fasteners to assemble the protein chains and sticky tack to affix the hydrogen bonds.

Model Key Hydrophilic amino acids (L) Hydrophobic amino acids (H) Links in amino acid chain Hydrogen Bonds Bond Score (B)

Blue Circles Red Circles White Bars Yellow Bars # of Yellow Bars

Solutions: (1) H

H

H

H

H

H

H

H

L

H

H

L

Figure 91. Hydrogren bonds for the example chain.

The Bond score is B = 6. (2) (a) (b) (c) (d) (3) (a)

H −L−H −L−H −L−L−H H −H −L−L−H −H −L−H L−H −L−L−H −L−L−H −L−L−H −L H −L−H −L−H −L−L−H −L−H −L−L−H −H Figure 92 shows the optimal configuration for the amino acid chain in problem 2a, which has a bond score of 3.

188

AMANDA SERENEVY

L

H

H

H

L

L

H

L

Figure 92. The optimal configuration for the amino acid chain in problem (2a). There is only one configuration with 3 hydrogen bonds (not counting rotations and reflections as different). One way to see this is to use a parity argument. In the chain H −L−H −L−H −L−L−H, it is not possible to arrange the chain so that any of the first three H’s are adjacent in the grid. To see why this is impossible, we first mark the vertices of a grid with alternating open circles and open squares as shown in Figure 93.

Figure 93. A parity argument. If we number the positions along any protein chain, starting with 1 and then embed it into a grid, all of the odd numbers will fall into the same shape of vertex and all of the even numbers will fall into vertices with the opposite shape. See Figure 94 for an example of a numbered protein chain embedded in a grid. In the chain, H − L − H − L − H − L − L − H, since the first three H’s are in odd positions in the chain, they cannot be adjacent to each other in the grid. However, the last H is in an even position.

THE H − L PROTEIN FOLDING MODEL

189

7

8

9

10

6

3

2

11

5

4

1

12

Figure 94. A numbered chain embedded in a grid. Thus, the last H is the only one which can be adjacent to any of the other three. Since we found a configuration where the first three H’s are each adjacent to the final H, we know that we have an optimal configuration. To show that there is not another optimal configuration (up to symmetry), note that the last four amino acids in the chain must be arranged into a square so that the third H can be adjacent to the last H. Up to symmetry, there is only one way to do this. After making the square, the front part of the chain can only be wrapped around in one direction to align the other two H’s with the last H. Important Note: It is important that this protein has a unique optimal configuration (up to symmetry) that maximizes the hydrogen bonds. Proteins automatically fold up so that the number of hydrogen bonds is maximized (so that the energy required to hold that configuration is minimal). Proteins do their jobs by attaching to parts of the cell and to other molecules, and they can only attach properly if they have the proper shape. If there is more than one shape that a given protein could fold into, then there is no guarantee that the protein will form the correct shape to do its job. The H − L model is not exactly the same as real proteins, but it is useful to study the conditions which allow each protein to have a unique optimal folding. After developing techniques for the simple model, we could then turn to more complicated models that better approximate real proteins and try to apply similar techniques in that case. The H −L protein folding model is a good example of a “toy model” that is too simplistic to capture much of the actual behavior of objects, but which still plays an important role in helping scientists and mathematicians to develop theories.

190

AMANDA SERENEVY

(b) The diagrams in Figure 95 show two optimal configurations for the amino acid chain in problem 2b. Both of these configurations have a bond score of 3. H

H

H

L

L

H

H

L

H

H

L

H

H

L

L

H

Figure 95. Two optimal configurations for the chain in problem (2b). To see that the bond score of 3 is the best possible, we again look at the parity of the positions of the H’s in the chain H − H − L − L − H − H − L − H. The H’s are in positions 1, 2, 5, 6, and 8. Since hydrogen bonds can only form between numbers of opposite parity that are not already adjacent in the chain, the 1st H could theoretically bond with the 6th and/or the 8th H’s. The 2nd H could theoretically bond with the 5th H. The 5th H could also bond with the 8th H. There are no other potential bond pairs. This means that we could theoretically have as many as 4 hydrogen bonds. However, when we try to make a configuration that has all 4 of these bonds, we find that it is not possible. Let’s begin by embedding the chain so that the 1st H is adjacent to the 8th and 6th as shown in Figure 96. Note that it will not be possible to make the 5th amino acid lie adjacent to the 8th one no matter what we do (since the 5th amino acid must be connected to the 6th one). So it is impossible to have a bond score greater than 3.

THE H − L PROTEIN FOLDING MODEL

191

H8

H1

L7

H6

Figure 96. It is not possible to arrange the amino acid chain from problem (2b) to form 4 hydrogen bonds. So for this protein chain, there are two different overall shapes which both have the best possible bond score. This amino acid chain would not make a very good protein — the protein might fold up in more than one way and might not do the job it is needed for. (c) Figure 97 shows the optimal configuration for the chain L − H − L − L − H − L − L − H − L − L − H − L. L

L

L

H

H

L

L

H

H

L

L

L

Figure 97. The optimal configuration for the chain in problem (2c). This is the only configuration (up to symmetry) with the optimal bond score of 4. We know that 4 is optimal because each of the H’s can form at most 2 hydrogen bonds since none of them are at the ends of the chain. Each of the 4 H’s has 2 potential bonds, but each bond would be shared by 2 different H’s. Thus, the total number of potential bonds is (4 × 2)/2 = 4. The proof above made use of a combinatorial argument based on how many bonds can be formed using each H. Thus, we now have another kind of proof technique for approaching H − L protein folding problems.

192

AMANDA SERENEVY

(d) The amino acid chain in (2d) is harder to fold optimally because of its length and its structure. It is a good challenge to try to find a configuration with a bond score greater than 2. Kami Atcitty and Aiden Pioche-Lee found the configuration in Figure 98 with a bond score of 3.

H

H

L

L

L

H

L

H

L

H

L

H

H

L

Figure 98. Kami Atcitty and Aiden Pioche-Lee’s embedding for the chain in (2d). Kamia Leano found a different configuration (not shown), which also has 3 hydrogen bonds. Anastasia Todacheeny found the configuration in Figure 99 with 4 hydrogen bonds. L

L

H

H

L

L

H

L

H

H

H

L

H

L

Figure 99. Anastasia Todacheeny’s embedding for the chain in (2d).

THE H − L PROTEIN FOLDING MODEL

193

Devon Lynch and Kayleigh Paddock found two different configurations which also have a bond score of 4. Mandi Wheeler found the configuration in Figure 100 with 5 hydrogen bonds. L

L

H

H

L

H

L

L

H

H

H

H

L

L

Figure 100. Mandi Wheeler’s embedding for the chain in (2d). Irvilinda Bahe, Amber Cly, Leyonah Endischee, and Sky Harper independently found the configuration in Figure 101 with 6 hydrogen bonds. L6

L7

L4

H5

H8

L9

H3

H14

H13

H10

L2

H1

L12

L11

Figure 101. The embedding for the chain in (2d) found by Irvilinda Bahe, Amber Cly, Leyonah Endischee, and Sky Harper. Note that in the chain in problem 3d, H − L − H − L − H − L − L − H − L − H − L − L − H − H, there are H’s in the 1st, 3rd,

194

AMANDA SERENEVY

5th, 8th, 10th, 13th, and 14th positions. The two H’s at the ends could potentially bond with 3 other H’s, while the other 5 H’s could each potentially bond with 2 other H’s. Totaling these and dividing by 2 (to fix the double counting that happens when we count each bond from both ends), we find an upper bound on the number of hydrogen bonds of (2 × 3 + 5 × 2)/2 = 8. Thus, it seems as though it might be possible to find a configuration of this chain with a higher bond score. On the other hand, it may be that we can prove that 6 is actually optimal due to spatial constraints. Suppose that the 1st H actually connects with three other H’s. (Note that this did not happen in the configuration found above.) The only three that the 1st H could bond to for parity reasons are the 8th, 10th, and 14th H’s. If the 1st H bonded with all three of these, we would begin with the diagram shown in Figure 102. H

H

H1

H

L2

Figure 102. The starting configuration if we try to make the 1st H bond with three others. The top H cannot be H8 because if the chain wrapped around from the L2 to the top position in either direction, there would be no way for the chain to get to the encircled H10 or H14 later without crossing itself. Thus, H8 must be on one side or the other (say, left, without loss of generality), and H10 must be on the top to be close enough to H8 to connect as shown in Figure 103. We also know where L9 and H14 must be. It is now clear that, with this starting configuration, while H14 could still bond with H3 , it cannot bond with H5 – the positions adjacent to H14 are simply too far away from H8 . Thus, H14 can have at most 2 bonds with this starting configuration. H5 cannot bond with H10 for the same reason. Thus, H5 ’s only potential bond is with H8 . H8 cannot bond with H13 because of the distance in this configuration between H8 and H14 . Thus,

THE H − L PROTEIN FOLDING MODEL

L9

H10

H8

H1

195

H14

L2

Figure 103. Filling in more details in the candidate starting configuration. H13 ’s only potential bond is with H10 . H3 can bond either with H14 or H8 , but not both. However, if H3 bonds with either one, it will be impossible for H5 to bond with H8 . Thus, among those three potential bonds, only one can actually exist. Thus, at best, with this starting configuration, we can have the following bonds: H1 − H8 , H1 − H10 , H1 − H14 , H10 − H13 , and one out of H3 − H14 , H3 − H8 , H5 − H8 , for a total of at most 5 hydrogen bonds. This is lower than the best bond score we found. Thus, we can conclude that it is not possible for this protein to realize all 8 of its potential hydrogen bonds, and we now know that not all three potential bonds for H1 will be realized in an optimal solution. Thus, we now have 7 bonds as a theoretical upper bound and we have found one way to make 6 bonds. Your Math Circle may be able to improve on this result. (4) Most of the questions listed in part (4) on the handout are open-ended research questions. If your students become interested in these questions, I recommend reading [2], which is available online. Here are a few notes about some of the questions. (a) Answers to this question will vary depending on the specific chains created. (b) The H’s which are not at the endpoints have at most two potential bonds and if there are H’s at the ends, they have at most 3 potential bonds. Thus, there are at most [2(h − 2) + 3(2)]/2 = h − 2 + 3 = h + 1 bonds in a chain with h H’s. Here is a table showing results for a few h values.

196

AMANDA SERENEVY

h Maximum B 1 0 0 2 0 3 4 1 1 5 2 6 7 2 3 8 4 9 (c) Given a chain configuration, reflected and rotated versions of that image also represent valid embeddings of the chain. Reversing the order of an H − L string does not impact the ways it can be embedded. This suggests that we should not count rotated or reflected versions of a configuration as different. (d) Many answers are possible. (e) These chains have a maximum of 4n hydrogen bonds. See [2] for more details. (f) Many answers are possible. (g) One interesting example related to this question is the chain H − L − H − L − L − H − L. Since there are H’s in the 1st, 3rd, and 6th positions, the maximum potential bond score will be 2 and will happen when the 6th H bonds to the 1st and 3rd H’s. Up to symmetry there is a unique configuration for this protein, which is shown in Figure 104.

L

H

L

H

H

L

L

Figure 104. This amino acid chain has a unique configuration (up to symmetry) that maximizes the bond score. To count the number of chains with the same length that have bond score B, we assume that we are not counting the reverse of any chain as different. As long as the three H’s do not change to L’s, the bond score will still be B for this configuration. There are four amino acids (the L’s in the diagram) that could be either

THE H − L PROTEIN FOLDING MODEL

197

H’s or L’s, and so there are 24 = 16 different chains with the same configuration which have the same bond score. For chains of length 7, B = 2 is the maximal bond score possible. However, some chains of length 7 with H’s in the key locations do have other optimal configurations. For example, the chain with all H’s has several configurations with 2 hydrogen bonds.

References [1] Hue Sun Chan and Ken A. Dill. “The protein folding problem.” Physics Today, pages 24–32, February 1993. [2] Oswin Aichholzer, David Bremner, Erik D. Demaine, Henk Meijer, Vera Sacrist´ an, and Michael Soss, Long proteins with unique optimal foldings in the H-P model, Comput. Geom. 25 (2003), no. 1-2, 139–159, DOI 10.1016/S0925-7721(02)00134-7. Special issue on the European Workshop on Computational Geometry—CG01 (Berlin). MR1971267

SOMA and Friends Contributed by Dave Auckly and Stan Isaacs Short Description: The SOMA cube is a puzzle with seven different pieces. The goal is to create a 3 × 3 × 3 cube using all of the pieces. It is difficult to solve by random tinkering, but using a few observations from some of the other tiling sessions makes the puzzle easier to solve. In this session, students will make and then solve a fun puzzle. A shorter version would just have students solve the puzzle. Materials: Collect a number of variants of the SOMA puzzle before the session. In addition to the original SOMA puzzle, Coffin’s Quartet, Bram Cohen’s Best 5, Best 6 and Best 7 puzzles are reasonable to include. A collection of 9/16 inch wood cubes or 1 inch wood cubes is a nice addition. Such cubes may be found at many craft stores. Tape is the fastest way to put the cubes together, glue is more permanent. A good solution is to have the students tape the cubes for the math session, then glue them later. It is also possible to make cubes using origami and connect them together. With 27 cubes for each student, each student can leave with their own puzzle. If you are going to prepare puzzles ahead of time for the students to analyze, Coffin’s Quartet would be a good choice. It has a unique solution and a very clean analysis. In fact all of the puzzles listed here have a unique solution (with the exception of the original SOMA puzzle). All of the grid cube dissection problems may be analyzed via questions (9) – (15) on the handout. If you are having students make their own, there is a nice logic to the pieces of the SOMA puzzle. Of course, you may have the students try all the puzzles here, and make up their own via dissection, tilting and coloring. Send us descriptions of any particularly nice puzzles your students invent. References/Authorship: The SOMA Cube was invented by Piet Hein in the 1930’s. The tiles in the SOMA puzzle are three-dimensional generalizations of a type of tiles known as polyominoes. The name polyomino was invented by S. Golomb, [3]. There are many variants that lead to other puzzles. The original SOMA puzzle has many solutions. Bram Cohen set out to make the most difficult variants of the puzzle. He created the BEST 5, Best 6, and BEST 7 versions. Stan Isaacs is an avid puzzle collector, and has 199

200

DAVE AUCKLY AND STAN ISAACS

shared puzzles with many people over the years. In particular he has taken puzzles to many Julia Robinson Math Festivals, together with hand-outs describing the puzzles. The book Winning Ways for Your Mathematical Plays contains an extensive discussion of all solutions to the SOMA puzzle [1]. This is a classic book with many wonderful ideas to explore. Mathematics Beneath & Beyond: Counting is surprisingly useful. Determining the number of corners each piece will fill goes a long way to solving such puzzles. Determining the number of edge centers, face centers as well is very powerful. In fact, the method used in conjunction with this counting — checkerboard coloring of the cells in a large cube — has many applications, in particular, in the mathematical field of combinatorics. (You can see instances of this and other types of coloring in the chapter on Parity and Other Invariants, page 149.) The notion of symmetry which plays an important role in analyzing the puzzles is one of most fundamental ones in mathematics. Broadly speaking, a symmetry or invariance preserves a certain property (e.g., geometric similarity) of an object under some operation applied to the object. This notion of invariance is formalized in an elegant branch of mathematics called group theory, which is one of the powerful tools of modern mathematics. Thus familiarizing students with the idea of symmetry and using it in the playful but useful way which helps to solve concrete puzzles would go a long way for budding mathematicians. One more interesting piece of mathematics in the script is the use of mathematical induction in the solution to problem 3. This proof can enliven the discussion with more advanced audience (but of course it can be omitted with younger or not so well prepared students).

SOMA AND FRIENDS

201

SOMA and Friends Student Handout In this session you will get to make your own puzzle. Of course you will then be a bit tempted to solve it, and this session will provide hints to help. (1) A polyomino is made by attaching a finite number of squares together, so that each square shares at least one full edge with a neighbor and the whole group of squares makes a single piece. The pieces are named by the number of squares in the piece: • a domino has 2 squares; • a triomino has 3 squares; • a tetromino has 4 squares; • a pentomino has 5 squares; • ... Notice that for larger numbers, it is possible to have holes in the pieces. Also notice that any two squares of a polyomino are connected by a chain of squares where adjacent squares share a full edge. Now look at Figure 105. Is the left configuration a polyomino? Is the right one? Why or why not?

Figure 105. Polyomino and Non-polyomino (2) There are two different ways to consider polyominoes equivalent. One way is to declare two polyominoes to be the same if one may be moved in the plane to match the other without using reflections. This is called orientation-preserving equivalence. On the other hand, the notion of equivalence allows reflections. It is called congruence. Use four squares to make a polyomino that looks like an L. Use four squares to make a polyomino that looks like the Greek letter Γ. Use four squares to make a polyomino that looks like a 7. Which of these are orientationpreserving equivalent? Which of these are congruent? (3) Make a list of all polyominoes that may be made out of three squares. How many are there? Now convert these into polyominoes made out of four squares. Make each possibility up to orientation-preserving equivalence, so your list will include two out of the three L, Γ and 7 tiles. How many polyominoes made out of four squares are there? You could continue to ask how many could be made out of five, six, etc. Note: It is possible to make many interesting shapes out of several polyominoes. Given any such shape, you may trace the outline, and it becomes

202

DAVE AUCKLY AND STAN ISAACS

a puzzle to figure out how to make the outlined shape from the given set of polyominoes. You may make several such and share with your friends. (4) A solid polyomino is the analogue of the polyomino, except it is made out of cubes. Make a 2 × 3 × 4 “brick” (a rectangular prism). Is it a 3D polyomino? A “brick” is any 3D polyomino which consists of multiple layers, where all layers are the same rectangular shape. (5) Make all 3-cube solid polyominoes that are not bricks. How many are there? (6) Make all 4-cube solid polyominoes that are not bricks. Consider two the same if there is an orientation-preserving equivalence. How many are there? (7) Which pieces are symmetric? Which have distinct mirror images? (8) If done right you will need exactly 27 wooden cubes to make all of the non-brick 3-cube and 4-cube solid polyominoes up to orientationpreserving equivalence. These are the SOMA pieces. Make these with wood cubes and tape. (You may glue them together later.) The SOMA puzzle simply asks you to assemble these into a 3 × 3 × 3 cube. Spend some time and try to solve the SOMA puzzle before you continue. (The next page contains a number of hints.) Note: When talking about the SOMA cube (and other 3 × 3 × 3 cube puzzles), we will sometimes call the 3 × 3 × 3 cube the large cube (to distinguish it from the small cubes, which we will call cells, and which make up pieces). The cells at the vertices of the large cube we will call corners, the cells between two corners we will call edges, the six cells in the center of each face we will call face centers, and the very center of the large cube we will call the cube center (or sometimes SOMA center ). (9) How many corners does the large cube have? (10) Look at each of the SOMA pieces. Write the maximal number of corners of the large cube the piece can fill. Write the minimal number of corners of the large cube the piece can fill. (11) What is the largest number of corners the SOMA pieces can fill if they all fill the most possible? (12) Where must the T go? Why? (13) It is worth remembering (or solving) the classical checkerboard-domino problem: Is it possible to remove two diagonally opposite corner squares from a checkerboard and then tile the rest using dominoes? Hints: (a) How many black squares are on the checkerboard after you remove the opposite corners? How many white squares? (b) When a domino is placed over two squares on a checkerboard, how many black squares will it cover? How many white squares will it cover? (c) Explain to someone why this classical checkerboard-domino challenge has no solution.

SOMA AND FRIENDS

203

(14) Now imagine the large cube is colored like a 3-dimensional checkerboard with each cell either black or white and no two adjacent cells the same color. Are all the corners the same color? Edges? Face centers? Cube center? How many black cells will you have? How many white cells? (15) Look at each piece. If you checkerboard the pieces with two colors, which pieces have an equal number of cells of each color, which have different numbers? (16) If the T piece fills two corners, to get the correct number of black and white cells and the correct number of corners what can you say about the placement of the Y piece? (The Y piece is the one that looks like three cells meeting at a corner, and is one of the pieces that require two layers.) (17) Make a table listing the possible placements of each piece. (18) Can you guess how many solutions there are in total for the SOMA cube? It would be possible to write a computer program to list all of the solutions. There are MANY ways to dissect a cube into pieces like the SOMA cube, giving very many possible puzzles. Some are more difficult than SOMA, and some are easier. Some have a single solution, while others (like the SOMA) have many solutions. And, of course, you could make pieces for a 4 × 4 × 4 cube or even larger. To describe some of these we introduce a bit of terminology. When the set of n3 cells in a n × n × n cube are partitioned into subsets (i.e., disjoint collections of cells that fill the entire n × n × n cube) in such a way that each subset is a solid polyomino, we call the result a grid dissection of the cube. Several interesting grid dissection puzzles to analyze in the same way are • Coffin’s Quartet • Cohen’s BEST 5 • Cohen’s BEST 6 • Cohen’s BEST 7 • SOMA cube with replacements • Tilted variants • Colored/labeled variants Coffin’s Quartet: This variant was designed by Stuart Coffin around 1975, [2]. The pieces interlock and they must be added in the correct order. Figure 106 displays the pieces in the Coffin cube. In this figure (as in the following figures of cube grid dissection puzzles and variants) we put the cells of each piece in a 3 × 3 × 3 cube (though possibly not in the position it would be in a solution to the puzzle) and then display three layers (Bottom, Middle, and Top) of the large cube.

204

DAVE AUCKLY AND STAN ISAACS

Figure 106. Coffin’s Quartet Cohen’s BEST 5, 6, and 7: The most mathematically pleasing puzzles would not only have a solution, they would also have a unique solution for a given number of pieces. Furthermore, the solution would be a bit surprising. Bram Cohen set out to make the most difficult grid dissection puzzles of the large cube. Each has a unique solution. The piece placements tend to be surprising — not where the pieces seem to fit well.

Figure 107. Cohen’s BEST 5

SOMA AND FRIENDS

205

Figure 108. Cohen’s BEST 6

Figure 109. Cohen’s BEST 7 SOMA with replacements: The SOMA Cube uses only non-convex pieces. But you can replace either the triomino with a straight 3 × 1 × 1 brick (i.e., a straight row of three cells), or one of the tetromino pieces with a square

206

DAVE AUCKLY AND STAN ISAACS

2 × 2 × 1 brick (four cells arranged in a square), or both the triomino and one of the tetrominos at the same time. Most of those variations can still be solved, but not all. Which ones? Tilted variants: You can tilt the pieces and the whole cube, so the pieces fit in fewer ways. In general pick three distances a, b, and c. Then make two parallelograms with side lengths a and b, two with lengths b and c, and two with lengths c and a. These six parallelograms may be used as the faces of a parallelopiped, and these parallelopipeds may be used in the say way the cells are used in the other grid dissection puzzles. These may be made out of card stock. The same arguments about corners and checkerboard coloring parity apply, but many of the possibilities are eliminated. It is less clear how to obtain extra information from the tilt to include in the analysis. You can color the cells or apply other patterns, to limit the solutions. This will not change the versions that already have a unique solution, but it will change the versions that have multiple solutions by limiting the number of possibilities. Here are several fun examples. Number like a die: Put numbers on each exposed cell face of each piece so that (1) Each small face on the same large face has the same number, which is one of 1, . . ., 6. (2) The numbers associated to opposite faces of the large cube add up to seven and the numbers 1, 2, 3, cycle counter-clockwise around their common vertex, just like the numbers on a regular die. Number in a SUDUKO pattern: Put the numbers 1, . . ., 9 on each exposed cell face so that: (1) Each face of the large cube contains the numbers 1, . . ., 9. (2) The numbers in any row through any face, and the continuation of this row into the two adjacent faces contains the number 1, . . ., 9. Checkerboard version: Color the faces of each cell, so that when assembled, the colors will make a 3D checkerboard pattern.

SOMA AND FRIENDS

207

SOMA and Friends Teacher Guide Solutions: (1) Square SD is only connected to square SE . Similarly square SE is only connected to square SD . Thus there is no way to connect SD to SF by a chain each meeting the other along full edges. Hence the tile on the right is not a polyomino. (2) The L is orientation-preserving equivalent to the 7 as one can see by rotating it 180◦ . The only way to make the Γ match with one of the others would be to make a reflection. This is possible via horizontal reflection to make the L or vertical reflection to make the 7. (3) There are only two triominoes: a (tilted) V such as the one shown in the left of Figure 105, or the 1 × 3 rectangle. To see this, work up from the bottom. There is only one 1-square polyomino — the square. To make a 2-square polyomino, there must be one square to begin, and the second square must join it along an edge. This gives a 1 × 2 rectangle. Now build up to the 3-square polyominoes. Any such must have two squares that meet along an edge, so that makes a 1 × 2 rectangle. The third square must then meet along one of the edges on the long side to give a V or along a short edge to give a 1 × 3 rectangle. We can construct any (n + 1)-polyomino by adding a square to an edge of some n-polyomino. This process is demonstrated for n = 3 in Figure 110 and the proof of this fact is below.

Figure 110. Making 4-polyominoes. Note that Figure 110 does not include reflected versions of the polyominoes. Therefore, there are 7 different tetrominoes up to orientationpreserving equivalence. Proposition: Any (n + 1)-polyomino contains a k-polyomino for every k ≤ n + 1.

208

DAVE AUCKLY AND STAN ISAACS

Proof: We proceed by induction. It clearly contains a square, i.e. a 1polyomino. Now assume it contains a k-polyomino, Pk , and consider a square S in Pk and a square S  that is not part of this k-polyomino. There must be a chain of squares connecting S  to S. The square in this chain closest to Pk without being in Pk must meet Pk in an edge. Adding this closest square gives a (k + 1)-polyomino. (4) A 2 × 3 × 4 brick is a polyomino. Every pair of cubes are joined by a chain of cubes meeting along full faces. (5) There is just one. It is a 3-dimensional version of the (tilted) V shown in the left of Figure 105. (6) There are six. We make them by considering ways to add a cube to the 3-cube solid polyominoes, just as we did to make the 4-square polyominoes above. The answer is displayed in the Figure 111.

Figure 111. Making 4-cube SOMA blocks (7) The types of symmetry we’re concerned with are rotational symmetry and mirror (or reflection) symmetry. A piece has a rotational symmetry of order n (also called n-fold rotational symmetry, or discrete rotational symmetry of the nth order), with respect to a particular axis if rotation by an angle of 360◦ /n (180◦ , 120◦ , 90◦ , 72◦ , 60◦ , 360/7 = 51 3/7◦ , etc.) around this axis does not change the object. A piece has a mirror symmetry if there is a plane going through it which divides it into two parts which are mirror images of one another. (a) The L piece has mirror symmetry relative to the plane through the centers of all 4 cells. It does not have any rotational symmetry. (b) The V piece has two mirror symmetries: one relative to the plane through the centers of all 3 cells, and another one relative to the plane which cuts the middle cell along the diagonal of its top square and the edge which is the point of the V . It has 180◦ rotational symmetry about the line orthogonal to the edge common to all 3 cells and going through the center of the middle cell. (c) The Z piece has mirror symmetry relative to the plane through the centers of all 4 cells and 180◦ rotational symmetry about the line orthogonal to that plane through the center of symmetry of the cross-section. (d) The T piece has a mirror symmetry relative to the plane through centers of all 4 cells and another one relative to the plane that

SOMA AND FRIENDS

209

bisects the 2 middle cells and is orthogonal to the line through centers of 3 connected cells. It has 180◦ rotational symmetry about the line through centers of the 2 middle cells. (e) The Y piece has three planes of reflection and therefore three mirror symmetries. A plane of reflection bisects any two adjacent cells through the diagonal of the square between them and goes through the point of intersection of all four cells. The piece has 60◦ rotational symmetry about the central diagonal of the cell adjacent to 3 others that goes through the point of intersection of all 4 cells. (f) The W and RW pieces do not have mirror symmetry (but they are mirror images of one another). Each of them has 180◦ rotational symmetry about the diagonal line that bisects the square between the 2 pairs of adjacent cells and goes through the point of intersection of all 4 cells. (8) Find one solution from https://en.wikipedia.org/wiki/Soma_cube in Figure 112.

Figure 112. One SOMA Solution (9) There are 8 corners.

210

DAVE AUCKLY AND STAN ISAACS

(10) The following table lists the possibilities. Piece

V

Z L T

Y

W

RW

Max corners

1

1

2

2

1

1

1

Min corners

0

0

0

0

0

0

0

(11) The most corners is the total of the maximum each piece can fill, i.e. 9. (12) Since the T can only fill 2 or 0 corners, we know it must fill 2 otherwise at most 7 corners could be filled. Thus the T lies along an edge filling two corners! (13) (a) The checkerboard has 8 × 8 = 64 squares. Half of these or 32 are black. If one of the corners is black, the opposite would also have to be black. Removing these would leave 30 black squares and 32 white squares. we could also switch the numbers by removing a white corner and its opposite. (b) A domino would have to cover one white square and one black square. (c) The number of black squares covered would have to be the same as the number of white squares covered by part (b), but the number of black squares is different from the number of white squares, so they can’t all be covered by dominoes. (14) All of the corners would have the same color. Let’s pick black. Then all the edges would be white, all the face centers would be black, and the center of the cube would be white. This gives 8 corners plus 6 face centers for a total of 14 black cells. Since there are 3 × 3 × 3 = 27 cells, the remaining 13 must be white. (15) The Z, L, W and RW pieces will all have two black and two white cells. Thus these contribute a total of 4 × 2 = 8 black cells. Given the position of the T , we know it must have 3 black cells and one white cell. We are now up to a total of 8 + 3 = 11 black cells. The Y may either have 3 black, or 1 black cell. The V will have 2 black cells or 1 black cell. (16) If the Y piece fills 3 black cells, it would bring the total to 14, but the V would have to contribute at least one more and we would have too many black cells. Thus the Y must fill 1 black cell. This means the V would have to fill 2 black cells. It follows that the tip of the V cannot fill a corner. Since the V does not fill its maximal number of corners, every other block must, so the Y must fill a corner, and the corner must be the only black cell of the Y , so the center of the Y .

SOMA AND FRIENDS

211

(17) Given any possible position for a piece, one could get other possible positions by rotating or reflecting the entire 3 × 3 × 3 cube. We will only list one placement of each piece up to rotation. T : The only placement for a T is on a face of the cube with the “top” of the T along an edge. Y : The Y has two possible placements — the center of the Y may be in the corner of the cube, or an arm of the Y may be in the center of the cube (filling one less than its maximal number of corners). L: The L has two possible placements — on a face of the cube with the “side” of the L along an edge (such could be reflected and then counted as a third type of placement), or on a face of the cube with just the foot in a corner (filling one less than its maximal number of corners). The L cannot be in a position filling no corners. Can you see why? Z: The Z has two possible placements — on a face of the cube (which could be reflected and then counted as a third type of placement), or passing through the center of the cube (filling one less than its maximal number of corners). V : The V has three possible placements — on a face with the “point” on an edge of the cube, or in a central slice with the “point” on an edge of the cube (filling no corners), or with the “point” in the center of the entire cube (filling no corners). W : The W has four possible placements — two that hit a corner are displayed in Figure 113, two that do not hit a corner are displayed in Figure 114.

Figure 113. Placements for W Hitting a Corner

Figure 114. Placements for W Missing a Corner RW : The RW has four possible placements obtained by reflecting the placements displayed in Figure 113 and Figure 114.

212

DAVE AUCKLY AND STAN ISAACS

To list the possible positions of each piece, we can use a matrix with 0, 1 and 2 entries. Each column of the matrix will represent the coordinates of the centers of one of the cells of the piece. Thus we could say that the T piece could sit at ⎛ ⎞ 0 0 0 0 ⎝0 0 0 1⎠ . 0 1 2 1 Reading the columns, this would put one cube with center at (0, 0, 0), one with center (0, 0, 1), one with center (0, 0, 2) and one with center at (0, 1, 1). Using symmetry we could list many more positions for the T piece. If we ignored the argument showing that the T had to fill two corners we could also have a position such as ⎞ ⎛ 1 1 1 1 ⎝0 0 0 1⎠ . 0 1 2 1 As stated above, we interpret the problem to mean, “describe up to symmetry all possible placements of each piece.” The only placement for the T is to hit two corners: ⎛ ⎞ 0 0 0 0 ⎝0 0 0 1⎠ . 0 1 2 1 The Y must have its center ⎛ 0 ⎝0 0

in a 0 0 1

corner: ⎞ 0 1 1 0⎠ . 0 0

The L must hit two corners: ⎛ ⎞ 0 0 0 0 ⎝0 0 0 1⎠ . 0 1 2 0 The Z must hit one corner: ⎛ ⎞ 0 0 0 0 ⎝0 0 1 1⎠ . 0 1 2 1 The W must hit one corner, ⎛ 0 ⎝0 0 and

but there are two ways to do this: ⎞ 0 0 1 0 1 1⎠ , 1 0 0

⎛ ⎞ 1 1 2 1 ⎝1 1 2 2⎠ . 1 0 0 0

SOMA AND FRIENDS

213

The possibilities for the RW block are just the mirror images of the possible positions of the W block. The corner of the V block must hit a face center or the cube center: ⎛ ⎞ 1 1 1 ⎝0 0 1⎠ , 0 1 0 or ⎛ ⎞ 1 1 1 ⎝1 1 2⎠ . 1 1 1 (18) As we already noticed, once one solution is found, the cube may be rotated to get other solutions. Of course these solutions are not very different. One may count the rotational symmetries of a cube as follows: There is a 90◦ , a 180◦ , and a 270◦ rotation about a line through the centers of each pair of opposite faces, for a total of 3 × 3 = 9 rotations. There is a 120◦ and 240◦ rotation about a line through each pair of the opposite vertices for 4 × 2 = 8 more. There is also a 180◦ rotation about a line through the midpoints of each pair of opposite edges for additional 6 × 1 = 6 rotations. There is also the trivial 0◦ rotation. Thus the grand total is 9 + 8 + 6 + 1 = 24 rotations. Since the reflection of any SOMA piece is also a SOMA piece (check it for every SOMA piece!), reflecting a solution will give another solution. One way to figure out the number of possible solutions to the SOMA puzzle is to write a computer program that checks all possible placements of all of the pieces including changes by rotations and reflections. To record the position of a piece one could start by using Cartesian coordinates to label the centers of the cells of the 3 × 3 × 3 cube, say (x, y, z), with each variable equal to 0, 1, or 2. In this case a list of three or four points would record the placement of a given piece. We have had students write such programs in the past, however this requires a high level of programming ability. Many puzzles may be encoded and solved with a computer in this way. There is a bit more discussion on this point at the end of this section. Solving the problem on a computer is perhaps a bit unsatisfying. To analyze all solutions by hand is a huge task, but is possible. Indeed John Conway generated a list of all possible solutions by hand and recorded it in a graphic he named the SOMAP. The keys to this analysis were three observations. First, exactly one piece will cover the center of the cube. Second, exactly one piece will fail to cover its maximal number of corners. Third, given one solution (with just one exception) it is possible to find a new solution by finding subconfiguration that may be made in two different ways and switching it. Such a configuration is displayed in Figure 115.

214

DAVE AUCKLY AND STAN ISAACS

Figure 115. An Interesting Sub-configuration Given this, the procedure is to use the first and second observations to find one solution, then use the third observation to generate many more solutions. Once a list of solutions is generated, return to the first two observations to argue that all solutions have been found. In the end there are 240 solutions not counting those generated by reflection or rotation. After reflecting all of these there will be 480. After rotating all of these there will be 480 × 24 = 11,520. This is all mapped out in Winning Ways [1]. Repeating this is more than all but the most ardent puzzler would wish to do. However, it is very reasonable to use this method to analyze all solutions to the variants of SOMA that have a unique solution. Note: For a better understanding of the explanation given for Coffin’s Quartet and all Cohen’s BEST puzzles it is advisable to have and make use of actual models of the required pieces. Building such models would be possible using Figures 106–109. Coffin’s Quartet: One of the four pieces consists of precisely seven unit cubes. (It is the piece in the lower left of Figure 106.) The rest have an even number of unit cubes. Following the checkerboard coloring of problem (14), we see that each piece will fill the same number of black cubes as white cubes, except the piece with seven units. Since there are more black cubes than white cubes, this means the seven unit piece must fill four black and three white spots. Thus the bulk of it (the portion represented by the center column in the “Bottom,” “Middle,” and “Top” slice) is on a face. Now count the number of corners filled by the piece with 7 unit cubes and the piece with 8 unit cubes. The answer is six. Since the pieces with 6 unit cubes can fill at most one corner, we see that each must fill one corner. This implies that the piece on the bottom right of Figure 106 must also fill the center unit cube. There are two unit cubes in this piece that can be the center. Each cube can be rotated in 24 different ways. This leaves a total of 48 cases to check. Putting this piece next to the piece with 8 unit cubes, all but one case can be ruled out. In many cases, the two would not even fit. In a few cases they do fit, but the piece with 7 cubes could then not fit. Of the very few cases that the 8 unit-cube piece, the bottom right piece and 7 unit-cube all fit, only one will leave room for the final piece.

SOMA AND FRIENDS

215

There is one more trick hidden in Coffin’s Quartet. There is only one order that the pieces may be added. The easiest way to see this is to assemble the cube, then see which piece can be removed (only the 7 unit piece). After this only the non-central 6 unit piece may be removed, then the remaining pieces may be separated.

Figure 116. The Final Coffin

Cohen’s BEST 5: To describe the solution, it helps to name the pieces. We name the Pieces P1, P2, P3, P4 and P5. Referring to Figure 107, the piece names correspond to the following positions: P1 P2 P3 P4 P5 Analysis follows problems (9) – (15) from the SOMA hand-out. Start with a count of the number of corners each piece can fill: Piece

P1

P2

P3

P4

P5

Possible corners 0, 1, 2 0, 1, 2 0, 1 0, 1, 2 0, 2 Since the maximum number of corners that could be filled is 9 and a cube has 8, we conclude that P5 must fill two corners. Now Assume that P1 and P2 each occupy two corners. Then there are three cases, the edges containing the corners these pieces touch are: a) in the same plane, b) parallel, but not in the same plane, c) perpendicular or skew. Case a) is impossible because all such configurations leave a single hole. Case b) is impossible because once the center of the cube is filled, there will be a hole that is too small for any other piece. Case c) is impossible because after adding piece P4 to each such configuration in every possible way, we see that there is no way to fit in the other two pieces. We conclude that the assumption is false and one of P1 or P2 must touch just one corner. These pieces are mirror images of each other. Assume that the unit cube that is not in the plane of the other five unit cubes is in the corner of the 3 × 3 × 3 cube. Now count the number of black cubes (when colored as in problem (14) that are filled. Pieces

216

DAVE AUCKLY AND STAN ISAACS

P1 and P2 will fill 8 black cubes. Piece P4 would fill 3 black cubes, and the remaining two pieces would have to fill at least two black cubes, leading to a total of 15 black cubes, but there are only 14 black cubes. This means that this second assumption is also false. We conclude that one of P1 or P2 must fill exactly one corner via a unit cube in the same plane as five unit cubes. (There is only one such unit cube on each.) Now count the number of black cubes that are will be filled. In total P1 and P2 fill 6 black cubes, and P4 fills 3 black cubes. Each of the other two pieces must fill at least 2 black cubes, so must fill exactly 2 black cubes. It follows that P5 must fill 2 corners and 3 edge spots. (This position is unique up to symmetry.) Only P3 can fill the center cube, and there are only two unit cubes in P3 that could be in the center and still have only 2 black unit cubes on P3. Each of these may be rotated in 24 different orientations. One finds that the cube in P3 that shares a face with three other cubes cannot be in the center. From this point only one of the remaining 24 orientations can work. The solution is displayed in Figure 117.

Figure 117. The Best 5

Cohen’s BEST 6: Once again, counting corners, and considering the checkerboard restricts the possibilities sufficiently to allow one to find the unique solution. Writing out all of the possibilities is a bit tedious. The solution is below.

Figure 118. The Best 6

SOMA AND FRIENDS

217

Cohen’s BEST 7: Finally, counting corners, and considering the checkerboard restricts the possibilities sufficiently to allow one to find the unique solution. Label the pieces in Figure 109 as follows: P1 P2 P3 P4 . P5 P6 P7 Since pieces P1 — P5 can each fill at most one corner, and P6 and P7 each fill two or zero, we see that P6 and P7 must each fill 2 corners. This means that combined they fill 4 black cubes. Pieces P2 — P4 each fill exactly two black cubes. Thus, P1 must fill three black cubes and P2 must fill exactly one. It follows that the corner of P2 must be black. If P1 and P2 both fill a corner, then One of P6 or P7 must share a face with P2. There are two ways this can happen up to symmetry. There are then a number possible placements of P1 that fill a corner that must be checked. None can be completed. It follows that P1 cannot fill a corner. This will force it to fill the center cube. In fact, the unit cube in P1 that meets exactly two other unit cubes of P1 must be in the very center, or there would be too many white cubes filled. This in turn forces each of P3 — P5 to fill a corner with a unit cube that meets two other unit cubes in the piece. There are still a number of possibilities to check, but the number is low enough that a bit of experimentation will yield the solution. Proving that there is only one solution still requires a tedious check of possibilities. The solution is below.

Figure 119. The Best 7

Mathematics Beneath & Beyond: The SOMA cube and its variants are all generalized (into 3 dimensions) tiling problems. There is a vast literature on the mathematics of tiling, [4]. There are also lovely mathematical treatments of puzzles and games. The SOMA puzzle and variants represent a small fraction of the puzzles addressed in

218

DAVE AUCKLY AND STAN ISAACS

[1]. Another way to consider the problem of this type is via computational complexity. One class of decision problems is the class NP. This stands for non deterministic polynomial time. Roughly this means that there is a polynomial time algorithm to check if some configuration is a solution. It is a major unsolved math problem to determine if there is a polynomial time algorithm that can solve any given decision problem of class NP. An important subclass of NP problems are the NP complete problems. Any algorithm that will solve all instances of one NP complete problem may be used to solve any problem in NP. Karp has a list of NP complete problems, [6]. Computational complexity is discussed in the context of puzzles and games in [5]. One particular example of an NP complete problem is the Exact Cover Problem. Given a set X and a collection of subsets of X, one asks if it is possible to select some of the sets in the collection of subsets so that every element of X is in exactly one. Any generalized grid tiling problem is an instance of an Exact Cover Problem. This is easy to see. Let X be the set of all cells in the region to be tiled. (For example, X could be the unit cells in the 3 × 3 × 3 cube.) Now consider all subsets of X that could be covered by exactly one tile. Knuth wrote an efficient program to solve Exact Cover problems [7]. It may in particular be used to solve the grid dissection puzzles described in this section. It is reasonable to conjecture that one could create a n × n × n grid dissection puzzle representing any given size n Exact Cover problem. Doing so would prove that grid dissection puzzles are NP complete.

References [1] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning ways for your mathematical plays. Vol. 4, 2nd ed., A K Peters, Ltd., Wellesley, MA, 2004. MR2051076 [2] Stewart. Coffin. The puzzling world of polyhedral dissections. Oxford University Press, Oxford, 1991. [3] Solomon W. Golomb, Polyominoes, 2nd ed., Princeton University Press, Princeton, NJ, 1994. Puzzles, patterns, problems, and packings; With diagrams by Warren Lushbaugh; With an appendix by Andy Liu. MR1291821 [4] Branko Gr¨ unbaum and G. C. Shephard, Tilings and patterns, A Series of Books in the Mathematical Sciences, W. H. Freeman and Company, New York, 1989. An introduction. MR992195 [5] Joseph O’Rourke, Games, puzzles, and computation [book review of MR2537584], SIAM Rev. 52 (2010), no. 4, 782–785, DOI 10.1137/siread000048000004000785000001. MR2743249 [6] Richard M. Karp, Reducibility among combinatorial problems, Complexity of computer computations (Proc. Sympos., IBM Thomas J. Watson Res. Center, Yorktown Heights, N.Y., 1972), Plenum, New York, 1972, pp. 85–103. MR0378476 [7] Donald Knuth, Dancing Links, https://arxiv.org/abs/cs/0011047, 2000.

How To Fold A Tie Into Sevenths Contributed by James Tanton Materials: • Strips of paper cut from photocopy paper, enough for about five strips per participant. • Markers • A tie • One’s wits

219

220

JAMES TANTON

How to Fold a Tie Into Sevenths Student Handout A Personal Problem I travel a great deal and I often have to pack a tie in my suitcase. I can’t lay the tie out flat in the case, nor can I fold the tie in half and lay out the folded tie, as the case is too short. So I fold the tie into quarters and lay this out instead. But then, during my travels, my tie develops crease marks at the quarter positions and one of those creases shows right at my mid-chest when I later wear my tie. I figure that if I could fold my tie into thirds, then no crease would show and the folded tie would fit along the length of my case. But how do I fold a tie into thirds? Luckily a friend happened to share with me a good trick for doing that. (See Figure 120 for a pictorial version of the instructions.) She said: Make a guess as to where you think the one-third mark is along the tie and hold the tie there between two fingers of your left hand. Lift the tie up and let it drape down either side of your hand. Pick up the end of the longer of the two sections that are draping down and hold that end too with your left fingers. This leaves a loop material hanging down on just one side. Grab the center of that loop with the thumb and forefinger of your right hand and then let go of all you are holding with your left. You now have the tie up in the air draped either side of your right hand. Now repeat this procedure again: pick up the end of the longer section that is draping down, hold that end between your right fingers, grab the center of the loop now formed with two fingers of your left hand, and then drop the contents of your right hand. You are now in the exact same configuration as the starting position of the trick, except this time you are now holding the tie with your left hand and at a position along the tie slightly different from where you started. My friend continued: Here’s the wild thing: That new position is a much better approximation to the one-third mark along the tie than your initial guess. And to improve the approximation even more, simply repeat this procedure. If you do it over and over again the fingers of your left hand will move to positions closer and closer to the true one-third mark along the tie. You’ll soon be so close that your eye can’t detect the error. I tried this demonstration and it did seem to work! I first started with a reasonable initial guess as to the location of the one-third mark along the

HOW TO FOLD A TIE INTO SEVENTHS

221

Figure 120. Folding a tie into thirds. tie, and then I tried the demonstration a couple more times with absurdly inaccurate initial guesses. It looked like it worked in those cases too. But then I wondered, just for the fun of it, is there a procedure for folding a tie into sevenths?

222

JAMES TANTON

Mathematical Exploration Let’s understand the mathematics at play. Let’s start with a strip of paper and explore some beginning questions about folding to work our way up to fully understanding — and answering — the tie question. Grab a strip of paper! Warm-up Pick up the right end of the strip, bring it over to the left end, crease, and unfold. This produces a crease halfway along the strip as shown in Figure 121.

Figure 121. Making a crease at position 12 . If we declare the strip to be one unit long with the left end of the strip at position 0 and the right end at position 1, then we can say that this crease is at position 12 . Now pick up the left end of the strip and bring to the crease we just produced, crease again and unfold. This gives a new crease at position 14 as shown in Figure 122.

HOW TO FOLD A TIE INTO SEVENTHS

223

Figure 122. A crease a position 14 . Now pick up the right end of the strip and bring it to this new crease just produced, crease again and unfold. This gives a third crease. (See Figure 123.)

Figure 123. The mystery crease. Question 1: What is the position of this third new crease? Now pick up the right end again and bring it to this new crease just produced. (See Figure 124.)

224

JAMES TANTON

Figure 124. Another mystery crease. Question 2: What is the position of this fourth new crease? Question 3: Now perform a fifth new crease by performing a “left fold” (that is, pick up the left end of the strip and bring it to the fourth crease just produced). What is the position of this new crease? Question 4: On a fresh strip of paper, create a crease at position 38 by performing a series of left and right folds. Can it be done? If so, what sequence of left and right folds did you follow? Question 5 (HARD AND OPTIONAL!): Find a sequence of left and right folds that produces a crease at the 23 32 mark. Question 6: All the fractional positions mentioned thus far are quantities of halves, quarters, eighths, sixteenths, and thirty-seconds. Must every fraction produced along the strip through a series of left folds and right folds be a fraction with denominator a power of two? What do you think? Coding Folds There is some ambiguity as to how we start our folding: we can either pick up the right end of the strip and fold it all the way left to produce a crease at positron 12 , or we can pick up the left end and fold it all the way right to produce a crease at position 12 . For the sake of consistency, let’s agree to always start our folding actions the first way, that is, with a right fold. And just to be clear, once one crease is created, let’s define our next folding actions as follows. Left Fold (L): Pick up the left end of the strip of paper, bring it to the crease just created, and produce a new crease to its left. (See Figure 125.) Right Fold (R): Pick up the right end of the strip of paper, bring it to the crease just created, and produce a new crease to its right.

HOW TO FOLD A TIE INTO SEVENTHS

225

Figure 125. The Left Fold (L) and the Right Fold (R). Now we can encode our folding instructions as a series of Rs and Ls, to be read left to right. For example, we can create a crease at position 38 by performing a right fold, then a right fold, and then a left fold. Let’s write 3 ↔ RRL. 8 By our convention, the folding code for a half is 1 ↔ R. 2 Question 7: Write the coding instructions for each of the fractions we have 13 23 folded thus far, namely, for 14 , 58 , 13 16 , 32 , and 32 . Question 8: Find the coding instructions for the remaining eighths, namely 1 7 2 8 , and 8 . (What are the folding instructions for 8 ?) Care to find the coding instructions all the remaining sixteenths as well? (The answer can be no!) Question 9: What fractions are produced by the coding instructions RL, RLRL, RLRLRL, and RLRLRLRL? Hint: This might be too tedious to compute, but we can at least carry out these actions on a fresh strip of paper. Do a right fold followed by a left fold and color the newest crease made with a marker. Now do another right fold followed by a left fold and color the newest crease. Do this several times. What do you notice about the colored creases? Question 10: Do you think there is any hope of folding a crease at position 1 1 3 ? How about one at position 7 ?

226

JAMES TANTON

The Powers of a Half Since we are working with fractions given in terms of halves, quarters, eighths, and the like, it might be fruitful to think about writing all fractions in terms of the powers of a half, that is, in terms of halves, quarters, and eighths, and so on.   1 1 = 2 2  2 1 1 = 2 4  3 1 1 = 2 8  4 1 1 = 2 16 1 1 1 For instance, we have that 58 equals 12 + 18 , and that 13 16 = 2 + 4 + 16 . There is no need to repeat a particular power of a half in a sum that 1 appears twice in a sum simply represents a fraction. For example, if 16 1 1 1 replace 16 + 16 with 8 .

Question 11: What is 38 as a sum of distinct powers of a half? What is 23 32 98 as such a sum? What is 256 as such a sum? Is there a relatively easy way to compute these sums? In everyday life we work with powers of ten and powers of a tenth: we have the decimal system of arithmetic, base ten. When we write 2783.75, for instance, we mean 2 × 1000 + 7 × 100 + 8 × 10 + 3 × 1 + 7 ×

1 1 +5× , 10 100

a sum of powers of ten and powers of a tenth. But we could write numbers in terms of the powers of two and powers of a half to create a base two system of arithmetic (binary arithmetic). For example, the number 1101.011, in binary, is 1×8+1×4+0×2+1×1+0×

1 1 1 +1× +1× , 2 4 8

which equals thirteen and three-eighths. Only the digits 0 and 1 are ever needed in binary. For example, two 1 , can be replaced by one eighth, 1× 81 , and three thirty-twos, sixteenths, 2× 16 3 × 32, can be replaced by one sixty-four and one thirty-two, 1 × 64 + 1 × 32. We can certainly write our folding fractions in binary. For example, from 5 1 1 5 13 1 1 1 = 8 2 + 8 we see that 8 is .101 in binary. From 16 = 2 + 4 + 16 , we get that 13 16 is .1101 in binary.

HOW TO FOLD A TIE INTO SEVENTHS

227

Shocker This next question is the heart of this activity. There is lots of mulling and deep reflection to be had here. Question 12: Here are all the folding fractions we have examined so far: 1 1 3 5 13 13 23 2 , 4 , 8 , 8 , 16 , 32 , and 32 . Write each fraction in binary. Compare each binary expression with the folding code for that fraction. What do you notice? WHOA! Can you explain what you notice? Carrying on. Question 13: I sometimes like to wear my tie so that 33 64 of its length show down my front. Give me a set of instructions I can follow so that I can find the thirty-three-sixty-fourth mark along my tie. Question 14: What is the fraction 13 in binary? Hint: We want to find values a, b, c, d, . . . so that 13 equals d + . . . with .abcd . . . in binary. This means 13 = a2 + 4b + 8c + 16 1 each value in a numerator either 0 or 1. Since 3 is smaller than a half, we need a = 0. Since 13 is bigger than a quarter, we can take b = 1. Perhaps continue reasoning this way to make a good guess as to what the binary expansion of 13 could be. (Alternatively, maybe question 9 suggests a good guess?) Prove that your guess is correct. Question 15: Following question 9, explain why repeatedly performing a right fold followed by a left fold gives a sequence of creases that converges to the position 13 . Can you now explain the opening tie demonstration? Question 16: Explain how to perform a series of actions on a tie to give the one-seventh mark along the tie to any degree of accuracy you desire.

228

JAMES TANTON

How to Fold a Tie Into Sevenths Teacher Guide Question 1: Question 2: Question 3: Question 4: position

This third crease is at position 58 . This fourth crease is at position 13 16 . This fifth crease is at position 13 . 32 Start with either a right fold or a left fold to get a crease at 1 2 . Then perform a right fold and then perform a left fold.

Question 5: Start with either a right fold or a left fold to get a crease at position 12 . Then perform two more right folds, then a left fold, and then a right fold.

Question 6: This seems to be the case. Each action of folding halves a section of the paper — either halves the section to the left of a crease or halves the section to its right. Thus, if one crease is at a position described in a quantity of eighths, say, it doesn’t seem surprising that the next crease will be described in terms of sixteenths. Comment: The intuition of this argument is correct. As the mathematics of this activity is developed, a rigorous argument for it will become clear. Question 7: 5 13 13 23 1 ↔ RL; ↔ RLR; ↔ RLRR; ↔ RLRRL; ↔ RRRLR 4 8 16 32 32 Question 8: 7 1 ↔ RLL; ↔ RRR 8 8 (And we already have the code for 28 = 14 .) 3 5 7 1 ↔ RLLL; ↔ RRLL; ↔ RLRL; ↔ RRRL; 16 16 16 16 9 11 13 15 ↔ RLLR; ↔ RRLR; ↔ RLRR; ↔ RRRR 16 16 16 16 Question 9: If one has the patience, we find 5 21 85 1 . RL ↔ ; RLRL ↔ ; RLRLRL ↔ ; RLRLRLRL ↔ 4 16 64 256 It looks like we have fractions with denominator one more than triple the numerator. If this is so, then these fractions approach the

HOW TO FOLD A TIE INTO SEVENTHS

229

value one-third. This seems to be the case if you physically check with a strip of paper. Question 10: The previous question suggests it is possible to produce creases as close as we please to the one-third mark. Is there sequence of folds we could repeat to give creases that converge to the one-seventh mark? 1 1 A1
1 1 Question 11: We have 38 = 2+1 8 = 4 + 8 . If we think of this as
2A + 4 + 8 we 23 16+4+2+1 1 1 1 see it matches .011 in binary. We have 32 = = 2 + 18 + 16 + 32 32 98 1 and so is .10111 in binary and 256 = 64+32+2 = 14 + 18 + 128 and so is 256 .0110001 in binary. Comment: Many people find it easiest to first write each number in the numerator as a sum of powers of two. HARD OPTIONAL CHALLENGE. A fraction between 0 and 1 has denominator a power of two (2, 4, 8, 16, 32, 64, . . .). Could that fraction be written as sum of distinct powers of a half in more than one way? Question 12: Here is a table of all the fractions mentioned, their folding codes, and their binary representations. Fraction Code Binary 1 R .1 2 1 RL .01 4 3 RRL .011 8 5 RLR .101 8 13 RLRR .1101 16 13 RLRRL .01101 32 23 RRRLR .10111 32 It appears that there is a relationship between folding codes and binary representations. Read the binary representation of a fraction backwards (that is, from right to left) interpreting each 1 as an R and each 0 as an L. Then you have the folding code for the fraction! 16+8+1 1 = 12 + 14 + 32 , and so has For example, the fraction 25 32 equals 32 binary representation .11001. And one can check that performing the instructions RLLRR does indeed produce a crease mark at position 25 32 . Why do matters work this way? Let’s first examine the mathematical effect of performing a left fold or a right fold. Suppose we have a crease at position X along the strip as showin in Figure 126.

230

JAMES TANTON

Figure 126. A crease at position X. As we are declaring the length of the whole strip to be one unit, the length of the section of the strip to the left of the crease is x units and the length of the section to its right is 1 − x units. A left fold produces a crease halfway along the left section of the strip and so gives a new crease is at position x2 . A right fold produces a crease halfway along the right section of the strip and so produces a new crease at position x + 1−x 2 . (See Figure 127.)

Figure 127. New creases from a given crease.

1 2

Some algebra shows that x + + x2 .

1−x 2

equals x +

1 2

− x2 , which equals

HOW TO FOLD A TIE INTO SEVENTHS

231

What effects do these operations have on binary representations? Suppose, in binary, we have x = .abcd . . .. This means that x = a b c d + 2 4 + 8 + 16 + . . .. c + A left fold gives a crease at position x2 , which equals a4 + 8b + 16 d x 32 + . . ., and in binary this reads 2 = .0abcd . . .. A right fold gives a c d + 32 + . . ., and crease at position 12 + x2 , which equals 12 + a4 + 8b + 16 1 x in binary this reads 2 + 2 = .1abcd . . ..

We see that, in a given binary representation of a folding fraction, a left fold has the effect of inserting a zero just right of the point and a right fold the effect of inserting one. That is, the digits 0 do indeed correspond to left folds and the digits 1 to right folds! Moreover, as each fold inserts a digit just to the right of the point, all the early folds appear at the right end of the binary representation and all the latter folds to the left, just after the point. This explains the need to read the binary representations backwards. right fold R .1= 12 left fold RL .01= 14 left fold RLL .001= 18 9 right fold RLLR .1001= 16 right fold RLLRR .11001= 25 32 32+1 1 Question 13: We have that 33 = 12 + 64 , and so is .100001 in 64 = 64 binary. Performing a right fold, followed by four left folds, and then a final right fold will produce a crease thirty-three sixty-fourths along the tie. (Be sure to regard the thick end of the tie as the left end.) Question 14: It turns out that 13 has an infinitely long expansion in binary. It equals .01010101. . . = .01. How might we see this? The number .01010101. . . equals

1 C1 1 1 1 C1 A1

S1  + + ... C2C + 4 + 8CC + 16 +
32 AA + 64 +  S 128 S 256

1 1 1 1 + + + + .... 4 16 64 256 Is there any reason to believe this sum equals one third? Stare at Figure 128. It has infinitely many layers with each layer containing three same-sized triangles, one of them shaded. =

232

JAMES TANTON

Figure 128. What is the sum of this series of fractions? As a third of each layer is shaded, one third of the whole entire triangle is shaded. But look at the largest shaded triangle. It is one quarter of the size of the whole triangle. And the next largest shaded triangle is one quarter of one quarter of the whole triangle, one sixteenth. And the next shaded triangle is one quarter of one quarter of one quarter, that is, one sixty-fourth of the whole triangle. And so on. So the total amount of shaded region of the triangle can also be expressed as 1 1 1 4 + 16 + 64 + . . .. It’s the same shaded region, so this quantity must also equal one third. So, yes, .01010101. . . in binary is the fraction 13 . Question 15: We have that 13 = .01010101 . . . in binary. What does this all mean for tie folding? In an attempt to find the one-third position along a tie, start by making one guess as to where it might be. Suppose you guess the position .abcd . . .. This guess might not at all look like a third. Performing a right fold and then a left fold inserts a 1 and a 0. You now have a crease at position .01abcd . . .. This looks more like a third. (Can you see that the actions described for a tie match right and left folds?) Now perform another right fold/left fold pair. This gives you a crease at position .0101abcd . . . This looks even more like a third. Do this another three times and you have a crease at a position that agrees with the true one third mark in the first ten binary places after the point: .0101010101abcd . . ..

HOW TO FOLD A TIE INTO SEVENTHS

233

As the tenth power of a half is .0009765625, this means that your current crease is accurate within a thousandth of a unit. Whoa! Question 16: We need to know what 17 is in binary. Like one third, it probably has an infinitely long repeating pattern. Now as one seventh is a little more than an eighth, we might guess that its expansion begins .001. . .. Some calculation shows that adding a sixteenth or a thirty-second to an eighth takes us beyond 17 , but adding a sixty-fourth does not. This suggests its expansion begins .001001. . .. At this point we might guess that 17 = .001001001 . . . = .001. Does it? 1 + Here’s one way to compute the value of .001001001. . . = 18 + 64 1 + . . . (the sum of the powers of an eighth). Give this quantity a 512 name. Call it A, for Angelique, say. 1 1 1 + + + . . . = A. 8 64 512 Multiplying through by 8 feels natural. 1 1 + + . . . = 8A. 8 64 We see a copy of Angelique in the left side. 1+

1 + A = 8A. A little algebra now shows that A = 17 . So yes, .001 is the binary representation of 17 . So what does this mean for tie folding? Well, make a guess as to where the one-seventh mark is along the tie. Say you guess the position .abcd . . .. Performing a right fold, then left fold, and then another left fold, gives you a crease at position .001abcd . . ., which is closer to being one-seventh. Repeating this sequence of folds, RLL, multiple times gives you better and better approximations to one-seventh. Try this: From a very large pile of dried beans or coins or pebbles pull out to the side what you think looks like one seventh of the pile. Now split the large pile that remains in half (just eye-balling it will do) and add one half to your estimation pile. Now split your estimation pile in half and in half again (again, eye balling suffices). You now have a count of objects in your estimation pile very close to being one-seventh of the original pile. Check it!

Boomerang Fractions Contributed by Amanda Serenevy Short Description: This session allowed students in our 2015 Baa H´ ozh´o Math Camp at Din´e College to explore some intriguing questions relating to fractions which had just been proposed at the Banff International Research Station (BIRS) Conference on Integer Sequences in late February of that year. One delightful aspect of this topic is that many of the initial conjectures of the mathematicians exploring this topic for the first time turned out to be wrong. On the other hand, there are lots of theorems relating to specific cases that students can easily discover and prove for themselves within one or two sessions. While it is not necessary for students to use a computer to enjoy exploring this topic, those who are familiar with any basic programming language will likely wish to use a computer in order to formulate and test conjectures. Materials: Board and writing implements for introducing the topic and sharing ideas, pencils, paper, handouts if desired. Students familiar with programming may wish to have access to computers. Problem Background: At the request of Gordon Hamilton, who organized the BIRS Conference on Integer Sequences in late February 2015, David Wilson suggested the original question. Gordon Hamilton, Joshua Zucker, Richard Guy, Amanda Serenevy and others at the conference participated in the original exploration of this question. Gordon Hamilton compiled the initial results to create an entry for the On-Line Encyclopedia of Integer Sequences [1]. Bob Klein contributed to the development of this Math Circle script by proposing terminology, and by assisting with initial presentations of this topic during the summer of 2015 to the Southeastern Ohio Math Teachers’ Circle and the students at Baa H´ozh´o Math Camp.

235

236

AMANDA SERENEVY

Boomerang Fractions Student Handout Choose a fraction m/n. Begin a sequence of numbers with 1. To get the next number of the sequence, add m/n. On subsequent steps either add m/n or take the reciprocal. We say that the longevity of the fraction is the minimum number of steps needed to return to 1. For example, consider the fraction 1/2. The sequence below shows the quickest way to return to 1 using this fraction, and so the longevity of 1/2 is 4. 1 3 →2 →1 2 2 Investigate some of the questions below or create your own questions about boomerang fractions. (1) What is the longevity of 1/3? Can you find more than one way to get back to 1 using the minimum number of steps? (2) What is the longevity of 1/4? (3) What is the longevity of 1/5? (4) For fractions of the form 1/n, find the longevities for n = 2, 3, 4, . . . , 16. (5) Prove that a fraction of the form 1/n has finite longevity. (6) What is the formula for the longevity of a fraction of the form 1/n? (Alternatively, can you find any bounds on this longevity?) (7) Are there some fractions with infinite longevity (meaning it is impossible to return to 1)? (8) Prove that fractions of the form n−1 n have finite longevity. (9) Find the longevity of some fractions of the form n−1 n . (10) If you make a tree of all of the possible sequences which can be generated using the two allowed operations, how many branches of each length will the tree have? (11) Is it possible to return to 1 if the fraction 9/11 is used? 1→

BOOMERANG FRACTIONS

237

Boomerang Fractions Teacher Guide Solutions: (1) The longevity of 1/3 is 9. Here are two different ways to demonstrate this longevity: 1→

5 7 8 1 2 4 → →2→ → →3 → →1 3 3 3 3 3 3

5 1 5 7 3 2 4 → →2 → → →  →1 3 3 2 6 6 2 3 The method of exhaustion (possibly aided by a computer) can be used to verify that there is no shorter longevity by generating a tree with every number that can be obtained in nine steps or fewer. Figure 129 shows an example of a tree which generates every pos1 sible number that can be obtained in four steps or fewer with m n = 2: 1→

Figure 129. A tree showing all numbers that can be ob1 tained in four steps or fewer if m n = 2. Let K1/2 (j) denote the set of rational numbers which can be obtained after j steps when using 1/2 as the amount that may be added (compare with the sets L(j) in the More Mathematics for a Curi  ous Reader section). Using the tree we see that K1/2 (1) = 32 ,  2 5 1 7  2  K1/2 (2) = 2, 3 , K1/2 (3) = 2 , 2 , 6 , K1/2 (4) = 3, 5 , 1, 53 , 67 . In particular, this tree demonstrates that the longevity of 12 is indeed 4. (2) The longevity of 1/4 is 7. The sequence below demonstrates this longevity. 1→

3 7 1 3 5 → → →2 → →1 4 2 4 2 4

238

AMANDA SERENEVY

Using the method of exhaustion, we can verify that a shorter route back to 1 does not exist. (3) The longevity of 1/5 is 20. Both of the sequences below demonstrate this longevity. 1→

7 8 9 1 7 9 11 13 6 → → → →2 → → → → → 5 5 5 5 2 10 10 10 10 2 13 16 19 22 5 3 4 3  → → → → →  → →1 2 3 15 15 15 15 3 5 5

7 8 9 1 7 9 11 13 6 → → → →2 → → → → → 5 5 5 5 2 10 10 10 10 17 19 21 23 5 2 3 4 15 → → → → →  → → →1 10 10 10 10 10 2 5 5 5 Using the method of exhaustion or a computer program, we can verify that a shorter route back to 1 does not exist. (4) The first 15 longevities for fractions of the form 1/n (starting with n = 2) are shown in the table below [1]. 1→

n Longevity

2 4

3 9

4 7

5 20

6 6

7 33

8 13

9 23

10 16

11 62

12 8

13 75

14 18

15 17

16 25

(5) A fraction of the form 1/n always has finite longevity. At the worst, we can simply add 1/n until we get up to n itself, then take the reciprocal to get 1/n, and then add 1/n until we get to 1. (6) One bound for the longevity of 1/n can be obtained by simply quantifying the technique described above. To get from 1 up to n, it is necessary to add (n2 − n) copies of 1/n. (To see this, note that 2 1 + n n−n = 1 + n − 1 = n.) Next, it takes one step to get from n to 1/n. Finally, we must add n − 1 copies of 1/n to get up to 1 from 1/n. Thus, one upper bound for the longevity is n2 − n + 1 + n − 1 = n2 . However, for the first 15 entries in the sequence, this bound on the longevity is only realized for n = 2 and n = 3. Thus, it may be possible to achieve a better bound on the longevities for this family of fractions. To the best of my knowledge, no one has yet found a formula in terms of n that gives the longevity of 1/n. (7) Any number m/n that is greater than or equal to 1 will have infinite longevity. This is because to land back at 1 again, we must add m/n to something (call it x) and then obtain 1. So x + m/n = 1. However, if m/n is greater than 1, then x must be negative or zero. It is not possible to reach any non-positive numbers via the operations of adding a positive number and taking reciprocals. Numbers less than -1 will also have infinite longevity. n−1 , the route back to 1 must pass through (8) For fractions of the form n 1 in the step before reaching 1. Following the route backwards one n

BOOMERANG FRACTIONS

239

 n−1 = more step, the reciprocal is n. Starting at 1, note that 1+n n 1 + n − 1 = n. Thus, there is always a route back to 1 and so the longevity is finite. (9) As hinted in the previous answer, the shortest route back to 1 must end 1 1 → ··· → n  → 1 n It is clear that the quickest route from 1 to n is to add n copies of n−1 n−1 . Thus, the longevity of any fraction of the form is n + 2. n n Here are a few examples for fractions of this form: 2 m = : Longevity 5 n 3 5 7 1 1→ → →3 →1 3 3 3 3 m = : Longevity 6 n 4 10 13 1 7 → →4 →1 1→ → 4 4 4 4 4 m = : Longevity 7 n 5 13 17 21 1 9 → → →5 →1 1→ → 5 5 5 5 5 (10) It is not effective to take a reciprocal twice in a row. Thus, when listing the next steps possible in a given path, if the last step added m/n, then there are two alternatives for the next step. If the last step took a reciprocal, then there is only one possibility for the next step. For the first step in the path, there is only one option: adding m/n. For the second step, there are two options: adding another m/n or taking a reciprocal. For the third step in the tree of all possible paths, there are three options – two of the options will add m/n, and one will involve taking a reciprocal. In general, at a given step there will be a paths that add m/n at that step and b paths that take a reciprocal at that step, for a total of a + b paths. At the next step, there will be a + b paths that add m/n and a paths that take a reciprocal, for a total of 2a + b paths. The step after that will include 2a + b paths that add m/n and a + b paths that take a reciprocal. Notice that the sum of the number of paths in the first two steps is equal to the number of paths in the third step, and that this will always be true. Because this holds true for all steps, and because the sequence begins with a 1 and a 2, we can see that the 

240

AMANDA SERENEVY

Fibonacci sequence gives the number of paths that are possible at each step. (11) It is surprisingly difficult to decide whether the fraction 9/11 has a finite longevity, even with a computer program. Will Orrick noted in an email exchange in August 2015 that the longevities of the fractions 7/9, 7/11, 9/11, 11/13, 11/14, 13/15, 13/16 are all rather difficult to determine. Using an optimized computer algorithm, Will was able to find that the longevities of 7/9, 7/11, and 11/14 are all finite. His algorithm was not able to determine whether the other fractions on the list above have finite longevities. This example illustrates that it is not always obvious whether or not returning to 1 is possible by adding m/n and by taking reciprocals.

BOOMERANG FRACTIONS

241

More Mathematics For a Curious Reader Generators It is interesting to note that a different choice of two operations (generators) could allow us to be certain that there is a path from any rational number to any other. For example, this is possible if we instead choose adding one and taking the negative reciprocal as the two operations. Is it always best to reciprocate down? Several of the mathematicians who played with this problem during the BIRS Conference on Integer Sequences in February 2015 conjectured that optimal paths would always use reciprocals to move from numbers above one to numbers below one. However, Joshua Zucker noted that the sequence for 1/21 is surprising in that the optimal path requires taking the reciprocal of a fraction below one to a fraction that is greater than 1. Common Factors Conjecture When exploring the longevities of fractions of the form 1p , where p is prime, it seems as though the shortest path back to 1 only takes reciprocals when the numerator and denominator have a factor in common. For example, in question (3) when we investigated the longevity of 15 , the most efficient paths 15 25 back to 1 took reciprocals at either 10 5 , 10 , and 15 (in the first solution given), 25 or at 10 5 and 10 (in the second solution) rather than occurring at fractions which did not simplify. Are these locations always the best places to take reciprocals? Is it possible to find an efficient route back to 1 that does not lead through taking a reciprocal of a simplifying fraction? Note on Some Computational Approaches Will Orrick sent the following notes about computational approaches. An obvious thing to try is to start with 1+m/n and trace through the binary search tree exhaustively, but this runs out of steam very quickly. A second idea is to observe that most of the paths one finds in playing around by hand use the reciprocal step relatively infrequently. So one can fix the number of times the reciprocal step gets used to some small number, 1, 2, 3, 4, 5, . . ., and go much deeper into the search tree. I did manage to turn up a path from 1 + m/n to 1 when m/n = 5/7, which used the reciprocal step four times. I had hoped that some of the difficult values of m/n would yield to this approach, but in retrospect it is clear that this was doomed. The other day I tried something different. Let A be the operation “add m/n” and let B be the operation “take the reciprocal

242

AMANDA SERENEVY

and then add m/n”. Define Lm/n (j) to be the set of positive rational numbers m/n such that some sequence of the operations A and B, using B exactly j times, when applied to m/n  yields  1. Example: Fix m/n = 7/9. Then L7/9 (0) = 29 , 1 since   A1 ( 29 ) = A0 (1) = 1. Also L7/9 (1) = 92 − 7i 9 | i = 0, 1, 2, 3, 4, 5 since the only way to reach 1 using B exactly once is to apply A some number of times to get 9/2, and then to apply B to get 1. (In this case there can’t be any applications of A after the B step since B includes the “add m/n” operation and any additional “add m/n” steps add a number greater than 1, and will therefore never result in 1.) One can continue this and can show that Lm/n (j) is always a finite set. It tends to grow with j, but the size isn’t monotonic. The growth is probably exponential, but the exponent is much higher for some values of m/n than for others. Fortunately, growth tends to be slower for the difficult values of m/n than it is for the easier values, which allows one, in some cases, to reach high enough j to find an Lm/n (j) that contains 1.

References [1] Sequence A256174 “Boomerang Fractions”. The On-Line Encyclopedia of Integer Sequences https://oeis.org

From Hats to Codes Contributed by Joe Buhler and Bob Klein Short Description: This session arose out of the 2016 Baa H´ ozh´o Math Camp at Din´e College, where a puzzle involving hat colors was used as a device to gently introduce students to some combinatorial ideas that lead to one of the most basic ideas in error-correcting and covering codes. The material has numerous connections to combinatorics, probability, information theory, logic, algebra, and computer science. The use of hats as a setting for mathematical ideas goes back (at least) to George Gamow. In 1998 Todd Ebert gave an amusing n-player “hats game” as an illustrative example in his Ph.D. thesis, and our story starts with the n = 3 case of that game. A similar puzzle about voting appeared a few years earlier in a paper by Aspnes, Beigel, Furst, and Rudich (see references at the end of this chapter), where successive generalizations of the problem culminate in important results in theoretical computer science. Objectives/Motivation: The first goal is to enable the students to fully understand the solution to the 3-player hats game, especially in terms of its formulation in terms of graphs. This should allow them to visualize and think about the n-player problem for n = 4, 5, . . ., and to understand the connection with the idea of covering codes. A more ambitious goal is to enable the students to also understand a little about perfect codes (that are covering and also single-error correcting), and why they can exist only for dimensions n = 2k − 1. We recommend taking the time, at some point during the session, to describe the fundamental importance of these error correcting codes and the astonishing range of their applications. They are a truly seminal idea in information theory, which has turned out to be an extraordinarily exciting and fertile area of research. The idea that “information” can be quantified and analyzed mathematically first arose in the late 1940s and 1950s, motivated then by concrete problems such as devising reliable telephone transmission over noisy phone lines. The Wikipedia page captures the extraordinary breadth of the field, mentioning numerous applications, including in:

243

244

JOE BUHLER AND BOB KLEIN

data compression, mobile phones, linguistics, electrical engineering, statistics, quantum computing, plagiarism detection, computer science, physics, neurobiology, cryptography, human vision, etc. Computers as we know them (including the smartphones in many students’ pockets!) could not exist without the use of information theory to efficiently and resiliently store information in memory and transmit it along wires or through the air. Preparation/Materials: It can be fun (but is by no means essential!) to simulate the hats games. This requires a way for students to “wear” a color that is visible to everyone but themselves. One way to do this is to paperclip a white paper strip around a student’s head, and attach a vertical strip of colored paper in the back. We once made the mistake at a Navajo school of buying dollar-store bowls as hats, but this didn’t work well as it bumped into a cultural taboo. The 5-player hats game may draw sustained attention from some of the students who do the first earlier problems easily. It may be useful to have scratch-paper copies of the 5-dimensional hypercube (explained below!), so that students can experiment. It is useful to read through the teacher discussion carefully, since the flow of the material is important. Some of the problem statements are intentionally terse, and this provides opportunities for teachers to clarify and elaborate as appropriate. The problems should be grouped as is convenient; occasionally statements of later problems give hints to earlier problems. The “probability interlude” material given at the end of the first session can be introduced as needed, depending on the background of the audience. We find it is usually best if done early, and reasonably quickly, since the ideas can be picked up from the examples. For some audiences, this interlude could be omitted since the concepts could be discussed and absorbed as they arise. References/Authorship: The use of hats as a device for expressing puzzles goes back (at least) to George Gamow. The 3-player hats game was first given by Todd Ebert in 1998 in his Ph.D. thesis. The Apnes, Beigel, Furst, and Rudich paper “The Expressive Power of Voting Polynomials” appeared in Combinatorica, 135-148, 1994. Both J. Buhler (2002), “Hat Tricks,” Mathematical Intelligencer, 44-49, and E. Brown & J. Tanton, (Apr. 2009) “A Dozen Hat Problems” give many related problems and have numerous references. Hamming codes, covering codes, and error-correcting codes are covered in innumerable books; the Wikipedia pages Covering code and Error correction and detection are reasonable entry points to the literature. More recently Lionel Levine formulated an especially devilish hat puzzle that is largely unsolved (reference: the internet).

FROM HATS TO CODES

245

Mathematics Beneath & Beyond: A graph is a set of points, called “vertices,” together with a set of pairs of vertices, called “edges.” The vertices are often adorned with a label or name in order to keep track of them. Consider a graph whose vertices are (or represent) four students, Natanii, Irvilinda, Charmayne, and Antonio, and whose edges join two students if they were in a math camp together. The graph in Figure 130 suggests that Natanii, Irvilinda, and Charmayne have all been to camp in the same year as each other, but Antonio has only been to camp during some other year where Natanii and Irvilinda were also present. Antonio

Charmayne

Natanii

Irvilinda

Figure 130. The “attended the same math camp” graph A set of vertices in a graph is an independent set if no two elements of the set are adjacent (connected by an edge). For instance, in the graph above, the set {Antonio, Charmayne} is independent since those two vertices are not connected (Antonio and Charmayne never attended the same math camp) but the set {Natanii, Irvilinda} is not an independent set since they are connected by an edge (there is at least one year when they both attended math camp). Finding large independent sets in the hypercube graphs that arise later is equivalent to finding good error-correcting codes. A set of vertices is a covering set if every vertex is either in the set or adjacent to an element of the set. We will see that good strategies in a “hats game,” to be introduced shortly, can be interpreted in terms of small covering sets in the cube graph (and, later, hypercube graphs). In the cube graph in Figure 131, a vertex “covers” 4 vertices: itself and its 3 neighbors. We will call that set a tripod; a set covers the cube if the tripods centered at its elements include all vertices. The set S = {001, 011, 101, 100} is a covering set since the tripods centered in those 4 points cover the graph, but the set {001, 110} is a particularly efficient covering set since it has only two vertices. In fact, this is obviously the smallest possible covering set in the cube, since the cube has 8 vertices and a tripod covers 4 vertices. Note that the two covering tripods do not intersect, so {001, 110} is also an independent set. A graph is a set of vertices and edges, and can be represented in many different ways. The two graphs in Figure 131 have the same connections

246

JOE BUHLER AND BOB KLEIN

011 111 111 001

101 010 110

000

100 000 Figure 131. Two views of the cube graph H3

and (at least when all of the vertices are labeled appropriately on the right) are the same graph. This graph is sometimes called the cube graph, which we will denote by H3 . The two-dimensional analogue, H2 , is a square with vertices that might be labeled {00, 01, 10, 11}; two vertices are connected by an edge if their labels differ in exactly one coordinate. There is an analogous “n-dimensional hypercube” graph for any dimension n, which we denote as Hn . It might seem impossible to visualize Hn for n > 3. However, they are easy to describe: there are 2n vertices of Hn that correspond to (i.e., will be labeled below by) strings of of 0’s and 1’s of length n. Moreover, two vertices are adjacent (i.e., have an edge between them) if and only if their labels differ in exactly one position. Although the graph H3 can be represented in the plane without any spurious edge crossings (i.e., not at vertices), it is impossible to represent Hn as a planar graph for n > 3. An independent set in a graph is a collection of points, no two of which share an edge; error-correcting codes begin with a study of maximal (or at least large) independent sets in Hn . The problem of finding maximal independent sets is somehow the opposite (or dual of) the problems of finding minimal covering sets. For some n there are sets of vertices in Hn that are both covering and independent. Such sets C have the property that each vertex is either in C or adjacent to an element of C, and no vertex is adjacent to two different elements of C. It is easy to see that the square H2 does not have such a set, whereas antipodal vertices of H3 do have this property. Such sets are called perfect codes, and they play a vivid and important role in coding theory. As will gradually emerge, much of this session is summarized by Figure 132.

FROM HATS TO CODES

247

Figure 132. A covering set in H3

From Hats to Codes Session 1: Hats We start with a simple game. A teacher places a red or blue hat on Alice and Bob. Each player can see the other’s hat color but can’t see his or her own hat color. No communication or signaling is allowed between them at any time. After 20 seconds, Alice and Bob each write a guess as to their own hat color on a piece of paper, and hand their pieces of paper to the teacher. If one of them is right and the other is wrong, they both get an ’A’ on the test, and otherwise (i.e, they were both right or both wrong) they both get an ’F.’ To clarify the intent of the rules, one should perhaps imagine that Alice and Bob are complete strangers, sitting in separate rooms where their smartphone shows their partner’s hat color, and after 20 seconds asks them to choose one of two options on their screen: red or blue. (1) What is the probability that Alice and Bob will pass the test? (2) The same test is given on the next day. Unlike the first day where Alice and Bob didn’t know about the test ahead of time, Alice and Bob are now friends and are smart enough to have had a strategy session before class. What is their best strategy, and what is its probability of success? The teacher then announces that there will be a test given the next day, this time for Alice, Bob, and Charmayne. The teacher is kind enough to describe how this test will work, and to allow time for a strategy session between the three friends, before class. The next day, red or blue hats are placed on all three students’ heads. This can be done in 8 different ways, and all of those placements are equally likely. After the strategy session no further communication is allowed between the players. Each player can see the other two players’ hat colors but not his or her own, and 20 seconds after the hats are placed, a bell rings. At that point each player writes exactly one of the following three words on a piece of paper — Red, Blue, or Pass — and hands the piece of paper to

248

JOE BUHLER AND BOB KLEIN

the teacher. A pass means that the student declines to make a prediction of his or her hat color, and otherwise the answer is interpreted as a prediction of the color of that student’s hat. The students win if a correct prediction is made, and there are no incorrect predictions. In other words, they lose if any of them makes a false prediction or if everyone passes. (3) What is the probability of success if they all just guess? (4) What is the probability of success if Alice guesses, and Bob and Charmayne pass? (5) Can you find a better strategy? Since any individual non-pass prediction has probability 1/2 of being correct, it might seem clear that this is the best possible winning probability, and it is attained by the strategy of having one player guess and the other two pass. Amazingly, the students can do better. Somehow, as in the game above, with two players, the players must coordinate their strategies so as to maximize their chance of success. The next few problems develop a systematic way to think about finding an optimal strategy. The eight possible hat placements, which will be called colorings or positions of the game, can be visualized in a natural way as the vertices of a cube. Name the colors 0 and 1 (after all, we’re mathematicians and those are the only names that we can think of. . .). If the students are ordered (say, Alice, Bob, Charmayne) then a placement of hats can be represented by a 3-tuple of 0’s and 1’s where, for instance, 011 denotes the placement in which Alice has a hat of color 0, and Bob and Charmayne have hats of color 1. In the computer world, a location that can store one of two values is called a bit, so that coloring of the 3 hats is a string of 3 bits. There are 8 possible hat colorings and it is (surprisingly) useful to visualize them as the vertices of a cube, as shown in Figure 133.

Figure 133. Hats placements = Cube vertices = Bit strings The key idea is that two positions are adjacent in the cube if one can be obtained from the other by changing the color of a single hat.

FROM HATS TO CODES

249

(6) Label the vertices in the diagram with binary strings so that the labels of adjacent vertices differ in exactly one coordinate. In how many ways can this be done? (7) In terms of this graph, what does each player know about the actual position when they see the other two colors? (8) Is there a strategy that never wins? Is it possible to have a strategy that always wins? (9) How can a “strategy” be described? A strategy tells players what they should do in any configuration that they see. During the game a player sees all hats colors except their own. In terms of the cube graph, this means that a player knows that there are two possibilities for the actual hat coloring, differing only in their hat color. In other words, after the hats have been placed, a player is “on” an edge of the graph. For instance if the coloring is v = 010, then Alice is on the edge between v and 110, Bob is on the edge from v to 000, and Charmayne is on the edge from v to 011. So a strategy is an assignment of either an arrow or Pass to every edge of the cube. If a player is on edge e joining vertices v and w, then the strategy assigns an element of {v, w, Pass} to e. More visually, one could put an arrow (pointing to one or the other endpoint) on the edge, or leave the edge unlabeled (meaning that a player on that edge passes). (10) How many possible strategies are there? These strategies are usually called “deterministic” strategies because the players’ actions are completely determined by the position. Several of the strategies are probabilistic in the sense that the player consults a source of randomness (e.g., the results of flipping a coin) and at least partially bases their action on the output of the source. Under the assumption that the colorings are chosen uniformly randomly, it is not too hard to see that no probabilistic strategy is better than the best deterministic strategy. So it suffices to consider deterministic strategies, which we will (mostly) do from now on. Given a (deterministic) strategy, there is a set W of colorings (vertices on the graph) for which the strategy wins, and a complementary set L of colorings for which it loses. For instance, if the strategy is “Alice says 0, and everyone else passes” then W is the set of 4 vertices whose first coordinate is 0 (which might be visualized as the left side of the cube in Figure 131), and L is the complementary set of 4 vertices with first coordinate 1, and its probability of success is 4/8 = 1/2. Our goal is to find a strategy for which the losing set is as small as possible. (11) If v is a losing position, then explain why we might as well, for the purposes of finding the best strategy, assume that every edge incident on v is labeled with an arrow pointing in the other direction. From now on we will assume strategies have this property, i.e., that students will always “vote against” losing positions rather than just passing.

250

JOE BUHLER AND BOB KLEIN

(12) Explain why (under this assumption) every vertex in the cube is either in or adjacent to an element of the losing set L. (13) Given a set L that has this property — every vertex in the cube is either in L or adjacent to an element of L — describe a strategy for which L is the exact losing set of the strategy. (14) Give a description (in terms of the cube graph!) of an optimal strategy, and explain why your strategy is optimal. Can you find a “memorable description” of the strategy? (If some students finish all of the problems before the end of the first session, they should be asked to think about the hats game with 4 or 5 people and try to find the best strategies for those games.) Probability Interlude. Probabilistic ideas arise in two ways above. First, the teacher chooses colorings in a way so that each coloring has the same chance of occurring. Second, strategies could themselves be probabilistic. Since it suffices to consider deterministic strategies, we only really need to work with the teacher’s “uniformly random” choice of hat placement. Even so, it may be useful to briefly formalize the idea of the probability of an event, in the case in which there is an underlying set of disjoint atomic events that are equally likely. If S is a finite set of equally likely “outcomes” then a subset E is called an event. The “probability” of the event E is |E|/|S|, where |A| denotes the number of elements in a set A. For each of the problems below, describe S and E and then calculate the probability of E. a. What is probability of heads when flipping a fair coin? b. What is probability of two heads when flipping a fair coin twice? c. What is probability of two heads when flipping a fair coin three times? d. What is probability of two heads when flipping a fair coin four times? e. What do you say to someone who says that the five equally likely outcomes when flipping a coin four times are ”0 heads, 1 head, 2 heads, 3 heads, 4 heads”?

FROM HATS TO CODES

251

Session 2: More players Another student, Danny, comes to the class the next day and the teacher announces that the four students will play the hats game! The teacher allows time for the hats game to be explained to Danny and for the four students to discuss strategy. The rules are the same: hats are placed (after which no communication is allowed), the students all choose from among three options, (pass, 0, 1) and they win only if there is at least one correct guess and no incorrect (non-pass) guesses. (15) How many colorings (hat placements) are possible? (16) The set of colorings is a graph if two colorings are deemed to be adjacent (i.e., have an edge connecting them) if and only if the colorings are the same except for exactly one player. Draw this graph, H4 , labeling all vertices. (The graph H4 is not planar so some edges will cross at points that are not vertices.) The graph H4 has 32 edges and is crowded no matter how it is drawn. One way to represent this graph is to draw two adjacent copies of H3 , with extra edges connecting matching vertices on the two copies. The drawing could be simplified somewhat by agreeing to only lightly draw, or even omit, the “crossing” edges that connect matching vertices in the two copies. (17) Label the vertices of the graph in Figure 134 with 4-bit strings so that two vertices are adjacent if and only if the labels differ in one bit. (Bonus question: how many such labelings are possible?)

Figure 134. H4 (18) For k = 1, 2, or 3, what is the optimal success probability if k students always pass? (19) What is the smallest possible size of a losing set C in H4 ? Name such a C. What strategy does it correspond to? Now we turn to the 5-player version of this game, and of course one can also imagine the n-player game for any positive integer n. Let pn denote the

252

JOE BUHLER AND BOB KLEIN

probability of success of an optimal strategy in the n-player game, so that |C| pn = 1 − n 2 where C is the smallest possible covering code. (20) Find an upper bound on p5 . (This is of course equivalent to finding a lower bound on the size of a covering set.) Explain why p5 ≥ p4 , and find a lower bound on p5 . (21) What is p5 ? (22) Find an upper bound on pn . The Hamming distance between two binary strings of the same length is the number of coordinates in which they differ. We let Hn denote the graph of (i.e., with vertices labeled by) all 2n binary strings of length n, where two vertices are joined by an edge if they differ in exactly one coordinate. The Hamming ball of radius r centered at x is the set of points within a distance of at most r of x. Note that the Hamming ball of radius 1 centered at x consists of x and all of its neighbors. In 3-dimensions this looks a bit like a “tripod,” and one might call the Hamming ball of radius 1 in Hn an n-pod, (23) How many vertices are in a Hamming ball of radius 2 in H5 ? What is the smallest number of Hamming balls of radius 2 that cover H5 ? (24) A covering set C is perfect if it is also an independent set. This means that the n-pods centered at the elements of the set cover the vertices, and are also disjoint. Explain why there is no perfect covering in H4 . Perfect covering sets C will be called perfect codes here. Here “code” does not refer to a secret message but just a “dictionary” that associates to each vertex the center of the n-pod on which it resides. In the larger world of error-correcting codes (roughly speaking, codes for which Hamming balls of some fixed radius r ≥ 1 cover Hk ) the perfect codes here are one of a small handful of examples. Note that a perfect code C satisfies all of the following: • Any v in Hn is either in C or is adjacent to an element of C. • Any v not in C is adjacent to a unique element of C. • The vertices of Hn are a disjoint union of the n-pods centered at elements of C. (25) Prove that if there is a perfect code in Hn then n has the form n = 2k − 1. (In particular, the next largest n after 3 for which a perfect code could possibly exist is n = 7.) The final problem asks you to prove that a set B, that will be described shortly, is a perfect code in H7 . It is convenient to fix some notation. As usual, let H7 denote the graph whose vertices are labeled by the 128 different 7-bit strings. Let H3∗ denote the set of 7 nonzero (not all-zero) bit strings of length 3. If v = v1 v2 v3 v4 v5 v6 v7 is a vertex in H7 then the subscripts on the individual bits will be identified with elements of H3∗ . In other words, we will

FROM HATS TO CODES

253

freely make the translation {1, 2, 3, 4, 5, 6, 7}  {001, 010, 011, 100, 101, 110, 111} in which an integer between 1 and 7 is replaced by a nonzero 3-bit string obtained by writing down the 3-bit expansion of the integer. (You may be feeling sorry about the omitted bit string 000, but it is about to play a starring role.) If v = v1 v2 v3 v4 v5 v6 v7 is in H7 then f (v) will denote the 3-bit string obtained by the following curious recipe: for each bit vi that is equal to 1, write the 3-bit alias for i in a row, and assemble the rows into a 3 by m array, where m is the number of nonzero bits in v. For instance, the element v = 1100110 has nonzero bits in positions 1, 2, 5, 6, so the array has 4 rows and 3 columns: 0 0 1 0 1 0 1 0 1 1 1 0 Then f (v) is obtained from this array by “summing the columns modulo 2,” which means that f (v) is the 3-bit string in which a bit is 0 (resp. 1) if the number of nonzero entries in the corresponding column is even (resp, odd). Thus f (v) = 000 for v = 1100110, since each column has an even number of 1’s. A vertex v will be said to be balanced if f (v) = 000, and B will denote the set of all balanced vertices in H7 . It should be remarked that if v = 0000000 then the array in the definition has 0 rows, and all of its empty columns have an even number of elemenets, so f (v) = 000 thus the 7-bit all-zero string is balanced. The delicate dance between bit strings of lengths 3 and 7, together with the various identifications used in the definition of f , may seem to be a bit mysterious but it is at the heart of the proof of the existence of perfect codes in H7 . (26) Show that the set of balanced strings in H7 is a perfect code. This question is hard in that it requires a full understanding of ideas that are only briefly sketched. It shows the existence of a (single errorcorrecting) perfect Hamming code in dimension 7, and the argument actually is the same in all dimensions of the form n = 2k − 1, so the existence of perfect codes in all dimensions of this form follows. If v is a vertex in H7 and i is an integer between 1 and 7, let v.i denote the vertex obtained by reversing bit i, i.e., by converting that bit to a 1 if it is 0, and to 0 if it is 1. By checking two cases it is easy to show that f (v.i) = f (v) + i, where the addition on the right is the “mod 2 addition of 3-bit strings.” I.e., f (v.i) can be calculated by appending the (3-bit alias of) i to the array in the definition of f (v). If i is nonzero then v.i is, in effect, obtained by reversing the ith bit of v. This means that v.i is adjacent to v in the graph H7 .

254

JOE BUHLER AND BOB KLEIN

To show that B is a perfect code, we have to show that any v in H7 is either in B, or else is adjacent to a unique element of B. Suppose that v is not in B, i.e., that f (v) is a nonzero bit string. Let i = f (v) and w = v.i. Then w is adjacent to (and not equal to) v and a magical fact follows: f (w) = f (v.i) = f (v) + i = f (v) + f (v) = 000, which means that w is balanced, and v is adjacent to the element w of B. It remains to show that no element of H7 is adjacent to two elements of B (i.e., B is a perfect cover). Assume that v is not in B, but that it is adjacent to two elements of B, i.e., v = u.i = w.j where u and w are balanced, and not equal. Then i = f (u) + i = f (u.i) = f (v) = f (w.j) = f (w) + j = j which implies that u = v.i = v.j = w, which contradicts the fact that they are distinct.

FROM HATS TO CODES

255

From Hats to Codes Teacher Guide (1) This is the first occurrence of “probability,” and stating the puzzle (or listening to their solutions) gives an opportunity to gauge student’s background with this idea. In fact, the problem makes no sense until the teacher’s method of choosing hat colors is specified. The natural choice is that the teacher flips a fair coin for each color, so that all colorings are equally likely. Nonetheless, other things are possible, and it is of course interesting to see how students respond given no precise specification. On an intuitive level, both Alice and Bob lack any knowledge about what the other sees, or what the other person is going to say, so their actions are complete guesses. This is equivalent to saying that they both flip a fair coin. If the hat colors are 0 and 1, then the space of equally likely outcomes is {00, 01, 10, 11} and the winning set is {01, 10} so the probability of success is one-half. (2) If they have a strategy then their actions can exploit the fact that they know what their partner will do! There are four possible colorings, and the students might be encouraged to make a table as below, where their job is to fill in Red or Blue for each of the four needed entries x, y, z, w below, and then calculate their probability of winning, given that they win exactly when one of them is correct. In the table, for instance, y is the color that Alice will say if she sees a Blue hat on Bob. Red Blue Alice x y Bob z w (One bonus question, once this table is up and perhaps before they’ve found the winning strategy, is “how many strategies are there?”) The strategy Red Blue Alice Red Blue Bob Blue Red can be verified to always succeed! I.e., it succeeds with probability 1. One memorable non-tabular way to remember this strategy is: Alice predicts that the two of them will have the same hat color, and Bob predicts that they will have opposite colors. Exactly one of them will be right. Of course once that “memorable” solution is enunciated, the result is obvious. If someone in the group gets that quickly that’s great, and you should move on to the next question. (3) There are 8 equally likely positions, and if no player passes then they win if and only if all three of their guesses are correct, so the success probability of the all-guess strategy is 1/8. (4) The players win if and only if Alice guesses correctly, so the success probability is 1/2.

256

JOE BUHLER AND BOB KLEIN

(5) Since the players lose if no one guesses, and any individual guess has probability exactly 1/2 of being correct, it might seem obvious that the highest possible winning probability is 1/2. The students can do better, by devising a strategy that coordinates their answers in a sufficiently clever way. It turns out that there is a strategy that succeeds with probability 3/4, and this is the best possible probability. It is conceivable that some student will have heard of the problem and blurt out an answer, or that a group will solve the problem. This is fine, and in any case the problems give a slow and steady development of a solution in visual terms. You might want to start the discussion of the next few problems by drawing a “stick graph” of a cube and asking the students to help you label it with 0/1 triples. (6) The vertex with label 000 can be chosen arbitrarily (from among the 8 choices), and its 3 neighbors have to have the labels 001, 010, 100. Those labels can be assigned in 6 ways — 3 choices for which vertex adjacent to 000 will be labeled with 001, times 2 possible choices for the remaining two labelings. It may be surprising, but a little experimenting with a graph will show the the other labelings are forced. For instance, the unique vertex adjacent to 001 and 100 has to be at a distance one from both (i.e., have a 1-bit change from each), and the only label with that property is 101. The upshot is that the total number of legal labelings of the graph is 8 × 6 = 48. (7) Once a player sees the two other hat colors, there are only two possible positions. For instance if Alice sees 0 on Bob and 1 on Charmayne then the only possible positions are 001 and 101, depending on Alice’s hat color. These positions differ only in Alice’s hat color and are therefore adjacent, as is easily visualized in, e.g., Figure 131. In some sense, Alice is “on” an edge in that she is undecided about which of the two vertices of the edge is the actual position. The three player’s views correspond to the respective edges that emanate, in 3 directions, from the actual position. Figure 131 might be useful in helping to visualize these ideas. (8) The strategy of everyone passing never wins. Any other strategy directs (at least) one player to make a non-pass statement; that statement is wrong half the time, so no strategy will always win. (This makes it sound like the optimal probability can’t be larger than 1/2, but this turns out to be wrong; it may be worth dwelling on this when discussing the answers.) (9) As discussed after the problem, a strategy consists of assigning to each edge of the graph either the word “pass” or one of the endpoints of the edge. Another answer is that a strategy is just a table in which each of the three players has a statement “my hat is 0,” “my hat is 1,” or “I pass” associated to each of the four edges of the graph that that player could be on. More visually, a strategy consists of placing one

FROM HATS TO CODES

257

of three things on each edge of the cube graph: an arrow pointing to one end (or vertex) of the edge, an arrow pointing to the other end of the edge, or “pass.” (It can be convenient to make the convention that an unlabeled edge denotes the “pass” option.) One might say that a strategy is equivalent to putting arrows on some subset of the edges of the cube graph. (10) There are 12 edges on a cube, and 3 choices for each edge, so the total number of possible strategies is 312 = 531,441.

(11)

(12) (13)

(14)

Remark: It is unlikely to be useful to cover the details of this in the session, but the justification of restricting to deterministic strategies is, as follows. Any (notion of a) probabilistic strategy gives a probabilistic strategy of the most general kind: an assignment of probabilities ps to each deterministic strategy s, where s ranges over all N = 531,441 deterministic strategies. If P (s) is the probability that s succeeds (when the hats are placed randomly) then the probability that the probabilis tic strategy succeeds is ps P (s), which is clearly no larger than the maximum of the P (s). If v is a losing position, then either all three edges meeting at v are unlabeled, or some edge is oriented “away” from v. The success probability is unchanged or improved if all edges are oriented away from v. If v is a winning position, then some arrow points toward it, and the tail of that arrow is a position adjacent to v, so v is a losing position. To define a strategy with losing set L we have to specify, for every edge of the cube, whether the player “on” that edge should pass or point to one endpoint of the edge or the other. Suppose L is a set of vertices with the stated property (i.e. every vertex of the cube is either in L or adjacent to an element of L. Define a strategy as follows: if an edge does not have an element of L as an endpoint, then pass; if an edge has one endpoint in L, then point to the other endpoint. Finally, if both endpoints are in L, point to either, as you wish. (Technically, there are problems if both endpoints have all of their neighbors in L; this would be a silly choice for L, since then one of the endpoints could be removed from L without changing the covering property, so we are implicitly assuming that L is not silly in this sense.) Now you can check that L is the losing set for this strategy. Any two antipodal elements of the cube graph give an optimal strategy: every element of the cube graph is in C or adjacent to an element of C. Clearly no one-element set has this property. One memorable way to describe this strategy is to choose the antipodal points to explicitly be the two monochromatic colorings (all red or all blue), so that the optimal strategy in that case can be stated as saying that the players should “avoid monochromaticity.” This means that if the two hats that they see are the same color then they bet against monochromaticity, and otherwise pass.

258

JOE BUHLER AND BOB KLEIN

Note the striking progression of ideas about what a strategy “is”: first, a table (or set of instructions), then a visualizable labeling of arrows on a graph, then merely a set of losing positions — vertices on the graph — that covers the graph in the sense that any winning position is adjacent to an element of C and finally a memorable rule that is easy to act on. The covering code idea suggests another view of what we should have been trying to do. All of the non-pass options will be right half the time. So the goal is to make all of the wrong answers happen on the same positions, i.e., “overlap,” and have the correct answers be given only once for any given position. In the 3-cube case there are 2 colorings for which all 3 students make incorrect statements, and 6 positions where 1 student makes a correct statement and the other 2 pass. Thus the team wins with probability 3/4 despite the fact that exactly half of their non-pass statements are true. The Probability Interlude problems can be solved by explicitly describing the set of equally likely outcomes, and the subset corresponding to the given event. a. What is probability of heads when flipping a fair coin? S = {H, T }, E = {H}, p = 1/2. b. What is probability of two heads when flipping a fair coin twice? S = {(H, H), (H, T ), (T, H), (T, T )}, E = {(H, H)}, p = 1/4. c. What is probability of two heads when flipping a fair coin three times? S = all ordered triples of H/T , E = {HHT, HT H, T HH}, p = 3/8. d. What is probability of two heads when flipping a fair coin four times? S = all ordered quadruples of H/T , E = {HHT T, HT HT, HT T H, T HHT, T HT H, T T HH}, p = 6/16 = 3/8. e. What do you say to someone who says that the five equally likely outcomes when flipping a coin four times are ”0 heads, 1 head, 2 heads, 3 heads, 4 heads”? Many answers are possible, e.g., starting with “There are 16 ordered strings of length four of heads and tails, and they are all equally likely . . .” or starting with “OK! Let’s test this by playing for money . . .” (15) There are 16 = 24 ways to choose between a hat color 0 or a hat color 1 for each of four players. (16) There are many ways to draw the graph in the plane (if “spurious” edge intersections are allowed); the key idea is that edges connect bit strings that differ in exactly one position. The students should be encouraged to see that many different representations in fact describe the “same” graph. (17) There are 4! ways to assign the position of the attributes left/right for the outer squares, inner/outer, up/down, and left/right for the smaller

FROM HATS TO CODES

(18)

(19)

(20)

(21)

259

squares, and 24 ways to assign 0 and 1 to each choice, and therefore 4! 24 = 384 ways to label the graph. If k players pass in the 4-player game, then the remaining players should play the optimal game for 4 − k players. So for j = 1, 2, 3 the optimal probabilities are 3/4, 1/2, 1/2. The problem of finding minimal covering sets in H4 has an unexciting answer. Think of H4 as two copies of H3 , and choose antipodal points in each of those subcubes. This gives a 4-element covering code. The corresponding strategy is to have one player pass and the other three play the 3-player optimal strategy. Is this minimal? Each point covers 5 vertices, so a 3-element set can cover only 15 vertices, and so no 3-element covering set exists. A 5-pod in H5 covers 6 points (itself, and its 5 neighbors). A code C that covers all 32 points in H5 must therefore have at least 6 points and 6 13 |Cmin | ≤1− = p5 = 1 − 32 32 16 where Cmin denotes a covering code of smallest possible size. To find a lower bound on p5 we need a covering set. We can think of H5 as the union of two copies of H4 and take a covering code in each copy. There are 4-point covering sets in H4 and therefore there is an obvious 8-point covering set in H5 , and an obvious corresponding strategy (have two players pass and the other three follow the optimal 3-player strategy) has success probability 3 8 = ≤ p5 . p=1− 32 4 It would be impressive if a group solved the optimal covering code in H5 problem completely in class (i.e., showed that 7-element covering codes exist, but 6-element covering codes do not). But it is a good question to capture the attention of students who have gotten through the earlier material quickly. We have sometimes found it handy to draw H5 as shown in Figure 135 on a piece of paper, and then xerox copies of it for students to experiment on. In H5 a 5-pod covers 6 points, so it takes at least 6 points to cover the 32 elements in H5 . There is a fairly obvious way to cover with 8 points: Write H5 as a union of four 3-cubes, and pick a 2-element covering set in each 3-cube. For instance in Figure 135, the NW, NE, SW, and SE subcubes could be, respectively, the 5-bit strings starting with 00, 01, 10, and 11 respectively. In that case, there are implicit edges going vertically and horizontally from points to their siblings in other subcubes. Each vertex has 5 outgoing edges, and covers 6 vertices. Suppose that there was a 6-element covering set C in H5 . How many points could be in each of the subcubes? We have to cover 8

260

JOE BUHLER AND BOB KLEIN

Figure 135. One view of H5 points in each subcube. Each point in C covers 4 points in its subcube and 1 point in each of its (horizontally and vertically) adjacent subcubes. Could there be a covering with 2 points in each of the NW and NE cubes, and 1 point in the SW and SE cubes? Well, no! Consider the SW cube. The unique point in the SW covers 4 vertices in the SW, and the points in the NW and SE cover 2 + 1 = 3 further points in the SW. This isn’t sufficient. The only possible subcube population numbers that can’t be immediately ruled out by this kind of reasoning is the following (or something equivalent to it): 2 points in each of the NE and SW cubes, and 1 point in each of the NW and SE cubes. Up to symmetry, an arbitrary point can be chosen in SE, and we take it to be A in the lower right hand corner, which covers all of the a points (see Figure 136). (Upper case letters are the elements of the covering set, and corresponding lower case letters are the vertices covered by those elements.) The uncovered points in the SE are a tripod, and have to be covered by the 2 points in the NE and 2 points in the SW. Without loss of generality, the central point of that tripod is covered by B in the SW. The three adjacent points are all equivalent, so we arbitrarily choose C next to B, and are forced to choose then the points D and E in the NE. The rest of the cover is forced: The two remaining uncovered points in the SE must be covered by D and E in the NE, and F in the SW. You can check that the 6 points A through F don’t quite form a covering set because they miss one point in NW and one point in NE. So no 6-point covering set exists in H5 . However, a quick check shows that if a seventh point is added that is antipodal to F in NW (at a point covered by C) the result is a 7-point covering set! So the optimal winning probability in the 5-player game is 1 − 7/32 = 25/32. Bonus questions: Depending on your audience, you could take a diversion into the geometry of hypercubes: how many vertices does

FROM HATS TO CODES

261

D b

c E a

F c

b B

C

b

c

a b

c a

a

a

A

Figure 136. H5 has no 6-point covering set!

(22)

(23)

(24)

(25)

H5 have? How many edges? How many “two-dimensional faces” (i.e., sets of 4 vertices in which 3 of the 5 coordinates are the same for all 4 elements of the set). Similarly for H6 , . . . In order for a code C in Hn to cover, the number of points in |C| n-pods must be at least 2n , so |C| 1 ≥ (n + 1)|C| ≥ 2n , i.e., 2n n+1 which implies that n 1 = . pn ≤ 1 − n+1 n+1 The points at a distance at most 2 from 00000 are the vertex itself, the 5 vertices in which 1 coordinate is nonzero, and the 10 vertices in which a pair of coordinates have been changed. So a Hamming ball of radius 2 in H5 has 16 points. Any point in H5 has a majority of 0 or a majority of 1 coordinates, and is therefore within a distance at most two of 00000 or 11111. Thus two Hamming balls of radius 2 are sufficient to (exactly) cover H5 . A 4-pod centered at v covers 5 points (itself, and its 4 neighbors). It is impossible for a set of 4-pods to cover the 16 points of H4 without intersecting. If a perfect code C exists in Hn the n-pods centered at the elements of C exactly cover Hn , so (n + 1)|C| = 2n .

This says that n+1 is a power of 2, i.e., there is a k such that n = 2k −1. The next such n larger than 3 is n = 7 = 23 − 1. (26) This question is quite hard in that it requires a full understanding of some ideas that are only briefly sketched. It shows the existence of (single error-correcting) perfect Hamming code in dimension 7, and

262

JOE BUHLER AND BOB KLEIN

the argument trivially extends to any dimension n = 2k − 1 in which Hamming codes exist. If v is a subset of S and s an element of S then let v.s denote the subset of S obtained by adding s to v, if s is not already in v, or taking s out of v if s is in v. Then it is easy to check that f (v.s) = f (v) + s, where the addition on the right is the mod-2 addition of two 3-bit strings. Moreover, the neighbors of v are exactly of this form, for a suitable choice of s, as described below. To prove that the set B of balanced 7-bit strings is a perfect code we have to show two things: any 7-bit string is in or adjacent to an element of B, and no element v not in B is adjacent to two elements of B. Suppose that v is an element of H7 , i.e., a subset of S. Let f (v) = s. If s is the all-zero bit string then v is in B. If s is a nonzero 3-bit string, i.e., s is in S, then f (v.s) = f (v) + s = s + s = 000, and v.s is in B. Moreover, v is adjacent to (i.e., has one bit changed in the 7-bit representation of) v.s. On the other hand, suppose that f (v) = s, where s is in S so that s is a nonzero bit string and v is not in B. Moreover, assume that v is adjacent to two elements of B, so that v = b.s = c.t where b and c are balanced and s and t are distinct elements of S. Then s = f (b) + s = f (b.s) = f (v) = f (c.t) = f (c) + t = t which is impossible since s and t are distinct. One bonus question would be to ask how many elements there are in B. Another would be to assume the existence of Hamming codes for all n = 2k − 1 and prove to prove that n n−1 < pn ≤ . n+1 n+1

From a Magic Card Trick to Hall’s Theorem Contributed by M. Kawski and H. A. Kierstead Short Description: A magic card trick is used to explore encoding sets by ordered sequences. Extended hands-on activities involving variations of this trick motivate the introduction of (combinatorial) graphs and Hall’s Theorem11 , and methods that students have developed for showing that these tricks are possible are used to prove Hall’s theorem. This activity is suitable for students who are comfortable with permun! n! and k!(n−k)! for tations and combinations, especially the formulas (n−k)! the numbers of k-permutations and k-combinations of n objects, and who are interested in problem solving. It is intended for grades 11 and 12, but Problem 1, and parts of Problems 2 and 3 could be used for grade 7. Materials: Each group of (say) three students needs: a standard deck of 52 playing cards (4 suits, 13 ranks per suit) with two additional jokers; 112 index cards and 56 paper clips and extras; colored markers (two colors, Sharpies or similar); a large table on which to organize cards and a board with colored markers for collaboration. Preparation: The session leader and an assistant need to practice the magic card trick. If time is limited, the sessions can be sped up if the index cards are prepared beforehand (or cards from a previous session are used). Objectives: Understand a magic card trick, and develop strategies for performing variations of it; understand these tricks in the context of encoding sets by ordered sequences; discover natural limitations in terms of numbers of permutations and combinations; try out and develop different strategies; model the tricks in terms of graph theory; learn about Hall’s theorem and its application to these tricks; generalize strategies developed for card tricks to proving Hall’s theorem. Students move from small examples to generalizations and abstractions, with emphasis on asking new questions at every stage; they participate in 11

This is a fundamental combinatorial theorem with wide mathematical applications. 263

264

M. KAWSKI AND H. A. KIERSTEAD

making definitions that allow them to more precisely communicate their findings, conjectures, and arguments. References/Authorship: The card trick has a long history dating back at least to the 1950s [10] and it is commonly known as the Fitch Cheney Trick. Several recent articles in MAA magazines such as [14] revisit this trick and discuss related items. Searches of the WWW bring up an abundance of further blog-spots and related discussions. In the math circles community a description may be found here [1]. Hall’s (Marriage) Theorem was proven by Philip Hall in 1935 [6]. Four years earlier, K¨onig [9] and Egerv´ary [4] independently proved an equivalent theorem. The authors first developed and class-tested the activity presented here in 2011 in a math circle [13] with high school age students.

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

265

From a Magic Card Trick to Hall’s Theorem Student Handout Problem 1. With the master magician (the session leader) outside, and the assistant magician in the classroom, the students select a hand of five cards from a standard deck of 52 playing cards, and give it to the assistant magician. The assistant carefully studies the cards, and then places them face-down in a stack on a table and leaves the room. The master magician then enters the room without communicating with the assistant. The master magician flips over the first four cards in the stack, and announces what the fifth card is.

Figure 137. Seeing the first four cards, what is the fifth? This trick is based on a secret code arranged ahead of time by the master and assistant magicians. Invent your own code for performing this trick. Practice the trick repeatedly so that you can perform it quickly and accurately. As an extra challenge add a joker to the deck and create a strategy to perform the analogous trick with five cards from the expanded deck of 53 cards. Suppose you add both a red and a black   joker? Problem 2. We call the first trick in Problem 1 the 52 5 because the deck has 52 cards and the hand has 5 cards. The trick with one  54 n joker added 53 is the 5 ; adding two jokers makes the 5 . The k uses an n  card deck and k-card hands. Use your strategy for the 52 5 to derive 15 3 6 strategies for the 2 , 3 and 4 . (Hint: Design your own deck. How many suits and ranks should you use  in each case?) Find a number n, as large as you can, so that the n6 is possible. Find numbers m, n, as

266

M. KAWSKI AND H. A. KIERSTEAD

    small as you can, so that the m and the n5 are impossible. Which 2    of the 73 , 83 , and 93 are possible? Why? Problem 3. For this problem, we abandon the suits and ranks of the deck of cards, and instead just consider a deck D whose cards arenumbered  (more by natural numbers. We can still, for example, do the 52 5 slowly) by identifying integers with card values. Here is one way: 1 → A♣, 2 → A♦, 3 → A♥, 4 → A♠, 5 → 2♣, . . . , 51 → K♥, 52 → K♠.  We will concentrate now on the 83 . Step 1: Break into teams of about three, each team at a separate table. Step 2: Label one set of index cards with all subsets of {1, 2, . . . , 8} that have three elements (use a green pen, write near the top of the vertical card). Here, for example, the combinations {2, 4, 7} and {4, 7, 2} are considered equal. How many are there? Step 3: Label a second set of index cards with all 2-card permutations of {1, 2, . . . , 8} (use a red pen, write near the top of the vertical card). Here, for example, the pairs (2, 3) and (3, 2) are considered different. How many are there? Step 4: Sort and lay out the index cards in a systematic pattern on the table. Use this pattern to check for duplicates. Check that your set of cards is complete by counting, and comparing results n! n! and n Ck = k!(n−k)! . with the formulas n Pk = (n−k)!

Figure 138. Checking for duplications and missing cards. Step 5: Find a pair of index cards such that one is a green combination {i, j, } and the other is a red permutation (r, s) with {r, s} ⊆

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

267

{i, j, }. For example {1, 3, 7} could form a pair with (3, 7) but not with (2, 7). Use a paper clip to attach these index cards so that both the combination and permutation are visible. We call this a clipped pair. Form as many non-overlapping clipped pairs as you can. Step 6: As the available permutations become scarce, discover methods to make changes to the pairs that result in more clipped pairs. The goal is to use every index card in a clipped pair (why?). Step 7: Reflect on whether it is always possible, as long as there are unclipped index cards, to increase the number of clipped pairs. Focus on strategies for breaking clipped pairs and clipping new pairs that result in more clipped pairs. Before removing any paperclips, plan which pairs to change in order to increase the number of clipped pairs. Step 8: When you succeed in clipping every index card to another, use the result to perform the trick (not necessarily from memory). Note: Before stating the next problem let us agree on some technical language which facilitates more precise descriptions of situations and arguments. We write n Ck for the set of k-combinations (hands) of the n-card deck D; then the cardinality of n Ck is |n Ck | = n C k . Similarly, we write n Pk for the set of k-permutations of D; then the cardinality of n Pk is |n Pk | = n P k .   In order to more effectively analyze the nk it is natural to visualize the possible   pairings in a graph as shown in Figure 139 for the special case of the 43 . {1, 2, 3}

{1, 2, 4}

{1, 3, 4}

{2, 3, 4}

u u u P a !H  u a ! QH  P    aa    Ha AHa ! Q P       !

A@ P

A H H aP a !    a Q A @ P H A H a A    ! P a H
!H H
a Q  aP A P! Ha A a! a P Q   
A@
H H a P !a   A @ a Q A H a P H    ! P H aa
 
Q H A !  a a PP @ AAu P @ a Qu HH u  u AA! u! u u AAu a u u HHu a 
 a
u Q P

(1, 2) (2, 1) (1, 3) (3, 1) (1, 4) (4, 1) (2, 3) (3, 2) (2, 4) (4, 2) (3, 4) (4, 3) Figure 139. Possible pairings of 4 C3 with 4 P2 displayed as the graph G4,3 Each possible hand is a combination in 4 C3 represented by a ‘top’ vertex drawn as a dot. Each permutation in 4 P2 is represented by a ‘bottom’ vertex drawn as a dot. There is an edge linking a hand to a permutation, drawn as a line, if it is possible to code the hand with the permutation. This means that all the cards in the permutation are also in the hand. For example, since (4, 2) can be formed from some of the elements of {1, 2, 4}, the combination {1, 2, 4} is linked to the permutation (4, 2) by an edge in the graph. We write {1, 2, 4}(4, 2) or (4, 2){1, 2, 4} for this edge.

268

M. KAWSKI AND H. A. KIERSTEAD

More generally, the set of edges E consists of the linked pairs of vertices of the form {c1 , c2 , c3 }(p1 , p2 ), where {p1 , p2 } ⊆ {c1 , c2 , c3 } ⊆ {1, 2, 3, 4}. We formalize such pictorial notions in modern mathematical language. Definition 1. A graph G = (V, E) consists of a set V of vertices and a set E of edges. Each edge e ∈ E consists of two linked vertices x and y, and is denoted by e = xy. The order of the vertices does not matter, so the edge e can also be written as e = yx. The vertices x and y are called the ends of e. A graph is finite if it has finitely many vertices. The graphs considered in our problems have a distinguishing structure that deserves its own name. Definition 2. A graph G = (V, E) is said to be bipartite if the vertex set V can be divided into a ‘top’ part C and a ‘bottom’ part P so that no two vertices of C are linked by an edge in E, and no two vertices of P are linked by an edge in E. In this case we say that G is a (C, P )-bipartite graph. The graph in Figure 139 is a (4 C3 , 4 P2 )-bipartite graph. We call it G4,3 . More generally, let Gn,k be the (n Ck , n P k−1 )-bipartite graph, whose edges are pairs HP where H is a hand of k cards and P is a (k − 1)-permutation of the cards of H. Definition 3. A matching in a graph G = (V, E) is a subset M ⊆ E of edges such that no vertex is in two different edges in M . A vertex is said to be matched if it is the end of an edge in M ; otherwise it is unmatched. A set of vertices X ⊆ V is said to be matched if every x ∈ X is matched. A maximum matching is a matching that has at least as many edges as any other matching.   Problem 4. Explain clearly why the nk has a solution if and only if the  graph Gn,k has a matching that matches n Ck . Illustrate this for the 43 using Figure 139 and explain your work in Problem 3 in these terms. Note: While struggling with Problem 3, you likely wondered whether there really was a matching that included all index cards. 4For example, Figure 140 shows the graph G4,2 that is associated with the 2 . It is apparent (why?) that this trick is impossible. Hall’s Theorem is a general result for bipartite graphs stating that it is always possible to find such a matching provided one necessary condition is satisfied. To precisely state this condition, we need some more notation. For a vertex v ∈ V in a graph G = (V, E) we write N (v) for the set of vertices, called neighbors of v, that are linked to v by an edge in E. For a set of vertices X ⊆ V we write N (X) for the set of vertices that are neighbors of some vertex in X. For example, in Figure 140, if v = {1, 2} then N (v) = {(1), (2)} and if X = {{1, 2}, {1, 3}} then N (X) = {(1), (2), (3)}.

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

{1, 2}

{1, 3}

{1, 4}

{2, 3}

{2, 4}

(1)

(2)

(3)

(4)

269

{3, 4}

u u u u HH HH u u HH   @ HH HH   @ HH HH HH   @ HH   HH  HH  HH @ H H  @ HH HH  HH   @u  Hu u u

Figure 140. Possible pairings of 4 C2 with 4 P1 as the bipartite graph G4,2 Hall’s Theorem. [6]. A finite12 (C, P )-bipartite graph has a matching such that C is matched if and only if |X| ≤ |N (X)| for all X ⊆ C.

(2)

Problem 5. Prove Hall’s theorem using methods developed in Problem 3. Here are some hints. Let G be a finite (C, P )-bipartite graph. First suppose G has a matching that matches C. To check that Eq.(2) holds, consider any subset X ⊆ C. Why must the vertices of X have at least |X| distinct neighbors? You may want to think about the special case of Problem 3 where X is a set of ‘red’ index cards that have been clipped to ‘green’ index cards. Now suppose G has no matching such that C is matched. The goal is to prove Eq. (2) fails for at least one subset X ⊆ C. Consider a maximum matching M in G; it exists because G is finite. Try to improve M using methods generalized from Problem 3. As M is maximum, this must fail. Construct such a set X by analyzing the reason for this failure. For inspiration, consider your work on Problem 3. In particular, recall a situation like the one sketched in Figure 141. The blue vertices on the top correspond to hands, the red ones on the bottom to permutations. The orange edges correspond to clipped pairs. The gray edges stand for other possible, but unused pairings. In this case there is an unclipped hand u = v0 , all of whose possible permutations may be in use. But by putting together a sequence of cards, we can find an alternating path of hands and permutations that connects u with an unclipped permutation vn . The key step was to break the pairs represented by the orange edges, and clip together new pairs corresponding to the violet edges. This process increased the number of clipped pairs. 12

It is possible to construct examples of infinite graphs for which the theorem fails. See the third comment in the Mathematics Beneath & Beyond section at the end of this article.

270

M. KAWSKI AND H. A. KIERSTEAD

uv= v0

vv 2

vv 4



A A

A

A A

A A A

A

A A

A

A

A

A

A

v A
v A
v



v1 uv= v0

v3 vv 2

...

v5

v3

v5




A


A A


A


A A


Av A

v


vn−2

vv 4


AA
AA AAA
AA
AA AAA

AA AA AAA

AA AA AA AA
AA
AA
AA
v AA
v AA
v

v1

vn−1 v

vn

vn−1 v

...


AA
AA AA
AA
AA AAA
AAv AAA
v

vn−2

vn

Figure 141. An augmenting path. Before extending this key trick to a proof of Hall’s Theorem, we formalize the situation illustrated in Figure 141 with more definitions of formal terminology. Definition 4. Let G be a graph. A path Q = v1 v2 . . . vn in G is a sequence of distinct vertices of G such that each pair vi vi+1 is an edge of G. The set of vertices of Q is V (Q) = {v1 , . . . , vn } and the set of edges of Q is E(Q) = {v1 v2 , v2 v3 , . . . , vn−1 vn }. For a matching M , the path Q is an M -alternating path if each vertex v ∈ V (Q) is an end of at most one edge in E(Q)  M ; in this case Q is M -augmenting if its ends v1 and vn are unmatched. Using this language, the sets of orange edges M = {v1 v2 , v3 v4 , . . . vn−2 vn−1 } and purple edges L = {v0 v1 , v2 v3 , . . . vn−1 vn } in Figure 141 form matchings, and Q = v0 v1 . . . vn is both an M -alternating path and an L-alternating path; also Q is an M -augmenting path, but not an L-augmenting path. Replacing M with L results in a larger matching. We conclude: Lemma 1. If a graph with a matching M has an M -augmenting path, then M is not a maximum matching. Now you can use this lemma to prove Hall’s Theorem. Problem n 6. Use your previous work and Hall’s Theorem to show that the k is possible if and only if n ≤ k! + k − 1.

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

271

From a Magic Card Trick to Hall’s Theorem Teacher Guide Solutions:   Problem 1. We start with a quick explanation of why the 52 5 is possible. The assistant and magician will use permutations of three cards, here called low, mid, and high to code the integers 1, 2, 3, 4, 5, 6. There are 6! such codes. The magicians could agree to use the following code, since it is easy to compute. (high, mid, low) → 1 (mid, high, low) → 2 (mid, low, high) → 3 (high, low, mid) → 4 (low, high, mid) → 5 (low, mid, high) →

6 The rule is: add the position (1, 2 or 3) of the high card to 0 if the mid card comes before the low card, and to 3 if mid card comes after the low card. The magicians agree to order the deck by ranks ordered by A, 2, . . . , 10, J, Q, K and break ties between ranks by suits ordered as ♣, ♦, ♥, ♠ (alphabetical order). So A♠ precedes K♦ and 5♦ precedes 5♥. Performing the 52 5 . The hand the assistant receives from the students has 5 cards. As there are only 4 suits in the deck, it must contain two cards c1 and c2 from the same suit. The assistant chooses c1 and c2 so that the number of steps t from the rank of c1 to the rank of c2 in the clockwise order A, 2, . . . , 10, J, Q, K, A is as small as possible. Then c1 will be put in the first position and c2 in the last position (c2 is the card the magician must determine). As they have the same suit, these cards have different ranks, so there are 11 remaining ranks. Thus 1 ≤ t ≤ 6. Order the remaining three cards to code t. After turning over the first four cards, the magician (i) knows that the suit of the first card is the same as the suit of the last card, (ii) can determine t from the pattern of the next three cards, and (iii) can calculate the rank of the last card by counting t clockwise steps from the rank of the first card. For example, in Figure 137, the fifth card is a heart, because the first card is. Since the next three cards have the pattern (high, low, mid), we have t = 4. Four steps from 10 in clockwise order yields  so the fifth card is the ace of hearts. 53A, Performing the 5 . Now the deck has a joker. There are three cases. Case 1 : The hand does not contain   the joker. Then the assistant uses the method designed for the 52 5 . Case 2: The hand contains no two cards of the same suit (so it contains a joker). The assistant chooses a pair of cards c1 and c2 so that it is possible to move clockwise from the rank of c1 to the rank of c2 in 0 ≤ t ≤ 3 steps and places c2 last. Then the assistant places the joker in position t + 1 and c1 in the first position not occupied by

272

M. KAWSKI AND H. A. KIERSTEAD

the joker. So if t = 0 then the joker comes first and c1 comes second; else c1 comes first. The order of the remaining two cards c3 and c4 does not matter (for now) but the smaller is put first. Case 3 : The hand contains a joker and at least two cards from the same suit. The assistant chooses c1 and c2 and calculates t as in Case 1, puts c1 first and the joker last, and orders the remaining three cards (including c2 !) to code t. After turning over the first four cards, the magician checks to see if a joker is showing. If so, the hand must satisfy Case 2. Then the other three face-up cards have different suits, and the suit of the last card is the only suit that is not showing. Moreover the magician can determine the rank of the last card from the position of the joker and the rank of the first-face-up non-joker. Now suppose the joker is not showing. Then the hand could be in Case 1 or Case 3. The assumes that Case 1 holds, and using   magician , calculates what the last card c should be. If the methods for the 52 5 c is showing the assumption is false, so the hand is in Case 3 and the last card is the joker. Else, the assumption is true, and the last card is c.   Performing the 54 5 . Now the deck has a red and a black joker. The coding must be slightly modified from the previous trick in case the hand has two jokers and also to distinguish the color of the joker in Case 3. If there are two jokers the assistant will show one of them and put the other last. In Case 2, the magician is notified that the last card is the other joker by reversing the order of c3 and c4 . In Case 3 the magician will see two cards in the same suit and a joker, so he will also know the last card is the other joker. If there are at least two cards in the same suit and a joker, then the assistant picks c1 and c2 as before, puts c1 in the first position and the joker in the last position, but codes 5 for the red joker or 6 for the black joker. Now the magician knows the hand is in Case 3, because in Case 1 the the last card would be in the same suit as c1 and c2 , and so t would be at most 4.  Problem 2. Impossibilities. For the 42 , consider Figure 140. As there are more hands (6) than permutations (4), any plan must assign different hands to the same permutation, and so the magician will not be able  to identify the last card with certainty. More generally, the nk is    impossible if n Ck > n Pk−1 . This is the case for the 93 and the 125 5 . Strategies. Put n = (2(k − 2)! + 1)(k − 1). The strategy for the 52   5   can be extended to the nk . In particular, we can do the 32 , 63 and 15 4 . To see this, start with a deck with k − 1 suits and 2(k − 2)! + 1 ranks. Again we use orderings of the ranks and the suits to order the cards, and we create a circular ordering of the ranks. Again there are

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

273

two cards c1 and c2 with the same suit, and we can choose the names c1 and c2 so that we can count from the rank of c1 to the rank of c2 in at most 1 ≤ t ≤ (k − 2)! clockwise steps. The assistant puts c1 first, c2 last, and orders the remaining k − 2 cards to code t using some agreed method. Again the magician can determine the suit of c2 from the suit of c1 , decode t from the order of the middle k − 2 cards,  and  calculate , including the rank of c2 from t and the rank of c1 . Similarly, the (n+1) k   7 the 3 , can be performed by generalizing the strategy for the 53 5 , but 8 54 our solution of the 5 fails for the 3 (why?). Here are some examples. For k = 2, there are three cards and one suit. The set of k − 2 = 0 middle cards is empty, and has 0! = 1 orderings (the empty ordering). The cards c1 and c2 are chosen so that c2 follows 1 step after c1 in the circular order. For k = 3 there are six cards and two suits, but still only one ordering of the middle elements. For k = 4 there are 15 cards, three   suits, and finally two orderings of the middle elements. For the n3 , one readily computes that n = 8 is the largest deck for which the trick might work, since n ≥ 9 implies n − 2 ≥ 7 > 6 = 3!, and so n(n − 1)(n − 2) > n(n − 1) = n P2 . n C3 = 3! Problem 3. For Step 4, one possible way to lay out the cards systematically is shown in the Math Circle photo in Figure 138. The obvious necessary condition that 8 C3 ≤ 8 P3 is satisfied, but at this stage it is unclear whether a coding is possible, due to the constraint that the cards in the 2-permutation that codes a hand must be chosen from the cards of that hand. Suppose we have started clipping pairs as in (1–4) of Problem 3, and eventually there are no more unclipped index cards that can be paired to extend the existing clipped pairs. Focus on one unclipped ‘blue’ card’ (hand, combination) c. Find all pairs whose ‘red’ card (permutation) could be matched to c and arrange them above c with their ‘red’ card below their ‘blue’ card as in Figure 142. There will be six of these pairs above c. Now focus on the ‘blue’ cards in these lower pairs. Find all the ‘red’ cards (either clipped or unclipped) that are eligible to be clipped with these ‘blue’ cards and put them and any clipped mate above one of the corresponding ‘blue’ cards as before. Continue this process. Eventually you are guaranteed to have an unclipped ‘red’ card p at the top of this structure.13 Now there is a path from the unclipped card p to the unclipped card c such that pairs alternate between being unclipped and clipped. Rearranging the clips yields more clipped pairs. Repeat this process 8 until all ‘blue’ cards are clipped. At this point it is clear that the 3 can be performed: The assistant and magician agree to 13

You do not need to prove this until Problem 5; for now reflect on how many ‘blue’ and how many ‘red’ cards there would be in the structure if it were false.

274

M. KAWSKI AND H. A. KIERSTEAD H

{1,5,7} (7, 5)

H

{2,4,6} (2, 4)

{4,5,7} (5, 4)

...

HH      HH   H H H  H

{1,2,3} (2, 3)

{2,4,5} (2, 5)

(8, 7)

 HH     HH     H   H H H

H

{4,5,6} (4, 5)

H

{1,7,8} (7, 8)

{2, 3,6} (3, 2)

{3,5,7} (3, 5)

{5,7,8} (5, 8)

{5,6,8} (8, 5)

HH H H HH H HH H HH H H

{2,5,7} (5, 2)

{3,5,8} (5, 3)

` ``` P  PP  ``` PP @  `` @  P`  `` P` P @

{2,3,5}

Figure 142. Trying to increase the number of clipped pairs of cards. code each hand with the permutation that is clipped to it. Given a hand the assistant arranges the cards to reveal the permutation clipped to that hand, so the magician will know the original hand and the hidden card. Problem 4. First suppose Gn,k has a matching such that n Ck is matched. Then the assistant and magician agree that the code for a hand H is the permutation P to which H is matched. Since HP ∈ E(Gn,k ) the assistant can arrange the first k − 1 cards of H in the order of P . Since no other edge of the matching has  P for an end, when the magician sees P he will know H. So the nk can be performed. On the other hand, if this trick can be performed, the assistant can agree that for any hand H she will always code H with the same permutation P . Moreover, she cannot code two hands with the same permutation. So all such pairs HP will form a matching of Gn,k such that n Ck is matched. Problem 5. Proof of Hall’s Theorem. For necessity, suppose G has a matching such that C is matched, and consider any X ⊆ C. Since every vertex x ∈ X is matched to a vertex of N (x) ⊆ N (X), and no two vertices of C are matched to the same vertex, |X| ≤ |N (X)|.

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

275

For sufficiency, let G = (V, E) be a finite (C, P )-bipartite graph such that C is not matched by any matching. This means that for every matching there is a vertex in C that is not matched. Our goal is to show that Eq. (2) fails. We do this by finding a subset X ⊆ C with |X| > |N (X)|. Let M be a maximum matching in G; it exists since G is finite. By hypothesis, there is an unmatched vertex u. Let A ⊆ V be the set of all vertices v such that there exists an M -alternating path starting with u and ending with v. Set X = A ∩ C and Y = A ∩ P . Then u ∈ X (witnessed by the trivial path that begins and ends with u). Since G is a (C, P )-bipartite graph we observe: For every M -alternating path uv1 . . . vn and for each index i, if i is odd then vi ∈ Y , and if i is even then vi ∈ X. Moreover, vi−1 vi ∈ M if and only if i is even. We first show that N (X) ⊆ Y , that is, every neighbor of a vertex in X is in Y . Suppose z ∈ N (X). Then there is a vertex w ∈ X with wz ∈ E. Then w is the end of an M -alternating path Q = v0 . . . vn with v0 = u and w = vn ∈ X. By our observation, the index n is even, and so vn−1 vn ∈ M . Either z ∈ V (Q) or Qwz is an M -alternating path. In either case, z ∈ Y . So N (X) ⊆ Y , and therefore |N (X)| ≤ |Y |. To finish, it suffices to show that |Y | < |X|. Consider a vertex ym ∈ Y . By our observation, it is the end of an M -alternating path Q = uy1 . . . ym , where m is odd. As M is a maximum matching, Q is not an M -augmenting path by Lemma 1. So ym is matched. Thus there is an x ∈ C such ym x ∈ M . As u is unmatched, x ∈ C  {u} and Qym x = uy1 . . . ym−1 ym x is an M -alternating path. Thus x ∈ X {u}. We have shown that: Every vertex of Y is matched to a vertex of X  {u}. Consequently |Y | ≤ |X  {u}|. As u ∈ X and X is finite, |X  {u}| < |X|.  The following corollary is often sufficient for applying Hall’s Theorem. Corollary 1. If G = (V, E) is a finite (C, P )-bipartite graph that satisfies (3)

|N (x)| ≥ |N (y)| for all vertices x ∈ C and y ∈ P , then G has a matching such that C is matched. Proof. Suppose the hypothesis holds. Then there exists an integer n such that |N (x)| ≥ n ≥ |N (y)| for all vertices x ∈ C and y ∈ P . Using Hall’s Theorem, it suffices to check (2). We will need the notation: for X, Y ⊆ V , the set of edges of G with one end in X and one end in Y is denoted by E(X, Y ) = {xy ∈ E : x ∈ X and y ∈ Y }.

276

M. KAWSKI AND H. A. KIERSTEAD

For Eq. (2), consider any subset X ⊆ C. Then n|X| ≤ |E(X, P )| = |E(X, N (X))| ≤ n|N (X)|. Dividing by n we have |X| ≤ |N (X)|, so (2) holds.  n Problem 6. We claimthat  the k is possible if and only if n ≤ k! + k − 1. Suppose the nk is possible. From Problem 2 we know n Ck ≤ n P k−1 . Thus, n! n! = n C k ≤ n Pk−1 = . k!(n − k)! (n − k + 1)! Multiplying both sides by k!(n − k + 1)!/n! yields n ≤ nk! + k − 1. Now suppose n ≤ k! + k − 1. We show that the k is possible by proving that the graph Gn,k has a matching such that n Ck is matched. By Corollary 1, it suffices to check that |N (H)| ≥ |N (p)| for all hands H ∈ n C k and permutations p = (c1 , . . . , ck−1 ) ∈ n P k−1 . The neighbors k! = k!. The of H are (k − 1)-permutations of H, so |N (H)| = (k−(k−1))! neighbors of p have the form {c1 , . . . , ck−1 , ck } where ck is a card not used in p. Thus |N (p)| = n − (k − 1) = n − k + 1. Since n ≤ k! + k − 1, we have |N (H)| = k! ≥ n − k + 1 = |N (p)|.

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

277

Presentation Suggestions: This project is aimed at strong students in grades 11 and 12, but by concentrating on combinations, permutations and coding, the first three problems would be appropriate for younger or less advanced students. Problem 1. Do not give away too much after performing the trick. If help is needed, suggest to separately look at the suit and the rank of the hidden card. The discussion should eventually focus on using the order (permutation) three cards for the rank of the hidden card. 54middle  53 of the The 5 and 5 problems are quite a bit harder and make for challenging homework problems. A reasonable suggestion to the students is to distinguish different cases. Problem 2. Do not miss empowering the students by asking them to make up their own card decks with fewer or more suits, and corresponding numbers of ranks. It is fun to perform the trick with, say, just 3 suits and ranks from ace to 5.   Point out to the students that a strategy for the nk assigns to each hand a permutation of its cards so that no two hands are assigned the same permutation. Problem 3. It may take students quite a bit of time to make their own index cards. But this provides ownership, and familiarity with the cards. If several students work together, expect it to take some time until they agree how to divvy up the task of labeling 112 cards. If time is at a premium you can give them decks, but in that case it is a good idea to at least ask them to check whether the provided decks are complete, suggesting to lay them out in some systematic arrangement.

278

M. KAWSKI AND H. A. KIERSTEAD

Figure 143. Identifying an augmenting path As unclipped cards become scarce, students will naturally start unclipping pairs of cards. We suggest to intervene as deemed appropriate: Try to discourage randomly break pairs and clipping them back together in different pairs. Instead ask for strategies, to work together, make plans, lay out a path between unclipped cards which eventually leads to fewer unclipped index cards as captured live in a math circle in the photo shown in Figure 143. If students are trying to code an unclipped hand {i, j, k}, challenge them to at the same time find a partner for a very distant unused code (m, n), e.g., where the sets {i, j, k} and {m, n} have no common members. This leads to be more strategic planning how to select a sequence of pairs that should be broken, and clipped together in a new way. Note between Problem 3 and Problem 4. We like to engage the students in the process of making definitions which is an important part of the mathematical enterprise, as are generalizations. Emphasize that technical definitions make it easier to communicate with each other, avoid misunderstandings. Abstracting from tangible playing cards and paper clips to model these by graphs and matchings is a key stage of this sequence of activities. Problem 5. After visualizing the trick in terms of graphs in Problem 4, an important theorem of graph theory is proved in Problem 5. The

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

279

necessity of the condition Eq. (2) in Hall’s theorem should be very clear by now, and provides a good start, by quickly getting the first part of the proof done. The organization of the sufficiency argument, using the contrapositive, likely needs some help by the session leader. But most of the remaining argument can certainly be done by teamwork of students with the session leaders. Key are the experiences with the paper clips, and the pictorial illustration of augmenting paths Figure 141. Regarding the contrapositive of the statement in Hall’s Theorem, rather than showing that if the inequality is satisfied for all subsets X, then one always can find an augmenting path, the suggested argument assumes that no augmenting path exists and then constructs one subset X of vertices that does not meet the inequality in Hall’s theorem. Problem 6. After proving Hall’s Theorem in Problem 5, the final problem applies graph theory to give a conclusive answer to our central question, first raised in Problem 2: For any fixed size k of a hand, what is the  largest number n of cards in a deck such that the nk is possible? Mathematics Beneath & Beyond:   is possible, but have not given (1) We have proved that the (k!+k−1) k a good way to perform it. There is a relatively easy algorithm; the hardest part of the calculation is coding and decoding {1, . . . , k!} with k Pk−1 . For k = 3 the algorithm can be performed with mental calculation; for k = 5 it can certainly be done in real time with pencil and paper. (2) Consider a variation of our trick. Start with the deck {1, . . . , 2k + 1}. The assistant receives a hand of k + 1 cards, and turns all but one face-up. The students are allowed to mix-up the order of the faceup cards. Again it is the magician’s task to determine the face-down card. The ideas from Problem 6 can be modified to prove that this trick is possible. Reading between the lines of [3, 8], there are many algorithms for performing it. Here is one. The assistant calculates the sum i of the cards in the hand modulo k + 1. She then removes the (i + 1)st -smallest card in numerical order. The magician calculates the sum j of the remaining cards modulo k + 1. He then guesses the (j + 1)st -largest missing card. For example with k = 6, if the hand is {1, 2, 5, 6, 9, 10, 13} then i = 4, so the assistant removes 9. Now j = 2, so the magician correctly guesses the third largest card 9 from the missing cards 12, 11, 9, 8, 7, 4, 3. (3) Here is an infinite C, P -bipartite graph G = (V, E) that satisfies Eq.(2), but has no matching for which C is matched: Set C = {x1 , x2 , x3 , . . . }, P = {y1 , y2 , y3 , . . . } and E = {x1 y1 , x1 y2 , x1 y3 , . . . } ∪ {x2 y1 , x3 y2 , x4 y3 , . . . }.

280

M. KAWSKI AND H. A. KIERSTEAD

For all X ⊆ C, we have |X| = |N (X)| if x1 ∈ / X and N (X) = P if x1 ∈ X, so Eq. (2) holds, but C is not matched by any matching M , since if x1 yi ∈ M then xi+1 is not matched. Infinite versions of Hall’s Theorem are explored in [5, 7, 12]. (4) We have introduced matchings in graph theory in the playful context of card tricks, but the general theory of matchings has great importance in graph theory, computer science and operations research. Here is a more serious, but still easily understood, application. A university must assign dorm roommates. Among the collection of incoming students there are many pairs of students who are incompatible with each other. Is it possible to match the students into roommate pairs so that there are no incompatibilities? This question goes beyond Hall’s Theorem as the associated graph whose vertices are students and whose edges are compatible pairs of students is most likely not bipartite. Even if we know there is a solution, can we find it effectively? For an even deeper problem, we rate the compatibility of any pair of students, and try to find a matching that maximizes the sum of the compatibilities of the matched pairs. For a comprehensive reference on matchings see [11].

References [1] Davis, T., Mathematical Card Tricks (2008). http://www.geometer.org/mathcircles/CardTricks.pdf [2] Reinhard Diestel, Graph theory, 5th ed., Graduate Texts in Mathematics, vol. 173, Springer, Berlin, 2017. MR3644391 [3] D. A. Duffus, H. A. Kierstead, and H. S. Snevily, An explicit 1-factorization in the middle of the Boolean lattice, J. Combin. Theory Ser. A 65 (1994), no. 2, 334–342, DOI 10.1016/0097-3165(94)90030-2. MR1268348 [4] Egerv´ ary, E., On combinatorial properties of matrices (Hungarian with German summary), Mat. Lapok, 38 (1931), 16–28. [5] Marshall Hall Jr., Combinatorial theory, 2nd ed., Wiley-Interscience Series in Discrete Mathematics, John Wiley & Sons, Inc., New York, 1986. A Wiley-Interscience Publication. MR840216 [6] Hall, P., On Representatives of Subsets, J. London Math. Soc., 10 no.11 (1935), 26–30. [7] Henry A. Kierstead, An effective version of Hall’s theorem, Proc. Amer. Math. Soc. 88 (1983), no. 1, 124–128, DOI 10.2307/2045123. MR691291 [8] H. A. Kierstead and W. T. Trotter, Explicit matchings in the middle levels of the Boolean lattice, Order 5 (1988), no. 2, 163–171, DOI 10.1007/BF00337621. MR962224 [9] K¨ onig, D., Graphen und Matrizen, Mat. Lapok, 38 (1931), 116–119. [10] Lee, W., Math Miracles, (1951) Micky Hades International. [11] L. Lov´ asz and M. D. Plummer, Matching theory, North-Holland Mathematics Studies, vol. 121, North-Holland Publishing Co., Amsterdam; North-Holland Publishing Co., Amsterdam, 1986. Annals of Discrete Mathematics, 29. MR859549 [12] Alfred B. Manaster and Joseph G. Rosenstein, Effective matchmaking (recursion theoretic aspects of a theorem of Philip Hall), Proc. London Math. Soc. (3) 25 (1972), 615–654, DOI 10.1112/plms/s3-25.4.615. MR0314610

FROM A MAGIC CARD TRICK TO HALL’S THEOREM

281

[13] Math Circle at ASU Tempe, calendar of events, https://math.la.asu.edu/ ~mathcircle/events.php#events. [14] Colm Mulcahy, Mathematical card magic, CRC Press, Boca Raton, FL, 2013. Fiftytwo new effects; With a foreword by Max Maven. MR3113665

Published Titles in This Series 24 Dave Auckly, Bob Klein, Amanda Serenevy, and Tatiana Shubin, Editors, Inspiring Mathematics, 2019 23 James Tanton, How Round Is a Cube?, 2019 22 Mel Currie, Mathematics: Rhyme and Reason, 2018 21 Laura Givental, Maria Nemirovskaya, and Ilya Zakharevich, Math Circle by the Bay, 2018 20 19 18 17

Anna Burago, Mathematical Circle Diaries, Year 2, 2018 Titu Andreescu and Mark Saul, Algebraic Inequalities: New Vistas, 2016 Alexander Shen, Geometry in Problems, 2016 V. I. Arnold, Lectures and Problems: A Gift to Young Mathematicians, 2015

16 V. I. Arnold, Experimental Mathematics, 2015 15 Thomas Kilkelly, The ARML Power Contest, 2014 14 Zvezdelina Stankova and Tom Rike, Editors, A Decade of the Berkeley Math Circle, 2014 13 12 11 10

Natasha Rozhkovskaya, Math Circles for Elementary School Students, 2014 Ivan Yashchenko, Invitation to a Mathematical Festival, 2013 Anna Burago, Mathematical Circle Diaries, Year 1, 2012 Judith D. Sally and Paul J. Sally, Jr., Integers, Fractions and Arithmetic, 2012

9 David M. Clark, Euclidean Geometry, 2012 8 Sergey Dorichenko, A Moscow Math Circle, 2012 7 Roman Fedorov, Alexei Belov, Alexander Kovaldzhi, and Ivan Yashchenko, Editors, Moscow Mathematical Olympiads, 2000–2005, 2011 6 Costas Efthimiou, Introduction to Functional Equations, 2011 5 Alexander Zvonkin, Math from Three to Seven, 2011 4 Roman Fedorov, Alexei Belov, Alexander Kovaldzhi, and Ivan Yashchenko, Editors, Moscow Mathematical Olympiads, 1993–1999, 2011 3 Judith D. Sally and Paul J. Sally, Jr., Geometry, 2011 2 Sam Vandervelde, Circle in a Box, 2009 1 Zvezdelina Stankova and Tom Rike, Editors, A Decade of the Berkeley Math Circle, 2008

The people of the Navajo Nation know mathematics education for their children is essential. They were joined by mathematicians familiar with ways to deliver problems and a pedagogy that, through exploration, shows the art, joy and beauty in mathematics. This combined effort produced a series of Navajo Math Circles—interactive mathematical explorations—across the Navajo Reservation. This book contains the mathematical details of that effort. Between its covers is a thematic rainbow of problem sets that were used in Math Circle sessions on the Reservation. The problem sets are good for puzzling over and exploring the mathematical ideas within. They will help nurture curiosity and confidence in students. The problems come with suggestions for pacing, for adjusting the problems to be more or less challenging, and for different approaches to solving them. This book is a wonderful resource for any teacher wanting to enrich the mathematical lives of students and for anyone curious about mathematical thinking outside the box. In the interest of fostering a greater awareness and appreciation of mathematics and its connections to other disciplines and everyday life, MSRI and the AMS are publishing books in the Mathematical Circles Library series as a service to young people, their parents and teachers, and the mathematics profession.

For additional information and updates on this book, visit www.ams.org/bookpages/mcl-24

MCL/24