No title

Cover 2 8/11/10 7:54 PM Page 2 To Our Readers: Although nearly every submission we receive at Math Horizons comes in...

Author: Steve Abbott | Bruce Torrence (Editors in Chief)

9 downloads 337 Views 22MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form

DOWNLOAD PDF

Cover 2

8/11/10

7:54 PM

Page 2

To Our Readers: Although nearly every submission we receive at Math Horizons comes in electronically, last spring a short story called “Superstrings and Thelma” arrived in our old-fashioned mailbox with a simple note attached: “Is this short story something you can use? I wrote the math column in Scientific American for 25 years. If my piece is not right for Math Horizons, there is no need to send it back. All best, Martin.” Just seeing the name on the page caused a palpable stir. Martin Gardner was well into his nineties, yet here we were, in possession of a charming piece of original fiction that had come directly from the typewriter of the most celebrated voice in recreational mathematics. What is the proper editorial response to someone whose life’s work arguably made the existence of a magazine like Math Horizons a possibility? Sadly, we did not get the chance to craft a reply. As a means of giving our newer readers a broader picture of Gardner’s influence, we decided to run the short story alongside a report from the ninth biennial “Gathering for Gardner” held in March of this year. Then, on May 22, Martin Gardner passed away, and our plans evolved into fashioning an appropriate tribute. We’ve added a previously published piece by Gardner that is more typical of his mathematical writing and selected several companion articles where his influence is unmistakable. Special thanks go to former MH editor Don Albers for penning an opening essay on short notice. In one way or another, all of us who love mathematics have received something from Martin Gardner over the years—and we have no plans to send it back. Bruce Torrence and Stephen Abbott Co-Editors, Math Horizons

Letters: Dr. Stephens’s Living Legacy Dear Math Horizons, I recently received a copy of your February 2010 issue with the article titled “The Best Undergraduate Math Program You’ve Never Heard Of” by Reuben Hersh. My godfather, Sylvester Reese (pictured, far right), sent me the article because I had always known my father, Delegate Howard P. Rawlings (pictured, far left) was a protégé of Dr. Clarence Stephens when he attended Morgan. In fact, Dr. Stephens had such a profound impact on my father that my younger sister Stephanie was named after him. Now we know why. Four years ago, I came to work at a community college where the students are likely similar to those Dr. Stephens taught at Morgan . . . poorly prepared, black, with low confidence. I have become frustrated by our inability to provide students with the nurturing environment to learn and EXCEL in math. I clearly got this from my father. Because of your article, I now understand how my father learned this from Dr. Stephens.

Rusczyk Is Right On Dear Editors,

Thank you for highlighting Dr. Stephens’s work at Morgan and Potsdam! Know his work was furthered by his students like Delegate Rawlings, former chair of the Maryland State Appropriations Committee and advocate for higher education; his namesake, Mayor of the City of Baltimore Stephanie Rawlings-Blake; and me.

In the “Conversation with Richard Rusczyk” (February 2010), Mr. Rusczyk presents one of the best-stated arguments for both the why and the how of the teaching and learning of math that I have ever read. His criticism of the “standard curriculum” seems more pointed at elementary school than high school math, at least in my experience, but his concise answer to the question “What should we train our students for?” was both compelling and inspiring.

Sincerely, Lisa Rawlings

Bob Nortillo Niles West High School

pp.03-04

8/15/10

6:51 PM

Page 3

Contents

p22 p18 p26 5

p7

Martin Gardner, 1914–2010: Magical Man of Numbers and Letters Don Albers The founding editor of Math Horizons offers some personal reflections on the life and impact of the undisputed champion of popular mathematics.

6

Talkative Eve / Superstrings and Thelma Martin Gardner A rerun of a 1996 Math Horizons piece followed by a previously unpublished short story provide a glimpse into Martin Gardner's creative endeavors.

9

Gathering for Gardner Bruce Torrence The author finds himself in the middle of a puzzler’s paradise where mathematics is king, probability rules, and brevity is not only the soul of wit but a way to get rich.

13

The View from Here: Confronting Analysis Tina Rapke An unfiltered firsthand account of one student’s struggle to stay rational in her attempts to understand the real numbers.

16

Infinite Bottles of Beer on the Wall

p9

Donald Byrd A classic ditty for long car rides goes transfinite in the service of illustrating the principles of cardinal arithmetic.

18

Congo Bongo Hsin-Po Wang A mathematical jungle drumming challenge has legendary problem-solvers Jacob Ecco and Justin Scarlet standing on their heads—hopefully at the same time.

22

Port-and-Sweep Solitaire Jacob Siehler If you have exhausted all the challenges of traditional peg solitaire, a new variation called “Port-and-Sweep” will take you in some imaginary new directions.

26

The Mathematics behind Lifesaving Kidney Exchange Programs Olivia M. Carducci What happens when a willing organ donor turns out to be incompatible with a loved one’s blood type? The tools of graph theory may have vital implications for the evolving formulation of a national kidney exchange network.

30

p13

The Playground The Math Horizons problems section, edited by Derek Smith

34

Aftermath: Facebook and Texting vs. Textbooks and Faces Susan D’Agostino What is really happening on the laptops of note-taking students during class—and to what effect?

SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS

3

pp.03-04

8/15/10

6:51 PM

Page 4

Math Horizons is for undergraduates and others who are interested in mathematics. Its purpose is to expand both the career and intellectual horizons of students. STEPHEN ABBOTT Middlebury College BRUCE TORRENCE Randolph-Macon College Editors LOIS M. BARON Managing Editor & Art Director KARA KELLER Assistant Managing Editor DEREK SMITH Lafayette College Problems Editor Math Horizons (ISSN 1072-4117) is published four times a year; September, November, February, and April by the Mathematical Association of America, 1529 Eighteenth Street, NW, Washington, DC 20036. September 2010, Volume XVIII, Issue 1. Periodicals postage paid at Washington, DC, and additional mailing offices. Annual subscription rates are $29.00 for MAA members, $38.00 for nonmembers, and $49.00 for libraries. Bulk subscriptions sent to a single address are encouraged. The minimum order is 20 copies ($200.00); additional subscriptions may be ordered in units of 10. To order, call (800) 331-1622. For advertising information, call (877) 622-2373. Printed in the United States of America. Copyright ©2010. The Mathematical Association of America. POSTMASTER: Send address changes to Math Horizons, MAA Service Center, P.O. Box 90973, Washington, DC 20090-1112.

EDITORIAL BOARD Ezra Brown Virginia Tech Edward Burger Williams College Nathan Carter Bentley University Timothy Chartier Davidson College Katherine Crowley Washington & Lee University William Dunham Muhlenberg College Joseph Gallian University of Minnesota Duluth Deanna Haunsperger Carleton College Dan Kalman American University Stephen Kennedy Carleton College Mark McClure University of North Carolina at Asheville

John Allen Paulos Temple University Loren Pitt University of Virginia Tommy Ratliff Wheaton College Adrian Rice Randolph-Macon College Marian Robbins California Polytechnic State University Karen Saxe Macalester College Peter Schumer Middlebury College Laura Taalman James Madison University James Tanton St. Mark's School Tom Tucker Colgate University Robin Wilson The Open University

Instructions for Authors Math Horizons is intended primarily for undergraduates interested in mathematics. Thus, while we especially value and desire to publish high-quality exposition of beautiful mathematics, we also wish to publish lively articles about the culture of mathematics. We interpret this quite broadly—we welcome stories of mathematical people, the history of an idea or circle of ideas, applications, fiction, folklore, traditions, institutions, humor, puzzles, games, book reviews, student math club activities, and career opportunities and advice. Manuscripts may be submitted electronically to Editors Stephen Abbott ([email protected]) and Bruce Torrence ([email protected]). If submitting by mail, please send two copies to Bruce Torrence, Math Department, Randolph-Macon College, P.O. Box 5005, Ashland, VA 23005-5505. Subscription Inquiries email: [email protected] Web: www.maa.org Fax: (301) 206-9789 Call: (800) 331-1622 or (301) 617-7800 Write: Math Horizons, MAA Service Center, P. O. Box 90973, Washington, DC 20090-1112 THE MATHEMATICAL ASSOCIATION OF AMERICA 1529 Eighteenth Street, NW Washington, DC 20036

On the cover: The portrait of Martin Gardner was made by coloring the individual tiles on a kite and dart Penrose tiling. This particular tiling exhibits fivefold rotational symmetry. Aperiodic tilings were introduced to the public by Gardner in his oft-cited January 1977 Scientific American column. Artwork by Bruce Torrence.

4


Albers

8/11/10

9:30 PM

Page 5

Martin Gardner, 1914-2010 Magical Man of Numbers and Letters Don Albers

ay 22 Martin Gardner died. For 25 of his 95 years, he wrote “Mathematical Games and Recreations,” a monthly column for Scientific American magazine, which was without a doubt the most popular and influential column on mathematics that has ever existed. His columns inspired thousands of readers to delve more deeply into the large world of mathematics that he loved to explore and explain. His readers included amateur and professional mathematicians and a host of people from many other fields. Among his correspondents were many famous mathematicians and scientists, including John Horton Conway, Persi Diaconis, Ron Graham, Douglas Hofstadter, Richard Guy, Don Knuth, Sol Golomb, and Roger Penrose. His extensive network of friends often enabled him to bring new developments in mathematics in clear English to his legions of fans.

M

What is astounding is that he never took a single college course in mathematics. He excelled in mathematics as

a high school student and wanted to attend Caltech, but at that time (1932), Caltech required two years of liberal arts at a college before transferring. So he went to the University of Chicago, where he became hooked on philosophy. After service in World War II, his writing career began to blossom with a position at Humpty Dumpty’s Magazine for Little Children. In 1957 he began writing his monthly column for Scientific American. Most mortals find the idea of writing a significant column on mathematics every month for 25 years to be mind-boggling. Although many of his columns dealt with topics from computer science, Gardner always composed at a typewriter, and the manuscripts that he sent to me when I was editor of Math Horizons and the College Mathematics Journal contained his revisions in pencil. Given that he was the author, I would have been happy if he had submitted his manuscripts written with crayons on lined paper. It’s very pleasing that today the MAA is the publisher of nearly all of his books on mathematics and a CD containing the 15 books of all his Scientific American columns. Mathematics, however, was only one of Gardner’s interests. He wrote extensively about close-up magic and counted some of the best-known magicians among his friends. His book Fads and Fallacies in the Name of

Science, a brilliant debunking of pseudoscience, remains a best-seller 58 years after publication. His Annotated Alice has sold over 400,000 copies, and the first printing of the most recent revision sold out a few days after publication. His works on philosophy, especially the philosophy of science, continue to be very popular. Over the course of his life, Gardner wrote more than 70 books, of which half are still in print. As a young editor, I was greatly impressed by his kindness, generosity, and gentle manner. We talked frequently on the phone, and finally in 2000 I met him in person for the first time at his home in Hendersonville, North Carolina. For most of the day we chatted in his library about his life and books. I was also treated to a demonstration of some of his magic tricks and a tour of his extensive files. By late afternoon, he said that it was time to go upstairs and join our wives in the living room where he proudly demonstrated his ability to prepare martinis. I knew Martin Gardner for 32 of his 96 years. As he was an inspiration to the end, I regret that it could not have been longer. About the author: Don Albers is currently the editorial director of the books program at the MAA. He is also the founder of Math Horizons, serving as its first editor from 1993 through 1998. He is co-editor of a number of books, including Mathematical People: Profiles and Interviews, which contains an in-depth conversation with Martin Gardner. email: [email protected] DOI: 10.4169/194762110X525511


5

Gardners

8/15/10

2:16 PM

Page 6

Talkative Eve Martin Gardnere

This article first appeared in the April 1996 issue of Math Horizons. Talkative Eve his cryptarithm (or alphametic, as some puzzlers prefer to call them) is an old one of unknown origin, surely one of the best and, I hope, unfamiliar to most readers:

T

EVE = .TALKTALKTALK... DID The same letters stand for the same digits, zero included. The fraction EVE/DID has been reduced to its lowest terms. Its decimal form has a repeating period of four digits. The solution is unique. To solve it, recall that the standard way to obtain the simplest fraction equivalent to a decimal of n repeating digits is to put the repeating period over n 9’s and reduce the fraction to its lowest terms.

Three Squares Using only elementary geometry (not even trigonometry), prove that angle C in Figure 1 equals the sum of angles A and B. I am grateful to Lyber Katz for this charmingly simple problem. He writes that as a child he went to school in Moscow, where the problem was given to his fourth-grade geometry class for extra credit to those who solved it. “The number of blind alleys the problem leads to,” he adds, “is extraordinary.”

Red, White, and Blue Weights Problems involving weights and balance scales have been popular during the past few decades. Here is an unusual one invented by Paul Curry, who is well known in conjuring circles as an amateur magician.

6

You have six weights. One pair is red, one pair white, one pair blue. In each pair, one weight is a trifle heavier than the other but otherwise appears to be exactly like its mate. The three heavier weights (one of each color) all weigh the same. This is also true of the three lighter weights. In two separate weighings on a balance scale, how can you identify which is the heavier weight of each pair?

ANSWERS Talkative Eve As stated earlier, to obtain the simplest fraction equal to a decimal of n repeated digits, put the repeating period over n 9’s and reduce to its lowest terms. In this instance TALK/9,999, reduced to its lowest terms, must equal EVE/DID. Consequently, DID is a factor of 9,999. Only three such factors fit DID: 101, 303, 909. If DID = 101, then EVE/101 = TALK/9,999, and EVE = TALK/99. Rearranging terms, TALK = (99)(EVE). EVE cannot be 101 (since we assumed 101 to be DID), and anything larger than 101, when multiplied by 99, has a five-digit product. And so DID = 101 is ruled out.

DID. Because EVE must be smaller than 303, E is 1 or 2. Of the 14 possibilities (121, 141, … , 292) only 242 produces a decimal fitting .TALKTALK…, in which all the digits differ from those in EVE and DID. The unique answer is 242/303 = .798679867986… If EVE/DID is not assumed to be in lowest terms, there is one other solution, 212/606 = .349834983498…, proving as Joseph Madachy has remarked, that EVE double-talked.

Three Squares There are many ways to prove that angle C in the figure is the sum of angles A and B. Here is one (see Figure 2). Construct the squares indicated by red lines. Angle B equals angle D because they are corresponding angles of similar right triangles. Since angles A and D add to angle C, B can be substituted for D, and it follows immediately that C is the sum of A and B. This little problem produced a flood of letters from readers who sent dozens of other proofs. Scores of correspondents avoided construction lines by making the diagonals equal to the square roots of 2, 5, and 10, then using ratios to find two similar triangles from which the desired proof would follow. Others generalized the problem in unusual ways.

If DID = 909, then EVE/909 = TALK/9,999, and EVE = TALK/11. Rearranging terms, TALK = (11)(EVE). In that case, the last digit of TALK would have to be E. Since it is not E, 909 is also ruled out. Only 303 remains as a possibility for Figure 1. Prove that angle A plus angle B equals angle C.


Gardners

8/15/10

2:16 PM

Page 7

Charles Trigg published 54 different proofs in the Journal of Recreational Mathematics Vol. 4, April 1971, pages 90–99. A proof using paper cutting, by Ali R. Amir-Moéz, appeared in the same journal, Vol. 5, Winter 1973, pages 8–9. For other proofs, see Roger North’s contribution to The Mathematical Gazette, December 1973, pages 334–36, and its continuation in the same journal, October 1974, pages 212–15. For a generalization of the problem to a row of n squares, see Trigg’s “Geometrical Proof of a Result of Lehmer’s,” in The Fibonacci Quarterly, Vol. 11, December 1973, pages 539-40.

Red, White, and Blue Weights One way to solve the problem of six weights—two red, two white, and two blue—is first to balance a red and a white weight against a blue and a white weight. If the scales balance, you know there are a heavy and a light weight on each pan. Remove both colored weights, leaving the white weights, one on each side. This establishes which white

weight is the heavier. At the same time it tells you which of the other two weights used before (one red, one blue) is heavy and which is light. This in turn tells you which is heavy and which is light in the red-blue pair not yet used. If the scales do not balance on the first weighing, you know that the white weight on the side that went down must be the heavier of the two whites, but you are still in the dark about the red Figure 2. Construction for proof of the three-square and the blue. Weigh the theorem. original red against the mate of the original blue (or dentist and amateur magician, devised the original blue against the mate of the the following variation. The six weights original red). As C. B. Chandler (who are alike in all respects (including color) sent this simple solution) put it, the except that three are heavy and three result of the second weighing, plus the light. The heavy weights weigh the memory of which side was heavier in same and the light weights weigh the the first weighing, is now sufficient to same. Identify each in three separate identify the six weights. weighings on a balance scale. For readers who liked working on this DOI: 10.4169/194762110X524684 problem, Ben Braude, a New York City

Short Fiction

Superstrings and Thelma Martin Gardner everal years ago I was a graduate student at the University of Chicago. I was working on my doctorate in physics, about possible ways to test superstring theory, when my brother in Tulsa died suddenly from a heart attack. Both parents had earlier passed away. After the funeral I drove past my past, marveling at the enormous changes that had taken place since I grew up there. I drove by the red brick building, now an enormous warehouse, that had

S

once been Tulsa Central High. My grades in history, Latin, and English lit were low, but I was good in math and had a great physics teacher. He was mainly responsible for my majoring in physics after a scholarship took me to the University of Chicago. While I was having dinner at a popular restaurant on the corner of Main and Sixth streets, the waitress stared at me with a look of surprise. “Are you Michael Brown?”

“That’s me,” I said. She smiled and held out a hand. “I’m Thelma O’Keefe. We were in the same algebra 101 class.” We shook hands. “You won’t remember me,” she said. “I was fat in those days, and shy, and not very pretty.” “That’s hard to believe,” I said. “You look gorgeous now.” “Well, thank you, kind sir,” she said,


7

Gardners

8/15/10

2:16 PM

Page 8

smiling. “You were a whiz at algebra. Do you remember when you caught Mr. Miller in a mistake he made on the blackboard, and how embarrassed he was?” “I remember. He was a miserable teacher. I think he hated math.” “I know I hated it,” Thelma said. “I’m sorry to hear that. Math can be exciting and beautiful when it’s taught properly.” After Thelma brought my receipt and credit card, I said, “Any chance I could see you after work? Maybe you could steer me to a late night bar where we could chat about old times?” “I’m free at eleven,” she said.

membranes, or branes for short. A superstring is a brane of one dimension. Other branes have higher dimensions. Our universe, I said, has ten or eleven dimensions, of which six or seven are squeezed into compact tiny spheres that are attached to every point in our spacetime. “I didn’t understand a word you spoke,” Thelma said. “It sounds nutty to me. Do you believe all that?”

P h o to g ra

ph co u rt

e sy o f A

I followed Thelma’s car to a pleasant little bar on the outskirts of town, near where she lived. She was divorced, she “Mostly. I think strings are for real, told me, and had a boy of ten who was but I’m not so sure about Witten’s probably asleep in her apartment. The membranes.” bar served only beer. She said she “Is everything made of strings?” didn’t drink anything with more alcohol Thelma asked. than beer. Her ex, she added, had an alcohol problem. I didn’t press her for “Everything.” details. Instead, I fear I talked too much “And what are the strings made of?” about myself, and mainly about I was unable to get Thelma out of my head. superstrings.

de li ne C.,

h t tp://w

w w.f lick

r.com /ph

o to s/ad

o ll in n /

Like a fool I failed to ask for her address and phone number. We shook hands. She said goodbye, then startled me with a quick kiss on the mouth.

Almost a year drifted by. My thesis was published as a book by the I kept thinking University of Chicago Press. My of her wonderful smile, and how good she smelled. I did my best to suggestions for testexplain that strings ing string theory were taken seriously “Nothing. They’re just pure were inconceivably tiny loops, like by most stringers. There was hope that mathematical structures.” rubber bands, that vibrated at different some of the tests might actually be “If the universe isn’t made of anything,” rates. Their frequencies generated all made by a new atom cruncher under she said, “how come it exists?” the properties of the fundamental construction in Switzerland. There were particles, such as electrons and quarks. “Good question. Nobody knows.” vague rumors about a Nobel prize. The simplest string vibration produces “Well, maybe God knows,” she said. I was unable to get Thelma out of my gravitons, conjectured particles that head. I kept thinking of her wonderful carry gravity waves. Outside the bar, standing by our cars, smile, and how good she smelled. It Thelma invited me to her apartment for “You have nice dark eyes,” she wasn’t perfume. Was it her hair? I some coffee. interrupted. thought about her more than I thought “No,” I said. “I really can’t stay another “Thanks,” I said. “Your eyes aren’t so about superstrings! minute. I have to be up early to catch a bad either.” The University of Oklahoma, at plane to Chicago. It was great getting I tried hard to explain how a famous Norman, hired me as an assistant to know you.” physicist named Ed Witten had professor. A suburb of Oklahoma City, “Will I see you again?” generalized strings to what he called Norman is only a few hours drive from M-theory. The M stands for Tulsa. “It would be a pleasure,” I said. 8


Gardners

8/15/10

2:16 PM

Page 9

None of the waitresses at the restaurant where Thelma had worked knew what had happened to her. She left her job six months before, and they hadn’t heard from her since.

Gathering for Gardner Bruce Torrence

No Thelma O’Keefe was in the Tulsa phone book. I drove back to Norman feeling sad and frustrated. Should I hire a detective? The Norman yellow book had a long list of “investigators” and two detective agencies. I was planning to call one of the agencies when my telephone rang. It was Thelma! “I heard you were asking about me,” she said. “Yes. How did you get my phone number?” “It’s on the Internet. How are strings?” “Not so good. It didn’t predict dark matter. It didn’t predict dark energy. It even failed to pass one of my tests. Lots of stringers are starting to have doubts, including me.” “If we meet again,” said Thelma, “don’t tell me about it.”

Further Reading Two recent books attacking string/M theory as pseudoscience are Not Even Wrong, by mathematician Peter Woit (Basic Books, 2006), and The Trouble with Physics, by Lee Smolin (Mariner Books, 2007). See Chapter 18, “Is String Theory in Trouble?” in my book, The Jinn from Hyperspace (Prometheus Books, 2008). Editor's note: A provocative and widely discussed article in which string theory is used to suggest that gravity is not a fundamental force, but is rather a consequence of entropy, is Erik Verlinde's “On the Origin of Gravity and the Laws of Newton” (2010), which can be found at http://arxiv.org/abs/1001.0785. DOI: 10.4169/194762110X525557

Photographs courtesy of Bruce Torrence

t’s a chaotic scene in the lobby of the Ritz-Carlton. There’s a dental convention getting under way, and as travel-weary orthodontic professionals trickle into the packed room, they encounter a bewildering display. Every tabletop holds a collection of fascinating objects—puzzles of every imaginable shape and design—around which magicians, mathematicians, and puzzle masters gather to discuss their latest inventions. The conversations are animated, and there is a tangible feeling that something important is unfolding. And indeed there is: unbeknownst to the dentists, the ninth Gathering for Gardner has just begun.

I

Atlanta has been host to this unlikely convention of tinkerers and free thinkers for almost two decades. They assemble to pay homage to Martin Gardner, the prolific and magnetic

author whose interests spanned the seemingly disparate disciplines of mathematics, puzzles, magic, and the spirited debunking of pseudoscience. This invitation-only affair attracts luminaries in all of these fields, and despite their obvious differences, there is a fertile and dynamic common ground and a deep mutual respect among the participants.

What Has Tuesday Got to Do with It? The organization of the conference is simple: there is one grand conference room with a single stage. Speakers give brief presentations in turn and are awarded a dollar coin for each minute that they finish ahead of their allotted time—an innovative and surprisingly inexpensive management tactic. Only one speaker really cashed in: Gary Foshee, a mechanical puzzle collector


9

Gardners

8/15/10

2:17 PM

Page 10

the week and that births are independent of one another. With these assumptions in hand, the puzzle succumbs easily to elementary probability theory. If one denotes a single birth by a twotuple such as (boy, Tuesday) or (girl, Sunday), then there are 14 equally likely scenarios for a single birth. Foshee has two children, and with no other information, it follows that there are 142 = 196 equally likely possibilities for two children. But at least one of his children is a (boy, Tuesday), and some simple counting reveals that just 27 of the 196 outcomes satisfy this criterion. To see this, simply note that there are 14 cases where the Tuesday boy is the first born, and 14 where he is the second born, and subtract the single case that was counted twice. Among these 27 equally likely possibilities, how many include two boys? Exactly 13—there are seven with (boy, Tuesday) as the first child, and seven with (boy, Tuesday) as the second child, from which we subtract the one we counted twice. Hence, the answer to Foshee’s riddle is 13/27, close to, but not exactly, 1/2.

and designer, faced the audience and spoke slowly, “I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?” After a pause, he continued. “The first thing you think is ‘What has Tuesday got to do with it?’ Well, it has everything to do with it.” And with that, he stepped down (and collected a hefty stack of coins from the organizers). Other talks focused on space-filling curves and genome folding, origami mazes, Lewis Carroll’s mathematics, dancing tessellations, psychological explanations for children’s magic theory, the history of Rubik’s cube, the relation between computer hacking and invention, self-replicating machines,

10

illusion and perception, and on and on. The topics bounced so completely from one idea to the next that the effect was both intoxicating and refreshing. Spontaneous discussion arose in and out of the conference room. “What does Tuesday have to do with it?” Foshee’s puzzle proved a perfect catalyst for discussion, and it is both fun and instructive to reason through it. One first has to make a few assumptions, and most of the people I spoke with agreed that it’s best to keep it simple. Let’s assume that there are no multiple births, and that any single birth is equally likely to be a boy or a girl. Assume further that births are uniformly distributed among the seven days of


Of course, the riddle leads naturally to a host of other questions. Does “one is a boy born on a Tuesday” mean that the other was not born on a Tuesday? I don’t read it this way. After all, the question makes clear that there is a possibility that the second child is a boy, so why shouldn’t the child be allowed to arrive on a Tuesday also? More glaring is the counterintuitive nature of the result itself. Why should something like the day of the week affect the outcome? In thinking about this, it’s important to understand that had the day of the week not been mentioned, the answer would be 1/3, not 1/2, for only one of the three equally likely gender scenarios BB, BG, GB yields two boys. It’s all about counting the possible outcomes. If you are still wondering what Tuesday has to do with it, you may wish to consult the Further Reading section at the end of the article.

Gardners

8/15/10

2:17 PM

Page 11

Just as Foshee’s riddle yields to elementary probability theory, the vast majority of questions and puzzles posed at the gathering can be approached using basic principles. Each question has some clever twist, and of course this is the hallmark of a good brainteaser. Not all yield so easily, however. Bill Gosper, discoverer of the “glider gun” in Conway’s game of life and considered by many to be the founder of the hacker community, proudly shared his “Dozenegger” puzzle. This is a physical puzzle in which twelve circles, each a different size meticulously laser-cut from a sheet of acrylic, must be snuggly packed (but not forced) into a larger elliptical cavity. (You can find it on his website—see the Further Reading section.) Magic and illusion played a big part in the proceedings. Individual presentations in this area fell squarely in the field of cognitive psychology, highlighting peculiarities in human perception. A cube with a corner removed can be interpreted at least three ways (a large cube with a smaller cube cut out of it; a large cube with a smaller cube jutting out from it; and a large three-sided “room” with a cube sitting in its back corner). Verses of Led Zepplin’s “Stairway to Heaven,” played backwards can be made to say pretty much whatever you want provided the

listener reads those words as it plays. Cinematic scenes in which actors changed costumes out of frame and returned in their new garb went unnoticed by the entire audience (until we were prompted to look for the change in a second showing). Performance artists and magicians also took to the stage and presented stunning illusions with amazing skill. Attendees were treated to close-up magic after hours by some of the best in the business. The net effect was one of fascination with a subtly disturbing aftertaste. Everyone came away with a similar feeling: how is it that we can be deceived so easily? Seeing and believing will never again be the same.

Abstract Structures Another highlight of the gathering was an afternoon dedicated to socializing and sculpture building. Attendees were invited to participate in the construction and installation of several mathematical sculptures at the home of Tom Rodgers, one of the main organizers of the event. On the lighter side, literally, Vi Hart led a group that created geometric balloon art. At the other extreme, Chaim Goodman-Strauss collaborated on a weighty steel sculpture that suggested a space-filling curve packed neatly into a cube. Other sculptures were created using materials such as aluminum, wood, bamboo, and plastic. Under the direction of their designers—George Hart, Carlo Séquin, Akio Hizume, Rinus Roelofs, among others—it was a tour de force of master craftsmen at the top of their game. The collaborative enterprise also emphasized to the pure mathematicians among us some of the practical difficulties that arise when an abstract idea is realized as a solid structure. Every sculpture presented unique challenges, and their successful resolution led to a deep sense of fulfillment as the day came to a close. Back in the grand conference room, the onslaught of ideas continued. Princeton mathematician John H. Conway gave a spirited presentation on the arithmetic


11

Gardners

8/15/10

2:17 PM

Page 12

of lexicographic codes, or lexicodes as he calls them. Pure mathematics par excellence! Later, Scott Morris and Bruce Oberg gave back-to-back presentations heralding and bemoaning, respectively, the number nine (this being the ninth Gathering for Gardner). Pure nonsense, with a nod to Gardner’s fictitious numerologist and polymath, Dr. Matrix. As the barrage of topics flowed from the podium, what at first seemed a disparate hodgepodge of individually fascinating ideas began to gel into a coherent form. There was mathematics, there were puzzles, there was sleight of hand and deception, and there was mathematical artwork—the realization of abstract ideas into concrete form. There is beauty in all these, of course. But together the topics celebrate nothing less than our capacity to reason, and they demonstrate the supremacy of that endeavor. It is reason, after all, that enables us to solve mathematical problems and puzzles, and it is reason that enables us to test our often flawed perceptions and arrive at the truth. Collectively, the topics also pay homage to Martin Gardner himself, as they are precisely the themes that arose again and again in his writing. In the end, it is deeply inspiring to witness the extent to which his legacy lives on.

http://www.stat.columbia.edu/~co ok/movabletype/archives/2010/05/ hype_about_cond.html. To feel the sting of Bill Gosper’s twelve-circle problem, download a (free) playable computer version of the puzzle at http://demonstrations.wolfram.co m/TheTroublesomeTwelveCircleProb lem/. View the “Dozenegger” at http://gosper.org/Dozenegger3.pdf. To test your skill at spotting a sleight of hand, check out “The Color-Changing Card Trick” at http://www.youtube.com/watch?v=v oAntzB7EwE.

Further Reading

The Ambiguous Corner Cube was first discussed in C. L. Strong’s “The Amateur Scientist” column in the November 1974 issue of Scientific American, p. 126. A template for constructing a physical model can be found at http://www.bu.edu/lite/inkjetscience/pdfs/ProjectLITECornerCu beThirdCut.pdf.

For more on Foshee’s Tuesday birthday problem, see Andrew Gelman’s Statistical Modeling, Causal Inference, and Social Science blog entry for May 27, 2010, titled “Hype about conditional probability puzzles” at

A brief introduction to Conway’s lexicode theorem is available at http://www.dpmms.cam.ac.uk/semin ars/Kuwait/abstracts/L25.pdf, and a more complete treatment can be

found in MASS Selecta: Teaching and Learning Advanced Undergraduate Mathematics, edited by S. Katok, A. Sossinsky, and S. Tabachnikov (AMS, 2003). For an introduction to the mysterious Dr. Matrix, see Martin Gardner’s The Magic Numbers of Dr. Matrix (Promethius Books, 1985). You might also enjoy the “Ask Dr. Matrix” online tool at http://www.iread.it/ask_matrix.php. About the author: Bruce Torrence is the Garnett Professor of Mathematics at Randolph-Macon College, and a co-editor of Math Horizons. email: [email protected] DOI: 10.4169/194762110X525593

12


Rapke

8/11/10

9:51 PM

Page 13

The View from Here

Confronting Analysis Tina Rapkee

hen I think of my experience course be interesting? Will I catch on quickly. Before this class, I never learning analysis, there is a quickly? Will the professor be understood my friends’ reactions when mixture of emotions. There impressed with my abilities? I had I told them that I was majoring in mathare great feelings of accomplishment taken a course from this professor ematics: “I’m not a math person,” “It’s and perseverance that are seeded in before, so that was not causing me too too hard,” “I have math dyslexia.” feelings of anxiousness, discouragemuch worry, but the text, Principles of Now here I was, reconsidering my own ment, and choice of major, inadequacy. Before about to give up on It was as if I was taking the math ride; it was fun, analysis, I found my hopes of math enjoyable and becoming a matheit was good, but now I had to get off the bus. interesting. Never matician. It was as before had it left a if I was taking the bad taste in my mouth. I took math math ride; it was fun, it was good, but Mathematical Analysis, by Walter courses because I found them easy now I had to get off the bus. I had Rudin, didn’t look like anything I had and they were good GPA boosters. reached the end of my abilities. seen before. Analysis would change all of that. It was unlike any other math I had ever Not being the type of person to give up Then the professor walked in and seen—to be honest, I wasn’t even sure easily, and with graduation so close, I began to talk about Dedekind cuts and if it was math. knew I had no choice but to complete other obscure-sounding mathematical the course. And I did—with hard work notions—density, orderings, countabiliAs I walked into the first lecture of my and extraordinary effort. I’ve since gone ty. It didn’t take long for me to realize I first analysis class, I was excited but on to graduate school and even passed was in way over my head. I had jumped had those first-day butterflies: Will the a candidacy examine in analysis! I am into the deep end, and I was drowning

W


13

Rapke

8/11/10

9:51 PM

Page 14

sharing my story because I believe that my initial reaction to analysis is not unique, and the strategies I used to make it through this experience could be helpful to others.

Irrational Thoughts My struggles began with the very first assignment. Looking back at my notes from that term, I see that we were asked to show that the set of positive rationals with squares greater than 2 has no least element. I stumbled through the proof. It was clear that I was trying to mimic a similar proof from the class notes. Did I really understand this argument? Apparently not! The difficulties I was having are glaringly obvious when I look at the second assignment. There we were asked to do the following problem from page 44 of Rudin: “Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = | p – q | . Let E be the set of all p 僆 Q such that 2 < p2 < 3. Show that E is closed and bounded in Q but that E is not compact.” The first line in my response was, “Let s be the least rational such that s2 > 2.” Wow, that’s amazing because I had just proved that no such least element exists! This demonstrates how confused I really was during the course—proving something one day and then claiming the opposite on the very next assignment. I remember a conversation with the professor after receiving a poor mark on the assignment. We talked about how I thought I could arrange the rationals in order. Part of my confusion came with the introduction of the concept of “countability.” I knew that a set was countable if there was a oneto-one correspondence between the set and the natural numbers, and we had learned that the rationals were a countable set, which meant that they

14

could be arranged in a sequence: r1, r2, r3, …. Now, given any collection of natural numbers, it is always possible to go through and pick out the smallest one, so it seemed to me that given a rational number rn on my list, I could go through the list of remaining rationals and let rn+1 be the “next largest one.” At the professor’s request, I experimented with trying to name the “next largest” rational, and as you can guess, I wasn’t able to do so. I remember thinking in terms of “thickness,” which turned out to be a temporary stand-in for density. Given a rational, I could not name the next one. I related this to the same property of the real numbers, so the rational numbers and the real numbers were similar in terms of “thickness.” But then this blurred into my understanding of cardinality and suggested to me that the rationals had a cardinality close to that of the reals. But the real numbers are uncountable and the rationals are countable. Why? Looking back at all the course assignments, I remember feeling tense and anxious. My stomach was constantly tied in knots. I wanted to run away and pretend the assignments never existed. I felt uneasy because I


knew I wouldn’t be able to complete them. I didn’t understand what was really going on. If I didn’t grasp the concepts, how could I complete the assignments?

Making Progress Rudin’s definition says a set is countable if there is a one-to-one correspondence with the natural numbers, which for me meant that to show A is countable I had to find an explicit function from the naturals to the set A. But Rudin explains that the rationals are countable in a less direct manner. He first provides the following result: Let A be a countable set, and let Bn be the set of all n-tuples (a1, a2, …, an) where each ak 僆 A. Then Bn is countable. After establishing this, he notes that we can apply this theorem with n = 2 by observing that the rational numbers are of the form a/b where a and b are integers. This is all well and good, but when first learning about countability, I was confused because I wanted to see a correspondence with the natural numbers via some explicit function. Many years later, I found the following theorem in Principles in Real Analysis by Aliprantis: For an infinite set A the following statements are equivalent:

Rapke

8/11/10

9:51 PM

Page 15

(i) A is countable; (ii) There exists a subset B 債 N (where N is the set of natural numbers) and a function f : B → A that is onto; (iii) There exists a function g: A → N that is one-to-one. Currently, I like justifying that the rationals are countable by using the following steps: 1. First consider the set P of positive rationals. 2. Let B = {2n3m : n, m 僆 N} which is a subset of N. 3. Construct the function f : B → P by letting f (2n3m) = n/m. 4. Appeal to the above theorem. 5. Finally, bring in the negative rationals by noting that the union of two countable sets is countable.

My Story Now I completed my first analysis course, did fairly well, and received a scholarship as a consequence. I’ve written my Ph.D. candidacy exam in analysis and passed. One of my favorite memories of studying was one night when analysis crept into my dreams. I woke up in a panicky cold sweat. In my dream I was being chased by some analysis monster. My only defense was to use the “blancmange function” (a continuous but nowhere differentiable function discussed in Hairer and Wanner’s book Analysis by Its History) as a boomerang. I took it as a good sign at the time that analysis concepts were finding their way into my subconscious.

The blancmange function

Photographs courtesy of Stephen Abbott

While studying for my candidacy exam, I used several different textbooks. I can’t say with certainty that my initial misconceptions with density and countability would have been resolved by seeing the theorems from Aliprantis’s Principles in Real Analysis sooner, but this book fundamentally changed my understanding of these ideas—which brings up an important point about textbooks: Before encountering analysis, I never consulted different sources. If you’re having trouble in math, you should consider finding alternate textbooks and other resources. I like Aliprantis and Burkinshaw because of the way that they lay things out and because they have a very conversational style. They use the word “we” a lot and make it seem that the proofs are a team effort. Understanding Analysis by Stephen Abbott is another good resource. It is also very well laid out, and the author doesn’t give things up; he makes you work, but he guides you and gives many strong hints. I really enjoy the discussions at the beginning of each section, which include some history and motivation. I also like Analysis by Its History because it provides some well-known counterexamples, and I like learning through history. As for a first text in analysis, I recommend A Friendly Introduction to Analysis by Witold

Kosmala because it has many figures and examples. However, over all of these, I find Rudin to be the most useful reference text. Rudin goes straight to the point with very few fillers. I like to think that Rudin provides a good struggle—you need to read between the lines, fill in the gaps, and try really hard. Since that first course, I’ve gone on to take many other analysis courses, including measure theory and functional analysis, and I’ve done quite well. It took hard work, but I’ve come to appreciate the beauty and proof techniques of analysis, and I now find it fascinating. In the end, if you’re having trouble with analysis and find yourself in this story, YOU ARE NOT ALONE! I’ve heard many other successful graduate students say that they had similar issues when they first confronted analysis. So keep with it. Some struggles reap big rewards! About the author: Tina Rapke is a Ph.D. candidate at the University of Calgary. She is pursuing an interdisciplinary degree in mathematics and education. email: [email protected] DOI: 10.4169/194762110X525566


15

Byrd

8/11/10

10:06 PM

Page 16

Infinite Bottles of Beer A Cantorial Approach to Cantorian Arithmetic and other Mathematical Melodies Donald Byrd

A

classic summer-camp song—at least that’s the way I think of it—goes like this:

100 bottles of beer on the wall, 100 bottles of beer; If one of those bottles should happen to fall, 99 bottles of beer on the wall. 99 bottles of beer on the wall, 99 bottles of beer; If one of those bottles should happen to fall, 98 bottles of beer on the wall. (etc.) And so on until you get to zero. The idea, of course, is to entertain kids on a long trip with a long song. For a longer trip, make a longer song by starting with 200 or 300 bottles. Some years ago, the following playful variation—for really long trips!—occurred to me: Infinite bottles of beer on the wall, Infinite bottles of beer; If one of those bottles should happen to fall, Infinite bottles of beer on the wall. Infinite bottles of beer on the wall, Infinite bottles of beer; If one of those bottles should happen to fall, Infinite bottles of beer on the wall. (etc.)

16

Basic transfinite version: Infinite bottles of beer on the wall, Infinite bottles of beer; If finite bottles should happen to fall, Infinite bottles of beer on the wall. Infinite bottles of beer on the wall, Infinite bottles of beer; If finite bottles should happen to fall, Infinite bottles of beer on the wall. (etc.)

Larger infinity version: Uncountable bottles of beer on the wall, Uncountable bottles of beer; If countable bottles should happen to fall, Uncountable bottles of beer on the wall.

General transfinite version: Aleph-n bottles of beer on the wall, Aleph-n bottles of beer; If, where m < n, Aleph-m bottles should happen to fall, Aleph-n bottles of beer on the wall. (etc.)

Indeterminate version: Infinite bottles of beer on the wall, Infinite bottles of beer; If infinite bottles should happen to fall, Indeterminate bottles of beer on the wall. (The End)

Geometric progression version: 2100 bottles of beer on the wall, 2100 bottles of beer;


Cartoon by John Johnson

And then I thought, why stop with this basic fact about transfinite arithmetic? So I started creating other versions and, once word got out, began receiving different verses from friends. A sampling of this collection appears below where most, but by no means all, are by me. The vast majority are mathematically inspired, and collectively they

actually carry some pedagogical value. I have no doubt many more entertaining and illuminating variations are possible, so please pass along your contributions.

Byrd

8/11/10

10:06 PM

Page 17

If half of those bottles should happen to fall, 299 bottles of beer on the wall. 299 bottles of beer on the wall, 299 bottles of beer; If half of those bottles should happen to fall, 298 bottles of beer on the wall. … (optional ending) One bottle of beer on the wall, One bottle of beer; If half of those bottles should happen to fall, Half of a bottle of beer on the wall. (etc.)

Euler’s identity version: 100 bottles of beer on the wall, 100 bottles of beer; If e2πi of those bottles should happen to fall, 99 bottles of beer on the wall. (etc.)

Differential version: 100 bottles of beer on the wall, 100 bottles of beer; If dbeer/dt = –1, 99 bottles of beer on the wall. (etc.)

Infinitesimal version: 100 bottles of beer on the wall, 100 bottles of beer; If db bottles should happen to fall, 100 bottles of beer on the wall. (etc.)

Identity version: 100 bottles of beer on the wall, 100 bottles of beer; If none of those bottles should happen to fall, 100 bottles of beer on the wall. (etc.)

Topological dimension version: 100 bottles of beer in a square, 100 bottles of beer; If one of the dimensions should happen to fall, 10 bottles of beer in a line.

If one of those dimensions should happen to fall, One bottle of beer on the wall. (The End)

Fractal dimension version: 1 bottle of beer on the wall, 1 bottle of beer; If the middle third of that bottle should happen to fall, (1/3 + 1/3) bottles of beer on the wall. (1/3 + 1/3) bottles of beer on the wall, (1/3 + 1/3) bottles of beer; If each middle third of those bottles should fall, (1/9 + 1/9 + 1/9 + 1/9) bottles of beer on the wall. (etc.)

Russell’s Paradox version: Four lines of lyrics on the wall, Four lines of lyrics; If one of the lines should happen to fall, (The End)

Base 2 version: 100 bottles of beer on the wall, 100 bottles of beer; If one of those bottles should happen to fall, 11 bottles of beer on the wall. (etc.)

Continuum hypothesis version: Beth-1 bottles of beer on the wall, Beth-1 bottles of beer; Take Aleph-1 down, and pass them around: There may be no bottles of beer on the wall. (The End)

About the author: Donald Byrd is Woodrow Wilson Indiana Teaching Fellow and Adjunct Associate Professor of Informatics and Music at Indiana University. Lacking the discipline to work hard at mathematics, he took up computing as the path of least resistance, but he still enjoys the ideas of math greatly. email: [email protected] DOI: 10.4169/194762110X524657

10 bottles of beer in a line, 10 bottles of beer;


17

Wang

8/15/10

2:34 PM

Page 18

Congo Bongo Hsin-Po Wang

Illustration courtesy of Greg Nemec

Dedicated to the memory of Martin Gardner n expedition into Congo uncovered a treasure chest in the shape of a regular octagon. At each corner was a bongo drum. A scroll attached to the chest, written in French, explained that there was a genie inside each bongo drum. A genie is either standing upright or doing a handstand. One may strike a number of bongo drums at the same time. When a bongo drum is struck, the genie inside will change its posture from right side up to upside down, or vice versa. The treasure chest will open if and only if all genies are right side up, or all are upside down. However, each time some bongo drums are hit the treasure chest will spin rapidly on its vertical axis. As the bongo drums are all identical in appearance, after the rotation it is impossible to tell which of them had just been hit.

A

Unfortunately, the scroll did not record the exact procedure by which the treasure chest might be opened. However, it mentioned that such a procedure had been documented.

18


This document was considered so valuable that it was put inside the treasure chest for safekeeping. It was a lot safer than was originally thought. The sponsor of the expedition was definitely not pleased with the current state of affairs, so she hired a team of mathematicians to try to open it. The task was seemingly hopeless, and many gave up, until Dr. Jacob Ecco arrived with his sidekick, Professor Justin Scarlet. “My dear professor,” said Dr. Ecco, “we must approach this challenge systematically. What is the most basic principle in problem solving?” “Downsizing,” replied the professor, who was well versed with the methods of Dr. Ecco. “Suppose there is only one bongo drum. Then the treasure chest will open automatically.” “What if there are two bongo drums?” asked Dr. Ecco.

Wang

8/15/10

2:34 PM

Page 19

“Well, if the treasure chest is not already open, hitting either of the bongo drums will do it.” “Excellent! As usual, my dear friend, you have provided me with the inspiration that leads to the solution.” “How?” cried Professor Scarlet. “I have made but the most trivial observation.” “Nevertheless, a solution to the case with two bongo drums leads to a solution to the case with four bongo drums, which in turn leads to a solution to the case with eight bongo drums,” Dr. Ecco proclaimed. “I can smell that you are going to use induction, but l still do not see the connection from one case to the next,” replied Professor Scarlet. “Be patient. Obviously, we will be treating pairs of bongo drums as a single bongo drum. How can we identify pairs of bongo drums? Adjacency is unsatisfactory because after the treasure chest spins around, it is impossible to tell whether a bongo drum is to be paired with its neighbor to the left or its neighbor to the right.” “The obvious solution is to identify opposite pairs of bongo drums,” pronounced Professor Scarlet. “What insight!” said Dr. Ecco with a slight smile.

“Yes!” exclaimed Professor Scarlet excitedly. “In state 1, the genies in each opposite pair of bongo drums are both right side up or both upside down. In the case with two bongo drums, we hit one of them. So here we hit an opposite pair of bongo drums, and the treasure chest will open.” “We will call hitting an opposite pair of bongo drums operation A, and as you keenly observed, performing this operation from state 1 will unlock the chest. What happens when I perform operation A if we are in state 2 or state 3?” “Let me see,” Professor Scarlet said. “Ah, we will remain in the same state as before.” “How can I move from state 2 to state 0 or 1?” Dr. Ecco asked. “Well, we may hit an adjacent pair of bongo drums. I suppose this will be operation B. If we hit both zeros, or both ones, we will be in state 0 immediately. If we hit one 0 and one 1, we will be in state 1.” “Indeed, you can check that state 3 is left unchanged by operation B, just as it was unchanged by operation A,” Dr. Ecco agreed. “The rest is easy now. We perform operation C by striking just one bongo drum. This will change state 3 into state 0, state 1, or state 2,” Professor Scarlet said triumphantly.

“This is fine if the genies in each opposite pair of bongo drums are both right side up or both upside down. But there is no guarantee that they start off like that,” his companion pointed out.

“Not so fast,” Dr. Ecco cauntioned. “After operation B, we may be in state 1, and if we rush into operation C now, state 1 will become state 3, and we will be going in circles.”

“True, but there is a way to make them end up that way. Let 0 or 1 indicate whether a genie is right side up or upside down, and consider a four-sided chest. If we are really lucky, the initial state may already be (0,0,0,0) or (1,1,1,1).”

“I was hasty,” admitted Professor Scarlet. “We must perform operation A once more to clear state 1 before taking care of state 3. After operation C, we will clear the lower states with the sequence ABA again. So the overall procedure for the four-sided chest is ABACABA.”

“You may be good enough to be that lucky, but that never works with me,” complained Professor Scarlet.

“Excellent. Here, l have drawn a state transition diagram for you,” Dr. Ecco said.

“I am not counting on luck. I am just singling out the most favorable scenario. We may regard (0,0,0,0) and (1,1,1,1) as the same state. Call it state 0. It is also referred to as an absorbing state, in that once we enter it we do not leave (for the problem is then solved). How many other states are there?” asked Dr. Ecco. Professor Scarlet paused. “Well, we have three other states, namely, (0,1,1,1), (0,0,1,1), and (0,0,0,1).” “You are correct as far as the number of other states is concerned, but wrong about their composition,” Dr. Ecco said. “By symmetry, we may consider (0,1,1,1) and (0,0,0,1) as the same state. We call it state 3. Your (0,0,1,1) is state 2, which is not the same as (0,1,0,1). This I call state 1.”

Here is the same drawing, with an emphasis on the structure of the transition operations:


19

Wang

8/15/10

2:34 PM

Page 20

“That explains everything clearly!” said the professor. “Now I am confident we are onto something that will crack the eight-sided treasure chest. Let’s see. We wish to treat each pair of opposing bongo drums as a single entity. As before, the simplest states are those in which the genies in each opposite pair of bongo drums are both right side up or both upside down. Here is my drawing of this part of the state transition diagram. It is really the same as yours, except that operation A means hitting every other pair of opposing bongo drums; operation B means hitting any two adjacent pairs of opposing bongo drums; operation C means hitting any pair of opposing bongo drums. By performing the sequence ABACABA when starting from any such simple state, the treasure chest will open.”

“The states with 2 matching pairs are classified according to whether those matching pairs are alternating or adjacent,” Dr. Ecco said. “The former are grouped under state 2 while the latter are grouped under state 2’. Operation D hits any four adjacent bongo drums. Operation E hits any two bongo drums separated by one other drum. Operation F hits any two adjacent bongo drums. Operation G hits any one bongo drum.” “Let me see. If we denote the sequence ABACABA by X, then the sequence that will open the treasure chest is XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX,” summed up Professor Scarlet. “We keep repeating X so that whenever we enter state 4, we will not return to another state. Whatever the initial state of the treasure chest may be, it will open by the end of this sequence.” The sponsor of the expedition was delighted with their analysis. She hired a team of Congo natives to strike the bongo drums in the prescribed fashion. Lo and behold, the treasure chest opened right before her eyes. Unable to contain her excitement, she reached inside, but found only another scroll. It was written in Kituba, but the most prominent line read something like XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX.

“Excellent, my dear professor,” exclaimed Dr. Ecco. “These states together form an expanded absorbing state in the overall diagram below. It is the box marked 4, where the box marked m contains all states with m matching opposite pairs, for 0 ≤ m ≤ 4.” “This diagram looks suspiciously like the ones before, except that states 4 and 0 are no longer equivalent. Why do you have a state 2 and a state 2’? And what are the operations D, E, F, and G?”

20


Later, back in his New York apartment, Dr. Ecco said to Professor Scarlet, “That was hilarious, wasn’t it? By the way, have you thought of the obvious question?” “Yes, l have,” replied the professor. “For what number of bongo drums can such a treasure chest be opened? From our analysis, the general procedure when the number is a power of two is clear. We treat each opposing pair as a single entity, thereby reducing to the preceding case. Then we progressively move all states into the expanded absorbing state. But what happens if the number of drums is not a power of two?”

Wang

8/15/10

2:34 PM

Page 21

“Then unlocking the chest is not always possible,” Dr. Ecco replied. “For in this case the number of bongo drums has an odd prime factor p. Suppose the pagan god Hemba is having fun with us. He will choose p evenly spaced bongo drums and make sure that the genies inside are not all right side up and not all upside down. Now ignore all other bongo drums except these p. The two types of bongo drums are not equal in number since p is odd. In order for us to succeed, we must strike precisely the bongo drums of one type. But Hemba will spin the treasure chest so that the bongo drums we plan to hit include at least one from the opposite group. This way, he can keep us from opening the treasure chest forever.” See problem 249 in the Playground on page 30 for more fun with these treasure chests.

Further Reading This problem was posed in the Senior A-Level paper in the 2009 Fall Round of the International Mathematics Tournament of the Towns. It is related to a problem posed in Martin Gardner’s famous “Mathematical Games” column in Scientific American (Feb. 1979), which is reproduced in his anthology Fractal Music, Hypercards and More Mathematical Recreations (W.H. Freeman, 1992). Gardner’s version of the problem is also treated in three other sources:

Ted Lewis and Steve Willard, The rotating table, Mathematics Magazine, 53 (1980): 174–179. William Laaser and Lyle Ramshaw, Probing the rotating table, in The Mathematical Gardner, edited by David Klarner, (Wadsworth, 1981) 288–307. Albert Stanger, Variations on the rotating table problem, Journal of Recreational Mathematics, 19 (1987): 307–308 and 20 (1988): 312–314. Jacob Ecco and Justin Scarlet are fictitional characters created by Dennis Shasha of the Courant Institute, New York. He is a leading creator of mathematical puzzles. Shasha has written a number of puzzle books featuring these clever detectives. A good place to start is his book The Puzzling Adventures of Doctor Ecco (Dover, 1998). About the author: Hsin-Po Wang is enrolled in Taipei Municipal Jianguo High School. He won a gold medal in the 2009 International Mathematical Olympiad and several awards in the International Mathematics Tournament of the Towns. He has also participated in science fair competitions in Taiwan and in the United States. email: [email protected] DOI: 10.4169/194762110X525601

xkcd

Cartoon courtesy of xkcd.com


21

Siehler

8/15/10

2:49 PM

Page 22

Port-and-Sweep Solitaire Jacob Siehler with the peg solitaire “problem space” sets in, and one abandons the game, abandons solving individual puzzles in favor of proving theorems, or seeks a twist on the rules to make the game new again. Opting for the latter, I propose a variation that I call “Port-and-Sweep Solitaire” (PaSS). PaSS is also played on a grid, and the First Rule of PaSS is that each square may hold 0, 1, or 2 counters—no more, no less. I usually play on my computer, but checkers on a checkerboard work just as well (and are pleasingly tactile). PaSS permits two types of move: 1. Sweep move: Add squares on the board, and

Image courtesy of Cristóbal Vila - etereaestudios.com

ow does this happen? I just wanted a nice game where I didn’t have to count higher than two, and I ended up dealing with imaginary numbers. But let me back up: I’ve been a little obsessed with a puzzle lately, and I would like to explain what’s puzzling me and how the square root of –1 can sneak in where you least expect it. The puzzle in question is a relative of the classic peg solitaire game pictured above, and before I introduce it properly, we can have a quick brush-up on the traditional version. You could write a whole book on the subject [1] but as you probably know, the game is quite simple: it is played on a grid, where each square may hold a peg (or a marble, or some other marker). The only allowable move is to jump one peg over an adjacent peg into an empty square, removing the jumped peg from the board:

H

This can be done in any direction on the grid—up, down, left, or right. An equivalent way of looking at the move is that it consists of adding to the “peg count” of three consecutive squares on the grid—subject to the rule that a square can’t contain more than one peg, or a negative number of pegs. The best-known problem in ordinary peg solitaire (OPS) asks the player to reduce an almost-full board of 33 squares to a single peg in the center square (see photo), but of course other problems are possible—easier problems to warm up on, or fresh challenges for those who have solved the classic configuration [8]. Eventually, though, a feeling of familiarity 22


2. Port move: Add on the board.

to four consecutive

to three consecutive squares

Both moves are subject to the First Rule, and as in OPS, they may be performed in all four directions. Lacking several hundred years of tradition, PaSS does not yet boast a single defining problem, but it does admit a wealth of varied and interesting ones. Figure 1 is a small, but typical, puzzle: we’re asked to reduce a given arrangement of sixteen counters on a 5 5 board to a single counter in the center. Up to symmetry, there are just three opening moves possible: port a 2 to the center, port a 2 to one of the corners, or sweep three 1’s to the edge.

Figure 1. Left: A small puzzle in its initial state. Right: The desired final state. Figure 2 shows a partial solution to this problem, using the third option to start. See if you can carry it to completion. This is a very forgiving problem; there are many ways to complete the solution I’ve begun, and my opening isn’t too special. Any two moves at the start can be carried to a successful conclusion. But if you arrive at one of the positions in Figure 3 on your third move, you will inevitably get stuck with no legal moves and at least three counters on the board. Figure 4 offers three more problems on the 5 5 board, which are somewhat less forgiving. In one case you can go

Siehler

8/15/10

2:49 PM

Page 23

An elegant way is to use an invariant—a function whose value is determined by the state of the board and will not vary when we change the board by legal moves. If such a function can be found that takes one value on the pristine state of the problem and a different value on the target state, then no sequence of legal moves can ever connect the two. PaSS has a natural affinity for mod 3 arithmetic, and we can take advantage of that to define a valuable pair of invariants for the game. These invariants will assign to each board a value in a charming and petite arithmetic universe of just nine elements; namely, the set F9 = {0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i}.

Figure 2. A promising opening sequence of moves.

F9 is a discrete, mod 3 analogue of the complex number system. Elements have the form a + bi; we can refer to a as the real part and b as the imaginary part. Addition uses the simple rule (a + bi) + (c + di) = (a + c) + (b + d)i, reducing both parts mod 3 to stay in the range {0, 1, 2}. To multiply, treat i as a square root of –1, but in the mod 3 world, –1 is the same as +2, so you can avoid minus signs by declaring i2 = 2. That leads to the peculiar, but totally legitimate, multiplication formula (a + bi) + (c + di) = (ac + 2bd) + (ad + bc)i.

Figure 3. Dead-end configurations. astray on the first move! Despite the modest length of the problems and the small board size, I think you’ll find their solutions satisfying. See Problem 250 in the Playground on page 30.

F9 is an abelian group under addition, and the nonzero elements form an abelian group under multiplication—that is,

F9 is a field of nine elements. We don’t need much of the multiplicative structure for the present purposes, but note that i(2i) = 1, that is, i–1 = 2i. Now, coordinatize your board in the usual Cartesian manner, letting the lower left square be (0,0). Let S(a,b) denote the number of counters on square (a,b), and define

π (S ) =

∑ S(a, b) i a + b and μ(S ) = ∑ S(a, b) i a − b ,

( a, b )

Figure 4. Three more games to try on a 5 5 board.

( a, b )

where all arithmetic—sums, products, and powers—takes place in F9, so each of the sums above, no matter how lengthy, will reduce to one of the nine elements.

Sorry, You Can’t Get There from Here Trying to reduce the problem in Figure 5 to a single counter will be less satisfying, however, and after a certain number of failed attempts, you will begin to suspect that it can’t be done. How can you prove that no solution exists, without examining every possible sequence of moves?

Suppose a position S is derived from S by a sweep to the right, taking one counter from each of (a, b), (a + 1, b), (a + 2, b) and adding two to (a + 3, b). Then

π (S ') = π (S ) − i a + b − i a + b +1 − i a + b + 2 + 2 i a + b + 3 = π (S ) − i a + b (1 + i + i 2 + i 3 ) = π (S ), since the sum in parentheses is zero in F9. Similar calculations show that the value of π is unchanged by sweeps in the other three directions. And if S is derived from S by porting a 2 from (a, b) to (a + 2, b), then

π (S ') = π (S ) − 2 i a + b + i a + b + 2 Figure 5. An impossible game.

= π (S ) + i a + b (1 + i 2 ) = π (S ), SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS

23

Siehler

8/15/10

2:50 PM

Page 24

and likewise for ports in other directions. This shows that the value of π remains unchanged under all legal moves in the game, and it’s routine to check that has the same property. Thus, the values of π and separate the possible positions on a given board into 81 distinct classes, and no play between positions in different classes is ever possible. A 5 5 board with a single counter in the center has π = 1 and = 1.

reduces the possibilities by a factor of 34, so at most 35 = 729 of these could plausibly be played to four corners. Among those, a large number share the disappointing property exhibited by Figure 7: one can easily reduce the red cells to a single 1 in the upper left corner, using ports alone, without affecting the rest of the board.

What of the board in Figure 5? Its invariants are π = 2 + i and = 1 + 2i, so it can never be played to a 1 in the center. In fact, it could never be played to a single 1 or single 2 on any square, since no single term in the sums for π and can contribute both a real and imaginary part. Figure 7. A four-corners game where no sweeping is needed. I should admit that I did not have to overexert myself to discover π and , because I knew of an analogous pair of invariants for ordinary peg solitaire, described by de Bruijn [6], using mod 2 arithmetic and a field of four elements. PaSS can therefore be seen as a plausible answer to the analogy problem, “What game is to the field of nine elements as peg solitaire is to the field of four elements?”

More Problems: Four in the Corners Back to problems you can solve: Figure 6 offers a few additional problems on a 6 6 board. Boards with even sides have no center square, so as an attractive, symmetric alternative, these are designed to play down to four 1’s, one in each of the four corners.

Figure 6. Some four-corners games, where the winning configuration has a 1 in each corner. How big is the “problem space” of potentially interesting four-corners puzzles? Not very big, if we’re picky about what we consider an interesting problem. I prefer problems with a lot of symmetry, and there are 39 = 19,683 six-by-six boards that have 90-degree rotational symmetry (including the four-corners configuration itself, and some dull things like the all-zero board). Taking the π and invariants into account

24


Due to symmetry, this can be repeated three more times, verbatim, to solve the problem. Boring! A port-and-sweep problem should require both ports and sweeps, I’m sure you agree. Eliminating the boring problems leaves a list of 273 candidates to be investigated—some that can’t be solved, some too small or too obvious to be interesting, some too large to be fun, and just a few in the Goldilocks region: tantalizing, but elusive. Of course, if we’re open to problems with less symmetry, there will be many more possibilities, and there are many tactics yet to be discovered which will aid in their construction and solution. As a parting shot, the 7 7 board in Figure 8 was generated by a computer working backward from a single 1 in the center and trying to obtain a symmetric configuration (in this case, just a reflective symmetry). I did not have the computer save its steps, so I know the problem can be solved, but I don’t know how! I believe it’s large enough to vex attempts to solve it by sheer computing power—but perhaps a reader will discover a more ingenious, tactical approach. See Problem 250 in the Playground on page 30.

Figure 8. Can you play this to a single 1 in the center?

Mutatis Mutandis Ordinary peg solitaire has been studied extensively, and while Beasley’s book [1] is the most comprehensive single work, other stimulating articles can be found. Anyone who is keen to explore will find that most of the problems and techniques that have been developed for OPS can be successfully,

Siehler

8/15/10

2:50 PM

Page 25

ahem, ported over to PaSS—but there are surprises to be found in the process. For example, “resource counts” can be constructed which provide an additional tool for proving certain PaSS problems unsolvable, but they are governed by more restrictive inequalities than in OPS. PaSS seems to be more complicated than OPS on a one-dimensional (1 n) board. And an investigation of the “Solitaire Army” problem ([2], [4]) will show that PaSS counters have much more forward mobility than their ordinary peg counterparts. That problem might lead you to wonder, “What is to PaSS as the Golden Ratio is to OPS?”—but I don’t believe any works of art will be inspired by the answer. Now, I’m going to have just one more try at the game in Figure 8…

Further Reading [1] John Beasley, The Ins and Outs of Peg Solitaire, Oxford University Press, 1985. [2] George I. Bell, George Bell’s Peg Solitaire Page, http://home.comcast.net/ ~gibell/pegsolitaire/index.html. [3] George I. Bell, A fresh look at peg solitaire, Mathematics Magazine, 80(1) 2007, 16–28. [4] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for Your Mathematical Plays, volume 4, 2nd edition, AK Peters, 2004. See chapter 23, pages 803–842. [5] Arie Bialostocki, An application of elementary group theory to central solitaire. The College Mathematics Journal, 29(3) 1998, 208–212. [6] N. G. de Bruijn, A solitaire game and its relation to a finite field, Journal of Recreational Mathematics, 5 1972, 133–137. [7] Christopher Moore and David Eppstein, One-dimensional peg solitaire, and duotaire, In Richard J. Nowakowski, editor, More Games of No Chance, Cambridge University Press, 2002. [8] Jacob A. Siehler, Problems in peg solitaire, http://demonstrations.wolfram.com/ProblemsInPegSo litaire/. About the author: Jacob Siehler is an assistant professor of mathematics at Washington and Lee University, who grew up reading Martin Gardner and who continues to grow by reading Martin Gardner. email: [email protected] DOI: 10.4169/194762110X525575


25

Carducci

8/16/10

9:07 PM

Page 26

The Mathematics behind Lifesaving Kidney Exchange Programs Olivia M. Carducci

Three kidneys that would not otherwise be available for transplant have been made available, and three people’s lives have been saved.

ary sighs deeply. She is sitting next to her husband, Bob, who has fallen asleep. Bob is hooked up to a dialysis machine as he has been three times a week for the last year. Mary thinks about all the traveling she and Bob had planned to do in their retirement years. That’s all on hold now; Bob can’t be away from the hospital for more than three days at a time. He needs a kidney transplant. Mary would be happy to donate one of her kidneys, but they are not compatible with Bob’s blood type. Bob is on the list for a cadaver transplant, but the wait for someone with his blood type averages three years, and he must get much sicker to advance to the top of the list. In the meantime, their lives are on hold.

M

Nearby, Kathy holds her breath as she waits for the doctor to tell her the results of her test. She is about to find out if her husband, Jim, can donate a kidney to her. The doctor says “positive crossmatch.” Kathy and Jim are not sure what a positive crossmatch is, but they know it means Jim’s kidney cannot be transplanted into Kathy’s

26

body. The two couples do not know each other and they have different doctors, but they will soon transform each other’s lives. It turns out that Jim can donate his kidney to Bob and Mary can donate her kidney to Kathy. The

identify “multiple patient exchanges” like the one described for these two couples. The doctors get all the glory, but mathematics makes the exchange work.

Background The only cure for end-stage renal failure is a kidney transplant. Dialysis can extend a patient’s life while the patient waits for a kidney to become available but does nothing to cure the disease. According to the National Kidney Foundation, as of April 2007 there were 70,870 patients on the United Network for Organ Sharing (UNOS) wait list for a kidney transplant. In 2006, 18,016 patients received a kidney transplant while 3,916 died waiting for a kidney to become available. Without an increase in the number of kidneys available for transplant, the UNOS wait list will continue to grow. two families are beneficiaries of a kidney exchange clearinghouse. A kidney exchange clearinghouse gathers information about incompatible patient-donor pairs and attempts to


One potential source of kidneys is living donors. A living donor is a person who is willing to donate one of his kidneys. People are born with two kidneys, but it is possible to live a long, healthy, and

Carducci

8/16/10

9:07 PM

Page 27

active life with only one. In 2006, more than a third of all transplanted kidneys came from living donors. Unfortunately, significant numbers of willing donors are unable to donate due to blood or tissue incompatibilities. This is the impetus for the creation of kidney exchange programs. In a kidney exchange, a group of donors who are unable to give a kidney directly to their loved ones donate instead to another member of the group. For example, consider three donor-recipient pairs: (Da, Ra), (Db, Rb), and (Dc, Rc). If Da donates a kidney to Rb, and Db’s kidney goes to Rc, and Dc’s kidney goes to Ra, then this is a three-way kidney exchange. Three kidneys that would not otherwise be available for transplant have been made available, and three people’s lives have been saved. There are several regional kidney exchange programs, most notably in New England, Ohio, and Baltimore. As we will see, mathematics, and graph theory in particular, is behind the algorithms that are used to identify potential exchanges.

The Top Trading Cycles and Chains Algorithm One of the first algorithms utilized for exchanges is the Top Trading Cycles and Chains (TTCC) algorithm. Consider a set of donor-recipient pairs (Di, Ri). Donor Di will often be referred to as kidney Di and recipient Ri will often be referred to as patient Ri. Assume that kidney Di is incompatible with recipient Ri, so that an exchange of some sort is required. We will use w to represent the cadaver wait list. Each patient ranks the available kidneys (and possibly the wait list w), with the most preferred kidney ranked first. For example, consider a kidney exchange problem with eight pairs (D1, R1), …, (D8, R8), and with preferences as follows:

R1 : D4

D5

w

R5 : D3

D1

R2 : D1

D3

D7

R6 : D7

w

R3 : D4

D5

w

R7 : D3

D7

R4 : D8

D2

w

R8 : D2

D8

w

Note that some lists end with w while others end with the recipient’s donor kidney. A recipient whose list ends with w is willing to have his donor’s kidney transplanted in exchange for the recipient being moved near the top of the wait list. A recipient whose list ends with his donor is only interested in an exchange that matches him with a living donor. If the recipient is matched with his donor’s kidney, the recipient does not receive a kidney in this round of the exchange. The TTCC algorithm consists of several rounds, or stages. At each stage, the exchange problem will be modeled as a directed graph. Recall that a directed graph consists of a set of points, called vertices, and a set of ordered pairs of vertices, called edges. At the end of each stage, a subset of the available kidneys will be assigned so the exchange problem for the next stage will be smaller. Consider a directed graph with a vertex for each patient Ri, each donor Di, and the cadaver wait list w. There is an edge from Di to Ri and from Ri to the patient’s most preferred kidney (or to w).

TTCC Algorithm Step 1: Formulate the directed graph

D7

w

with vertices for all available donorrecipient pairs. Step 2: If there are no cycles in the graph, skip to step 3. Otherwise, identify all cycles, and carry out the corresponding exchanges. (If edge (Ri, Dj) is in the cycle, then patient Ri receives kidney Dj. If i = j, then patient Ri does not receive a kidney in this exchange.) Remove all donor-recipient pairs that were in a cycle from the list of available donor-recipient pairs. If the available donor-recipient list is empty, stop; otherwise, go to step 1. Step 3: All patients are in a chain that ends with w, but patients can be in more than one such chain. Choose a chain ending with (Rp, w). If edge (Ri, Dj) is in the chosen chain, then patient Ri receives kidney Dj, and Rp will receive priority on the wait list. Remove the donor-recipient pairs in the chain from the list of available donor-recipient pairs, but the kidney at the start of the chain is still available. If the available donor-recipient list is empty, stop; otherwise, go to step 1. Let’s return to the example above. For step 1 in the initial stage, we construct the directed graph as shown in Figure 1. (We don’t show vertex w as no edges are incident with it.)

Figure 1. Directed graph for the initial stage of step 1.


27

Carducci

8/16/10

9:07 PM

Page 28

Step 2: The only cycle is (D1, R1, D4, R4, D8, R8, D2, R2, D1). Make the corresponding assignments, and remove the donor-recipient pairs from the list of available pairs. Step 1: The new graph is shown in Figure 2. Note that the edge (R3, D5) has been added since kidney D4 is no longer available.

Figure 2. Directed graph for the second stage of step 1. Step 2: The only cycle is (D5, R5, D3, R3, D5). Make the corresponding assignments, and adjust the list of available donor-recipient pairs. Step 1: The new graph is shown in Figure 3.

The final assignments are:

R1

R2

R3

R4

R5

R6

R7

R8

D4

D1

D5

D8

D3

w

D7

D2

and D6 is available to the cadaver queue. The cycle found in stage 1 illustrates the main problem with the TTCC algorithm: this cycle involves four patients and kidneys. The logistics of coordinating four transplants makes this impractical. Each transplant requires two operating rooms and their staffs. Some incompatibilities between recipients and donors can only be identified by mixing their blood. Only 10 percent of pairwise matches will exhibit such incompatibilities, but with a cycle of three donor-recipient pairs, such an incompatibility will occur 27 percent of the time, and with a cycle of four pairs such an incompatibility will occur 47 percent of the time. The TTCC algorithm can generate arbitrarily long cycles.

Pairwise Exchanges Figure 3. Directed graph for the third stage of step 1. Step 2: Again, there is one cycle, (D7, R7, D7). This assignment means patient 7 does not receive a kidney in this run of the algorithm. Step 1: The new graph is shown in Figure 4.

Figure 4. Directed graph for the final stage of step 1. Step 2: Go to step 3. Step 3: The only path is (D6, R6, w), which means patient 6 is given priority on the cadaver queue and kidney 6 is made available to a patient on the cadaver queue. Stop. 28

Exchanges that involve only two donor-recipient pairs (like the one in the introduction) will be called pairwise exchanges. Consider an undirected graph G with a vertex for each donorrecipient pair and an edge between two vertices if each of the two donor kidneys is compatible with the recipient from the other vertex. In this case, all compatible kidneys are viewed as equally desirable. This may seem problematic, but since success rates are comparable for all compatible live kidney donations, many transplant surgeons consider it reasonable. Let V denote the vertices of G and E the edges. A matching is a subset of E such that each vertex in V is in at

R1 : D2

D4

D7

R2 : D1

D3

D4

R3 : D2

D4

D7

R4 : D3

D5

D6


D7

most one edge of . Each edge of represents a pairwise exchange. An optimal set of pairwise exchanges is a matching containing the most edges, called a maximum cardinality matching. A patient whose vertex is not included in the maximum cardinality matching does not receive a donor kidney in this exchange. An augmenting path for is a path from a vertex not in an edge of to another vertex not in an edge of whose edges are alternately not in and in . An augmenting path can be used to increase the number of edges in a matching by one. The following theorem can be found in most undergraduate graph theory texts. Theorem: A matching in a graph G is a maximum cardinality matching if and only if there is not an augmenting path for in G. Maximum Cardinality Matching Algorithm: Step 1: Formulate the graph G with a vertex for each donor-recipient pair and an edge between compatible pairs. Set the matching = . Step 2: If there is no augmenting path in G, then output the maximum cardinality matching . Otherwise, find an augmenting path P in G. Step 3: Form the new matching by reversing the roles of the edges in P; edges in P 傽 are removed from and edges in P – are added to . Go to step 2. As an example, consider seven donorrecipient pairs, (D1, R1), (D2, R2), …, (D7, R7) represented by vertices v1, v2, …, v7, respectively, with the recipients’ compatible donors as listed below.

R5 : D4

D6

R6 : D4

D5

D7

R7 : D1

D2

D3

D4

D6

Carducci

8/16/10

9:07 PM

Page 29

Since D2 appears on R1’s list and D1 appears on R2’s list, there is an edge from v1 to v2; although D4 appears on R1’s list, D1 does not appear on R4’s list, so there is no edge between v1 and v4. Step 1: Formulate the graph representing the compatibilities. This is shown in Figure 5.

Obviously, considering only pairwise exchanges avoids the issue of long cycles. However, if we apply the pairwise exchange algorithm to the example used to illustrate the TTCC algorithm, we see that there is only one pairwise exchange possible: patients 1 and 5. Permitting longer cycles clearly has the potential to help more patients. Evidence from simulations suggests

doctors must have faith that the allocation is fair, or the system will fail. Mathematics is key to this lifesaving endeavor.

Acknowledgment The author would like to thank Anthony Nance for many helpful discussions and much encouragement on this project.

Further Reading

Figure 5: Graph for step 1. Step 2: Augmenting path: (1, 2). (Any edge could be chosen.) Step 3: = {(1, 2)} Step 2: Augmenting path: (3, 2), (2, 1), (1, 7) Step 3: = {(3, 2), (1, 7)} Step 2: Augmenting path: (4, 3), (3, 2), (2, 1), (1, 7), (7, 6) Step 3: = {(4, 3), (2, 1), (7, 6)} Step 2: There are no augmenting paths left, so the matching {(4, 3), (2, 1), (7, 6)} is a maximum cardinality matching. Clearly there are other matchings that also include three edges, for example {(1, 7), (3, 4), (5, 6)}, but no matching includes more than three edges. The medical community can prioritize patients for living-donor transplants similar to the prioritizing on the cadaver waiting list, and the matching that includes the highest priority patients can be chosen as the exchange to be carried out.

that permitting three-way exchanges helps significantly more patients than allowing only pairwise exchanges, but that any gains from permitting longer cycles are minimal.

Current Work Integer programming has been used to generate pairwise and three-way exchanges. Large integer programming problems can require an unreasonable amount of time to solve, even on the fastest computers, so this solution was limited to regional exchanges. Recently, Tuomas Sandholm of Carnegie Mellon University has developed a method of breaking down the large integer programming problem into manageable pieces. His approach is currently being tested on a regional level, but the hope is that one day there will be a national database for living donors similar to the cadaver queue. How does the story end? It remains to be seen. The effort to organize a fair and efficient kidney exchange at the national level is ongoing. One thing is certain: The algorithm that determines how the kidneys are allocated is essential. The donors, recipients, and

A compendium of information on organ donation is published online by the National Kidney Foundation at www.kidney.org/news/newsroom/fs_ new/25factsorgdon&trans.cfm. See www.unos.org for an up-to-the-minute count on the number of people waiting for some type of transplant. For information on the New England Kidney Exchange Program, see Alvin E. Roth, Tayfun Sönmez, and M. Utku Ünver, “A Kidney Exchange Clearing house in New England,” American Economic Review, Papers and Proceedings, 95 (2), May 2005, 376–380. A good source of information on pairwise matchings is Alvin E. Roth, Sönmez Tayfun, and M. Utku Ünver, “Pairwise Kidney Exchange,” Journal of Economic Theory, 125, 2005, pp. 151–188. To better understand why three-way exchanges may be optimal, see Susan L. Saidman et al., “Increasing the Opportunity of Live Kidney Donation by Matching for Two- and Three-Way Exchanges,” Transplantation, 81 (5), March 15, 2006, pp. 773–782. The integer programming approach being developed at Carnegie Mellon University is outlined in an article appearing online at www.carnegiemellontoday.com/arti cle.asp?aid=514. About the author: Olivia Carducci teaches all levels of mathematics at East Stroudsburg University. She is the proud mother of four children, both a Boy Scout and Girl Scout leader, and a soccer coach. email: [email protected] DOI: 10.4169/194762110X525539 29


Playground

8/16/10

11:08 PM

Page 30

THE PLAYGROUND! Derek Smith

Welcome to the Playground! Playground rules are posted at the bottom of page 33, except for the most important one: Have fun!

THE SANDBOX In this section, we highlight problems that anyone can play with, regardless of mathematical background. But just because these problems are easy to approach doesn’t necessarily mean that they are easy to solve! This issue’s Sandbox problem comes from Joseph Ward, a high school student at Silver Creek Academy (AZ). Joseph was playing around with pencils one day when he discovered that some numbers of pencils could be stacked lengthwise to form the lowest k > 1 rows of an equilateral triangle when viewed from the end. Two different ways to stack n = 9 pencils are shown below:

He also found that, as hard as he tried, there were some numbers of pencils, such as n = 8, that could not be stacked in this way. Problem 248, Pencil Stack, asks you to determine the positive integers n for which there is at least one way to form the lowest k > 1 rows of an equilateral triangle.

THE ZIP-LINE

sequence of drum strikes that leaves all n genies oriented in the same way; the complicating factors are that you can’t see the genies’ orientations inside the drums and that the chest rapidly spins a random amount between each set of simultaneous drum strikes. The end of the article explains why no sequence of drum strikes is guaranteed to open an n-drum chest unless n is a power of 2. However, for other n it may be possible to have a pretty good chance of opening the chest during a certain sequence of drum strikes, assuming that the spinning between strikes is truly random. This leads us to Problem 249, Congo Bongo not so Wrongo: What is the smallest number of drum strikes needed on a three-drum treasure chest to give you at least a 90 percent chance of opening it? (As a separate further challenge, which may require computer assistance, try to answer the same question for a five-drum chest.)

The other Zip-Line problem, Problem 250, is simple to state: Win at Port-and-Sweep! The rules of the game are described in the first paragraphs of “Port-and-Sweep Solitaire” on pages 22–25: Each square of the board must have a value of 0, 1, or 2 at all times; a “port” move subtracts 2 from one cell and adds 1 to a cell distance 2 away; a “sweep” move subtracts 1 from three consecutive cells and adds 2 to a cell next in line. The goal is to end with a board where every cell is 0 except for a single 1 in the middle. (a) Start with the 3rd board in Figure 4 from the article:

This section offers problems with connections to articles that appear in this issue. Not all of the problems in this section require you to read the corresponding articles, but doing so can never hurt, of course. The first Zip-Line problem is motivated by the article “Congo Bongo” on pages 18–21. As described on the first page of the article, we attempt to open a treasure chest in the shape of a regular n-gon with a bongo drum at each corner. Inside each drum there is a genie standing either upright or on its hands, and striking any subset of the drums at the same time will cause all of the genies inside of those drums to invert themselves. The treasure chest will open if you can devise a

30


(b) Start with the challenge board in Figure 8:

Playground

8/16/10

11:08 PM

Page 31

THE JUNGLE GYM Any type of problem may appear in the Jungle Gym—climb on! Matthew McMullen of Otterbein University (OH) provides the Jungle Gym problem, Twice Sliced. Polya’s Pizzeria produces pizzas that are perfectly round, 12 inches in diameter, and cut into six congruent slices. You and a friend decide to share a slice by making a straight cut that divides the slice into equal areas, and you would like to consider cuts besides the “boring” cut that goes through the midpoint of the crust. Part (a) of Problem 251 asks you to describe all possible cuts, like cut a in the figure below, that don’t go through the curved outer crust of the slice, while part (b) asks you to describe all of the cuts (like b) that do.

the per-class mean class size is

C=

c1 + c2 + + ck , k

and the per-student mean class size is

S=

c12 + c22 + + ck2 . c1 + c2 + + ck

Since all of the numerators and denominators are positive, cross multiplying or dividing shows that S ≥ C if and only if

k (c12 + c22 + + ck2 ) ≥ (c1 + c2 + + ck )2 . At this point, the group from Taylor invoked the famous Cauchy-Schwartz inequality, 2

⎛ k 2⎞ ⎛ k 2⎞ ⎛ k ⎞ ⎜ ∑ xi ⎟ ⎜ ∑ yi ⎟ ≥ ⎜ ∑ xi yi ⎟ , ⎝ i =1 ⎠ ⎝ i =1 ⎠ ⎝ i =1 ⎠ which holds for all positive quantities x1, x2, …, xk and y1, y2, …, yk. Simply letting xi = 1 and yi = ci for each i then turns the Cauchy-Schwartz inequality into exactly what we wanted to show!

FEBRUARY WRAP-UP Larry Lesser of the University of Texas at El Paso led off February’s Sandbox by asking you to compare two types of averages: if a 9-student school consists of two 2-student classrooms and one 5-student classroom, then you can check that the mean class size on a per-class basis is 3, while the mean class size on a per-student basis is 33/9 = 11/3. Problem 240, Meaningful Means, asks if 11/3 > 3 is just one case of a more general phenomenon: if n students are distributed among k non-empty classes, must the per-student mean class size always be at least as large as the per-class mean class size? Correct solutions were received from student Xie Jun (University of Michigan–Shanghai Jiao Tong University Joint Institute) and a student group consisting of Seth Foote, Megan Frantz, and Audrey Nice (Taylor University, IN), as well as from Allen Fuller (Gordon College). Also, Alexander Bogomolny’s solution to this problem can be found at his “Cut The Knot” website. Here, we highlight the solution given by the group from Taylor. Let ci represent the number of students in class i. Then

n = c1 + c2 + + ck ,

The companion problem to Problem 240 asked if the per-class standard deviation for class size must always be at least as large as the per-student standard deviation for class size; for the example above, we get (sample) standard deviations of 3 and 2.5 , respectively. The answer is no: check this for 5 students distributed into one class of size 2 and three classes of size 1.

September’s second Playground problem came from Tuan Le of Fairmont High School (CA). Problem 241, As Easy as a, b, c, was another inequality: show that

L = a 4 (a − b )(a − c ) + b 4 (b − a )(b − c ) + c 4 (c − a )(c − b ) is greater than or equal to

R = k ((a − b )(b − c )(c − a ))2 when k = 1 and a, b, and c are non-negative real numbers. As a further challenge problem, Tuan asked you to find the largest value of k for which this relationship is guaranteed to hold. No correct “by hand” solutions were received by the submission deadline; there were two incorrect submissions. However, Stan Wagon of Macalester College offers computer-based solutions to both the regular problem (without the need for the non-negativity condition on a, b, and c) and the challenge problem in the form of a Mathematica notebook which can be found at http://www.maa.org/mathhorizons/supplemental.htm. SEPTEMBER 2010 • MATH HORIZONS • WWW.MAA.ORG/MATHHORIZONS

31

Playground

8/16/10

11:08 PM

Page 32

For comparison, we present Tuan’s original solution to the challenge problem below and ask which method of proof you prefer. We begin with the arithmetic-geometric mean inequality for two non-negative real variables x and y, which states that

x+y ≥ xy . 2 Setting x = a3b and y = ab3 lets us conclude that a3b + ab3 ≥ 2a2b2. Similarly, we deduce that a4 + b4 ≥ 2a2b2, so we know that

a 4 + a 3b + a 2 b 2 + ab 3 + b 4 ≥ 5 a 2 b 2 for any non-negative values a and b. Multiplying both sides of this inequality by the non-negative quantity (a – b)2 results in

a 5 (a − b ) + b 5 (b − a ) = (a − b )(a 5 − b 5 ) ≥ 5(a − b )2 a 2 b 2 , an inequality that we’ll return to soon. Looking back at the original problem statement, note that since both L and R are symmetric expressions in a, b, and c, we may without loss of generality suppose that c is the minimum value of the three variables. Also, notice what happens to L and R if we subtract from each of a, b, and c a common value t satisfying 0 < t ≤ c: the value of R doesn’t change, and the value of L is reduced because the fourth powers get smaller. Thus, to solve the problem it suffices to prove that L ≥ R in the case when c = 0, which you can check gives exactly the inequality presented at the end of the previous paragraph when k = 5.

The Zip-Line problem in February was attached to the article “Bacterial Computing: Using E. coli to Solve the Burnt Pancake Problem.” A sloppy chef has prepared the stack of six pancakes shown, with each pancake burnt on one side and with the two burntside-up pancakes shown with solid borders:

32


Holding two spatulas, you may alter the stack with the following type of move: the top spatula is inserted anywhere to lift all pancakes above it, while the bottom spatula is inserted in a lower position and used to flip over as a group all of the pancakes resting on it, after which the top spatula simply lowers any suspended pancakes back onto the stack. Problem 242, Flipjacks, asked you to turn the stack above into the standard smallest-to-largest all-burnt-sides-down stack in the fewest number of moves. For your flipping convenience, a Mathematica notebook was provided at http://www.maa.org/mathhorizons/supplemental.htm. Correct 5-move solutions were received by the University of Minnesota–Duluth Math Club, the “Why So Series?” student group from Mountain Lakes High School (NJ), and students Kim Moorehead, Ben Nelson, and Chris Vroon from Taylor University. Two other submissions provided 6-move solutions. All of the correct solutions mentioned the following sequence of 5 moves, as presented by the group from Mountain Lakes:

The group from Taylor also described a different 5-move sequence that has the same first, second, and fifth moves as the one above—can you find it? The group from Duluth wrote

Playground

8/16/10

11:08 PM

Page 33

a computer program to calculate all solutions that use no more than 6 moves; they found exactly 20 different 5-move solutions, and none with fewer moves.

“FIVE MORE MINUTES, KIDS!” The final new problem in February was given by Paolo Perfetti of Universita degli Studi di Roma “Tor Vergata.” Problem 243, Not Quite π2/6, asked about the convergence of the series ∞

1

∑ n2

Harold and the Purple Crayon. No correct solutions were received by the deadline, so the problem remained open, with a hint given in April’s column. Correct solutions have now been received by students Sabrina Dangc (Westmont College), Xie Jun, and Bao Yong (St. Cloud State University), as well as by Jeff VanderKam (Princeton, NJ) and Stan Wagon.

f ′(1 / n ),

n =1

where f : [0, 1] →R is an increasing function on [0,1] that is differentiable on (0,1] and where f : (0, 1] → R is decreasing. Correct solutions were received by student Jean Huang (Harvard University) and the Northwestern University Math Problem Solving Group; five incorrect solutions were also received. The group from Northwestern tackled the problem by first noting that for any positive integer n, the Mean Value Theorem says that there is some xn in the interval (1/ ( n + 1), 1/n ) satisfying

1 1 1 ⎞ f ′ ( xn ) ⎛1 . f( )− f( ) = f ′ ( xn ) ⎜ − = ⎟ ⎝ n n + 1 ⎠ n(n + 1) n n +1 Since xn < 1/n and f is decreasing, we know that f (xn) > f (1/n) and thus ∞

⎛

1

1

⎞

∞

f ′(1 / n )

∑ ⎜⎝ f ( n ) − f ( n + 1)⎟⎠ > ∑ n(n + 1) .

n =1

n =1

The series on the left is a telescoping series equal to f (1) − lim n →∞ f (1 / n ), which exists because f is increasing and bounded below by f (0). Also, since f is increasing we know that f (1/n) > 0 for all n, so the terms of the series on the right are positive—thus, the series converges by the comparison test. The convergence of the series in the problem statement then follows by the limit comparison test. Jean solved the problem using a different method and offered a generalization: ∞

1

∑ nα +1 f ′(1 / nα )

n =1

is convergent for any real number > 0.

CLEANING UP Problem 233 from last September was A Slice of Purple Pi, which asked for a proof of a result that appeared in a mathematics article by Crockett Johnson, the creator of

Solution techniques varied, but all showed that if = ⬔ABF, then x = cos() satisfies the cubic equation 8x3 – 4x2 – 4x + 1 = 0. Clever application of trigonometric identities could then show that cos(7) = –1 (not 1, as erroneously suggested in April’s hint), although a more streamlined derivation is available through the use of

cos(α ) =

eiα + e− iα . 2

SUBMISSION & CONTACT INFORMATION The Playground features problems for students at the undergraduate and (challenging) high school levels. All problems and/or solutions may be submitted to Derek Smith, Mathematics Department, Lafayette College, Easton, PA 18042. Electronic submissions (preferred) may also be sent to [email protected]. Please include your name, email address, and school affiliation, and indicate if you are a student. If a problem has multiple parts, solutions for individual parts will be accepted. Unless otherwise stated, problems have been solved by their proposers. The deadline for submitting solutions to problems in this issue is November 10, 2010. DOI: 10.4169/194762110X525584


33

Aftermath

8/12/10

9:02 PM

Page 34

Aftermath

Facebook and Texting vs. Textbooks and Faces Susan D’Agostino ast semester, my business statistics students were not exactly thrilled when I announced an in-class ban on electronic devices, including laptops, phones, and digital music devices.

L

“But I use my cell phone as a calculator!” one student protested. “Can’t I use my MP3 player to help focus during exams?” another pleaded. “I found a cool app that gives p-values for the standard normal distribution!” another offered hopefully, as if using statistical jargon would entice me to cave. “Humor me,” I responded. “Let this class be the one hour and fifteen minutes of your day in which you are completely unplugged.” I felt like a counselor at an outpatient program for recovering addicts. Halfway through the semester, I did what any self-respecting statistics instructor would have done: I surveyed my 67 students and used the tools I was teaching—confidence intervals for means and proportions—to compile the data. My results provide estimates—with a 95 percent confidence level—for the in-class, electronic multitasking habits of business majors at midsized, regional universities. Every student in this category has, at some point, used a laptop, phone, or digital music device in class. In a seventy-five-minute class that permits students to be “plugged in,” a student with an open laptop takes electronic notes just as much as he social networks: 34 minutes with a margin of error of 5 minutes. Looking at websites that are relevant to class is only slightly more common than looking at websites that are irrelevant to class: 36 as opposed to 32 minutes. A student with an open laptop spends, on average, 27 minutes sending and receiving email and 11 minutes reading an electronic newspaper. That these numbers sum to more than the seventy-five class minutes hints at the prevalence of in-class, electronic multitasking. Overall, when electronic devices are permitted in class, a majority of students using the devices—58 percent— multitask at least half the time. Students self-reported on the number of multitasking activities they engaged in beyond listening to the lecture or participating in class discussion: 52 percent of the examples involved one activity, including social networking or texting. Forty-six percent of the examples cited two, three, or four activities, including social networking, emailing, and doing homework. An intrepid 2 percent of the examples involved five multitasking activities: social networking, instant messaging, searching online, playing games, and texting. 34

To my surprise, the vast majority of students—94 percent—expressed either a favorable or neutral opinion of my policy. Were these the same students who originally made me feel like a counselor for substance abusers? “Knowing I can’t text allows me to pay better attention,” wrote one student. “Not having my computer out means that I can’t find myself on Facebook,” wrote another student. “I like the reduced noise distractions from [the absence of] electronic devices,” wrote a third. “It’s a good policy. I always see the students with laptops looking at Facebook or playing games,” another offered. So what about the responses from students who did not appreciate my policy? One commented that he “miss[ed] the unlimited amount of information that a computer has.” Another was put off by having to “carry notebooks and pens for note taking.” Another mentioned his concern about being unreachable in an emergency. Of course, I had informed my students that the university’s security office would deliver an emergency message to a student in class if needed. The Kaiser Family Foundation recently reported that the average 18-year-old spends over seven hours daily using electronic media devices for recreational purposes outside of the classroom. Based on my study, this statistic would likely increase dramatically if recreational use of electronics inside of the classroom were counted. College students should not sell their in-class time short. Class should be a time and place devoted to wrestling with ambiguity, not deferring to online encyclopedias edited by anyone with an inclination to blog. Currently, this assistant professor of math is wrestling with whether the anonymous student who wrote the following comment on my survey intended to be ironic: “I think [the in-class ban on electronics] is a good policy…. In this age of technology, people need to stay connected at all times. It absolutely gets in the way during class. Unfortunately, I really do not know how to fix the issue. I guess you could Google it?” Aftermath essays are intended to be editorials and do not necessarily reflect the views of the MAA. To respond, go to Aftermath at www.maa.org/mathhorizons. About the author: Susan D’Agostino is an assistant professor of mathematics at Southern New Hampshire University. email: [email protected]


DOI: 10.4169/194762110X525548

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

No title

Recommend Documents