MARTIN GARDNER KNOTTED DOUGHNUTS
KNOTTED DOUGHNUTS AND OTHER MATHEMATICAL ENTERTAINMENTS
MARTIN GARDNER
KNOTTED DOU...
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MARTIN GARDNER KNOTTED DOUGHNUTS
KNOTTED DOUGHNUTS AND OTHER MATHEMATICAL ENTERTAINMENTS
MARTIN GARDNER
KNOTTED DOUGHNUTS AND OTHER MATHEMATICAL ENTERTAINMENTS
W. H. Freeman and Company New York
Library of Congress CataloginginPublication Data Gardner. Martin, 1914 Knotted doughnuts and other mathematical entertainments Includes bibliographies and index. 1. Mathematical recreations. I. Title. QA95.GZ7 1986 793.7'4 ISBN 07 16717948 ISBN 07 16717999 (pbk.)
8531134
Copyright O 1986 by W. H. Freeman and Company No part of this book may be reproduced by any mechanical, photographic, or electronic process, or in the form of a phonographic recording, nor may it be stored in a retrieval system, transmitted, or otherwise copied for public or private use, without written permission from the publisher. Printed in the United States of America
To Gerry Pie1 and Dennis Flanagan and all my other good friends at Scientific American during the 25 years that I had the great privilege of writing the magazine's Mathematical Games column
contents Preface xi CHAPTER ONE
Coincidence 1 CHAPTER T W O
The Binary Gray Code 11 CHAPTER THREE
Polycubes 28 CHAPTER FOUR
Bacon's Cipher 45 CHAPTER FIVE
Doughnuts: Linked and Knotted 55
CONTENTS
x
CHAPTER SIX
The Tour of the Arrows and Other Problems
68 CHAPTER SEVEN
Napier's Bones
85
Napier's Abacus
CHAPTER NINE
Sim, Chomp and Racetrack
CHAPTER TEN
Elevators
123 CHAPTER ELEVEN
Crossing Numbers
133 CHAPTER TWELVE
Point Sets on the Sphere
145
CONTENTS
xi
Newcomb's Paradox
155 CHAPTER FOURTEEN
Reflections on Newcomb's Paradox
CHAPTER FIFTEEN
Reverse the Fish and Other Problems
176
LookSee Proofs
192 CHAPTER SEVENTEEN
Worm Paths
205
Waring's Problems
CHAPTER NINETEEN
Cram, Bynum and Quadraphage
232
CONTENTS
xii
CHAPTER TWENTY
The I Ching
244 CHAPTER TWENTYONE
The La$er Curve
257 Index of Names
273
Because this is the eleventh collection of my Scientijc American columns, there is little to say in a preface that I have not said before. As in earlier volumes, I have made corrections and additions throughout and included addendurns to material sent by readers and to update chapters in ways that were not easy to squeeze into the earlier text. References cited in the chapters are given more fully in the bibliographies that follow the chapters. Martin Gardner
KNOTTED DOUGHNUTS AND OTHER MATHEMATICAL ENTERTAINMENTS
CHAPTER ONE
Coincidence Don't worry. Lightning never strikes twice in the same
BILLY BEE
Since the beginning of history, unusual coincidences have strengthened belief in the influence on life of occult forces. Events that seemed to miraculously violate the laws of probability were attributed to the will of gods or devils, God or Satan, or at the very least to mysterious laws unknown to science and mathematics. O n the other hand, skeptics have argued that in the unthinkably intricate snarls of human history, with billions on billions of events unfolding every second around the globe, the situation is really the other way around. It is surprising that more strange coincidences are not publicized. "Life," wrote G. K. Chesterton in Alarms and Discursions, "is full of a ceaseless shower of small coincidences. . . . It is this that lends a frightful plausibility to all false doctrines and evil fads. There are always such props of accidental arguments upon anything. If I said suddenly that historical truth is generally told by redhaired men, I have no doubt that ten minutes' reflection (in which I decline to indulge) would provide me with a handsome list of instances in support of it." " W e trip over these trivial repetitions and exactitudes at every turn," Chesterton continued, "only they are too trivial even for conversation. A man named Williams did walk into a strange house and murder a man named Williamson. . . . A journalist of my acquaintance did move quite unconsciously from a place called Overstrand to a place called Overroads."
CHAPTER ONE
2
In his Poetics, Aristotle attributes to Agathon the remark that it is probable that the improbable will sometimes happen. All the same, most coincidences surely go unrecognized. For instance, would you notice it if the license plate of a car just ahead of you bore digits that, read backward, gave your telephone number? W h o except a numerologist or logophile would see the letters U , S , A symmetrically placed in LOUISIANA or at the end O ~ J O H NPHILIP SOUSA, the name ofthe composer of our greatest patriotic marches? It takes an odd sort of mind to discover that Newton was born the same year that Galileo died, or that Bobby Fischer was born under the sign of Pisces (the Fish). That's not all. "Fish" is chess slang for a mediocre player. In 1972, when Bobby Fischer's blunder cost him the first game in his famous match in Iceland with Boris Spassky, he said afterward, "I'm a fish! I played like a fish!" There are two other reasons why strange coincidences are seldom recorded. W h e n trivial ones are noticed, it is easy to forget them, and when they are remarkable enough to be remembered, one may hesitate to speak about them for fear of being thought superstitious. Skeptics maintain that with all of this in mind the number of astonishing coincidences that continually occur as the result of ordinary statistical laws is far greater than even occultists realize. T h e ancient view that many coincidences are too improbable to be explained by known laws has recently been revived by Arthur Koestler. In his book The Roots of Coincidence, he devotes many pages to a theory developed by Paul Kammerer, an eccentric Austrian biologist, whose Lamarckian convictions were much admired by T. D. Lysenko and who was the hero of Koestler's previous book, The Case of the Midwife Toad. Kammerer wrote a book, Das Gesetz der Serie (1919), about his theory of coincidences. It describes exactly 100 coincidences concerning words, numbers, people, dreams and so on that he had collected over a period of 20 years. Kammerer's seventh coincidence is typical. O n September 1 8 , 1 916, his wife was in a doctor's waiting room admiring magazine reproductions of ~ a i n t i n g s by a man named Schwalbach. A door opened and the receptionist asked i f ~ r i u Schwalbach was in the room. Kammerer's 10th coincidence is even more impressive. Two soldiers were separately admitted to the same hospital. They were 19, had pneumonia, were born in Silesia, were volunteers in the Transport Corps and were named Franz Richter. Kammerer was ~ e r s u a d e dthat such oddities could be accounted for only by assuming a universal law, independent of physical causality, that brought "like and like together." Koestler is sympathetic to this view. He suggests that some of the results of parapsychology, such as the tendency of falling dice to show a certain number more often than expected, can be explained not as the influence
COINCIDENCE
3
of mind on matter but as coincidences produced by a transcendent "integrative tendency. " Estimating the probability that a hidden law is at work behind a series of apparent coincidences is a difficult task, and statisticians have developed sophisticated techniques for doing so. How easy it is for our intuitions to go astray is illustrated by many familiar paradoxes. If 23 students are in a classroom and you pick two at random, the probability that their birthdays (month and day) match is about 11365. T h e probability that at least two ofthe 23 have the same birth date, however, is a trifle better than 112. T h e reason is that now there are 1 2 3 . . . 22 = 253 possible matching pairs, and figuring the exact probability of coincidence is a bit tricky. In a class of 35 students the probability of a birthday coincidence rises to about 85 percent. If students call out their birth dates one at a time until someone raises a hand to indicate that his birthday matches the one just called, you can expect a hand to go up after about nine calls (see "Note on the 'Birthday Problem,' " by Edmund A. Gehan in The American Statistician, April, 1968, page 28). William Moser has pointed out that the chances are better than even that two people in a group of 14 will have birth dates that either are identical or fall on consecutive days of the year. Among seven people, he calculates, the probability is about 60 percent that two will have birthdays within a week of each other, and among four people the probability is about 70 percent that two will have birthdays within 30 days of each other. Variants of the basic idea are endless. T h e next time you are in a gathering of a dozen or more people try checking on such things as the exact amount of change each person has, the first names of his parents, the street numbers of his home, the playing card each writes secretly on a slip of paper and so on. T h e number of coincidences may be scary. Another simple demonstration of an event that seems improbable but actually is not can be given with a deck of playing cards. Shuffle the cards, then deal them while you recite their names in a predetermined order, say ace to king of spades followed by the same sequence for hearts, clubs and diamonds. T h e probability that a card named in advance, such as the queen of hearts, will be dealt when it is named is 1/52, but the probability that at least one card will be dealt when named is almost 213. Ifyou name only the values, the probability of a "hit" rises to 98 percent, or very close to certain. In the foregoing instances the probabilities can be calculated precisely. For most events in daily life, however, probability estimates of coincidences are necessarily vague. For example, a great deal of research has been done on the "smallworld problem." What is the probability that if you meet a stranger on
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CHAPTER ONE
4
an airplane, the two of you will have at least one acquaintance in common? Not only are accurate statistics hard to come by but also the very terms of the problem are impossible to define precisely. Who, for instance, is an "acquaintance"? In spite of such formidable difficulties there is strong evidence that it is indeed a smaller world than most people imagine. Suppose a person is given a document and asked to transmit it to someone he does not know who lives in another city in another part of the U.S. T h e procedure is to send the document to a friend whom he knows on a firstname basis and who seems the most likely to know the "target" person. T h e friend in turn then sends the document to one of his friends with the same instructions, and the chain continues until the document reaches the target. How many intermediate links will the chain have? Most people guess about 100. When psychologist Stanley Milgram made actual tests, he found that the links varied from two to 10 and that the median was five. Pick two women at random. T h e s rob ability that both are wearing green shoes is low, but if you consider 20 ways the women can matchcolor of eyes, first names, type of hairdo and so on the probability of a coincidence is close to certainty. It is hard to believe, but gross miscarriages ofjustice have resulted from a failure to understandjust such trivial truths. In 1964 a black man and his white wife were convicted of a muggins in San Pedro, Calif., mainly because they were the only couple in the area who matched the reports of witnesses on five counts: the girl was a blonde, she had a ponytail, her companion was black, he had a beard, they drove a yellow car. T h e prosecutor estimated each probability separately 1/10 for a yellow car, 111,000 that a couple are black and white, and so on then he multiplied the five fractions and convinced the jury that the probability was 1112,000,000 that a matching couple lived in the vicinity. Not until four years later (see Time, April 26, 1968, page 41) did the California Supreme Court reverse the decision after a judge less ignorant of mathematics persuaded the court that the estimate should have been about 411100. Anyone who watches carefully for coincidences involving himself can easily find them. "Did you ever notice that remarkable coincidence?" F. Scott Fitzgerald wrote in 1928 to the British writer Shane Leslie. "Bernard Shaw is 61 years old, H. G. Wells is 5 1,G. K. Chesterton is 41, you're 31, and I'm 21 all the great authors of the world in arithmetical progression." Carl Sandburg was quoted in The New York Times, January 6, 1967, as saying that having completed his 89th birthday he confidently expected to live to 99. He had two greatgrandfathers and a grandfather who had died in years that were multiples of 11. Having got safely past 88, Sandburg expected to go on to 99. Unfortu
COINCIDENCE
5
nately he died six months later. Lewis Carroll recorded in his diary that most good things that happened to him, of which the best were meeting new and comely little girls, occurred on Tuesdays. Surely the strangest coincidence involving a major U.S. magazine was the case ofthe "deadly double" ads in The New Yorker, November 22,1941, which generated rumors about Japanese undercover agents for many years after. The longsubmerged rumors surfaced in 1967 when a former U.S. naval intelligence agent, Ladislas Farago, told the story in a press release for his book The Broken Seal, an account of American and Japanese intelligence operations before World W a r 11. Sixteen days before Pearl Harbor The New Yorker ran two advertisements (pages 32 and 86) for a new dice game called T h e Deadly Double [see Figure 11. Were these advertisements placed by the Japanese to inform their undercover agents of the planned attack on Pearl Harbor? Farago's press release pointed out the following correlations. T h e attack was on December 7. In the smaller first advertisement, note the 12 (for December) on one die and the 7 on the other. Above the dice are the words "Achtung. Warning. Alerte!" T h e numbers 5 and 0, Farago said, could have indicated the planned time for the attack, which did not start until 7:00 A.M. T h e XX, or 20. is the approximate latitude of Pearl Harbor. Farago admitted that he did not know what the 24 stood for. T h e second advertisement shows two people playing the dice game during an air raid, with the XX repeated on the symbol ofthe doubleheaded eagle. A Times story of March 12, 1967, based on Farago's press release, stated that the mysterious dice game had never existed. Farago told the Times that he had first learned of the ads from his friend A1 Hirschfeld, the newspaper's theatrical caricaturist. When Farago questioned officials at The New Yorker, he said, "They were very closemouthed about it." These fantastic allegations were quickly dissipated by the Time's followup story on March 14. T h e dice game did exist. Mrs. E. Shaw Cole, widow of the man who invented it, had been found in Montclair, N.J. She had helped her late husband, Roger Paul Craig, write the ads. Several New York department stores were selling the game in 1941. Agents of the Federal Bureau of Investigation, Mrs. Cole said, actually had visited them after the Pearl Harbor attack, but any relation between the attack and the ads was just "one big coincidence." "What can I say?" said Farago. Several years ago I asked Dr. Matrix, the famous numerologist, for his opinion on the advertisements. T h e XX, he told me, indicates that two X's are to be appended to the alphabet. The first number on the die, 12, instructs us to count to the 12th letter, L. T h e second number 24, tells us to count 24 letters forward from L, including of course the extra X's, and carrying the count back
1 @a1 See Advertisement Page
I
86
MONARCH PUBLISHING CO. Now York
I
W e hope )ou'll never have to spend a t a n alrraid shelter, long winter's n ~ g h in but we were just thinking . . . it's only common sense to be prepared. If you're not too busy between now and Christmas, why not sit d o w n and plan a list of the things you'll w a n t to have on hand. . . . Canned goods, of course, and candles, Sterno, bottled water, sugar, coffee or tea, brand), and plenty of cigarettes, sweaters and blankets, books or magaand though zines, vitamin capsules it's no time, really, to be thinking of whnt's fashionable, w e bet that most of your friends will remember to include those intriguing dice and chips which make Chicago's favorite game
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THE DEADLY DOUBLE
$2.50 at leading Sporting Coodo and Department Stores Everywhere
Figure 1
Two advertisements that appeared in The N e w Yorker for November 22. 1941.
COINCIDENCE 7
to the beginning. This second count ends on H. T h e 7 on the die at the right tells us to count seven letters forward from H to 0. T h e three letters found in this straightforward manner are L, H and 0, the initials of Lee Harvey Oswald. T h e advertisements in The New Yorker appeared in the November 22, 1941, issue. November 22 was the date of President John F. Kennedy's assassination, and 22 added to 1941 is 1963, the year of the assassination. It is easy to understand how anyone personally involved in a remarkable coincidence will believe that occult forces are at work. You can hardly blame the winner of the Irish Sweepstakes for thinking that Providence has smiled on him even though he knows it is absolutely certain that someone will win. Gamblers are particularly susceptible to this belief, and they tend to be more superstitious than most. In every big city in the U.S. there are thousands of policy "hunch players" who like to bet on numbers prominent in the news. It is hardly surprising that now and then such hunches pay off. In 1958, for example, 48 people died when a Jersey Central commuter train plunged into Newark Bay. T h e last car taken from the water was shown in newspapers and on television with its number 932 clearly visible. Thousands ofManhattan policy players bet on 932 and won. A similar coincidence was reported in The New York Times for January 24, 1967. T h e President's daughter, Luci Johnson Nugent, had just given birth to a boy weighing eight pounds 10 ounces. All over Brooklyn bets were made on various permutations of these three digits. When 081 won, Brooklyn policy banks were closed for days because of the losses. In science, as in daily life, it is not always easy to know if an observed correlation of "like and like" is pure coincidence or evidence of underlying structure. It was coincidence (plus some fudging) that the planetary orbits fitted Kepler's pattern of nested Platonic solids but not coincidence that data on their orbits fitted his patterns of ellipses. It is undoubtedly coincidental that the disks ofthe sun and moon, seen from the earth, are almost exactly the same size. T h e sun's diameter is 400 times that ofthe moon, but incredibly it is just 400 times as far away, as though nature planned it that way to give us a spectacular display of the sun's corona during a total eclipse. O n the other hand, for half a century most geologists were convinced that the fit of the edges of the land masses on each side of the Atlantic was sheer coincidence. Alfred L. Wegener's theory that the two land masses had once been a supercontinent that had split and drifted apart (a notion that had been advanced by Francis Bacon) was considered crankish until about 10 years ago. Now it is the preferred hypothesis. There are similar difficulties in mathematics. The curious repetition of 1828 in the first nine decimals of e (2.718281828 . . .) is almost certainly coincidental. Consider now the square roots of ,999 and .9999999. They are respectively .9994 . . . and .99999994. . . . Is it accidental that in each case the
Figure 2
Benson Ho's answer.
irrational square root of a decimal fraction consisting of n 9's begins with n 9's followed by a 4? No, as Richard G. Gould has pointed out in a letter; it can be shown to be true of all such "rep9" decimal fractions. You have only to express their square roots as (1 expand the expression by the binomial theorem and interpret the results properly to establish the theorem. T h e number 4 is a square number, and if YOU append to it the next consecutive square number, 9, the result is 49, another square. Is it a coincidence or a special case of a general law? One more curious question (both will be answered next month): An old brainteaser asks for the ordering principle behind the sequence 8549176320, which contains all 10 digits. T h e answer is that they are in the alphabetical order of their names. When the Massachusetts Institute of Technology's Technology Review printed this answer in its issue of July, 1967, page 10, it added a second answer that had been supplied by a reader named Benson P. Ho. His solution is best explained by his diagram [see Figure 21. T h e digit above the right arm of each V is subtracted from the digit above the left arm. Ifthe result is negative, add 10. T h e result goes under each V. Arrow pairs point to digits that are the sum of the two digits at the back of each arrow. If the sum is greater than 10, subtract 10. Note that the diagonal series of digits, when they are read upward, repeats the original series. It is a remarkable coincidence. O r is it?
ANSWERS Neither of the two numerical oddities are coincidences. S. N. Collings, in The Mathematical Gazette (December, 1971, page 418), generalizes the fact that joining consecutive squares 4 and 9 produces the
COINCIDENCE 9
square 49 as follows: Let (n  1)2and n2 be two consecutive squares. Join them to form a twodigit number in a notation with a base of n2 1. (In the case of 3* the base is 32 1 = 10.) T h e new number will be (n  1)' X (n2 1) n2, which equals the square number (n2  n I)~. Philip G. Smith, Jr., discovered that a reverse procedure always yields the same square. Interpret each of the squares in a base equal to the smaller square plus 1,put the larger of the two squares in front of the smaller and interpret the result in a base equal to the smaller square plus 1.In decimal notation: consecutive squares 9 and 16 join to produce square number 169. If the opposite procedure is followed, the result is 9 followed by 16,with 16 regarded as a single symbol of base17 notation. T h e number's decimal equivalent is (9 X 17) 16 = 169, the same square that was obtained before. O n the surface it seems surprising that both procedures always give the same result, but, as Smith showed, it is merely a special case of the following general theorem. Let x and y be any positive real numbers. If both are expressed in base x 1,andx is appended toy, the value is the same as expressing the numbers in base y 1 and appending y to x. In the first case the value is y(x 1) x, and in the second x(y 1) y. T h e two expressions are clearly equivalent. T h e pattern that Benson P. Ho found for the series 8549176320 is a ho, ho, ho hoax. It is not hard to show that any series of digits ending in 0 , subjected to Ho's procedure, will give the same result.
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ADDENDUM Judith Bronowski wrote to correct my statement that Francis Bacon anticipated continental drift. It is true that in the second book of Novum Organum (Section 28), Bacon spoke of the remarkably similar shapes of the Atlantic coasts of South America and Africa, but his only explanation was that this could "not be attributed to mere accident." T h e earliest known record of explaining this seeming coincidence by assuming that a continent split and the two parts drifted from each other is, according to Bronowski, in a book called The Creation and Its Mysteries Revealed, by Antonio SniderPelligrini (Paris, 1858). T h e book had no influence on geologists. Wegener was apparently the first to suggest continental drift as a serious scientific theory.
CHAPTER ONE
10
BIBLIOGRAPHY "Coincidences." William S. Walsh in HandyBook of Literary Curiosities, pages 170  174. Lippincott, 1892. "The Small World Problem." Stanley Milgram in Psychology Today, May, 1967, pages 61 67.
The Roots ofcoincidence. Arthur Koestler. Random House, 1972. My review of this book is reprinted in my Science: Good, Bad and Bogus, Prometheus Books, 1981, Chapter 22, along with Koestler's rebuttal and my reply to Koestler. See also N. T. Gridgeman's more detailed review in Philosophy Forum, Vol. 14,1975,pages 307  316. The Challenge of Chance. Alister Hardy, Robert Harvie and Arthur Koestler. Random House, 1973.
Incredible Coincidences. Alan Vaughan. Lippincott, 1979. An outstanding instance of a naive book by an occultist with no comprehension of statistics or any awareness ofthe danger of taking anecdotes as scientific evidence. "On Coincidence." Ruma Falk in The Skeptical Inquirer, Vol. 6, 1981 1982, pages 1831. "Mere Coincidence?" Robert A. Wilson in Science Digest, January, 1982, pages 8485, 95. "Against All Odds." Richard Blodgett in Games, November, 1983, pages 14 18. "The Powers of Coincidence." Rudy Rucker in Science 85, February, 1985, pages 5457.
The Magic Numbers of Dr. Matrix. Martin Gardner. Prometheus Books, 1985.
CHAPTER TWO
The Binary Gray Code T h e binary Gray code is fun, For in it strange things can be done. Fifteen, as you know, Is one, oh, oh, oh, And ten is one, one, one and one.
Although the decimal system is now in common use throughout the world, mathematicians and computers often manipulate integers by using other systems, some with such exotic features as mixed bases, negative bases, irrational bases or floating points. One of the most useful of these systemsone with surprising puzzle applications is the Gray code. T h e first puzzle application of a Gray code, which I shall describe below, was in 1872, when a binary version provided an elegant solution to a much older mechanical puzzle. T h e term "Gray," however, derives from Frank Gray, a research physicist at the Bell Telephone Laboratories, who died in 1969. His contributions to modern communication technology were immense. T h e . method now in use for compatible color television broadcasting was developed by Gray (numerologists note!) in the 1930's. In the 1940's he devised what was soon to be called the binary Gray code to avoid the large errors that could arise in transmitting signals by pulse code modulation (PCM). T h e first publication ofthis code was in his U.S. Patent 2632058 (March 17,1953) for a Gray coder tube that eliminated the quantizing grid wires used in early PCM transmission tubes.
CHAPTER TWO
12
Exactly what is a Gray code? It is a way of symbolizing the counting numbers in a positional notation so that when the numbers are in counting order, any adjacent pair will differ in their digits at one position only, and the absolute difference at that position will be 1.For instance, 193 and 183 could be adjacent counting numbers in a decimal Gray code (the middle digits differ by I), but not 193 and 173, nor 134 and 143. There is an infinity ofGray codes, since they apply to any base system and for each base there are many different ways to construct the code. T o appreciate the value of such a system, consider what happens when the odometer of a car reads 9,999 miles. T o register the next mile, five wheels must rotate to show 10,000. Because the wheels move slowly, there is little chance of error. But if counting is recorded electronically at enormously high speeds, when two or more digits change simultaneously the likelihood of producing a false number zooms upward. The probability is greatly reduced if the counting procedure requires only one decision whenever the magnitude to be coded is halfway between two adjacent quantized steps, regardless of whether the magnitude is increasing or decreasing. If the counting is by Gray code, only one digit of the counter changes by only one unit at each step. T h e mileage meter is a familiar example ofwhat are called analogtodigital (AID) converters. A continuous (in this case always increasing) variable, the mileage (or, if you prefer, the number of times the car wheels have rotated) is given a digital output. There are many other control systems in which analogtodigital conversion must proceed at enormously high speed while the variable being measured fluctuates rapidly. Examples include windtunnel simulations of airplanes and guided missiles, and PCM applications where voltages, shaft positions, wave amplitudes of sounds, colors and so on must be translated almost instantly to a digital output signal. At one time a human observer would take pointer readings or inspect a curve on a graph, record the magnitude in digital form and feed this information to a computer. Today the slow and errorprone middleman is eliminated by analogtodigital converters connected directly to the computer. A great increase in accuracy and often a considerable saving in hardware result from counting scales in Gray codes. Binary Gray codes are the simplest. If we limit the code to one digit, there are only 2 l = 2 numbers, O and 1. Disregarding reversals, there is only one Gray code: 0, 1. W e can graph this as a straight line, its ends labeled 0 and 1 [see Figure 3, IeH. T h e Gray code is obtained by moving along the line in either direction. A Gray code for two binary digits has 2' = 4 numbers: 00, 01, 10 and 11.T h e corners of a square can be labeled with these numbers [see Figure 3, middle]. T h e labeling is such that the binary numbers at any pair of adjacent corners differ in only one place. W e can start at any corner and visit all four
THE BINARY GRAY CODE
Figure 3
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Graphs for one (left),two (middle) and threedigit (right) binary Gray codes.
corners by going clockwise or counterclockwise around the square. Ifwe ignore reversals, this produces four Gray codes. T h e line starting at 00 yields the Gray code 00, 01, 11, 10. T h e code is cyclic because the path can return to 00. A Gray code for threedigit binary numbers has Z3 = 8 numbers that can be placed on the corners of a cube [see Figure 3, right]. Adjacent corners have binary triplets that differ in only one place. Any continuous path that visits every corner once only generates a Gray code. For example, the path shown by the dashed line starting at 000 produces 000, 001, 011, 010, 110, 111, 101, 100. This is a cyclic code because the path can return from 100 to 000 in one step. Such paths are called Hamiltonian paths after the Irish mathematician William Rowan Hamilton. As the reader has probably guessed, binary Gray codes correspond to Hamiltonian paths on cubes of n dimensions. A Gray code for fourdigit binary numbers has Z4 = 16 numbers that fit the corners of a hypercube in 4space, for five digits a hypercube in 5space and so on. Interested readers will find this covered in detail in E. N. Gilbert's paper (see the bibliography). Gray codes for other bases correspond to Hamiltonian paths on more complicated ndimensional graphs. T h e number of Gray codes for any base increases explosively as the number of digits increases. The number of Gray codes, even for the binary system, is known only for four or fewer digits. An illfated attempt to obtain the number for five binary digits is recounted in Graph Theory and Its Applications, by Ronald C. Read, who wrote a BFI program for finding the number of Hamiltonian paths on the fivedimensional cube. BFI is Read's acronym for brute force and ignorance. ("It should be BFBI," he has since remarked, "the second B standing for 'Bloody,' but one has to preserve a measure of decorum in published papers.") "These are algorithms," he explains, "devoid of any subtlety whatever, which simply keep thumping the problem on the back until it disgorges an answer." After the program ran for a short time (on a computer in Kingston, Jamaica), a sample of the output was
CHAPTER TWO
14
examined in order to estimate how long the run would be. The guess was 10 hours, and so the computer was set to run unattended overnight. During the night a tropical thunderstorm cut the power supply, and the computer stopped. "Idle curiosity," Read continues, "prompted us to look to see where the program had got to before being so abruptly terminated, and in doing so we discovered that we had made a rather serious error in calculating our previous estimate of the running time. Our revised estimate turned out to be more like ten years!" Read sensibly abandoned the project. T h e problem was not solved until 1980 (see the addendum). For practical purposes it is important to select a Gray code with two desiderata: ( I ) rules for its formation should apply to the entire set of counting numbers; (2) it should have simple conversion rules for translating a standard number to its Gray code equivalent and vice versa. T h e simplest Gray code with both features is called a reflected Gray code. For most mathematicians it is the Gray code. T o convert a standard binary number to its reflected Gray equivalent, start with the digit at the right and consider each digit in turn. Ifthe next digit to the left is even (0), let the former digit stand. If the next digit to the left is odd ( I ) ,change the former digit. (The digit at the extreme left is assumed to have a 0 on its left and therefore remains unchanged.) For example, applying this procedure to binary number 110111 gives the Gray number 101100. T o convert back again, consider each digit in turn starting at the right. If the sum of all digits to the left is even, let the digit stay as it is. If the sum is odd, change the digit. Applying this procedure to 101100 restores the original binary number 110111. Inspection of the numbers from 0 through 42 and their reflected binary Gray code equivalents will show that every two adjacent Gray numbers differ at only one place, and of course the difference is necessarily 1[see Figure 41. It is called a reflected code because the series can be generated rapidly by the following algorithm. Start with 0, 1 as a onedigit Gray code, then reflect (reverse) and append the digits to get 0 , 1,1 , O . Next put 0's in front ofthe first two numbers and 1's in front ofthe last two numbers. T h e result is a twodigit Gray code: 00, 01, 11, 10. T o extend the series to threedigit Gray numbers, reflect the twodigit code 00, 01, 11, 10, 10, 11, 01, 00. As before, put 0's in front ofthe first half of these numbers and 1's in front of the last half: 000, 001, 01 1, 010, 110, 111, 101, 100. This corresponds to a Hamiltonian path starting at 000 on a cube. Proceeding in this fashion, first reflecting the entire series, then adding 0's and 1's on the left, one can quickly generate the reflected binary Gray code to
THE BINARY GRAY CODE
15
FEDCBA
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Figure 4
FEDCBA
21 22 1l l1 11 l0 11 1 23 11100 24 25 1 0° l1o0O1 1 1 10111 26 27 101 10110 100 28 10010 29 1 1 00 01 } 30 1 00 °010 l1 1 10000 1111 31 1110 32 1010 33 110001 110011 34 l o0 l0 l1} 35 110010 1 1000 36 110110 37 1 1 O0 0 o O1 1 38 110101 110100 11011 39 11010 40 11110 41 111101 42 111111 Reflected binary Gray code for 0 through 42.
O} 1 11 10 110
'''1
00001
1101111 1111001
any desired counting number. Note that for each set of ntuplets the code is cyclic in that the first and last ntuplets also differ at only one spot. If the code is used by a counter consisting of wheels, such as the usual mileage meter, the meter can go from its highest number back to 0's with a final unit change ofonly one wheel. In 1872 Louis Gros published in Lyon a brochure on Thiorie du Baguenodier. "Baguenodier" (more commonly spelled "baguenaudier") is the French name for a classic puzzle known in the Englishspeaking world as Chinese rings, although any connection between the puzzle and China is unknown to me. In his brochure Gros applied a binary notation to this puzzle for the first time. The puzzle had been first described in 1550 by Girolamo Cardano in his De Subtilitate Rerum, and it was later analyzed at considerable length by John Wallis in his Algebra in 1693. Many versions of the Chinese rings (the number of rings can vary) are currently on sale around the world. If you are handy with tools, the puzzle can be made with curtain rings, stiff wire and a strip of wood with holes drilled through it [see Figure 51.
CHAPTER TWO
Figure 5
16
Chinese ring puzzle.
T h e object of the puzzle is to free all the rings from the double bar. For a first move the two end rings can be dropped either individually or both at once. T o simplify the solution, we shall assume that only one of the two end rings is dropped at a time. With the exception of those two rings (which can always be taken off or put on simultaneously), a ring will come off or go on only when its immediate neighbor closer to the end is on and all the other rings beyond are off. This is the peculiar feature of the puzzle that makes it so frustrating and repetitious. Let each ring be represented by a binary digit: 1 for on, O for off. The binary Gray number for 42 [see Figure 61 is 111111.Ifwe let this represent the six rings on the upper rod, each Gray number going from 42 back to 0 shows which ring is to be removed or put on to solve the puzzle in a minimum number of moves! For n rings it is apparent that to determine the number of moves required, we simply write n as a Gray number of n units, convert it to standard binary and so obtain the answer. In this case the Gray number 111111 corresponds to 1010 10 in standard binary, which is 42 in decimal notation. (Gros explained all this in a slightly different way, but it amounts to the same thing.) T o find the number by formula, use %(2"  2) when n is even and l/11(2"+'  1)when n is odd. W e have assumed that for each move only one ring is removed or put on. T h e braces in Figure 4 indicate pairs of moves that can be made simultaneously with the two end rings. If these are counted as single moves, the sixring puzzle can be solved in 3 1moves instead of 42. T h e formulas for this "fast way" of solving an nring puzzle are 2"'  1 if n is even and 2"' if n is odd. With a sixring puzzle the slowtofast ratio is 42131 = 1.355;for seven rings it is 85164 = 1.328. T h e ratios continue as 1.338, 1.332, 1.334,. . . . N. S. Mendelsohn has shown that this oscillating series converges rapidly to 1%.
+'
THE BINARY GRAY CODE
Figure 6
17
First six positions for solving ring puzzle using the Gray code.
Twentyfive rings require 22, 369, 621 steps. Assuming that a skilled operator can do 50 steps a minute, he could solve the puzzle the slow way, working 10 hours per day, in a little more than two years. By doing it the fast way, however, he could cut the time by about half a year. Jesse R. Watson of Altadena, Calif., headed a firm called Watson Products that manufactured a handsome, sixring, aluminum version of the rings in the early 1970's. In his instructions he asked the following question: Suppose the initial position for an nring puzzle has the last ring (the one nearest the handle) on and all other rings off. Watson calls this the position of "maximum effort" because it requires more moves than any other position to take all the rings off. Assuming that the slow method is used, what simple formula gives the required minimum number of moves? T h e binary Gray code also solves the wellknown Tower of Hanoi puzzle, in which n disks of diminishing sizes are stacked in a pyramid. T h e problem is to transfer them one at a time to a second spot, using a third spot as a temporary resting place with the proviso that no disk be placed on top of a smaller disk. (See Chapter 6 of The Scientijc American Book of Mathematical Puzzles €3 Diversions.) T o solve this puzzle for five disks, label the disks of the initial pyramid, starting with the smallest, from A to E. Label the columns of Figure 4 from A to F as shown. Take the Gray numbers in sequence. At each step move the disk that corresponds to the column in which there is a change of digit. T h e sequence begins ABACABAD. . . . O n every move a disk can be transferred to only one spot. T h e sequence solves the puzzle in 2n  1moves, which in this case is 31. Rules for converting numbers in other base systems to reflected Gray numbers are simple generalizations of the rules for binary numbers. (There are several general conversion procedures, but I give the simplest here.) If the base is even, the rules are the same as for the binary system, except that when a digit is altered it is changed to its "complement" with respect to n  1 when n is the base, that is, to its difference from n  1. In the binary system, n  1 = 1, so
CHAPTER TWO
18
GRAY
GRAY
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Figure 7
13 12 11 10 20 21 22 23 24 25 26 27 28 29 39
Reflected decimal Gray code
that this means a simple change of 0 to 1 or 1 to 0 . In the decimal system, numbers are complemented with respect to 9 (that is, subtracted from 9). Therefore to convert a decimal number to a Gray number take each digit in turn beginning at the right. If the next digit to the left is even, leave the former digit unchanged. If the left digit is odd, complement the former digit. For example, 1972 becomes 1027. T o convert back to the decimal system, work with sums. If all digits to the left have an even sum, let the digit stand. If the sum is odd, subtract the digit from 9. Only a slight modification of rules is required for numeral systems with an odd base. In such cases the sum rule applies to conversion in either direction. In  the ternary system, for instance, complementation is with respect to 2. Regardless of which way you convert, complement when the sum on the left is odd; otherwise let the digit stand. Ternary Gray numbers, in counting order, are 0 , l . 2 , 12, 11, 10, 20, 21, 22, 122, 121, 120, . . . . In all bases, Gray counting numbers of the reflecting type (unless otherwise specified, these are considered the Gray numbers for a given base) are quickly determined by generalizing the procedure given for binary numbers. This is best explained by using the decimal systemis an example [see Figure 71. Note that the unit's column begins with the sequence 0 through 9; then it proceeds from 9 through 0 , then from 0 through 9 and so on. In the 10's column, ten 0's (not shown) are followed by ten l ' s , then by ten 2's, ten 3's and so on through
THE BINARY GRAY CODE
19
ten 9's until 99 is reached. Now the doublets are reflected after every 100 steps, and in the third column from the right a hundred 0's are followed by a hundred l ' s , then by a hundred 2's and so on until 999 is reached. T h e reader should have little difficulty applying this procedure to other base systems. In the ternary system, for example, reflections occur in the right column every third step, in the next column every ninth step, in the next column every 27th step and so on through increasing powers of 3. Because Gray codes are relatively unknown to students of recreational mathematics, I suspect they have many puzzle applications other than the ones given here. I would be glad to hear from readers who know of recreational uses for Gray codes with bases greater than 2.
ANSWER
zn
From a "maximum effort" position (only the last ring is on the bar), 1 moves are required to remove all the rings by the slow method. Numbers of this form are called Mersenne numbers. The same formula gives the number of moves required for transferring n disks in the Tower of Hanoi puzzle. Henry E. Dudeney, in his discussion ofthe puzzle (see the bibliography), has this to say about a "maximum position" task. "If there are seven rings and you take offthe first six, and then wish to remove the seventh ring, there is no course open to you but to reverse all those 42 moves that never ought to have been made. In other words, you must replace all the seven rings on the loop and start afresh!"
ADDENDUM T h e limerick at the head of the chapter is only half anonymous. It is my variation on the following anonymous tribute to the binary system: T h e binary system is fun, For with it strange things can be done. Two as you know Is a one and an oh, And five is one hundred and one.
Like so many mathematical ideas, the origin of the Gray code fades into history. George R. Stibitz, a physiologist at Dartmouth Medical School, sent me a copy of his 1943 patent (No. 2,307,868),applied for in 1941 when he was with Bell Laboratories. It describes a counting device using elastic balls and magnets. Electric pulses shift the balls back and forth, varying their positions in
CHAPTER TWO
20
accord with the cyclic Gray code. Recalling this patent prompted Stibitz to write: An ingenious fellow one day Wrote numbers a newfangled way. As earlier had Stibitz, But that name inhibits Historians who call the code "Gray."
So far as I have been able to learn, the earliest technical use of the Gray code was by mile Baudot (1845  1903), a French engineer who applied the cyclic code to telegraphy. For details and references see "Origins ofthe Binary code," by G. G. Heath in Scientijc American, August, 1972, pages 7683. T h e term "reflected code" was first used by Gray in his 1953 patent. "Because this code in its primary form may be built up from the conventional binary code by a sort of reflection process and because other forms may in turn be built u p from the primary form in similar fashion, the code in question, which has as yet no recognized name, is designated in this specification and in the claims as a 'reflected binary code.' " Sydney N . Afriat, an economist at the University of Ottawa, has written an entire book about the Chinese rings, The Ring of Linked Rings, published in London in 1982 by Duckworth and Company. Afriat also discusses the Tower of Hanoi and how the two puzzles are solved by the Gray code. The book includes computer programs for both puzzles and an extensive bibliography. In recent years several mechanical puzzles have been marketed that use a Gray code for their solution. A notable example is T h e Brain, invented by computer scientist Marvin H . Allison, Jr., and made in the 1970's by a company called MagNif. It consists of a tower of eight transparent plastic disks that rotate horizontally around their centers. The disks are slotted, with eight upright rods going through the slots. T h e rods can be moved to two positions, in or out, and the task is to rotate the disks to positions that permit all the rods to be moved out. T h e Gray code supplies a solution in 170 moves. A curious puzzle called Loony Loop its complicated history would require a chapter consists of four intertwined steel loops and a ring of nylon cord that seems permanently captured by the loops. T h e task is to free the nylon cord. T h e puzzle generalizes to n metal loops and is solved by applying a ternary Gray code to a sequence of moves. Many readers reminded me of the similarity of Gray codes to a word puzzle invented by Lewis Carroll, called Doublets, better known today as Word Ladders. T h e idea is to change a word to one of the same length by altering one
THE BINARY GRAY CODE
21
letter at a time, forming a different word at each step, in a minimum number of steps. (See the chapter on Carroll in my New Mathematical Diversions from Scientific American.) As often noted, word ladders resemble the way the genetic code is altered by evolutionary mutations. On the relation of the Gray code to wordladder problems, see “The Arithmetic of Word Ladders,” by Rudolph W. Castown, in the quarterly journal Word Ways, Vol. 1, August, 1968, pages 165–169. The Gray code solves many brainteasers that appear from time to time. A typical example is the switching puzzle on page 26 of The Surprise Attack in Mathematical Problems, by L. A. Graham (Dover, 1968). An earlier instance is Problem 319, solved in American Mathematical Monthly, December, 1938, pages 694–696. Imagine a light bulb connected to n switches in such a way that it lights only when all the switches are closed. A push button opens and closes each switch, but you have no way of knowing which push opens and which closes. What is the smallest number of pushes required to be certain you will turn on the light regardless of how the switches are set at the outset? This device, by the way, is the basis of an amusing trick, unpatented and inventor unknown, that is currently sold in magic stores under various trade names. Louis Tannen’s Magic Studio in Manhattan sells it under the name Electronic Monte. There are three push buttons and one light. The magician demonstrates how a single button seems to control the light, but the control mysteriously changes from one push button to another, like the pea in a threeshell game. Many legends tell how the rings puzzle was invented in ancient China, but the world’s expert on early Chinese inventions, Joseph Needham, finds no evidence of its Asian origin. (See his Science and Civilization in China, Vol. 3, page 111.) The Japanese became so intrigued by the puzzle in the 17th century that they wrote Haiku poems about it, and symbols of the linked rings appeared on heraldic emblems. There is a large literature on the puzzle in both China and Japan, but I know of no published bibliography. The rings were sometimes used in Europe as a whimsical locking device for bags and chests. In England the puzzle was usually called the “tiring irons,” probably because it is tiring to solve, especially if the rings are large and heavy. According to the Oxford English Dictionary, it was earlier called “tarrying irons,” perhaps because one is long delayed in solving it. Among its quotations, the OED gives the following 1782 doggerel: Have you not known a small machine Which brazen rings environ, In many a country chimney seen Yclep’d a tarringiron?
CHAPTER TWO
22
T h e puzzle has been sold around the world in hundreds offorms. A beautiful handcarved ivory version with nine rings (from the puzzle collection of Tom Ransom of Toronto) provided the cover for the May 1977 issue of Computer. T h e puzzle illustrates the lastin  firstout principle of stack machines, the topic of five articles in the issue. At the other extreme is a small sevenring version I found advertised in a 1936 Johnson Smith and Company catalog, where it is called the "Chinese Ringbar Puzzle"; price: 15 cents. "This is an extremely difficult puzzle," the description reads, "yet very simple when you are familiar with the method. . . . You may try forever and not be able to remove the bar from the rings. Just as you think you are getting it done, you are further off the solution than ever, and you have to give u p in despair." An elaborate electronic version of the puzzle, with eight lights and eight push buttons, is the topic of "The Princeps Puzzle," by James W. Cuccia, in Popular Electronics. T h e article gives detailed instructions on how to make the thing. (I am indebted to Dr. Burton J. Bacher for calling this article to my attention.) Science and Invention, September, 1927, page 397, describes a "marvelous escape trick" in which a "fair damsel" is shackled on the stage in the manner shown in Figure 8. Diagrams show how the lady is released by manipulating the rings around her arms and legs in the manner of the Chinese puzzle. This preposterous stage trick, the article says, was invented by one Theodore P. Brunner of Los Angeles, who has it protected by U.S. Patent 1,625,452. Since my column on Gray codes was published in Scientijic American in 1972, the number of 5bit codesthe same as the number of Hamiltonian paths on a fivedimensional cube has been determined. A good upper bound has been established for the 6bit Gray code. At the 1980 IEEE International Conference on Circuits and Computers, at Port Chester. N.Y., a paper was presented titled "Gray Codes: Improved Upper Bounds and Statistical Estimates for n > 4 bits," and published in 1983 (see the bibliography). T h e authors were Jerry Silverman, Virgil E. Vickers and John L. Sampson, electrical engineers at the Rome Air Development Center, Hanscom Air Force Base, Chicopee, Mass. Their estimates for the 5 and 6bit codes were based on Monte Carlo techniques. T h e authors open with a succinct definition of an nbit Gray code as "a list of all the 2" binary ntuples ordered so that adjacent elements differ by a change in only one bit." They point out that such codes are widely used in AID conversion, shaft encoding, codes for data retrieval, control mechanisms, switching and network theory and experimental design. Although the reflected binary code is the most widely used, other types of Gray codes are preferred for special purposes. Finding a formula for the number of nbit Gray codes as a function of n remains a difficult unsolved combinatorial problem.
THE BINARY GRAY CODE 23
Figure 8
An absurd stage trick using the Chinese rings.
Their statistical estimates agreed with the exact value for the 4bit code to within 0.06 percent. T o test their estimate for the 5bit code they made a precise calculation on a PDP11 computer. At first they feared the running time would be about 11years, but by taking advantage of symmetries and a "look ahead" method that predicts deadend branches, they were able to reduce the running time to 750 hours. "We are happy to report," the three researchers told me in a 1980 letter, "that a fivedimensional fly can walk along the edges of a fivedimensional cube in exactly 187,499,658,240 ways." W e need to make clear just what this number counts. It allows the fly to start at any corner of the hypercube and trace a Hamiltonian path that ends at any other corner. Reversals of each path are included. (Ifreversals are excluded, the number must be halved.) The number of Hamiltonian circuits paths starting anywhere but ending on a corner adjacent to the starting corneris 58,O18,928,640.This, too, includes reversals. If you want the corresponding figures for paths and circuits that start only at the corner taken as zero, then each of the above figures must be divided by 2" = 32, where n is the dimension number.
CHAPTER TWO
24
Because my column did not disclose the numbers for Hamiltonian paths and circuits (including reversals) on lowerorder cubes, I give them below: n
Hamiltonian circuits
Hamiltonian (noncyclic) paths
T h e Hanscom researchers were the first to publish the figures for the 5bit Gray code, but they were not the first to find them. After my column appeared, the same results were sent to me in the fall of 1972 by David Vanderschel of Houston, Alex G . Bell and Peter Hallowell of the Rutherford High Energy Laboratory in Chilton, England, and Steve Winker of Naperville, Ill. None published their results, but I reported them in my column for April, 1973. The fact that all four programs agreed is strong evidence that the numbers are accurate. T h e number of 6bit Gray codes remains unknown. T h e Hanscom researchers estimate it as close to 2.4 X a number so large that it is probably not possible to determine it precisely in any reasonable computer time, unless, of course, someone discovers a formula or some new algorithm shortcuts. I don't know if anyone has noticed this before, but it occurred to me that the number of Gray ternary codes of n digits is equal to the number of Hamiltonian paths on ncubical lattices with three points on each edge and the faces toroidally joined. This is best explained with examples. There are six onedigit ternary Gray codes. W e represent them on the single edge of a onedimensional "cube" by three points, then close the ends of the line to make a circle, as shown in Figure 9a. By starting at any point and counting reversals, we see that the six Hamiltonian paths provide the onedigit codes 012, 120, 201 and their reversals; 210, 021 and 102. T h e twodigit ternary Gray codes are obtained from the square lattice shown in Figure 9b. Its nine points are labeled with the nine twodigit combinations of 0 , l and 2. Points on each side ofthe square are connected to the three points on the opposite side. T h e graph can, of course, be drawn on a torus with three parallel lines going around it one way and three circling it the other way. T h e number of Gray codes is the number of Hamiltonian paths on this graph. Already it is not easy to see how to count the paths systematically, and I have made no attempt to do so. For threedigit ternary codes we go to the cubical lattice shown in Figure 9c, its 27 points labeled with the 27 threedigit combinations of 0, 1 and 2. As
THE BINARY GRAY CODE
Figure 9
25
Ternary Gray codes as Hamiltonian paths.
before, imagine each point connected by a line (not shown) to the corresponding point on the opposite face. T h e procedure clearly generalizes to ndimensional hypertoruses. Gray codes with bases higher than 3 can be similarly generated by Hamiltonian paths on more complicated hyperlattices.
CHAPTER TWO
26
BIBLIOGRAPHY Gray Codes "Reflected Number Systems." Ivan Flores in IRE Transactions on Electronic Computers, Vol. EC5, 1956, pages 7982. "Affine Mary Gray Codes." Martin Cohn in Information and Control, Vol. 6, 1963, pages 7078. "Digital Transmission of Analog Signals." William R. Bennett in Introduction to Signal Transmission. McGrawHill, 1970. "Using the Decimal Gray Code." N. Darwood in Electronic Engineering, February. 1972, pages 2829. "On the Use of Binary and Gray Code Schemes for ContinuousTone Picture Transmission." E. S. Deutsch in Pattern Recognition, Vol. 5 , 1973, pages 121  132. "Distance2 Cycle Chaining of Constant Weight Codes." D. T. Tang and C . N. Liu in IEEE Transactions on Computers, Vol. 22, 1973, pages 176 180. "Efficient Generation ofthe Binary Reflected Gray Code and Its Applications." J. R. Bitner, G. Ehrlich and E. M . Reingold in Communications A C M , Vol. 19, 1976, pages 517521. "Gray Codes." Edward M . Reingold, Jurg Nievergelt and Narsingh Deo in Combinatorial Algorithms, pages 173  188. PrenticeHall, 1977. "A Technique for Generating Gray Codes." J. E. Ludman and J. L. Sampson in Journal of Statistical Planning and Inference, Vol. 5. 1981, pages 171  180.
The Chinese Rings "Le Jeu du Baguenaudier." Edouard Lucas in Re'cre'ations MathPmatiques, Vol. 1, Chapter 7. Paris, 1883. "Der Baguenaudier." W .Ahrens in Mathematische Unterhaltungen und Spiele, Vol. 1. Berlin: Druck und Verlag von B. G . Teubner, 1910. "The Tiring Irons." H . E. Dudeney in Amusements in Mathematics, Problem 417. Nelson. 1917. "Some Binary Games." R. S. Scorer, P. M . Grundy and C. A. B. Smith in Mathematical Gazette, Vol. 28, 1944, pages 96  103. T h e rings puzzle is generalized to a tier ofk rods. "Chinese Rings." Maurice Kraitchik in Mathematical Recreations. Dover, 1953 "The Icosian Game and the Tower of Hanoi." Martin Gardner in The Scient$c American Book of Mathematical Puzzles and Diversions. Simon and Schuster, 1959.
THE BINARY GRAY CODE
27
"Problems and Puzzles." Joseph Needham, Science and Civilization in China, Vol. 3 , Section 19. Cambridge University Press, 1959. "An Old Puzzle." E. H. Lockwood in Mathematical Gazette, Vol. 53, 1969, pages 362  364. "The Princeps Puzzle." James W. Cuccia in Popular Electronics, Vol. 34, 1971, pages 2632. "Chinese Rings." W. W. Rouse Ball in Mathematical Recreations and Essays, 12th edition, edited by H. S. M. Coxeter. University of Toronto Press, 1974. Hamiltonian Paths on the nCube
"Gray Codes and Paths on the nCube." E. N. Gilbert in The Bell System Technical Iournal, Vol. 37, 1958, pages 815826. "Graph Theory Algorithms." Ronald C. Read in Graph Theory and Its Applications, edited by Bernard Harris. Academic, 1970. "Cyclic Codes in AnalogtoDigital Encoders." Charles F. Cole, Jr., in Computer Design, May, 1971, pages 107  112. Shows how Gray codes can be counted by tracing Hamiltonian paths on twodimensional matrices known in network theory as Karnaugh maps. "A Technique for Generating specialized Gray Codes." Virgil E. Vickers and John
L. Silverman in lEEE Transactions on Computers, Vol. C29, 1980, pages 329331. "Statistical Estimates of the nBit Gray Codes by Restricted Random Generations of Permutations of 1 to 2"." Jerry Silverman, Virgil E. Vickers and John L. Sampson in IEEE Transactions on Information Theory, Vol. IT29, 1983, pages 894901. "A CubeFilling Hilbert Curve." William J. Gilbert in Mathematical Intelligencer, Vol. 6, 1984, page 78. Shows how the 3bit reflective binary Gray code can be used to start a sequence that generates a Hilbert curve that will at the limit completely fill an ndimensional cube.
CHAPTER THREE
Polycubes In 1958 Piet Hein’s Soma cube was first introduced to U.S. puzzle buffs in my September Scientific American column. (The column is reprinted in The 2nd Scientific American Book of Mathematical Puzzles & Diversion.) The puzzle has since been sold around the world under a variety of trade names. The only authorized version is marketed in the U.S. by Parker Brothers, with an informative booklet written and illustrated by Piet Hein. Three issues of Soma Addict, a newsletter edited by Thomas V. Atwater, were published, as well as many articles on Soma in mathematical journals. The Soma pieces are a subset of what have been called polycubes. These are solid figures created by joining unit cubes at their faces. Like their flat cousins the polyominoes, they pose an extraordinarily difficult combinatorial problem: Given n cubes, is there a formula for calculating the number of distinct polycubes of order n? If so, it has not yet been found, although there are, of course, recursive procedures by which all polycubes of order n can be constructed: Simply add a cube in all possible ways to each polycube of order n – 1 and eliminate duplicates. Since there is no way to “turn over” an asymmetric polycube in 4space analogous to the way an asymmetric polyomino can be reversed in 3space, mirrorimage pairs of polycubes are considered different. It is obvious that for orders 1 and 2 only one polycube is possible for each and that three unit cubes can form two polycubes. It also is easy to determine that there are eight tetracubes and 29 pentacubes. Several computer programs have verified a hand computation, first made by David Klarner, that there are 166 hexacubes. As far as I am aware, the number of heptacubes is still undetermined. The Soma cube consists of the seven irregular shapes [See Figure 10] that can be formed by combining three or four unit cubes — all nonconvex polycubes of orders 1 through 4. There are 240 distinct ways (not counting rotations and reflections) that the seven pieces will form a 3by3by3 cube. This was first
POLYCUBES
29
Figure 10
The seven Soma pieces.
determined by John Horton Conway and M. J. T. Guy and has since been verified by many computer programs. Parker's Soma booklet states that Conway and Guy used a computer for their work an error I am now happy to correct. As Conway puts it in a letter, he and Guy, both mathematicians at the University of Cambridge, obtained the 240 solutions by hand "one wet afternoon" when they had no morepressing chores. "I think for a puzzle the size of Soma," Conway adds, "it's an admission of defeat to use a computer. If you find the right way of organizing the material, it should take less time to do the whole thing by hand than it does to program the machine." By first establishing a few ingenious theorems (some of which were found by Guy's father, R. K. Guy) and using a parity coloring technique, they were able to check all possibilities with great efficiency. Conway and Guy later discovered that if you begin with any of 239 solutions (one solution is an anomaly), all of the others can be obtained in 238 steps by altering the position of no more than three pieces at each step. Conway has drawn a large graph (which he calls the Somap) showing how the 239 solutions are linked to one another and giving each solution a concise notation, called its "somatype." T h e map does not give any one solution, but once you have built the cube in any of the 239 ways, the map enables you to transform it to all the others by moving two or three pieces at a time. T h e map is too complex to reproduce here, but you will find it on pages 802  803 of Winning Ways, Vol. 2, by Elwyn R. Berlekamp, John H. Conway and Richard Guy (Academic, 1982). T h e Soma cube's popularity flows from the enormous variety of pleasing shapes that can be made with its pieces and from the many clever ways of proving that certain 27cube shapes are impossible. It is not, however, the first
CHAPTER THREE
Figure 1 1
30
Polycube pieces for the Diabolical cube.
polycube dissection of the order3 cube to be marketed as a puzzle. A sixpiece set was sold in Victorian England under the name of the Diabolical cube [see Figure 11, top]. (Its pieces are reproduced on page 108 of Puzzles Old and New, by "Professor Hoffmann," published in London in 1893.) I do not know how many basic solutions the Diabolical cube has, but perhaps a reader can tell me. I found only eight. T h e pieces can be cut from wood or made by gluing together alphabet blocks. As Piet Hein has noted, the unknown inventor surely intended a dissection of the cube into a set of "flat" polycubes containing one each of orders 2 through 7. Another dissection of the cube into six polycubes was made by the Polish mathematician J. G. Mikusiriski [see Figure 12, middle]. It appears in Hugo Steinhaus's Mathematical Snapshots (Oxford University Press, 1950). These pieces are currently on sale here and abroad under several trade names. There are just two solutions, both difficult to find. Still another interesting cube dissection, suggested by Thomas H. O'Beirne of Glasgow, is to cut the order3 cube into nine tricubes, all shaped like the 3piece of the Diabolical cube. Random attempts to build a cube with the nine tricubes are likely to be very frustrating unless you hit on a systematic procedure.
Figure 12 Polycube pieces for J. G. Mikusinski's cube
Gorilla
Camel
Swan
Goose
Tyrannosaurus
Bird
Duck
Giraffe
Figure 13 Soma animals created by Rev. John W. M. Morgan Nine animals from a zoo of several dozen Soma figures created by Rev. John
W. M . Morgan, vicar ofSt. Matthew's Church in Luton, England, are shown in Figure 13.T h e animals all have bilateral symmetry except for the giraffe, whose head leans to one side (he is thinking), and the dog, whose hidden rear portion violates symmetry. T h e bird actually will perch on one leg as shown.
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Figure 14
32
Soma structures with hidden holes: (a) penthouse, (b) staircase and (c) tower
Three Soma structures of a delightful new type were created by Benjamin L. Schwartz of McLean, Va. [see Figure 14]. The penthouse has a cubical hole at its center and is not hard to construct. The tower is flat on its two hidden sides and has three interior holes. The stairway also has three interior holes. The last two are difficult to build. In both cases the three holes are inside, invisible from all angles. Another pleasant exercise is to construct Schwartz’s three figures with the six Diabolicalcube pieces. None is possible with the holes on the interior, but each can be made with one or more holes at the back, so that the structures appear as shown in the illustration. The notorious wall in Piet Hein’s instruction booklet is an insoluble Soma problem [see Figure 15]. Many impossibility proofs have now been found, but
Figure 15
The impossible wall
POLYCUBES 33
the simplest (discovered independently by many Soma addicts) is based on the wall's 10 corner cubes, shown shaded in the illustration. If each Soma piece is considered in turn, it is apparent that five of the pieces can provide only one corner cell each and that the other two can provide no more than two each. All together, therefore, the pieces can supply a maximum of nine corners. Since there are 10 corners, the wall is impossible. It is possible, however, to build a wall that from the front looks exactly like the one in the illustration. Ifthe wall is viewed from behind, however, the hidden corner (indicated by the arrow) is missing, and an extra cube protrudes at some other spot. The corner proofof impossibilityapplies also to the six pieces of Mikusiriski's cube but not to the Diabolical pieces. Unfortunately, they will not make a genuine wall either, and readers may enjoy proving it by a different technique. The Diabolical pieces will, however, like the Soma pieces (but not Mikusiriski's), make an ersatz wall that appears genuine from the front. This is a harder task than forming the Diabolical cube. There are several ways to do it with one hole hidden below the top center corner and one backprojecting cube at the base, where it is hard to see even when looking downward from the front. A notsofunny joke to play on a victim is to let him see a false wall from the front (formed by either Soma or Diabolical pieces), knock the wall apart and then offer him $50 if he can rebuild the structure (with no holes, of course) within three hours. The building of fake structures opens up numerous amusing possibilities. One can build Soma bricks that are 3 by 3 by 4 or 2 by 3 by 6, that appear solid but are actually hollow in back like the fa~adesof buildings on a movie set. A spurious 2by2by8 tower can even display two extra cubes on top. A 1by4by6 Soma wall, standing on edge, has three invisible cubes projecting from the back. Of course, gravity must be taken into account in problems of this type because the structures should be capable of standing alone, without the aid of adhesive or concealed supports. Many people have worked on structures formed from larger sets of polycubes. The eight tetracubes were manufactured in Hong Kong in 1967 (by E. S. Lowe Co., Inc.) and marketed as the Wit's End puzzle. The set came boxed as a 2by2by8solid. A 2by4by4 solid also is possible. Indeed, a group in the Artificial Intelligence Laboratory at the Massachusetts Institute of Technology used a computer to show that it had 1,390 basic solutions. Both of these solids are enlarged replicas of two of the tetracubes, and enlarged replicas of the remaining six tetracubes can also be made. The 29 pentacubes are the subject of U.S. Patent 3,065,970,November 27, 1962, issued to Serena Sutton Besley. Unfortunately, no rectangular solid has 5 X 29 = 145 unit cubes, but by adding a duplicate pentacube, Mrs. Besley
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34
Figure 16
The solid pentominoes.
obtained 150 unit cubes. The 30 pieces will form bricks of 5 by 5 by 6, 3 by 5 by 10, 2 by 5 by 15 and 2 by 3 by 25. Klarner had earlier found that if the 1by1by5 piece is omitted, the remaining 28 pentacubes will form two separate 2by5by7 solids. Two solutions are given by Solomon W. Golomb in his Polyominoes, page 118. Other problems devised by Klarner, using 28 or fewer pentacubes, are in Golomb’s book on pages 159–160. If the 12 pentominoes are given a unit thickness, the set is known as the solid pentominoes [see Figure 16]. Golomb introduces this popular set of polycubes on page 116 of his book and gives additional problems with them on pages 158–159. The set will form enlarged replicas of ten of the pieces. When Golomb’s book appeared, the W and X pieces had been proved impossible, but replicating the F piece (sometimes called the R piece) remained undecided until 1970. It was solved by J. M. M. Verbakel of the Philips Research Laboratories in the Netherlands. It is not known if his solution (see the bibliography) is unique. C. J. Bouwkamp, associated with the same laboratory, reported in 1969 (see the bibliography) on his computer programs that produced all the solutions for packing the 12 solid pentominoes in boxes of 2 by 3 by 10, 2 by 5 by 6 and 3 by 4
POLYCUBES
35
by 5. T h e number of basic solutions is 12, 264 and 3,940. respectively. Bouwkamp's paper gives the 12 solutions for the 2 by 3 by 10 and comments on some of their unusual properties. In July, 1967, the Technological University of Eindhoven published Bouwkamp's 310page Catalogue of Solutions ofthe Rectangular 3 X 4 X 5 Solid Pentomino Problem. An unusual task that links the solid pentominoes with the Soma puzzle has been proposed by J. Edward Hanrahan of La Mesa, Calif. He reports that it is possible to form 4by4by2 solids with the Soma pieces so that on the upper 4by4 layer there will be five cubical holes joined to form hollow molds for each of the 12 solid pentominoes except the I pentomino, which is obviously too long to fit. Working with cubical holes suggests many curious and unsolved polycube questions. What, for example, is the largest volume of empty space that can be put inside a solid formed with a specified set of polycubes? "Inside" can be defined in various ways. What is the maximum number of unit holes that do not touch one another or touch the outside surface (under various definitions of "touch")? Here is an intriguing, unpublished and unsolved hole problem that can be worked on with a set of either the flat pentominoes or the solid ones. Stephen Barr of New York (not Stephen Barr the writer, who lives in Woodstock, N.Y.) recently set himself the task of creating a flatpentomino pattern having the maximum number of unit holes that do not in any way touch the perimeter or one another. (Each hole must be surrounded by eight squares.) His best result, 12 holes, is shown in Figure 17 in one of several solutions. It can be proved that 14 holes are impossible. I leave it to readers to settle the question of whether or not a pattern with 13 holes can be achieved.
Figure 1 7 T h e maximumhole problem
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Figure 18
36
Solutions to the Soma problems.
ANSWERS Solutions to the Soma tasks of building the penthouse (with one interior hole) and the tower and stairsteps (each with three inside holes) are shown in Figure 18. Numerals indicate the pieces as they are numbered in Figure 10. Thomas H. O’Beirne’s simple procedure for building the 3by3by3 cube with nine bent tricubes is to use six of them to make three 1by2by3 slabs. The remaining three tricubes are piled into a stack of height 3; then the slabs are placed vertically [see Figure 19]. The picture is a view of the cube from above. I said that the number of heptacubes had not been calculated. I have since learned from David Klarner and C. J. Bouwkamp that an ALG0L60 program, written in 1969 by A. J. Dekkers at the Philips Research Laboratories in the Netherlands, found 210 – 1 = 1,023 heptacubes. This was confirmed in 1972 with a program written by Timothy L. Bock, of Oberlin, Ohio. The results of an earlier program were proved faulty by Klarner’s father, who had built a set of wooden heptacubes that included several the program had missed. Klarner assures me that the complete set of heptacubes will pack a 2by6by83 box, but whether it packs a 3by4by83 box is not yet known. Bouwkamp, who also works at the Philips Laboratories, informs me that he wrote a program in 1970, proving that J. M. M. Verbakel’s way of replicating
POLYCUBES 37
Figure 19 Solution to the tricube problem.
the Fpentacube with the 12 solid pentominoes is unique. "It is understandable," Bouwkamp comments, "that in Golomb's book the replication of this ~entacubewas left undecided, and most remarkable that Verbakel hit on it by trial and error."
ADDENDUM Wade E. Philpott, of Lima, Ohio, was the only reader who sent all 13 solutions to the Diabolical cube. I once had occasion to show this puzzle to John Horton Conway of the University of Cambridge. He mentally labeled the pieces with a checkerboard coloring; he then began testing the pieces rapidly, talking out loud and occasionally scribbling a note. It was like watching Bobby Fischer play blitzkrieg chess. About 15 minutes later he announced that there were just 13 solutions. T o distinguish them, designate each piece of the Diabolical cube by the number of unit cubes it contains. There are three ways in which the two largest pieces, 6 and 7, can go: 1. Parallel and side by side. When properly placed, with the 5piece wrapped around a projecting cube of 6, the 4piece can go in three places. There are five solutions. 2.
Parallel but on opposite sides of the cube. There are two solutions.
3. Perpendicular to each other. Crossing in one way yields four solutions, another way two, or six solutions in all.
Philpott also sent a proof that a pattern of 14 unit holes, each surrounded by eight cells, cannot be achieved with the 12 pentominoes. The proof establishes that at least 59 squares are needed to surround 14 holes. On all such patterns
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38
a
b
Figure 20 Symmetrical solutions to the 13hole problem.
each pentomino, except the P and W pieces, will fit. Adding a 60th cell will accommodate only one of the two pieces, proving that the 60 cells of the pentomino set are not enough. Essentially the same proof had earlier been formulated by Joseph Madachy. Readers too numerous to mention sent 13hole solutions to the problem. The beautifully symmetrical one shown in Figure 20a was found only by Andrew L. Clarke of Freshfield, England. It so intrigued C. J. Bouwkamp that he wrote a computer program to see if the pattern was unique. He found just one other solution [see Figure 20b], except for rotations and reflections. If the conditions allow holes to touch the border and also one another at their corners, how many holes are possible? The maximum is 18. The pattern shown in Figure 21, first discovered by Christer Lindstedt of Sweden, may be unique except for a trivial shift of the straight pentomino. If the holes are not restricted
Figure 21
18unit holes
to unit squares, there are many 18hole solutions. (See "Pentomino Problem," in Journal of Recreational Mathematics, Vol. 17, No. 3, 1984  1985, pages 220  224.) T h e solution I gave for the Soma penthouse with the interior hole is not very stable. John Conway informed me that pieces 4 , 5 , 6 and 7 can each be used to make the projecting "penthouse," and that the most stable configuration is obtained by forming the cube shown in Figure 22, removing the 7piece, inverting it, and replacing it. The structure is so stable that if you turn it upside down and put a book on top, it will balance on its projecting cube. This solution was also sent to me by Geof D. Clayton, of Beaverton, Ore.
Figure 22
How to make a stable penthouse
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40
Figure 23 The Lesk cube.
I was wrong in saying that the dog in the Reverend John Morgan’s Soma zoo had an asymmetrical rear portion. Peter Neuret of West Germany, sent a symmetrical solution. “My dog was infuriated to read that his hidden rear portion violates symmetry,” Morgan wrote. “Just come over here and say that again to his face.” Morgan sent two symmetrical solutions. David Bird of England, raised the interesting question, What is the lowestorder polycube that contains a unit hole completely surrounded? The answer is an order11 polycube. Six unit cubes are needed to cover the hole’s six sides, and five more are required to join them. Note that even if we exclude higherorder polycubes with no interior holes, there are surely structures interlocked in such a way that they cannot be built without going through a fourth dimension. I have no idea what the simplest example would be. Mathematical Digest, a school periodical in Christchurch, New Zealand, in issue No. 58 (1978) introduced the six polycubes shown in Figure 23. The editors call it the Lesk cube after its designer Lesk Kokay, who had been seeking a dissection of the 3by3by3 cube into six polycubes that would form the cube in only one way. Unfortunately, it has at least three solutions. Is there a six or sevenpiece dissection with a unique solution? If so, it has not come to my attention. In 1973 an order3 cube with, as I recall, seven pieces was on sale in the U.S. under the trade name Qube. The box stated that it had only one solution, but this was achieved by a blackandwhite checkerboard coloring of tsshe pieces and the requirement that the cube be similarly colored. Tom Marlow wrote from England to say that the number of hexacubes and heptacubes were known as far back as 1948. In The Fairy Chess Review, Vol. 7, 1948, page 8, Dr. J. Niemann gave the number of heptacubes as 1,023, along with a neat system for classifying them. His figure for the hexacubes was 167, but this was corrected to 166 in a later issue. Sets of the 12 solid pentominoes have been marketed both here and abroad
POLYCUBES 41
under various trade names. In the U.S. a handsome polished hardwood set is available from Kadon Enterprises, 1227 Lorene Drive, Pasadena, Md. 21122. It is called Quintillions and comes with a 9by12 checkerboard on which games can be played with the pieces. T h e other 17 pentacubes (those that are not "flat") are also available from Kadon as Super Quintillions. T h e company also sells "QuintArt" sculptures, produced by bonding together Quintillion pieces. A fourpage QuintGram, issued twice a year since 1981, is devoted to puzzles based on the solid pentominoes. See the bibliography for other references to pentomino problems. Joseph Dorrie of Madison Heights, Mich., proposed another subset of the pentacubes those pieces that are no longer than three units wide along any of the three coordinate directions. There are 25 such pieces, and they form what Dorrie calls the "Dorian cube," a term he has copyrighted. Another set of polycubes suggested for puzzle purposes consists of all the polycubes in orders 1 through 5 . Scott L. Forseth, in "Solid Polyomino Constructions," in Mathematics Magazine, Vol. 19, 1976, pages 137  139, shows how these 41 polycubes will pack a 2by3by31 = 186 box. H e found two solutions and thinks there are many others. In 1979 a Los Altos, Calif., firm called Lemmel Associates introduced a puzzle game called Putzl. Invented by L. E. Minnick, it uses two sets ofthe eight tetracubes, each a different color. One player uses one set; his opponent, the other. They take turns placing one of their pieces on the table. The object is to build an order4 cube. When a visible face of this cube is completed, the face is won by the player who has the most squares of his color on the face. (The game can be played in reverse, the win going to the person with the least of his color on the face.) A played piece must fit snugly on the previously placed pieces without creating any holes or extending beyond the imagined order4 cube. If no such play is possible, the player passes. T h e game ends when no play can be made, and the winner is the person who has captured the most faces. T h e cube's top face is rarely completed, although the 16 pieces will form the cube. Minnick has prepared a handbook for the game. Lakeside Industries, a division of Leisure Dynamics of Minneapolis, marketed in 1969 a series of six polycube puzzles under the name Impuzzables. Each consisted of a set of five, six or seven plastic polycubes, each set a different color, that fitted together to make a 3by3by3 cube. T h e colors were assigned in order of difficulty from the easiest (yellow) to the hardest (blue). Gerard D'Arcey, a California game inventor, designed the puzzles. Without knowing any of the polycube shapes, or even how many there are to each impuzzable, how quickly can you prove that all the polycubes from all six sets will build a 3by6by9 brick?
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BIBLIOGRAPHY Polycubes in General "Solid Polyominoes." S. W. Golomb in Polyominoes. Scribner, 1965. "Packing Boxes with Congruent Figures." D. A. Klarner and F. Gobel in Koninklijke, Nederlandse Akademie van Wetenschappen. Proceedings, Series A, Vol. 72, 1969, pages 465  472. "Symmetry of Cubical and General Polyominoes." W. F. Lunnon in Graph Theory and Computing, edited by Ronald C. Read. Academic, 1972. "Tiling Space with the Aid of the Holomorph." James P. Conlan in Journal of Combinatorial Theory, Vol. 14, 1973, pages 167  172. "Packing Boxes with Congruent Polycubes." Andrew L. Clarke inJournal ofRecreational Mathematics, Vol. 10, 1977  1978, pages 177  182. Polycubes. J. Meeus and P. J. Torbijn. Paris: CEDIC, 1917. A marvelous survey of the topic in a book of 176 pages.
Tetracubes "Tetracubes." Jean Meeus injournal of Recreational Ll/lathematics,Vol. 6,1973, pages 257  265.
Pentacubes "Constructions with Pentacubes." N. R. Wagner inlournal of Recreational Mathematics, Vol. 5 , 1972. pages 266268. "Constructions with Pentacubes 2." Iournal of Recreational Mathematics, Vol. 6, 1973, pages 211214. Pentacubes. Sivy Farhi, published by the author in 1977. The fifth edition (1981) is obtainable from Pentacube Puzzles, Ltd., Box 308, Auckland 1,New Zealand. This is a 70page booklet of pentacube problems to accompany a set ofthe 29 pieces. Mr. Farhi also published (1982) a booklet titled Soma World that contains more than 2,000 Soma constructions. The author's address is 19 Yogelsang Place, Flynn, Canberra, Australia. "A Search for NPentacube Prime Boxes." David Klarner inlournu1 of Recreational Mathematics, Vol. 12, No. 4, 19791980, pages 252257. "Packing Handed Pentacubes." C. J. Bouwkamp in The Mathematical Gardner, edited by David Klarner. Prindle, Weber and Schmidt, 1981.
POLYCUBES
43
The Soma Cube "The Soma Cube." Martin Gardner in The Second Scient$c American Book of Mathematical Puzzles and Diversions, Chapter 6. Simon and Schuster, 1961.
The Solid Pentominoes Catalog of Solutions of the Rectangular 3 X 4 X 5 Solid Pentomino Problem. C. J . Bouwkamp. Department of Mathematics, Technische Hogeschool Eindhoven, Netherlands, 1967. "Packing a Rectangular Box with the Twelve Solid Pentominoes." C. J. Bouwkamp in Journal ofCombinatoria1 Theory, Vol. 7, 1969, pages 278280. "Packing a Box with YPentacubes." C. J. Bouwkamp and D. A. Klarner injournal of Recreational Mathematics, Vol. 3 , 1970, pages 1026. See also Klarner's followup letter in the October 1970 issue, page 258.
"A New Solid Pentomino Problem." C. J. Bouwkamp and D. A. Klarner inJourna1of Recreational Mathematics, Vol. 4 , 1971, pages 179 186. "The FPentacube Problem." J. M . M . Verbakel injournal ofRecreationa1 Mathematics, Vol. 5 , 1972, pages 2021. "Solid Pentomino Multiplications." Ad Mank inJourna1ofRecreational Mathematics, Vol. 7 , 1974, pages 279282.
Packing the Steps with Solid Pentominoes. C .J . Bouwkamp. Department of Mathematics, 'Technische Hogeschool Eindhoven, Netherlands, 1979. Gives the 137 solutions to problem 44 on page 158 of S. W. Golomb's book Polyominoes.
CHAPTER FOUR
Bacon's Cipher Cryptography is a science of deduction and controlled experiment; hypotheses are formed, tested and often discarded. But the residue which passes the test grows and grows until finally there comes a point when the experimenter feels solid ground beneath his feet: his hypotheses cohere, and fragments of sense emerge from their camouflage. The code "breaks." Perhaps this is best defined as the point when the likely leads appear faster than they can be followed up. It is like the initiation of a chainreaction in atomic physics; once the critical threshold is passed, the reaction propagates itself.
JOHNCHADWICK, The Decipherment of Linear B
It is not hard to understand why philosophers and historians of science are so divided in their opinions about Sir Francis Bacon, the Elizabethan writer, philosopher and Lord Chancellor. O n the one hand, his insights into scientific method were primitive and defective. O n the other, he had a prophetic vision of science as a vast, collective and systematic enterprise that could provide humanity with undreamedof knowledge. And knowledge, he insisted, is power. For the first time man would have the power to master nature and control his own destiny. Although Bacon had little skill in mathematics, he did invent an ingenious cipher system of considerable interest to students of both recreational mathematics and word play. T h e "biliteral cipher," as Bacon called it, was one ofthe earliest demonstrations of how easily information can be transmitted by a simple binary code. T h e system is related to a fascinating combinatorial problem that has practical applications to errorcorrecting codes. Not least, Bacon's cipher has been responsible for the funniestand most bizarre claims ever
BACON’S CIPHER
45
propounded by the Baconians—those nevergiveup pseudoscholars who still labor mightily to convince the world that Bacon wrote the plays of Shakespeare. There are hints about the biliteral cipher in Bacon’s Advancement of Learning (1605), but he did not fully disclose the method until he expanded his brief remarks on ciphers for the later encyclopedic edition of this work in Latin, De Augmentis Scientiarum (1623). In Book 6 he repeats his earlier summary of the three virtues every good cipher should have: (1) “Easy and not laborious to write”; (2) “Safe and impossible to decipher”; (3) “If possible, such as not to raise suspicion.” A cipher with the third merit, known as a “concealment cipher,” is one in which the very existence of the true cipher text is not suspected. Bacon first explains a whimsical concealment dodge using two cipher alphabets. The genuine message is written with one set of symbols, then a false message is written with a second set. The two ciphers are interwoven to make a single cipher text. If this is intercepted and a translation demanded of the sender, he strikes out the symbols of the true text, explaining that they are what cryptographers today call “nulls,” meaningless symbols inserted only to make the cipher harder to break. He then reveals the key to the remaining symbols. Because an intelligible message now emerges, Bacon writes, who would suspect that the apparent nulls actually conceal another message? “But for avoiding suspicion altogether,” Bacon continues, “I will add another contrivance, which I developed myself when I was at Paris in my early youth.” The contrivance, the biliteral cipher, is based on a key that assigns to each letter of the alphabet a different sequence of two symbols in groups of five. As Bacon explains, there are 32 such sequences, more than enough for the English alphabet, which in Bacon’s day consisted of 24 letters. (I and J were interchangeable, as were U and V.) Bacon used a and b for the two symbols, assigning aaaaa to A, aaaab to B, aaaba to C and so on. “Nor is it a slight thing which is thus by the way affected,” Bacon writes. “For instance we see how thoughts may be communicated at any distance of place by means of any object perceptible either to the eye or ear, provided only that those objects are capable of two differences, as by balls, trumpets, torches, gunshots, and the like.” Indeed, the Morse telegraphic code is essentially a biliteral sound cipher, although pauses are used as a kind of third symbol so that no more than four dots and dashes are needed for each letter. Bacon’s plan was to use this cipher for concealing the plaintext (message to be enciphered) in an innocentlooking “cover text.” One has only to distinguish between two different ways of printing each letter. A crude method would be to let italicized letters stand for a and roman letters for b. The word “Bacon,” with only the first letter italicized, would represent the permutation abbbb, which in
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Bacon's alphabet means Q. It is obvious that any cover text, provided it is five times the length of the plaintext, can be printed so that it carries the secret message. T h e difference between roman and italicized letters is, ofcourse, too obvious. Bacon proposed using two type fonts that differed in minute ways. Only someone aware of these subtle differences would know how to scan the printing, label each letter a or b, divide the letters into quintuplets and read the hidden message. Bacon gave two examples of how these fonts could conceal a message. A short Latin cover text meaning "Do not go until I come" deciphers as a message of opposite advice: "Flee." A longer example of how "anything can be written by anything" is a passage Bacon took from a letter of Cicero [see Figure 241. When the letters are labeled a and b (according to the two fonts), the concealed Latin message (copied from one the Spartans had once sent by a cylindrical ciphering device called a scytale) translates into English as "All is lost. Mindarus is killed. The soldiers want food. W e can neither get hence, nor stay longer here." Elizabethan printing was so crude by modern standards that no two appearances of the same letter on a page, when examined under a strong magnifying glass, are exactly alike. Lead molds were imperfect, type was often damaged, ink dried irregularly on rough and dampened paper, and printers often mixed fonts on the same page. It is not surprising that anyone persuaded that Bacon wrote the plays of Shakespeare would suspect that Bacon might have used his own cipher to state the fact in early folios, perhaps even pepper the pages with other secret revelations. Elizabethan printing has provided Baconians with a marvelous arena for the unhampered play of unconscious impulses. With a magnifying glass in hand and flexible biliteral rules allowing a and b forms of each letter to be distinguished in any possible way (and in more than one way for each letter), a clever Baconian can extract from a long passage of Shakespeare's almost any short message he likes. T h e first appearance of a T may be labeled a because it has a slightly thinner upright line than other T's; the next T may be labeled a because it has a tiny curl at the end of the crossbar, and so on. Cipher keys are allowed to vary from passage to passage. If a Baconian is not a mountebank, the secret messages he finds will spring from deep within his subconscious, like the messages spelled on Ouija boards or by automatic handwriting or transmitted by mediums from the Great Beyond. Strangely enough, the first major effort to decipher Shakespeare's plays did not exploit Bacon's cipher. T h e flamboyant Populist politician from Minnesota, Ignatius Donnelly, used a different system, even more farfetched, for his 1,000page crank work The Great Cryptogram (1888). (This tome and Don
A letter of Cicero's in which the two type fonts conceal a secret war dispatch.
Figure 24
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nelly's Atlantis and Ragnarok form the most impressive set of crackpot works written by an American before 1900.) It remained for Mrs. Elizabeth Wells Gallup (1846 19341, a Michigan teacher and high school principal, to apply Bacon's own cipher with unflagging persistence to Shakespeare's plays, producing the best and most hilarious plaintexts in the history of Baconiana. Like Donnelly, Mrs. Gallup is a splendid specimen of the intelligent, learned, honest and thoroughly selfdeluded crank. Her opus The Biliteral
Cipher of Sir Francis Bacon Discovered in His Works and Deciphered by Mrs. Elizabeth Wells Gallup (1899) had a shattering impact on fellow Baconians. She found secret messages not only in the Shakespeare folios but also in the writings of Marlowe, Spenser, Burton and other writers whose books she believed had also been written by Bacon. "Queene Elizabeth is my true mother," one message read, "and I am the lawful1 heire to the throne. Find the Cipher storie my bookes containe; it tells great secrets, every one ofwhich, if imparted openly, would forfeit my life." Many of the great secrets turned out to be bawdy details of Elizabethan court life. "Surprise followed surprise," wrote Mrs. Gallup, "as the hidden messages were disclosed, and disappointment as well was not infrequently encountered. Some of the disclosures are of a nature repugnant, in many respects, to my very soul. . . . As a decipherer I had no choice, and I am in no way responsible for the disclosures, except as to the correctness of the transcription." "Colonel" George Fabyan (the military title was honorary), a wealthy textile manufacturer, became Mrs. Gallup's convert and major benefactor. He brought her to Riverbank Laboratories on his 500acre estate in Geneva, Ill., where he established a staff of cryptanalysts to work under Mrs. Gallup's supervision. She remained there for 20 years, studying photographic enlargements of Elizabethan manuscripts and trying to teach her bewildered staff how to decipher them. Ironically, as David Kahn observes in his book The Codebreakers, it was at Riverbank that young William F. Friedman was first introduced to the art of codebreaking. Later he became one of the world's greatest cryptanalysts. (It was his team that cracked the Japanese "purple code" of World W a r 11.)While he was at Riverbank, he met and married another of Mrs. Gallup's assistants, Elizabeth Smith. T h e two eventually became the most illustrious husbandandwife team in the history of cryptanalysis. Both, I hasten to add, quickly caught on to how Mrs. Gallup was deceiving herself. Indeed, the chapters devoted to Mrs. Gallup in their book The Shakespearean Ciphers Examined totally demolish Mrs. Gallup's monumental and pathetic lifetime labors. Back to mathematical reality. In recent decades mathematicians have developed many ingenious procedures for forming cyclic chains in which all possible
BACON’S CIPHER
49
Figure 25 A concise key for a bilateral cipher.
sequences of n symbols, taken k at a time, are given once only by each set of k adjacent symbols. For example, consider the 32symbol chain aaaaabbbbbabbbaabbababbaaababaab If you view the chain as cyclic (end joined to beginning), every group of five adjacent symbols is one of the 25 = 32 sequences of a and b in sets of five. There are 2,048 ways to construct such a chain, if reversals are considered different. For two symbols the formula giving the number of chains is k–1 2(2 –k) where k is the number of symbols in a group. Any of the 2,048 chains provides a convenient way of recording the key to a biliteral cipher. Simply print the alphabet, with the first six digits appended to bring the number of symbols to 32, in a circle and add the chain of a’s and b’s inside the circle [see Figure 25]. To obtain the sequence for, say, R, check the set of five symbols that start at R and go clockwise (or the other way if you prefer) around the circle. The cipher has many unusual applications. A deck of 52 playing cards, for instance, can be arranged so that the colors (or odd and even values, or high and low cards or any other binary division) will encipher a 10letter word or phrase.
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Of course, threesymbol chains provide triliteral ciphers, four symbols provide quadriliteral ciphers (the genetic code!) and so on. Although it is a defect of Bacon's system that a cipher text must be five times as long as the plaintext, a remarkable merit of the system is that more than one message can be hidden in the same cipher text. One has only to choose letters carefully so that they can be divided into a's and b's in more than one way. Consider, for example,
GkwRt ceUya porrE O u r cipher key will again be the concentric circles in Figure 25, reading clockwise. If a stands for letters whose positions in the alphabet are odd (a, c, e, . . .), and b for evenpositioned letters (b, d, f, . . .j, the text deciphers as aaabb aaaaa babba, which spells CAT. If a refers to a letter in the first half of the alphabet and b to letters in the second half, the same text deciphers as aabbb aabba bbbba, which spells DOG. And if a means uppercase and b lowercase, the translation is abbab bbabb bbbba, or PIG. Here is an exercise for readers: QUZGF MTXYX J L U N XNEEN WLREW TSNJE
Using the same key as before, can you determine three ways ofbifurcating the alphabet so that the above cipher text can be translated in three ways, each giving a sixletter last name of a famous mathematician? (Hints: T h e three divisions have to do with the name of a poet, legs and topology.) Although Bacon himself did not make the metaphor explicit, his cipher may be taken as symbolic of the curious way he viewed scientific knowledge. It is an attitude still held today by many philosophers and scientists. Bacon did not believe that the laws of science were infinite in number. Like his fellow Anglicans, he was convinced that God had created a natural world that was sharply  . cut off from the supernatural. In this world a finite number of simple principles combine, like the variables of an nliteral cipher, to form all the laws of nature. T h e 19thcentury English logician John Venn made this point in his Empirical Logic (page 357), where he described Bacon's position as an "alphabetical view of the Universe, in its extremest form. . . . W e find [the universe] all broken up, partitioned, and duly labeled in every direction; so that, enormously great as is the possible number of combinations which these elements can produce, they are neverthelessfinite in number, and will therefore yield up their secrets to plodding patience when it is supplied with proper rules."
BACON'S CIPHER
51
Science, to pursue the metaphor, is one stupendous task of cryptanalysis. Bacon was persuaded that eventually, and not far in the future either, all the ciphers would be broken and mankind would know not all truth by any means, but all the basic natural laws. T h e future of science would then be merely a filling in of details and the exploitation of laws by new inventions. Although few scientists today would venture such a prediction, more limited Baconian sentiments are often expressed with reference to a particular science. Nigel Calder, in his vivid survey of the new astronomy, Violent Universe (Viking, 1969),suggests that our century may turn out to be unique in the history of astronomy as the century in which astronomers first became "knowalls," omniscient in the sense of having mapped the fundamental outlines of the entire cosmos. "Or," Calder adds, "will our descendants smirk about our ideas as we do about those of our ancestors?" W h o can be sure, even with reference to a single science, whether in the long run (whatever that means) Bacon will be proved right or wrong? W e can say that at the moment nature appears to be far shaggier and more complicated than the Lord Chancellor suspected. There are ciphers within ciphers within ciphers, and there is not a clue in sight about whether any ofthese regresses has an end.
ANSWERS T h e three translations ofthe Baconian cipher are Fermat, Galois, Newton. T h e three biliteral keys respectively are
1. Any letter in WILLIAM SHAKESPEARE is a; all others are b.
2.
Any letter with one or more legs when printed as acapital is a (A, F, H , I, K,
M, N, P, Q, R, T, X, Y). Noleg letters are b. 3. Any letter that in simplest capital form is topologically equivalent to a line segment is a (C, I, L, M , N, S, U, V, W, Z). All others are b.
ADDENDUM I received a fascinating letter from Marguerite Gerstell, then an instructor at the Florida Institute of Technology in Jensen Beach. Using the same circular key that I used for my puzzle, she encoded the names o f j u e eminent mathematicians in the following cipher text: HUUSN IUUII YPDAW WVALP EZRWZ TISOS
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52
"Four of them are easy to find," she wrote. "Anyway, a smart gal can help you with the fifth." Here is how she did it:
1. NAPIER is encoded by replacing all vowels (including Y) in the cipher text with a and all other letters with b. 2. EUCLID is encoded by replacing with a all letters whose ordinals (position in the alphabet) are multiples of a square greater than 1. 3. KUMMER is encoded by substituting a for letters that are among the first 15 of the alphabet.
4.
CAUCHY
is encoded by replacing leftright symmetrical letters by a.
5 . CANTOR is encoded by substituting a for letters not in the phrase "anyway a smart gal. " A surprising thought occurred to Gerstell. Why not use the name itself as a basis for distinguishing a and b? She sent four examples of cipher texts, each concealing the names of three mathematicians, by using this curious selfreference technique. Here is one of them: ZYMWL EIGAI UMBOI JULRY MYEGA IXYZM LOSUL
T h e three names are Zermelo, Galileo and Fourier. In each case the name is encoded by letting a stand for letters in the name being concealed. As Gerstell pointed out, it is not easy to accomplish this with more than three names. It would be an interesting challenge, she wrote, to try to maximize the number of names that could be simultaneously encoded in this way. Gerstell's cipher texts all use the same cyclic chain that I suggested for a biliteral cipher. Such chains are now known as de Bruijn sequences, after the Dutch mathematician N. G. de Bruijn. For a fascinating history of such chains see "Memory Wheels," by Sherman K. Stein, in the second edition of his Mczthematics: The ManMade Universe (Freeman, 1969).In recent years mathematically minded magicians have invented a variety of bewildering card tricks based on de Bruijn sequences. References to where you can find some of them are in the answer section of Chapter 12 in my Magic Numbers of Dr. Matrix (Prometheus, 1985). For a recent article, with a good bibliography, on de Bruijn sequences see "De Bruijn SequencesA Modern Example of the Interaction of Discrete Mathematics and Computer Science," by Anthony Ralston in Mathematics Magazine, Vol. 55, 1982, pages 131 143.
BACON’S CIPHER
53
Someone ought to write a book about the sad life of Mrs. Gallup. Little seems to be on record about her. Apparently she taught at various public schools in Michigan (at Wayne, Flint, Fenton and Holly) and was a principal of the Holly high school. Friedman says she died in 1934, but an obit in the British periodical Baconia (October, 1935, page 106), called to my attention by David Shulman, gives the date of her death as April 1933 and her age as 87. She was born February 4, 1846, near Waterville, N.Y., educated at State Normal College of Michigan and was later a graduate student at the University of Marburg and the Sorbonne. I have been unable to determine what subject she taught or who Mr. Gallup was. All her tomes were published by Howard Publishing Company, Detroit, which I take to be her own company. The first edition of her opus (1899) was a mere 246 pages, but the second edition (1900) expanded it to 480 pages. The third edition (1901) is even larger—two volumes. In 1902 she issued a booklet titled Biliteral Cipher of Francis Bacon: Replies to Criticisms. Concerning the Biliteral Cipher of Francis Bacon, Discovered in His Works: Pros and Cons of the Controversy was a 1910 book of 229 pages. She also published (1901) a 147page work titled The Tragedy of Anne Boleyn: A Drama in Cipher Found in the Works of Sir Francis Bacon. A bibliography of articles about Mrs. Gallup’s obsessions would run to many pages. Here are the few references I was able to track down. “Mrs. Gallup’s Cipher.” Blackwood’s Magazine, Vol. 171, 1902, pages 267–269. “Mrs. Gallup and Francis Bacon.” Andrew Lang in The Monthly Review, Vol. 2, 1902, pages 146–162. “Mrs. Gallup’s Bad History.” Robert S. Rait in Fortnightly Review, Vol. 77, 1902, pages 328–334. Studies in the Biliteral Cipher of Francis Bacon. Gertrude Horsford Fiske. J. W. Luce, 1913. “The Encyclopedia Britannica and Mrs. Gallup.” B. Wright in Baconia, No. 132, 1949, pages 154–160. A picture of Mrs. Gallup can be found in Friedman’s book, cited earlier, and a different photograph appears in all her books. Georg Cantor, by the way, the genius who founded modern set theory, was a passionate believer in the Bacon–Shakespeare theory. During his later years of manic depression, when he was dabbling in theosophy and other occult mat
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54
ters, he wasted enormous amounts of time trying to prove the theory, lecturing on the topic and writing many articles. Cantor believed that his set theory had been directly inspired by God and was therefore flawless. His biblical studies convinced him that Jesus was the natural son of Joseph of Arimathea, and he wrote the pamphlet Ex Oriente Lux to prove it. (See "Georg Cantor's Creation of Transfinite Set Theory: Personality and Psychology in the History of Mathematics," by Joseph W. Dauben in Annals ofthe New York Academy ofSciences, Vol. 32 1 , 1979, pages 27  44, a volume titled Papers in Mathematics, edited by Paul Meyer.) I closed my column by expressing doubts that science was near discovering that everything in physics could be explained, as Bacon suggested, by a finite set of laws. At the moment this hope has sprung up again among many top physicists, who believe they are on the verge of constructing a grand unifiedfield theory that will cover all the forces of nature and explain why all the particles are just what they are. See my review of two recent books expressing this euphoria: "Physics: T h e End of the Road?", in The New York Review of Books, June 13, 1985, pages 31  34.
BIBLIOGRAPHY The Philosophy of Francis Bacon. Fulton H . Anderson. University of Chicago Press, 1948. The Shakespearean Ciphers Examined. William F. and Elizabeth S. Friedman. Cambridge University Press, 1957. The Codebreakers. David Kahn. Macmillan, 1967. "Origins of the Binary Code." F. G. Heath in Scienttjic American, August, 1972.
CHAPTER FIVE
Doughnuts: Linked and Knotted As you ramble on through life, brother, Whatever be your goal, Keep your eye upon the doughnut And not upon the hole!
A torus is a doughnutshaped surface generated by rotating a circle around an axis that lies on the plane of the circle but does not intersect the circle. Small circles, called meridians, can be drawn around the torus with radii equal to that of the generating circle. Circles of varying radii that go around the hole or center of the torus on parallel planes are called parallels [see Figure 261. Both meridians and parallels on a torus are infinite in number. There are two other less obvious infinite sets of "oblique" circles with radii equal to the distance from the center of the generating circle to the center of the torus's hole. Can you find them? Members of one set do not intersect one another, whereas any member of one set twice intersects any member of the other. T o a topologist, concerned only with properties that do not alter when a figure is elastically deformed, a torus is topologically equivalent to the surface of such objects as a ring, a bagel, a life preserver, a button with one hole, a coffee cup, a soda straw, a rubber band, a sphere with one handle, a cube with one hole through it and so on. Think of these surfaces as a thin membrane that can be stretched or compressed as much as one wishes. Each can be deformed until it
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56
Axis J.
Figure 26
Parallel
The torus
becomes a perfect toroidal surface. In what follows, "torus" will mean any surface topologically equivalent to a torus. A common misunderstanding about topology is the belief that a rubber model of a surface can always be deformed in threedimensional space to make any topologically equivalent model. This often is not the case. A Mobius strip, for example, has a handedness in 3space that cannot be altered by twisting and stretching. Handedness is an extrinsic property it acquires only when embedded in 3space. Intrinsically it has no handedness. A 4space creature could pick u p a lefthanded strip, turn it over in 4space and drop it back in our space as a righthanded model. A similar dichotomy applies to knots in closed curves. Tie a single overhand (or trefoil) knot in a piece of rope and join the ends. T h e surface of the rope is equivalent to a knotted torus. It has a handedness, and no amount of fiddling with the rope can change the parity. Intrinsically the rope is not even knotted. A 4space creature could take from us an unknotted closed piece of rope and, without cutting it, return it to us as knotted in either left or right form. All the properties of knots are extrinsic properties of toruses (or, if you prefer, onedimensional curves that may be thought of as toruses whose meridians have shrunk to points) that are embedded in %space. It is not always easy to decide intuitively if a given surface in 3space can be elastically deformed to a different but topologically equivalent surface. A striking instance, discussed more than 20 years ago [see "Topology," by Albert W. Tucker and Herbert S. Bailey, Jr., in Scientific American, January, 19501, concerns a rubber torus with a hole in its surface. Can it be turned inside out to
DOUGHNUTS: LINKED AND KNOTTED
Figure 27
57
Reversible cloth torus.
make a torus of identical shape? T h e answer is yes. It is hard to do with a rubber model (such as an inner tube), but a model made of wool reverses readily. Stephen Barr, in his Second Miscellany of Puzzles (Macmillan, 1969), recommends making it from a square piece of cloth. Fold the cloth in half and sew together opposite edges to make a tube. Now sew the ends ofthe tube together to make a torus that is square shaped when flattened. For ease in reversing, the surface hole is a slot cut in the outer layer of cloth [shown by the broken line in Figure 2 71. After the cloth torus is turned inside out, it is exactly the same shape as before, except that what were formerly meridians have become parallels, and vice versa. T o make the switch visible, sew or ink on the model a meridian of one color and a parallel of another so that both colors are visible from either side of the cloth. In 1958 Mrs. Eunice Hakala sent me a model she had made by cutting off the ribbed top of a sock and joining the tube's ends. T h e ribbing provides a neat set of parallels that turn into meridians afier the torus is reversed. Let us complicate matters by considering a torus tied in a trefoil knot. If we ignore handedness, there are only two such toruses: one with an external knot and one with an internal knot [see Figure 28 a,b]. A way to visualize the
Figure 28
Torus with outside knot (a), inside knot (b) and pseudoknots (c)
internally knotted torus is to imagine that the externally knotted torus on the lefi is sliced open along a meridian outside the knot. One end is turned back, as though reversing a sock; then the tube is expanded and drawn over the entire knot, and its ends are joined once more. O r imagine a solid wood cube with a hole bored through it that, instead of going straight, ties a knot before it emerges on the opposite side. T h e surface of such a cube is topologically equivalent to an internally knotted torus. You might suppose that a torus could be simultaneously knotted externally and internally, but it can't be done. One kind of torus seems to have both an outside and an inside knot [see Figure 2 8 ~ 1 Actually . both knots are humbugs. Untying the outer knot simultaneously unties the inner one, proving that the model is topologically the same as an unknotted torus its hole elongated like the hole of a garden hose. Although an outsideknotted torus is intrinsically identical with an insideknotted one, it is not possible to deform one to the other when it is embedded in 3space. Ifthere is a hole in the side of an outsideknotted torus, can the torus be reversed in 3space to put the knot inside? In the answer section I shall show how R. H . Bing, a topologist at the University of Wisconsin, answers this question with a simple sketch. A similar but harder problem is solved by Bing in his paper "Mapping a 3Sphere onto a Homotopy 3Sphere," in Topology Seminar, Wisconsin, 1965, edited by Bing and R. J. Bean (Princeton University Press, 1966). Imagine a cube with two straight holes [see Figure 2 9 ~ 1Its . surface is topologically the same as a twohole doughnut. W e can also have a cube with two holes, one straight,
DOUGHNUTS: LINKED AND KNOTTED
Figure 29
59
Three varieties of a twohole torus.
one knotted [Figure 29b1. It is not possible in 3space to deform the second cube so that the knot dissolves and the model looks like the first one. A third cube has one straight hole and one knotted hole with the knot around the straight hole [Figure 2 9 ~ 1 .Can this cube be elastically deformed until it becomes the first model? It is hard to believe, but the answer is yes. Bing's proof is so elegant and simple that the diagrams for it are almost selfexplanatory [see Figure 301. In elastic deformation a hole can be moved any distance over a surface without altering the surface's topology. As the hole moves, the surface merely stretches in back and shrinks in front. In Bing's proofthe knotted tube is drawn as a single
Figure 30
R. H. Bing's proof.
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60
line to make the proofeasier to follow. T h e hole, at the base ofthis tube is moved over the cube's surface, as indicated by the arrows, dragging the tube along with it. It goes left to the base of the other tube, climbs that tube's side, moves to the right across the top of the cube, circles its top hole counterclockwise, continues left around the other hole, over the cube's front edge,  down the front face, around the lower edge to the cube's bottom face and then across that face to the position it formerly occupied. It is easy to see that the tube attached to this hole has been untied. Naturally the procedure is reversible. If you had a sufficiently pliable doughnut surface with two holes, you could manipulate it until one hole became a knot tied around the other. Topologists worried for decades about whether two separate knots side by side on a closed rope could cancel each other; that is, could the rope be manipulated until both knots dissolved? No pair of canceling knots had been found, but proving the impossibility of such a pair was another matter. It was not even possible to show that two trefoil knots of opposite handedness could not cancel. Proofs ofthe general case were not found until the early 1950's. One way of proving it is explained by Ralph H . Fox in "A Quick Trip through Knot Theory," in Topology of 3Manijolds and Related Topics, edited by M . K. Fort, Jr. (PrenticeHall, 1963). It is a reductio ad absurdum proof that unfortunately involves the sophisticated concept of an infinity of knots on a closed curve and certain assumptions about infinite sets that must be carefully specified to make the proof rigorous. When John Horton Conway, the University of Cambridge mathematician, was in high school, he hit on a simpler proof that completely avoids infinite sets of knots. Later he learned that essentially the same proof had been formulated earlier, but I have not been able to determine by whom. Here is Conway's version as he explained it years ago in a letter. It is a marvelous example ofhow a knotted torus can play an unexpected role in proving a fundamental theorem of modern knot theory. Conway's proof, like the one for the infinite knots, is a reductio ad absurdum. W e begin by imagining that a closed string passes through the opposite walls of a room [see Figure 311. Since we shall be concerned only with what happens inside the room, we can forget about the string outside and regard it as being attached to the side walls. O n the string;  are knots A and B. Each is assumed to be genuine in the sense that it cannot be removed by manipulating the string if it is the only knot on the string. It also is assumed that the two knots will cancel each other when both are on the same closed curve. T h e proof applies to pairs of knots of any kind whatever, but here we show the knots as simple trefoils of opposite parity. If the knots can cancel, it means that the string can be manipu
DOUGHNUTS: LINKED AND KNOTTED
Figure 31
61
John Horton Conway1sproof.
lated until it stretches straight from wall to wall. Think of the string as being elastic to provide all the needed slack for such an operation. In the center figure we introduce an elastic torus around the string. Note that the tube "swallows" knot A but "circumnavigates" knot B (Conway's terminology). Any parallel drawn on this tube, on the section between the walls, obviously must be knotted in the same way as knot B. Indeed, it can be shown that any line on the tube's
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62
surface, stretching from wall to wall and never crossing itself at any spot on the tube's surface, will be knotted like knot B. "Now," writes Conway, "comes the crunch." Perform on the string the operation that we assumed would dissolve both knots. This can be done without breaking the tube. Because the string is never allowed to pass through itself during the deformation, we can always push the tube's wall aside if it gets in the way. T h e third drawing in Figure 31 shows the final result. T h e string is unknotted. T h e tube may have reached a horribly complicated shape impossible to draw. Consider a vertical plane passing through the straight string and cutting the twisted tube. W e can suppose that the tube's cross section will look something like what is shown with the possibility ofvarious "islands," but there will necessarily be two lines, XY and MN,from wall to wall that do not cross themselves at any point on the vertical plane. Each line will be unknotted. Moreover, each line also is a curve that does not cross itselfon the tube's surface. As we have seen, all such lines were (before the deformation) knotted like knot B. T h e deformation has therefore removed a knot equivalent to knot B from each of these two lines. Therefore knot B, alone on aline, can be removed by manipulating that line. But knot B, by definition, is a genuine knot that cannot be so removed. W e have contradicted an assumption. If two knots on a string can cancel, neither knot (since the same proof can be applied to knot A) can be genuine. Both must really hcve been pseudoknots. Although a onehole torus can be embedded in 3space in only three ways (outside knot, inside knot, no knot), a twohole torus has so many bizarre forms that the number is, I believe, not yet known. In some cases it can be reduced to a simpler form by deformation. For example, a tubethroughhole is equivalent to an ordinary twohole doughnut [see Figure 321, but what about the other two figures [b and c]? They are among several dozen monstrosities sketched by Piet Hein in a moment of meditation on twohole toruses. In b an inside knot goes through an outside one, and in c an outside knot goes through a hole. 1s it possible, by deformation, to dissolve the inside knot of b and the outside knot of c? With more complicated pairs of twoholers embedded in 3space, proofs that one can be deformed to the other are not so easy. As one of Piet Hein's "grooks" puts it: There are doughnuts and doughnuts with knots and with no knots and many a doughnut so nuts that we know not.
Figure 32
Twohole toruses.
Here are three more toroidally knotty questions
1. How many closed curves can be drawn on a torus, each a trefoil knot of the same handedness, so that no two curves cross each other at any point?
2. If two closed curves are drawn on a torus so that each forms a trefoil knot but the knots are of opposite parity, what is the minimum number of points at which the two curves will intersect each other?
3.
Show how to cut a solid twohole doughnut with one slice of a knife so that the result is a solid outsideknotted torus. T h e "slice" is not, of course, planar. More technically, show how to remove from a twohole doughnut a section topologically equivalent to a disk so that what remains is a solid knotted torus. (This amusing result was discovered by John Stallings in 1957 and communicated to me by James Stasheff.)
CHAPTER FIVE
Figure 33
64
Solution to torusreversed problem
ANSWERS R. H . Bing shows how an internally knotted torus can be reversed through a hole to produce an externally knotted torus [see Figure 331. A small hole, h, is enlarged to cover almost the entire side of the cylinder, leaving only the shaded strip on the right. T h e top and bottom disks ofthe cylinder are flipped over, and the hole is shrunk to its original size. As in reversing the unknotted torus through a hole, the deformation interchanges meridians and parallels. You might not at first think so because the circle, m , appears the same in all three pictures. T h e fact is, however, that initially it is a parallel circling the torus's elongated hole, whereas after the reversal it is a meridian. Moreover, after the reversal the torus's original hole is no longer through the knotted tube, which is now closed at both ends. As indicated by the arrow, the hole is now surrounded by the knotted tube. Piet Hein's twohole torus, with an internal knot passing through an external one, is easily shown to be the same as a twoholer with only an external knot. Simply slide one end of the inside knot around the outside knot (in the manner explained earlier) and back to its starting point. This unties the internal knot. Piet Hein's twoholer, with the external knot going through a hole, can be unknotted by the deformation shown in Figure 34.
DOUGHNUTS: LINKED AND KNOTTED
Figure 34
65
Unknotting a twohole torus.
Answers to the final three toroidal questions are as follows:
1. An infinity of noncrossing closed curves, each knotted with the same handedness, can be drawn on a torus [see Figure 35 top]. If a torus surface is cut along any of these curves, the result is a twosided, knotted band.
2. Two closed curves on a torus, knotted with opposite handedness, will intersect each other at least 12 times.
3. A rotating slice through a solid twohole doughnut is used to produce a solid that is topologically equivalent to a solid, knotted torus [see Figure 35, bottom]. Think of a short blade as moving downward and rotatingone and a half turns as it descends. If the blade does not turn at all, two solid toruses result. A halfturn produces one solid, unknotted torus. One turn produces two solid, unknotted, linked toruses. Readers may enjoy investigating the general case of n halfturns.
Figure 35 Knotted, nonintersecting curves on a torus (a) and rotating slice through a twohole torus (b)
ADDENDUM In studying the properties oftopological surfaces, one must always keep in mind the distinction between intrinsic properties, independent of the space in which the surface is embedded, and properties that arise from the embedding. T h e "complement" of a surface consists of all the points in the embedding space that are not in the surface. For example, a torus with no knot, one with an outside knot and one with an inside knot all have identical intrinsic properties. No two have topologically identical complements; hence, no two are equivalent in their extrinsic topological properties. John Stillwell, a mathematician at Monash University, Australia, sent several fascinating letters, in which he showed how an unknotted torus with any number of holes such toruses are equivalent to the sufaces of spheres with
DOUGHNUTS: LINKED AND KNOTTED
Figure 36 the right.
67
The surface on the left can be continuously deformed to the surface on
handles could be turned inside out through a surface hole. He was not sure if a knotted torus, even with only one hole, can be turned inside out through a hole in its surface. I leave this as a problem for the reader. Stillwell also posed the following question. Suppose two ordinary doughnut surfaces are linked, and one has a hole in its surface. Can the torus with the surface hole "swallow" the other torus so that at the finish the eaten torus is completely inside the cannibal? T h e answer is yes. I gave this problem in my April 1977 column in Scientijic American; the answer appeared the following month. Many beautiful, counterintuitive problems involving links and knots in toruses have been published. See Rolfsen's book, cited in the bibliography, especially the startling problem on page 95, where he shows that the surface on the left of Figure 36 is topologically equivalent to the surface shown on the right. For other curious equivalences ofthis sort see Herbert Taylor's torus problem in my Scientijic American column for December 1979, and "The Toroids of Dr. Klonefake," Problem 9, in my Science Fiction Puzzle Tales (Clarkson Potter, 1981).
BIBLIOGRAPHY "Elementary Point Set Topology." R. H. Bing in The American Mathematical Monthly, Part 11, Vol. 67, 1960, special supplement to Part I.
lntuitive Concepts in Elementary Topology. Bradford Henry Arnold. PrenticeHall, 1962. introduction to Knot Theory. Richard H. Crowell and Ralph H. Fox. SpringerVerlag, 1963.
First Concepts of Topology. W . G. Chinn and N. E. Steenrod. Random House New Mathematical Library, 1966. Knots and Links. Dale Rolfsen. Publish or Perish, 1976
CHAPTER SIX
The Tour of the Arrows and Other Problems 1. THE TOUR OF THE ARROWS Sketch a large 4by4 checkerboard on a sheet of paper, obtain 16 paper matches, and you are set to work on this new solitaire puzzle. T h e matches represent arrows that point in the direction of the match head. Put a single spot on both sides of one match, two spots on both sides of eight matches and three spots on both sides of seven matches. When a match is placed on a square ofthe board pointing north, south, east or west, the single spot means it points to the immediately adjacent cell, two spots mean it points to the second cell and three spots mean it points to the third cell. Seven matches can be placed to map a closed tour [see Figure 371. Start at any match of the seven and place your finger on the cell to which it points. T h e arrow on that cell gives the next "move." Follow the arrows until you return (in seven moves) to where you started. T h e problem is to place all 16 matches, one to a cell so that they map a closed tour that visits every cell. There are just two solutions, not counting rotations and reflections. T h e tour will have a length of 1 (2 X 8) ( 3 X 7) = 38. It is not hard to prove that this is the longest closed tour that can be made on the board by using any combination of the three types of arrows. Brian R. Barwell, a British engineer who introduced the problem in theJournal ofRecreational Mathematics (October, 19691, found that only one other maximumlength tour is possible. It requires six 3arrows, ten 2arrows and no 1arrow. Readers are invited to search for all three patterns. T h e arrows are, of course, merely a convenient way to map a maximumlength, closed tour by a chess rook, which lands on each cell exactly once. (Queen tours of this type are less interesting because there are so many of them;
+
+
THE TOUR OF THE ARROWS AND OTHER PROBLEMS 69
Figure 37
A closed arrow tour
bishop tours cannot close and cannot visit all cells; and knight tours cannot vary in length.) T h e 2by2 field is trivial, and the 3by3 is easily analyzed. (Its maximum tour has a length of 14.) As far as I know, the 5by5 and all higher squares have yet to be investigated.
2. FIVE COUPLES My wife and I recently attended a party at which there were four other married couples. Various handshakes took place. No one shook hands with himself (or herself) or with his (or her) spouse, and no one shook hands with the same person more than once. After all the handshakes were over, I asked each person, including my wife, how many hands he (or she) had shaken. T o my surprise each gave a different answer. How many hands did my wife shake? (From Lars Bertil Owe of Lund, Sweden.)
3. SQUARETRIANGLE POLYGONS An unlimited number of cardboard squares and equilateral triangles, each with unit sides, are assumed to be available. With these pieces it is easy to form convex polygons with from 3 to 10 sides [see Figure 381. Can you make an
CHAPTER SIX
Figure 38
70
Convex polygons with from 3 to 10 sides
11sided convex polygon with the pieces? And what is the largest number of sides a convex polygon formed by the pieces can have?
4. TEN STATEMENTS Evaluate each of the 10 statements as to its truth or falsity:
1. Exactly one statement on this list is false.
2.
Exactly two statements on this list are false.
3.
Exactly three statements on this list are false.
4.
Exactly four statements on this list are false.
5.
Exactly five statements on this list are false.
6.
Exactly six statements on this list are false.
7. Exactly seven statements on this list are false. 8.
Exactly eight statements on this list are false.
9. Exactly nine statements on this list are false. 10. Exactly ten statements on this list are false.
5. PENTOMINO FARMS Victor G. Feser of Saint Louis University has proposed four maximumarea problems, each using the full set of 12 pentominoes. Three have been solved, and the fourth is probably solved.
THE TOUR OF THE ARROWS AND OTHER PROBLEMS
Figure 39
71
Pentomino fence problems
1. Form a rectangular "fence" around the largest rectangular field. T h e 4by7 has been proved maximum [see Figure 39a]. 2. Form a rectangular fence around the largest field of any shape. The maximum is 6 1 unit squares [see Figure 39b]. 3. Form a fence of any shape around the largest rectangular field. The 9by10 is maximum [see Figure 39c].
4. Form a fence of any shape around the largest field of any shape. (As in the preceding problems, the fence must be at least one unit thick at all points.) This is the most difficult of the four. In Figure 39d you see a solution of 127 squares. This was believed to be maximum until Donald E. Knuth, the Stanford computer scientist, recently raised it to 128. Knuth has an informal proof that 128 cannot be exceeded. Readers will find it a pleasant and difficult task to find a 128 solution.
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6. THE UNEVEN FLOOR A kitchen has an uneven floor. There are no "steps," but the continuous random waviness of the linoleum is such that when one tries to place on it a small square table with four legs, one leg is usually offthe floor, causing the table to wobble. If one does not mind the table top being on a slant, is it always possible to find a place where all four legs are firmly on the floor? O r can a floor wave in such a way that no such spot is available? T h e problem can be answered by a simple, elegant proof.
7. THE CHICKENWIRE TRICK This strange parlor trick comes from T a n Hock Chuan, a Chinese professional magician who lives in Singapore. He described it in a letter to Johnnie Murray, an amateur conjuror of Portland, Maine, who passed it on to me. A blank sheet of paper about eight by five inches (half a sheet of typewriter paper works nicely) is initialed by an onlooker so that later it can be identified. T h e magician holds it behind his back (or under a table) for about 30 seconds. When he brings it back into view, it is covered with creases that form a regular hexagonal tessellation [see Figure 401. How is it done? T h e performer is usually accused of pressing it against a piece of chicken wire, but the creasing actually is done without using anything except the hands.
Figure 40
Chickenwire folds
Figure 41
Where was the white king?
8. WHERE WAS THE KING? T h e philosophermathematicianlogician Raymond Smullyan invented this elegant chess problem when he was a student at the University of Chicago in 1957. He showed it to his friend William Browder, now a distinguished mathematician at the university, who passed it on to his father, Earl Browder, former head of the Communist Party in the U.S. and an ardent chess player. T h e father sent it to the Manchester Guardian, where it was inadvertently published without mentioning Smullyan. A later issue gave proper credit for the problem, and other retrograde problems by Smullyan ran in subsequent issues. A retrograde chess problem is one that can be solved only by deducing the moves that precede the position shown. In this case we see in Figure 41 a position in a legal game just after the white king has been knocked offthe board. Where was the king standing? and what was White's last move?
9. POLYPOWERS By convention, the value of a ladder of exponents such as
is computed by starting at the top and working down. T h e highest pair equals 4, then z4 = 16, and 216 = 65,536. How large is 265,536?A few years ago Geoffrey W. Hoffmann of West Germany sent me a computer printout of this number. It starts 20035 . . . and has 19,729 digits. Adding another 2 to the ladder gives a number that will never be calculated because the answer, as Hoffmann put it, would require the age of the universe in computer time and the space of the universe to hold the printout. Even a ladder as short as three 9's is 9387,420,489, a number of more than 360 million digits. In 1933 S. Skewes published a paper in which he showed that if
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n(x) is the number of primes less than x, and li(x) is the logarithmic integral function, then n(x)  li(x) is positive for some x less than
an integer said to be the largest known to play a role in a nontrivial theorem. In 1971 Aristid V. Grosse, one of the pioneer atomic chemists at Columbia University in 1940 (he is now president of Germantown Laboratories, Inc., affiliated with the Franklin Institute), began an investigation of exponential ladders of identical numbers that are calculated in the opposite direction (up) and their relations to down ladders. He coined the term "polypowers" for ladders of both types. Ladders of two x's are called "dipowers," of three x's "tripowers" and so on, according to the Greek prefixes. T h e value of x can be rational or irrational, transcendental, complex or entirely imaginary. In most cases the polypowers are single valued, continuous and differentiable. Since 1to any polypower of 1is 1,all these functions and their derivatives, when graphed against x, cross one another at x = 1, and their values at 0 are the limits as x approaches 0. Grosse's notes, which already fill many volumes, lead into a lush jungle of unusual theorems as well as new classes of numbers. U p and down dipowers obviously are identical, but for all higher polypowers the two directions give different numbers. T h e triplet of 9's, for example, when calculated upward is a number of only 77 digits. Except for the triplet of 2's, going "up all the way" on a ladder of identical integers gives the minimum number, and "down all the way" gives the maximum. In what follows, the arrows indicate these maximum and minimum numbers. What happens when u p and down ladders of different lengths are equated? If an up triplet of x's equals a down triplet of x's, x = 2. ( W e exclude x = 1 as being trivial.) Each additional x on the u p ladder increases the value ofx by 1.If three down x's equal four u p x's, x = 3; if three down equals five up, x = 4 and SO on. As an introduction to polypowers, readers are asked to solve the three . equations below, which begin a series with down tetrapowers on the left: ~
Readers may enjoy investigating ladders of fractional x's, reciprocals ofx and more exotic forms. Grosse has also developed the concept of a perfect polypower, that is, x to the xth power (up or down) an x number oftimes. (Example:
THE TOUR OF THE ARROWS AND OTHER PROBLEMS
75
Figure 42 Answer to arrow tours
π to the πth power, π times up, is 588,916.33+.) The reverse operation to polypowers he calls “polyroots.” Have these fields been investigated before? In spite of considerable effort, neither he nor I have uncovered references. ANSWERS 1. The three ways of forming maximumlength arrow tours on the 4by4 field are shown in Figure 42. Edward N. Peters, on the faculty of the University of Rochester Medical School, discovered a general procedure for constructing maximumlength rooktours on square boards of any size. See his paper “Rooks Roaming Round Regular Rectangles,” in Journal of Recreational Mathematics, Vol 6, 1973, pages 169–173. Frederick Hartmann of Rolling Hills Estates, Calif., extended the analysis to nonsquare rectangular boards, but so far as I know, his results remain unpublished. When the board is n × 1, it reduces to the “worstroute” problem for a postman delivering mail to a row of n houses (see my Sixth Book of Mathematical Games from Scientific American, W. H. Freeman, 1971, Chapter 23). Maximumlength rook tours on these linear boards are unique from n = 1 through 4, then increase in number steadily as n exceeds 4. For n = 7, for example, there are 18 such tours. Hartmann gave an algorithm for constructing at least one maximumlength tour on any rectangular board. If m and n are the lengths of the sides, with m equal to or greater than n, and C is obtained from the table shown in Figure 43, the formula for the length of the tour is
n( 3m 2 + n 2 − 10 ) +C . 6 For square boards of side n the formula reduces to 2n 3 − 5n +C 3 with C = 1 for odd n and C= 2 for even n.
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m
n
{n/2)
C
even odd even even odd odd
even odd odd odd even even

2 1 312 112
Figure 43

even odd even odd
(n/2) indicates the greatest integer contained in the ( ).
0 1
Table for the value of C
Figure 44 (from Hartmann) gives the maximumlength rook tours for values of m and n through 12. Neither Peters nor Hartmann attempted the much more difficult task of finding a formula for the number of distinct tours on a given board.
Figure 44 12 X 12
Maximumlength rook tours for m X n boards from 1 X 2 through
THE TOUR OF THE ARROWS AND OTHER PROBLEMS
77
MY WIFE
Figure 45
Answer to the handshaking problem
2. Among the five married couples no one shook more than eight hands. Therefore if nine people each shake a different number of hands, the numbers must be 0, 1 , 2 , 3 , 4 , 5 , 6 , 7and 8. T h e person who shook eight hands has to be married to whoever shook no hands (otherwise he could have shaken only seven hands). Similarly, the person who shook seven hands must be married to the person who shook only one hand (the hand ofthe person who shook hands only with the person who shook eight hands). T h e person who shook six must be married to the person who shook two, and the person who shook five must be married to the person who shook three. T h e only person left, who shook hands with four, is my wife. T h e above reasoning, which makes use of the familiar "pigeonhole principle." can be clarified by diagramming the problem [see Figure 451. Every graph that lacks loops and multiple edges must contain at least two points that have the
Figure 46 Elevensided and 12sided convex polygons and three other polygons of 11 sides
same number of lines attached to them. In this case the graph has only two such points, those representing me and my wife.
3. ,4n 11sided convex polygon can be formed with unitsided squares and equilateral triangles, as shown in Figure 46, [top left].T h e angles possible for a convex polygon formed with the pieces are 6 0 , 9 0 , 120 and 150 degrees. For a polygon with the maximum number of sides, all angles must be 150 degrees. T h e number of sides will then be 12. Figure 40 [top right] shows the smallest example. Several readers "proved" that an 11sided polygon could not be formed with squares and equilateral triangles of unit sides. T h e flaw, of course, was failing to realize that a side could be more than one unit long. Wade Philpott pointed out that any convex pentagon formed with unit equilateral triangles can be used as the core of an 11sided polygon. Simply place unit squares next to each triangle and complete the perimeter with six triangles. T h e solution I gave leads to an infinite family of 11sided polygons, shown in the middle of Figure 46. At the bottom are two other examples with different inner pentagons. The problem derives from one posed by Joseph Malkewitch in Mathematics Magazine and answered by Michael Goldberg in the May 1969 issue, page 158.
THE TOUR OF THE ARROWS AND OTHER PROBLEMS
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4. Only the ninth statement is true. David L. Silverman contributed the problem to the Journal of Recreational Mathematics, January, 1969, page 29, presenting it in the form of 1,969 statements. Underwood Dudley answered it in the October issue, page 231, as follows: "At most one of the statements can be true because any two contradict each other. All the statements cannot be false, because this implies that the list contains exactly zero false statements. Thus exactly one statement can be true. Thus exactly n  1 are false, and the ( n  1)st (the 1,968th)statement is true." Alan Brown pointed out that if the word "exactly" is removed from each of the 10 statements in the logic problem, there is a different and unique solution: T h e first five statements are true; the last five, false. T h e problem obviously generalizes to as many statements as you care to add. What happens if you decrease the number to just one? 1. Exactly one statement on this list is false
Norman Pos wrote to point out that the problem then reduces to the traditional liar paradox: "This sentence is false." T o circumvent the paradox, Pos added a zero statement at the beginning: 0. Exactly none of the statements on this list is false
Pos was surprised to discover that this shifts the one true statement from position n  1 to position n. That adding such a statement at the top of, say, 1,000 numbered statements would shift the unique true sentence from nexttolast to last he found an amusing case of syntactical "action at a distance."
5. A solution to the farm problem, enclosing 128 square units, is shown in Figure 47. I learned later that this problem had been proposed by R. J. French in The Fairy Chess Review, Vol. 4 , 1939, page 43. French said the area was more than 120. 1 have not been able to determine if the problem was answered in subsequent issues. After I published Knuth's 128 solution, Yoichi Kotani sent a proof, along with 1,440 solutions, that 128 is the maximum. Robert Reid Dalmau of Lima, Peru, sent the same set of solutions. In 1978 Takakazu Shimauchi published in Japanese a proof that 128 is the maximum (Sugaku Seminar, March, 1978, pages 1116). For references on pentomino farm problems in the Journal of Recreational Mathematics, see the issues for January, 1968, pages 55 61; October, 1968, pages 234235; July, 1969, pages 187 188; andVol. 17, No. 1 , 1 9 8 4  1985,
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Figure 47
The largest pentomino "farm"
pages 75  77. If the 12 pieces are allowed to touch only at corners and all edges are required to be horizontal and vertical, a farm of 160 square units is the largest known. If the pieces are allowed any orientation and corner touching, the area can be raised to slightly more than 161. 6. A square table can always be placed somewhere on a wavy floor with all four legs touching the floor. T o prove this, put the table anywhere. Assume that only three legs, A, B, C, are on the floor and D is off[see Figure 481. It is always possible for three legs to touch the floor because three points, anywhere in space, mark the corners of a triangle. Rotate the table 90 degrees around its center, keeping legs A and B always on the floor. This brings the table to a position where C is now the only leg that does not touch the floor. During the rotation D has moved to the floor and C has left it. But D must have touched the floor before C left, otherwise there would be a position at which only A and B would touch the floor, and we know that it is always possible for three legs to touch. At some point in the rotation, therefore, all four
Figure 48
The wobblytable proof
T H E TOUR O F T H E ARROWS AND OTHER PROBLEMS
81
legs must have been in contact with the floor. A similar argument can be applied to wobbly rectangular tables by giving them 180degree rotations. Many readers called attention to two tacit assumptions that are necessary to make this proof valid:
1. T h e table, like all normal tables, has four legs of equal length, their lower ends at the corners of a square.
2. T h e legs of the table are sufficiently long and the unevenness of the floor is sufficiently mild, so while the table is rotated, there is never a moment at which three legs cannot be made to touch the floor. The theorem is actually useful. Suppose you have a circular table with four legs that wobbles a bit when you move it to a porch. Ifyou don't mind the table's surface being on a slight slant, you don't have to search for something to slip under a leg: Just rotate the table to a stable position. If you have to stand on a fourlegged stool or chair to replace a light bulb and the floor is uneven, you can always rotate the stool or chair to make it steady.
7. T o put a chickenwire pattern of creases into a small sheet of paper, first roll the sheet into a tube about half an inch in diameter. With the thumb and forefinger of your left hand, pinch one end of the tube flat. Keeping pressure on the pinch with your left hand, your right thumb and forefinger, pinch the tube flat at a spot as close as possible to the first pinch, making the pinch at right angles to the first one. Press firmly with both hands, at the same time pushing the two pinches tightly against each other to make the creases as sharp as possible. Now the right hand retains its pinch while the left hand makes a third pinch adjacent to and perpendicular to the second one. Continue in this way, alternating hands as you move along the tube, until the entire tube has been pinched. (Children often do this with soda straws to make "chains.") Unroll the paper. You will find it hexagonally tessellated in a manner that is most puzzling to the uninitiated. John H. Coker wrote to say that when he was a child in Yugoslavia in the early 1930's, his schoolteacher rolled and pinched notes to other teachers in this manner. Because it is extremely difficult to unroll such a tube and then reroll it exactly as before, the tube provided security from the eyes of children asked to transmit the notes.
8. Place the pieces as shown in Figure 49, and make the following moves: White
Black
1
BQ4 (check)
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Figure 49
Retrograde chess
White
Black
2 PB4
P takes P en passant (double check)
3 K takes P (check)
Removing the White king will now leave the position given with the statement of the problem. In addition to books of philosophical essays and logic problems, Raymond Smullyan published two collections of his marvelous chess problems: The Chess Mysteries of Sherlock Holmes (Knopf, 1979) and The Chess Mysteries of the Arabian Knights (Knopf, 1981). 9. T h e key to simplifying the three polypower equations is the basic law = .(bXc)
Applying this to the first equation gives
T h e two bottom x's are equal; therefore their parenthetical exponents are equal. Cancel the bottom x's and repeat the procedure:
T h e bottom x's again drop out, leaving x X= 2, which gives x the value 1.55961k.
THE TOUR OF THE ARROWS AND OTHER PROBLEMS
83
The same procedure simplifies the second equation (down4 equals up4) to 1.82545t. Each succeeding equation increases the value ofxX
X~ = 3, and x =
by 1. T h e third equation reduces to x X= 4, or x = 2. The general procedure is to replace the up ladder by a number one less than the number of its x's and remove two x's from the down ladder. (Example: Down5 = up5 reduces to down3 = 4.) Correspondence about polypowers was unusually heavy, and readers raised many interesting questions. Several readers pointed out that parentheses could be placed on a ladder in a variety ofways. In fact, the number ofways is given by the sequence known as the Catalan numbers. However, not all ways of parenthesizing give distinct values for the ladder. Determining the number of such values is a difficult problem, and I do not know the solution. Many readers called attention to unusual, littleknown theorems about infinite ladders of exponents. Consider, for example, a ladder of x's that grows steadily upward to infinity. I would have thought that if x is greater than 1 the ladder's value (working from top down) would diverge as the ladder grows. This is not true. If x is an integer, the value diverges only if x exceeds elie = 1.4446. . . . If x is a real number, it converges only if it is equal to or greater than eFe = 0.0659 . . . and equal to or less than elle. I found this amazing. A delightful paradox is related to the above theorem. Assume that an infinite ladder ofx's has a value of 2. What is the value ofx? Because all the x's above the bottomx form an infinite chain, we can assume that the value ofthis chain is also 2. Substituting 2 for this chain gives the equation x2 = 2, for which x = fi. All well and good. Now apply the same dodge to an infinite ladder of x's that equals 4. This leads to x4 = 4, so again x = &. How can an infinite ladder converge to two different numbers? Actually, an infinite ladder of square roots of 2 cannot converge to 4, and in this case the dodge is not applicable. T o show this exactly is complicated. You will find it explained in "A Matter of Definition," by M. C. Mitchelmore in American Mathematical Monthly, Vol. 81, 1974, pages 643  647. For general discussions of infinite ladders see "Infinite Exponentials," by D. F. Barrow in American Mathematical Monthly, Vol. 43, 1936, pages 150160; "Exponentials Reiterated," by R. A. Knoebel, ibid., Vol. 88, 1981, pages 235252; and "Infinite Exponentials," by P. J. Rippon in Mathematical Gazette, Vol. 67, 1983, pages 189196. Knoebel gives a long bibliography of earlier references. Several readers sent references relevant to Grosse's labors, but unfortunately they were all in German or French. I still know of no good references in English to the sort of problems Grosse has been investigating. Some comments on big numbers may be of interest. I mentioned that the largest number that can be written in conventional notation with no symbols
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84 9
other than three digits is 9' . In the nexttolast chapter of Ulysses,Joyce reveals that Leopold Bloom was once fascinated by this number, and a paragraph is devoted to describing how big it is. Skewes is pronounced Skewease. T h e large number that bears his name was based on the assumption that the Riemann hypothesis is true. What if it isn't? In 1955 Skewes published a proofthat the number would then be the much larger
For an entertaining account of all this see "Skewered!" by Isaac Asimov in Fantasy and Science Fiction, November, 1974. Skewes made his calculations at the request of J. E. Littlewood, who tells about it in the chapter titled "Large Numbers" in A Mathematician's Miscellany (Methuen, 1953). Even Skewes's second number is very tiny and no longer the largest ever involved in a legitimate proof. T h e record is now held by Ronald L.Graham, of Bell Laboratories. Graham's number arose in connection with a problem in a branch of graph theory called Ramsey theory. (See my Scientijic American column for November 1977.) T h e number can be expressed compactly only in a special notation devised by Donald E. Knuth for handling numbers of such unimaginable magnitude.
CHAPTER SEVEN
Napier's Bones In his celebrated Budget ofParadoxes Augustus De Morgan defines a "graphomath" as a person ignorant of mathematics who tries to describe a mathematician. As an example, he quotes from the second chapter of Sir Walter Scott's novel The Fortunes of Nigel, in which David Ramsay, a whimsical clockmaker and amateur mathematician, swears "by the bones of the immortal Napier!" It is hard to tell from the passage whether Scott actually was uninformed or whether he merely intended Ramsay to make an ignorant or ajoking remark. In any case, "Napier's bones" have nothing to do with the skeletal remains of Baron John Napier (1550 1617), the Scottish mathematician who discovered logarithms and who was the first important mathematician of Britain. The phrase refers to a set of numbered rods that Napier invented for doing multiplication. W e shall discuss his method later, but first some remarks about Napier himself. His father, Sir Archibald Napier, master of the Scottish mint, was just 16 when John was born. And John was a mere 13 when he entered the University of St. Andrews. He left the university without getting a degree, took over the family castle and estates at Merchiston (now part of Edinburgh), married and had one son and one daughter, was widowered, remarried and continued the symmetry with five sons and five daughters. T h e Protestant Reformation in Scotland had started at about the time John was born, and while a youth at St. Andrews he became a passionate Calvinist with a compulsion to explicate biblical prophecy. In 1593 he published what he always considered his masterpiece (much more important than logarithms), the full title of which was "A Plane Discovery of the whole Revelation of Saint Iohn: set downe in two treatises: T h e one searching and proving the true interpretaiion thereof: T h e other applying the same paraphrastically and historically to the text. Set foorth by. John Napier L, of Marchistoun younger. Whereunto are annexed certaine Oracles of Sibylla, agreeing with the Revelation and other places of Scripture.
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Edinburgh, printed by Robert Waldegrave, printer to the King's Majestie, 1593. Cum privilegio Regali." It was the first major Scottish work on the Bible and one ofthe most thorough attempts ever made before or since to explore the symbolism of the Apocalypse. It is ironic that today, when many college students seem more interested in the Second Coming than in current politics, there is no available reprint ofNapier's treatise. It was enormously influential in its day, with 21 English editions and numerous European translations. Perhaps the main reason the book is out of print is that Napier made a slight miscalculation about the end of the world. He had been strongly influenced by the religious speculations of Michael Stifel, a German algebraist who proved that Pope Leo X was the Antichrist by rearranging the Roman numerals in LEO DECIMVS to make DCLXVI, or 666, the notorious "mark ofthe Beast." Where did Stifel get the x? From Leo X and because LEO DECIMVS has 10 letters. What happened to the M? He left that out because it stood for mysterium. Stifel predicted that the world would end on October 3, 1533. Napier perceived that this was a mistake. He decided that it was the Pope of 1593 who was the actual Antichrist. God had ordained that exactly 6,000 years would elapse between the earth's creation and its destruction. Since there was some uncertainty about the exact date ofcreation, Napier set the end ofthe world as being between 1688 and 1700. Napier begins his book by apologizing for having written it in a language so base as English, and he concludes it by appealing to the pope as follows: "In summar conclusion, if thou o Rome aledges thyselfe reformed, and to beleeue true Christianisme, then beleeue SaintJohn the Disciple, whome Christ loued, publikely here in this Reuelation proclaiming thy wracke, but if thou remain Ethnick in thy priuate thoghts, beleeuing the old Oracles of the Sibyls reuerently keeped somtime in thy Capitol: then doth here this Sibyl1 proclame also thy wracke. Repent therefore alwayes, in this thy latter breath, as thou louest thine Eternal1 salvation. Amen. " "Strange," comments De Morgan in his Budget, "that Napier should not have seen that this appeal could not succeed, unless the prophecies of the Apocalypse were no true prophecies at all." After clearing up the apocalyptic mysteries, Napier turned his ingenuity toward ways of defending Scotland against a threatened invasion by Catholic Spain. His 1596 document was titled Secrett lnventionis, profitabill and necessary
in theis dayesfor defense ofthis Iland, and withstanding ofstrangers, enemies ofGod's truth and religion. It describes three inventions: mirrors for setting fire to enemy ships (shades of Archimedes!), a machine gun and a metal chariot (that is, a tank) housing soldiers who could fire through holes in the sides.
NAPIER’S BONES
87
Napier’s next book, the Latin title of which begins Mirifici Logarithmorum Canonis Descriptio... (A Description of the Marvelous Rule of Logarithms...), appeared in 1614. This was the book in which Napier explained logarithms, called them logarithms (a term he coined), and gave the world its first log table. It has often been pointed out that if exponents had then been in use, logarithms would have immediately been recognized as a great toil saver but Napier conceived of them without reference to exponents at all. This is not the place to explain how he arrived at logs the hard way by considering the relation of an arithmetic series to a geometric series. The London geometer Henry Briggs quickly realized that 10 was the most convenient base for logarithmic calculations in the decimal system, and Napier at once agreed. It is said that when the two men first met at Merchiston Castle (where Briggs remained for a month), they admired each other for 15 minutes before either spoke a word. Navigators and astronomers, notably Johannes Kepler, found the base10 logs (or common logarithms as they are now called) invaluable, and years of drudgery were devoted by Briggs and others to preparing better and better log. tables. (Today it is faster to compute a log all over again on a pocket electronic calculator—it takes less than a second—than to look it up in a book!) In Napier’s posthumous work Mirifici Logarithmorum Canonis Constructio…(1619) he explained how he calculated his original logs. In doing so he made systematic use of a decimal point, placing it above the baseline and using it exactly as it is used today in England. Two of the most amusing of many anecdotes about Napier are recounted by Howard W. Eves in his delightful In Mathematical Circles. Because a neighbor’s pigeons were flying onto Napier’s estate and eating grain, Napier told his neighbor that he would impound the birds as payment. The neighbor replied hat Napier was welcome to any pigeon he could catch alive. Napier scattered brandysoaked peas over his grounds and the pigeons were soon staggering about in such a stupor that he had no trouble collecting all of them in a sack. It was a time when almost everyone in Scotland (including Napier) believed in astrology and black magic. One day Napier called his servants together and told them that his black rooster had the occult power to tell him which servant had been stealing from the estate. One at a time each servant was asked to enter a dimly lighted room and stroke the bird’s back. As Napier had anticipated, only the guilty person, fearing exposure, would not do as asked. Napier had covered the rooster’s black feathers with soot, and so only the guilty servant emerged with clean hands. The age was also one of intense interest in calculating. The average person did arithmetic on his fingers, but more skillful mathematicians took great
CHAPTER SEVEN
Figure 50
88
Rabdoiogy, or "Napier1sbones"
delight in completing tedious computations. Napier's hobby was to find ways to simplify such work. Logarithms were, of course, his best invention, but in 1617 (the year he died) he brought out a little book called Rabdologia that explained three other methods of calculating. T h e book's title was his name for the first method, one that soon became known as "Napier's bones" because it used rods that often were made of animal bone. T h e reader is urged to make a set of Napier's bones by labeling 11 strips of heavy cardboard (or Popsicle sticks, tongue depressors or any other available wooden strips) as shown in Figure 50. T h e index rod is not essential, but it makes it easier to locate desired rows. Each of the rods has a digit at the top. Below the digit, from the top down, are the products when that digit is multiplied successively by numbers 1 through 9. The set of bones obviously is nothing more than a multiplication table cut into strips so that it can be manipulated manually, with a zero strip added to serve as a placeholder. T h e procedure is ridiculously simple. Suppose you wish to multiply 4,896 by 7. Rods topped with 4, 8 , 9 and 6 are placed side by side with the index rod on the left [see Figure 511. Only row 7 (the multiplier) is considered. Write down 2, the last digit of the row, as the final digit ofthe product. T h e product's next digit (working to the left on both rods and paper) is obtained by adding the next pair of digits (the diagonally adjacent digits inside the little parallelogram) of the row. They are 4 3, so put down 7 as the second digit from the end of your product. T h e sum of the next pair (6 6 = 12) is more than 9, therefore write 2 as the third digit ofthe product and carry 1.T h e next pair, 5 and 8, add to 13, but you are carrying 1,so the sum is 14. Put down 4 and again carry 1.The last digit of the row is 2. Two plus 1 is 3, so 3 is the final digit (on the left) of your product.
+
+
NAPIER'S BONES
89
Figure 51
4,896 X 7
= 34,272
You have now obtained the correct answer, 34,272, by using only simple addition. Of course, if you know your multiplication table through the 9's, you can do it just as easily without the rods. In Napier's day, however, the ordinary person's ability to calculate was feeble, so the rods became an instant success throughout Great Britain and continental Europe. T o multiply 4,896 by a larger number, say 327, it is necessary to obtain three partial products and add them in the usual way. In other words, write down 34,272 (the product of 4,896 and 7 ) ; then put below it the products obtained from rows 2 and 3, jogging them to the left in the standard manner,
then add to obtain the final product. T h e rods are of little use unless you have more than one set because a multiplicand may contain duplicate digits. Napier's rods had square cross sections, each face of a rod corresponding to one of the strips in our cardboard set. He arranged the four columns so that the top digits on opposite sides of each rod added to 9. T h e following are the quadruplets of Napier's set of 10 bones:
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90
It is clear that such a set of 10 rods can handle all multiplicands of 10 digits or fewer that are possible to form with the rods, but many multiplicands cannot be formed, so it was advisable to own more than one set. As a little puzzle in combinatorics, can the reader determine the largest multiplicand one set of Napier's bones will form such that all smaller multiplicands can also be formed by the set? As a second exercise, find the corresponding largest multiplicand for two sets of Napier's bones. Napierls rods can be used for division too, but the process is more trouble than it is worth. In short division you must select rods that form the desired dividend on the row for the digit divisor and read offthe quotient from the top.
g rods
NAPIER’S BONES
91
Figure 53 673 × 8 = 5,384
If the dividend cannot be formed, form the largest number you can that is less than the dividend and subtract that number from the dividend to obtain the remainder. In long division the rods can be used for determining the successive products of the divisor and each digit in the quotient. The charm of Napier’s rods lies in their simplicity. If we are willing to complicate them a bit, however, we can eliminate the bother of having to carry 1’s in our head. The cleverest way of doing this was invented about 1890 by Henri Genaille, a French civil engineer. The picture of these rods is almost selfexplanatory [see Figure 52]. They work exactly like Napier’s except that the product is read directly from right to left. Start with the digit at the top right of the desired row. The next digit is the one to which the shaded triangle (at the left of the previous digit) points. From now on move from each digit into the shaded triangle directly at its left and go to the digit to which it points. For example, to multiply 673 by 8, start with 4 at the top right [see Figure 53], and see how easily you can move to the left through the chain of triangles to obtain the product 5,384. Both Napier’s bones and Genaille’s rods are marvelous teaching devices because it is not hard to see why they work, and when you do, you obtain valuable insight into the multiplication procedure. If you have difficulty understanding why Genaille’s rods operate, you can find it explained in the article by B. R. Jones (see the bibliography), from which our illustrations were taken. The second calculating method in Rabdologia had to do with arranging metal plates inside a box. It is too complicated and impractical to explain here. But Napier’s third method, which he regarded as being primarily an amusement, requires only a chessboard and a supply of counters. By moving the counters as you would rooks or bishops, you can do addition, subtraction, multiplication, division and square roots all in the binary system, as explained in the next chapter.
CHAPTER SEVEN
92
ANSWERS A single set of Napier's original 10 rods will form every multiplicand of 11,110 or less. Two sets will form every multiplicand of 111,111,110or less. For n sets of bones the number is 471 1's followed by a 0.
ADDENDUM Napier had no notion of a "base" for his logarithms. The matter is complicated, but, as Carl Boyer explains in his History of Mathematics, if you divide all Napier's numbers and logarithms by lo7, you have a system practically the same as one based on lie. Natural logs based on e later came to be known as Napierian logarithms, even though Napier never had such a system. Boyer does a good job clarifying the confusing details. Napier's bones were based on an ancient way of multiplying that came to be called the Gelosia system, because its lattice lines looked like the gratings on Italian windows. There is a good account of this, together with a survey of the curious mechanical devices (some with rotating cylindrical rods) that came after Napier's bones, in the paper by M. R. Williams cited in the bibliography. I had assumed that David Ramsay, mentioned in the first paragraph of the chapter, was invented by Scott. Not so. He actually lived and made and sold clocks and watches for a living. He served James I as an astrologer, as did his son William. In 1652 William published a book on astrology with a curious dedication to his father that reads in part: "It's true your carelessness in laying up while the sun shone for the tempests of a stormy day hath given occasion to some inferiorspirited people not to value you according to what you are by nature and in yourself, for such look not to a man longer than he is in prosperity. . ." William Lilly, a famous British astrologer of the time, wrote an autobiography in which he gives a hilarious account of how he, David Ramsay and others tried to locate a treasure reportedly buried in the cloisters of Westminster Abbey. It was late at night, and a great wind developed that prevented their dowsing rods from turning. Lilly writes that he "dismissed the demons," but that the real cause of their failure was that they were surrounded by more than 30 people who kept laughing and deriding them. Lilly's autobiography is also the source of the anecdote I gave about the first meeting of Napier and Briggs. Where today are Napier's body bones? Nobody seems to know. As Williams discloses, there are reports of his having been buried in at least two different spots in Edinburgh.
NAPIER'S BONES
93
BIBLIOGRAPHY T h e Art of Numbering by SpeakingRods: Vulgarly Termed Napier's Bones. W . Leybourn. London, 1667. "John Napier" and "Logarithms." J. W . L. Glaisher in The Encyclopedia Britannica, 11th edition, 1911.
Napier Tercentenary Memorial Volume. Edited by Cargill Gilston Knott. Longmans, 1915. "Lord Napier First Scottish Expositor of the Revelation." Leroy Edwin Frgom in T h e Prophetic Faith of Our Fathers, Vol. 2. Review and Herald, 1948. Froom is a SeventhDay Adventist. His section on Napier is the best account of Napier's eschatology that I know. "Genaille's Rods: An Ingenious Improvement on Napier's." B. R. Jones in The Mathematical Gazette, Vol. 48, 1964, pages 1722. "John Napier and the History of Logarithms." N. T. Gridgeman in Scripta Mathematica, Vol. 29, 1969, pages 4965. "From Napier to Lucas: The Use of Napier's Bones in Calculating Instruments."
M. R. Williams in Annals ofthe History ofComputing, Vol. 5 , 1983, pages 279 296; see also comments in Vol. 6, 1984, pages 403  404. "Napier's Bones." Michael R. Williams in A History of Computing Technology. PrenticeHall, 1985.
CHAPTER EIGHT
Napier's Abacus "Napierls bones" (the topic of the previous chapter) are the calculating rods that were invented by John Napier, the 16thcentury Scottish mathematician who discovered logarithms. In Rabdologia, the book in which Napier first explained his "bones," he also described a curious method of calculating by moving counters across a chessboard. This method, which seems to have been completely forgotten, deserves to be remembered for several reasons: It is not only a pleasant recreation but also a valuable teaching device and of considerable historic interest. It is the world's first binary computer, and it came almost 100 years before Leibniz explained how to calculate with binary numbers! Although Napier did not express numbers explicitly in binary notation, we shall see how his counting board is equivalent to doing so. T h e use of checkered boards and cloths for calculating was widespread in Europe during the Middle Ages and the Renaissance. English words such as "exchequer," "check" and "counter" derive from these boards; even "bank" comes from the German word for counting board, Rechenbank. T h e algorithms for calculating on these boards, however, were clumsy. By adopting a binary system and basing his algorithms on old methods ofmultiplying by "doubling," Napier created a remarkably efficient counting board unlike any that had been in use before. Napier's counting board is a chessboard of arbitrary size, with columns and rows labeled by the doubling series l , 2 , 4 , 8 ,1 6 , 3 2 , . . . . These numbers are, of course, successive powers of 2. Before explaining Napier's methods for multiplying, dividing and extracting square roots, let us see how his board can be used for addition and subtraction. Suppose we want to add 89 41 52 14. Each number is expressed by placing counters on a row of the board [see Figure 54a]. A counter has the value of its column. (Ignore the row numbers on the right margin.) Thus the fourth row shows 8 9 as the sum of 64 16 8 1. If you think of each counter as 1 and the empty cells as 0, then 89 is
+
+
+ +
+ +
NAPIER'S ABACUS 95
Figure 54
Binary addition: 89
+ 41 + 52 + 1 4
represented in binary notation as 1011001, and similar notations can be made for the other three numbers. T h e counters can be positioned rapidly because any positive integer is uniquely represented as a sum of the powers of 2. Start at the left and put a counter on the largest power less than the number to be represented; then move right and place a counter on the next larger power that, when added to the previous power, will not exceed the desired number. Continue in this way until the unique binary representation is obtained. T o add the four numbers, first move all the counters down like rooks in chess to the bottom row [see Figure 54b]. Adding the values of all these counters will give the correct sum, but we want to express the sum in binary notation. T o do this, "clear" the row of multiple counters on a cell by the following procedure. Start at the right, taking each cell in turn. Remove every pair of counters on a cell and replace them with a single counter on the adjacent cell to the left. W e shall call this "halving up." Clearly it will not affect the sum of the counters's values because every pair of counters of value n is replaced by one counter of value Zn. T h e final result after clearing is the binary number 11000100, or 196 in decimal notation [see Figure 54~1. Subtraction is almost as simple. Suppose you want to take 83 from 108. Represent the larger number on the second row and the smaller on the bottom row, as shown in Figure 55a. You can now do subtraction in the usual manner, starting at the right and borrowing as you go, but I prefer to alter the entire second row (preserving the total value of its counters) until each counter on the bottom row has one or two counters above it, and no empty cell on the bottom
CHAPTER EIGHT 96
Figure 55
Binary subtraction: 108  83
row has more than one counter above it. This is done by "doubling down" on the second row removing a counter and replacing it with two counters on the next cell to the right. How the top row looks after it is transformed to meet the two specified conditions is shown in b. T h e next step is to "king" (as in checkers) each counter on the bottom row by moving a counter on top of it from the cell directly above. After this is done, the second row shows in binary notation the difference between the two numbers. In this case it is 11001,or 25, as shown in Figure 55c. A different subtraction method is to "complement" the smaller number and add. A number is complemented by placing a counter on each of its empty cells and removing all counters originally there. In other words, each 0 is changed to 1 , and each 1to 0. (If the subtrahend has fewer digits than the minuend, before complementing you must add zeros to the left of the subtrahend until it is the same length as the minuend.) Add the two numbers, clear the row by halving up and transfer the counter at the extreme left to the extreme right. Clear again if necessary. T o use the preceding example, we change 1010011 to its complement 101100. Adding and clearing produces 10011000. Shifting the counter from left to right gives 11001, or 25, the correct difference. Multiplication is delightfully easy. Napier explains it with the example 19 X 13 = 247. One number, say 19, is indicated below the board by marking the proper columns; the other number, 13, by marking the proper rows. A counter goes on each cell at the intersection of a marked row and column [see Figure 5 6 ~ 1Every . counter not on the column at the extreme right is moved diagonally up and right (like a chess bishop) until it is on the rightmost column. T h e result is shown in b. T h e sum of the values of these counters (as indicated on the right margin) is 247, the desired product, but we wish to express it in binary notation. That is quickly done by halving up until the column is cleared. T h e final result is 11110111, or 247, as shown in c. It is easy to see why it works. Counters on the first row keep their values when moved to the right, counters on the second row double in value, counters on the
I I I I*] Figure 56
1J
Binary multiplication: 19 X 13
third row quadruple in value and so on. The procedure is equivalent to multiplying with logarithms to base 2. In our example, 19 is expressed as 2' 2', and 1 3 as z3 22 2'. Cross multiplying in the familiar manner (remembering the basic law of exponents: x n X x rn = xn+rn) yields z7 2 . z4 2 . 23 2' 2' 2'. This corresponds exactly to Napier's procedure. Indeed, moving the counters is equivalent to cross multiplying. W e are, in effect, multiplying by adding exponents. Napier was not the first to recognize that powers of 2 can be multiplied by adding their exponents. As early as 1500 it had been clearly explained with exponential notation by Nicolas Chuquet, a physician oflyons, in the algebraic part of his Triparty en la sciences des nombres. It is Napier, however, who gets the credit for the first mechanical device operating with logs based on 2. Napier next explains how to do long division on his abacus, using the example 250 + 13. T h e procedure, as one would expect. is the reverse of
z6 +
+ + + + + +
z4 +
+
+
CHAPTER EIGHT 98
.l.I.
v REMAINDER
Figure 57
Binary division: 250 + 13
multiplication. Complications arise that make it difficult to explain, although in practice one soon learns to do it quickly. T h e divisor, 13, is marked at the bottom ofthe board, and the dividend is indicated by counters on the column at the extreme right [see Figure 57a]. You must now move the dividend counters like chess bishops, down and left, to produce a pattern that has counters (one to a cell) only on marked columns, and each marked column must have its counters on the same rows. Only one such pattern can be formed, but to do so it is necessary at times to double down on the right column, that is, remove single counters, replacing each with a pair of counters on the next lower cell. Start with the top counter and move it diagonally to the leftmost marked column. If you see that you cannot proceed to form the desired pattern, return the counter to its original cell, double down and try again. If the first attempt fails, the second will succeed in beginning the required pattern, although more doubling down may be necessary. Continue in this manner, doubling down whenever you see that you must, gradually filling in the pattern by extending it down and right until finally the unique pattern is constructed [see Figure 5 7b]. After the final counter at the bottom right corner of the pattern is in place, you will have three counters left over. They represent the remainder. T h e rows containing counters are marked on the right margin, symbolizing 10011,or 19, the correct quotient. T h e three extra counters give the fraction 3/13.
NAPIER'S ABACUS 99
...
.a. .a. ..a
4/
REMAINDER
Figure 58
Binary extraction of square root:
J1,238
A similar procedure is used to find integral square roots. If the root is not integral, the procedure gives the root of the largest square less than the original number. Counters left over then represent the difference between that number and the original. Napier demonstrates by finding the square root of the largest square less than 1,238. This requires a board extended higher than the standard chessboard. As in division, the number is represented by counters on the rightmost column [see Figure 58a]. Since no divisor is marked on the bottom, how do we form a pattern? W e must move counters diagonally down to produce a pattern with two properties: (1)Every column with counters must have its counters on the same rows, and (2) the pattern must have bilateral symmetry along the diagonal passing through the board's lower right corner. This ensures, of course, that multiplier and multiplicand are identical. As before, start with the top counter and see if you can move it to the diagonal of symmetry. If you can, that is the correct first move. If you cannot, double down and move one of the counters to the diagonal of symmetry. Continue in this fashion, doubling down when necessary, until the required symmetrical pattern is achieved. T h e result is 35 X 35 = 1,225, with 1 3 leftover counters that represent the difference between the square and 1,238.
CHAPTER EIGHT
Figure 59
100
Patterns for squaring 1 through 15
T h e 15 patterns that generate all squares from 1through 225 are shown in Figure 59. Studying them will familiarize you with the kind of pattern that must be formed for square roots. Note that in every pattern each row and column has a counter on the diagonal of symmetry. Napier's device will operate with any base notation, but above base 2 it is necessary to work with multiple counters on single cells. As the base increases, the system becomes progressively more cumbersome and uninteresting, and more multiplying must be done in the head. For example, to multiply 77 by 77 in decimal notation each of the four cells at the lower right corner must hold 7 X 7 = 49 counters. After moving them to the right column you have 49 counters on the bottom cell, 98 on the next and 49 on the next. Then every set of 10 counters on a cell is replaced by a single counter immediately above it, resulting finally in counters on four cells that signify the product, 5,929. T h e most interesting extension of Napier's board was suggested by Donald E. Knuth, the Stanford computer scientist. A checkered board can be used very efficiently for calculating in the "negabinary system." Because this remarkable notation is based on powers of  2 , the rows and columns of the board are labeled with the series 1,2, 4,  8, 16,  32, . . . , in which alternate powers are negative. The main virtue of negabinary is that every positive and every negative integer can now be uniquely represented in binary notation without the use of signs. Examples are 13 = 11101 (16  8 4 1) and 13 = 110111 (32 16 4  2 k 1).
+
+
+ +
+
+ +
NAPIER'S ABACUS
101
Figure 60
Negabinaq notation of integers
T h e negabinary forms of positive and negative integers from 1 through 20 are shown in Figure 60. Note that every positive number has an odd number of negabinary digits and every negative number has an even number of negabinary digits. Every odd number, regardless of sign, ends in 1; every even number, regardless of sign, ends in 0. Many other basic theorems are easily discovered. For example, a negabinary number is divisible by 3 if and only if its number of 1's is a multiple of 3. Observe that every palindromic negabinary number on the list (a number that is the same in both directions) is a positive or a negative prime. Is this true in general? If not, what is the first exception? I know of no better way to become acquainted with this extraordinary notation (so rich in recreational possibilities) than to calculate with it on Napier's board. Addition is handled exactly as before except that in clearing the sum the following two rules are observed:
1. A pair of counters on one cell and a single counter on the next higher cell cancel one another. Remove all three. 2. If any cells still have double counters, remove each pair and put single counters on each of the two next higher cells.
CHAPTER EIGHT
102
Figure 61 Clearing rules for negabinary and Fibonacci notations
The clearing procedure, thanks to the cancellation rule, is unusually rapid [see Figure 61, left]. The fastest way to do subtraction is to change the sign of the subtrahend and add! Changing the sign is the same as multiplying by –1, or 11 in negabinary. Since multiplying by 11 is the same as adding a number to itself, with one replica shifted one cell to the left, we can reverse the sign of any negabinary number by the following simple algorithm: Add a new counter to every cell that is immediately to the left of a counter originally there, then clear the row as explained. For example, 11(–1 in decimal notation) becomes 121, but the first two digits cancel (by rule 1), leaving 1, which is positive. Applying the algorithm again restores 11, or –1. When this algorithm is used on standard binary numbers, by the way, it is the same as multiplying by 3. (Do you see why?) Any two negabinary numbers can be multiplied by using Napier’s procedure and clearing the result according to negabinary rules. The product will have the correct sign when translated into decimal notation. Try multiplying –4 and –6. They are 1100 and 1110 in negabinary [see Figure 62]. After multiplying and clearing, you get 1101000, or +24. If you had multiplied –4 and +6, or +4 and –6, the result would have been 111000, or –24. Division and squareroot procedures are much trickier, although interested readers should be able to devise them. In square roots both positive and negative roots appear as solutions. Are there ways to use Napier’s board efficiently for converting a signed binary number to negabinary, and vice versa? Yes; we can exploit two simple algorithms given by Knuth as the answer to Exercise 12 on page 177 of his Seminumerical Algorithms [see bibliography]. Readers are encouraged to work them out before checking the answers section. It is hard to believe, but the idea of negativebase notation (it applies to any radix) did not occur to anyone until the 1950’s, when many people independently thought of it. In 1955, when Knuth was a high school senior, he wrote a short paper on it for a science talent search, but the first published account (at least in English) seems to be a short letter by Louis B. Wadel in IRE Transactions on Electronic Computers (Vol. EC6, 1957, page 123). The term “negabi
NAPIER'S ABACUS
Figure 6 2
103
Negabinary multiplication:
4
X 6
nary" was coined by Maurits P. de Regt, wk ose series of pioneering articles on negative radix arithmetic is listed in the bibliography. Knuth also suggests the Fibonacci labeling, 1, 2 , 3, 5, 8, 13, . . . , for Napier's board. It is difficult to multiply or divide with it, but addition and subtraction can be handled by representing each integer as the sum of the fewest possible Fibonacci numbers. Start by putting a counter on the column with the highest value less than the number to be represented; then work downward until the desired sum is obtained. (This method of representing a number uniquely in Fibonacci notation is known as Zeckendorf's theorem.) For example, 19 is uniquely indicated by 101001, or 13 5 1. T h e adding procedure is the same as Napier's except that a row is cleared by the following two rules:
+ +
1. If single counters are on adjacent cells of the board, remove them and put one counter on the next higher cell. 2. For every pair of counters on the same cell, remove them and put one counter on the next higher cell and one on the second lower cell. For example, two counters on cell 1 3 are replaced by one on cell 21 and one on cell 5 [see Figure 61, right].
CHAPTER EIGHT
104
If you imagine the row extended two more cells to the right, with values of 1 and 0 (or, alternatively, that the columns are labeled 0 , 1, 1, 2, 3, 5 , . . . ), then the above two rules suffice. Otherwise there are two exceptions. A pair of counters on 2 is replaced by one on 3 and one on 1,and a pair on 1is replaced by one on 2. T o subtract, I know of no better way than the "kinging" procedure explained for binary subtraction. You must, of course, first change the minuend to the required pattern by applying the two clearing rules in reverse. There may be a better method. Indeed, there may be all kinds of clever algorithms for calculating on Napierls board, in various notations, that no one has yet discovered.
ANSWERS T o change a signed binary number to negabinary
1. Express the number in binary on row 2. 2. If the number is positive, move all counters that have negative values (in negabinary) down like rooks to the first row. ( O n a standard chessboard this means moving down all counters on white squares.) If the number is negative, move down all counters of positive value (those on black squares).
3 . Regard both rows as negabinary numbers. Subtract the first row from the second, using the procedure explained in the previous chapter for negabinary subtraction.
4.
Clear the bottom row by negabinary rules T o convert a negabinary number to a signed binary
1. Express the number in negabinary on row 2.
2.
If the number is positive (an odd nurnber of digits), move down all the negative counters (white squares). Ifthe nurnber is negative (an even number of digits), move down all positive counters (black squares).
3. Regard both rows as binary numbers. Subtract the first row from the second, using a binary procedure.
4. Clear the answer by binary rules and prefix the proper sign (plus if the original number was positive, minus if it was negative). T h e answer to the question about negabinary palindromes is that the smallest composite number that is palindromic in negabinary is 21. Its positive form is 10101; its negative form is 111111.
Figure 63 Fibonacci notation for 7 X 7
ADDENDUM John Harris of Santa Barbara, Calif., discovered an ingenious way to multiply numbers in Fibonacci notation, using the Napier counting board. He added an extra 1row and Icolumn outside the heavy line to the counting board [see Figure 631. Suppose you want to multiply 7 by 7. Place the counters according to
CHAPTER EIGHT
106
Napier's rules [see a]. More counters are now positioned according to the following rule: O n the diagonal that extends down and to the right from each counter, n, put a counter on every alternate cell, starting with the cell two cells away from counter n [b]. Each counter outside the heavy line is moved to the nearest cell inside the line [c]. Now move all counters u p and to the right along their diagonals to the heavy line [dl. Clear the column according to the Fibonacci clearing rules given earlier [el. T h e counters, reading from the top down, give the correct product in Fibonacci notation. Readers familiar with the Fibonacci series will enjoy proving that Harris's algorithm works. Division by this method, however, seems to be hopelessly complicated. Napier's abacus furnishes insights into many important combinatorial formulas. For example, in how many ways can you make a selection from n different objects? T h e answer 2"  1 is apparent from the way the columns (or rows) are labeled. Let the eight columns of the standard chessboard be eight objects. Each selection of columns corresponds to a binary number from 1 to I 1 111111,or 255. That 255 = 28  1 is obvious, because adding 1to it makes the binary number 100000000, or 28 = 2.56. Assuming one counter to a cell, we ca.n ask several questions about the number of patterns of a specified kind that can be placed on an n X n chessboard. How many patterns can be formed in which each nonempty column has its counters on the same rows? Clearly this is the same as asking how many products can be made by multiplying two numbers, each from 1 through 2"  1. How many of these patterns have bilateral symmetry along the main diagonal that passes through the board's lower right corner? This is the same as asking how many squares can be made by squaring a number from 1through 2"  1. How many patterns can be made with no restrictions whatever? Think of the rows as joined to form one long chain of n X n cells. Every pattern will be expressed by a binary number from 1 through 2(nXn) 1. If we count the absence of all counters as a pattern, the number of patterns possible is 2"'. Donald Knuth called my attention to the entertaining article "Binary Notation," by E. William Phillips in the British publicationJourna1 of the Institute of Actuaries, Vol. 67,1936,pages 187 221. T h e purpose ofthe paper is to defend a notation based on 8 as superior to decimal notation. T o show how easily numbers can be multiplied when given in binary notation, the author reinvents Napier's abacus without realizing it. Christopher J. Schultz wrote to propose the following procedures for changing a signed binary number to a negabinary number, and vice versa. In many ways they are simpler than the algorithms I gave.
NAPIER'S ABACUS
107
1. If the number is positive, check the nexttorightmost column; if negative, check the rightmost column.
2. If the column contains a counter, consider the columns to the left of it a complete binary number and add 1 to it, using binary arithmetic and clearing rules. 3. Move two columns to the left and repeat step 2. Continue in this way through the entire number. T o change a negabinary number to a signed binary number: 1.
Starting at the left, check the nexttofirst column.
2.
If the column contains a counter, consider the columns to the left of it a complete binary number and subtract 1 from it, using binary arithmetic and clearing rules.
3. Move two columns to the right and repeat step 2. Ifthe last column checked is the rightmost column, sign the number negative; otherwise sign it positive. Many readers suggested ways, which they considered better than the one I gave, for performing division on Napier's abacus and also for dividing and doing square roots in Fibonacci notation. Craige Schensted was inspired by Napier's device to invent a chessboard computer on which many astonishing calculations can be made. T h e basic idea is to allow the columns to be labeled with the powers of one base and the rows to be labeled with the powers of a different base. Each cell represents the product of its row and column numbers. It would require a long chapter to do justice to the elegant ways Schensted found for using such a board to solve problems that otherwise would be difficult. I gave 1950 as the date on which papers about negativebase number systems first began to appear. In his History ofBinary and other Nondecimal Numeration [see the bibliography], Anton Glaser disclosed that in 1885 Vittorio Griinwald published an article in which he covered all the basic arithmetical operations in a negative10 system. This is the only reference to negativebase notation known to me prior to 1950.
BIBLIOGRAPHY On Negative Bases " O n Bases for the Sets of Integers." N. G. De Bruijn in Publication Mathematics Debrecen, Vol. 1 , 1950, pages 232242.
CHAPTER EIGHT 108
"A Look at Base Negative Ten." Richard D. Twaddle in Mathematics Teacher, Vol. 56, 1963, pages 8890. "Using a Negative Base for Nuniber Notation." Chauncy H. Wells, Jr., in Mathematics Teacher, Vol. 56. 1963, pages 91 93.
"Negative Radix Arithmetic." Maurits de Regt in Computer Design, Vols. 6 and 7. 1967, 1968.
The Art of Computer Programming, Vol. 2, Seminumerical Algorithms. Donald E. Knuth. '4ddisonWesley, 1969, page 171, and exercises on pages 176, 177, and 179. Histor?, of B i n a ~and Other Nondecimal Numeration. Anton Glaser, privately published, 1971. "Negative Based Number Systems." W . J. Grlbert and R. James Green in Mathematics Magazine, Vol. 52. 1979, pages 240  244.
On Fibonacci Notation
Fibonacci and Lucas Numbers. Verner E. Hoggatt. Jr. Houghton Mifflin, 1969, pages 7071. "Zeckendorf's Theorem and Some Applications." J. L. Brown, Jr., in Fibonacci Quarterly, Vol. 2, 1964, pages 162  168. "Representations of Natural Numbers as Sums of Generalized Fibonacci Numbers." D . E. Daykin injournal ofthe London Mathematical Society, Vol. 35, 1960, pages 143 160. "Generalizations of Zeckendorf's Theorem." Timothy J. Keller in Fibonacci Quarterly, Vol. 10, 1972, pages 95 112.
CHAPTER
NINE
Sim, Chomp and Race track New mathematical games of a competitive type, demanding more intellectual skill than luck, continue to proliferate both in the U.S. and abroad. In Britain they have become so popular that a monthly periodical called Games and Puzzles was started in 1972 just to keep devotees informed. st rate^ and Tactics (a bimonthly with offices in New York City) is primarily concerned with games that simulate political or military conflicts, but a column by Sidney Sackson reports on new mathematical games of all kinds. Sackson's book A Gamut of Games (Random House, 1969) has a bibliography of more than 200 of the best mathematical board games now on the market. Simulation games are games that model some aspect of human conflict: war, population growth, pollution, marriage, sex, the stock market, elections, racism, gangsterismalmost anything at all. They are being used as teaching devices, and some notion of how widely can be gained from the fact that a 1973 catalogue, The Guide to Simulation Gamesfor Education and Training, by David W . Zuckerman and Robert E. Horn, runs to 500 pages. We will take a look at three unusual new mathematical games. None requires a special board or equipment; all that is needed are pencil and paper (graph paper for the first game) and (for the third) a supply of counters. Race Track, virtually unknown in this country, is a truly remarkable simulation of automobile racing. I do not know who invented it. It was called to my attention by Jurg Nievergelt, a computer scientist at the University of Illinois, who picked it up on a recent trip to Switzerland. T h e game is played on graph paper. A racetrack wide enough to accommodate a car for each player is drawn on the sheet. T h e track may be of any length or shape, but to make the game interesting it should be strongly curved [see Figure 641. Each contestant should have a pencil or pen of a different color. T o line up the cars, each player draws a tiny box just below a grid point on the starting line. In the example illustrated the track will take three cars, but for
CHAPTER NINE
Figure 64
110
The Race Track game
simplicity a race of two cars is shown. Lots can be drawn to decide the order of moving. In the sample game, provided by Nievergelt, Black moves first. You might suppose that a randomizing device now comes into play to determine how the cars move, but such is not the case. At each turn a player
SIM, CHOMP AND RACE TRACK
111
simply moves his car ahead along the track to a new grid point, subject to the following three rules:
1. T h e new grid point and the straight line segmentjoining it to the preceding grid point must lie entirely within the track.
2. No two cars may simultaneously occupy the same grid point. In other words, no collisions are allowed. For instance, consider move 22. Gray, the second player, would probably have preferred to go to the spot taken by Black on his 22nd move, but the nocollision rule prevented it. 3. Acceleration and deceleration are simulated in the following ingenious way. Assume that your previous move was k units vertically and m units horizontally and that your present move is k' vertically and m' horizontally. The absolute difference between k and k' must be either 0 or 1, and the absolute difference between m and m' must be either 0 or 1.In effect, a car can maintain its speed in either direction, or it can change its speed by only one unit distance per move. T h e first move, following this rule, is one unit horizontally or vertically, or both.
The first car to cross the finish line wins. A car that collides with another car or leaves the track is out of the race. In the sample game Gray slows too late to make the first turn efficiently. He narrowly avoids a crash, and the bad turn forces him to fall behind in the middle of the race. He takes the last curve superbly, however, and he wins by crossing the finish line one move ahead of Black. Neither driver, I should add, always makes his best moves. Nievergelt programmed Race Track for the University of Illinois's Plato IV computerassisted instruction system, which uses a new type of graphic display called a plasma panel. Two or three people can play against one another, or one person can play alone. The game became so popular that the authorities made it inaccessible for a week to prevent students from wasting too much time on it. O u r second pencilandpaper game is called Sim, after Gustavus J. Simmons, a mathematician at the Sandia Corporation laboratories in Albuquerque, who invented it when he was working on his Ph.D, thesis on graph theory. He was not the first to think of it (the idea occurred independently to a number of mathematicians), but he was the first to publish it and to analyze it completely with a computer program. In his note titled "On the Game of Sim" (see the bibliography) he says that one of his colleagues picked the name as short for S I M SIMmOnS, ~ ~ and because the game resembles the familiar game of nim. Six points are placed on a sheet of paper to mark the vertexes of a regular hexagon. There are 15 ways to draw straight lines connecting a pair of points, producing what is called the complete graph for six points [see Figure 651. Two
Figure 65 T h e game of Sim Sim players take turns drawing one of the 15 edges of the graph, each using a different color. T h e first player to be forced to form a triangle of his own color (only triangles whose vertexes are among .the six starting points count) is the loser. If only two colors are used for the edges of a chromatic graph, it is not hard to prove that six is the smallest number of points whose complete chromatic graph is certain to contain a triangle with sides all the same color. Simmons gives the proof as follows: "Consider any vertex in a completely filledin game. Since five lines originate there, at least three must be the same color say blue. No one of the three lines joining the end points of these lines can be blue ifthe player is not to form a blue triangle, but then the three interconnecting lines form a red triangle. Hence at least one monochromatic (all one color) triangle must exist, and a drawn game is impossible." With a bit more work a stronger theorem can be established. There must be at least two monochromatic triangles. A detailed proof of this is given by Frank Harary, a University of Michigan graph theorist, in his paper "The TwoTriangle Case of the Acquaintance GraphH[see the bibliography]. Harary calls it an acquaintance graph because it provides the solution to an old brainteaser: Of any six people, prove that at least three are mutual acquaintances or at least three are mutual strangers. Harary not onjy proves that there are at least two such sets but also shows that if there are exactly two, they are of opposite types (colors on the graph) if and only if the two sets have just one person (point) in common. Because Sim cannot be a draw, it follouis that either the first or the second player can always win if he plays correctly. When Simmons wrote his note in 1969, he did not know which player had the win, and in actual play among equally skillful players wins are about equally divided. Later he made an exhaustive computer analysis showing that the second player could always win. Because of symmetry, all first moves are alike. T h e computer results showed
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that the second player could respond by coloring any of the remaining 14 edges and still guarantee himself a win. (Actually, for symmetry reasons, there are only two fundamentally different second moves: one that connects with the first move, and one that does not.) After the first player has made his second move, exactly half of the remaining plays lead to a sure win for the second player and halfto a sure loss, assuming, of course, that both sides play rationally. If 14 moves are made without a win, the last move, by the first player, will always produce two monochromatic triangles of his color. This 14move pattern is unique in the sense that all such patterns are topologically the same. Can you find a way of coloring 14 edges of the Sim graph, seven in one color and seven in another, so that there is no monochromatic triangle on the field? T h e most interesting unanswered question about Sim is whether there is a relatively simple strategy by which the second player can win without having to memorize all the correct responses. Even if he has at hand a computer printout ofthe total game tree, it is of little practical use because it is enormously difficult to locate on the printout a position isomorphic to the one on the board. Simmons's computer results have been verified by programs written by Michael Beeler at the Artificial Intelligence Laboratory of the Massachusetts Institute of Technology and, more recently by Jesse W. Croach, Jr., of West Grove, Pa., but no one has been able to extract from the game tree a useful mnemonic for the second player. Sim can, of course, be played on other graphs. O n complete graphs for three and four points the game is trivial, and for more than six points it becomes too complicated. The pentagonal fivepoint graph, however, is playable. Although a draw is possible, I am not aware of any proof that a draw is inevitable if both sides make their best moves. Our third game, which I call Chomp, is a nimtype game invented by David Gale, a mathematician and economist at the University of California at Berkeley. Gale is the inventor of Bridgit, a popular topological board game still on the market. What follows is based entirely on results provided by Gale. Chomp can be played with a supply of counters [see Figure 661 or with 0 ' s or X ' s on a sheet of paper. The counters are arranged in a rectangular formation. Two players take turns removing counters as follows. Any counter is selected. Imagine that this counter is inside the vertex of a right angle through the field, the base of the angle extending east below the counter's row and its other side extending vertically north along the left side of the counter's column. All counters inside the right angle are removed. This constitutes a move. It is as though the field were a cracker and a rightangled bite were taken from it by jaws approaching the cracker from the northeast. 

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Figure 66 Chomp on a 5by6 field T h e object ofthe game is to force your opponent to chomp the poison counter at the lower left corner of the array [black counter]. The reverse form of Chomp winning by taking this counter is trivial because the first player can always win on his first move by swallowing the entire rectangle. What is known about this game? First, we dispose of two special cases for which winning strategies have been found.
1. When the field is square, the first player wins by taking a square bite whose side is one less than that ofthe original squ