David Gale*

For the general philosophy of this section see Vol. 13, no. 1 (1991). Contributors to this column who wish an acknowledgment of their contributions should enclose a self-addressed postcard.

We All Make Mistakes II

In response to my request for examples of of mistakes by famous mathematicians several people mentioned a well-known error of Lebesgue. The following nice de- scription was supplied by Doug Lind.

"Lebesgue was trying to prove that the projection of a Borel set in the plane is a Borel subset of the line [Sur les fonctions repr6sentable analytiquement, Journal de Math~matiques, Series 6, Volume 1 (1905), page 195]. His argument uses the 'fact' that if you have a decreas- ing sequence of sets in the plane, then the projection of their intersection is the intersection of their projections (this is contained in the third paragraph, starting with "Supposons que e soit F de classe 1 ou 2"). Of course this is wrong (no first-year graduate student should make such a mistake!), and led to the Souslin-Lusin theory of analytic sets." [20-second time-out while the reader finds the obvious counterexample.]

"Lusin even asked Lebesgue to write a preface to his book on analytic sets [Lefons sur les Ensembles Analy- tiques et leurs Applications]. There Lebesgue says that this was the most fruitful error that he had ever com- mitted!"

Careful Card-Shuffling and Cutting Can Create Chaos

Imagine (if you can) a countably infinite deck of cards. Each card is marked with a different natural number and initially the cards are arranged in their natural

order with card 1 on the top and placed face down on a (finite) table.

Definition A perfect n-shuffle consists in picking up the top n cards of the deck and interlacing them with the next n. Thus, if one executes a 5-shuffle on the deck in its initial ordering, the resulting ordering will be

6,1,7,2,8,3,9,4,10,5,11,12,13 . . . .

Consider now executing a sequence of shuffles, first a 1-shuffle, then a 2-shuffle, then a 3-shuffle etc.

CONJECTURE. In the course of this sequence of shuffles every card will come to the top of the deck infinitely often.

This conjecture is a mild modification of a conjecture of Richard Guy, as will be explained later, so I will refer to it from now on as Guy's conjecture. Here are the orderings given by the first eight shuffles:

0 1,2,3 . . . . 1 2,1,3,4,5 . . . . 2 3,2,4,1,5,6 . . . . 3 1,3,5,2,6,4,7,8,9 . . . . 4 6,1,4,3,7,5,8,2,9,10 . . . . 5 5,6,8,1,2,4,9,3,10,7,11,12 . . . . 6 9,5,3,6,10,8,7,1,11,2,12,4,13,14 . . . . 7 1,9,11,5,2,3,12,6,4,10,13,8,14,7,15,16 . . . . 8 4,1,10,9,13,11,8,5,14,2,7,3,15,12,16,6,17,18 . . . .

* Column editor's address: Depar tment of Mathematics, University of California, Berkeley, CA 94720 USA.

Note that at this point cards 1 through 6 have made their way to the top of the deck, 1 having been there already three times. However card 7 won' t get there

54 THE MATHEMATICAL INTELLIGENCER VOL. 14, NO. 1 �9 1992 Springer Verlag New York

until shuffle 78, a first indication of the sort of erratic behavior to be described shortly.

Getting back to Guy's conjecture, what basis is there for believing it may be correct? There are two argu- ments for it, the first "empirical," the second "proba- bilistic" (quotation marks here mean that what we are about to say is, rigorously speaking, nonsense). We consider first the probabilistic argument suggested by Raphael Robinson. Note first that card c stays in its place until the [c/2]th shuffle after which it may bounce around in a seemingly turbulent manner, but it can never reach a position further than 2n from the top of the deck on the nth shuffle, so the position of c on the nth shuffle is some number between 1 and 2n. Now assuming (here's where the nonsense sets in) that the position of c is random, the probability that it is not at the top of the deck is 1 - 1/2n, and (more nonsense) assuming successive positions are independent, we see that the probability that c never reaches the top of the deck is

1 - [ ( 1 - 1/2n), n = l

which is zero because of the divergence of the har- monic series.

As for "empirical evidence," Ilan Adler has checked all cards from I to 5000, listing for each card the num- ber of the shuffle at which it reaches the top of the deck, and the results are interesting. Things go along comparatively quietly until we get to card 39, which takes 13,932 shuffles to reach the top, after which things settle down again, although card 43 requires 30,452 shuffles; but then there is a major explosion. After card 53 takes a mere 30 shuffles, card 54 goes on a wild rampage (card 54, where are you?), finally mak- ing it to the top on shuffle 252,992,198.

Collecting all the data took about 80 hours of com- puter time on a NeXT Work Station, but most of that time was taken up with three "monsters," 4546, 3729, and the current world champion 3464, which took re- spect ively 2,263,846,432 and 15,009,146,841 and 21,879,255,397 shuffles. Of course, this sort of behavior is what one would expect from the fallacious probabil- ity argument. The longer a card stays away from the top, the longer it is likely to continue to do so, i.e., the chance of "choosing" a I on the millionth shuffle is one in a million.

Since perfect shuffles behave so wildly, perhaps one should look at something simpler. Instead of shuffling one might try simply cutting the cards. The traditional cut takes the top, say, n cards and places them on the bottom of the deck. In our model, however, the bottom is too far away, so instead let us define an n-cut to interchange the top n with the next n cards. Thus a 5-cut on the original order produces

6,7,8,9,10,1,2,3,4,5,11,12,13 . . . .

If we now perform n-cuts in consecutive order starting with a 1-cut then a 2-cut, etc., it is trivial to show that Guy's conjecture is true, for card c remains in place until the [c/2]th cut, after which it moves up the deck by one on every other cut until it hits the top, where- upon it jumps down and then again proceeds to work its way back up to the top. So to make things interest- ing, instead of merely cutting, we will cut, but after each cut we discard the top card. The question is then whether every card is eventually discarded. The statis- tical behavior here is somewhat more restrained than for the shuffles, although there are occasional spurts. For example, card 752 survives over nineteen million cuts before being discarded. It seems though that this problem may be tractable. The orbit of a given card has a clear pattern which the reader will easily find by working a few examples, and the question of how long a card will survive boils down to a question in number theory--which, however, as of this writing has not been settled.

A little bit about the origin of these questions: It all began with a problem proposed by Clark Kimberling which appeared in Crux Mathematicorum volume 7, number 2 (Feb. 1991). Kimberling considers the follow- ing array:

1 2 3 4 5 6 7 8 9 1 0 . . . 2 3 4 5 6 7 8 9 1 0 1 1 . . . 4 2 5 6 7 8 9 1 0 1 1 1 2 . . . 6 2 7 4 8 9 1 0 1 1 1 2 1 3 . . . 8 7 9 2 1 0 6 1 1 1 2 1 3 1 4 . . . 6 2 1 1 9 1 2 7 1 3 8 1 4 1 5 . . .

Here each row is obtained from the previous one by a sort of leap-frog procedure. Start with the number to the right of the diagonal term, which is underlined. Then go to the number to the left of the diagonal, then back to the 2nd number to the right, then the 2nd number to the left, etc., until you reach the first num- ber in the row. Then jump back to the right and leave the remaining numbers in their natural order. Once a number appears on the diagonal it is expelled. Kim- berling now asks, "(a) Is 2 eventually expelled? (b) Is every number eventually expelled?" The procedure is easily interpreted as a shuffle, which I will call the Kimberling Shuffle (sounds like the name of a nine- teen-thirties dance craze), in which on the nth round one discards card n, then reverses the order of the first n - 1 cards and interlaces them with the next n - 1.

Richard Guy noticed right away that the answer to (a) was yes, and in fact 2 is expelled on row (shuffle) 25, as a fairly easy hand calculation shows. Guy then conjectured that (b) is also true, and with the help of his grandson Andy Guy, who is studying computer science at Cambridge, verified the conjecture for all

THE MATHEMATICAL INTELLIGENCER VOL. 14, NO. 1, 1992 5 5

numbers up to 1200. Their table shows the same sort of wild behavior as the one for the perfect n-shuffles. In a private communication Guy has written, "I 'd guess all numbers are expelled, but I also guess that no one's going to prove it." So he has actually made two con- jectures, with the interesting property that if either one is confirmed the other one won' t be.

A Spanish Self-Descriptor

Readers may recall that last spring Lee Sallows gave a recipe for constructing self-descriptive sentences. Ob- viously the procedure is language-independent, so here is a Spanish version constructed by Miguel A. Lerma of the computer science department of the Uni- versidad Politecnica of Madrid.

ESTA FRASE CONTIENE EXACTAMENTE DOSCIENTAS TREINTA Y CINCO LETRAS: VEINTE A'S, UNA B, DIECISEIS C'S, TRECE D'S, TREINTA E'S, DOS F'S, UNA G, UNA H, DIECINUEVE I'S, UNA J, UNA K, DOS L'S, DOS M'S, VEINTIDOS N'S, CATORCE O'S, UNA P, UNA Q, DIEZ R'S, TREINTA Y TRES S'S, DIECINUEVE T'S, DOCE U'S, CINCO V'S, UNA W, DOS X'S, CUATRO Y'S, Y DOS Z'S.

I understand Sallows also has a Dutch example.

Problems

Products of two cycles (91-6) by column editor (These are three problems in one, a qu ick ie , a not-so-quickie, and an unsolved)

The first thing one learns about permutations is that any permutation is the product of disjoint cycles. (a) (quickie) Prove that if disjointness is not required then any permutation is the product of at most two cycles. (All group theorists seem to know this but I have been unable to find any reference to it in the literature.) (b) (unsolved, as far as I know) Is every bijection of a countable set the product of at most two cycles? A cycle here means a bijection which has exactly one orbit which is not a fixed point (so infinite cycles are included, as in infinite cyclic groups). (c) Prove that the bijection on 77 which maps n to n + 3 is the product of two cycles.

5 6 THE MATHEMATICAL INTELLIGENCER VOL. 14, NO. I, 1992

A Re-view of Some Reviews

Among his many noteworthy accomplishments, Paul Erd6s may well hold the all-time world record for the number of papers which he has co-authored. It is in- teresting, therefore, to note that there is at least one paper of which he is the sole author which was, nev- ertheless, in some sense a collaboration. Irving Kap- lansky had just finished writing a review of a (joint)

paper by Erd6s w h e n he encountered the author him- self and ment ioned that he had admired the result but wondered whether the proof of the main theorem, which ran over a page and a half, couldn' t be substan- tially shortened. Erd6s took another look and quickly found that indeed it could. The excerpt from Math. Reviews [Vol. 7, 1946, page 164] reproduced below pro- vides the full story.

So here is a rare example of a paper published in its entirety in two different journals - -and now this makes it three (perhaps another world record).

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