# The Lumber Room

"Consign them to dust and damp by way of preserving them"

## Eugene Curtain and Max Washauer

If you came here because you were reading Peter Winkler’s “7 Puzzles You Think You Must Not Have Heard Correctly”, the names are supposed to be Eugene Curtin and Max Warshauer, and the paper is called “The locker puzzle”, published in The Mathematical Intelligencer, Volume 28, Number 1 (March 2006), pages 28–31.
[If not, you should read the amazing “7 Puzzles You Think You Must Not Have Heard Correctly”, spend several days trying the first problem, read the brilliant solution, and then come back here if you’re interested in learning why no other solution can do better.]

The paper is available here if your institution has access. If not, here’s a sketch of the proof that the strategy cannot be improved upon. [Update 2010-01-06: Oliver Nash has a post about the puzzle, explaining both the original solution and the proof of optimality, here. Just the original solution is also worked out by David MacKay here.]

First, let us modify the rules slightly so that each prisoner must continue looking in boxes until he finds the box containing his name. The prisoners win if no prisoner opens more than 50 (i.e., n/2) boxes. This change obviously makes no difference to the outcome. Let’s call this (modified) game Game 1.

A different game involves all the prisoners being in the room at the same time, and is played as follows. The first prisoner opens boxes until he finds his name (i.e., the number “1”). Then, the lowest-numbered prisoner whose name hasn’t been revealed starts opening boxes until he finds his name. Then the next lowest-numbered whose name hasn’t been revealed opens boxes, and so on. The prisoners win if no one opens more than 50 boxes. Call this Game 2.

Let’s say we observe the prisoners as they play Game 2, and record the order in which boxes were revealed. This completely specifies what happened. For example, (with 10 prisoners instead of 100) if we record the list 2,6,1,4,9,7,10,8,3,5, we know that the first prisoner revealed boxes containing 2, 6, 1, then the third (lowest unrevealed) prisoner opened boxes with 4,9,7,10,8,3, then prisoner 5 opened 5, and they lost because the third prisoner opened 6 > 5 boxes.

• Prove: No matter what strategy the prisoners follow, each permutation has the same probability (1/n!) of being the list recorded.
• Prove: “The classical records-to-cycles bijection”. It sends 2,6,1,4,9,7,10,8,3,5 to (2 6 1)(4 9 7 10 8 3)(5), for example.
• So the probability of the prisoners winning Game 2 (no matter what strategy they follow) is exactly the probability that a random permutation has no cycle of length greater than n/2.
• Prove: Any strategy for Game 1 corresponds to a strategy for Game 2 with the same probability. (Trivial: the only change is that you don’t have to open boxes you’ve already seen opened.)

This proves that the pointer-chasing strategy is optimal for Game 1.

Here’s the puzzle as it was originally considered, still open: suppose there are n prisoners and 2n boxes, half of them empty. The prisoners can each examine n lockers. The pointer-chasing strategy doesn’t work as empty boxes point nowhere. Does the probability of winning go to 0 as n→∞?

Written by S

Mon, 2009-08-10 at 23:45:25

## “Every good theorem must have a good counterexample”

Abhyankar[1] attributes the quote to Severi.

[1]: Historical Ramblings in Algebraic Geometry and Related Algebra, Shreeram S. Abhyankar, The American Mathematical Monthly, Vol. 83, No. 6 (Jun. – Jul., 1976), pp. 409-448. Also available here, because it won a Lester R. Ford Award (“articles of expository excellence”) and also a Chauvenet Prize (“the highest award for mathematical expository writing”).

Abhyankar, after distinguishing between the flavours of “high-school algebra” (polynomials, power series), “college algebra” (rings, fields, ideals) and “university algebra” (functors, homological algebra) goes on to present his fundamental thesis (“obviously a partisan claim”):

The method of high-school algebra is powerful, beautiful and accessible. So let us not be overwhelmed by the groups-ring-fields or the functorial arrows of the other two algebras and thereby lose sight of the power of the explicit algorithmic processes given to us by Newton, Tschirnhausen, Kronecker, and Sylvester.

Perhaps for this reason, Dr. Z calls Abhyankar (“one of my great heroes”) “the modern prophet of high-school-algebra”.

Anyway, enough rambling. Back to “Every good theorem must have a good counterexample”. Discuss.

[Edited to add: The statement in its original context was referring to a phenomenon where a pleasing conjecture is found to have counterexamples, until it is resolved by realising that we must, say, count multiplicities the “right” way—the right way turning out to be whatever makes the conjecture true. Thus Bezout’s theorem, etc., and the quote just means, as he paraphrases, “don’t be deterred if your formula is presently invalid in some cases; it only means that you have not yet completely deciphered god’s mind”. On the other hand, what I (mis?)remembered was that one must know “counterexamples” to a theorem in the sense that one must know why the conclusion is not true if the hypotheses are weakened: thus one doesn’t really understand a theorem till one knows at least one “counterexample” (and at least two proofs).]

Written by S

Tue, 2009-03-10 at 06:34:18

## Mathematics and notation: the Hindu-Arabic numeral system

Quick: What is CCXXXVII × CCCXXIX?

From page 15 of The Life of Pi by Jonathan Borwein:

The Indo-Arabic system came to Europe around 1000 CE. Resistance ranged from accountants who didn’t want their livelihood upset to clerics who saw the system as ‘diabolical,’ since they incorrectly assumed its origin was Islamic. European commerce resisted until the 18th century, and even in scientific circles usage was limited into the 17th century.

The prior difficulty of doing arithmetic is indicated by college placement advice given a wealthy German merchant in the 16th century: “If you only want him to be able to cope with addition and subtraction, then any French or German university will do. But if you are intent on your son going on to multiplication and division — assuming that he has sufficient gifts — then you will have to send him to Italy.” (George Ifrah, p. 577)

[The rest of the pages of the slides are also great and worth reading!]

Just to give some context of the time: The Hindu-Arabic system was introduced into Europe by Leonardo of Pisa (Fibonacci) — an Italian — in his Liber Abaci, written in 1202. Gutenberg (in Germany) invented the printing press around 1450. In Italy, Tartaglia lived 1500-1557, Cardano 1501-1576, Sturm 1507-1589, Giodano Bruno (1548-1600), and Ludovico Ferrari (1522-1565). (And outside Italy, Robert Recorde (as we’re talking about notation) (1510-1558) in Wales, François Viète (1540-1603) in France, etc. See this image.) Of course Galileo Galilei (1564-1642) was Italian too, but came later, as did Newton, Fermat, the Bernoullis, and all the others.

While on the topic of mathematics and notation, see also this post: Visual Clarity in the Naming of Variables.
[And while not exactly notation, Donald Knuth’s Calculus via O notation]

Written by S

Sun, 2008-12-14 at 05:19:19

## A “Möbius” inversion formula

[Not the Möbius inversion formula, but something similar.]
As usual, define the Möbius function μ on the natural numbers as
$\mu(n) = \begin{cases}0 \text{ if n is not square free,}\\ -1\text{ if n is squarefree with an odd number of prime factors,}\\ 1\text{ if n is squarefree with an even number of prime factors.}\\\end{cases}$
Let $f$ be any function defined on the natural numbers, and let $F$ be the function defined as $\displaystyle F(n) = \sum_{j=1}^{n}{f(\left\lfloor{n/j}\right\rfloor)}$.
Then it is true that $\displaystyle f(n) = \sum_{k=1}^{n}{\mu(k)F(\left\lfloor{n/k}\right\rfloor)}$.

Note that f need not be multiplicative; it can be any arbitrarily defined function. I have no idea why it is true. Help?

Written by S

Fri, 2008-11-28 at 18:55:44

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## Lattice points visible from the origin

[A test of LaTeX-to-Wordpress conversion. Bugs remain, point them out. Original PDF]

Here is a problem I love. It is simple to state, and it has a solution that is not trivial, but is easy to understand. The solution also goes through some beautiful parts, so I can promise it’s worth reading :-)

[The solution is not mine. Also, the question is just an excuse for the ideas in the solution. :P]

Question. Suppose you are standing on an infinite grid in the plane. You can see infinitely in all directions, but you cannot see through grid points: a point is hidden from view if some other grid point lies in your line of sight. What fraction of the grid points can you see?

Let us first imagine that we are standing at the origin, and that the grid is that of the lattice (integer) points.

The blue points are visible; the grey points are not

Read the rest of this entry »

Written by S

Fri, 2008-11-07 at 15:00:30

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## A theorem by Euler on partitions

There are so many beautiful facts in mathematics that are not as well known as they deserve to be. [Just today I couldn’t find an online reference for the Robinson-Schensted(-Knuth) correspondence.] Some efforts in the direction of communicating these to an online audience have been collected in the Carnival of Mathematics (41 so far) posts on various blogs. Another good idea I recently found is Theorem of the Day. [EtA: Found at Theorem of the Day: RSK correspondence.]

Anyway… today I attended a lecture on Euler by William Dunham (great talk!) where after a brief biography of Euler and a tour through some of his results, he showed Euler’s proof of a particular theorem about partitions. It will be familiar to anyone who has read about integer partitions, but it deserves to be familiar to everyone! So here is a sketch of the proof, for your reading pleasure. [It doesn’t do justice to present only a “sketch”, but I have assignments due tomorrowtoday… feel free to take it and turn it into some form worthy of the content. :-)]

Now when I say “a theorem by Euler”, it is very ambiguous which one I mean, and even with the “on partitions” restriction, there are several. So let me state it:

The number of partitions of a number into odd parts is the same as the number of partitions of the number into distinct parts.

[If I remember correctly what was narrated in the talk, one of the Bernoullis wrote to Euler asking him a question about partitions, and within a few days (think of the postal system then) Euler replied with a proof of the theorem, along with a apology for the delay caused by the poor eyesight he had recently been suffering from. This is an outline of his proof, ask me about the details.]

Let D(n) denote the number of partitions into distinct parts, and O(n) denote the number of partitions into odd parts. Then we have:
$\displaystyle \sum_{n\ge0}D(n)x^n$
$\displaystyle = (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)\dots$
$\displaystyle = \frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^6}{1-x^3}\frac{1-x^8}{1-x^4}\frac{1-x^{10}}{1-x^5}\dots$
$\displaystyle = \frac{1}{(1-x)(1-x^3)(1-x^5)\dots}$
$\displaystyle = (1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+x^{5+5}+\dots)$
$\displaystyle = \sum_{n\ge0}O(n)x^n$
which proves the theorem.
The first equality, which gives the generating function for the D(n)s, can be seen as follows: when you “expand” and write out $(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)\dots$, you get $1 + x + x^2 + (x^3+xx^2) + (x^4+xx^3) + (x^5+xx^4+x^2x^3) + \dots$, where the coefficient of any term $x^n$ is exactly the number of ways of writing $x^n$ as a product of distinct factors of the form $x^k$, which is $D(n)$. Similarly, the last equality, about the generating function for the O(n)s, is because the coefficient of any term $x^n$ in $(1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+x^{5+5}+\dots)$ is exactly the number of ways of writing $n$ as a product of (not necessarily distinct) $x^k$ for odd $k$.

The result might not be terribly important, but this idea was the starting point for proving several observations about partitions, and the method has led to the discovery of several facts and is more widely applicable than you think!

If you want a combinatorial proof by bijection (“For Entertainment Only”) here is an outline of one:
Given a partition of n into odd parts, say
$n = a_11 + a_33 + a_55 + \dots$,
write each $a_i$ “in binary”, by which I mean as a sum of distinct powers of two. So you have
$n = (2^{b_{11}}+2^{b_{12}}+...)1 + (2^{b_{31}}+2^{b_{32}}+...)3 + \dots$.
Now just get rid of the brackets, and note that all terms are distinct.
E.g. for $19 = 5+3+3+3+1+1+1+1+1$,
i.e. $19 = (1)5 + (2+1)3 + (4+1)1$, you get the partition
$19 = 5 + 6 + 3 + 4 + 1$.
[Exercise: Why are all parts distinct?]

Conversely, given a partition into distinct parts, we can separate the “even part” from the “odd part” in each term, so for example with $19 = 5+6+3+4+1$, we write
$19 = 5 + (2)3 + 3 + (4)1 + 1$, and collect the coefficients of each odd number, so $19= 5 + (2+1)3 + (4+1)1$, which was our original odd partition.
[Exercise: Why can’t we do the same thing starting with any partition?]

[Question: Is the bijective proof related to the algebraic one?]

Written by S

Wed, 2008-10-15 at 06:26:39

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