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Tag: puzzle

exotic chess positions (1)

Published March 4, 2008 by lievenlb

Ever tried a chess problem like : White to move, mate in two! Of course you have, and these are pretty easy to solve : you only have to work through the finite list of white first moves and decide whether or not black has a move left preventing mate on the next white move. This is even a (non-optimal) fool-proof algorithm to find the solution to this kind of chess problems. Right?

Wrong! There exist concrete positions, provable mate in two in which it is NOT possible to determine the winning first move for white! So, what’s wrong with the argument above? We did assume that, given the position, it is possible to determine all legal moves for the two players. So?

Well, some moves are legal only depending on the history of the game. For example, you are only allowed to do a castling if your king nor your rook made a prior move. Further, you can only make an en-passant-capture on the next move.

But surely all this is just theoretical? No-one ever constructed a provable 2-mate with impossible winning move. Wrong again. The logician Raymond Smullyan did precisely that in his retro-chess puzzle book Chess mysteries of Sherlock Holmes. Here’s the position :



Presumably every chess player goes for the mate : 1. Kf5-e6 2. g7-g8D But what if black counters your first move with a castling 1. … 0-0-0 ? Surely he isnt allowed to do this. Why not, is there any clue in the position to prove that either the king or the rook must have moved before?

Well, what was black previous move then? It cannot be the pawn move e6-e5 as before that move the white King would be in check, so what was it? Just one possibility left : it must have been e7-e5.

This offers then another winning strategy for white, as white can capture en-passant. 1. d5xe6 e.p. and then if black castles 1. … 0.0.0, 2. b6-b7 or is black does any other move : 2. g7-g8.

Hence, whatever the games’ history, white has a mate in two! However, looking ONLY at the given position, it is impossible for him to judge whether Kf5-e6 will do the trick!

Anyone seen similar constructions?

UPDATE

According to the wikipedia page on Retrograde chess analysis, the Smullyan-idea is an adaptation of a much older problem due to W. Langstaff in the Chess Amateur of 1922. Here’s the situation (the solution is the same as above)



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the King’s problem on MUBs

Published February 28, 2008 by lievenlb

MUBs (for Mutually Unbiased Bases) are quite popular at the moment. Kea is running a mini-series Mutual Unbias as is Carl Brannen. Further, the Perimeter Institute has a good website for its seminars where they offer streaming video (I like their MacromediaFlash format giving video and slides/blackboard shots simultaneously, in distinct windows) including a talk on MUBs (as well as an old talk by Wootters).

So what are MUBs to mathematicians? Recall that a d-state quantum system is just the vectorspace $\mathbb{C}^d $ equipped with the usual Hermitian inproduct $\vec{v}.\vec{w} = \sum \overline{v_i} w_i $. An observable $E $ is a choice of orthonormal basis ${ \vec{e_i} } $ consisting of eigenvectors of the self-adjoint matrix $E $. $E $ together with another observable $F $ (with orthonormal basis ${ \vec{f_j} } $) are said to be mutally unbiased if the norms of all inproducts $\vec{f_j}.\vec{e_i} $ are equal to $1/\sqrt{d} $. This definition extends to a collection of pairwise mutually unbiased observables. In a d-state quantum system there can be at most d+1 mutually unbiased bases and such a collection of observables is then called a MUB of the system. Using properties of finite fields one has shown that MUBs exists whenever d is a prime-power. On the other hand, existence of a MUB for d=6 still seems to be open…

The King’s Problem (( actually a misnomer, it’s more the poor physicists’ problem… )) is the following : A physicist is trapped on an island ruled by a mean
king who promises to set her free if she can give him the answer to the following puzzle. The
physicist is asked to prepare a d−state quantum system in any state of her choosing and give it
to the king, who measures one of several mutually unbiased observables on it. Following this, the physicist is allowed to make a control measurement
on the system, as well as any other systems it may have been coupled to in the preparation
phase. The king then reveals which observable he measured and the physicist is required
to predict correctly all the eigenvalues he found.

The Solution to the King’s problem in prime power dimension by P. K. Aravind, say for $d=p^k $, consists in taking a system of k object qupits (when $p=2l+1 $ one qupit is a spin l particle) which she will give to the King together with k ancilla qupits that she retains in her possession. These 2k qupits are diligently entangled and prepared is a well chosen state. The final step in finding a suitable state is the solution to a pure combinatorial problem :

She must use the numbers 1 to d to form $d^2 $ ordered sets of d+1 numbers each, with repetitions of numbers within a set allowed, such that any two sets have exactly one identical number in the same place in both. Here’s an example of 16 such strings for d=4 :

11432, 12341, 13214, 14123, 21324, 22413, 23142, 24231, 31243, 32134, 33421, 34312, 41111, 42222, 43333, 44444

Here again, finite fields are used in the solution. When $d=p^k $, identify the elements of $\mathbb{F}_{p^k} $ with the numbers from 1 to d in some fixed way. Then, the $d^2 $ of number-strings are found as follows : let $k_0,k_1 \in \mathbb{F}_{p^k} $ and take as the first 2 numbers the ones corresponding to these field-elements. The remaning d-2 numbers in the string are those corresponding to the field element $k_m $ (with $2 \leq m \leq d $) determined from $k_0,k_1 $ by the equation

$k_m = l_{m} * k_0+k_1 $

where $l_i $ is the field-element corresponding to the integer i ($l_1 $ corresponds to the zero element). It is easy to see that these $d^2 $ strings satisfy the conditions of the combinatorial problem. Indeed, any two of its digits determine $k_0,k_1 $ (and hence the whole string) as it follows from
$k_m = l_m k_0 + k_1 $ and $k_r = l_r k_0 + k_1 $ that $k_0 = \frac{k_m-k_r}{l_m-l_r} $.

In the special case when d=3 (that is, one spin 1 particle is given to the King), we recover the tetracode : the nine codewords

0000, 0+++, 0—, +0+-, ++-0, +-0+, -0-+, -+0-, –+0

encode the strings (with +=1,-=2,0=3)

3333, 3111, 3222, 1312, 1123, 1231, 2321, 2132, 2213

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Archimedes’ stomachion

Published February 11, 2008 by lievenlb

The Archimedes codex is a good read, especially when you are (like me) a failed archeologist. The palimpsest (Greek for ‘scraped again’) is the worlds first Kyoto-approved ‘sustainable writing’. Isn’t it great to realize that one of the few surviving texts by Archimedes only made it because some monks recycled an old medieval parchment by scraping off most of the text, cutting the pages in half, rebinding them and writing a song-book on them…

The Archimedes-text is barely visible as vertical lines running through the song-lyrics. There is a great website telling the story in all its detail.

Contrary to what the books claims I don’t think we will have to rewrite maths history. Didn’t we already know that the Greek were able to compute areas and volumes by approximating them with polygons resp. polytopes? Of course one might view this as a precursor to integral calculus… And then the claim that Archimedes invented ordinal calculus. Sure the Greek knew that there were ‘as many’ even integers than integers… No, for me the major surprise was their theory about the genesis of mathematical notation.

The Greek were pure ASCII mathematicians : they wrote their proofs out in full text. Now, here’s an interesting theory how symbols got into maths… pure laziness of the medieval monks transcribing the old works! Copying a text was a dull undertaking so instead of repeating ‘has the same ratio as’ for the 1001th time, these monks introduced abbreviations like $\Sigma $ instead… and from then on things got slightly out of hand.

Another great chapter is on the stomachion, perhaps the oldest mathematical puzzle. Just a few pages made in into the palimpsest so we do not really know what (if anything) Archimedes had to say about it, but the conjecture is that he was after the number of different ways one could make a square with the following 14 pieces

People used computers to show that the total number is $17152=2^8 \times 67 $. The 2-power is hardly surprising in view of symmetries of the square (giving $8 $) and the fact that one can flip one of the two vertical or diagonal parts in the alternative description of the square

but I sure would like to know where the factor 67 is coming from… The MAA and UCSD have some good pages related to the stomachion puzzle. Finally, the book also views the problema bovinum as an authentic Archimedes, so maybe I should stick to my promise to blog about it, after all…

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