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Mathieu’s blackjack (2)

Published July 17, 2007 by lievenlb

(continued from part one). Take twelve cards and give them values 0,1,2,…,11 (for example, take the jack to have value 11 and the queen to have value 0). The hexads are 6-tuples of cards having the following properties. When we star their values by the scheme on the left below and write a 0 below a column if it has just one star at the first row or two stars on rows two and three (a + if the unique star is at the first row or two stars in the other columns, and a – if the unique star in on the second row or two stars in rows one and two) or a ? if the column has 3 or 0 stars, we get a tetracodeword where we are allowed to replace a ? by any digit. Moreover, we want that the stars are NOT distributed over the four columns such that all of the possible outcomes 0,1,2,3 appear once. For example, the card-pile { queen, 3, 4, 7, 9, jack } is an hexad as is indicated on the right below and has column-distributions (1,1,2,2).

$\begin{array}{|c|ccc|} \hline 6 & 3 & 0 & 9 \\ 5 & 2 & 7 & 10 \\ 4 & 1 & 8 & 11 \\ \hline & & & \end{array} $ $\begin{array}{|c|ccc|} \hline & \ast & \ast & \ast \\ & & \ast & \\ \ast & & & \ast \\ \hline – & 0 & – & + \end{array} $

The hexads form a Steiner-system S(5,6,12), meaning that every 5-pile of cards is part of a unique hexad. The permutations on these twelve cards, having the property that they send every hexad to another hexad, form the sporadic simple group $M_{12} $, the _Mathieu group_ of order 95040. For now, we assume these facts and deduce from them the Conway-Ryba winning strategy for Mathieu’s blackjack : the hexads are exactly the winning positions and from a non-hexad pile of total value at least 21 there is always a legal (that is, total value decreasing) move to an hexad by replacing one card in the pile by a card from the complement.

It seems that the first proof of this strategy consisted in calculating the Grundy values of all 905 legal positions in Mathieu’s blackjack. Later Joseph Kahane and Alex Ryba gave a more conceptual proof, that we will try to understand.

Take a non-hexad 6-pile such that the total value of its cards is at least 21, then removing any one of the six cards gives a 5-pile and is by the Steiner-property contained in a unique hexad. Hence we get 6 different hexads replacing one card from the non-hexad pile by a card not contained in it. We claim that at least one of these operations is a legal move, meaning that the total value of the cards decreases. Let us call a counterexample a misfit and record some of its properties until we can prove its non-existence.

A misfit is a non-hexad with total value at least 21 such that all 6 hexads, obtained from it by replacing one card by a card from its complement, have greater total value

A misfit must contain the queen-card. If not, we could get an hexad replacing one misfit-card (value > 0) by the queen (value zero) so this would be a legal move. Further, the misfit cannot contain the jack-card for otherwise replacing it by a lower-valued card to obtain an hexad is a legal move.

A misfit contains at least three cards from {queen,1,2,3,4}. If not, three of these cards are the replacements of misfit-cards to get an hexad, but then at least one of the replaced cards has a greater value than the replacement, giving a legal move to an hexad.

A misfit contains more than three cards from {queen=0, 1,2,3,4}. Assume there are precisely three $\{ c_1,c_2,c_3 \} $ from this set, then the complement of the misfit in the hexad {queen,1,2,3,4,jack} consists of three elements $\{ d_1,d_2,d_3 \} $ (a misfit cannot contain the jack). The two leftmost columns of the value-scheme (left above) form the hexad {1,2,3,4,5,6} and because the Mathieu group acts 5-transitively there is an element of $M_{12} $ taking $\{ 0,1,2,3,4,11 \} \rightarrow \{ 1,2,3,4,5,6 \} $ and we may even assume that it takes $\{ c_1,c_2,c_3 \} \rightarrow \{ 4,5,6 \} $. But then, in the new value-scheme (determined by that $M_{12} $-element) the two leftmost columns of the misfit look like

$\begin{array}{|c|ccc|} \hline \ast & . & ? & ? \\ \ast & . & ? & ? \\ \ast & . & ? & ? \\ \hline ? & ? & & \end{array} $

and the column-distribution of the misfit must be either (3,0,2,1) or (3,0,1,2) (it cannot be (3,0,3,0) or (3,0,0,3) otherwise the (image of the) misfit would be an hexad). Let {i,j} be the two misfit-values in the 2-starred column. Replacing either of them to get an hexad must have the replacement lying in the second column (in order to get a valid column distribution (3,1,1,1)). Now, the second column consists of two small values (from {0,1,2,3,4}) and the large jack-value (11). So, at least one of {i,j} is replaced by a smaller valued card to get an hexad, which cannot happen by the misfit-property.

Now, if the misfit shares four cards with {queen,1,2,3,4} then it cannot contain the 10-card. Otherwise, the replacement to get an hexad of the 10-card must be the 11-card (by the misfit-property) but then there would be another hexads containing five cards from {queen,0,1,2,3,jack} which cannot happen by the Steiner-property. Right, let’s summarize what we know so far about our misfit. Its value-scheme looks like

$\begin{array}{|c|ccc|} \hline 6 & III & \ast & 9 \\ 5 & II & 7 & . \\ IV & I & 8 & . \\ \hline & & & \end{array} $ and it must contain three of the four Romans. At this point Kahane and Ryba claim that the two remaining cards (apart from the queen and the three romans) must be such that there is exactly one from {5,6} and exactly one from {7,8,9}. They argue this follows from duality where the dual pile of a card-pile $\{ x_1,x_2,\ldots,x_6 \} $ is the pile $\{ 11-x_1,11-x_2,\ldots,11-x_6 \} $. This duality acts on the hexads as the permutation $~(0,11)(1,10)(2,9)(3,8)(4,7)(5,6) \in M_{12} $. Still, it is unclear to me how they deduce from it the above claim (lines 13-15 of page 4 of their paper). I’d better have some coffee and work around this (to be continued…)

If you want to play around a bit with hexads and the blackjack game, you’d better first download SAGE (if you haven’t done so already) and then get David Joyner’s hexad.sage file and put it in a folder under your sage installation (David suggests ‘spam’ himself…). You can load the routines into sage by typing from the sage-prompt attach ‘spam/hexad.sage’. Now, you can find the hexad from a 5-pile via the command find_hexad([a1,a2,a3,a4,a5],minimog_shuffle) and you can get the winning move for a blackjack-position via blackjack_move([a1,a2,a3,a4,a5,a6],minimog_shuffle). More details are in the Joyner-Casey(Luers) paper referenced last time.

Reference

Joseph Kahane and Alexander J. Ryba, ‘The hexad game‘

Mathieu’s blackjack (1)

Published July 16, 2007 by lievenlb

Mathieu’s blackjack is a two-person combinatorial game played with 12 cards of values 0,1,2,…,11. For example take from any deck the numbered cards together with the jack (value 11) and the queen (value 0) (btw. if you find this PI by all means replace the queen by a zero-valued king). Shuffle the cards and divide them into two piles of 6 cards (all of them face up on the table) : the main-pile and the other-pile. The rules of the game are

players alternate moves
a move consists of exchanging a card of the main-pile with a lower-valued card from the other-pile
the player whose move makes the sum of all cards in the main-pile under 21 looses the game

For example, the starting main-pile might consist of the six cards

This pile has total value 3+4+7+8+9+11=42. A move replaces one of these cards with a lowever vlued one not in the pile. So for example, replacing 8 with 5 or 1 or 2 or the queen are all valid moves. A winning move from this situation is for example replacing 8 by the queen (value 0) decreasing the value from 42 to 34

But there are otthers, such as replacing 11 by 5, 9 by 1 or 4 by 2. To win this game you need to know the secrets of the tetracode and the MINIMOG.

The tetracode is a one-error correcting code consisting of the following nine words of length four over $\mathbb{F}_3 = { 0,+,- } $

$~\begin{matrix} 0~0 0 0 & 0~+ + + & 0~- – – \\ +~0 + – & +~+ – 0 & +~- 0 + \\ -~0 – + & -~+ 0 – & -~- + 0 \end{matrix} $

The first element (which is slightly offset from the rest) is the slope s of the words, and the other three digits cyclically increase by s (in the field $\mathbb{F}_3 $). Because the Hamming-distance is 3 (the minimal number of different digits between two codewords), the tetracode can correct one error, meaning that if at most one of the four digits gets distorted by the channel one can detect and correct this. For example, if you would receive the word $+~++- $ (which is not a codeword) and if you would know that at most one digits went wrong, you can deduce that the word $+~0+- $ was sent. Thus, one can solve the 4-problem for the tetracode : correctt a tetracodeword given all 4 of its digits, one of which may be mistaken.

Another easy puzzle is the 2-problem for the tertracode : complete a tetracodeword from any 2 of its digits. For example, given the incomplete word $?~?0+ $ you can decide that the slope should be + and hence that the complete word must be $+~-0+ $.

We will use the MINIMOG here as a way to record the blackjack-position. It is a $4 \times 3 $ array where the 12 boxes correspond to the card-values by the following scheme

$\begin{array}{|c|ccc|} \hline 6 & 3 & 0 & 9 \\ 5 & 2 & 7 & 10 \\ 4 & 1 & 8 & 11 \\ \hline \end{array} $

and given a blackjack-position we place a star in the corresponding box, so the above start-position (resp. after the first move) corresponds to

$~\begin{array}{|c|ccc|} \hline & \ast & & \ast \\ & & \ast & \\ \ast & & \ast & \ast \\ \hline – & 0 & 0 & + \end{array}~ $ respectively $\begin{array}{|c|ccc|} \hline & \ast & \ast & \ast \\ & & \ast & \\ \ast & & & \ast \\ \hline – & 0 & – & + \end{array} $

In the final row we have added elements of $\mathbb{F}_3 $ indicating wher ethe stars are placed in that column (if there is just one star, we write the row-number of the star (ordered 0,+,- from top to bottom), if there are two stars we record the row-number of the empty spot. If we would have three or no stars in a column we would record a wild-card character : ?

Observe that the final row of the start position is $-~00+ $ which is NOT a tetracodeword, whereas that of the winning position $-~0-+ $ IS a tetracodeword! This is the essence of the _Conway-Ryba winning strategy_ for Mathieu’s blackjack. There are precisely 132 winning positions forming the Steiner-system S(5,6,12). By an S(5,6,12) we mean a collection of 6-element subsets (our card-piles) from a 12-element set (the deck minus the king) having the amazing property that for EVERY 5-tuple from the 12-set there is a UNIQUE 6-element set containing this 5-tuple. Hence, there are exactly $\begin{pmatrix} 12 \\\ 5 \end{pmatrix}/6 = 132 $ elements in a Steiner S(5,6,12) system. The winning positions are exactly those MINOMOGs having 6 stars such that the final row is a tetracodeword (or can be extended to a tetracodeword replacing the wildcards ? by suitable digits) and such that the distribution of the stars over the columns is NOT (3,2,1,0) in any order.

Provided the given blackjack-position is not in this Steiner-system (and there is only a 1/7 chance that it is), the strategy is clear : remove one of the stars to get a 5-tuple and determine the unique 6-set of the Steiner-system containing this 5-tuple. If the required extra star corresponds to a value less than the removed star you have a legal and winning move (if not, repeat this for another star). Finding these winning positions means solving 2- and 4-problems for the tetracode. _Another time_ we will say more about this Steiner system and indicate the relation with the Mathieu group $M_{12} $.

References

J.H. Conway and N.J.A. Sloane, ‘The Golay codes and the Mathieu groups’, chp. 10 of “Sphere Packings, Lattices and Groups“

David Joyner and Ann Casey-Luers, ‘Kittens, S(5,6,12) and Mathematical blackjack in SAGE‘

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Hexagonal Moonshine (3)

Published July 13, 2007 by lievenlb

Hexagons keep on popping up in the representation theory of the modular group and its close associates. We have seen before that singularities in 2-dimensional representation varieties of the three string braid group $B_3 $ are ‘clanned together’ in hexagons and last time Ive mentioned (in passing) that the representation theory of the modular group is controlled by the double quiver of the extended Dynkin diagram $\tilde{A_5} $, which is an hexagon…

Today we’re off to find representations of the extended modular group $\tilde{\Gamma} = PGL_2(\mathbb{Z}) $, which is obtained by adding to the modular group (see this post for a proof of generation)

$\Gamma = \langle U=\begin{bmatrix} 0 & -1 \\\ 1 & 0 \end{bmatrix},V=\begin{bmatrix} 0 & 1 \\\ -1 & 1 \end{bmatrix} \rangle $ the matrix $R=\begin{bmatrix} 0 & 1 \\\ 1 & 0 \end{bmatrix} $

In terms of generators and relations, one easily verfifies that

$\tilde{\Gamma} = \langle~U,V,R~|~U^2=R^2=V^3=(RU)^2=(RV)^2=1~\rangle $

and therefore $\tilde{\Gamma} $ is the amalgamated free product of the dihedral groups $D_2 $ and $D_3 $ over their common subgroup $C_2 = \langle~R~\rangle $, that is

$\tilde{\Gamma} = \langle U,R | U^2=R^2=(RU)^2=1 \rangle \ast_{\langle R | R^2=1 \rangle} \langle V,R | V^3=R^2=(RV)^2=1 \rangle = D_2 \ast_{C_2} D_3 $

From this description it is easy to find all n-dimensional $\tilde{\Gamma} $-representations $V $ and relate them to quiver-representations. $D_2 = C_2 \times C_2 $ and hence has 4 1-dimensonal simples $S_1,S_2,S_3,S_4 $. Restricting $V\downarrow_{D_2} $ to the subgroup $D_2 $ it decomposes as

$V \downarrow_{D_2} \simeq S_1^{\oplus a_1} \oplus S_2^{\oplus a_2} \oplus S_3^{\oplus a_3} \oplus S_4^{\oplus a_4} $ with $a_1+a_2+a_3+a_4=n $

Similarly, because $D_3=S_3 $ has two one-dimensional representations $T,S $ (the trivial and the sign representation) and one simple 2-dimensional representation $W $, restricting $V $ to this subgroup gives a decomposition

$V \downarrow_{D_3} \simeq T^{b_1} \oplus S^{\oplus b_2} \oplus W^{\oplus b_3} $, this time with $b_1+b_2+2b_3=n $

Restricting both decompositions further down to the common subgroup $C_2 $ one obtains a $C_2 $-isomorphism $V \downarrow_{D_2} \rightarrow^{\phi} V \downarrow_{D_3} $ which implies also that the above numbers must be chosen such that $a_1+a_3=b_1+b_3 $ and $a_2+a_4=b_2+b_3 $. We can summarize all this info about $V $ in a representation of the quiver

Here, the vertex spaces on the left are the iso-typical factors of $V \downarrow_{D_2} $ and those on the right those of $V \downarrow_{D_3} $ and the arrows give the block-components of the $C_2 $-isomorphism $\phi $. The nice things is that one can also reverse this process to get all $\tilde{\Gamma} $-representations from $\theta $-semistable representations of this quiver (having the additional condition that the square matrix made of the arrows is invertible) and isomorphisms of group-representation correspond to those of quiver-representations!

This proves that for all n the varieties of n-dimensional representations $\mathbf{rep}_n~\tilde{\Gamma} $ are smooth (but have several components corresponding to the different dimension vectors $~(a_1,a_2,a_3,a_4;b_1,b_2,b_3) $ such that $\sum a_i = n = b_1+b_2+2b_3 $.

The basic principle of _M-geometry_ is that a lot of the representation theory follows from the ‘clan’ (see this post) determined by the simples of smallest dimensions. In the case of the extended modular group $\tilde{\Gamma} $ it follows that there are exactly 4 one-dimensional simples and exactly 4 2-dimensional simples, corresponding to the dimension vectors

$\begin{cases} a=(0,0,0,1;0,1,0) \\\ b=(0,1,0,0;0,1,0) \\\ c=(1,0,0,0;1,0,0) \\\ d=(0,0,1,0;1,0,0) \end{cases} $ resp. $\begin{cases} e=(0,1,1,0;0,0,1) \\\ f=(1,0,0,1;0,0,1) \\\ g=(0,0,1,1;0,0,1) \\\ h=(1,1,0,0;0,0,1) \end{cases} $

If one calculates the ‘clan’ of these 8 simples one obtains the double quiver of the graph on the left. Note that a and b appear twice, so one should glue the left and right hand sides together as a Moebius-strip. That is, the clan determining the representation theory of the extended modular group is a Moebius strip made of two hexagons!

However, one should not focuss too much on the hexagons (that is, the extended Dynkin diagram $\tilde{A_5} $) here. The two ‘backbones’ (e–f and g–h) have their vertices corresponding to 2-dimensional simples whereas the topand bottom vertices correspond to one-dimensional simples. Hence, the correct way to look at this clan is as two copies of the double quiver of the extended Dynkin diagram $\tilde{D_5} $ glued over their leaf vertices to form a Moebius strip. Remark that the components of the sotropic root of $\tilde{D_5} $ give the dimensions of the corresponding $\tilde{\Gamma} $ simples.

The remarkable ubiquity of (extended) Dynkins never ceases to amaze!

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Mathieu’s blackjack (2)

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