Tag: Mathieu

Mathieu’s blackjack (1)

Published July 16, 2007 by lieven

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‘

Generators of modular subgroups

Published July 12, 2007 by lieven

In older NeverEndingBooks-posts (and here) proofs were given that the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is the group free product $C_2 \ast C_3 $, so let’s just skim over details here. First one observes that $\Gamma $ is generated by (the images of) the invertible 2×2 matrices

$U= \begin{bmatrix} 0 & -1 \\\ 1 & 0 \end{bmatrix} $ and $V= \begin{bmatrix} 0 & 1 \\\ -1 & 1 \end{bmatrix} $

A way to see this is to consider X=U.V and Y=V.U and notice that multiplying with powers of X adds multiples of the second row to the first (multiply on the left) or multiples of the first column to the second (multiply on the right) and the other cases are handled by taking multiples with powers of Y. Use this together with the fact that matrices in $GL_2(\mathbb{Z}) $ have their rows and columns made of coprime numbers to get any such matrix by multiplication on the left or right by powers of X and Y into the form

$\begin{bmatrix} \pm 1 & 0 \\\ 0 & \pm 1 \end{bmatrix} $ and because $U^2=V^3=\begin{bmatrix} -1 & 0 \\\ 0 & -1 \end{bmatrix} $

we see that $\Gamma $ is an epimorphic image of $C_2 \ast C_3 $. To prove isomorphism one can use the elegant argument due to Roger Alperin considering the action of the Moebius transformations $u(z) = -\frac{1}{z} $ and $v(z) = \frac{1}{1-z} $ (with $v^{-1}(z) = 1-\frac{1}{z} $) induced by the generators U and V on the sets $\mathcal{P} $ and $\mathcal{N} $ of all positive (resp. negative) irrational real numbers. Observe that

$u(\mathcal{P}) \subset \mathcal{N} $ and $v^{\pm}(\mathcal{N}) \subset \mathcal{P} $

Hence, if $w $ is a word in $u $ and $v^{\pm} $ of off length we either have $w(\mathcal{P}) \subset \mathcal{N} $ or $w(\mathcal{N}) \subset \mathcal{P} $ so $w $ can never be the identity. If the length is even we can conjugate $w $ such that it starts with $v^{\pm} $. If it starts with $v $ then $w(\mathcal{P}) \subset v(\mathcal{N}) $ is a subset of positive rationals less than 1 whereas if it starts with $v^{-1} $ then $w(\mathcal{P}) \subset v^{-1}(\mathcal{N}) $ is a subset of positive rationals greater than 1, so again it cannot be the identity. Done!

By a result of Aleksandr Kurosh it follows that every modular subgroup is the group free product op copies of $C_2, C_3 $ or $C_{\infty} $ and we would like to determine the free generators explicitly for a cofinite subgroup starting from its associated Farey code associated to a special polygon corresponding to the subgroup.

To every even interval [tex]\xymatrix{x_i = \frac{a_i}{b_i} \ar@{-}[r]_{\circ} & x_{i+1}= \frac{a_{i+1}}{b_{i+1}}}[/tex] in the Farey code one associates the generator of a $C_2 $ component

$A_i = \begin{bmatrix} a_{i+1}b_{i+1}+ a_ib_i & -a_i^2-a_{i+1}^2 \\\ b_i^2+b_{i+1}^2 & -a_{i+1}b_{i+1}-a_ib_i \end{bmatrix} $

to every odd interval [tex]\xymatrix{x_i = \frac{a_i}{b_i} \ar@{-}[r]_{\bullet} & x_{i+1} = \frac{a_{i+1}}{b_{i+1}}}[/tex] in the Farey code we associate the generator of a $C_3 $ component

$B_i = \begin{bmatrix} a_{i+1}b_{i+1}+a_ib_{i+1}+a_ib_i & -a_i^2-a_ia_{i+1}-a_{i+1}^2 \\\ b_i^2+b_ib_{i+1} + b_{i+1}^2 & -a_{i+1}b_{i+1} – a_{i+1}b_i – a_i b_i \end{bmatrix} $

and finally, to every pair of free intervals [tex]\xymatrix{x_k \ar@{-}[r]_{a} & x_{k+1}} \ldots \xymatrix{x_l \ar@{-}[r]_{a} & x_{l+1}}[/tex] we associate the generator of a $C_{\infty} $ component

$C_{k,l} = \begin{bmatrix} a_l & -a_{l+1} \\\ b_l & – b_{l+1} \end{bmatrix} \begin{bmatrix} a_{k+1} & a_k \\\ b_{k+1} & b_k \end{bmatrix}^{-1} $

Kulkarni’s result states that these matrices are free generators of the cofiniite modular subgroup determined by the Farey code. For example, for the M(12) special polygon on the left (bounded by the thick black geodesics), the Farey-code for this Mathieu polygon is

[tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{\bullet} & \frac{1}{3} \ar@{-}[r]_{\bullet} & \frac{1}{2} \ar@{-}[r]_{\bullet} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

Therefore, the structure of the subgroup must be $C_{\infty} \ast C_3 \ast C_3 \ast C_3 $ with the generator of the infinite factor being

$\begin{bmatrix} -1 & 1 \\\ -1 & 0 \end{bmatrix} $ and those of the cyclic factors of order three

$\begin{bmatrix} 3 & -1 \\\ 13 & -4 \end{bmatrix}, \begin{bmatrix} 7 & -3 \\\ 19 & 8 \end{bmatrix} $ and $\begin{bmatrix} 4 & -3 \\\ 7 & -5 \end{bmatrix} $

This approach also gives another proof of the fact that $\Gamma = C_2 \ast C_3 $ because the Farey code to the subgroup of index 1 is [tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] corresponding to the fundamental domain on the left. This finishes (for now) this thread on Kulkarni’s paper (or rather, part of it). On the Lost? page I will try to list threads in a logical ordering when they materialize.

Reference

Ravi S. Kulkarni, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1133

Farey codes

Published July 1, 2007 by lieven

John Farey (1766-1826) was a geologist of sorts. Eyles, quoted on the math-biographies site described his geological work as

“As a geologist Farey is entitled to respect for the work which he carried out himself, although it has scarcely been noticed in the standard histories of geology.”

That we still remember his name after 200 years is due to a short letter he wrote in 1816 to the editor of the Philosophical Magazine

“On a curious Property of vulgar Fractions.
By Mr. J. Farey, Sen. To Mr. Tilloch

Sir. – On examining lately, some very curious and elaborate Tables of “Complete decimal Quotients,” calculated by Henry Goodwyn, Esq. of Blackheath, of which he has printed a copious specimen, for private circulation among curious and practical calculators, preparatory to the printing of the whole of these useful Tables, if sufficient encouragement, either public or individual, should appear to warrant such a step: I was fortunate while so doing, to deduce from them the following general property; viz.

If all the possible vulgar fractions of different values, whose greatest denominator (when in their lowest terms) does not exceed any given number, be arranged in the order of their values, or quotients; then if both the numerator and the denominator of any fraction therein, be added to the numerator and the denominator, respectively, of the fraction next but one to it (on either side), the sums will give the fraction next to it; although, perhaps, not in its lowest terms.

For example, if 5 be the greatest denominator given; then are all the possible fractions, when arranged, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, and 4/5; taking 1/3, as the given fraction, we have (1+1)/(5+3) = 2/8 = 1/4 the next smaller fraction than 1/3; or (1+1)/(3+2) = 2/5, the next larger fraction to 1/3. Again, if 99 be the largest denominator, then, in a part of the arranged Table, we should have 15/52, 28/97, 13/45, 24/83, 11/38, &c.; and if the third of these fractions be given, we have (15+13)/(52+45) = 28/97 the second: or (13+11)/(45+38) = 24/83 the fourth of them: and so in all the other cases.

I am not acquainted, whether this curious property of vulgar fractions has been before pointed out?; or whether it may admit of any easy or general demonstration?; which are points on which I should be glad to learn the sentiments of some of your mathematical readers; and am

Sir, Your obedient humble servant,
J. Farey. Howland-street.”

So, if we interpolate “childish addition of fractions” $\frac{a}{b} \oplus \frac{c}{d} = \frac{a+c}{b+d} $ and start with the numbers $0 = \frac{0}{1} $ and $\infty = \frac{1}{0} $ we get the binary Farey-tree above. For a fixed natural number n, if we stop the interpolation whenever the denominator of the fraction would become larger than n and order the obtained fractions (smaller or equal to one) we get the Farey sequence F(n). For example, if n=3 we start with the sequence $ \frac{0}{1},\frac{1}{1} $. The next step we get $\frac{0}{1},\frac{1}{2},\frac{1}{1} $ and the next step gives

$\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{1}{1} $

and as all the denomnators of childish addition on two consecutive fractions will be larger than 3, the above sequence is F(3). A remarkable feature of the series F(n) is that if $\frac{a}{b} $ and $\frac{c}{d} $ are consecutive terms in F(n), then

$det \begin{bmatrix} a & c \\\ b & d \end{bmatrix} = -1 $

and so these two fractions are the endpoints of an even geodesic in the Dedekind tessellation.

A generalized Farey series is an ordered collection of fractions $\infty,x_0,x_1,\cdots,x_n,\infty $ such that $x_0 $ and $x_n $ are integers and some $x_i=0 $. Moreover, writing $x_i = \frac{a_i}{b_i} $ we have that

$det \begin{bmatrix} a_i & a_{i+1} \\\ b_i & b_{i+1} \end{bmatrix} = -1 $

A Farey code is a generalized Farey sequence consisting of all the vertices of a special polygon that lie in $\mathbb{R} \cup \{ \infty \} $ together with side-pairing information. If two consecutive terms are such that the complete geodesic between $x_i $ and $x_{i+1} $ consists of two sides of the polygon which are paired we denote this fact by
[tex]\xymatrix{x_i \ar@{-}[r]_{\circ} & x_{i+1}}[/tex]. If they are the endpoints of two odd sides of the polygon which are paired we denote this by [tex]\xymatrix{x_i \ar@{-}[r]_{\bullet} & x_{i+1}}[/tex]. Finally, if they are the endpoints of a free side which is paired to another free side determined by $x_j $ and $x_{j+1} $ we denote this fact by marking both edges [tex]\xymatrix{x_i \ar@{-}[r]_{k} & x_{i+1}}[/tex] and [tex]\xymatrix{x_j \ar@{-}[r]_{k} & x_{j+1}}[/tex] with the same number.

For example, for the M(12) special polygon on the left (bounded by the thick black geodesics), the only vertices in $\mathbb{R} \cup \{ \infty \} $ are $\infty,0,\frac{1}{3},\frac{1}{2},1 $. The two vertical lines are free sides and are paired, whereas all other sides of the polygon are odd. Therefore the Farey-code for this Mathieu polygon is

[tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{\bullet} & \frac{1}{3} \ar@{-}[r]_{\bullet} & \frac{1}{2} \ar@{-}[r]_{\bullet} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

Conversely, to a Farey-code we can associate a special polygon by first taking the hyperbolic convex hull of all the terms in the sequence (the region bounded by the vertical lines and the bottom red circles in the picture on the left) and adding to it for each odd interval [tex]\xymatrix{x_i \ar@{-}[r]_{\bullet} & x_{i+1}}[/tex] the triangle just outside the convex hull consisting of two odd edges in the Dedekind tessellation (then we obtain the region bounded by the black geodesics). Again, the side-pairing of the obained special polygon can be obtained from that of the Farey-code.

This correspondence gives a natural one-to-one correspondence special polygons <---> Farey-codes . _Later_ we will see how the Farey-code determines the group structure of the corresponding finite index subgroup of the modular group $\Gamma = PSL_2(\mathbb{Z}) $.

Reference

Ravi S. Kulkarni, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1133