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sporadic simple games

About a year ago I did a series of posts on games associated to the Mathieu sporadic group $M_{12} $, starting with a post on Conway’s puzzle M(13), and, continuing with a discussion of mathematical blackjack. The idea at the time was to write a book for a general audience, as discussed at the start of the M(13)-post, ending with a series of new challenging mathematical games. I asked : “What kind of puzzles should we promote for mathematical thinking to have a fighting chance to survive in the near future?”

Now, Scientific American has (no doubt independently) taken up this lead. Their July 2008 issue features the article Rubik’s Cube Inspired Puzzles Demonstrate Math’s “Simple Groups” written by Igor Kriz and Paul Siegel.

By far the nicest thing about this article is that it comes with three online games based on the sporadic simple groups, the Mathieu groups $M_{12} $, $M_{24} $ and the Conway group $.0 $.

the M(12) game

Scrambles to an arbitrary permutation in $M_{12} $ and need to use the two generators $INVERT=(1,12)(2,11)(3,10)(4,9)(5,8)(6,7) $ and $MERGE=(2,12,7,4,11,6,10,8,9,5,3) $ to return to starting position.



Here is the help-screen :



They promise the solution by july 27th, but a few-line GAP-program cracks the puzzle instantly.

the M(24) game

Similar in nature, again using two generators of $M_{24} $. GAP-solution as before.



This time, they offer this help-screen :



the .0 game

Their most original game is based on Conway’s $.0 $ (dotto) group. Unfortunately, they offer only a Windows-executable version, so I had to install Bootcamp and struggle a bit with taking screenshots on a MacBook to show you the game’s starting position :



Dotto:

Dotto, our final puzzle, represents the Conway group Co0, published in 1968 by mathematician John H. Conway of Princeton University. Co0 contains the sporadic simple group Co1 and has exactly twice as many members as Co1. Conway is too modest to name Co0 after himself, so he denotes the group โ€œ.0โ€ (hence the pronunciation โ€œdottoโ€).

In Dotto, there are four moves. This puzzle includes the M24 puzzle. Look at the yellow/blue row in the bottom. This is, in fact, M24, but the numbers are arranged in a row instead of a circle. The R move is the “circle rotation to the right”: the column above the number 0 stays put, but the column above the number 1 moves to the column over the number 2 etc. up to the column over the number 23, which moves to the column over the number 1. You may also click on a column number and then on another column number in the bottom row, and the “circle rotation” moving the first column to the second occurs. The M move is the switch, in each group of 4 columns separated by vertical lines (called tetrads) the “yellow” columns switch and the “blue” columns switch. The sign change move (S) changes signs of the first 8 columns (first two tetrads). The tetrad move (T) is the most complicated: Subtract in each row from each tetrad 1/2 times the sum of the numbers in that tetrad. Then in addition to that, reverse the signs of the columns in the first tetrad.

Strategy hints: Notice that the sum of squares of the numbers in each row doesn’t change. (This sum of squares is 64 in the first row, 32 in every other row.) If you manage to get an “8”in the first row, you have almost reduced the game to M24 except those signs. To have the original position, signs of all numbers on the diagonal must be +. Hint on signs: if the only thing wrong are signs on the diagonal, and only 8 signs are wrong, those 8 columns can be moved to the first 8 columns by using only the M24 moves (M,R).

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the Kingโ€™s problem on MUBs

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 dโˆ’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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iTouch as network sniffer

In the iTouch warwalking post I was considering trying to gain access to closed networks for innocent purposes such as checking mail, rather than stealing secret passwords from people allowing you free access to their wireless network, but still, I should have thought of the following possibility

Here’s a walk-through :

  • type the following command into your iTouch Terminal.app (assuming you’ve installed the BSD subsystem) :

tcpdump -v -s 65535 -w log.txt

  • once you’ve collected enough packets, cancel the command (ctrl c), AFPd the file from the iTouch to your Mac and open it with Wireshark (this is the most convenient way to install binaries under Leopard as well as an updated version of X11. For other platforms, or source code, see here)

  • do whatever black magic you feel you have to perform using Wireshark (the new name for Ethereal) or other password crackers

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