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 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 . The 2-power is hardly surprising in view of symmetries of the square (giving ) 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…

Ibrahim Belkadi, one of my first-year group theory students invented the micro-sudokube, that is, a cube having a solution to a micro-sudoku on all its sides such that these solutions share one row along an edge. For example, here are all the solutions for a given central solution. There are 4 of them with and

The problem is : how many micro-sudokubes are there? Ibrahim handed in his paper and claims that there are exactly 32 of them, up to relabeling , so in all there are micro-sudokubes.

The proof-strategy is as follows. Fix one side and use relabeling to put the solution on that side to be one of 12 canonical forms (see for example this post. Next, work out as above for each of these standard forms in how many ways it can be extended. A nice idea of Ibrahim was to develop a much better notation for micro-sudokubes than the above flattenet-out cube. He uses the fact that a micro-sudokube is entirely determined by the solutions on two opposite sides (check this for yourself). Moreover, fixing one side determines one-half of all the neighboring sides. His notation for the 4 solutions above then becomes

and he can then use these solutions also in other standard form (the extra notation using the names A,B,C 1-4 for the 12 canonical forms).

Via the Arcadian functor I learned of the existence of the Sudokube (picture on the left).

Sudokube is a variation on a Rubik’s Cube in which each face resembles one-ninth of a Sudoku grid: the numbers from one to nine. This makes solving the cube slightly more difficult than a conventional Rubik’s Cube because each number must be in the right place and the centre cubies must also be in the correct orientation.

A much more interesting Sudoku-variation of the cube was invented two weeks ago by one of my freshmen-grouptheory students and was dubbed the mini-sudokube by him. Here’s the story.

At the end of my grouptheory course I let the students write a paper to check whether they have research potential. This year the assignment was to read through the paper mini-sudokus and groups by Carlos Arcos, Gary Brookfield and Mike Krebs, and do something original with it.

Mini-Sudoku is the baby version of Sudoku. Below a trivial problem and its solution

Of course most of them took the safe road and explained in more detail a result of the paper. Two of them were more original. One used the mini-sudoku solutions to find solutions for 4×4 sudokus, but the most original contribution came from Ibrahim Belkadi who wanted to count all mini-sudokubes. A mini-sudokube is a cube with a mini-sudoku solution on all 6 of its sides BUT NUMBERS CARRY OVER CUBE-EDGES. That is, if we have as the mini-sudoku given by the central square below on the top-face of the cube, then on the 4 side-faces we have already one row filled in.

The problem then is to find out how many compatible solutions there are. It is pretty easy to see that top- and bottom-faces determine all squares of the cube, but clearly not all choices are permitted. For example, with the above top-face fixed there are exactly 4 solutions. Let and then these four solutions are given below

Putting one of these solutions (or any other) on a -Rubik cube would make a more interesting puzzle, I think. I’ve excused Ibrahim from having to take examination on condition that he writes down what he can prove on his mini-sudokubes by that time. Looking forward to it!

Conway’s puzzle M(13) involves the 13 points and 13 lines of . On all but one point numbered counters are placed holding the numbers 1,…,12 and a move involves interchanging one counter and the ‘hole’ (the unique point having no counter) and interchanging the counters on the two other points of the line determined by the first two points. In the picture on the left, the lines are respresented by dashes around the circle in between two counters and the points lying on this line are those that connect to the dash either via a direct line or directly via the circle. In the first part we saw that the group of all reachable positions in Conway’s M(13) puzzle having the hole at the top positions contains the sporadic simple Mathieu group as a subgroup. To see the reverse inclusion we have to recall the definition of the ternary Golay code named in honour of the Swiss engineer Marcel Golay who discovered in 1949 the binary Golay code that we will encounter later on.

The ternary Golay code is a six-dimenional subspace in and is spanned by its codewords of weight six (the Hamming distance of whence it is a two-error correcting code). There are weight six codewords and they can be obtained from the 132 hexads, we encountered before as the winning positions of Mathieu’s blackjack, by replacing the stars by signs + or - using the following rules. By a tet (from tetracodeword) we mean a 3×4 array having 4 +-signs indicating the row-positions of a tetracodeword. For example

is the tet corresponding to the bottom-tetracodeword. A col is an array having +-signs along one of the four columns. The signed hexads will now be the hexads that can be written as vectors as (depending on the column-distributions of the stars in the hexad indicated between brackets)

For example, the hexad on the right has column-distribution so its signed versions are of the form tet-tet. The two tetracodewords must have the same digit (-) at place four (so that they cancel and leave an empty column). It is then easy to determine these two tetracodewords giving the signed hexad (together with its negative, obtained by replacing the order of the two codewords)

signed as

and similarly for the other cases. As Conway&Sloane remark ‘This is one of many cases when the process is easier performed than described’.

We have an order two operation mapping a signed hexad to its negative and as these codewords span the Golay code, this determines an order two automorphism of . Further, forgetting about signs, we get the Steiner-system S(5,6,12) of hexads for which the automorphism group is hence the automorphism group op the ternary Golay code is , the unique nonsplit central extension of .

Right, but what is the connection between the Golay code and Conway’s M(13)-puzzle which is played with points and lines in the projective plane ? There are 13 points so let us consider a 13-dimensional vectorspace with basis . That is a vector in X is of the form and consider the ‘usual’ scalar product on X. Next, we bring in the lines in .

For each of the 13 lines l consider the vector with support the four points lying on l and let be the subspace (code) of X spanned by the thirteen vectors . Vectors satisfy the remarkable identity . Indeed, both sides are bilinear in so it suffices to check teh identity for two line-vectors . The right hand side is then 4.4=16=1 mod 3 which equals the left hand side as two lines either intersect in one point or are equal (and hence have 4 points in common). The identity applied to gives us (note that the squares in are {0,1}) information about the weight (that is, the number of non-zero digits) of codewords in

Let be the collection of of weight zero (modulo 3) then one can verify that is the orthogonal complement of with respect to the scalar product and that the dimension of is seven whereas that of is six. Now, let for a point p be the restriction of

to the coordinates of , then is clearly a six dimensional code in a 12-dimensional space. A bit more work shows that is a self-dual code with minimal weight greater or equal to six, whence it must be the ternary Golay code! Now we are nearly done. Next time we will introduce a reversi-version of M(13) and use the above facts to deduce that the basic group of the Mathieu-groupoid indeed is the sporadic simple group .

References

Robert L. Griess, “ Twelve sporadic groups” chp. 7 ‘The ternary Golay code and ‘

John H. Conway and N. J.A. Sloane, “ Sphere packings, lattices and groups” chp 11 ‘The Golay codes and the Mathieu groups’

John H. Conway, Noam D. Elkies and Jeremy L. Martin, ‘The Mathieu group and its pseudogroup extension ‘ arXiv:math.GR/0508630

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If you only tune in now, you might want to have a look at the definition of Mathieu’s blackjack and the first part of the proof of the Conway-Ryba winning strategy involving the Steiner system S(5,6,12) and the Mathieu sporadic group .

We’re trying to disprove the existence of misfits, that is, of non-hexad positions having a total value of at least 21 such that every move to a hexad would increase the total value. So far, we succeeded in showing that such a misfit must have the patern

That is, a misfit must contain the 0-card (queen) and cannot contain the 10 or 11(jack) and must contain 3 of the four Romans. Now we will see that a misfit also contains precisely one of {5,6} (and consequently also exactly one card from {7,8,9}). To start, it is clear that it cannot contain BOTH 5 and 6 (then its total value can be at most 20). So we have to disprove that a misfit can miss {5,6} entirely (and so the two remaining cards (apart from the zero and the three Romans) must all belong to {7,8,9}).

Lets assume the misfit misses 5 and 6 and does not contain 9. Then, it must contain 4 (otherwise, its column-distribution would be (0,3,3,0) and it would be a hexad). There are just three such positions possible

Neither of these can be misfits though. In the first one, there is an 8->5 move to a hexad of smaller total value (in the second a 7->5 move and in the third a 7->6 move). Right, so the 9 card must belong to a misfit. Assume it does not contain the 4-card, then part of the misfit looks like (with either a 7- or an 8-card added)

contained in the unique hexad

Either way the moves 7->6 or 8->6 decrease the total value, so it cannot be a misfit. Therefore, a misfit must contain both the 4- and 9-card. So it is of the form on the left below

If this is a genuine misfit only the move 9->10 to a hexad is possible (the move 9->11 is not possible as all BUT ONE of {0,1,2,3,4} is contained in the misfit). Now, the only hexad containing 0,4,10 and 2 from {1,2,3} is in the middle, giving us what the misfit must look like before the move, on the right. Finally, this cannot be a misfit as the move 7->5 decreases the total value.

That is, we have proved the claim that a misfit must contain one of {5,6} and one of {7,8,9}. Right, now we can deliver the elegant finishing line of the Kahane-Ryba proof. A misfit must contain 0 and three among {1,2,3,4} (let us call the missing card s), one of with and one of with . Then, the total value of the misfit is

So, if this value is strictly greater than 21 (and we will see in a moment is has to be if it is at least 21) then we deduce that . Therefore belongs to the misfit. But then the move moves the misfit to a 6-tuple with total value 21 and hence (as we see in a moment) must be a hexad and hence this is a decreasing move! So, finally, there are no misfits!

Hence, from every non-hexad pile of total value at least 21 we have a legal move to a hexad. Because the other player cannot move from an hexad to another hexad we are done with our strategy provided we can show (a) that the total value of any hexad is at least 21 and (b) that ALL 6-piles of total value 21 are hexads. As there are only 132 hexads it is easy enough to have their sum-distribution. Here it is

That is, (a) is proved by inspection and we see that there are 11 hexads of sum 21 (the light hexads in Conway-speak) and there are only 11 ways to get 21 as a sum of 6 distinct numbers from {0,1,..,11} so (b) follows. Btw. the obvious symmetry of the sum-distribution is another consequence of the duality t->11-t discussed briefly at the end of part 2.

Clearly, I’d rather have conceptual proofs for all these facts and briefly tried my hand. Luckily I did spot the following phrase on page 326 of Conway-Sloane (discussing the above distribution) :

“It will not be easy to explain all the above observations. They are certainly connected with hyperbolic geometry and with the ‘hole’ structure of the Leech lattice.”

So, I’d better leave it at this…

References

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

John H. Conway and N. J.A. Sloane, “Sphere packings, Lattices and Groups” chp. 11 ‘The Golay codes and the Mathieu groups’

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