Suppose n greater/equal 5. Look at the entry in position (0, …, 0, 1, 2) with d – 3 zero’s. There is no value possible that doesn’t violate the combination of supersymmetry and rigidity.

Of course there is still some question about what happens if n = 4, but the problem is not as interesting anymore as I thought yesterday, I’m sorry.

]]>if d odd there are either one or zero solutions depending on whether d is divisible by 3 or not. (Not counting solutions with n = 1 or n = 0). The proof rests on 4 more general but easy results:

1) For a rigid supersymmertic latin (d-1)-cube, we have that n is at most 1 + the smallest prime divisor of (d – 1).

2) For n = 2 and n = 3 every supersymmetric latin hypercube for d greater than or equal to 3 is of group type. (This can be shown by induction on d)

3) For n = 3 the ‘zero section’ group type sslhc is rigid if and only if d is divisible by 3 and the other two are never rigid.

4) n can not be a divisor of d-1

Now the question is what happens for d even. Some additional results:

5) For n = 2 both group type supersymmetric latin hypercubes are rigid if d is even and neither one is when d is odd;

6) For n = 4 the zero section group type sslhc coming from the Klein four group is rigid if and only if d is even

So for d = 4 we expect at least 3 solutions and for d = 6 we expect at least 4 solutions. I checked by hand that there are no others for these values of d. In particular all examples I saw so far are of group type. However,I don’t know what to think of that. For the cyclic group of four elements and for every group of five or more elements it is easy to see that no group type sslhc is rigid.

In semi-related news: statement 2) does not extend to n = 4 and higher; an example of a (non-rigid) non-group type supersymmetric latin cube (d = 4, n =4) is given by the supersymmetric orbits of

(AAAA),

(AABB), (AACC),

(BBDD), (CCDD)

(AADD), (ABCD)

(BBBC), (CCCB)

(DDDD)

So… I don’t really see why rigid sslhc’s should be group type, but as I said, I know no counterexamples. Do you?

Hapy holidays,

Vincent

]]>The above shows I think that there are to many sslss to handle. Hence a different question. Let’s call a ssls ‘rigid’ if it is its own orbit under your equivalence relation. One might also call them hypersymmetric since they have even bigger symmetry groups than ‘ordinary’ supersymmetric latin squares. In d=3 (so latin squares) the only example is in your origional post (n = 3) and it is not hard to show that there are no others. My question is: can we classify all rigid supersymmetric latin (d-1)-cubes for all d?

]]>Via some detours I found out that these special sslss (or actually the orbits of length 6) already exist under a different name: steiner triple systems. Apparently in the 19th century it has been proven that such systems and hence sslss with numbers 1 to n on the diagonal exist if and only if n is congruent 1 or 3 mod 6.

Their number growth rather fast: http://oeis.org/A030129

In particular there are apparently 2 sslss of this type of order 13, although I was not able to find them. I get the impression that the number of Steiner triple systems of order 21 is yet unknown, although we can construct one from the sslss of order 3 and 7 using your construction.

]]>0654321

6132540

5321604

4213065

3560412

2406153

1045236

I find it a bit disturbing how easy it is to find these examples. There is no clever reasoning behind it, I just wrote the first few numbers and then the puzzle essentially solved itself, like an extra simple sudoku-puzzle. Still I like to believe that these sslss (due to their rich symmetry) ‘come from’ some other ‘special’ object in some unified way…

]]>Here is an example with n = 7 and 4 elements of the form (x, x, x) (unless I made a mistake and this thing is not supersymmetric after all):

0654321

6145230

5426103

4563012

3210654

2301546

1032465

So it seems the role of group-type sslss is not as big as the n=3 case suggests. Pity, but still interesting!

]]>Your non-group-type example fascinates me. How did you find it?

Here is a new question inspired by your theorems:

The statement that the ssls contains a unique element of the form (x, x, x) (i.e. entry x in position (x, x)) iff n is not divisible by 3

is true for group-type squares. (Similarly for elements of the form (x, x, …, x) in the d-dimensional case if n is not divisible by d). Is it true for all sslss?

Regarding supersymmetric Latin squares and your construction, I do have some new results. I’m going to do another post on this topic in the next couple days, and I’ll include the proofs there, but I wanted to get these facts to your attention without delay.

Let S(G,g) be the square which arises as the set of triples of elements of G which sum to g, and call a supersymmetric Latin square “group-type” if it is equivalent to some S(G,g).

THEOREM. Suppose 3 does not divide n. Then there is a one-to-one correspondence between equivalence classes of group-type ssls’s of order n and isomorphism classes of abelian groups of order n. That is, S(G,g) and S(H,h) are equivalent iff G is isomorphic to H.

THEOREM. Not all ssls are group-type. In particular, the square

AFEDCB

FBDCEA

EDCBAF

DCBAFE

CEAFBD

BAFEDC

is not group-type. Up to equivalence, this is the only ssls of order at most 6 which is not group-type.

]]>