Red Ed wrote:King Roger? Pah. Nothing but a jester in the court of King Andy!

Perhaps your "King Andy" is from the Roddick family instead of the Murray family? The latter one crashed out yesterday. But anyway the former one should also be eliminated later today by "King Novak". However, I suspect all these "Kings" at the end would all be perished under the horrific inhuman power of "Emperor Rafa"...

Back to the topic...

Red Ed wrote:Here's what I would need from you:

- You give me a list of "nice" symmetries. By "nice", I mean simple stuff like rotation, transposition, band/row cycling, flipping the whole grid horizontally or vertically; that sort of thing. The list must be sufficient to generate (by combining) any possible symmetry. It doesn't matter if there are some redundant symmetries in there -- in fact, it should be expected that there'll be quite a bit of redundancy. eleven's list from the first post would be a good starting point:

- You tell me the cost of each of those "nice" symmetries. The cost is a positive integer. Really nice symmetries have cost 1. Slightly less nice ones have cost up to, oh I don't know, 5?

- You give me a combination cost. This tells me how bad it is to write a symmetry as a combination of two others.

In return, I could express each of the 122 aut groups as a list of generator symmetries such that the overall cost of the each list was minimal...

Interested? Or do you want to go ahead and do this manually yourself?

My group of "nice symmetries" should be the 13 basic ones and 13 merged ones, repeatedly specified in the last few pages. I keep saying I'll formulate them nicely later but keep struggling to find time. Now that the tennis schedule is less packed perhaps I can find more time to finally get some work done. Just give me a couple more days...

In the mean time, here is something a little bit different (but still on topic of this thread). On the

last page JPF cited the

Maximum Minlex grid found by you:

- Code: Select all
`123456789`

457893612

986217354

274538196

531964827

698721435

342685971

715349268

869172543

He mentioned there are 72 automorphisms, I was deeply interested and studied this grid in details during my limited free time among work/meals/sleep/tennis. Here are the early results:

This grid has Jumping Rows and Jumping Columns (hence implied Jumping Diagonals on both directions). It also has Row Sticks, Column Sticks, Diagonal Mirrors (both directions) as well as Half Turn (which is implied by double Diagonal Mirrors):

- Code: Select all
`r132465897`

c132465879

d123456789 -> d563819427

536891247

792645138

841237596

684123759

927456381

153789624

415378962

279564813

368912475

JR: (176)(258)(394)

JC: (138)(279)(456)

J\: (195)(263)(487)

J/: (142)(357)(698)

RS: (4)(5)(6)(17)(28)(39)

CS: (2)(5)(8)(13)(46)(79)

\M: (1)(5)(9)(24)(37)(68)

/M: (3)(5)(7)(19)(26)(48)

HT: (5)(19)(28)(37)(46)

Note there are merged symmetries too (CS+JR, \M+J\ etc) which enable us to see some additional symmetries "on the fly" (e.g. if we shift r789 to the top, then there is CS: (3)(4)(9)(18)(27)(56)

).

Anyway, after some pondering, I figure that to count the number "72" by trying to identify the "independent factors" of the symmetry groups and then multiplying their orders together is a pretty naive approach, as demonstrated by

eleven and me on several posts on the last 2 pages.

Here is a more sensible approach to verify the number "72" here:

Firstly, fix the centre cell r5c5. Now how many isomorphs can we generate?

Obviously, the grid as it is (identity) is the very 1st one.

\M (transpose) gives us another.

/M (anti-transpose) gives us the 3rd.

Applying both simutaneously gives us a 4th, which is essentially the same one obtained by HT (180 rotation).

Now, applying CS (column sticks) on each of these 4 can give us 4 more isomorphs. (For the unfamiliar, the CS transformation is essentially swapping each mini-column to the opposite one in the local stack, e.g. r123c1 is swapped with r789c3, r789c4 is swapped with r123c6).

Note that we don't need RS (row sticks) now since \M+CS is essentially the identical transformation.

Hence with r5c5 fixed, we get 8 isomorphs.

The next step would be using JR & JC, we can freely move each centre of the 3x3 blocks to r5c5 via whole band/whole stack shifting. And we can be sure each time we'll get new isomorphs because each of r258c258 has a different unique value.

From each of these 9 centres at r5c5 we will get 8 isomorphs. So the total number of isomorphs is 9x8=72. Voila!

From there we can freely select the independent factors:

\M (2), /M (2), CS (2), JR (3), JC (3)

or

HT (2), \M (2), CS (2), JR (3), J\ (3)

The fact that \M+JR implies JC etc should not be bothered too much. Apparently it's much easier to analyse if we seperate the order-2 & order-3 (and possibly order-9) operations and don't mix them together.

I think with some effort all 122 groups can be analysed like that. But as always it requires substantial time and effort!

(I'll work hard at your request to give you a nice specification of each symmetries, and hope that the help of your program can speed up things considerably.)

In the mean time, I'm also very interested in the problem of working out this (Max Minlex) grid manually, which I think I have a very nice fully logical solution that doesn't need any machine help. But it probably belong to another thread when I finally get to write it up nicely.

[Edited: some minor corrections]