From:

~~~ Subject:

In his message of Fri, 17 Dec 1993 00:54:00 EST, Jerry Bryan

<BRYAN%WVNVM.BITNET@mitvma.mit.edu> makes some observations on the

distances between the following positions in the edge group:

I = Solved,

P = Pons Asinorum (or Mirror),

E = All edges flipped, and

PE = P E = Pons Asinorum with all edges flipped.

[I _will_ continue to use permutation multiplication as we have done

so in this group since its inception. I realize that this agrees with

some textbooks and is backwards from others, but it would be far more

confusing to write these functionally all the time.] Jerry's

brute-force search has shown that d(I,PE)=15, and he notes that

conjugation by E shows us that d(P,E)=15 as well. He concludes:

I have the sensation in describing this that the Edge group is

square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees

apart, and Mirror-Image-of-Start and Edges-Flipped at the other

two corners of the square.

Well, it's not quite a square, since d(I,P)=12 and d(I,E)=9, according

to Jerry's message of Wed, 8 Dec 1993 10:02:15 EST. Conjugation will

similarly show that d(E,PE)=12 and d(P,PE)=9. So we are dealing with

a rectangle. The sides of the rectangle are 9 and 12, and the

diagonal is 15: a most fortuitous set of numbers, in that we can

actually embed such a rectangle in the Euclidean plane!

We can map the positions of the edge group to 4-tuples of distances.

For any position X, let

f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).

If f(X)=(a,b,c,d), then conjugation shows us that f(X E)=(b,a,d,c),

f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). So the set of quadruples has

the symmetries of the rectangle.

We know f(I)=(0,9,12,15). What is more, the earlier results on

symmetry show us that I is at a local maximum distance from E, P, and

PE. So, letting I1 be the unique (up to M-conjugacy) position

adjacent to I, we have F(I1)=(1,8,11,14). (This destroys Euclidean

embeddability.) An analogous result holds for the unique neighbor of

each corner of the rectangle.

We also have Jerry's results of Wed, 8 Dec 1993 22:41:28 EST and

23:16:50 EST that H (the 6-H pattern) and HE=H E are at distances 8

and 13 from start, respectively. Since H is an M-conjugate of P H,

this gives us f(H)=(8,13,8,13). [Note: there are two distinct

M-conjugates of H, call them H and Hbar. This distinction is

important when we compose permutations: H H = I, but H Hbar = P. So

we have to be careful when conflating M-conjugates.] We can by

symmetry find f(H1)=(7,12,7,12) for H's unique neighbor H1.

What quadruples are possible? If f(X)=(a,b,c,d), and X is not one of

the eight corners and neighbors, we have

max(2,9-b,12-c,15-d) <= a <= min(14,9+b,12+c)

with constraints on b, c, and d from symmetry. A quick hack tells me

there are 7836 such quadruples. I wonder how many of them are

realized? If it's fairly few, I would like to see a diagram of

quadruples, with lines between those quadruples that represent

adjacent positions (adjacent quadruples differ by at most one in each

coordinate). Maybe with the number of positions for each quadruple,

too. I have an idea that such a diagram might tell us something about

the problem, or at least look pretty.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil