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