Date: Tue, 28 Dec 93 14:16:49 -0500 (EST)
From: Dan Hoey <hoey@aic.nrl.navy.mil >
~~~ Subject: Re: Some Additional Distances in the Edge Group

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