From:

Subject:

"Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu> writes:

I spoke too quickly when I said the antipodal was simply

Start with the edges flipped. I stared at it, flipped the edges in

my mind, and it "looked" solved, so I assumed it was Start.

It's interesting to note that this is All-Edges-Flipped composed with

a mirror reflection of Start. Begging the question: *which* mirror

reflection? The answer is, it doesn't matter: since these are the

edges of a cube without centers, all reflections are the same

position. As long as we get to choose which reflection, the canonical

one would be the central reflection. When composed with All-Edges-

Flipped, it makes the following antipode. (I think using BFTDLR

notation instead of 123456 makes these diagrams a lot easier to read).

+ T + + F + T T R L + T + + B ++ L + + F + + R + + D + + D + + D + L L F F R R => F B R L B F + L + + F + + R + + T + + T + + T ++ D + + B + D D R L + D + + F + + B + + T + B B R L + B + + D +

I am not yet for sure what they look like, but of the other two states

with order-24 equivalence classes, one is at level 9 and the other

is at level 12. Since the only one at an even level is at level 12,

I am assuming that will be the one which is Start with the edges all

flipped. The one at level 9 will probably be the mirror image of Start.

If an order-24 equivalence class means what I think it does, I'm

pretty sure those two states have to be Mirror-Start and All-Edges-

Flipped, being the only sufficiently symmetric positions. But as to

their depth, the parity argument (which Chris Worrell also cited) is

not valid here. Remember that the cube has no face centers, so the

position is not changed by rotating the assemblage of edges in space

(i.e., with respect to the absent face centers). But a quarter-turn

of the cube in space is an odd permutation of the edges. So permuta-

tion parity is not an intrinsic property of edge positions. We can

show that there is no sort of parity here by explicitly constructing

an odd cycle. Just use a process that would permute the edges of a

cube with faces as (FR,FT,FL,FD)(BR,BT,BL,BD)(RT,TL,LD,DR). This has

to be an odd process, but it's an identity on the faceless cube.

My (very cheap) guess for where we will find the other two M-symmetric

positions is opposite to Jerry Bryan's. On a cube with faces, the

central reflection of the edges with respect to the faces is Pons

Asinorum, which has the easy 12-qt tight lower bound we've seen before

(or if not, you can of course get it from me with email). I'm

guessing that this bound happens to be tight on the cube without

faces, as well. But I have no proof of this guess, and I'm very

grateful we won't have to settle for guesses for very long.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil