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Date: Sat, 16 Aug 1997 22:35:13 -0400
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From: Dan Hoey
To: reid@math.brown.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199708152344.TAA05912@life.ai.mit.edu> (message from michael
reid on Fri, 15 Aug 1997 19:49:54 -0400)
Subject: Re: isoglyphs
> perhaps i am missing something, but doesn't
> D2 R2 U2 D2 R2 U D (12q, 7f)
> produce an isoglyph of type 4D ?
Oops, you're right. I goofed because I half-remembered a different
result, that there are no 6-bar patterns of the nice symmetric sort,
with three mutually perpendicular pairs of parallel bars.
> ...
> i hadn't even considered chiral versus achiral isoglyphs. indeed,
> all the "continuous" isoglyphs given by herbert are chiral.
> achiral isoglyphs certainly exist, for example
> D2 R2 U' B' L B U B L F2 R D' L2 U2 B2 D (22q, 16f)
> of type 11; pattern
> *..
> ***
> ***
> and others can be derived from this. i suspect that there is no
> chiral form of this isoglyph, but i'm not absolutely certain.
Modulo some oversight, I think this is true, and not hard to
demonstrate.
Recall that a "ground" facet is one that is not on its home face.
First note that a corner cubie will have 0, 2, or 3 ground facets. So
on any isoglyph of corner type 1, there are a total of 6 ground corner
facets, and these ground facets must appear on two corner cubies
(three ground facets each) or three corner cubies (two ground facets
each). If two corner cubies, those cubies must be antipodes, and they
are either rotated (forming a meson, FTR+ BLD- or equivalent) or
exchanged ((FTR,BLD) or equivalent, implying odd edge permutation
parity). If ground facets appear on three corner cubies, the cubies
must be a three-cycle of cubies on nonadjacent corners ((FTR,FDL,BTL)
or equivalent).
I've done some analysis by facets on these three cases, which is too
messy to describe, but which leads me to the conclusion that the above
position is the only isoglyph of its pattern, implying the conclusion
that there is no chiral form.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil