Message #2154
From: David Vanderschel <DvdS@Austin.RR.com>
Subject: Re: [MC4D] Faulty Logic Counting States for 4D Center-Edge Cubies
Date: Fri, 11 May 2012 21:48:49 -0500
Andrew Gould wrote (about me):
>He claimed that an even length cubie permutation cycle (of
>3C pieces) implies 3 even length permutation cycles of
>stickers. Counterexample: a 2-cycle of the 3C pieces, but
>a single 6-cycle of their stickers.
Andy is absolutely correct. I had figured out long ago,
using the proof he provides below, based on the effect of
quarter turn twists, the fact that the sticker permutation
parity was equal to the cubie permutation parity. For some
reason, when I thought to provide a proof for the fact in my
post, I did not remember how I had done it before and I came
up with the incomplete proof based on the permutation cycles
in the whole pile’s actual state. I did not think about it
enough because I knew I was getting the right result. :-(
(The proof using the state cycles can be resuscitated by
observing that the length of any sticker cycle corresponding
to a cubie cycle must be the cubie cycle length (for 3 of
them) or 3 times the cubie cycle length (for just 1), but it
is not worth pursuing this approach because the proof based
incrementally on the twists that generate the state is much
easier. (The tricky part is proving that you cannot get a
sticker cycle of length twice that of the cubie cycle.))
>I believe you would still be able to arrive at the last
>sentence of that paragraph ("Thus the parity of the sticker
>permutation is equal to the parity of the cubie
>permutation."), but I believe this approach may be easier
>(it seems to be what David Smith was getting at in Roice’s
>last link):
I did not even see where David Smith attempted to prove
equality of the permutation parities for the cubies and
their stickers. He just seemed to take it as given.
>Note that an element of the generating set (of twists)
>performs an odd permutation on the 3C pieces iff (if and
>only if) it performs an odd permutation on the 3C
>stickers. Thus, any composition of generators performs
>an odd permutation on the 3C pieces iff it performs an
>odd permutation on the 3C stickers.
And, yes, it is pretty clear based on the twisting argument.
Thus I offer the following revision to my (hopefully really
now) valid argument:
There are 96 sticker positions in the pile occupied by<br>
stickers from these 32 cubies. When we scramble the<br>
puzzle, we are permuting not only the cubies but the<br>
individual stickers themselves. It may be observed that<br>
the permutation parity of the stickers equals the<br>
permutation parity of the cubies: It suffices to check<br>
the effect of a quarter turn of an external slice, since<br>
any twist can be generated by composing such quarter<br>
turns. For a quarter turn, there are 3 4-cycles of the<br>
cubies and 9 4-cycles of the stickers. Thus permutation<br>
parity always changes for both cubies and stickers. (If<br>
you want to consider twisting a center slice, there are<br>
2 4-cycles of cubies and 6 of stickers for no change in<br>
permutation parity of either.)
Now we can come back to the placement of 31 of the<br>
cubies, each with any of 6 possible orientations. The<br>
permutation parity of the cubies themselves is now<br>
determined and so is that of the stickers. Of the six<br>
possible orientations for the 32nd cubie only half of<br>
them will lead to the one possible overall sticker<br>
permutation parity value that is possible for whatever<br>
the cubie permutation parity is. (Without working out<br>
the actual parity of the cubie permutation, we don't<br>
know what that parity is. Indeed, it can go either way;<br>
but, whichever way that is, only half of the 6 possible<br>
orientations will lead to the correct corresponding<br>
overall sticker permutation parity.)
Thank you, Andy, for the correction.
Regards,
David V.