Message #635

From: rev_16_4 <>
Subject: [MC4D] Re: Parity on MC m^n
Date: Mon, 02 Feb 2009 00:34:47 -0000

Ok, one last reply today. I’m going to trim this down to the points
I’m commenting on (apologies if this is in bad form):

>> On a 3^3, a single swapped pair has odd parity. This cannot happen
>> due to the even (aka double odd) parity of a single quarter twist
>> on a 3^3. However, using reduction, the "single swapped edge pair"
>> (with correct orientation) parity err on a 4^3 can seem to occur.
>> This will alway take an even number of quarter twists to solve.

> Furthermore, we can say that some of those twists will have to be
> of the inner slice. In fact, both inner and outer twists will be
> required, an even number of both. I can provide my reasoning for
> this conclusion if desired.

I agree completely.

>> You can see a similar phenomenon with a 3^3….

> It is interesting to note you can have an odd number of corner pair
> swaps with an even number of edge pair swaps if the orientations of
> the corners also come into play, the most simple example being a
> cube that is solved except for two corners. In this case I guess
> the double odd parity is shared with the orientations instead of
> the edges.

Are you refering to two corners that aren’t oriented correctly? If
not I’m not sure what you mean. On a 3^3, you cannot swap a single
pair of corners, regardless of orientation, unless an odd # of pairs
of edges are swapped as well.

> I think we can deduce the 4C case is impossible because corners are
> only permuted/oriented by outer twists (that is, by the same twist
> set as the 3^4), and this is not a possible configuration on the
> 3^4. The 3C case I’m unsure of, but I plan to experiment with it
> some.

I’m not so sure. The same thing is possible on the 4^3, yet not on
the 3^3….

>>[…"Wait a minute! This position is unsolvable!"]

> That is exactly what happened to me though ;) But using a mental
> parity model of a 4^3, not a 3^4. Considering two adjacent,
> identically colored and oriented pieces as behaving like a single
> edge piece on a 4^3, I reached this state and did not know how to
> solve it (because the solution necessitated breaking these already
> solved, artificially joined pieces back in two). Since I had
> already mentally recategorized those sets of two as single merged
> pieces, the puzzle model I was trying to use to my advantage left
> me stuck.

Interesting. Essentially you went 4^4, to 4*3^3, to 3^4? This parity
case happened between 4*3^3 and 3^4? Is this close to accurate?

> Overall, I’d say the different usages of the word parity is still
> clouding the discussion here. I am conceding that my use of the
> phrase "parity problem" is perhaps too general. After all, a
> problem is simply a configuration you don’t know how to solve using
> a tool set of sequences.
> And parity just means even or odd, which can be applied in a number
> of ways. I promise it’s clear in my mind though :) I like the
> reductionist approach of analyzing parities and deriving what that
> says about the solution. I like the concept of "parity problem"
> defined in the context of mismatches between the different parity
> characteristics of various puzzles, that they are mental surprises
> (perceived impossibilities) when one tries to use solution methods
> across multiple puzzles.

I agree, the terminology is annoying. I was typing some of these
posts thinking "Do I really need to say ‘single edge pair-swap parity
error’ every time?" I figured it was too ambiguous otherwise. I like
your definition for traditional "parity problems" above. It’s exactly
what’s happening, regardless of how you solve the puzzle.

This was the question I kept having while trying to develop a
solution across all n^d. How do you address a possible parity
condition that might only happen on d>10, but has no relevence on
realistic puzzles? 4_4Roice2.log is a good example. This can’t happen
on a 4^3 (or 3^4). Are there similar issues on the 4^5? I don’t know.
Not with the caging method I settled on. And not with even n, d>3 if
the only limitation you self impose is "only conjugators and
commutators are legal." I suppose this is the spirit of the no parity
problem comment I made. I’m also confident that given a 4^100, and
time was no issue, I could solve any given solvable parity case with
the solution I have. The only reason time is an issue is because each
algorith would be somewhere around 14 * 2^97 quarter-twists!

Thanks for the lively discussion.