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Hi, folks.

Having read Vanderschel's msg of Aug. 6, it appears to me that the

explanations of the orientation parities are unnecessarily complex,

though the material on permutation parities is presented in a way

that should be immediately convincing to anyone familiar with the

notion of even and odd permutations from elementary group theory (and

anyone who isn't should be!). Here's my attempt at a more elegant

demonstration.

In what follows, I will use the term "facelet" to denote any (visible)

face of a cubie. Thus, each edge cubie has two facelets and each

corner cubie has three facelets.

I will address Edge Orientation Parity (EOP) first. Consider the

following diagram:

+-------+ | 1 | |0 U 0| | 1 | +-------+-------+-------+-------+ | 1 | 0 | 1 | 0 | |0 L 0|1 F 1|0 R 0|1 B 1| | 1 | 0 | 1 | 0 | +-------+-------+-------+-------+ | 1 | |0 D 0| | 1 | +-------+

The numbers label absolute positions, not facelets, and therefore

remain in the same configuration when the cube is manipulated.

Imagine that a mark is placed on each facelet that occupies a

position labeled "0" when the cube is in the solved configuration.

Thus, each edge cubie will have one marked and one unmarked facelet.

(Unlike the numbers, the marks are attached to the facelets and will

move as the cube is manipulated). The parity of an edge cubie in an

arbitrary configuration is defined as the number labelling the

position occupied by its marked face, and the EOP is defined as the

sum of the parities of all edge cubies modulo 2. A quarter turn of

any face reverses the parity of 4 edge cubies, thus leaving the EOP

fixed. By induction, no sequence of manipulations starting from the

solved configuration can produce a configuration EOP = 1. So much

for EOP.

[Just for grins, here's a cute way to determine the parity of an edge

cubie without consulting (or reconstructing) the diagram: Assign

each of the cubie's two facelets a number in {0,1,2} according as it

is oriented parallel to its home position ("Self" -> 0), parallel

to the other facelet's home position ("Other" -> 2) or perpendicular

to both home positions ("Neither" -> 1); add these two numbers and

reduce modulo 3 (yes, three!) giving the parity of the cubie. The

sum can never be 2 because that would imply that the two facelets

were parallel to each other. Here's an addition table with

double-ended arrows showing the possible transitions:

(S) 0 (N) 1 (O) 2 +-------+-------+-------+ | | |no | (S) 0 | 0 <-+-> 1 | + | | ^ | ^ | way| +---+---+---+---+-------+ | v |no | | | (N) 1 | 1 <-+---+---+-> 0 | | | |way| ^ | +-------+---+---+---+---+ |no | v | v | (O) 2 | + | 0 <-+-> 1 | | way| | | +-------+-------+-------+

]

Now for Corner Orientation Parity (COP). Consider the diagram below:

+-------+ |0 0| | U | |0 0| +-------+-------+-------+-------+ |2 1|2 1|2 1|2 1| | L | F | R | B | |1 2|1 2|1 2|1 2| +-------+-------+-------+-------+ |0 0| | D | |0 0| +-------+

Once again, imagine that we mark all (corner) facelets which occupy

positions labeled "0" when the cube is in the solved configuration.

The parity of a corner cubie in an arbitrary configuration is defined

as the number which labels the position of the cubie's marked face.

The COP is the sum of all the parities of all corner cubies modulo 3.

Inspection of the diagram will reveal that the twists U, u, D, and d

leave the parities of all corner cubies unchanged. Any of the other

possible quarter twists will increment (modulo 3) the parities of two

corner cubies and decrement the parities of two others, thereby

leaving the COP unchanged.

[As Vanderschel pointed out, one way to compute the parity of a

corner cubie by looking at it is to note the number of clockwise (as

viewed from outside the cube) 120 degree twists of the cubie that it

would take to bring the marked facelet parallel to its home position.

Note that the parity of a corner cubie, unlike that of an edge cubie,

depends on the selection of a particular pair of opposing colors for

the marked facelets. While this lack of symmetry may be considered

unfortunate, it is an inevitable result of the fact that four is not

divisible by three. It is easy to show that the COP as a whole is

independent of the choice of distinguished faces.]

Vanderschel also mentions the extended problem, but does not quite

make it clear that the FOP changes every time a qtw is done. This

constrains all three of {FOP, EPP, CPP} to be the same, so that only

two of the eight plausible states of the <FOP, EPP, CPP> vector are

actually achievable by twisting.

Jim