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I am having difficulty understanding the third stage of Thistlethwaite's algorithm.

I am using http://www.jaapsch.net/puzzles/thistle.htm.

The third stage has a factor of [8!/(4!4!)]2 *2*3. This corresponds to the fact that the edges in the L and R faces are placed in their correct slices, the corners are put into their correct tetrads, the parity of the edge permutation (and hence the corners too) is made even, and the total twist of each tetrad is fixed.

If I understand correctly, a tetrad is formed from the path a corner travels using only moves in <L2,R2,F2,B2,U2,D2>. This forms two tetrads, each consisting of four corners, and each corner can then be placed into its respective tetrad.

Then the edge slices can be arranged and the parity determined. This counts up to

C(8, 4)2 × 2 = 9,800

The size of stage three, however, is 29,400 which means I am off by a factor of 3.

What I don't understand is the meaning of tetrad twist and how it generates a factor of 3.

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  • $\begingroup$ It would appear this question was also asked on math.SE, but nobody knew the answer. There is a link there to a Thistlethwaite-based solver that resolves stage 3 with a condition stronger than "total twist fixed", FWIW. $\endgroup$ – COTO Nov 22 '14 at 6:31
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Tricky question! The extra factor of 3 comes from how the permutation of the corners in one tetrad affects the permutation of the corners in the other one.

As I'm sure you know, the tetrads are each made up of 4 corners in an orbit like in the following image. I colored the corners in one tetrad in shades of pink and the corners in the other tetrad in shades of teal:

highlight the two tetrads

It should be easy to see that any permutation is possible within a tetrad. For example U2 swaps a pair of corners in each tetrad and if a pair of corners can be swapped in a tetrad then any permutation (4!) in each tetrad is possible.

The trick comes when you try to perform a 3-cycle in one of the tetrads. For example the commutator [R2, F2] (the sequence R2 F2 R'2 F'2) performs a 3-cycle of the UFR and DRB and DFL corners. But, and this is where the factor of 3 comes in, it also performs a 3-cycle in the other tetrad.

So instead of the total permutations of the corners being (4! * 4!) / 2 (where the 2 comes from the fixed parity) it is actually (4! * 4!) / (2 * 3) because a lone 3-cycle is disallowed.

Here is an example of the sequence R'2 F'2 R'2 F'2 D'&2 F'2 D&2 F'2 (where D&2 is the equatorial slice turned the same direction as D) that shows two 3-cycles at once:

tetrad 3-cycle pair

In the group generated by <L2,R2,F2,B2,U2,D2> you just can't overcome this restriction so when you move from <L,R,F2,B2,U2,D2> to <L2,R2,F2,B2,U2,D2> there is an unintuitive extra factor of 3 in the reduction.

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  • $\begingroup$ Thanks for the clear diagrams. It took me a while to wrap my head around this, but I finally solved a cube using this algorithm. Now I'm just having trouble recognizing when the corners are twisted incorrectly. $\endgroup$ – resyst Nov 28 '14 at 0:29

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