Disjoint Cycles and Twist/Flip Parity Rules

I've been fooling around writing a virtual Rubik's cube program, initially simply as an exercise for teaching myself the Open GL 3D rendering API. In representing the puzzle I was led to assign each cublet an X,Y,Z coordinate specifying its starting position in the cube.
(-1,-1,-1) being the coordinate for the (left,down,back) cublet through to (1,1,1) being the coordinate for the (right,up,front) cublet. The transformed position and orientation of each cublet is then specified as an element of the O symmetry point group, there being a one to one correspondence between the 24 elements of the O point group and the 24 states a cublet may assume via Rubik cube face turns: 12 edge positions with two flip states each or 8 corner positions with 3 twist states each.

In due course I was led to looking at Rubik group elements in terms of disjoint cycles, i.e

Cycle Analysis
Corner cycle: (0) X 3 (C3xyz2)
Corner cycle: (2,19,17) X 3 (C3x'y'z)
Corner cycle: (5,14,12,7) X 3 (C3x'yz')
Edge cycle: (1,15) X 2
Edge cycle: (3,18,4,10,9,16)
Edge cycle: (6,11)
Edge cycle: (8) X 2
Edge cycle: (13)

In my representation these cycles are the consequence of applying a series of O point group operations to a cublet. For example the above cycle, (2,19,17), results from applying the operations, [C2yz][C2xy][C3xy'z'2] = [C3x'y'z], to the left,down,front cublet. This sequence returns the cublet to its home position rotated 120 deg about the C3 axis passing through that position. Similarly, the cycle, (1,15) = [C3x'yz'][C2yz] = [C2xy], returns the left,down edge cublet to its home position rotated 180 deg about the C2 axis passing through that position. This has led to the observation that there are always an even number of edge cycles equivalent to C2 operations, and either two (opposite sense) or three (same sense) corner cycles equivalent to C3 operations. It seems obvious that this is related to the oft cited parity rules concerning the total twist and the total flip of the cube being zero. However, it is not obvious to this non-mathematitian why this should be so. Why is it not possible to satisfy the edge flip parity rule with a single two cycle? Once through the cycle would afford two cublets flipped after all.

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Cubelet flip


I hadn't heard the term "O point group" before (googling seems to show it is used in chemistry), but I assume it means the group of 24 rotations of a cube (or cubelet).

It is possible to define what edge flip means within this group. When a cubelet is rotated, this can be considered as a permutation of its 6 facelets. Call a cubelet flipped if its facelets have undergone an odd permutation. The 24 rotations then break down into two sets of 12 - one set flips a cubelet, the other doesn't.

No flip:
the eight 60 degree rotations about a vertex
the three 180 degree rotations about a face centre
the six 180 degree rotations about an edge
the six 90 degree rotations about a face centre

Note that this is consistent - two flip rotations combine to give a no-flip rotation, two no-flip rotations combine to a no-flip rotation, and flip plus no-flip gives flip.

Any quarter turn move on the cube flips 4 edge cubelets. Before the move, 0,1,2,3 or 4 of those edge cubelets could have already been in a flipped state, and after the move there will then be 4,3,2,1, or 0 edge cubelets in a flipped state respectively.

Note that the number of flipped cubelets remains odd or remains even, regardless of the move. As the solved cube has by definition no flipped edge cubelets, there will always be an even number of flipped edge cubelets when it is mixed.

A 'twisted two-cycle' will indeed flip two edges when performed twice, but when done only once it will flip exactly one edge cubelet. As I just showed, that is not possible to do without also flipping other edge cubelets.

Corner twists are rather more messy. It is unfortunately not possible to define cubelet twist by looking only at the rotation without taking the cubelet location into account.

Jaap's Puzzle Page: http://www.geocities.com/jaapsch/puzzles/

Flip cycles

Thanks for your comment.

O is the Schoenflies designation for the group, commonly used by chemists of which I am one.

Ok. It makes sense. If there is a cycle, (a,b,c,d . . . ) = C2 the product ...[d][c][b][a] must have odd parity. This means an odd number of the terms have odd parity. Since a,b,c .. are the rotational states of the cubelets involved in the cycle, at this point in the cycle an odd number of them are flipped. Thus there must be another C2 cycle to even the total parity.

I hadn't realized it before but a cubelet's flip is given by the A2 representation of its rotational state. You've allowed me to simplify my program. I'll have to have another look at corner twist, there must be a simple way to pull it out of the character table.

Again, thank you.
Bruce MacKenzie