# Disjoint Cycles and Twist/Flip Parity Rules

Submitted by B MacKenzie on Wed, 05/16/2007 - 00:13.

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.

(-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.