Challenge with three faces

No, I do not need other tricks. A corner tri-swap in one face (without affecting edges) takes 10 face turns at most, and if it takes 10, then these 10 moves are in three adjacent faces. Indeed, I used cube explorer to compute this trick (it is a beauty, just as the 12 turns for twisting two opposite corners on a normal cube), but it is still useful for the general cube. I use (symmetric/inverse variants of) FURU'R'F' to get the edges of the last face right, and conjugations on the above corner trick to complete the last face.

Further, I conjugate on (R2U2)3 to get the edges of the slice between the first face D and the last face U right (using FURU'R'F' for edge orientation) and (R2U)2 R2U2R2 for the corners of the first face D (using FURU'R'F' for corner orientation). (R2U)2 R2U2R2 was also computed with cube explorer, but is again handy in general. However, if you do not agree, then you can just start with the first face before doing the slice between the first and the last face, and you do not need (R2U)2 R2U2R2.

Your edge tri-swap is only 11 face turns with only two faces. The one I use has 7 face/slice turns and 9 face turns of four faces. But there is more: it proves that any rearrangement of the edges in two faces is possible with those faces. This is not true for the corners: any tri-swap is impossible. Furthermore, it proves that parity is the only thing that connects the rearrangement of the edges and that of the corners. Since the edge orientation cannot be changed with two faces, the edges are now understood. The corners orientation can be changed, as (FUF'U')2 makes clear. By some conjugations on that, one can get any reorientation of the corners with only two faces (affecting the edges, but that was no problem).

In order to prove that a tri-swap of the corners is not possible, one can make a more symmetrical object. This object will have so many symmetries that the number of corner position configurations reduces to only three modulo symmetry on both sides. Then it is easy to see that a tri-swap does not belong to those positions.

We start with the 2x2x2-cube with the faces U and F. The object we make has an imaginary face I adjacent to U in addition to U and F. Turning I clockwise corresponds to U2FU2. The face I is not entirely adjacent to F, because F2UF2 = I' instead of I itself. Now what happens when we apply U2. It is a move, but not only that. It can also be seen as the symmetry that interchanges F and I. Similarly, F2 interchanges U' and I and I2 interchanges U' and F. So only quarter face turns may change the symmetry class, and one can verify that there are only three symmetry classes SPS.

Ok, now we know that tri-swap is impossible, but can all symmetries be arranged. Indeed, one can show that within the half face turns, the only missing symmetry is the one that inverts all faces, and that one is possible. With 12 symmetries, the class SU2S has size 12. The class SUS has size 36 and the class SUFS has size 72, which makes a total of 120 positions. Indeed, a single flip of two corners is impossible as well, so it is not surprising that only 1/6 of all corner positions can be made with two faces.

My macros are optimal in the q-turn metric. The swaps are chosen to preserve flip/twist. There may be shorter swap macros which also change the orientation. They are also chosen such that the involved cubies are visible as much as possible when looking at the UFR corner.

I have also done a fair bit of work with the two face group. You are correct that all position permutations of the edges are possible. However, edge flip is constrained, one cannot do a double edge flip with two faces. As for the corners, there are 6! = 720 position permutations of which only 120 may be solved. So in addition to tri-swaps certain double swaps are impossible. It's been a while, but I believe the cross-swap across one face is possible using two faces but the parallel swaps are not. I was never able to come up with a characterization of the permutation group such that I could inspect a permutation and tell whether it was a group member or not. The only way I could do it was to generate all 120 allowed permutations starting with the generators and make a list. If you have ideas relating to this I'd be interested in hearing them.

I already formulated a criterion: look if you can get U2, U or UF by applying symmetries on both sides. But when I computed all 12 cosets SPS, I discovered a more direct criterion.

Call two corner cubies vertically adjacent if either of the following holds on a clean cube:

1. they share one of the faces U,F and one of the faces R,L,
2. they are opposite, i.e. have distance sqrt(3).

Notice that 1. is for the faces U and F, and 2. is for the imaginary face I. Now for each of the three faces U,F,I, compute their adjacency score, by adding the following values for each pair of vertically adjacent corner cubies that is contained in the face at hand:

1. if the distance between the cubies is 1, then add 2 to the adjacency score,
2. if the distance between the cubies is sqrt(2), then add 1 to the adjacency score,
3. if the distance between the cubies is sqrt(3) (on face I), then add 2 to the adjacency score.

MIRACLE: the corpos position can be reached by way of U and F, if and only if all three adjacency scores are equal to 4.

Here are the sizes of the 12 cosets SPS ordered by their adjacency scores:

• 444a 12 (SU2S)
• 444b 36 (SUS)
• 444c 72 (SUFS)

Subtotal: 120

• 224 36
• 233 72
• 334a 72
• 334b 72
• 345 144
• 446 72
• 466 36
• 556 72
• 666 24

Total: 720

I am having a hard time following your arguments. I am unfamiliar with much of the terminology you are using.

Let's see if we can get on the same page here. We are dealing with the UF corner position permutations. We will lay out these cubies by projection on a plane:

UBL  URB

ULF  UFR

DFL DRF

and number the positions:

0     1
2     3
4     5

The two generators, U and F, may then be represented as the permutations, (2,0,3,1,4,5) and (0,1,4,2,5,3) respectively. Recursively forming products starting with these two generators forms a group of 120 permutations representing all the possible arrangements of these six cubies ignoring orientation. We will refer to this group as the UF permutation group. This group is a subgroup of the 720 element S6 group which is the group of all possible permutations of six objects.

The geometry of the two faces reveals four elements of symmetry; identity, a two-fold rotation, a plane of symmetry along the UF edge and a plane of symmetry perpendicular to the UF edge. These may be represented as the permutations:

(0,1,2,3,4,5) (5,4,3,2,1,0) (4,5,2,3,0,1) (1,0,3,2,5,4)

These four permutations form a group isomophic with the C2v symmetry point group (Schoenflies).

Now, by connecting position 0 to 5 and position 1 to 4 through a hyper dimension you form a third face and reveal a higher symmetry. Here I get lost. What is the nature of this higher order symmetry group? Your discussion suggests the three faces are symmetrically equivalent yet they have different distances for the length of the diagonal. How many elements does the symmetry group have? If I add U2 , F2, and I2 to the C2v group elements above, would that be sufficient to generate the full group?

When you refer to "applying symmetries on both sides" am I right in thinking you are talking about a conjugation: s' q s , where s is a symmetry element?

You use the term "cosets SPS". Cosets usually refers to cosets of a subgroup within a parent group. Within the S6 group there are five cosets of our UF permutation group each of a 120 elements. I thought you must be meaning symmetry equivalence classes but then your numbers don't make sense. If they are symmetry equivalence classes, where is the identity element. You ought to have one class containing only the identity element since s' [identity]s = identity for all s.

I am confused.

Indeed, you have the 4 symmetries you describe above. My symmetry set S contains 12 symmetries, however. One can describe S by taking one extra permutation so say that the symmetry set S is by definition the group generated by your 4 symmetries and the permutation U2 = (3,2,1,0,4,5). Now one can show that F2 and I2 are contained in S as well. I is equal to (1,5,2,3,0,4) by definition, and just as with U and F, the corner cubies of I are those that are moved when turning I, i.e. DRF, DFL, URB and UBL.

If two corner cubes in I have distant sqrt(3), then they become corner cubies of F after applying the symmetry U2, and will have distant 1 within F. So the distances 1 and sqrt(3) are similar in some sense, whence I count the same score increments (i.e. 2) for both distances. The distance sqrt(2) is different, but always preserved after applying the symmetry U2, which makes its score increment (i.e. 1) consistent.

With applying symmetries on both sides of q, I mean t q s, where t and s are arbitrary symmetries of S. With the coset S q S, I mean the set of all t q s. The identity element is 444a, which is just S U4 S = S and therefore contains 12 elements. By the way,

• U4 = (0,1,2,3,4,5)
• UFUFU = (5,4,3,2,1,0)
• U2F2U2 = I2 = (4,5,2,3,0,1)
• UFUF'U2F = (1,0,3,2,5,4)

so your 4 symmetries can be made. The other 8 are:

• U2 = (3,2,1,0,4,5)
• F2 = (0,1,5,4,3,2)
• U2F2 = (4,5,1,0,3,2)
• F2U2 = (3,2,5,4,0,1)
• U'FUFU = (5,4,0,1,2,3)
• UFUFU' = (2,3,4,5,1,0)
• F'UFUF' = (2,3,0,1,5,4)
• U'FUFU' = (1,0,4,5,2,3)

Ok, things are clearer now. I was puzzled by your root 3 distance and was trying to work in the diagonal of a cube. I see now that you get it from the plane projection diagram. Now the actual geometry is a figure in some abstract non-Euclidean space. The apparent root 3 distance is an artifact of the projection and the distance 0-5 is really 1, the same as the distance 0-2. The three faces are truly symmetry equivalent, face I is not different from face U or face F.

I am intrigued by the symmetry group closing off the third face generates and have done a shirt sleeve analysis in the table below. If the face were formed normally one would get a prism. The two fold rotation axis becomes three axes. This is carried through with our hyper prism, giving elements 3, 4 and 11. The same for the plane of symmetry along the edge, giving elements 1, 6 and 9. The square prism has a three fold axis and we have two elements of order 3 in our symmetry group, elements 7 and 8. Now the prism is not closed off normally but crosswise. This introduces a twist in the spacial relationships. Now, I would conceive of this not as twist in the figure like a Möbius strip, but rather as a twist in the space the object is embedded in. This to allow the three faces to remain "square and planer" and maintain the symmetry. This introduces an interesting feature to the symmetry group. We find two elements, elements 5 and 10, of order 6. The nature of these elements is revealed by looking at the way they cycle. Three applications of the operation gives element 2, which is the counterpart of the perpendicular plane of symmetry from the original C_2v group. In order for this to be, elements 5 and 10 must be mirrored elements. They seem to be something akin to six fold improper rotation elements. It's like the trigonal faces of the prism are not lined up, but rotated 60˚ wrt one another. Although I wouldn't stretch the analogy too far--these symmetries arise from a strange spacial relationship with properties all their own. The three fold elements are strange also. They rotate the vertexes around the diagonals of the faces rather than the edges.

Anyway, I've had fun trying to get my mind around the symmetry. I am still trying to understand your symmetry classes which puts the E element in with U2

0	1
2	3
4	5
Element 0: E (identity) Cycle: E 
Order: 1

0	1
5	4
3	2
Element 1: sigma1 Cycle: sigma1 E 
Order: 2

1	0
3	2
5	4
Element 2: sigma Cycle: sigma E
Order: 2

1	0
4	5
2	3
Element 3: C2a Cycle: C2a E 
Order: 2

2	3
0	1
5	4
Element 4: C2b Cycle: C2b E 
Order: 2

2	3
4	5
1	0
Element 5: S6 Cycle: S6 C3 sigma C3' S6' E 
Order: 6

3	2
1	0
4	5
Element 6: sigma2 Cycle: sigma2 E 
Order: 2

3	2
5	4
0	1
Element 7: C3' Cycle: C3' C3 E
Order: 3

4	5
1	0
3	2
Element 8: C3 Cycle: C3 C3' E 
Order: 3

4	5
2	3
0	1
Element 9: Sigma3 Cycle: Sigma3 E 
Order: 2

5	4
0	1
2	3
Element 10: S6' Cycle: S6' C3' sigma C3 S6 E
Order: 6

5	4
3	2
1	0
Element 11: C2c Cycle: C2c E 
Order: 2

The symmetry group is the dihedral group of the hexagon. Below, the hexagon is drawn along with the six cubie positions.

   4 *-----* 2
    /       \
   /         \
1 *           * 0
   \         /
    \       /
   3 *-----* 5

In order to determine the sets SqS, I did use Möbius strips. I drawed connections between vertically adjacent cubies on them. Vertically adjacent cubies with distance sqrt(3) on the cube give a line of length sqrt(5) on a regular strip, but the wrap-around of the Möbius strip makes the line length 1. Vertically adjacent cubies with distance sqrt(2) on the cube may give a line of length 2 on a regular strip, but the wrap-around of the Möbius strip makes the line length sqrt(2). With these drawings, I discovered the face score property of reachable positions with U and F.