Some 3-color cube results

The Rubik's Cube can be simplified by using only 3 colors instead of the usual six colors. Generally, opposite faces would share the same color, and that is the convention I assume here in talking about a 3-color cube.

Kunkle/Cooperman showed that a scrambled cube can always be brought to a position within the squares group within 16 moves. This puts an upper bound for God's number for the 3-color cube at 16. It is also well-known that the cube can be put into the <U,D,L2,R2,F2,B2> group in 12 moves. That puts a lower bound on God's number for the 3-color cube at 12. The superflip equivalent for the 3-color cube requires 14 moves according to an optimal 3-color cube solver program I have written. (From solving a million random positions, it appears that about 1.4% of positions of the 3-color cube require 14 moves to solve.) This raises the lower bound for the 3-color cube to 14.

I got the following distribution optimally solving a million randomly scrambled 3-color cubes.

moves    count
  7          1
  8         37
  9        357
 10      4,421
 11     49,818
 12    378,592
 13    552,604
 14     14,170

I note that the 3-color cube has 211*37*(12 choose 4)*(8 choose 4)2 = 10,863,756,288,000 configurations of corners and edges with respect to fixed centers. Because each center can not be distinguished from the opposite center, there are really only about 1/4 that many positions. I get the following counts for symmetry classes.

 size               count
   1                    2
   2                    8
   3                   14
   4                    6
   6                  222
   8                  352
  12                8,086
  16                8,566
  24            5,982,162
  48      226,325,259,954
total     226,331,259,372

With under a quarter trillion symmetry-reduced positions, a complete breadth-first search should be a doable task.

The centers of the 3-color 4x4x4 cube can be solved in 11 moves.

I have completed a breadth-first for solving the the center pieces of a 3-color 4x4x4 cube. The results are given below. As the 4x4x4 does not have any fixed centers to serve as a reference, positions are counted under the assumption that the orientation of the cube matters (24!/8!3 = 9,465,511,770 total positions).

distance      positions          distinct
    0                 6                 1
    1                18                 1
    2               432                 4
    3             7,722                45
    4           130,608               520
    5         2,041,152             7,404
    6        30,316,656           106,399
    7       365,166,768         1,271,430
    8     2,664,549,816         9,259,729
    9     5,689,455,696        19,762,729
   10       713,840,016         2,479,705
   11             2,880                12
          -------------        ----------
Total     9,465,511,770        32,887,979

I have also broken down the numbers by the sizes of the symmetry classes.

                                       class size
        6       12       18       24       36       48       72       96      144      288
dist 	-       --       --       --       --       --       --       --      ---      ---
 0      1
 1                        1
 2                                                            2                 2
 3                        1                 4                 7                17       16
 4                                          4        1       18        1       96      400
 5                        4                 6        2       38        6      547     6801
 6                                          8                82       14     2110   104185
 7                                 2        6       19      189       31     6609  1264574
 8               1                          7       13      150       87    15263  9244208
 9                                          2       12      119      222    14851 19747523
10                                                   6       10      158     1952  2477579
11                                                                              4        8
Total   1        1        6        2       37       53      615      519    41451 32845294
Pos     6       12      108       48     1332     2544    44280    49824  5968944  9459444672

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STM finished

STM took a little longer.

 0             1
 1             9
 2           126
 3          1851
 4         29501
 5        469428
 6       7470639
 7     118929033
 8    1878680252
 9   29058914384
10  416296633120
11 4001003671862
12 6315305304516
13  100086181328
14          1950

Great job, Tom!

Good work coming up with God's number (and distance distributions) for these tricolor cube metrics, Tom!

I note that I have verified that the 162 maneuvers you listed for the face turn metric do in fact generate 15f* positions according to my optimal solver.

"Positions" that differ only in the way the cube is oriented are generally not counted as distinct positions. With the tricolor cube, you can perform x2, y2, and z2 cube rotations without appearing to affect the centers. This means that each "actual" position has four different "looks" depending upon how the cube is oriented (without changing the 3 axes). So you have provided distance distributions in terms of the 10,863,756,288,000 configurations I mentioned, but you did not get a distribution in terms of the "actual" positions. (Well, I didn't either for my tricolor 4x4x4 centers analysis, so don't feel you have to.)

Also, I'm guessing you got a distribution in terms of the equivalence classes, but I don't see those figures.

You can get 192 "symmetries" similar to how you can get 1152 "symmetries" for the 2x2x2 cube. Just use h = m'*g*m*c where m is an element of M, and c is an element of { identity, x2, y2, z2 }. But as has been pointed out, you still only get at most 48 elements per equivalence class, and conjugating by the elements of M will suffice to find all elements of a symmetry class. The tricolor 4x4x4 centers puzzle has more "symmetries" because there are no pieces tied to a particular axis.


Thanks! I let my computer run long on them, rather than writing a new faster program. Specifically, I made relatively small changes to the coset solver I used for the full 3x3 cube and just let it run. Thus, I don't (easily) have classification into symmetry classes (just as for the 15f* result of the main cube, all I have is a full count, although *that* can be combined with Silviu Radu and Herbert Kociemba's solution of all symmetrical positions to break out the symmetry).

I don't get the 192 "symmetries", and I argue (as do others) that there are not such.

For any c as described above, m' g m c == c' m' g m c (since c' does not change the identity position; remember, we no longer have a group because many pieces are identical). Thus, there is m_2 = m c that's an element of M such that m_2' g m_2 == m' g m c.


My point is that we can generally talk about as many as 1152 "symmetries" for cubic puzzles. (I may very well be abusing the term symmetry here, hence I've used quotation marks.) These 1152 "symmetries" have been used to perform God's Algorithm calculations for such puzzles as the 2x2x2 cube and the edges-only (no centers) 3x3x3 cube, for example. So they are useful "symmetries." The 1152 symmetries allow us to classify "states" as equivalent if they differ by a rotation of the whole cube (and hence, intrinsically the same state), by a symmetrical equivalence (conjugating by elements of the symmmetry group of the cube), or a combination of both.

What you (Tom) have shown is that some of these symmetries are redundant with the usual 48 symmetries when dealing specifically with the tricolor 3x3x3 cube. And I never claimed in this thread that these "symmetries" weren't redundant with respect to the tricolor 3x3x3 cube. (5/6 of the 1152 "symmetries" always have a visible effect on the centers, so can quickly be deemed irrelevant. This leaves 192 potential symmetries, but as your argument shows, the extra 144 are redundant with the usual 48.)

In the centers-only 4x4x4 tricolor cube, however, we find that the usual 48 symmetries alone are not sufficient, at least if we want states (within the state space of the specified 9,465,511,770 states) that correspond to the same instrinsic state to be in the same equivalence class.

For example, consider the maneuver R' b' u F2 u' b u F2 u' R (with lower case representing inner layer turns) applied to a 4x4x4 supercube. This maneuver has the effect of performing a pure 3-cycle of center pieces. If we conjugate this by the elements of M, we find that this has 3-fold symmetry. Now rotate the whole cube 90 degrees in the same direction as a U move. We now conjugate by the elements of M and produce 48 different states (in the full state space). Without changing the intrinsic state of the cube (only applying a rotation to the whole cube), we have changed it from being 3-fold symmetrical to being asymmetrical! The same holds if we only assume the centers have stickers, and if we assume 3 colors for the centers as well. Clearly, we want a symmetry scheme that provides consistent amount of symmetry for states that correspond to the same intrinsic state of the cube. So simple M-conjugation will not work here.


Your use of the term "symmetry" for ( m' q m c) is entirely appropriate. For the G = < R , U , F , D , B , L > representation of the 2 x 2 x 2 cube:

m' G m c = G for all m ∈ M and c ∈ C

Thus the group is invariant under this transformation and it is a symmetry of the group. It may be productive rather to consider this symmetry as two different symmetries:

c G = G for all c ∈ C


m' G m = G for all m ∈ M

If one reduces the group by the rotation symmetries by picking the element with the Down-Left-Back cubie properly oriented in the Down-Left-Back cubical to represent the class the generators reduce to <R , U , F> and you end up with the fixed cubie representation of the puzzle. Thus by using a fixed cubie representation one may reduce the size of the group 24 fold from the get-go. Moreover, now there is a one to one correspondence between the distinct states of the cube and the representation. On the other hand when one proceeds to reduce the group by M conjugation one must apply a rotation to get the conjugates back in the group. So, perhaps it's six of one and half a dozen of the other.

1152 symmetries

Okay, I now understand that the 1152 symmetries are not for the tricolor 3x3x3 but instead some other puzzles, such as 2x2x2.

And of course you are correct.

Indeed, I will mention here that taking advantage of the 1152 symmetries of a subset of a puzzle (such as corners on a 3x3x3) permits you to get a relatively deep pruning table pretty cheaply.

Another similar trick is, for instance, in phase 2 of Kociemba's algorithm, which deals only with permutations, you can pretty much ignore the permutation of the middle edges (a factor of 12 reduction) without impacting the quality of the pruning table much.

For the centers-only 4x4x4, how many symmetry classes do you see?

Tri-color cube symmetry

Let me add my two cents to the discussion of the extra symmetry shown by the tri-color cube. The way I look at it there is no extra symmetry beyond the familiar 48 cubic group symmetries. Rather the D2 subgroup (the two fold rotations about the principal axes of the cube) of the cubic group is different in kind, giving by conjugation degenerate cube states rather than merely symmetry equivalent cube states.

Conjugations of a tri-color cube state by the D2 symmetries give states indistinguishable from the original (after a 180° whole cube rotation in most cases). Now the 48 members of the cubic symmetry group may be partitioned into cosets of the D2 symmetry group. Conjugation of a cube state by the elements of each of these cosets likewise give four states indistinguishable from each other. Thus a 48 member equivalence class is composed of only 12 distinct states since it will partition into 12 sets of 4 indistinguishable cubes. Smaller equivalence classes may reduce by a factor of 4, 3 or not at all depending on the symmetry shown by the elements of the class. Classes with elements showing D2 symmetry will not be degenerate, classes showing only one C2 symmetry will have three fold degeneracy and classes showing none of the D2 symmetries will have four fold degeneracy.

Tri-color cube symmetry

I finally got off my duff and did a symmetry analysis of the tri-color cube in line with my earlier comments:

Class Size Degeneracy    Class Count           Cosets       D2 Reduced Cosets
    48         4       226,325,259,954  10,863,612,477,792  2,715,903,119,448
    24         4             5,868,914         140,853,936         35,213,484
    24         2               113,248           2,717,952          1,358,976
    16         4                 8,566             137,056             34,264
    12         4                 1,466              17,592              4,398
    12         2                 6,620              79,440             39,720
     8         4                   352               2,816                704
     6         2                   184               1,104                552
     6         1                    38                 228                228
     4         4                     6                  24                  6
     3         1                    14                  42                 42
     2         1                     8                  16                 16
     1         1                     2                   2                  2
Total                  226,331,259,372  10,863,756,288,000  2,715,939,771,840

So the 43,252,003,274,489,856,000 elements of the six color cube partition into 10,863,756,288,000 cosets by the permutations of the indistinguishable cubies of the tri-color cube. Since the center facelets of opposite faces on the tricolor are indistinguishable this count includes positions which differ only by a 180° whole cube rotation. These duplicates may be eliminated by reduction by D2 symmetry to give 2,715,939,771,840 distinct positions. Reduction by the full 48 member cubic symmetry group gives 226,331,259,372 symmetry distinct positions.

Note that my previous post is in error when it states that a cubic group equivalence class showing only one C2 symmetry is three fold degenerate. Rather the degeneracy is two. Since C2x * C2y = C2z, if a element shows C2x symmetry then the C2y and C2z conjugates are exactly the same rather than needing a 180° whole cube rotation to be the same. Such exact duplicates are eliminated already in forming the equivalence class. Thus my supposed three fold degeneracy collapses into two fold degeneracy.

When the counts for classes of the same size but different D2 degeneracy are combined they agree with those found by Bruce Norskog above. Thus Bruce's counts are confirmed.

QTM finished

Here is the distance distribution:
 0             1
 1             6
 2            51
 3           468
 4          4278
 5         39660
 6        368421
 7       3414258
 8      31624673
 9     292626892
10    2701053152
11   24741079606
12  219528695052
13 1668009153892
14 6209687290956
15 2733247000902
16    5513935702
17            30
There turns out to be exactly three (unique mod M) antipodes:
24 D3F3R1B3L1F1U1L3B3F3U1B3R3B3F3F3U3
4 B3U1B3F1D1F3L3B1F1L3B3D1B1F3U1F3U3
3 F3R3D1U3F3F3L3R3B1L3R3D3R3R3F3U3U3

HTM finished

Here is the distance distribution:
 0             1
 1             6
 2            75
 3           888
 4         11082
 5        142230
 6       1828443
 7      23404284
 8     299243019
 9    3801549936
10   47276120420
11  540970449436
12 4110442428686
13 6007250783685
14  153690321787
15          4022
I also have a full set of antipodes, now. The counts by symmetry are as follows:
 1 22
 2 94
 3  8
 4 39
 6  7
 8  9
24  2
It's a bit surprising to me that there are more antipodes with 2-way symmetry than those with 1-way symmetry, considering how vastly the 1-way symmetrical positions outnumber the 2-way symmetrical positions.

Different metrics

In the QTM, there are probably only two antipodes, at a distance of 17:

24 D3F3R1B3L1F1U1L3B3F3U1B3R3B3F3F3U3
4 B3U1B3F1D1F3L3B1F1L3B3D1B1F3U1F3U3

(I have not yet run all positions so I don't say this with 100% confidence.)

It's interesting to note that the diameter in QTM is likely only two more than the diameter in HTM, unlike the full cube, where the diameter is almost certainly six more. But like the full cube, there are very few antipodes.

Note that the three-color cube is not bipartite in the QTM, as the six-color cube is.

In the slice turn metric, there are quite a few distance-14 likely antipodes (again, without a full exploration, I'm not 100% certain that 14 is the diameter). The more symmetric among them are:

24 I3B3R3I2R1B1I1F1R1J3D3R3F3U3
24 U3B1L1J3U3L1R2U1B3F2I3R3F3U3
8 L3B1D3U3L3J3L1U1K3F1D2R1F3U3
6 F3L3U3J3L1I2R3B2D1K1D3R3F3U3
4 K3U3R3J2U2B3K1J3L2B3R1D2F3U3

(My slice turn notation is unusual; I use I1 as a synonym for U1D3, J1 as a synonym for F1B3, and K1 as a synonym for R1L3; I do not yet have a program to convert this to the more common slice turn notation.)

Here's how I would transpose

Here's how I would transpose the turn sequences:

24 MU R' F' MU2 B L MU' B L MF' R' D B' L'
24 U B L MF R' U D2 R B' F2 MR F' U' R' 
8 L' B D' U' L MF U R MU' R F2 D R'
6 F' L' U' MF U MR2 U' F2 L MU' F' U' L' B'
4 MR B' R' MU' L2 D' MF MR U2 L' D F2 R' B'

And here's the routine I use for the transposition. The trick is to keep a running account of the rotation of the reference frame produced by the replacement of face turns with middle slice turns and transpose the turns accordingly. The code is in MacOS objective c which you would have to port. In my enumeration of the turns r is a clockwise turn and s is a counterclockwise turn.

-(NSData *)transposeToSingmasterPath: (NSData *)path
    NSMutableData    *result;
    const RM_Turn    *source;
    RM_Turn            op, op2,  *destination;
    unsigned            n, max, symTag1, symTag2;
    result = [NSMutableData dataWithLength: [path length]];
    source = [path bytes];
    destination = [result mutableBytes];
    max = [path length] / sizeof(RM_Turn);
    symTag1 = [cubicGroup identityTag];
    for( n = 0 ; n < max ; n++ )
        op = [self conjugateOfTurn: source[n] bySymTag: symTag1];
        switch (op) 
            case RLr:
                op2 = MRs;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4x3];    //90° cw rotation about the x axis
            case RLs:
                op2 = MRr;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4x];
            case RL2:
                op2 = MR2;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4x2];
            case UDr:
                op2 = MUs;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4y3];
            case UDs:
                op2 = MUr;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4y];
            case UD2:
                op2 = MU2;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4y2];
            case FBr:
                op2 = MFs;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4z3];
            case FBs:
                op2 = MFr;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4z];
            case FB2:
                op2 = MF2;
                symTag2 =  [cubicGroup tagForElementNamed: CS_C4z2];
                op2 = op;
                symTag2 = [cubicGroup identityTag];
        destination[n] = op2;
        // sym1 = sym1 * sym2  i.e. the right product with the last frame rotation
        symTag1 = [cubicGroup productOfOperatorTag: symTag1
                                          stateTag: symTag2 ];
    return result;

Coset space

The tricolor cube may be considered as a coset space of the standard six color cube. On the tricolor cube the cubies may be divided into five sets of four indistinguishable cubies. The permutations of these sets of indistinguishable cubies give ( 4! )5 / 2 = 3,981,312 standard cube states indistinguishable from identity on the tricolor cube. The distinct states of the tricolor cube may be mapped to the left cosets of this group of states. Thus the size of the tricolor coset space = 43,252,003,274,489,856,000 / 3,981,312 = 10,863,756,288,000. This agrees with the count you arrive at by different means above.

Now the D2 two fold symmetries are members of 3,981,312 element base set of identity states. So as I see it the four fold degeneracy you mention is already accounted for. Is there something I'm missing?

On further thought

I've been musing about this all afternoon and I believe I can answer my own question. The four fold degeneracy Bruce Norskog referred to arising from the center facelets of opposite faces being indistinguishable is not subsumed by the 3,981,312 fold degeneracy outlined above. If it were then d * q (where d is one of the three four spot permutations and q is an element of the standard cube group) would be in the same coset as q. This is not in general the case. However, in certain cases d * q is in the same coset. This is exemplified by the identity base set | E |. Since the four spot permutations are members of the base set, d * | E | = | E |. Thus the further reduction of the coset space is something less than 4.

What one can say is that the product d * q will be in same coset as the symmetry conjugate d * q * d since d is an element of |E|.

d * q * | E | = d * q * d * | E |

Thus the nearly four fold degeneracy Bruce mentioned is subsumed by the reduction by symmetry equivalence class performed by Tom Rokicki.

Three color cube: God's Number is 15 ?

I'm fairly certain God's Number for the three-color cube in the half-turn metric is 15. Here are a bunch of positions that I believe are at distance 15; would you verify them for me? I've listed the symmetry at the front. (I'm not sure I understand where the "extra" symmetry you mention above comes from; I'm using normal 48-way symmetry in my analysis.) I believe this is the majority of the distance-15 positions (but not all of them). I am fairly certain there are no distance-16 positions, but I have not proved this yet.
24 R3B3D1F3R2F3D3U2F3R3U1B3R2F3U3
24 R3U1F1D3U1R3B2F1R1D1B1F3L3F3U3
8 B3D2R1D2U3F3L3B1L3U3F2D1L1F3U3
8 B3U2L1D3U2F3R3B1R3D3F2U1R1F3U3
8 B3U2L1F3D2F3U2L3D1F3L3D3R3F3U3
8 F3D2U3F3R1B1D3R1D2R2B1D1U2F3U3
8 L3D3U1R1B3D1F2L3U3F2L1D2U2F1U3
8 R3B3U1B1F3D3F2L3B3F3L2F2R1F3U3
8 R3U2F3L2F2D1L3R1D2B1U1F2L3F3U3
8 R3U3F3D3F3R2B3L3F2U3L3F3D2F3U3
8 U3L3B3U1L2F3R3B3U3L3D1B2R1F3U3
6 L3D1L2D2L2F1R2B1F2R1U1B3R2F3U3
6 L3U1B2L2R3F2U2F3U1B3F2L3R1F3U3
6 R3B1F3D3B2U1F1U1L1D1B3R2U2F3U3
6 R3F1D2U1R3B2L2B2D3B3L2D3U1F3U3
6 U3F3U3B2L1U3R3B1U1F2U1B1R2F3U3
6 U3L2R2F1U1L1R3B1U1L1B3D3L2F3U3
4 B3R1U2F2R3D1R3F3U3L1D2F3L2F2U3
4 B3R2F3L2R1B2F1R1U3R2D3B3R2F3U3
4 B3R2U1B1F2L1R1F1D3L1U1B2R1F3U3
4 B3R3U1B3L2D1L2B1R3D3F2L1U3F3U3
4 D3B1D3L3R3B1R3U1B1D1L3F2R3F3U3
4 D3B3R1B2U1F3L3F2D3B2F1L2U3F3U3
4 D3B3R1U1R1F2L3D3F3L3D1U1R3F3U3
4 D3B3R2D1U2B1F3D1R1U3B1D2R1F3U3
4 D3F1U2L3U2B3L3D1B2U1R2D3U2F3U3
4 D3F3R2F3D3F2U3F1R1D3B3D3U3F3U3
4 D3L3F3D3R3U3B1U3L1B2U2L1U2F3U3
4 F3D1U3B1L2B1D1U2L3D1B2L1U3F3U3
4 F3D2R3D2L2U1B3L1D1B1L2F2R3F3U3
4 F3D3U3L3D1R2F1R1B1U1F3D2R1F2U3
4 F3L1D3L1R2U1F1L2D1R3B1D2U3F3U3
4 F3L2U1F1D2L3U3B2U2L3F2D1U2F3U3
4 F3L3D3R3F3L3U3R3U1L1D3B3R3F2U3
4 F3L3R2D3B2F3L2U1B1F2R1D2U3F3U3
4 F3R1B1F1L2D3L2U2R2B3L3F1U2F3U3
4 F3R2D1F2D2L2U1R1F3D2R2B3U2F3U3
4 F3U2L3D3U1F2L1F1R1F3D3F2D3F3U3
4 L3B2R3D1U1L3F1D3R2F3L3B3U1F3U3
4 L3B2U3F3U3L2U1F2D1U3R3D3U3F3U3
4 L3B3U1L1D3F2R1D3F3L3D3L3U3F3U3
4 L3F3D1B3U1F2U3L1U2R2F1L1U3F3U3
4 L3U2F1U3L1D2B3R3U1L2R2U1R2F3U3
4 R3B2U1L2U2F3L3B1R1D3L2B1U2R2U3
4 R3B3F1L3F2L3B3D3B3F1D1L3D2F3U3
4 R3D1L1B2D1B3F2L3U2R2B2F3R3F2U3
4 R3F1L1U1B1L2U2L3B1D3B1F2R1F2U3
4 R3F1L3D1U1R3F1L1D1B2L1R3B3F2U3
4 R3F2D3R2D1U2F1D2B3R3B1L1U3F1U3
4 R3U2B1R1D1B1D1B2F3L3R2B3F2R3U3
4 U3B1D3L3U3L1D1U2L2R3D3R3U2F3U3
4 U3F3D2L3U1B2L3D1U3B2L3B2U3F3U3
4 U3F3D3F3L3B2L3B1D3F1L1U3B3R3U3
4 U3L3B1R2U3F2L3B1L1B3D3L2R3F3U3
4 U3L3U3R3F1D1B1L3F2D3B1U3B2F3U3
4 U3R1F2L3F2D2B1U3F3D1R3B1U2F3U3
3 B3R1D1F3D3B1D3R1D2B3F2L3U1F3U3
3 B3R1F1D2L1U3R1D3F1L1U1F3U1F3U3
3 F3R1B3U3R3B1U1B3D3B3D3F2L1F3U3
3 L3F1D1R1B3R3F3U1B1D1F3U1L1F3U3
3 L3F3R2U1B3L2U3F1L2R1U2F3D1F1U3
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