Ideal Toy Company stated on the package of
the original Rubik cube that there were more than
three billion possible states the cube could attain.
It's analogous to Mac Donald's proudly announcing
that they've sold more than 120 hamburgers.
(J. A. Paulos, Innumeracy)

Analyzing Rubik's Cube with GAP.

For the example we consider the group of transformations of Rubik's magic cube. If we number the faces of this cube as follows

                     +--------------+

                     |  1    2    3 |

                     |  4  top    5 |

                     |  6    7    8 |

      +--------------+--------------+--------------+--------------+

      |  9   10   11 | 17   18   19 | 25   26   27 | 33   34   35 |

      | 12  left  13 | 20 front  21 | 28 right  29 | 36  rear  37 |

      | 14   15   16 | 22   23   24 | 30   31   32 | 38   39   40 |

      +--------------+--------------+--------------+--------------+

                     | 41   42   43 |

                     | 44 bottom 45 |

                     | 46   47   48 |

                     +--------------+

then the group is generated by the following generators, corresponding to the six faces of the cube (the two semicolons tell GAP not to print the result, which is identical to the input here).

  gap> cube := Group(

  >   ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19),

  >   ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35),

  >   (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11),

  >   (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24),

  >   (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27),

  >   (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40)

  > );;

First we want to know the size of this group.

  gap> Size( cube );

  43252003274489856000

Since this is a little bit unhandy, let us factorize this number.

  gap> Collected( Factors( last ) );

  [ [ 2, 27 ], [ 3, 14 ], [ 5, 3 ], [ 7, 2 ], [ 11, 1 ] ]

(The result tells us that the size is 2^27 3^14 5^3 7^2 11.)

Next let us investigate the operation of the group on the 48 points.


  gap> orbits := Orbits( cube, [1..48] );

  [ [ 1, 3, 17, 14, 8, 38, 9, 41, 19, 48, 22, 6, 30, 33, 43, 11, 46,

        40, 24, 27, 25, 35, 16, 32 ],

    [ 2, 5, 12, 7, 36, 10, 47, 4, 28, 45, 34, 13, 29, 44, 20, 42,

        26, 21, 37, 15, 31, 18, 23, 39 ] ]

The first orbit contains the points at the corners, the second those at the edges; clearly the group cannot move a point at a corner onto a point at an edge.

So to investigate the cube group we first investigate the operation on the corner points. Note that the constructed group that describes this operation will operate on the set [1..24], not on the original set [1,3,17,14,8,38,9,41,19,48,22,6,30,33,43,11,46,40,24,27,25,35,16,32].


  gap> cube1 := Action( cube, orbits[1] );

  <permutation group with 6 generators>

  gap> NrMovedPoints(cube1);

  24

  gap> Size( cube1 );

  88179840

Now this group obviously operates transitively, but let us test whether it is also primitive.

  gap> corners := Blocks( cube1, MovedPoints(cube1) );

  [ [ 1, 7, 22 ], [ 2, 14, 20 ], [ 3, 12, 16 ], [ 4, 17, 18 ],

    [ 5, 9, 21 ], [ 6, 10, 24 ], [ 8, 11, 23 ], [ 13, 15, 19 ] ]

Those eight blocks correspond to the eight corners of the cube; on the one hand the group permutes those and on the other hand it permutes the three points at each corner cyclically.

So the obvious thing to do is to investigate the operation of the group on the eight corners. The action gives a homomorphism to a permutation group on the corners:


  gap> blockhom1 := ActionHomomorphism( cube1,corners,OnSets);

  <action homomorphism>

  gap> cube1b := Image(blockhom1);

  Group([ (1,2,5,3), (1,3,7,4), (3,5,8,7), (2,6,8,5),

          (1,4,6,2), (4,7,8,6) ])

  gap> Size( cube1b );

  40320

Now a permutation group of degree 8 that has order 40320 must be the full symmetric group S(8) on eight points.

The next thing then is to investi the kernel of this operation on blocks, i.e., the subgroup of cube1 of those elements that fix the blocks setwise.


  gap> Factors( Size( Kernel( blockhom1 ) ) );

  [ 3, 3, 3, 3, 3, 3, 3 ]

  gap> IsElementaryAbelian( Kernel( blockhom1 ) );

  true

We can show that the product of this elementary abelian group 3^7 with the S(8) is semidirect by finding a complement, i.e., a subgroup that has trivial intersection with the kernel and that generates cube1 together with the kernel.

  gap> cmpl1:=Complementclasses(cube1,Kernel(blockhom1));

  [ Group([(1,3,5,2)(7,16,21,14)(9,20,22,12),(1,2,3,4,5,6,13)

           (7,14,16,17,21,10,15)(9,24,19,22,20,12,18), 

	   (1,2,3,4,5,8,13)(7,14,16,17,21,23,15)

	   (9,11,19,22,20,12,18)]) ]

  gap> cmpl1:=cmpl1[1];;

  gap> Size(cmpl1);

  40320

We verify the complement properties:

  gap> Size( Intersection( cmpl1, Kernel( blockhom1 ) ) );

  1

  gap> ClosureGroup( cmpl1, Kernel( blockhom1 ) ) = cube1;

  true

There is even a more elegant way to show that cmpl1 is a complement.

  gap> IsBijective(RestrictedMapping(blockhom1,cmpl1));

  true

Of course, theoretically it is clear that cmpl1 must indeed be a complement.

In fact we know that cube1 is a subgroup of index 3 in the wreath product of a cyclic 3 with S(8). This missing index 3 tells us that we do not have total freedom in turning the corners. The following tests show that whenever we turn one corner clockwise we must turn another corner counterclockwise.


  gap> (1,7,22) in cube1;

  false

  gap> (1,7,22)(2,20,14) in cube1;

  true

More or less the same things happen when we consider the operation of the cube group on the edges.

  gap> cube2 := Action( cube, orbits[2] );;

  gap> Size( cube2 );

  980995276800

  gap> edges := Blocks( cube2, MovedPoints(cube2) );

  [ [ 1, 11 ], [ 2, 17 ], [ 3, 19 ], [ 4, 22 ], [ 5, 13 ], [ 6, 8 ],

    [ 7, 24 ], [ 9, 18 ], [ 10, 21 ], [ 12, 15 ], [ 14, 20 ], [ 16, 23 ] ]

  gap> blockhom2 := ActionHomomorphism( cube2, edges, OnSets );;

  gap> cube2b := Image(blockhom2);;

  gap> Size( cube2b );

  479001600

  gap> Factors( Size( Kernel( blockhom2 ) ) );

  [ 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 ]

  gap> IsElementaryAbelian( Kernel( blockhom2 ) );

  true

  gap> cmpl2:=Complementclasses(cube2,Kernel(blockhom2));;

  gap> Length(cmpl2);

  4


So there are even 4 classes of complements here.



This time we get a semidirect  product  of a 2^11 with an S(12), namely a

subgroup of index 2 of the wreath product of a cyclic 2 with S(12).  Here

the missing index 2 tells us again that we  do not  have total freedom in

turning the edges.   The following tests show  that whenever  we flip one

edge we must also flip another edge.


  gap> (1,11) in cube2;

  false

  gap> (1,11)(2,17) in cube2;

  true

Since cube1 and cube2 are the groups describing the actions on the two orbits of cube, it is clear that cube is a subdirect product of those groups, i.e., a subgroup of the direct product. Comparing the sizes of cube1, cube2, and cube we see that cube must be a subgroup of index 2 in the direct product of those two groups.

  gap> Size( cube );

  43252003274489856000

  gap> Size( cube1 ) * Size( cube2 );

  86504006548979712000

This final missing index 2 tells us that we cannot operate on corners and edges totally independently. The following tests show that whenever we exchange a pair of corners we must also exchange a pair of edges (and vice versa).

  gap> (17,19)(11,8)(6,25) in cube;

  false

  gap> (7,28)(18,21) in cube;

  false

  gap> (17,19)(11,8)(6,25)(7,28)(18,21) in cube;

  true

Finally let us compute the centre of the cube group, i.e., the subgroup of those operations that can be performed either before or after any other operation with the same result.

  gap> Centre( cube );

  Group([(2,34)(4,10)(5,26)(7,18)(12,37)(13,20)(15,44)

         (21,28)(23,42)(29,36)(31,45)(39,47)])

We see that the centre contains one nontrivial element, namely the operation that flips all 12 edges simultaneously.

This concludes our example. Of course, GAP can do much more, but demonstrating them all would take too much room.

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Martin Schönert, 24 Apr 00