guzman
Regular
How many non isomorphic cube stickerings do you get
if you're allowed to sticker a 2x2x2 cube in any way you want ?
Just to be really clear:
Definition of isomorphic stickerings:
Two cubes A and B (both nxnxn) are said to be isomorphic
if there exists a bijection f
between the cubies of A and the cubies of B
such that, for any cubies a and a' of A we have:
1) the type of f(a) is the same type of a
(for instance, f maps corners to corners and edges to edges)
2) f(a) and f(a') are indistinguishable (that is they have the same stickerings) if and only if, a and a' are indistinguishable.
3) a and f(a) have the same number of orientations.
The question (in the 2x2x2 case) is not difficult to answer but I think some of you may enjoy solving the problem.
More tedious is the case of 3x3x3 non-isomorphic stickerings.
I didn't solve it precisely. My rough estimate is that there are more or less 38.000.000 non-isomorphic stickerings of the 3x3x3 cube.
[We could also think about modifying the definition in order to define when f is a stickering omomorphism and says therefore when A is a sort of subcube of B ... I dont know if this leads somewhere but it seems to me quite a natural topic ... ]
PS: Observe that with this defintion it is as if the cubes are considered disassembled; in other words the way the cubes are assembled doesn't play any role.
Edit: I substituted "cubes" with "cube stickrings" which is more clear
if you're allowed to sticker a 2x2x2 cube in any way you want ?
Just to be really clear:
Definition of isomorphic stickerings:
Two cubes A and B (both nxnxn) are said to be isomorphic
if there exists a bijection f
between the cubies of A and the cubies of B
such that, for any cubies a and a' of A we have:
1) the type of f(a) is the same type of a
(for instance, f maps corners to corners and edges to edges)
2) f(a) and f(a') are indistinguishable (that is they have the same stickerings) if and only if, a and a' are indistinguishable.
3) a and f(a) have the same number of orientations.
The question (in the 2x2x2 case) is not difficult to answer but I think some of you may enjoy solving the problem.
More tedious is the case of 3x3x3 non-isomorphic stickerings.
I didn't solve it precisely. My rough estimate is that there are more or less 38.000.000 non-isomorphic stickerings of the 3x3x3 cube.
[We could also think about modifying the definition in order to define when f is a stickering omomorphism and says therefore when A is a sort of subcube of B ... I dont know if this leads somewhere but it seems to me quite a natural topic ... ]
PS: Observe that with this defintion it is as if the cubes are considered disassembled; in other words the way the cubes are assembled doesn't play any role.
Edit: I substituted "cubes" with "cube stickrings" which is more clear
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