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Rubik can be solved in 27f
Submitted by silviu on Sat, 04/01/2006 - 16:39.In this paper we give a proof that Rubiks cube can be solved in 27f.
The idea is to eliminate the 476 cosets at distance 12 in the group H=< U,D,L2,F2,R2,B2 >.
In this way we never have to consider in the 2 phase algorithm that a coset is at distance 12.
So we only solve cosets at distance 11. Together with my earlier result of 28 this gives a proof of 27.
The same idea was used by Bruce Norskog in his 38q proof.
However we do not really need to compute all 476 cosets. In fact we only need to compute 7 cosets of the group
T = Intersection ( < U,D,L2,F2,R2,B2 > , < F,B,L2,U2,R2,D2 > , < L,R,F2,B2,U2,D2 > )
The group H is not invariant under all symmetries. But the group T is invariant under all 48.
The idea is to eliminate the 476 cosets at distance 12 in the group H=< U,D,L2,F2,R2,B2 >.
In this way we never have to consider in the 2 phase algorithm that a coset is at distance 12.
So we only solve cosets at distance 11. Together with my earlier result of 28 this gives a proof of 27.
The same idea was used by Bruce Norskog in his 38q proof.
However we do not really need to compute all 476 cosets. In fact we only need to compute 7 cosets of the group
T = Intersection ( < U,D,L2,F2,R2,B2 > , < F,B,L2,U2,R2,D2 > , < L,R,F2,B2,U2,D2 > )
The group H is not invariant under all symmetries. But the group T is invariant under all 48.
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