# 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.

Let xH be one of the 476 cosets at distance 12. Then this coset can be written as a union of cosets of T. That is each xH can be written be partioned as {x1T,x2T,x3T,...}. Doing this operation for each of the 476 cosets gives us 2332400 cosets of the group T.

Now if we take one of the 2332400 cosets and conjugate it by an element m of M i.e. take xT and map it to m'xmT. If m'xmT is not in one of the 476 distance 12 cosets then we can remove it because positions from this coset can be conjugated and their conjugates can be solved with the 2-phase algorithm in max 27 moves.

I have done this operation for each of the 2332400 cosets and after conjugating each by the 48 symmetries only 40 are still at distance 12. If we reduce this 40 cosets by symmetry we are left with only 7 cosets which I solved optimally with another modification of Reids solver. As seen below all were solvable in less than 21 moves.

The first column represents the number of positions and the second the number of positions unique with respect to the subgroup of M that leaves the coset in question invariant.

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.

Let xH be one of the 476 cosets at distance 12. Then this coset can be written as a union of cosets of T. That is each xH can be written be partioned as {x1T,x2T,x3T,...}. Doing this operation for each of the 476 cosets gives us 2332400 cosets of the group T.

Now if we take one of the 2332400 cosets and conjugate it by an element m of M i.e. take xT and map it to m'xmT. If m'xmT is not in one of the 476 distance 12 cosets then we can remove it because positions from this coset can be conjugated and their conjugates can be solved with the 2-phase algorithm in max 27 moves.

I have done this operation for each of the 2332400 cosets and after conjugating each by the 48 symmetries only 40 are still at distance 12. If we reduce this 40 cosets by symmetry we are left with only 7 cosets which I solved optimally with another modification of Reids solver. As seen below all were solvable in less than 21 moves.

The first column represents the number of positions and the second the number of positions unique with respect to the subgroup of M that leaves the coset in question invariant.

BD LD FD RD BU LU FU RU LB RB LF RF FRU BUR BLU FUL FDR FLD BDL BRD 15f 720 30 16f 14508 615 17f 226036 9646 18f 2120898 90394 19f 1618793 69802 20f 357 49 DL BL DR FR UR FL UL BR BD BU FU FD UFR URB UBL ULF DRF DFL DLB DBR 15f 552 24 16f 13722 585 17f 241910 10171 18f 2504080 105523 19f 1221024 51925 20f 24 4 UR DF UL DB DL UB DR UF LF RF LB RB FDR FRU BDL BLU BRD FUL FLD BUR 14f 64 8 15f 1204 151 16f 22362 2836 17f 354750 45087 18f 2610544 331516 19f 992332 127268 20f 56 14 UL UF UR UB DL DF DR DB RF LF RB LB FUL RUF BUR LUB FLD LBD BRD RFD 14f 16 2 15f 512 64 16f 12452 1567 17f 265113 33378 18f 2611642 329002 19f 1091498 138842 20f 79 25 DR RF DL LB UL RB UR LF DF UF UB DB UFR URB UBL ULF DRF DFL DLB DBR 14f 64 8 15f 1040 132 16f 18338 2337 17f 312168 39681 18f 2549550 322317 19f 1100104 140093 20f 48 8 FU RF BU LB FD RB BD LF RD LU RU LD ULF UFR URB UBL DBR DRF DFL DLB 15f 624 80 16f 16050 2050 17f 279938 35657 18f 2406252 305501 19f 1278378 163575 20f 70 17 UB RD UF LD FD LU BD RU LB RB LF RF FRU BUR BLU FUL LBD RDB RFD LDF 15f 496 124 16f 10896 2791 17f 197241 49936 18f 2218817 560150 19f 1553826 395614 20f 36 25I want to thank Michael Reid who's solver has been of great help.