# Rubik can be solved in 35q

Submitted by silviu on Wed, 03/22/2006 - 10:19.

Let H be the group < U,D,L2,F2,B2,R2 > and let N be the subgroup of H that contains
all even elements in H.
I have run an exhaustive search on the coset space G/N and got the following table:

0q 1 1q 9 2q 68 3q 624 4q 5544 5q 49992 6q 451898 7q 4034156 8q 35109780 9q 278265460 10q 1516294722 11q 2364757036 12q 235188806 13q 28144The group N contains no elements of odd length and the maximum length is 24. The position of length 24 is a local maxima. I have explained in my previous posts that when combining two solutions g an h and the phase 2 solution is a local maxima then the length of the total gh is less than or equal to L(g)+L(h)-2. Where L(g) means the length of g. This shows that two phase solutions based on the group N are maximum 35 quarter turns long.