arenol wrote:I think I'm getting the idea.
The other day I solved the entire cube except three corners in 25 moves. Unfortunately, solving the remaining corners would take another 10 moves (need a conjugate and commutator to solve it). That makes it a 35 turn solution.
However, If I track the three corners throughout the solve, I might come across a situation where I can cycle them in 8 moves (without conjugation). Such an insertion alone is two turns saved (but I didn't find any in this case).
Furthermore, I might even find a place where I can cycle them in a way that the first moves of the insertion is the reverse of the last few moves up to that point, or the last moves is the reverse of the subsequent moves after the insertion. That can also save a number of turns, I guess. (I'll look into that case now...)
Or am I on the wrong track here?
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