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what is the minumum number of algorithms needed such that for any last layer case, you can apply 2 algorithms and solve the cube?
it's probably some really silly thing that would be silly to recognize but it seems like an interesting question.
what about 3 look?
I can't really think of a way of finding such a method so i'll let the puzzle theory people find it instead
it's probably some really silly thing that would be silly to recognize but it seems like an interesting question.
what about 3 look?
I can't really think of a way of finding such a method so i'll let the puzzle theory people find it instead