New answers tagged solvability
3
Let's assume there is a sequence $A$, such that for any starting configuration $x$ there exists a number $n(x)$, such that applying $A^{n(x)}$ will solve the cube.
It follows that given two configurations $x$ and $y$, we can move from $x$ to $y$ with $A^{n(X) - n(Y)}$. In other words, any sequence can be replaced by repeating A (or its reverse) for a ...
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