# Tag Info

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 ...