It can be shown that if two permutations $\sigma, \tau\in S_n$ have exactly one common non-fixed point, then their commutator $\sigma\tau\sigma^{-1}\tau^{-1}$ is a $3$-cycle.

This fact can supposedly be applied to solving twisty puzzles, since if two different moves "intersect" in exactly one piece (of a certain type), then their commutator will result in a $3$-cycle of that piece type. It is often the case that only even permutations of pieces are possible in a twisty puzzle, so one could then potentially correctly permute all the pieces using basic commutators.

My question is about applying the fact about permutations in $S_n$ to twisty puzzles. The usual definition of a permutation involves moving the same points, no matter where they are. So, the $3$-cycle $(123)$ will cycle those three points no matter where they're located.

However, in a twisty puzzle, moves consist not of permuting the same pieces no matter their position, but of permuting the same positions no matter what pieces are in them. For example, say I have a "twisty puzzle" consisting of three numbers in a line:


And the basic moves are to switch the two numbers on the left (call this move $L$) or to switch the two on the right (call this move $R$).

Consider the move sequence $L$, then $R$, then $L$, then $R$. Since permutations are composed from right to left, I guess I should write the product of these four moves as $RLRL$. The first $L$ switches $1$ and $2$, resulting in:


The first $R$ switches $1$ and $3$, resulting in:


The second $L$ switches $2$ and $3$:


The second $R$ switches $2$ and $1$:


The final result of all four moves is indeed a $3$-cycle, namely $(132)$. However, the four permutations, in order of application, were $(12), (13), (23), (12)$, so we have $(132)=(12)(23)(13)(12)$.

Although the two moves of this "twisty puzzle" intersect in just one position (the middle one), the commutator of $L$ and $R$ being a $3$-cycle doesn't directly follow from the result about permutations in $S_n$, since the $4$-move sequence wasn't directly a commutator.

Now, it turns out that $(132)$ is also equal to the commutator $(13)(12)(13)(12)$, because $(12)(23)=(13)(12)$. I'm trying to find a simple way to translate permutations of numbers to permutations of positions on a twisty puzzle (which could contain any piece). I think a similar thing will happen if $L$ and $R$ have orders greater than $2$, while still intersecting in a single point, but is there a way to connect the fact that the commutator of two (for simplicity, single-cycle) permutations intersecting in a single non-fixed point is a $3$-cycle to the fact that the commutator of two cycles of positions in a twisty puzzle intersecting in one position is also a $3$-cycle?

  • $\begingroup$ I put everything I know about this on my webpage with links to an online solver to demonstrate the algorithms. math.utoledo.edu/~niverso/algs.html $\endgroup$ – N8tron May 23 '18 at 23:03

You are mistaken about permutation notation. The numbers in the notation are the positions being exchanged, not the contents of those positions. So $L = (12)$ and $R = (23)$, and the permutation you've performed is (when written right-to-left): $$RLRL = (23)(12)(23)(12)$$ not $(12)(23)(13)(12)$.

Since $L^{-1} = L$ and $R^{-1} = R$, this is indeed a commutator.

  • $\begingroup$ In the cubing community they write in left-to-right notation $\endgroup$ – N8tron May 23 '18 at 23:04
  • $\begingroup$ @N8tron - I am used to seeing that too, at least when expressed by permutations. But in mathematics, we have a longstanding tradition of writing functions as $f(x)$, not $(x)f$, which leads us to use the right-to-left notation for applying operators to objects. $\endgroup$ – Paul Sinclair May 23 '18 at 23:11
  • $\begingroup$ The math community is not 100% set on right-to-left. Mostly people in abstract algebra, especially in group theory. $\endgroup$ – N8tron May 23 '18 at 23:24
  • 1
    $\begingroup$ @N8tron - since I am a part of the math community, and as I just said, I am used to seeing that too, I was well aware that different notations are possible. I was merely noting why the OP thought operators should always be applied right-to-left. I chose in my answer not to cover this so that I could address the real issue. $\endgroup$ – Paul Sinclair May 23 '18 at 23:38

To answer the supposedly part. I've used intersections of permutations that only move one cubie in common to solve many twisty puzzles. I've moved on to slightly more efficient algorithms. But here's the key to getting 3-cycles using the theorem you stated. Conjugation!

Face twists and slices move too many elements in common. For instance the right face R twist and upper twist U move 2 corners and one edge in common so their commutators tend to move quite a bit more. Here are a couple examples with the names people sometimes call them.

Here's an example on how you can use conjugation to find a move that only moves one thing in common with $L$: U'R'U https://alg.cubing.net/?alg=U-R-U. Here are a couple of useful 3-cycles you can make in this fashion:

I've put what I know on my webpage along with many other commutator and block building techniques here http://math.utoledo.edu/~niverso/algs.html


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