Lesson 6

We now show how to realize the \(3\)-cycle \(a \to b \to c \to a\).

First, we move \(a\) to the top-left corner by shifting the pieces along the following cycle.

Next, if \(c\) is not in the bottom-left corner, we move it to the position directly below \(a\) by using the following cycle.

If, on the other hand, \(c\) is in the bottom-left corner, we use the following cycle to move it below \(a\).

Finally, using the following cycle, we move \(b\) to the position immediately to the right of \(a\).

As a result, we obtain a configuration of the following form.

We then move the empty cell to the right of \(c\) and perform the following four moves.

Finally, if we apply the reverse of the sequence of moves that brought \(a\), \(b\), \(c\), and the empty cell to the top-left corner, we obtain the required \(3\)-cycle \(a \to b \to c \to a\).