I have been studying the $8$-puzzle and have thus far managed to wrap my head around the following information:
The following illustrates the solved position of the $8$-puzzle, where $9$ is the empty tile:
$$ \begin{matrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{matrix} $$
Suppose that I have the following valid position of the $8$-puzzle:
$$ \begin{matrix} a & b & c\\ d & e & f\\ g & h & i \end{matrix} $$
If I enumerate its tiles, beginning at $1$, from left to right and from top to bottom, then applying the following permutation to it solves it:
$$ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ a & b & c & d & e & f & g & h & i \end{pmatrix} $$
Is decomposing this permutation into a sequence of valid $2$-cycles what I now have left to do? I deem a $2$-cycle $(p\;q)$ as valid if either $p$ or $q$ is the empty tile, and if the other tile is either vertically or horizontally adjacent to this empty tile.
Have I understood and laid this problem correctly? Thanks in advance.