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I've been working on The Fifteen puzzle, which is to transform the slide puzzle in configuration $A$ below into configuration $B$, but I'm unsure how to proceed. Can someone provide a small hint as to what the right direction to investigate would be?

$$ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 15 & 14 \end{bmatrix} $$

$$ B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 \end{bmatrix} $$

What I tried so far

My guess is that this is impossible, and the idea is to find an invariant s/t the value for $A$ is different from that of $B$. Some of the things that I tried:

  • Coloring the board with alternating colors but didn't find anything useful.
  • Looked at sums mod 2 for each row and each column. We can get $A$ and $B$ to match if we consider swapping sums in this case.
  • All items in the matrix mod 2, but wasn't sure where to go further with this.

I haven't been able to use the fact that we can only swap with the position of the blank square. Can someone provide a hint (not a solution) of what the right direction to go in is? This other question may be similar, but it seems like it discusses the solution, so I've not looked at it.

crossvalidateme
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    It should be useful to check out this Numberphile video: https://youtu.be/YI1WqYKHi78 – Yuta May 07 '20 at 01:29
  • Wikipedia gives you an invariant that should work. – user780985 May 07 '20 at 01:31
  • First hint: call the blank square a number as well, say $16$. Now, consider each game state as some permutation of the $16$ numbers with the original starting state as the identity permutation. How might you describe going from one game state to another then in the language of permutations? – JMoravitz May 07 '20 at 01:42
  • Second hint: If the blank tile started at the bottom right, after some number of moves it ended at the bottom right, what can you say about the total number of moves made? – JMoravitz May 07 '20 at 01:46

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