15 puzzle

The 15 puzzle (also called Gem Puzzle, Boss Puzzle, Game of Fifteen, Mystic Square and more) is a sliding puzzle. It has 15 square tiles numbered 1 to 15 in a frame that is 4 tile positions high and 4 tile positions wide, with one unoccupied position. Tiles in the same row or column of the open position can be moved by sliding them horizontally or vertically, respectively. The goal of the puzzle is to place the tiles in numerical order (from left to right, top to bottom).

Named after the number of tiles in the frame, the 15 puzzle may also be called a "16 puzzle", alluding to its total tile capacity. Similar names are used for different sized variants of the 15 puzzle, such as the 8 puzzle, which has 8 tiles in a 3×3 frame.

The n puzzle is a classical problem for modeling algorithms involving heuristics. Commonly used heuristics for this problem include counting the number of misplaced tiles and finding the sum of the taxicab distances between each block and its position in the goal configuration. Note that both are admissible. That is, they never overestimate the number of moves left, which ensures optimality for certain search algorithms such as A*.

Johnson & Story (1879) used a parity argument to show that half of the starting positions for the n puzzle are impossible to resolve, no matter how many moves are made. This is done by considering a binary function of the tile configuration that is invariant under any valid move and then using this to partition the space of all possible labelled states into two mutually inaccessible equivalence classes of the same size. This means that half of all positions are unsolvable, although it says nothing about the remaining half.

The invariant is the parity of the permutation of all 16 squares plus the parity of the taxicab distance (number of rows plus number of columns) of the empty square from the lower right corner. This is an invariant because each move changes both the parity of the permutation and the parity of the taxicab distance. In particular, if the empty square is in the lower right corner, then the puzzle is solvable only if the permutation of the remaining pieces is even.

Johnson & Story (1879) also showed that on boards of size m × n, where m and n are both larger or equal to 2, all even permutations are solvable. It can be proven by induction on m and n, starting with m = n = 2. This means that there are exactly two equivalency classes of mutually accessible arrangements, and that the parity described is the only non-trivial invariant, although equivalent descriptions exist.

Archer (1999) gave another proof, based on defining equivalence classes via a Hamiltonian path.

Wilson (1974) studied the generalization of the 15 puzzle to arbitrary finite graphs, the original problem being the case of a 4×4 grid graph. The problem has some degenerate cases where the answer is either trivial or a simple combination of the answers to the same problem on some subgraphs. Namely, for paths and polygons, the puzzle has no freedom; if the graph is disconnected, only the connected component of the vertex with the "empty space" is relevant; and if there is an articulation vertex, the problem reduces to the same puzzle on each of the biconnected components of that vertex. Excluding these cases, Wilson showed that other than one exceptional graph on 7 vertices, it is possible to obtain all permutations unless the graph is bipartite, in which case exactly the even permutations can be obtained. The exceptional graph is a regular hexagon with one diagonal and a vertex at the center added; only ⁠1/6⁠ of its permutations can be attained, which gives an instance of the exotic embedding of S₅ into S₆.