It is also NP-hard to approximate the fewest slides within an additive constant, but there is a polynomial-time constant-factor approximation. 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.įor larger versions of the n puzzle, finding a solution is easy, but the problem of finding the shortest solution is NP-hard. 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. 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. 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. 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. Archer (1999) gave another proof, based on defining equivalence classes via a hamiltonian path. This is straightforward but a little messy to prove by induction on m and n starting with m= n=2. Johnson & Story (1879) also showed that the converse holds on boards of size m× n provided m and n are both at least 2: all even permutations are solvable. In particular, if the empty square is in the lower right corner then the puzzle is solvable if and only if the permutation of the remaining pieces is even. This is an invariant because each move changes both the parity of the permutation and the parity of the taxicab distance. 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 done by considering a function of the tile configuration that is invariant under any valid move, and then using this to partition the space of all possible labeled states into two equivalence classes of reachable and unreachable states. 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. they never overestimate the number of moves left, which ensures optimality for certain search algorithms such as A*. 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. The n puzzle is a classical problem for modelling algorithms involving heuristics. Similar names are used for different sized variants of the 15 puzzle, such as the 8 puzzle that has 8 tiles in a 3×3 frame. Named for the number of tiles in the frame, the 15 puzzle may also be called a 16 puzzle, alluding to its total tile capacity. The goal of the puzzle is to place the tiles in numerical order. Tiles in the same row or column of the open position can be moved by sliding them horizontally or vertically, respectively. The 15 puzzle (also called Gem Puzzle, Boss Puzzle, Game of Fifteen, Mystic Square and many others) is a sliding puzzle having 15 square tiles numbered 1–15 in a frame that is 4 tiles high and 4 tiles wide, leaving one unoccupied tile position. To solve the puzzle, the numbers must be rearranged into order
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