How do you solve the 15-puzzle with A-Star or Dijkstra's Algorithm?

Question

I\'ve read in one of my AI books that popular algorithms (A-Star, Dijkstra) for path-finding in simulation or games is also used to solve the well-known \"15-puzzle\".

C

Answer 1

The graph theoretic way to solve the problem is to imagine every configuration of the board as a vertex of the graph and then use a breath-first search with pruning based on something like the Manhatten Distance of the board to derive a shortest path from the starting configuration to the solution.

One problem with this approach is that for any n x n board where n > 3 the game space becomes so large that it is not clear how you can efficiently mark the visited vertices. In other words there is no obvious way to assess if the current configuration of the board is identical to one that has previously been discovered through traversing some other path. Another problem is that the graph size grows so quickly with n (it's approximately (n^2)!) that it is just not suitable for a brue-force attack as the number of paths becomes computationally infeasible to traverse.

This paper by Ian Parberry A Real-Time Algorithm for the (n^2 − 1) - Puzzle describes a simple greedy algorithm that iteritively arrives at a solution by completing the first row, then the first column, then the second row... It arrives at a solution almost immediately, however the solution is far from optimal; essentially it solves the problem the way a human would without leveraging any computational muscle.

This problem is closely related to that of solving the Rubik's cube. The graph of all game states it too large to solve by brue force, but there is a fairly simple 7 step method that can be used to solve any cube in about 1 ~ 2 minutes by a dextrous human. This path is of course non-optimal. By learning to recognise patterns that define sequences of moves the speed can be brought down to 17 seconds. However, this feat by Jiri is somewhat superhuman!

The method Parberry describes moves only one tile at a time; one imagines that the algorithm could be made better up by employing Jiri's dexterity and moving multiple tiles at one time. This would not, as Parberry proves, reduce the path length from n^3, but it would reduce the coefficient of the leading term.