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预先计算A *的结果

[英]Precalculate Result of A*

原文 2017-09-21 18:49:58 6 2 algorithm/ search/ a-star/ 8-puzzle

Currently learning about the A* search algorithm and using it to find the quickest solution to the N-Puzzle . 当前正在学习A*搜索算法，并使用它来找到N-Puzzle的最快解决方案。 For some random seed of the initial starting state, the puzzle may be unsolvable which would result in extremely long wait times until the algorithm has search the entire search-space and determined there is not solution to the give start state. 对于初始启动状态的某些随机种子，难题可能无法解决，这将导致非常长的等待时间，直到算法搜索了整个搜索空间并确定没有给定启动状态的解决方案。

I was wondering if there is a method of precalculating whether the A* algorithm will fail to avoid such a scenario. 我想知道是否有一种方法可以预先计算A*算法是否无法避免这种情况。 I've read a bit about how it is possible but can't find a direct answer as to a method in which to do it. 我已经阅读了一些有关如何实现的信息，但找不到有关执行该方法的直接答案。

Any guidance or options are appreciated. 任何指导或选择，不胜感激。

2 个解决方案

I think A* does not offer you a mechanism to know whether or not a problem is solvable. 我认为A *无法为您提供一种机制来了解问题是否可以解决。 Specifically for N-Puzzle , I think this could help you to check if it can be solved or not: 专门针对N-Puzzle ，我认为这可以帮助您检查它是否可以解决：

http://www.geeksforgeeks.org/check-instance-8-puzzle-solvable/ http://www.geeksforgeeks.org/check-instance-8-puzzle-solvable/

It seems that if you are in a state where you have an odd amount inversion, you know for sure the problem for that permutation is infeasible. 看来，如果您处于数量倒置的状态，则可以肯定知道该排列的问题是不可行的。

For the N-puzzle specifically, there are only two possible parities, so you just need to check which parity the current puzzle is. 特别是对于N-puzzle，只有两个可能的奇偶校验，因此您只需要检查当前谜题是哪个奇偶校验即可。

There is an in-depth explanation on how to do this on the math stackexchange 有关如何在数学stackexchange上执行此操作的深入说明

For general A* problems, no, there is no way to pre-compute if the graph is solvable. 对于一般的A *问题，不可以，如果图形可解，则无法进行预先计算。