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dijkistra是否保证可允许的启发式

[英]does dijkistra guarantee an admissible heuristic

We know that dijkistra is a modified version of A* algorithm where in the estimated cost is set to zero.My question is the heuristic function used by dijkistra an admissible. 我们知道dijkistra是A *算法的修改版本,其中估计成本设置为零。我的问题是dijkistra使用的启发式函数是可以接受的。 Apologies if the question does not make sense.This was asked in an interview one week ago ,and I don't recall the exact wordings. 抱歉,这个问题没有道理。一个星期前在一次采访中问过这个问题,我不记得确切的措词。

In A* algorithm H-cost is a heuristic and it is admissible, otherwise we wouldn't be able to find the shortest path. 在A *算法中,H成本是一种启发式算法,可以接受,否则我们将无法找到最短路径。

Dijkstra uses heuristic which is constantly zero as H-cost is constantly zero. Dijkstra使用的启发式算法始终为零,因为H成本始终为零。 So it is also admissible as it doesn't overestimate the shortest path. 因此,它也可以接受,因为它不会高估最短路径。

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