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What is the difference between the heuristic function and the evaluation function in A*? The evaluation function, often denoted as $f$, is the function that you use to choose which node to expand during one iteration of A* (i.e. decide which node to take from the frontier, determine the next possible actions and which next nodes those actions lead to, and ...


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You can easily find a counterexample. Suppose that there are three nodes $s$, $p$, and $goal$ such that $s \rightarrow p \rightarrow goal$. The real cost of going from $s$ to $p$ is $c(s,p) = 10$ and $c(p, goal) = 10$. Also, $h_1(s) = 18$, $h_1(p) = 9$, $h_1(goal) = 0$, $h_2(s) = 17$, $h_2(p) = 1$.On the other hand, $h^*(s) = 19$ and $h^*(p) = 10$. Now, $...


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The sketch of the proof for your first question: for an open node $n$, if $f_1(n) = g(n) + h_1(n)$, in the same situation in using $h_2$, it will be $f_2(n) = g(n) + 3 h_1(n)$. Hence, all the time for any node $n$, $f_2(n) \leqslant 3f_1(n)$. On the other hand, we know that A* with the admissible husritic function $h_1$ will be admissible (from Theorem 2 of ...


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I'm not familiar with your game so I can't tell you what a good heuristic woul be in your specific case, but I can give you some advice on how to look for a good heuristic function. As a rule of thumb, the heuristic function for a MiniMax algorithm is best kept simple and efficient, so you can get deeper into the tree. But it depends on how costly it is to ...


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