Imagine that we have a set of heuristic functions $\{h_i\}_{i=1}^N$, where each $h_i$ is both admissible and consistent (monotonic). Is $\sum_{i=1}^N h_i$ still consistent or not?

Is there any proof or counterexample to show the contradiction?


No, it will not necessary be consistent or admissible. Consider this example, where $s$ is the start, $g$ is the goal, and the distance between them is 1.

s --1-- g

Assume that $h_0$ and $h_1$ are perfect heuristics. Then $h_0(s) = 1$ and $h_1(s) = 1$. In this case the heuristic is inadmissible because $h_0(s)+h_1(s) = 2 > d(s, g)$. Similarly, as an undirected graph the heuristic will be inconsistent because $|h(s)-h(g)| > d(s, g)$.

If you'd like to understand the conditions for the sum of heuristics to be consistent and admissible, I would look at the work on additive PDB heuristics.

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  • $\begingroup$ Would you please give an example and update your post? $\endgroup$ – Mostafa Ghadimi Feb 28 at 16:34
  • $\begingroup$ @MostafaGhadimi In my answer I described an example with two states, the goal state and a second state distance one from the goal. Is there something particular you are looking for more than this? $\endgroup$ – Nathan S. Feb 28 at 16:37
  • $\begingroup$ I didn't get what you mean by this example. $\endgroup$ – Mostafa Ghadimi Feb 28 at 16:38
  • $\begingroup$ I will add a simple drawing of this example. $\endgroup$ – Nathan S. Feb 28 at 16:39
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    $\begingroup$ For a more extreme version of this answer, consider taking a single admissible, consistent heuristic, and then adding up an infinite number of copies of them. For any base heuristic value $> 0$, this sum is going to end up being $\infty$, which is generally not admissible. $\endgroup$ – Dennis Soemers Feb 28 at 16:46

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