# Is the summation of consistent heuristic functions also consistent?

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.

• Would you please give an example and update your post? – Mostafa Ghadimi Feb 28 at 16:34
• @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? – Nathan S. Feb 28 at 16:37
• I didn't get what you mean by this example. – Mostafa Ghadimi Feb 28 at 16:38
• I will add a simple drawing of this example. – Nathan S. Feb 28 at 16:39
• 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. – Dennis Soemers Feb 28 at 16:46