# 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 $$|h(s)-h(g)| > d(s, g)$$.
• 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. Feb 28, 2020 at 16:46