If $h_1(s)$ is a consistent heuristic and $h_2(s)$ is a admissible heuristic, is $\min(h_1(s),\ h_2(s))$ consistent?


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, $h_1(s) \leqslant c(s,p) + h_1(p)$ and $h_1(s) \leqslant c(s, goal) + h_1(goal)$ that satisfies the consistency constraint of $h_1$. Also, $h_2(s) \leqslant h^*(s)$ and $h_2(s) \leqslant h^*(p)$ that approve $h_2$ is admissible.

However, for any node $n$, if we define $h_3(n) = \min(h_1(n), h_2(n))$:

$$h_3(s) = \min(h_1(s), h_2(s)) = 17 \not\leqslant c(s,p) + \min(h_1(p), h_2(p)) = $$ $$c(s,p) + h_3(p) = 10 + 1 = 11. $$ It means that $h_3$ as a heuristic function is not consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.