TP1 - Q3 - h not so admissible?

[loupwolf]

Hi, it seems to me that, in the TP1, the "h" function of the Q3 is not admissible because it overestimates the remaining distance from B to G2. The A* algorithm can't find the shortest path "('SBEG2', 13)". Instead it converges towards "('SACDG1', 14)".

Best regards,
Loup

[KenN7]

Hi :)
Yes indeed, h is not admissible in this case, the exercise does not make any assumptions on this.
In this case, A* does not find an optimal solution, congrats on finding the trick ;)
I agree this could be more clear in the exercise, either by adding a disclaimer on the possible "non-admissibility" of h, or by making h admissible by e.g. changing the B-1->E to a B-3->E