New answers tagged

1

There is an inherent assumption in heuristic search that the heuristic function points you in the right direction. A* largely depends on how good the heuristic function is. Two nice properties for the heuristic function are for it to be admissible and consistent. If the latter stands, I can't think of any case where BFS would outperform A*. However, this ...


0

I don't get why the (a).i. uses (𝑥,𝑦) coordinates whereas (b).i. uses boolean list. I guess they can be used interchangeablly right? No these cannot be used interchangeably. The situations really are different enough that completely different state representations are necessary. In the case of (a) you know the current bug's location, and nothing else ...


2

The only general situation that comes to my mind where BFS could be preferred over A* is when your graph is unweighted and the heuristic function is $h(n) = 0, \forall n \in V$. However, in that case, A* (which is equivalent to UCS) behaves like BFS (except for the goal test: see section 3.4.2 of this book), i.e. it will first expand nodes at level $l$, then ...


Top 50 recent answers are included