A* Pathfinding - Updating Parent

Question

I've read A* Pathfinding for Beginners and looked at several source code implementations in C++ and other languages. I understand most of what is happening, with the exception of one possible issue I think I have found, and none of the tutorials/implementations I've found cover this.

When you get to this part:

If an adjacent square is already on the open list [...], if the G cost of the new path is lower, change the parent of the adjacent square to the selected square. Finally, recalculate both the F and G scores of that square.

Changing the G score of a square should also change the G score of every child, right? That is, every square that already has this square as the parent, should get a new G score now also. So, shouldn't you find every child (and child of child) square in the open list and recalculate the G values? That will also change the F value, so if using a sorted list/queue, that also means a bunch of resorting.

Is this just not an actual problem, not worth the extra CPU for the extra calculations, and that is why the implementations I've seen just ignore this issue (do not update children)?

Answer 1

It depends on your heuristic.

For correctness, the basic A* algorithm requires that you have an admissible heuristic, that is, one that never overestimates the minimum cost of moving from a node to the goal. However, a search using an admissible heuristic may not always find the shortest path to intermediate nodes along the way. If that's the case with your heuristic, you might later find a shorter path to a node you've already visited and need to expand that node's children again. In this situation, you shouldn't use a closed list, as you need to be able to revisit nodes multiple times if you keep finding shorter routes.

However, if you use a consistent heuristic (meaning that the estimated cost of a node is never more than the estimated cost to one of its neighbors, plus the cost of moving from the node to that neighbor), you will only ever visit a node by the shortest path to it. That means that you can use a closed list and never revisit a node once you've expanded its children.

All consistent heuristics are admissible, but not all admissible heuristics are consistent. Most admissible heuristics are also consistent though, so you'll often seen descriptions and example code for A* that assumes the heuristic is consistent, even when it doesn't say so explicitly (or only mentions admissibility).

On the page you link to, the algorithm uses a closed list, so it requires a consistent heuristic to be guaranteed of finding an optimal path. However, the heuristic it uses (Manhattan distance) is not consistent (or admissible for that matter) given the way it handles diagonal moves. So while it might find the shortest path, it could also find some other path and incorrectly believe it is the shortest one. A more appropriate heuristic (Euclidean distance, for example) would be both admissible and consistent, and you'd be sure of not running into trouble.

Answer 2

@eselk : As the square, whose parent and G-score are to be updated, is still in OL, so this means that it has Not been expanded yet, and therefore there would be no child of the square in the OL. So updating G-scores of children and then their further children does not arise. Please let me know if this is not clear.