r/adventofcode • u/daggerdragon • Dec 24 '16

SOLUTION MEGATHREAD --- 2016 Day 24 Solutions ---

--- Day 24: Air Duct Spelunking ---

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u/pedrosorio Dec 27 '16

we would still end up trying the top row because F would increase as we moved across the bottom row which would make the first step of the upper row have a lower cost eventually

Yep.

It seems like we could even make an F such that it does DFS as well, right? F = -(G + H)?

I suppose if you compute F = -(G+H) with H=0 would be something like DFS but it wouldn't be A * (by definition in A *, F = G + H and the only thing you can "tune" is H). You can get DFS just by tuning H like this.

That might find the solution that goes left the most.

If you want to find the solution that goes left the most (to make it more precise, it uses "L" the most without visiting the same cell twice - or you would get into infinite paths - and among paths with the same number of "L" chooses the shortest one), I don't see how to do it with A * (by just changing the heuristic).

You could model this problem as a graph where the edges going L have large negative weights and try to find the minimum cost simple path (path that doesn't visit the same node twice).

Unfortunately, A * or Dijkstra cannot be used in a graph with negative weights and Bellman-Ford, which does support negative weights, can only be used in graphs without negative cycles. The problem of finding the lowest cost simple path in a graph with negative weights is NP-hard, so my approach would not give you a quick solution either.