c++,search,graph,tree,iterative-deepening

I'm almost certain that it's not the most efficient way to go about it, but I found a way that works. void Graph::IDS(int x, int required) { if(x == required) return; cout << "Iterated Deepening Search for " << required << ", starting from vertex " << x << "...

algorithm,time-complexity,depth-first-search,iterative-deepening

If the tree is not balanced and the answer lies nearer to the root than the deepest leaf, then the answer will be found by a depth-limit which is less than the maximum depth of the tree, whereas depth first search may have to search half of the tree to...

algorithm,breadth-first-search,iterative-deepening

Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b : https://mhesham.wordpress.com/tag/depth-first-search/ Iterative Deepening DFS worst case # of nodes: N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. +...

prolog,shortest-path,iterative-deepening

This is a classical case for iterative deepening. Just enter at the toplevel: ?- length(Path, N), path(0, 2, Path). The first answer will be of shortest length. And that's what you can do in Prolog very elegantly: You are starting to enumerate an infinite set, hoping that you will find...