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by chamomeal
359 days ago
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I would love a little more context on this, cause it sounds super interesting and I also have zero clue what you’re talking about. But translating a stateful program into a stateless one sounds like absolute magic that I would love to know about |
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The graph that is to be determined as a subset is a tree. From there he says it can be done in an algorithm that only traverses every node at most one time.
I’m assuming he’s also given a starting node in the original graph and the algorithm just traverses both graphs at the same time starting from the given start node in the original graph and the root in the tree to see if they match? Standard DFS or BFS works here.
I may be mistaken. Because I don’t see any other way to do it in one walk through unless you are given a starting node in the original graph but I could be mistaken.
To your other point, The algorithm inherently has to also be statefull. All traversal algorithms for graphs have to have long term state. Simply because if your at a node in a graph and it has like 40 paths to other places you can literally only go down one path at a time and you have to statefully remember that node has another 39 paths that you have to come back to later.