Scenario every possible path is also a word in the dictionary. This still leaves the problem NP-hard, because in the worst case Wordament graph We simply start with the first letter, and trie the The above gives enough infomation to enumerate all possible paths in our

If the path s ~~> t is valid, then the path s ~~> t ~> t' is valid iff.A path s ~~> t is valid iff the path occurs in the trie.We can now use our trie to navigate through our graph: A path to a lead node is always a valid word.Every node hold a letter, and a value indicating whether this path to.We load the entire dictionary into a a trie. Number of valid paths in our graph significantly. Luckilly we don't have to enumerate all the paths as we can restrict the Even counting the number of paths is a Sharp If every path would be valid (that is ever path is a word in theĭictionary), then the problem is reduced to finding every path betweenĮvery pair of vertices in the graph. Path between a starting vertex and an ending vertex. Where edges exist between neighboring nodes. The wordament problem can be understood as a graph where each vertex isĪ letter. The dictionary is now way too large (thanks.Solving this took 1.9 sec, on an MacBook Pro with Intel Core ghz Notice that the words are ordered by the corner letter onwards. Require 'wordament' game = Wordament:: Wordament.

Wordament and wordament snap attack