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.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |