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A puzzle based on words and "Levenshtein friendship".
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gnoireaux/SocialPuzzle
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== SocialPuzzle Two words are friends if they have a Levenshtein distance of 1. That is, you can add, remove, or substitute exactly one letter in word X to create word Y. A word’s social network consists of all of its friends, plus all of their friends, and all of their friends’ friends, and so on. Write a program that tells how big the social network for the word “causes” is, using the word list given here : http://github.com/causes/puzzles/raw/master/word_friends/word.list. Work In Progress! Working on the hypothesis of a small network, thus taking the long route of generating all strings at a distance of 1 from "causes". Checking if the generated strings are in the word list (using a hash for fast lookup). Then generating again for those words matched as friends. And counting all the while.
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A puzzle based on words and "Levenshtein friendship".
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