/SocialPuzzle

A puzzle based on words and "Levenshtein friendship".

Primary LanguageRubyOtherNOASSERTION

== 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.