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Re^3: Hangman Assistant

by blokhead (Monsignor)
on Jul 13, 2009 at 18:14 UTC ( #779698=note: print w/replies, xml ) Need Help??

in reply to Re^2: Hangman Assistant
in thread Hangman Assistant

That makes sense. I hadn't considered also taking into account the positions of the letters when you guess a correct letter. So in my example, when the word has a Q, and you guess U, you can still potentially get some useful information if many of the candidate words have U appearing in different places (i.e., you could distinguish QUEUE from QUEST). In this case, it would be "best" (if minimizing total # of guesses) to try to choose a letter whose absence / presence at all positions will partition the candidates into a large number of sets, each with size as small as possible.

Update: expanding with an example: Suppose the word to be guessed matches S T _ _ _. Then suppose we are considering E for our next guess. All of the candidate words will then fall into one of these 8 classifications:

S T _ _ _ (no E in the word) S T _ _ E (E in last position ONLY) S T _ E _ (etc..) S T _ E E S T E _ _ S T E _ E S T E E _ S T E E E
So we have 8 buckets, and we put all of the candidate words into the appropriate bucket. Suppose the bucket with most words has n words in it. Then in the worst case, after guessing E, we will have n remaining candidates. So you can take n to be the worst-case score of guessing E. Now compute this score for every letter, and take the letter with lowest score.

Note that there might be other ways to score each possible next-letter guess. Number of non-empty buckets comes to mind as an "average case" measure (to be maximized). Again, this is all assuming we're minimizing the total number of guesses. That way, all of the possible outcomes are (i.e., the guessed letter appears or doesn't appear in the word) are treated the same. To minimize the number of wrong guesses, you have to treat the "doesn't appear in the word" outcome differently and weight things in some better way.


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