in reply to Re: Challenge: Mystery Word Puzzle
in thread Challenge: Mystery Word Puzzle

Where each of the variables (a,b,d,e,...,u) can be 0 or 1.

Therein lies the problem. What if the answer was "shoes" ... you now have 2 S'es ... that's why this cannot really be solved algebraically. Now, if the requirement was that you didn't take unique letters but instead followed the rules of Jotto, then simultaneous equations would work just fine.

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Re^3: Challenge: Mystery Word Puzzle
by Solo (Deacon) on Jan 13, 2005 at 05:10 UTC
    Per the OP:
    The number of letters in common is the exact number of unique letters in common. A 't' would only be counted once even if it existed twice in both the mystery word and the hint word

    Update:That is why the final equation is LESS THAN or equal to. The letters may appear more than once, reducing the total number of unique letters, but no more than the number of letters in the secret word can be used.

    --Solo

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    You said you wanted to be around when I made a mistake; well, this could be it, sweetheart.
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