Michael Tobis wrote:
> The first piece of code that I ever voluntarily wrote was intended to
> solve this puzzle:
>
> Assign the number 2 to 'a', 3 to 'b' ... 27 to 'z'. To each word assign
> the value of the product of its characters. Find the English (or
> language of your choice) word whose product is closest to a million (or
> number of your choice).
>
> I didn't have a lexicon handy. So I just printed out the letter
> combinations closest to a million and manually played Jumble with them.
> This was on a PDP-11 using Fortran, so you should have no problem doing
> this in Python.
>
> The prize was a VCR, which was a big deal in those days, but I didn't
> have a Green Card yet, so I didn't submit my answer, for (probably
> unfounded) fear the immigration authorities would give me a hard time
> for unauthorized income. My answer returned
> <hack hack type> 999856 for its product. Can you do better?
dioxid = 1000000
ixodid = 1000000
>
> You can go further. If you have a unix-like system you already have a
> lexicon, probably at /usr/share/dict/words .
Also check out www.puzzlers.org. There, and other places, you
can find word lists, such as the one I used above to find words
that numify to exactly 1000000.
>
> Compare various ways to solve this problem for performance, for
> arbitrary ciphers, which you can pass in as a dictionary like
> {'a':2,'b':4,'c':5 .... etc.
>
> To get you started here's the calculation, which may be interesting in
> itself. I don't have the Fortran, but I'm sure it was MUCH longer than
> this!
>
> #####
> cipher0= {}
>
> for i in range(ord('a'),ord('z')+1):
> cipher0[chr(i)] = i - ord('a') + 2
>
> from operator import mul
>
> def numify(word,cipher=cipher0):
> return reduce(mul,[cipher[c] for c in word])
>
> if __name__ == "__main__":
> print numify("hello")
>
> # prints 146016
>
> ####
>
> mt
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