Explained clear solution for Vigenere Cipher by veky - python coding challenges

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Explained solution in Clear category for Vigenere Cipher by veky

import string, itertools

def diff(first, second):
    for c1, c2 in zip(first, second):
        yield string.ascii_uppercase[ord(c1) - ord(c2)]

def decode_vigenere(old, old_enc, enc):
    key_haystack = ''.join(diff(old_enc, old))
    for key in itertools.accumulate(key_haystack):
        long_key = key * len(enc)
        if long_key.startswith(key_haystack):
            return ''.join(diff(enc, long_key))

Sept. 11, 2015

Comments:

brubru777 on Jan. 1, 2017, 7:59 a.m.
Very interesting solution. The diff function is very nice. But the way you use to find the key is not so good. <ul> <li>if the new message is shorter than the old one, you don't need to compute the long key</li> <li>if the new message is longer, multiplying the key by len(enc) seems a bit excessive. `ceil(len(enc) / len(key))` would be much better.</li> </ul>

veky on Jan. 2, 2017, 11:26 a.m.
As said in the title, the point was to explain the algorithm. This was just a technical detail, necessary because generators don't have rich enough API - otherwise I'd be satisfied with `islice(cycle(key), len(enc))`. :-) And ceil is surely an overkill... singular one-extra of `1 + len(enc)//len(key)` is cheaper than importing math, surely.

brubru777 on Jan. 9, 2017, 8:33 a.m.
Thanks for your reply. Sorry that my comment sounded bad (it does when I read it again). I just thought that this solution missed a little something in order to be perfect. It's still better than mine though. :) <blockquote> singular one-extra of 1 + len(enc)//len(key) is cheaper than importing math </blockquote> That sounds like a reasonable compromise.

veky on Jan. 9, 2017, 11:34 a.m.
Not all of my solutions here are perfect. If they were, what would I do anymore? :-D

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