accumulate creative solution for Vigenere Cipher by veky - python coding challenges

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accumulate solution in Creative category for Vigenere Cipher by veky

def decode_vigenere(a, b, c, I=__import__):
 m=lambda*u:''.join(I("string").ascii_uppercase[ord(s)-ord(t)]for s,t in zip(*u))
 for p in I("itertools").accumulate(m(b, a)):
  if (p * len(c)).startswith(m(b, a)): return m(c, p * len(c))

Sept. 19, 2014

Comments:

maurice.makaay on Nov. 5, 2014, 10:10 p.m.
Hey, itertools.accumulate()! Didn't know that one yet. New in Python 3.3+ I see. Thanks.

veky on Nov. 5, 2014, 10:40 p.m.
I'm always glad to teach people something. ;-)

vananabu on Sept. 11, 2015, 3:25 p.m.
take likes. i'll puzzle out into your solution later:)

veky on Sept. 11, 2015, 4:31 p.m.
Argh. Have we really come to that, that people give me thumbs on credit? :-) It's not so hard to understand this solution. <a href="http://www.checkio.org/mission/vigenere-cipher/publications/veky/python-3/explained/">Here</a> it is in longer form.

vananabu on Sept. 11, 2015, 7:59 p.m.
It sound like a credit:) but it's more for me. Usually, I think over the solutions first go. I seen that solution is good. My comment is a reminder for me)

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