matasano Cryptochallenge (Part three)

Becoming more interesting.

80 Kggcafo(EK/{(dacm(i(xg}fl(gn(jikgf
82 Ieeacdm*GI-y*fcao*k*zedn*el*hkied
86 Maaeg`i.CM)}.bgek.o.~a{`j.ah.loma`
87 L``dfah/BL(|/cfdj/n/`zak/`i/mnl`a
88 Cooking MC's like a pound of bacon
90 Ammikle"OA%q"nkig"c"rmwlf"md"`caml

is resulty of our search, and key is definitely 88, but how to get there?

Somewhat sad string, contains apostrofe, and cant be fished with alnum() method of a stringe, even right string is wrong anyway.

Long story shortened, we have only 256 possible keys here, since keysize is one byte, because key is same long as chunks to encode. in our case, we have some chunks, we want to iterate from 1 to 256, xor all our chunks and see if result makes sence.

Quite good critery for such size would be some letter counting. In this case there were few unnecessary solution, which are easily seen due to compactness of output.

Lets count ‘a’s and if there is at least one ‘a present, than we gather this as possible candidate.

Here is challenge, and here is solution.