Single-byte XOR cipher

Challenge

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Script

from script02 import xor

if __name__ == '__main__':
    chal = '1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736'.decode('hex')

    for i in range(0x7f):
        print (str(i) +' : '+  xor(chal, len(chal) * chr(i)))

Click HERE to see script02.py

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FL4G

88 : Cooking MC's like a pound of bacon

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Advanced Method

Challenge says try to use a sort of 'scoring' method of alphabet frequency to find real plain text among the candidates.

I think that measuring e's frequency will not be working because the plain text is too short!

I want to apply 'Index of Coincidence' method, but not now.. 'cause it will be somewhat hard working!

COMING SOON..