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dabombguyman
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The result isn't a mono-alphabetic substitution because every 1 and 0 is given a random character from their list. If you don't know the list that each one and zero has then you'll have to find it, and I've shown how frequency analysis doesn't work, so that is not an option. Also I calculated the possible combinations correctly. In my encryptor each 1 and 0 has 26 possible characters, so there are 26! combinations for each, making the number of TOTAL combinations 26!^2. Put that in a calculator and see what you get.

My challenge still holds. If you think you can crack it so easily, solve the puzzle above.

BTW: Thanks for helping me remember to work out a few kinks.
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