I am playing around with Enigma Encryption, and especially solving it piece-wise, like in https://www.youtube.com/watch?v=RzWB5jL5RX0 shown. The idea is, that the text gets clearer, if you get more and more settings correct, but are still missing other settings.

For this, you need a fitness of text function, to decide the "englishness" of decrypted text. I found this one https://planetcalc.com/8038/#calculator8006 with single character frequencies, bigrams and trigams blended a quite nice approach.

Small numbers are "good", I guess. If I was to compare the corpus to itself, the formula in step 5 should (and will) deliver 0. Right?

However, the way I translated it into c# does not really work I think. All results come out as very small numbers, something like 1.0E-15 - even total gibberish before even attempting a decrypt.

Checking the formula in step 5, I wonder if this is missing abs() around the differences, or possibly squaring. Can you share your knowledge, if the formula in step 5 is missing operations?

  • $\begingroup$ Forgot to mention: Obviously, I just added abs() and tried it, and it looks quite reasonable now. Squaring did not look good. Am I missing something why without abs() could be correct? $\endgroup$
    – Ralf
    Apr 1 at 16:40


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