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I'm reading from Cornell , I know for a single permutation, I can find the ciphertext length (100), divide in some key lengths (10x10, 5x20), then try all permutation (10x9x8... 5x4x3x2x1). then find anything legible.

Is there any case the above method will not work, and is there any better method to crack this permutation cipher?

Also, what if I use two-permutation patterns, for example; for the 100 plaintext length:

  • I divide it into 10 characters per column.
  • Then for the first 4 characters - I use one permutation.
  • For the next 6 characters, I use another permutation.

Am I right that this makes no difference on the cracking step? i.e. I tried 10x10, then I tried all permutations 10! I will find anything legible.

kelalaka
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pianobegginer
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1 Answers1

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While transposition ciphers may move bits around, they are entirely linear; and effectively perform substitution... except the data positions are substituted, instead of the values.

Let's say I have the ciphertext

raspberry

and I encrypt it to get

bsapryrre

Statistical analysis becomes even easier than with a substitution cipher: I simply look for a word or set of words with the same count of each letter. Once I have a set of possible plaintexts (there won't necessarily be only 1 match) I effectively have a reduced set of possible plaintexts, one of which forms a known plaintext.

If I have the next ciphertext block as

ip

I can perform the same attack on the next block, effectively creating a series of anagrams. In fact, this attack is so simple that it's already been implemented on every online anagram solver.

I might find multiple choices choices with meaning:

  • raspberry pi
  • raspberry ip

Since one makes more sense than the other, you know what your message is. This principal can be applied indefinitely to break transposition ciphers without even needing to break the key.

Serpent27
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