(MeowChow) Re: CipherText

A message encrypted with this system can be broken easily in only 2 ke
+y periods
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That's not quite true. As you said, having only two key periods of message length means that when you XOR the two encrypted segments together, you are left with the XOR of the two plaintext segments. This is not usually enough information to reconstruct the original plaintext message, unless advanced statistical techniques enter the picture. To see why, consider that:

("case" ^ "hook") eq ("gone" ^ "lark")
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This is not an isolated case either, it's just the first I came up with after not more than a few minutes of randomly poking around with XOR.

What if you don't know the key length to start with? Some well-chosen +shifts and xors expose the redundancy in the N*(N-1) period, allowing + it to be attacked with frequency analysis as though it had a length +of only N+(N-1).
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I'm not sure what you mean by attacking as though length was N+(N-1). You can reduce a double XOR of lengths N and N-1 to solving for N+(N-1) if you have known plaintext of that length, but I'm not aware of any other attacks that reduce the problem in that way.

To find an unknown key length, however, is trivial. The standard technique for discovering an unknown key length is to try XORing the text against itself shifted at various distances, until you see a spike in the index of coincidence. The lowest common multiple of the spiked distances is then your key length.

I'm not a professional cryptographer either, but I like to play one on perlmonks.com ;-)

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

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Re: (MeowChow) Re: CipherText by no_slogan (Deacon) on Jul 24, 2001 at 00:31 UTC
This is not usually enough information to reconstruct the original plaintext message, unless advanced statistical techniques enter the picture. Fortunately, you don't have to reconstruct the entire plaintext message from the two chunks of xored text. About 2N characters, not necessarily contiguous, is enough to start attacking the two original xor keys. It's pretty easy (and kind of fun) to do that much by eye. Just write a little perl script to plug in some common words, and see if what comes out looks like English text. Sometimes you'll guess wrong, but once you get three or four words that form a sensible phrase, things start looking pretty certain. The standard technique for discovering an unknown key length is to try XORing the text against itself shifted at various distances, until you see a spike in the index of coincidence. The lowest common multiple of the spiked distances is then your key length. Your method is probably better than mine. Told you I was an amateur.	[reply]

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Re: (MeowChow) Re: CipherText
by no_slogan (Deacon) on Jul 24, 2001 at 00:31 UTC

This is not usually enough information to reconstruct the original plaintext message, unless advanced statistical techniques enter the picture.

The standard technique for discovering an unknown key length is to try XORing the text against itself shifted at various distances, until you see a spike in the index of coincidence. The lowest common multiple of the spiked distances is then your key length.

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