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

In reply to (MeowChow) Re: CipherText by MeowChow
in thread CipherText by NodeReaper

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