Delete comment from: Ken Shirriff's blog
In English text, it's pretty likely that there will be some spaces 3 characters apart, 4 characters apart, 5 characters apart, 6 characters apart, and so on. (For instance, every 5-letter word will be surrounded by spaces 6 characters apart.) Similarly, there are likely to be e's and t's spaced a variety of distances apart.
If the key is 6 characters long, each pair of spaces 6 characters apart will turn into a pair of something 6 characters apart. One pair might turn into a pair of Q's, another might turn into a pair of @'s. Likewise with pairs of any other plaintext character. Thus, you'll end up with as many 6-character-offset pairs as in the plaintext. Pairs at other distances, say, 5, will turn into different characters when the key is applied. The result is the ciphertext will have a bunch of cases with the same character 6 apart, and not very many for other distances. You'll get some matches at other distances just by chance, but this will be fairly small.
As bitshifter mentions, the statistics are important. English text tends to have a fair number of repeats. If the plaintext were random data, this technique wouldn't work.
Feb 13, 2009, 11:57:00 AM
Posted to Simple Cryptanalysis with Arc