Example 3: A Transposition System

In this system, we will assume every line of the message is 63 characters long. The key is a permutation of the numbers from 1 to 63, and each line of the plaintext is rearranged using the permutation to produce the corresponding ciphertext. For example if the key is

$$1\ 11\ 21\dots61\ 8\ 18\dots54$$

(we would really want to use a more complicated permutation) and we use the same plaintext as in the previous two examples, we obtain:

TTRNRT UHOMO SFECE HYSGEH REDEN E NHS E A LE I I CTCE O SI

FN ET AHBCT DNDO AOBTRA TALOO TY IW CBEO K SEV H AS TOE HE

C HNO OWOA S UMOGR TIWC RNK BOU S STIT O NF EDTN

We are using the version of the plaintext including blanks. The second line of the plaintext has 55 characters, so we add 8 blanks on the end.

One method of decoding looks at a column of the ciphertext and asks what other column could immediately follow it. For example, it is possible that the column following OBO (the tenth ciphertext column) is UAO (the 8th), but the column TFC would yield the improbable two-letter combination BF.

As always, a longer message is easier to decode. Unlike simple substitution, it seems that blanks make the decoding process more difficult.

What about a known-plaintext attack? Since there is only one Y in the first line of the plaintext, we can tell that column 12 of the plaintext is column 21 of the ciphertext, but there are other things we can't tell. In this example, there are 8 columns of three blanks at the end of the plaintext, and we can't be sure which of these corresponds to which of the all-blank ciphertext columns. (it doesn't matter for this message, but we would like to know the entire key to deal with longer plaintexts in the future) A carefully chosen plaintext can give us the entire key at once.