Breaking The Rectangular Transposition Encryption System

We shall view Rectangular Transposition in a slightly different manner than we did originally. We assume that the sender and receiver choose a secret period p and a secret permutation of length p, say a=(a1,a2,...,ap) where each ai is taken from the set {1,2,...,p} and no two are the same. The plaintext is broken up into blocks of length p and then each block is transposed by the permutation a. For example, if the (i+1)st block of plaintext is Xip+1,Xip+2,...,Xip+p, then the (i+1)st block of cyphertext would be Xip+a1,Xip+a2,...,Xip+ap. This encryption system is vulnerable to a biletter-frequency analysis, since letters that are close together in the plaintext message, tend to stay close together in the cyphertext message.

To break Rectangular Transposition encryption, one guesses a period p and then, uses the given matrix of statistics to determine the validity of p according to the following cases

  1. All the entries in the matrix are approximately the same
  2. Each row except one has a maximum that is much larger than all the other entries in the row. Similarly, each column except one has a maximum that is much larger than all the other entries in the column.
In case 1, the opponent concludes that p was a wrong guess for the period and proceeds to repeat the calculations with another value of p. In case 2, the opponent guessed that p is the right value and from the matrix, reconstructs the permutation by clicking on the cell that contains the maximum value.

The Applet below is programmed to illustrate this codebreaking process.

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