Revision as of 12:10, 16 September 2008 by Zcurosh (Talk)

Application of Linearity

1. How Can Bob Decrypt the Message?

Alice makes sure the first thing that she does is tell Bob what secret matrix she is going to use. This is critical since this matrix is the basis of the encrypted message. Alice then replaces each letter by its corresponding order in the alphabet and puts a zero for a space. Using this integer vector, she applies the secret matrix three entries at a time which yields a new integer vector. She then sends this new vector to Bob. Bob is easily able to decode this message. In order to do this, Bob first has to find the inverse of the secret matrix. Then he applies the inverse secret message to the encrypted vector 3 integers at a time. This produces an integer vector in which each number corresponds to the order of that letter in the alphabet. Bob can find the message by replacing each number with the corresponding letter in the alphabet, and if he encounters a 0, he simply needs to insert a space.

2. Can Eve Decrypt the Message Without Finding the Inverse of the Secret Matrix?

It is possible for Eve to decrypt the message. Assume M=message vector, E=encrypted vector, and S=secret matrix. The vectors and matrix are related in this way:


$ M*S=E\! $

So Eve can then find the Secret Matrix using the equation,

$ M=E*S^{-1}\! $

Now that she knows the secret matrix, it is a simple matter of finding the inverse secret matrix and applying it to the encrypted vector to find the message. Using MATLAB, the secret matrix is,

$ \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood