Introduction to Cryptography with Coding Theory, 3rd Edition

C.5 Examples for Chapter 6

Example 21

The ciphertext

$$22\text{,}09\text{,}00\text{,}12\text{,}03\text{,}01\text{,}10\text{,}03\text{,}04\text{,}08\text{,}01\text{,}17$$

was encrypted using a Hill cipher with matrix

$$\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 10\end{array}\right)\text{.}$$

Decrypt it.

SOLUTION

A matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is entered as $[a\text{,}b;c\text{,}d]$. Type $M\ast N$ to multiply matrices $M$ and $N$. Type $v\ast M$ to multiply a vector $v$ on the right by a matrix $M$.

First, we put the above matrix in the variable $M$.

>> M=[1 2 3; 4 5 6; 7 8 10]

M =

$$\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 10\hfill \end{array}$$

Next, we need to invert the matrix mod 26:

>> Minv=inv(M)

   Minv =

$$\begin{array}{ccc}\hfill -2/3\hfill & \hfill -4/3\hfill & \hfill 1\hfill \\ \hfill -2/3\hfill & \hfill 11/3\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 1\hfill \end{array}$$

Since we are working mod 26, we can’t stop with numbers like $2/3$. We need to get rid of the denominators and reduce mod 26. To do so, we multiply by 3 to extract the numerators of the fractions, then multiply by the inverse of 3 mod 26 to put the ...