Introduction to Cryptography with Coding Theory, 3rd Edition

A.6 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 $\{\{a\text{,}b\}\text{,}\{c\text{,}d\}\}$. Type $M\text{.}N$ to multiply matrices $M$ and $N$. Type $v\text{.}M$ to multiply a vector $v$ on the right by a matrix $M$.

First, we need to invert the matrix mod 26:

In[1]:= Inverse[{{ 1,2,3},{ 4,5,6},{7,8,10}}]

Out[1]= {{-${\displaystyle \frac{2}{3}}$, -${\displaystyle \frac{4}{3}}$, 1}, {${\displaystyle \frac{2}{3}}$,${\displaystyle \frac{11}{3}}$, -2}, {1, -2, 1}}

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 “denominators” back in (see Section 3.3 ...