Hill Ciphors Case

Introduction

The Hill Cipher was invented by Lester S. Hill, who was a member of the mathematics faculty of Hunter College in New York City, in a 1929 paper "Cryptography in an algebraic alphabet. It is a polygraphic cipher which is based on linear algebra. The general idea of a cypher in cryptology is to use an algorithm for encryption or decryption similar to a code. However a code uses a codebook, which is linked to a string of numbers and symbols, to form the meaning of a random codetext, while a ciphertext contains all the information of the plaintext, which is hidden with a key. This key has to be selected beforehand and is of crucial importance to decrypt a specific ciphertext. Hill's main achievement was the application of mathematics to design and analyses various cryptosystems, with the help of number theory. At his time Hill's system however was deemed to be overly complicated. To facilitate the process of encryption Hill therefore invented a cipher machine, which used gears and chains in order to quickly encrypt and decrypt messages using his system. This machine however did never really distribute well.

Method

The base of a hill cipher is a matrix. This matrix can have any size as long as it is a square (2x2, 3x3, 4x4….nxn). This matrix will later serve as the key for the decryption. The next step is to link the numbers to its symbols. For simplicity we can do this by assigning the number 0 to 26 to their corresponding letters in the alphabet. This means that A would be equivalent to 0, B equivalent to 1, and so forth. These numbers are then multiplied with the key in order to get the cipher text.

So for Example:

Let's take the following matrix as the key:

If our plain text consists of the letters DEF they would correspond to the numbers 345 ( In case of a long message this message has to be broken down into several blocks). This numbers are then arranged to a 3x1 matrix and multiplied with our key matrix in order to get the cyphertext:

In the case the plaintext does contain a whole number of blocks it can be "rigged" in order to have a whole number of blocks.

The next step is also the most difficult one: the decryption. In order to change our VOY back to DEF we need to find an inverse matrix with the modulo 26, that will then serve as decryption key.

Let K be the key matrix. Let d be the determinant of K. We wish to find K-1 (this is the matrix of x's above), such that K × K-1 = I (mod 26), where I is the identity matrix.

where d × d-1 = 1(mod 26), and adj(K) is the adjoint matrix of K.

d