Polygraphic Substitution Cipher - Part 2

1. Prof. Neeraj Bhargava
Mrs. Shubha Chaturvedi
Department of Computer Science, School of Engineering & System
Sciences
MDS University Ajmer, Rajasthan

2.  It is also known as Multiple groups substitution.
 It involves replacing a one group of character in the
plaintext message with another groups of character in
cipher text.
 In a Polygraphic Substitution cipher, plaintext letters are
substituted in larger groups, instead of substituting
letters individually.
 Here the plaintext is divided into groups of adjacent
letters of the same fixed length and than each such
group is transformed into a different group of same
length letters in cipher text.
 n playfair cipher unlike traditional cipher we encrypt a
pair of alphabets(digraphs) instead of a single alphabet.

3. 1. Playfair cipher
2. Hill cipher

4.  Hill cipher is a polygraphic substitution cipher based on
linear algebra.
 Each letter is represented by a number modulo 26. Often
the simple scheme A = 0, B = 1, …, Z = 25 is used, but
this is not an essential feature of the cipher.
 To encrypt a message, each block of n letters (considered
as an n-component vector) is multiplied by an invertible n
× n matrix, against modulus 26.
 To decrypt the message, each block is multiplied by the
inverse of the matrix used for encryption.
 The matrix used for encryption is the cipher key, and it
should be chosen randomly from the set of invertible n ×
n matrices (modulo 26).

5.  The hill cipher formula can be expressed in
terms of columns ,vectors and matrics.
Encryption =c=(k*p) mod 26
Decryption=p=(k*c)mod 26

7.  We have to encrypt the message ‘ACT’ (n=3).The key is ‘GYBNQKURP’ which
can be written as the nxn matrix:

8.  To decrypt the message, we turn the ciphertext back into a vector, then
simply multiply by the inverse matrix of the key matrix (IFKVIVVMI in
letters).The inverse of the matrix used in the previous example is:

9.  Given Plaintext is = “DOG”
 Step1-Put DOG in a matrix from according to
numbers. D= 3
O= 14
G= 6
 Choose a random key according to the size of
plain text. Random key is 3*3
 3 6 24 1
14 13 16 10
6 20 17 5

10.  Step3-multiple the 2 matrices
3 6 24 1
14 13 16 10
6 20 17 5
 3*6 18 + 14*24 336 + 6*1 6 = 360/26 = 22
3*13 39 + 14*16 224 + 6*10 60 = 323/26 = 11
3*20 60 + 14*17 238 + 6*15 90 = 388/26 = 24

11. Now we get a cipher text which is =
22
11
24
 Step4- we will convert this cipher text into
the plain text:
22 = W
11 = L
24 = Y