Affine Cipher by Java

By Shiddhant Gupta

This project is designed for encrypting and decrypting the text using Java Programming Language.

This project is designed for encrypting and decrypting the text using Java Programming Language.

The Affine cipher is a type of monoalphabetic substitution cipher, wherein each letter is an alphabet is mapped to its numeric equivalent and encrypted by using a simple mathematical function and converted back to a letter. It uses two keys for encryption and decryption.

Formulas

1.) For Encryption

E(x) = ( a.pt + b) mod m

2.) For Decryption

D(x) = a^-1( ct - b) mod m

where

a and b denote the keys.

m denotes the size of the alphabets. By default, it is 26.

pt denotes plaintext.

ct denotes ciphertext.

a^-1 denotes the modular multiplicative inverse of a mod 26.

E and D denote Encryption and Decryption.

NOTE: a and m should be coprime means a and m has only one common factor i.e. 1  

How to find a^-1?

1 = a.a^-1 mod 26

For Example Encrypt CodersPacket using Affine Cipher by using 5 and 2 as keys.

Step 1:  Check for coprime:- 5 and 26 are coprime because 1 is the only common factor between them.

Step 2: Write the respective key values of the plain text and then apply the encryption formula and get the ciphertext.

Plaintext	Input Key-Value	Formula : ( ax + b) mod 26	Output Key-Value	Ciphertext
C	02	( 5*2 + 2 ) mod 26	12	M
O	14	( 5*14 + 2 ) mod 26	20	U
D	03	( 5*03 + 2 ) mod 26	17	R
E	04	( 5*04 + 2 ) mod 26	22	W
R	17	( 5*17 + 2 ) mod 26	09	J
S	18	( 5*18 + 2 ) mod 26	14	O
P	15	( 5*15 + 2 ) mod 26	25	Z
A	00	( 5*00 + 2 ) mod 26	02	C
C	02	( 5*02 + 2 ) mod 26	12	M
K	10	( 5*10 + 2 ) mod 26	00	A
E	04	( 5*04 + 2 ) mod 26	22	W
T	19	( 5*19 + 2 ) mod 26	19	T

So the encrypted message is: MURWJOZCMAWT

For Example Decrypt MURWJOZCMAWT using Affine Cipher by using 5 and 2 as keys.

Step 1: Check for coprime:- 5 and 26 are coprime because 1 is the only common factor between them.

Step 2: Find a^-1 : 1=a.a^-1 mod 26

                          1=5.a^-1 mod 26 => put a^-1 = 1 then 5 mod 26 == 1 => 5!=1

                          1=5.a^-1 mod 26 => put a^-1 = 2 then 10 mod 26 == 1 => 10!=1

                          1=5.a^-1 mod 26 => put a^-1 = 3 then 15 mod 26 == 1 => 15!=1

                          1=5.a^-1 mod 26 => put a^-1 = 4 then 20 mod 26 == 1 => 20!=1

                          1=5.a^-1 mod 26 => put a^-1 = 5 then 25 mod 26 == 1 => 25!=1

                          1=5.a^-1 mod 26 => put a^-1 = 21 then 105 mod 26 == 1 => 1==1

So a^-1 = 21

Step 2: Write the respective key values of the ciphertext and then apply the decryption formula and get the plaintext.

Ciphertext	Input Key-Value	Formula : a-1( x - b) mod 26	Output Key-Value	Plaintext
M	12	21( 12 - 2) mod 26	02	C
U	20	21( 20 - 2) mod 26	14	O
R	17	21( 17 - 2) mod 26	03	D
W	22	21( 22 - 2) mod 26	04	E
J	09	21( 09 - 2) mod 26	17	R
O	14	21( 14 - 2) mod 26	18	S
Z	25	21( 25 - 2) mod 26	15	P
C	02	21( 02 - 2) mod 26	00	A
M	12	21( 12 - 2) mod 26	02	C
A	00	21( 00 - 2) mod 26	10	K
W	22	21( 22 - 2) mod 26	04	E
T	19	21( 19 - 2) mod 26	19	T

So the decrypted message is: CODERSPACKET

Possible Outputs :