Playfair Cipher Encryption Program In Python

In this post I'm going to explain you how to code playfair cipher to encrypt plaintext to cipher. So this post is all about encryption program and will soon add decryption program in this series also so stay tuned!

If you don't know how the algorithm works checkout previous post in this series.

So Without further ado, LET'S BEGIN!!👨‍💻👩‍💻

Can you recall what were the steps of algorithm?

Yes, You're right!
The Algorithm was consist of three steps:

1. Convert plaintext into digraphs (i.e., into pair of two letters)
2. Generate a Cipher Key Matrix
3. Encrypt plaintext using Cipher Key Matrix and get ciphertext.

 

Programming Playfair Cipher:

Step: 1 Creating function to convert plaintext into digraphs

	def convertPlainTextToDiagraphs (plainText):
	# append X if Two letters are being repeated
	for s in range(0,len(plainText)+1,2):
	if s<len(plainText)-1:
	if plainText[s]==plainText[s+1]:
	plainText=plainText[:s+1]+'X'+plainText[s+1:]
	# append X if the total letters are odd, to make plaintext even
	if len(plainText)%2 != 0:
	plainText = plainText[:]+'X'
	return plainText

view raw convertPlainTextToDiagraphs.py hosted with ❤ by GitHub

 

Step: 2 Function to generate a cipher key matrix

	def generateKeyMatrix (key):
	# Intially Create 5X5 matrix with all values as 0
	# [
	# [0, 0, 0, 0, 0],
	# [0, 0, 0, 0, 0],
	# [0, 0, 0, 0, 0],
	# [0, 0, 0, 0, 0],
	# [0, 0, 0, 0, 0]
	# ]
	matrix_5x5 = [[0 for i in range (5)] for j in range(5)]
	simpleKeyArr = []
	"""
	Generate SimpleKeyArray with key from user Input
	with following below condition:
	1. Character Should not be repeated again
	2. Replacing J as I (as per rule of playfair cipher)
	"""
	for c in key:
	if c not in simpleKeyArr:
	if c == 'J':
	simpleKeyArr.append('I')
	else:
	simpleKeyArr.append(c)
	"""
	Fill the remaining SimpleKeyArray with rest of unused letters from english alphabets
	"""
	is_I_exist = "I" in simpleKeyArr
	# A-Z's ASCII Value lies between 65 to 90 but as range's second parameter excludes that value we will use 65 to 91
	for i in range(65,91):
	if chr(i) not in simpleKeyArr:
	# I = 73
	# J = 74
	# We want I in simpleKeyArr not J
	if i==73 and not is_I_exist:
	simpleKeyArr.append("I")
	is_I_exist = True
	elif i==73 or i==74 and is_I_exist:
	pass
	else:
	simpleKeyArr.append(chr(i))
	"""
	Now map simpleKeyArr to matrix_5x5
	"""
	index = 0
	for i in range(0,5):
	for j in range(0,5):
	matrix_5x5[i][j] = simpleKeyArr[index]
	index+=1
	return matrix_5x5

view raw generateKeyMatrix.py hosted with ❤ by GitHub

 

Now we need to locate index of particular character inside cipher key matrix to perform encryption:

	def indexLocator (char,cipherKeyMatrix):
	indexOfChar = []
	# convert the character value from J to I
	if char=="J":
	char = "I"
	for i,j in enumerate(cipherKeyMatrix):
	# enumerate will return object like this:
	# [
	# (0, ['K', 'A', 'R', 'E', 'N']),
	# (1, ['D', 'B', 'C', 'F', 'G']),
	# (2, ['H', 'I', 'L', 'M', 'O']),
	# (3, ['P', 'Q', 'S', 'T', 'U']),
	# (4, ['V', 'W', 'X', 'Y', 'Z'])
	# ]
	# i,j will map to tupels of above array
	# j refers to inside matrix => ['K', 'A', 'R', 'E', 'N'],
	for k,l in enumerate(j):
	# again enumerate will return object that look like this in first iteration:
	# [(0,'K'),(1,'A'),(2,'R'),(3,'E'),(4,'N')]
	# k,l will map to tupels of above array
	if char == l:
	indexOfChar.append(i) #add 1st dimension of 5X5 matrix => i.e., indexOfChar = [i]
	indexOfChar.append(k) #add 2nd dimension of 5X5 matrix => i.e., indexOfChar = [i,k]
	return indexOfChar
	# Now with the help of indexOfChar = [i,k] we can pretty much locate every element,
	# inside our 5X5 matrix like this => cipherKeyMatrix[i][k]

view raw indexLocator.py hosted with ❤ by GitHub

 

Step: 3 Encrypt plaintext using cipher key matrix and get ciphertext

	def encryption (plainText,key):
	cipherText = []
	# 1. Generate Key Matrix
	keyMatrix = generateKeyMatrix(key)
	# 2. Encrypt According to Rules of playfair cipher
	i = 0
	while i < len(plainText):
	# 2.1 calculate two grouped characters indexes from keyMatrix
	n1 = indexLocator(plainText[i],keyMatrix)
	n2 = indexLocator(plainText[i+1],keyMatrix)
	# 2.2 if same column then look in below row so
	# format is [row,col]
	# now to see below row => increase the row in both item
	# (n1[0]+1,n1[1]) => (3+1,1) => (4,1)
	# (n2[0]+1,n2[1]) => (4+1,1) => (5,1)
	# but in our matrix we have 0 to 4 indexes only
	# so to make value bound under 0 to 4 we will do %5
	# i.e.,
	# (n1[0]+1 % 5,n1[1])
	# (n2[0]+1 % 5,n2[1])
	if n1[1] == n2[1]:
	i1 = (n1[0] + 1) % 5
	j1 = n1[1]
	i2 = (n2[0] + 1) % 5
	j2 = n2[1]
	cipherText.append(keyMatrix[i1][j1])
	cipherText.append(keyMatrix[i2][j2])
	cipherText.append(", ")
	# same row
	elif n1[0]==n2[0]:
	i1= n1[0]
	j1= (n1[1] + 1) % 5
	i2= n2[0]
	j2= (n2[1] + 1) % 5
	cipherText.append(keyMatrix[i1][j1])
	cipherText.append(keyMatrix[i2][j2])
	cipherText.append(", ")
	# if making rectangle then
	# [4,3] [1,2] => [4,2] [3,1]
	# exchange columns of both value
	else:
	i1 = n1[0]
	j1 = n1[1]
	i2 = n2[0]
	j2 = n2[1]
	cipherText.append(keyMatrix[i1][j2])
	cipherText.append(keyMatrix[i2][j1])
	cipherText.append(", ")
	i += 2
	return cipherText

view raw encryption.py hosted with ❤ by GitHub

 

Final Step! Running our code from main method:

	def main():
	#Getting user inputs Key (to make the 5x5 char matrix) and Plain Text (Message that is to be encripted)
	key = input("Enter key: ").replace(" ","").upper()
	plainText =input("Plain Text: ").replace(" ","").upper()
	convertedPlainText = convertPlainTextToDiagraphs(plainText)
	cipherText = " ".join(encryption(convertedPlainText,key))
	print(cipherText)
	if __name__ == "__main__":
	main()

view raw main.py hosted with ❤ by GitHub

 

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Thanks a lot for reading out!

I hope the code with comments was actually sufficient to understand the logic.

Here's the repository link to the whole code file GitHub Repo

But if you still have any doubts feel free to mention below in discussion section, I would love to answer them.

Also drop a heart ❤ if you gained some knowledge!

Comments

[Sophia Iroegbu]

This is genuis. I was watching NCIS Hawi'i and wanted to try this, glad I found this.

[Karan-Munjani]

Glad you find it useful..