Simple Vigenere cipher in Python

See: - Part 2/3 - Part 3/3

I am currently reading "The code book" by Simon Singh, and he just described how the Vigenere cipher works... I am not coding any Python lately, so I have decided to implement it (real quick), not using any algorithm but manually, as someone would have done 300 years ago, preparing a Vigenere square, and then looking up the values in the table.

The Python code is pretty simple:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 #!/usr/bin/env python # Simple Vigenere cipher implementation in Python import string mykey="WHITE" input_text="en un lugar de la mancha de cuyo nombre no quiero acordarme" ciphertext = [] matrix = [] encryption_tuple= [] row = 0 control = 0 # Alphabet used as reference source = string.ascii_uppercase # Creating the Vigenere Square. A 26x26 matrix. # In the example provided by the book, instead of using the regular alphabet as reference # we shift the items, so the column used as reference doesn't start in A, but in B for row in range(len(source)): matrix.append([ x for i,x in enumerate(source) if i > row ]) for i,x in enumerate(source): if i <= row: matrix[row].append(x) # Creating the tuple based on the letter and key. ie: # ('D', 'W'), ('I', 'H'), ('V', 'I'), ('E', 'T'), ('R', 'E'), ('T', 'W'), ... # In case special characters are not considered, this is cleaner: # import itertools # text=[ x for x in input_text.upper() if x in string.ascii_letters] # encryption_tuple = [(x,y) for x,y in zip(text, itertools.cycle(mykey))] for x,y in enumerate(input_text.upper()): control = 0 if control % len(mykey) == 0 else control if y in string.punctuation or y in string.whitespace: encryption_tuple.append((y,y)) else: encryption_tuple.append((y,mykey[control])) control += 1 # Each element y in the tuple is the key in the alphabet matrix for x,y in encryption_tuple: if source.find(x) == -1: ciphertext.append(x) else: ref_row = matrix.index(y) ciphertext.append(matrix[ref_row][source.index(x)]) # Print guide print("-> Reference:") print(" " + ' '.join([x for x in source])) # Printing Vigenere square print("-> Square:") for id,i in enumerate(matrix,1): print("{:02d} {}".format(id,' '.join(i))) # Print results print("-> Key: {0}".format(mykey)) print("-> Input text: {0}".format(input_text)) print("-> Output text: {0}".format(''.join(ciphertext)))

I stored the code in GITHUB.

In my case, as in the book, I have used "WHITE" as keyword, and the string to cipher is hardcoded in the script (input_text)

The main step is to map the input_text("en un lugar...") with the keyword("WHITE"), so we end up with something like this:

('E', 'W')
('N', 'H')
(' ', ' ')
('U', 'I')
('N', 'T')
(' ', ' ')
('L', 'E')
('U', 'W')
('G', 'H')
('A', 'I')
('R', 'T')
(' ', ' ')
('D', 'E')
[...]
• The keyword will be repeated over and over until input_text string finishes
• For the sake of clarity, I haven't stripped the white spaces or any other punctuation symbol from the input message, but it makes the cipher (even) weaker.

Those tuples are going to be used to look for the letter to replace the actual symbol. For example: In order to cipher "E", we will use the row 22, as is the one starting by "W", therefore, the letter "E" will be replaced by "A". Next letter is "N", and we will use row 7, so it will be replaced by "U", and so on...

Not very likely to replace opengpg, but it works:

\$ python vigenere_cipher.py
-> Reference:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
-> Square:
01 B C D E F G H I J K L M N O P Q R S T U V W X Y Z A
02 C D E F G H I J K L M N O P Q R S T U V W X Y Z A B
03 D E F G H I J K L M N O P Q R S T U V W X Y Z A B C
04 E F G H I J K L M N O P Q R S T U V W X Y Z A B C D
05 F G H I J K L M N O P Q R S T U V W X Y Z A B C D E
06 G H I J K L M N O P Q R S T U V W X Y Z A B C D E F
07 H I J K L M N O P Q R S T U V W X Y Z A B C D E F G
08 I J K L M N O P Q R S T U V W X Y Z A B C D E F G H
09 J K L M N O P Q R S T U V W X Y Z A B C D E F G H I
10 K L M N O P Q R S T U V W X Y Z A B C D E F G H I J
11 L M N O P Q R S T U V W X Y Z A B C D E F G H I J K
12 M N O P Q R S T U V W X Y Z A B C D E F G H I J K L
13 N O P Q R S T U V W X Y Z A B C D E F G H I J K L M
14 O P Q R S T U V W X Y Z A B C D E F G H I J K L M N
15 P Q R S T U V W X Y Z A B C D E F G H I J K L M N O
16 Q R S T U V W X Y Z A B C D E F G H I J K L M N O P
17 R S T U V W X Y Z A B C D E F G H I J K L M N O P Q
18 S T U V W X Y Z A B C D E F G H I J K L M N O P Q R
19 T U V W X Y Z A B C D E F G H I J K L M N O P Q R S
20 U V W X Y Z A B C D E F G H I J K L M N O P Q R S T
21 V W X Y Z A B C D E F G H I J K L M N O P Q R S T U
22 W X Y Z A B C D E F G H I J K L M N O P Q R S T U V
23 X Y Z A B C D E F G H I J K L M N O P Q R S T U V W
24 Y Z A B C D E F G H I J K L M N O P Q R S T U V W X
25 Z A B C D E F G H I J K L M N O P Q R S T U V W X Y
26 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

-> Key: WHITE
-> Input text: en un lugar de la mancha de cuyo nombre no quiero acordarme
-> Output text: AU CG PQNIK HA SI FEJJPT HA JCRS JVUUVA UW JYELZH EYVZWENTM

I'll try to find time to code the decryption part and also to try different ways of implementing it.

By the way, I totally recommend the book!

Later