Cryptography Tutorial - The Homophonic Cipher

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Cryptography > Polyalphabetic Ciphers > Homophonic Cipher (1/1) (30 min.)

Objective:

1) Understand how to encrypt, decrypt and break the Homophonic Cipher.

The Homophonic Cipher

Another straightforward option to disguise letter frequencies is realized in the Homophonic Cipher: Coding symbols are assigned to each plain letter based on their relative occurrences. In example, we may assign the hundred 2-digit numbers 00, 01, ..., 99 to the 26 plain letters based on their relative frequencies as shown below.

Plain letters with rounded %-frequency	Assigned cipher numbers	Plain letters with rounded %-frequency	Assigned cipher numbers
e - 12	00,06,13,32,52,53,71,72,83,93,94	m - 3	33,51,80
t - 10	14,16,30,31,43,58,73,79,84	p - 2	12,50
o - 8	11,15,25,41,42,57,78,85	y - 2	49,68
i - 8	03,10,34,35,54,56,77,86	f - 2	24,45
a - 8	18,19,20,36,55,62,76,87	g - 2	01, 96
n - 7	02,37,38,59,61,69,70	w - 2	81,98
r - 6	09,26,39,60,75,88	b - 2	48,97
s - 6	17,28,63,74,89	v - 1	99
h - 5	04,08,27,64	k - 1	67
l - 4	21,40,65,82	x - 1	47
d - 3	05,29,66	j - 1	95
u - 3	07,22,91	q - 1	90
c - 3	23,44,92	z - 1	46

Exercise 1:
Verify that the sender encodes cryptography to 440968123085962655122749. c=44, r=09, ....

Exercise 2:
Like any other polyalphabetic cipher, same plain letters are not encoded to the same cipher letter in the homophone cipher. Nevertheless, can the recipient decode the cipher in a unique manner? Or, does he have to choose among several possible plain texts? Explain. The decryption yields a unique plain text since there is exactly one letter for each 2-digit number.

Exercise 3:
The sender and the recipient share the above table as a key. As a recipient, decode 792793 087851785004116993 923512649360 922069 9772 66932342663205 . the homophonic cipher can be decoded

Cryptoanalysis of the Homophonic Cipher:
How can the Homophonic Cipher be broken? Can it at all? The letter frequency are definitely disguised? So, what else is there to attack such a cipher? Although the letter frequencies of individual letters were successfully disguised those of 2- or 3-letter combinations are not. Knowing that the most common 2-and 3-letter combinations within words are "th", "in", "he", "er" and "the", "ing", "and" an eavesdropper may find that the symbol 64 is often preceded by 31 and frequently followed by 93. Thus, 64 may well be letter "h", 31 be "t" and 93 "e". Using these three correspondences, the eavesdropper continues to infer on other correspondences. Although this is not a complete cryptanalysis it shows how an eavesdropper may gain information piece by piece to eventually break the whole cipher. This approach certainly requires the eavesdropper to intercept a relative large cipher text. In addition to using computer assistance to find common di- and trigrams, he may use traditional cryptographic virtues such as gut instinct, logic and detective work to break secret messages.

Conclusion
Both Polyalphabetic Ciphers offer a better privacy protection than Monoalphabetic Substitution Ciphers do. However, don't rely on them. You may want to use it to quickly disguise the content of a private message from somebody glancing at it. In fact, ...use the Vigenere Cipher. For better privacy, you could use the perfectly secure One Time Pad (remember that the one-time usage of the key makes it not very convenient to use) or even better, you know get ready to study the commonly used RSA cipher.

Related web sources:

Yahoo's Encryption & Security

Introduction to Cryptography

Enigma and the Codebreakers

Enigma History

Enigma Emulator

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