390 Codes, Ciphers, and Cryptography

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Presentation transcript:

390 Codes, Ciphers, and Cryptography Polygraphic Substitution Ciphers – Hill’s System

Hill’s System We now look at the system for enciphering blocks of text developed by Lester Hill. Matrices form the basis of this substitution cipher! We’ll work with blocks of size two letters – the idea can be generalized to larger blocks.

Steps to Encipher a Message 1. Choose a 2 x 2 matrix with entries in Z26 for a key. Make sure that (ad – bc)-1 (mod 26) exists, i.e. (ad – bc) = 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, or 25. This will guarantee that A-1 exists (mod 26).

Steps to Encipher a Message 2. Split the plaintext into pairs and assign numbers to each plaintext letter, with a = 1, b = 2, … , z = 26 = 0 (mod 26). Plaintext: p1p2|p3p4| … |pn-1pn If necessary, append an extra character to the plaintext to get an even number of plaintext characters.

Steps to Encipher a Message

Example 8 Use Hill’s scheme to encipher the message: “Meet me at the usual place at ten rather than eight o’clock.”

Example 8 Solution: For the key, choose a 2 x 2 matrix, with entries in Z26. Note that (ad – bc) (mod 26) = (9*7 – 4*5) (mod 26) = (63 – 20) (mod 26) = 43 (mod 26) = 17 (mod 26) and 17-1 (mod 26) exists! More on this later …

Example 8 Next convert the plaintext into pairs of numbers from Z26: me | et | me | at … cl | oc | kz. 13,5 | 5,20 | 13,5 | 1,20 | … 3,12 | 15,3 | 11,0 Now convert the plaintext to numbers to ciphertext numbers, using (*) above.

Example 8

Example 8 Thus, “me” is encrypted as “GV”. Try the next pair!

Example 8

Example 8 Thus, “et” is encrypted as “UI”. HW – Finish encrypting message! Note that for the word “meet”, the first “e” is encrypted as “G” and the second “e” is encrypted as “U”. Frequency analysis won’t work for this scheme!

Deciphering a Message To decipher a message encrypted with Hill’s Scheme, we can use the idea of matrix inverses! Since ciphertext (ck,ck+1) is obtained from plaintext (pk,pk+1) by multiplying key matrix A by plaintext (pk,pk+1), all we need to do is multiply matrix A-1 by ciphertext (ck,ck+1).

Deciphering a Message

Deciphering a Message

Deciphering a Message The same idea will work for matrices of numbers from Z26! Matrix A will be invertible, provided that (ad-bc)-1 (mod 26) exists! The only difference is that instead of 1/(ad-bc), we need to use (ad-bc)-1.

Deciphering a Message

Deciphering a Message

Deciphering a Message

Example 9 Decipher the ciphertext found above in Example 8! Write ciphertext as pairs of numbers in Z26: GV | UI 7,22 | 21,9 Use the inverse of the key matrix to decipher!

Example 9

Example 9 Thus, “GV” is deciphered as “me”. Repeat with “UI”.

Example 9

Example 9 Thus, “UI” is deciphered as “et”.

References Cryptological Mathematics by Robert Edward Lewand (section on matrices).