Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "390 Codes, Ciphers, and Cryptography Polygraphic Substitution Ciphers – Hill’s System."— Presentation transcript:

1 390 Codes, Ciphers, and Cryptography Polygraphic Substitution Ciphers – Hill’s System

2 2 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.

3 3 Steps to Encipher a Message 1. Choose a 2 x 2 matrix with entries in Z 26 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).

4 4 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: p 1 p 2 |p 3 p 4 | … |p n-1 p n If necessary, append an extra character to the plaintext to get an even number of plaintext characters.

5 5 Steps to Encipher a Message

6 6 Example 8 Use Hill’s scheme to encipher the message: “Meet me at the usual place at ten rather than eight o’clock.”

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

8 8 Example 8 Next convert the plaintext into pairs of numbers from Z 26 : 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.

9 9 Example 8

10 10 Example 8 Thus, “me” is encrypted as “GV”. Try the next pair!

11 11 Example 8

12 12 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!

13 13 Deciphering a Message To decipher a message encrypted with Hill’s Scheme, we can use the idea of matrix inverses! Since ciphertext (c k,c k+1 ) is obtained from plaintext (p k,p k+1 ) by multiplying key matrix A by plaintext (p k,p k+1 ), all we need to do is multiply matrix A -1 by ciphertext (c k,c k+1 ).

14 14 Deciphering a Message

15 15 Deciphering a Message

16 16 Deciphering a Message The same idea will work for matrices of numbers from Z 26 ! 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.

17 17 Deciphering a Message

18 18 Deciphering a Message

19 19 Deciphering a Message

20 20 Example 9 Decipher the ciphertext found above in Example 8! Write ciphertext as pairs of numbers in Z 26 : GV | UI 7,22 | 21,9 Use the inverse of the key matrix to decipher!

21 21 Example 9

22 22 Example 9 Thus, “GV” is deciphered as “me”. Repeat with “UI”.

23 23 Example 9

24 24 Example 9 Thus, “UI” is deciphered as “et”.

25 25 References Cryptological Mathematics by Robert Edward Lewand (section on matrices).


Download ppt "390 Codes, Ciphers, and Cryptography Polygraphic Substitution Ciphers – Hill’s System."

Similar presentations


Ads by Google