DTTF/NB479: Dszquphsbqiz Day 8 Announcements: Please use pencil on - PowerPoint PPT Presentation

DTTF/NB479: Dszquphsbqiz Day 8 Announcements:  Please use pencil on quizzes if possible Questions? Today:  Congruences  Chinese Remainder Theorem  Modular Exponents

Hill Cipher implementation Encryption  Easy to do in MATLAB.  Or find/write a matrix library for language X. Decryption  Uses matrix inverse.  How do we determine if a matrix is invertible mod 26?

How to break via known plaintext? Good work on last session’s quiz. Idea: Assume you know the matrix size, n. Then grab n sets of n plaintext chars  ciphertext This gives n 2 equations and n 2 unknowns. Then solve using basic linear algebra, but mod n. Caveat: sometimes it doesn’t give a unique solution, so you need to choose a different set of plaintext. Hmm. This could make a nice exam problem…

Substitution ciphers Each letter in the alphabet is always replaced by another one. Which ciphers have we seen are substitution ciphers?  Which aren’t and why?  Breaking ciphertext only uses linguistic structure. Frequencies of: Single letters  Digrams (2-letter combinations)  Trigrams  Where do T&W get their rules like “80% of letters preceding n are  vowels”? (p. 26) See http://keithbriggs.info/documents/english_latin.pdf Lots of trial and error when done by hand. Could automate with a dictionary.

Fairy Tales HTTP :// XKCD . COM /872/ Goldilocks’ discovery of Newton’s method of approximation required surprisingly few changes.

Basics 4: Congruence Def: a≡b (mod n) iff (a-b) = nk for some int k Properties ∈ ≠ ≡ ≡ , , , , 0 , (mod ), Consider a b c d Z n If a b c d n then ≡ ∃ ∈ = + + ≡ + (mod ) . . ( ) ( )(mod ) a b n if k Z s t a b nk a c b d n ≡ − ≡ − 0 (mod ) | ( ) ( )(mod ) a n iff n a a c b d n ≡ ≡ (mod ) (mod ) a a n ac bd n ≡ ≡ = ≡ (mod ) (mod ) gcd( , ) 1 (mod ), a b n iff b a n If a n and ab ac n then ≡ ≡ ⇒ ≡ ≡ , (mod ) (mod ) (mod ) a b b c n a c n b c n You can easily solve congruences ax≡b (mod n) if gcd(a,n) = 1.  For small numbers, do by hand  For larger numbers, compute a -1 using Euclid

1-2 Solving ax≡b (mod n) when gcd(a,n )≠1 Let gcd(a,n)=d  Example: 2x ≡ 7(mod 10 ) If d doesn’t divide b then no solution Else divide everything by d Example: and solve 3x ≡ 3 (mod 6) (a/d)x=(b/d)(mod (n/d)) Get solution x 0 Multiple solutions: x 0 , x 0 +n/d,x 0 +2n/d,…x 0 +(d-1)n/d Always write solution with the original modulus This is an easy program to code once you have Euclid…

How could we write x ≡ 16 (mod 35) as a system of congruences with smaller moduli?

Chinese Remainder Theorem Equivalence between a single congruence mod a composite number and a system of congruences mod its factors Two-factor form  Given gcd(m,n)=1. For integers a and b, there exists exactly 1 solution (mod mn) to the system: ≡ (mod ) x a m ≡ (mod ) x b n

3-4 CRT Equivalences let us use systems of congruences to solve problems ≡ 3 (mod 7 ) x Solve the system: ≡ 5 (mod 15 ) x How many solutions? 2 ≡ 1 (mod 35 ) x  Find them.

Chinese Remainder Theorem n-factor form  Let m 1 , m 2 ,… m k be integers such that gcd(m i , m j )=1 when i ≠ j. For integers a 1 , … a k , there exists exactly 1 solution (mod m 1 m 2 …m k ) to the system: ≡ (mod ) x a m 1 1 ≡ (mod ) x a m 2 2 ... ≡ (mod ) x a m k k

5-6 Modular Exponentiation Compute last digit of 3^2000 Compute 3^2000 (mod 19) Idea:  Get the powers of 3 by repeatedly squaring 3, BUT taking mod at each step.

Modular Exponentiation (All congruences are mod 19) ≡ 2 3 9 Compute 3^2000 = ≡ ≡ 4 2 3 9 81 5 (mod 19) = ≡ ≡ 8 2 3 5 25 6 = ≡ ≡ − 16 2 Technique: 3 6 36 17 ( 2 ) or = ≡ ≡  Repeatedly square 32 2 3 17 289 4 3, but take mod at = ≡ 64 2 3 4 16 each step . ≡ ≡ ≡ 128 2 3 16 256 9 ≡  Then multiply the 256 3 5 terms you need to ≡ 512 3 6 get the desired power. ≡ 1024 3 17 ≡ 2000 1024 512 256 128 64 16 3 ( 3 )( 3 )( 3 )( 3 )( 3 )( 3 ) Book’s ≡ 2000 3 ( 17 )( 6 )( 5 )( 9 )( 16 )( 17 ) powermod() ≡ 2000 3 ( 1248480 ) ≡ 2000 3 9 (mod 19 )

Modular Exponentiation ≡ 2 3 9 Compute 3^2000 = ≡ (mod 152) 4 2 3 9 81 = ≡ ≡ 8 2 3 81 6561 25 = ≡ ≡ 16 2 3 25 625 17 = ≡ ≡ 32 2 3 17 289 137 = ≡ ≡ 64 2 3 137 18769 73 ≡ 128 3 9 ≡ 256 3 81 ≡ 512 3 25 ≡ 1024 3 17 ≡ 2000 1024 512 256 128 64 16 3 ( 3 )( 3 )( 3 )( 3 )( 3 )( 3 ) ≡ 2000 3 ( 17 )( 25 )( 81 )( 9 )( 73 )( 17 ) ≡ 2000 3 ( 384492875 ) ≡ 2000 3 9 (mod 152 )