Lecture 4 Notes; April 7, 1997; Notetaker: Eddie Lo Equivocation * 8-bit integer H(x) = -SUM from i=1 to 2^8 (1/2^8) log (1/2^8) = log 2^8 = 8 If parity = 0 H(x| bit 8 is 0) = -SUM from i=1 to 2^7 (1/2^7) log (1/2^7) = log 2^7 = 7 Joint Probability / Uncertainty: X = die throw, Y = coin toss * Case 1: throw, toss simultaneously p(X=3,Y=head) = p(Y=head) p(X=3|Y=head) As p(X=3|Y=head) = p(X=3) because X, Y are independent, p(X=3,Y=head) = p(Y=head) p(X=3) = 1/12 * Case 2: Toss coin; on heads, roll 3-sided die; on tails, roll 6-sided die p(X=3,Y=head) = p(Y=head) p(X=3|Y=head) = (1/2)(1/3) = 1/6 p(X=3,Y=tail) = p(Y=tail) p(X=3|Y=tail) = (1/2)(1/6) = 1/12 In case 1, X=3, Y=head (independent) H(X|Y) = H(X) = log 6 In case 2, X=3,Y=head (dependent) H(X|Y) = SUM over Y p(Y) SUM over X p(X|Y) log p(X|Y) = p(Y=head) SUM over X p(X|Y=head) log p(X|Y=head) + p(Y=tail) SUM over X p(X|Y=tail) log p(X|Y=tail) = (1/2) log 18 H(M|C) = H(M) and p(M|C) = p(M) define perfect security Key Equivocation The uncertainty H(K|C) in the key given some ciphertext * Case 1: KHOOR ZRUOG as a Caesar cipher: H(K|C) = 0 (see handout) * Case 2: J as a Caesar cipher: H(K|C) = 1 (could be I or A) Formula: H(K|C) = SUM over C p(C) SUM over K p(K|C) log p(K|C) In case 1: p(C) = 26^(-10) and p(K|C) = 0 if K <> 23 and 1 if K = 23 H(K|C) = SUM over K p(K|C) log p(K|C) = SUM from i=1 to 26^10 of 26^(-10) * 0 = 0 In case 2: p(C) = 26^(-1) p(K|C) = 0 if K <> 9 and K <> 25; 1/2 if K = 9; 1/2 if K = 25 H(K|C) = - SUM over C p(C) SUM over K p(K|C) log p(K|C) = - SUM from i=1 to 26 (1/26) ((1/2) log (1/2) + (1/2) log (1/2)) = - log (1/2) = 1 Unicity Distance Smallest length of text N such that H(K|C) is approximately 0 N = H(K)/D implies N is the unicity distance when H(K) - DN = 0 This is impossible when # of Keys = # meaningful mesg So: H(K) = log 2^{RN} = RN so RN - DN = 0 so rN = 0 where R is absolute rate and r is the actual rate of the language Example: Caesar cipher H(K) = log 26 = 4.7, so N = 4.7/3.2 = 1.5. So you need 1.5 characters to solve a Caesar cipher. Number Theory 5 = 12 mod 7 12 congruent to 5 (Mathmatical notation) 12 = 5 + 7k a mod n = b Principle of modular arithmetic: mods can be moved into multiplication and addition Examples: 1) 7*9 mod 4 = (7 mod 4) (9 mod 4) mod 4 = 3 2) 2^{20} mod 3 = (2^{10})(2^{10}) mod 3 =(2^5)(2^5)(2^5)(2^5) mod 3 = (2^2)(2^3)(2^2)(2^3)(2^2)(2^3)(2^2)(2^3)(2^2)(2^3) mod 3 = (2^2 mod 3)(2^3 mod 3)(2^2 mod 3)(2^3 mod 3)(2^2 mod 3) (2^3 mod 3)(2^2 mod 3)(2^3 mod 3)(2^2 mod 3)(2^3 mod 3) mod 3 = [(1)(2)(1)(2)(1)(2)(1)(2)(1)(2)] mod 3 = 16 mod 3 = 1 Solve: ax mod n = 1 for any a and n (if possible) Consider 2x mod 5 = 1. 2x mod 5 generates a complete set of residues, as 2(0) mod 5 = 0 2(1) mod 5 = 2 2(2) mod 5 = 4 2(3) mod 5 = 1 2(4) mod 5 = 3 (and so 2x mod 5 = 1 has a solution -- x = 3). But 2x mod 4 will not: 2(0) mod 4 = 0 2(1) mod 4 = 2 2(2) mod 4 = 0 2(3) mod 4 = 2 (and so 2x mod 4 = 1 does not have a solut1on.) Solution requires gcd(a,n) = 1 or a, n relatively prime. 2x mod 5 gives a complete set of residues. After removing the zero get the reduced set of residues, which is the set for which ax mod n = 1 has a solution. Totient function PHI is the number of elements in the reduced set of residues; PHI(4) = 2, PHI(5) = 4, PHI(p) = p-1 when p is prime 15 = 3*5; showed that PHI(15) = 8 = 2*4 = (3-1)*(5-1) In general, PHI(pq) = (p-1)(q-1) when p, q are prime.