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Ch1 - Algorithms with numbers Basic arithmetic Addition Multiplication Division Modular arithmetic RSA –factoring is hard Primality testing Addition 53+35=88 Cost? (n – number of bits) O(n) Multiplication 13x11=143 Cost? O(n2) al-Khwārizmī Operations determining parity (even or odd) addition duplation (doubling a number, left shift) mediation (halving a number, rounding down, right shift) al-Khwārizmī Cost? O(n2) Can we do better? Division Cost? Modular arithmetic A system for dealing with restricted ranges of integers Addition x+y mod N, assuming x, y <N O(n), n - number of bits N has (size of input) (x+y mod N = x+y or x+y-N) Multiplication x*y mod N ? Modular arithmetic RSA Ron Rivest, Adi Shamir, Leonard Adleman (1977) Algorithm for public-key cryptography, based on the presumed difficulty of the factoring problem. 2002 A.M. Turing Award RSA is one of the most used cryptographic protocols on the net. Your browser uses it to establish a secure session with a site. Needed for implementing RSA: FLT (Fermat’s Little Theorem) Fast Exponentiation Extended Euclidean Algorithm Modular inverses CRT (Chinese Remainder Theorem) Turing Lecture on Early RSA Days, Ronald L. Rivest Turing Lecture on Early RSA Days, Ronald L. Rivest Turing Lecture on Early RSA Days, Ronald L. Rivest In April 2012, the factorization of 143 is achieved. RSA public-key cryptosystem In a public-key cryptosystem, everyone has a public key and a secret key. Suppose Alice and Bob are two participants. Alice PA , SA Bob PB , SB The keys specify 1-1 functions from message M to itself: M= SA (PA (M)) M= PA (SA (M)) Encryption: encrypt M PA Communication channel Bob decrypt SA Alice PA(M) M RSA Digital signatures: SA PA Communication channel M Alice M Bob SA(M) =? Accept RSA algorithm Select at random 2 large prime numbers p & q; (p & q might be, say, 100 decimal digits each.) Compute n: n = pq; Select an odd integer e that is relatively prime to (n) = (p-1)(q-1); Compute d as the multiplicative inverse of e, modulo (n); (de 1 mod (n)) Publish P = (e, n) as the RSA public key; Keep secret S = (d, n) as the RSA secret key. If M Zn ={0,1,…,n-1}, P(M) = Me mod n S(C) = Cd mod n, C=P(M). RSA example Pick p = 47, q=71. n=pq=3337. (n) = (p-1)(q-1)=46*70=3220, choose e=79 (at random). d =79-1 mod 3220 = 1019. PA=(79, 3337). SA=(1019, 3337). Message: M = 6882326879666683 = 688 232 687 966 668 3 M1 = 688 68879 mod 3337 = 1570 =C1 M2 = 232 23279 mod 3337 = 2756 =C2 … C = 1570 2756 2091 2276 2423 158 C1 = 1570 15701019 mod 3337 = 688 =M1 … C2 = 158 1581019 mod 3337 = 3 =M2 Another example m e mod n n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N A M E S M I L E Y 1201 1920 0014 0113 0500 1913 0912 0525 120113 mod 4559, 192013 mod 4559, … 1074 0116 1478 2150 3906 4256 1445 2462 RSA Bob receives the encrypted blocks c = m e mod n. He have a private decryption exponent d which when applied to c recovers the original blocks m : (m e mod n )d mod n = m For n = 4559, e = 13 the decryptor d = 3397. RSA n = 4559, d = 3397 1074 0116 1478 2150 3906 4256 1445 2462 1074 3397 mod 4559, 01163397 mod 4559, … 1201 1920 0014 0113 0500 1913 0912 0525 LA S T N A M E S M I L E Y RSA Technical difficulties: How do we know the algorithm works correctly? How to pick large prime numbers? Compute pq How to choose e Compute d How to compute Me, Cd Can any one break the code? RSA If I want to encrypt credit card numbers, how big my p and q should be? If I want to encrypt words of four random characters from ASCII set, how big my p and q should be? How to pick large prime numbers ? Primality testing Hard, but much easier than factoring. Fermat’s Little Theorem(~1640): If p is prime, then a, s.t. 1≤a<p, ap-11 (mod p). ? The numbers make us fail are called Fermat pseudoprime -extremely rare (ex. 2340=1mod341; Carmichael number 561, 2560=1mod561) Lagrange’s Prime Number Theorem Theorem: The number of prime numbers between 1 and x is “about” x/lnx . Not only are primes easy to detect, but they are also relatively abundant. Carmichael number A number c is a Carmichael number if it is not a prime, and still for all prime divisors d of c it so happens that d-1divides c-1. The smallest Carmichael number is 561 = 31117 . If c is a Carmichael number and a is relatively prime to c, then ac-1 1 mod c. Primality testing Primality testing Fermat's Last Theorem Fermat's Last Theorem states that xn + yn = zn has no non-zero integer solutions for x, y and z when n > 2. RSA Technical difficulties: How do we know the algorithm works correctly? How to pick large prime numbers? Compute pq How to choose e Compute d How to compute Me, Cd? Can any one break the code? How to compute e M, d C ? Modular exponentiation In order to implement RSA, exponentiation relative some modulo needs to be done a lot. So this operation better be doable, and fast. Q: How is it even possible to compute 28533397 mod 4559 ? After all, 28533397 has approximately 3397·4 digits! Modular exponentiation A: By taking the mod after each multiplication. For example: 233 mod 30 -73 (mod 30) (-7)2 ·(-7) (mod 30) 49 · (-7) (mod 30) 19·(-7) (mod 30) -133 (mod 30) 17 (mod 30) Modular exponentiation Therefore, 233 mod 30 = 17. Q: What if had to figure out 2316 mod 30. Same way tedious: need to multiply 15 times. Is there a better way? Modular exponentiation A: Better way. Notice that 16 = 2·2·2·2 so that 2316 = 232·2·2·2 = (((232)2)2)2 Therefore: 2316 mod 30 (((-72)2)2)2 (mod 30) (((49)2)2)2 (mod 30) (((-11)2)2)2 (mod 30) ((121)2)2 (mod 30) ((1)2 )2 (mod 30) (1)2 (mod 30) 1(mod 30) Which implies that 2316 mod 30 = 1. Q: How about 2325 mod 30 ? Modular exponentiation A: The previous method of repeated squaring works for any exponent that’s a power of 2. 25 isn’t. However, we can break 25 down as a sum of such powers: 25 = 16 + 8 + 1. Apply repeated squaring to each part, and multiply the results together. Previous calculation: 238 mod 30 = 2316 mod 30 = 1 Thus: 2325 mod 30 2316+8+1 (mod 30) Modular exponentiation x25 mod N Cost? – polynomial time (n=logN) Modular exponentiation How do we compute xy mod m , m>0? repeated squaring algorithm: mod-exp(x, y, m) if y = 0 then return(1) else z = mod-exp(x, y div 2, m) if y mod 2 = 0 then return(z * z mod m) else return(x * z * z mod m) Compute d ? Modular Inverse GCD Greatest common divisor Example: Euclid Algorithm If a,bZ+, apply division (mod) repeatedly as follows: a = q1b + r1, b = q2r1 + r2, r1 = q3r2 + r3, …… rk-2 = qkrk-1+ rk, rk-1 = qk+1rk Then, rk = GCD(a,b). Proof: (1) rk|a, rk|b (2) if d|a, d|b, then d| rk. where 0 < r1 < b where 0 < r2 < r1 where 0 < r3 < r2 where 0 < rk-1 < rk Recursion Theorem a,b N, b0, gcd(a,b) = gcd(b, a mod b). Proof : Let d = gcd(a,b). d|a, d|b. d|a-qb = a mod b d|b, d|a mod b d|gcd(b, a mod b). Let d = gcd(b, a mod b). d|b, d| a mod b. d|a-qb, d|b d|a d|gcd(a,b). gcd(a,b) = gcd(b, a mod b). Computing GCD Euclid gcd(x,y) { if y = 0 then return(x) else return(gcd(y,x mod y)) } Euclid Algorithm Example: Computing gcd(125, 87) 125 = 1*87 + 38 87 = 2*38 + 11 38 = 3*11 + 5 11 = 2*5 + 1 5 = 5*1 gcd(125,87)=1 gcd(125,87) = 1 11 - 2*5 =1 11 - 2*(38-3*11) = 1 - 2*38 + 7*11 = 1 - 2*38 + 7*(87 - 2*38) = 1 7*87 - 16*38 = 1 7*87 - 16*(125-1*87) = 1 - 16*125 + 23*87 = 1 1 = 125*(-16) + 87*23 1 = as + bt Extended Euclidean Algorithm obtain gcd(a,b) and x,y, s.t. gcd(a,b) = ax+by. Extended-Euclid (a,b) if (b==0) return (a,1,0); (d’,x’,y’)=Extended-Euclid(b, a mod b); (d,x,y)=(d’, y’, x’-a/by’); return (d,x,y); Ex: demo a 412 260 152 108 44 20 4 b 260 152 108 44 20 4 0 q 1 1 1 2 2 5 x 12 -7 5 -2 1 0 1 y -19 12 -7 5 -2 1 0 d 4 4 4 4 4 4 4 Cost? Theorem: The algorithm above correctly computes the gcd of x and y in time O(n), where n is the total number of bits in the input (x; y) Multiplicative Inverse Multiplicative inverse x of a, modulo n: ax = 1 mod n. ax = kn+1 If gcd(a,n)=1, ax-kn = gcd(a,n). ax+ny = gcd(a,n). Therefore, x can be found using extended Euclidean algorithm. Is the multiplicative inverse unique? Multiplicative Inverse Theorem: n>1, if gcd(a,n)=1, then ax=1 (mod n) has a unique positive solution, modulo n. Example: a = 79; n = 3220. x = 1019. ax = 80501 = 25*3220+1. x = -2201. ax = -173879 = -54*3220+1. RSA Technical difficulties: How do we know the algorithm works correctly? How to pick large prime numbers? Compute pq How to choose e Compute d How to compute Me, Cd? Can any one break the code? How do we know RSA works correctly? Chinese Remainder Theorem (~1700 old) http://en.wikipedia.org/wiki/RSA_Factoring_C hallenge#The_prizes_and_records