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6.4 Factoring

6.4 Factoring. 1. The Pollard’s p-1 algorithm input： an integer n , and a prespecified “bound” B output：factors of n. Why? Suppose p is a prime divisor of n, and suppose that q <= B for every prime power q|(p-1). Then (p-1)|B! At the end of for loop, we have

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6.4 Factoring

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  1. 6.4 Factoring

  2. 1. The Pollard’s p-1 algorithm input： an integer n , and a prespecified “bound” B output：factors of n

  3. Why? Suppose p is a prime divisor of n, and suppose that q <= B for every prime power q|(p-1). Then (p-1)|B! At the end of for loop, we have a=2B! mod n Now 2p-1=1 mod p (by Fermat’s little Thm) Since (p-1)|B!, it follows a=2B! =1 mod p and hence p|(a-1). Since we also have p|n, d=gcd(a-1, n) will be a non-trivial divisor of n (unless a=1).

  4. E.g. n=15770708441, B=180 a = 2180! = 11620221425 D = gcd(a-1, n) = 135979 In fact, the complete factorization of n into primes is 15770708441 = 135979 x 115979 The factorization succeeds because 135978 has only “small” prime factors: 135978 = 2 x 3 x 131 x 173

  5. 2. The Pollard’s rho algorithm input： an integer n output：factors of n (1) Selecting a “random” function f with integer coefficients , and any Begin with x=x0 and y=y0. (2) Repeat the two calculations until d=gcd(x-y,n)>1. (3) Do the following compare 3.1 If d<n, we have succeeded. 3.2 If d=n, the method is failed. Goto (1). (*) A typical choice of f(x)=x2+1, with a seed x0=2.

  6. 5 26 1 26 449 1 126 240 19 • Complexity of rho method We expect this method to use the function f at most • E.g：n=551, f(x)=x2+1 mod 551 and x0=2.

  7. 3. Dixon’s random squares algorithm • The idea is to locate with if gcd(x+y,n) is a nontrivial factor of n. (Why?) since n|(x-y)(x+y) but neither of x-y or x+y is divisible by n. • Eg. n=15, x=2, y=7 (22=72 mod 15) => gcd(2+7,15)=3 is a nontrivial factor of n. • Eg. n=77, x=10, y=32 (102=322 mod 77) => gcd(10+32,77)=7 is a nontrivial factor of n.

  8. factor base and pt-smooth • A factor base B={p1, p2,…,pt} consisting of the first t primes is selected. If b factors over B, b is said to be pt-smooth. • Eg：B={2,3,5}, b=23*56 is 5-smooth;b=23*76 is not 5-smooth. • We may include -1 in B to handle the negative b B={p0, p1, p2,…,pt}, with p0=-1.

  9. Algorithm input： a composite integer n and factor base B= {p1, p2,…,pt} output：factors of n (1) Suppose t+1 pairs (ai, bi=ai2 mod n) are obtained, where bi is pt-smooth over B and the factorizations are given by (2) A set S is to be selected so that has only even powers of primes appearing. (3) Let , and do the following compare 3.1 If 3.2 If

  10. 1 231 1018 2*509 (discard!) 1 105 968 23*112 2 115 3168 25*32*11 3 1006 6336 26*32*11 4 3010 8800 25*52*11 5 4014 882 2*32*72 6 4023 2816 28*11 • Eg ：n=10057, t=5, B={2,3,5,7,11} If S={4,5,6}, then x=3010*4014*4023 mod n=2748 y=27*3*5*7*11 mod n=7042 Since , we obtain a nontrivial factor gcd(x+y,n)=89, and 10057=89*113. If S={1,5}, then x=105*4014 mod n=9133 and y=22*3*7*11=924. Unfortunately, , and no useful information is obtained.

  11. Eg ：n=15770708441, t=6, B={2,3,5,7,11, 13} 83409341562 = 3*7 (mod n) 120449429442 = 2*7*13 (mod n) 27737000112 = 2*3*13 (mod n) (8340934156*12044942944*2773700011)2 = (2*3*7*13)2 (mod n) 95034357852 = 5462 (mod n) gcd(9503435785–546, 15770708441)=115759 to find the factor 115759 of n

  12. Improvements: • We may include -1 in B to handle the negative b B={p0, p1, p2,…,pt}, with p0=-1. • Define Let ai=z+m and bi= q(z) = ai2 - kn for z=0,1,-1,2,-2, … k=1,2, …

  13. Quadratic sieve algorithm (simple version) input： a composite integer n output：factors of n (1) choose a suitable P and construct a factor base (2) Define (3) Let ai=z+m and bi=q(z)=ai2-n for z=0,1,-1,2,-2,… A set S is to be selected so that has only even powers of primes appearing. (4) Let , and do the following

  14. 0 100 -57 -3*19 -1 99 -256 -28 1 101 144 24*32 -3 97 -648 -23*34 5 105 968 23*112 • Eg ：n=10057 If S={1}, then x=101 and y= =22*3. Since , we obtain a nontrivial factor gcd(x+y,n)=113, and 10057=89*113. If S={-1,-3, 5}, then x=99*97*105 and y=27*32*11. Unfortunately, , and no useful information is obtained.

  15. 4. Factoring algorithms in practice (Asymptotic running times) 1. Quadratic sieve 2. Elliptic curve (p is the smallest prime factor of n) 3. Number field sieve

  16. The RSA Challengehttp://www.rsa.com/rsalabs/node.asp?id=2092 • RSA-640 is factored! (11/2/2005) (193 digits) • RSA-200 is factored! (5/9/2005) (663 bits) • RSA-576 is factored! (12/3/2003) (174 digits) • RSA-160 is factored! (1/6/2003) (530 bits) • RSA-155 is factored! (8/22/1999) (512 bit) • RSA-140 is factored! (2/2/1999)

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