# I. Introduction. MPRI Cours Lecture IV: Integer factorization. What is the factorization of a random number? II. Smoothness testing. F.

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1 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 MPRI Cours I. Introduction Input: an integer N; logox F. Morain logocnrs logoinria Output: N = k i=1 pα i i with p i (proven) prime. Lecture IV: Integer factorization Major impact: estimate the security of RSA cryptosystems. Also: primitive for a lot of number theory problems. I. Introduction. II. Smoothness testing. III. Pollard s RHO method. IV. Pollard s p 1 method. 2013/10/14 How do we test and compare algorithms? Cunningham project, RSA Security (partitions, RSA keys) though abandoned? Decimals of π. F. Morain École polytechnique MPRI cours /22 What is the factorization of a random number? F. Morain École polytechnique MPRI cours /22 II. Smoothness testing N = N 1 N 2 N r with N i prime, N i N i+1. Prop. r log 2 N; r = log log N. Size of the factors: D k = lim N + log N k / log N exists and On average k D k N 1 N 0.62, N 2 N 0.21, N 3 N an integer has one large factor, a medium size one and a bunch of small ones. Def. a B-smooth number has all its prime factors B. B-smooth numbers are the heart of all efficient factorization or discrete logarithm algorithms. De Bruijn s function: ψ(x, y) = #{z x, z is y smooth}. Thm. (Candfield, Erdős, Pomerance) ε > 0, uniformly in y (log x) 1+ε, as x with u = log x/ log y. ψ(x, y) = x u u(1+o(1))

2 B-smooth numbers (cont d) A) Trial division Prop. Let L(x) = exp ( log x log log x ). For all real α > 0, β > 0, as x ψ(x α, L(x) β x α ) = L(x). α 2β +o(1) Ordinary interpretation: a number x α is L(x) β -smooth with probability ψ(x α, L(x) β ) x α = L(x) α 2β +o(1). Algorithm: divide x X by all p B, say {p 1, p 2,..., p m }, m = π(b) = O(B). Cost: m divisions or p B T(x, p) = O(m lg X lg B). Implementation: use any method to compute and store all primes 2 32 (one char per (p i+1 p i )/2; see Brent). Useful generalization: given x 1, x 2,..., x n X, can we find the B-smooth part of the x i s more rapidly than repeating the above in O(nm lg B lg X)? F. Morain École polytechnique MPRI cours /22 B) Product trees Algorithm: Franke/Kleinjung/FM/Wirth improved by Bernstein 1. [Product tree] Compute z = p 1 p m. p 1 p 2 p 3 p 4 p 1 p 2 p 3 p 4 p 1 p 2 p 3 p 4 2. [Remainder tree] Compute z mod x 1,..., z mod x n. z mod (x 1 x 2 x 3 x 4 ) z mod (x 1 x 2 ) z mod (x 3 x 4 ) z mod x 1 z mod x 2 z mod x 3 z mod x 4 3. [explode valuation] For each k {1,..., n}, compute y k = z 2e mod x k with e s.t. 2 2e x k ; print gcd(x k, y k ). F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 Validity and analysis Validity: let y k = z 2e mod x k. Suppose p x k. Then ν p (x k ) 2 e, since 2 ν p ν 2 2e. Therefore ν p (y k ) 2 e ν and the gcd will contain the right valuation. Division: If A has r + s digits and B has s digits, then plain division requires D(r + s, s) = O(rs) word operations. In case r s, break into r/s divisions of complexity M(s). Step 1: M((m/2) lg B). Step 2: D(m lg B, n lg X) (m/n)m(n) lg B lg X. Ex. B = 2 32, m , X = Rem. If space is an issue, do this by blocks. Rem. For more information see Bernstein s web page. F. Morain École polytechnique MPRI cours /22

3 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 III. Pollard s RHO method Epact Prop. Let f : E E, #E = m; X n+1 = f (X n ) with X 0 E. X 0 X 1 X 2 X µ 1 X µ X µ+1 X µ+λ 1 Thm. (Flajolet, Odlyzko, 1990) When m πm λ µ m. Prop. There exists a unique e > 0 (epact) s.t. µ e < λ + µ and X 2e = X e. It is the smallest non-zero multiple of λ that is µ: if µ = 0, e = λ and if µ > 0, e = µ λ λ. Floyd s algorithm: X <- X0; Y <- X0; e <- 0; repeat X <- f(x); Y <- f(f(y)); e <- e+1; until X = Y; Thm. e π 5 m m. F. Morain École polytechnique MPRI cours /22 Application to the factorization of N F. Morain École polytechnique MPRI cours /22 Practice Idea: suppose p N and we have a random f mod N s.t. f mod p is random. function f(x, N) return (x 2 + 1) mod N; end. function rho(n) 1. [initialization] x:=1; y:=1; g:=1; 2. [loop] while (g = 1) do x:=f(x, N); y:=f(f(y, N), N); g:=gcd(x-y, N); endwhile; 3. return g; Choosing f : some choices are bad, as x x 2 et x x 2 2. Tables exist for given f s. Trick: compute gcd( i (x 2i x i ), N), using backtrack whenever needed. Improvements: reducing the number of evaluations of f, the number of comparisons (see Brent, Montgomery). Conjecture. RHO finds p N using O( p) iterations.

4 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 Theoretical results (1/4) Bach (2/4) Thm. (Bach, 1991) Proba RHO with f (x) = x finding p N after k iterations is at least ( k 2) when p goes to infinity. Sketch of the proof: define p + O(p 3/2 ) f 0 (X, Y) = X, f i+1 (X, Y) = f 2 i + Y, f 1 (X, Y) = X 2 + Y, f 2 (X, Y) = (X 2 + Y 2 ) + Y,... Divisor of N found by inspecting gcd(f 2i+1 (x, y) f i (x, y), N) for i = 0, 1, 2,.... Prop. (a) deg X (f i ) = 2 i ; (b) f i X 2i mod Y; (c) f i Y 2i Y mod X; (d) f i+j (X, Y) = f i (f j (X, Y), Y). (e) f i is absolutely irreducible (Eisentein s criterion). (f) f j f i = fj 1 2 f i 1 2 = (f j 1 + f i 1 )(f j 2 + f i 2 ) (f j i + X)(f j i X). (g) For all k 1 f l+n ± f l f l+kn ± f l. F. Morain École polytechnique MPRI cours /22 Bach (3/4) For i < j, ρ i,j is the unique poly in Z[X, Y] s.t. (a) ρ i,j is a monic (in Y) irreducible divisor of f j f i. (b) Let ω i,j denote a primitive (2 j 2 i ) root of unity. Then ρ i,j (ω i,j, 0) = 0. Proof: and and for i 1: f j f i X 2i (X 2j 2 i 1) mod Y, X 2j 2 i 1 = µ 2 j 2 i Φ µ (X). f j f 0 ρ 0,j d j,d j ρ 0,d f j 1 + f i 1 ρ i,j d j i,d j i ρ. i,i+d Conj. we have in fact equalities instead of divisibility., F. Morain École polytechnique MPRI cours /22 Bach (4/4) Thm1. f k f l factor over Z[X, Y]. Moreover, they are squarefree. Proof uses projectivization of f. Weil s thm: if f Z[X, Y] is absolutely irreducible of degree d, then N p the number of projective zeroes of f is s.t. ( ) d 1 p. N p (p + 1) 2 2 Thm2. Fix k 1. Choose x and y at random s.t. 0 x, y < p. Then, proba for some i, j < k, i j, f i (x, y) = f j (x, y) mod p is at least ( k ) 2 /p + O(1/p 3/2 ) as p tends to infinity. Proof: same as i < j < k and ρ i,j (x, y) 0 mod p. Use inclusion-exclusion, Weil s inequality and Bézout s theorems. Same result for RHO to find p N.

5 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 IV. Pollard s p 1 method First phase Idea: assume p N and a is prime to p. Then Invented by Pollard in Williams: p + 1. Bach and Shallit: Φ k factoring methods. Shanks, Schnorr, Lenstra, etc.: quadratic forms. Lenstra (1985): ECM. Rem. Almost all the ideas invented for the classical p 1 can be transposed to the other methods. (p a p 1 1 and p N) p gcd(a p 1 1, N). Generalization: if R is known s.t. p 1 R, will yield a factor. gcd((a R mod N) 1, N) How do we find R? Only reasonable hope is that p 1 B! for some (small) B. In other words, p 1 is B-smooth. Algorithm: R = p α B 1 p α = lcm(2,..., B 1 ). Rem. (usual trick) we compute gcd( k ((ar k 1) mod N), N). F. Morain École polytechnique MPRI cours /22 Second phase: the classical one F. Morain École polytechnique MPRI cours /22 Second phase: BSGS Let b = a R mod N and gcd(b, N) = 1. Hyp. p 1 = Qs with Q R and s prime, B 1 < s B 2. Test: is gcd(b s 1, N) > 1 for some s. s j = j-th prime. In practice all s j+1 s j are small (Cramer s conjecture implies s j+1 s j (log B 2 ) 2 ). Precompute c δ b δ mod N for all possible δ (small); Compute next value with one multiplication b s j+1 = b s j c sj+1 s j mod N. Cost: O((log B 2 ) 2 ) + O(log s 1 ) + (π(b 2 ) π(b 1 )) multiplications +(π(b 2 ) π(b 1 )) gcd s. When B 2 B 1, π(b 2 ) dominates. Rem. We need a table of all primes < B 2 ; memory is O(B 2 ). Record. Nohara (66dd of , 2006; see Select w B 2, v 1 = B 1 /w, v 2 = B 2 /w. Write our prime s as s = vw u, with 0 u < w, v 1 v v 2. Lem. gcd(b s 1, N) > 1 iff gcd(b wv b u, N) > 1. Algorithm: 1. Precompute b u mod N for all 0 u < w. 2. Precompute all(b w ) v for all v 1 v v For all u and all v evaluate gcd(b vw b u, N). Number of multiplications: w + (v 2 v 1 ) + O(log 2 w) = O( B 2 ) Memory: O( B 2 ). Number of gcd: π(b 2 ) π(b 1 ).

6 Second phase: using fast polynomial arithmetic Algorithm: 1. Compute h(x) = 0 u<w (X bu ) Z/NZ[X] 2. Evaluate all h((b w ) v ) for all v 1 v v Evaluate all gcd(h(b wv ), N). Analysis: Step 1: O((log w)m pol (w)) operations (using a product tree). Step 2: O((log w)m int (log N)) for b w ; v 2 v 1 for (b w ) v ; multi-point evaluation on w points takes O((log w)m pol (w)). Rem. Evaluating h(x) along a geometric progression of length w takes O(w log w) operations (see Montgomery-Silverman). ( ) Total cost: O((log w)m pol (w)) = O. B 0.5+o(1) 2 Trick: use gcd(u, w) = 1 and w = F. Morain École polytechnique MPRI cours /22 Second phase: using the birthday paradox Consider B = b mod p ; s := #B. If we draw s elements at random in B, then we have a collision (birthday paradox). Algorithm: build (b i ) with b 0 = b, and { b 2 b i+1 = i mod N with proba 1/2 b 2 i b mod N with proba 1/2. We gather r s values and compute where r i=1 j i (b i b j ) = Disc(P(X)) = i P(X) = r (X b i ). i=1 use fast polynomial operations again. P (b i ) F. Morain École polytechnique MPRI cours /22

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