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**Abstract**• The Number Field Sieve is asymptotically the fastest algorithm for factoring a large integer N with no small prime factors, such as an RSA modulus. An early step in the algorithm selects two polynomials with a common root modulo N and “small” coefficients. We know ways to select two polynomials when one is linear, but that choice causes one polynomial norm to be much larger than the other. This talk says what is known about higher-degree selections, esp. a search for two cubic polynomials. Peter L. Montgomery Microsoft Research and CWI**Searching for Higher-Degree Polynomials for the General**Number Field Sieve Peter L. Montgomery Microsoft Research, USA, and CWI October, 2006**Number Field Sieve (NFS)**• Asymptotically best known algorithm for factoring large integers with no small prime factors. • Also best known algorithm for discrete logarithms modulo large primes. Peter L. Montgomery Microsoft Research and CWI**SNFS and GNFS**• Special Number Field Sieve (SNFS) • Number being factored has nice algebraic form. • Record 6353− 1 (274 digits, 2006). • General Number Field Sieve (GNFS) • No known nice algebraic form. • Record RSA200 (200 digits, 2005). Peter L. Montgomery Microsoft Research and CWI**NFS Stages – Part I**• Input: Composite integer N, no small factors. • Polynomial selection • Find polynomials f1, f2 with common root m modulo N. • Homogeneous form: Fk(a, b) = b deg(fk)fk(a/b) . • Sieving • Find many integer pairs (ai, bi) where both homogeneous polynomial values |Fk(ai, bi)| are smooth (k = 1, 2). • Normalized so gcd(ai, bi) = 1 and bi > 0. • Called relations. • Need one relation per prime ideal in your factor bases. Peter L. Montgomery Microsoft Research and CWI**NFS Stages – Part II**• Matrix construction and linear algebra • Let k be a (complex) root of fk. • Find nonempty set S of indices such that πjS (aj – bjk) is a square in Q(k), for each k. • Each aj – bjk has smooth norm. • Find square roots in Q(k). • Apply homomorphisms mapping each k to m mod N . • Get integer congruence A2≡B2 (mod N). Hope GCD(A + B, N) is nontrivial factor of N. Peter L. Montgomery Microsoft Research and CWI**NFS with Two Polynomials**• Given N, which we want to factor. • Also input desired degrees d1, d2 . • Find irreducible polynomials f1, f2 of degrees d1, d2 with common root m modulo N (but not in C). • resultant(f1, f2) will be a nonzero multiple of N, preferably a small multiple. • Determinant formula for resultant gives lower bound on coefficient sizes in f1, f2 . Peter L. Montgomery Microsoft Research and CWI**Sample SNFS Polynomial Selection**• N = (2512 + 1)/2424833 (148 digits). • 9th Fermat number made SNFS famous. • Guess to use degrees 5 and 1. • Common root m = 2103. • f1(X) = X−m and f2(X) = X5 + 8. • Resultant = ± (m5 + 8) or 19e6 N. • Homogeneous F1 (a, b) = a−mb, and F2 (a, b) = a5 + 8 b5. Peter L. Montgomery Microsoft Research and CWI**Norm Sizes**• Assume we sieve 2e12 points, in rectangle |a| 1e6 and 0 < b 1e6. • Approximate homogeneous sizes a− 1e31 b and a5 + 8b5. • Norm bounds approx 1e37 and 9e30. • Smaller norms more likely to be smooth. • Both norms must be smooth. Peter L. Montgomery Microsoft Research and CWI**Alternate Choices for 2512 + 1**• Degree 4, m = 2128≈ 3e38. f2(X) = X4 + 1. • a−mb and a4 + b4. • Bounds 3e44 and 2e24. • Degree 6, m = 285≈ 4e25. f2(X) = 4X6 + 1. • a−mb and 4a6 + b6. • Bounds 4e31 and 5e36. • Degree 5 bounds were 1e37 and 9e30. • Close call between degrees 5 and 6. • 1990 technology needed monic polynomials. Peter L. Montgomery Microsoft Research and CWI**Roots Modulo Small Primes**• X 4 + 1 • One root modulo 2, four modulo 17. • X 5 + 8 • One root modulo each of 2, 3, 5, 7, 13, 17, 19, 23. • 4X 6 + 1 • Projective root modulo 2. • Two roots modulo each of 5, 17. • This quintic norm has more prime divisors < 25 than the other norms, on average. Peter L. Montgomery Microsoft Research and CWI**Resultant Lower Bounds on Coefficient Sizes**• Assume fk has degree dk, coefficient bound Bk (k = 1, 2). • Determinant formula for resultant(f1, f2) has d2 rows with coefficients of f1 and d1 rows with coefficients of f2. • Need B1d2B2d1N (approx). • If rectangular sieving region is 2A ×A, we want both BkAdk small, about same size. Peter L. Montgomery Microsoft Research and CWI**Base-m Method**• Set m≈ N1/(d+1) if degree d wanted. • Write N = a0 + a1m + ... + ad md in base m. • Each ai is O(m), possibly negative. • f1(X) = X−m . • f2(X) = a0 + a1X + ... + ad Xd . • Let rectangular sieving region be 2A ×A. • |a| A and 0 < bA. • Norm bounds mA and (d+1)mAd . • Norms too far apart (ratio (d+1)Ad−1 ). Peter L. Montgomery Microsoft Research and CWI**Rating Polynomials**• Heuristics to increase density of smooth norms: • Try to make norm small on average. • Prefer real roots, so norm is near zero on parts of sieving region. • Try to have many roots modulo small primes and prime powers. • For example, X2 + 7 is divisible by 8 whenever it is even. • Brian Murphy (ANTS, 1998) confirmed that these properties improve yield when using two quadratic polynomials. Peter L. Montgomery Microsoft Research and CWI**Improved Base-m**• Assume degrees d and 1 wanted, with d 4. • Looking for f(m) = N where (if d = 5) f(X) = a5X5 + a4X4 + a3X3 + a2X2 + a1X+ a0. • Pick leading coefficient ad. • Prefer many small prime divisors. • Set m = round((N/ad)1/d). • Fill in initial a0 to ad−1 using arithmetic mod m. Usually |ad−1| d ad/ 2. • Reject unless |ad−2| << m. Peter L. Montgomery Microsoft Research and CWI**Skewed Sieving Region**• Let f0 be the initial f, with “small” ad to ad−2 and f0(m) = N. • Suppose the rectangular sieving region of area 2A2 is |a| Ar and 0 < bA/r. • If r = 1, norm bound is about a0Ad or mAd. • If r >> 1, big terms are ad−3(Ar)d−3 (A/r)3 and ad−2(Ar)d−2 (A/r)2 and ad (Ar)d. • Assuming first and last dominate, equate them • r = (ad−3 /ad)1/6 or (m/ad)1/6. • New norm bound ad−3(Ar)d−3 (A/r)3 is about mAdrd−6. • When d = 5, this is factor of r improvement over r = 1. • Linear X−m norm improves slightly too. Peter L. Montgomery Microsoft Research and CWI**Improved Modular Properties**• Try f(X) = f0(X) + C(X) (X − m) . • C(X) of degree d−4 to be determined • ad to ad−2 not affected. • ad−3 to a0 grow, but little effect on norm bound if C has small coefficients. • f(m) = f0(m) = N. • Sieve to find C(X) for which f has good modular properties. • Used for RSA140 and RSA155 (1999). • Brian Murphy’s PhD thesis. Peter L. Montgomery Microsoft Research and CWI**Two Quadratic Polynomials**• Suppose m is common root (mod N) of fk = ak X2 + bk X + ck (k = 1, 2) . • Assume O(N1/4) coefficients, coprime in Q[X]. • [m2, m, 1] orthogonal to both [ak , bk , ck ] (mod N) . • Let v = cross product of [ak , bk , ck ] over Z. • Coefficients of v are O(N1/2), not all zero. • vis multiple of [m2, m, 1] (mod N). • v is a geometric progression mod N. • Not a GP over Z if fk are irreducible. • Polynomials → Geometric progression mod N. Peter L. Montgomery Microsoft Research and CWI**GP → Quadratic Polynomials**• Let R = [r2, r1, r0] = O(N1/2) be geometric progression mod N, but not over Z. • Look at 2-D lattice in Z3 where R . v = 0. • Smallest basis vectors [ak, bk, ck] have typical size O(|R|1/2) = O(|N|1/4). • Resulting polynomials ak X2 + bk X + ck have common root r2 / r1≡r1 / r0 mod N . Peter L. Montgomery Microsoft Research and CWI**Constructing 3-term GP modulo N**• Choose prime q < N1/2 for which N is a quadratic residue. • Find x0 near N1/2 with x02≡N (mod q). • Return [q, x0, (x02 – N)/q]. • Different q lead to different GP and different pairs of quadratics. • Used for 3367 − 1 c105 in 1993-94. Peter L. Montgomery Microsoft Research and CWI**Two Quadratics for N = 2005**• Guess q = 23. 22≡ 2005 mod 23 • GP (23, 44, −3) since 442≡−3 ∙ 23. • Ratio 44/23 ≡−3/44 ≡ 1048 mod 2005. • [23, 44, −3] orthogonal to [−1, 1, 7] and [6, −3, 2]. • 7X2 + X −1 and 2X2−3X + 6 share root 1048 mod 2005. Peter L. Montgomery Microsoft Research and CWI**More than two Quadratics**• If f and g are two same-size quadratics with a common root, merge them with f±g. • Try ℓ quadratics, where ℓ 2, shared factor base bound. • Changes to rest of NFS straightforward. • ℓ (ℓ − 1)/2 chances per (a, b) for two norms to be smooth. • Need to produce ℓ / 2 times as many smooth relations. • Sieve 1 / (ℓ − 1) times as many points (hence smaller norms). • Sieving takes ℓ / 2 times as long per (a, b). • Estimate ℓ / 2(ℓ − 1) as much time as two quadratics. • Average ℓ /2 − 1 + 2−ℓ free ideals per prime. • Hard to find over two quadratics which excellent modular properties, so the ℓ (ℓ − 1)/2 above is unrealistic. Peter L. Montgomery Microsoft Research and CWI**Two Cubics → Five-term GP**• Suppose m is common root (mod N) of fk = ak X3 + bk X2 + ck X + dk (k = 1, 2) . • By resultant bound, O(N1/6) coefficients is smallest we can hope for. • Find vector v orthogonal over Z to both [ak, bk,ck , dk , 0] and both [0, ak, bk,ck,dk ]. • Simple determinant formula for v. • Components of v will be O(N2/3). • Multiple of [m4, m3, m2, m, 1] mod N. Peter L. Montgomery Microsoft Research and CWI**Five-term GP →Two Cubics**• Let R = [r4, r3, r2, r1, r0] = O(N2/3) be 5-term GP mod N, but not over Z. Ratio s = r1/r0 mod N. • Must avoid 2nd-order linear recurrence. • Look at 2-D lattice in Z4 orthogonal to both R ′ = [r3, r2, r1, r0] and ( [r4, r3, r2, r1] −s R ′ ) / N . • Smallest basis vectors [ak, bk, ck,dk] have typical size O((|R|2/N)1/2) = O(|N|1/6). • Resulting polynomials have common root s mod N . • For two degree-d polynomials, with O(N1/2d) coefficients, need 2d−1 terms of size O(N1−1/d). Peter L. Montgomery Microsoft Research and CWI**Desire a five-term GP mod N**• Exhaustive search finds many O(N2/3) solutions when N≈ 1e8. • Example: • [109, 151, 154, 11, 144] ratio 14 = 154/11 mod 2005 • Largest entry 154 vs. 20052/3≈ 159.0 . • X3− 4X2 + 3X + 3 and 3X3−X2−X− 2 share root 14 mod 2005. • Avoid (1st or) 2nd order linear recurrence. • Example: [39, 22, −39, −22, 39] mod 2005 = 392 + 222. • X3 + X and X2 + 1 share a quadratic factor. • Don’t know how to find quickly when N is large. Peter L. Montgomery Microsoft Research and CWI**Can we use Matrix Inverse?**• Matrix inverse is O(N1/3) / N (109 151 154 ) (−11 10 11) (151 154 11 ) ( 10 4 −11) = 2005 I3 (154 11 144 ) ( 11 −11 3) • Entries in second are bilinear forms evaluated at coefficients of f1 and f2 • (a1b2−b1a2a1c2−c1a2a1d2−d1a2) • (a1c2−c1a2a1d2+b1c2−c1b2−d1a2 b1d2−d1b2 ) • (a1d2−d1a2 b1d2−d1b2 c1d2−d1c2 ) • Related to coefficients of [f1(X)f2(Y) −f2(X)f1(Y)] / (X−Y) . • Second matrix O(N1/3), symmetric, determinant ±N. • First has constant backwards diagonals. Peter L. Montgomery Microsoft Research and CWI**O(1) Cubic and O(N1/3) Quadratic for Prime N**• Choose irreducible cubic f1 with known linear factor X− mod N and O(1) coefficients. • One of X3− (2, 3, 6, 12) will work. • Find quadratic f2 with O(N1/3) coefficients and root modulo N. • Can use LLL to choose f2 . • Follow construction of GP from two O(N1/6) cubics. Output satisfies 2nd order recurrence. • N is prime in discrete logarithm problem. Peter L. Montgomery Microsoft Research and CWI**Sizes when Factoring c200**• Assume 2e18 points sieved. • Two quadratics. • Coefficients 1e50. Norms 3e68. • Two cubics. • Coefficients 2e33. Norms 8e60. • Two degree 4. • Coefficients 1e25. Norms 5e61. • Degree 3 or 4 appears best if we use equal degrees. Peter L. Montgomery Microsoft Research and CWI**c200 Sizes for Original Base-m**• Assume degree d = 5. Sieving area 2e18. • m = (c200)1/6 = 2e33. • Coefficients (except leading) 1e33. • Norms (d+2)(1e33)(1e9)d =7e78 and m(1e9) = 2e42. • Norm bounds far apart, compared to equal degrees. Peter L. Montgomery Microsoft Research and CWI**c200 Sizes for Modified Base-m**• Assume degree d = 5. Sieving area 2e18. • Assume a5≈ 1e10 and m = (1e200/a5)1/5≈ 1e38. • Assume we can find a3 small enough. • r≈ (m/a5)1/6≈ 5e4. • Bounds (1e18)1/2r = 5e13 on a and 2e4 on b. • a5 (5e13)5 and m(5e13)2(2e4)3 both 2e78. • Norm bound around 1e79 (six summands). • Linear bound (2e4)(1e38) = 2e42. • Little different than original base m. • But improved modular properties. Peter L. Montgomery Microsoft Research and CWI**Non-monic Linear Polynomial – Part I**• Start with N, d, ad. • Instead of finding f0 with f0(m) = N, find a P for which the congruence adMd≡ N (mod P) has many solutions M. • P is product of primes, each ≡ 1 (mod d) with N / ad a d-th residue. • Size of M chosen so | M− (N / ad)1/d | P / 2. Peter L. Montgomery Microsoft Research and CWI**Non-monic Linear Polynomial – Part II**• For each such M, where adMd≡ N (mod P), find polynomial f0 (X)= Σj=0d aj Xj Z[X] with N = Pdf0(M/P). • As earlier, reject unless coefficient of Xd−2 is small. • Can perform this test quickly when same P is reused. • f2(X) = f0(X) + C(X)(PX−M) for some C(X). • f2(X) and f1(X) = PX−M share root m = M / P mod N. • Due to Thorsten Kleinjung, Math. Comp., Oct. 2006. • Used for RSA576 (2003) and RSA200 (2005). Peter L. Montgomery Microsoft Research and CWI**RSA200 polynomials**• Found by Kleinjung et al • f1(X) = 374029011720 X 5 + 2711065637795630118X 4 + 19400071943177513865892714 X 3 − 33803470609202413094680462360399 X 2 − 120887311888241287002580512992469303610 X + 38767203000799321189782959529938771195170960 • M = 37570227807001155896638712233675454511 • P = 12722245648421103686881 = 11 . 31 . 61 . 71 . 191 . 331 . 461 . 521 . 691 . 821 Peter L. Montgomery Microsoft Research and CWI**Norm Sizes for RSA200**• a5≈ 23 . 35 . 5 . 7 . 13 . 422861 ≈ 4e11. • r≈ 1800. • Linear PX −M≈ 1e22 X− 4e37. • On rectangle |a| < 1.8e12 and 0 < b < 5.6e5, of area 2e18, norm bounds about 1e74 (quintic) and 2e43 (linear). • Quintic much smaller than predicted. Peter L. Montgomery Microsoft Research and CWI