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Diffstat (limited to 'libgo/go/math/big/prime.go')
| -rw-r--r-- | libgo/go/math/big/prime.go | 320 |
1 files changed, 320 insertions, 0 deletions
diff --git a/libgo/go/math/big/prime.go b/libgo/go/math/big/prime.go new file mode 100644 index 0000000000..3e9690e55e --- /dev/null +++ b/libgo/go/math/big/prime.go @@ -0,0 +1,320 @@ +// Copyright 2016 The Go Authors. All rights reserved. +// Use of this source code is governed by a BSD-style +// license that can be found in the LICENSE file. + +package big + +import "math/rand" + +// ProbablyPrime reports whether x is probably prime, +// applying the Miller-Rabin test with n pseudorandomly chosen bases +// as well as a Baillie-PSW test. +// +// If x is prime, ProbablyPrime returns true. +// If x is chosen randomly and not prime, ProbablyPrime probably returns false. +// The probability of returning true for a randomly chosen non-prime is at most ¼ⁿ. +// +// ProbablyPrime is 100% accurate for inputs less than 2⁶⁴. +// See Menezes et al., Handbook of Applied Cryptography, 1997, pp. 145-149, +// and FIPS 186-4 Appendix F for further discussion of the error probabilities. +// +// ProbablyPrime is not suitable for judging primes that an adversary may +// have crafted to fool the test. +// +// As of Go 1.8, ProbablyPrime(0) is allowed and applies only a Baillie-PSW test. +// Before Go 1.8, ProbablyPrime applied only the Miller-Rabin tests, and ProbablyPrime(0) panicked. +func (x *Int) ProbablyPrime(n int) bool { + // Note regarding the doc comment above: + // It would be more precise to say that the Baillie-PSW test uses the + // extra strong Lucas test as its Lucas test, but since no one knows + // how to tell any of the Lucas tests apart inside a Baillie-PSW test + // (they all work equally well empirically), that detail need not be + // documented or implicitly guaranteed. + // The comment does avoid saying "the" Baillie-PSW test + // because of this general ambiguity. + + if n < 0 { + panic("negative n for ProbablyPrime") + } + if x.neg || len(x.abs) == 0 { + return false + } + + // primeBitMask records the primes < 64. + const primeBitMask uint64 = 1<<2 | 1<<3 | 1<<5 | 1<<7 | + 1<<11 | 1<<13 | 1<<17 | 1<<19 | 1<<23 | 1<<29 | 1<<31 | + 1<<37 | 1<<41 | 1<<43 | 1<<47 | 1<<53 | 1<<59 | 1<<61 + + w := x.abs[0] + if len(x.abs) == 1 && w < 64 { + return primeBitMask&(1<<w) != 0 + } + + if w&1 == 0 { + return false // n is even + } + + const primesA = 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 37 + const primesB = 29 * 31 * 41 * 43 * 47 * 53 + + var rA, rB uint32 + switch _W { + case 32: + rA = uint32(x.abs.modW(primesA)) + rB = uint32(x.abs.modW(primesB)) + case 64: + r := x.abs.modW((primesA * primesB) & _M) + rA = uint32(r % primesA) + rB = uint32(r % primesB) + default: + panic("math/big: invalid word size") + } + + if rA%3 == 0 || rA%5 == 0 || rA%7 == 0 || rA%11 == 0 || rA%13 == 0 || rA%17 == 0 || rA%19 == 0 || rA%23 == 0 || rA%37 == 0 || + rB%29 == 0 || rB%31 == 0 || rB%41 == 0 || rB%43 == 0 || rB%47 == 0 || rB%53 == 0 { + return false + } + + return x.abs.probablyPrimeMillerRabin(n+1, true) && x.abs.probablyPrimeLucas() +} + +// probablyPrimeMillerRabin reports whether n passes reps rounds of the +// Miller-Rabin primality test, using pseudo-randomly chosen bases. +// If force2 is true, one of the rounds is forced to use base 2. +// See Handbook of Applied Cryptography, p. 139, Algorithm 4.24. +// The number n is known to be non-zero. +func (n nat) probablyPrimeMillerRabin(reps int, force2 bool) bool { + nm1 := nat(nil).sub(n, natOne) + // determine q, k such that nm1 = q << k + k := nm1.trailingZeroBits() + q := nat(nil).shr(nm1, k) + + nm3 := nat(nil).sub(nm1, natTwo) + rand := rand.New(rand.NewSource(int64(n[0]))) + + var x, y, quotient nat + nm3Len := nm3.bitLen() + +NextRandom: + for i := 0; i < reps; i++ { + if i == reps-1 && force2 { + x = x.set(natTwo) + } else { + x = x.random(rand, nm3, nm3Len) + x = x.add(x, natTwo) + } + y = y.expNN(x, q, n) + if y.cmp(natOne) == 0 || y.cmp(nm1) == 0 { + continue + } + for j := uint(1); j < k; j++ { + y = y.mul(y, y) + quotient, y = quotient.div(y, y, n) + if y.cmp(nm1) == 0 { + continue NextRandom + } + if y.cmp(natOne) == 0 { + return false + } + } + return false + } + + return true +} + +// probablyPrimeLucas reports whether n passes the "almost extra strong" Lucas probable prime test, +// using Baillie-OEIS parameter selection. This corresponds to "AESLPSP" on Jacobsen's tables (link below). +// The combination of this test and a Miller-Rabin/Fermat test with base 2 gives a Baillie-PSW test. +// +// References: +// +// Baillie and Wagstaff, "Lucas Pseudoprimes", Mathematics of Computation 35(152), +// October 1980, pp. 1391-1417, especially page 1401. +// http://www.ams.org/journals/mcom/1980-35-152/S0025-5718-1980-0583518-6/S0025-5718-1980-0583518-6.pdf +// +// Grantham, "Frobenius Pseudoprimes", Mathematics of Computation 70(234), +// March 2000, pp. 873-891. +// http://www.ams.org/journals/mcom/2001-70-234/S0025-5718-00-01197-2/S0025-5718-00-01197-2.pdf +// +// Baillie, "Extra strong Lucas pseudoprimes", OEIS A217719, https://oeis.org/A217719. +// +// Jacobsen, "Pseudoprime Statistics, Tables, and Data", http://ntheory.org/pseudoprimes.html. +// +// Nicely, "The Baillie-PSW Primality Test", http://www.trnicely.net/misc/bpsw.html. +// (Note that Nicely's definition of the "extra strong" test gives the wrong Jacobi condition, +// as pointed out by Jacobsen.) +// +// Crandall and Pomerance, Prime Numbers: A Computational Perspective, 2nd ed. +// Springer, 2005. +func (n nat) probablyPrimeLucas() bool { + // Discard 0, 1. + if len(n) == 0 || n.cmp(natOne) == 0 { + return false + } + // Two is the only even prime. + // Already checked by caller, but here to allow testing in isolation. + if n[0]&1 == 0 { + return n.cmp(natTwo) == 0 + } + + // Baillie-OEIS "method C" for choosing D, P, Q, + // as in https://oeis.org/A217719/a217719.txt: + // try increasing P ≥ 3 such that D = P² - 4 (so Q = 1) + // until Jacobi(D, n) = -1. + // The search is expected to succeed for non-square n after just a few trials. + // After more than expected failures, check whether n is square + // (which would cause Jacobi(D, n) = 1 for all D not dividing n). + p := Word(3) + d := nat{1} + t1 := nat(nil) // temp + intD := &Int{abs: d} + intN := &Int{abs: n} + for ; ; p++ { + if p > 10000 { + // This is widely believed to be impossible. + // If we get a report, we'll want the exact number n. + panic("math/big: internal error: cannot find (D/n) = -1 for " + intN.String()) + } + d[0] = p*p - 4 + j := Jacobi(intD, intN) + if j == -1 { + break + } + if j == 0 { + // d = p²-4 = (p-2)(p+2). + // If (d/n) == 0 then d shares a prime factor with n. + // Since the loop proceeds in increasing p and starts with p-2==1, + // the shared prime factor must be p+2. + // If p+2 == n, then n is prime; otherwise p+2 is a proper factor of n. + return len(n) == 1 && n[0] == p+2 + } + if p == 40 { + // We'll never find (d/n) = -1 if n is a square. + // If n is a non-square we expect to find a d in just a few attempts on average. + // After 40 attempts, take a moment to check if n is indeed a square. + t1 = t1.sqrt(n) + t1 = t1.mul(t1, t1) + if t1.cmp(n) == 0 { + return false + } + } + } + + // Grantham definition of "extra strong Lucas pseudoprime", after Thm 2.3 on p. 876 + // (D, P, Q above have become Δ, b, 1): + // + // Let U_n = U_n(b, 1), V_n = V_n(b, 1), and Δ = b²-4. + // An extra strong Lucas pseudoprime to base b is a composite n = 2^r s + Jacobi(Δ, n), + // where s is odd and gcd(n, 2*Δ) = 1, such that either (i) U_s ≡ 0 mod n and V_s ≡ ±2 mod n, + // or (ii) V_{2^t s} ≡ 0 mod n for some 0 ≤ t < r-1. + // + // We know gcd(n, Δ) = 1 or else we'd have found Jacobi(d, n) == 0 above. + // We know gcd(n, 2) = 1 because n is odd. + // + // Arrange s = (n - Jacobi(Δ, n)) / 2^r = (n+1) / 2^r. + s := nat(nil).add(n, natOne) + r := int(s.trailingZeroBits()) + s = s.shr(s, uint(r)) + nm2 := nat(nil).sub(n, natTwo) // n-2 + + // We apply the "almost extra strong" test, which checks the above conditions + // except for U_s ≡ 0 mod n, which allows us to avoid computing any U_k values. + // Jacobsen points out that maybe we should just do the full extra strong test: + // "It is also possible to recover U_n using Crandall and Pomerance equation 3.13: + // U_n = D^-1 (2V_{n+1} - PV_n) allowing us to run the full extra-strong test + // at the cost of a single modular inversion. This computation is easy and fast in GMP, + // so we can get the full extra-strong test at essentially the same performance as the + // almost extra strong test." + + // Compute Lucas sequence V_s(b, 1), where: + // + // V(0) = 2 + // V(1) = P + // V(k) = P V(k-1) - Q V(k-2). + // + // (Remember that due to method C above, P = b, Q = 1.) + // + // In general V(k) = α^k + β^k, where α and β are roots of x² - Px + Q. + // Crandall and Pomerance (p.147) observe that for 0 ≤ j ≤ k, + // + // V(j+k) = V(j)V(k) - V(k-j). + // + // So in particular, to quickly double the subscript: + // + // V(2k) = V(k)² - 2 + // V(2k+1) = V(k) V(k+1) - P + // + // We can therefore start with k=0 and build up to k=s in log₂(s) steps. + natP := nat(nil).setWord(p) + vk := nat(nil).setWord(2) + vk1 := nat(nil).setWord(p) + t2 := nat(nil) // temp + for i := int(s.bitLen()); i >= 0; i-- { + if s.bit(uint(i)) != 0 { + // k' = 2k+1 + // V(k') = V(2k+1) = V(k) V(k+1) - P. + t1 = t1.mul(vk, vk1) + t1 = t1.add(t1, n) + t1 = t1.sub(t1, natP) + t2, vk = t2.div(vk, t1, n) + // V(k'+1) = V(2k+2) = V(k+1)² - 2. + t1 = t1.mul(vk1, vk1) + t1 = t1.add(t1, nm2) + t2, vk1 = t2.div(vk1, t1, n) + } else { + // k' = 2k + // V(k'+1) = V(2k+1) = V(k) V(k+1) - P. + t1 = t1.mul(vk, vk1) + t1 = t1.add(t1, n) + t1 = t1.sub(t1, natP) + t2, vk1 = t2.div(vk1, t1, n) + // V(k') = V(2k) = V(k)² - 2 + t1 = t1.mul(vk, vk) + t1 = t1.add(t1, nm2) + t2, vk = t2.div(vk, t1, n) + } + } + + // Now k=s, so vk = V(s). Check V(s) ≡ ±2 (mod n). + if vk.cmp(natTwo) == 0 || vk.cmp(nm2) == 0 { + // Check U(s) ≡ 0. + // As suggested by Jacobsen, apply Crandall and Pomerance equation 3.13: + // + // U(k) = D⁻¹ (2 V(k+1) - P V(k)) + // + // Since we are checking for U(k) == 0 it suffices to check 2 V(k+1) == P V(k) mod n, + // or P V(k) - 2 V(k+1) == 0 mod n. + t1 := t1.mul(vk, natP) + t2 := t2.shl(vk1, 1) + if t1.cmp(t2) < 0 { + t1, t2 = t2, t1 + } + t1 = t1.sub(t1, t2) + t3 := vk1 // steal vk1, no longer needed below + vk1 = nil + _ = vk1 + t2, t3 = t2.div(t3, t1, n) + if len(t3) == 0 { + return true + } + } + + // Check V(2^t s) ≡ 0 mod n for some 0 ≤ t < r-1. + for t := 0; t < r-1; t++ { + if len(vk) == 0 { // vk == 0 + return true + } + // Optimization: V(k) = 2 is a fixed point for V(k') = V(k)² - 2, + // so if V(k) = 2, we can stop: we will never find a future V(k) == 0. + if len(vk) == 1 && vk[0] == 2 { // vk == 2 + return false + } + // k' = 2k + // V(k') = V(2k) = V(k)² - 2 + t1 = t1.mul(vk, vk) + t1 = t1.sub(t1, natTwo) + t2, vk = t2.div(vk, t1, n) + } + return false +} |