| // 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 // x 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, false) |
| if y.cmp(natOne) == 0 || y.cmp(nm1) == 0 { |
| continue |
| } |
| for j := uint(1); j < k; j++ { |
| y = y.sqr(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. |
| // https://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. |
| // https://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", https://web.archive.org/web/20191121062007/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.sqr(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.sqr(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.sqr(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.sqr(vk) |
| t1 = t1.sub(t1, natTwo) |
| t2, vk = t2.div(vk, t1, n) |
| } |
| return false |
| } |