mirror of https://go.googlesource.com/go
321 lines
10 KiB
Go
321 lines
10 KiB
Go
// Copyright 2016 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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package big
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import "math/rand"
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// ProbablyPrime reports whether x is probably prime,
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// applying the Miller-Rabin test with n pseudorandomly chosen bases
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// as well as a Baillie-PSW test.
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//
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// If x is prime, ProbablyPrime returns true.
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// If x is chosen randomly and not prime, ProbablyPrime probably returns false.
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// The probability of returning true for a randomly chosen non-prime is at most ¼ⁿ.
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//
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// ProbablyPrime is 100% accurate for inputs less than 2⁶⁴.
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// See Menezes et al., Handbook of Applied Cryptography, 1997, pp. 145-149,
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// and FIPS 186-4 Appendix F for further discussion of the error probabilities.
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//
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// ProbablyPrime is not suitable for judging primes that an adversary may
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// have crafted to fool the test.
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//
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// As of Go 1.8, ProbablyPrime(0) is allowed and applies only a Baillie-PSW test.
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// Before Go 1.8, ProbablyPrime applied only the Miller-Rabin tests, and ProbablyPrime(0) panicked.
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func (x *Int) ProbablyPrime(n int) bool {
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// Note regarding the doc comment above:
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// It would be more precise to say that the Baillie-PSW test uses the
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// extra strong Lucas test as its Lucas test, but since no one knows
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// how to tell any of the Lucas tests apart inside a Baillie-PSW test
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// (they all work equally well empirically), that detail need not be
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// documented or implicitly guaranteed.
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// The comment does avoid saying "the" Baillie-PSW test
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// because of this general ambiguity.
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if n < 0 {
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panic("negative n for ProbablyPrime")
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}
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if x.neg || len(x.abs) == 0 {
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return false
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}
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// primeBitMask records the primes < 64.
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const primeBitMask uint64 = 1<<2 | 1<<3 | 1<<5 | 1<<7 |
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1<<11 | 1<<13 | 1<<17 | 1<<19 | 1<<23 | 1<<29 | 1<<31 |
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1<<37 | 1<<41 | 1<<43 | 1<<47 | 1<<53 | 1<<59 | 1<<61
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w := x.abs[0]
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if len(x.abs) == 1 && w < 64 {
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return primeBitMask&(1<<w) != 0
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}
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if w&1 == 0 {
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return false // x is even
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}
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const primesA = 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 37
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const primesB = 29 * 31 * 41 * 43 * 47 * 53
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var rA, rB uint32
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switch _W {
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case 32:
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rA = uint32(x.abs.modW(primesA))
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rB = uint32(x.abs.modW(primesB))
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case 64:
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r := x.abs.modW((primesA * primesB) & _M)
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rA = uint32(r % primesA)
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rB = uint32(r % primesB)
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default:
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panic("math/big: invalid word size")
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}
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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 ||
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rB%29 == 0 || rB%31 == 0 || rB%41 == 0 || rB%43 == 0 || rB%47 == 0 || rB%53 == 0 {
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return false
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}
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return x.abs.probablyPrimeMillerRabin(n+1, true) && x.abs.probablyPrimeLucas()
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}
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// probablyPrimeMillerRabin reports whether n passes reps rounds of the
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// Miller-Rabin primality test, using pseudo-randomly chosen bases.
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// If force2 is true, one of the rounds is forced to use base 2.
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// See Handbook of Applied Cryptography, p. 139, Algorithm 4.24.
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// The number n is known to be non-zero.
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func (n nat) probablyPrimeMillerRabin(reps int, force2 bool) bool {
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nm1 := nat(nil).sub(n, natOne)
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// determine q, k such that nm1 = q << k
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k := nm1.trailingZeroBits()
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q := nat(nil).shr(nm1, k)
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nm3 := nat(nil).sub(nm1, natTwo)
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rand := rand.New(rand.NewSource(int64(n[0])))
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var x, y, quotient nat
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nm3Len := nm3.bitLen()
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NextRandom:
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for i := 0; i < reps; i++ {
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if i == reps-1 && force2 {
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x = x.set(natTwo)
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} else {
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x = x.random(rand, nm3, nm3Len)
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x = x.add(x, natTwo)
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}
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y = y.expNN(x, q, n, false)
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if y.cmp(natOne) == 0 || y.cmp(nm1) == 0 {
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continue
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}
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for j := uint(1); j < k; j++ {
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y = y.sqr(y)
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quotient, y = quotient.div(y, y, n)
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if y.cmp(nm1) == 0 {
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continue NextRandom
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}
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if y.cmp(natOne) == 0 {
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return false
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}
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}
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return false
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}
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return true
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}
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// probablyPrimeLucas reports whether n passes the "almost extra strong" Lucas probable prime test,
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// using Baillie-OEIS parameter selection. This corresponds to "AESLPSP" on Jacobsen's tables (link below).
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// The combination of this test and a Miller-Rabin/Fermat test with base 2 gives a Baillie-PSW test.
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//
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// References:
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//
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// Baillie and Wagstaff, "Lucas Pseudoprimes", Mathematics of Computation 35(152),
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// October 1980, pp. 1391-1417, especially page 1401.
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// https://www.ams.org/journals/mcom/1980-35-152/S0025-5718-1980-0583518-6/S0025-5718-1980-0583518-6.pdf
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//
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// Grantham, "Frobenius Pseudoprimes", Mathematics of Computation 70(234),
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// March 2000, pp. 873-891.
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// https://www.ams.org/journals/mcom/2001-70-234/S0025-5718-00-01197-2/S0025-5718-00-01197-2.pdf
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//
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// Baillie, "Extra strong Lucas pseudoprimes", OEIS A217719, https://oeis.org/A217719.
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//
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// Jacobsen, "Pseudoprime Statistics, Tables, and Data", http://ntheory.org/pseudoprimes.html.
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//
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// Nicely, "The Baillie-PSW Primality Test", https://web.archive.org/web/20191121062007/http://www.trnicely.net/misc/bpsw.html.
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// (Note that Nicely's definition of the "extra strong" test gives the wrong Jacobi condition,
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// as pointed out by Jacobsen.)
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//
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// Crandall and Pomerance, Prime Numbers: A Computational Perspective, 2nd ed.
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// Springer, 2005.
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func (n nat) probablyPrimeLucas() bool {
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// Discard 0, 1.
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if len(n) == 0 || n.cmp(natOne) == 0 {
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return false
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}
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// Two is the only even prime.
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// Already checked by caller, but here to allow testing in isolation.
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if n[0]&1 == 0 {
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return n.cmp(natTwo) == 0
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}
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// Baillie-OEIS "method C" for choosing D, P, Q,
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// as in https://oeis.org/A217719/a217719.txt:
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// try increasing P ≥ 3 such that D = P² - 4 (so Q = 1)
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// until Jacobi(D, n) = -1.
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// The search is expected to succeed for non-square n after just a few trials.
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// After more than expected failures, check whether n is square
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// (which would cause Jacobi(D, n) = 1 for all D not dividing n).
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p := Word(3)
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d := nat{1}
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t1 := nat(nil) // temp
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intD := &Int{abs: d}
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intN := &Int{abs: n}
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for ; ; p++ {
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if p > 10000 {
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// This is widely believed to be impossible.
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// If we get a report, we'll want the exact number n.
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panic("math/big: internal error: cannot find (D/n) = -1 for " + intN.String())
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}
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d[0] = p*p - 4
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j := Jacobi(intD, intN)
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if j == -1 {
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break
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}
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if j == 0 {
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// d = p²-4 = (p-2)(p+2).
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// If (d/n) == 0 then d shares a prime factor with n.
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// Since the loop proceeds in increasing p and starts with p-2==1,
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// the shared prime factor must be p+2.
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// If p+2 == n, then n is prime; otherwise p+2 is a proper factor of n.
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return len(n) == 1 && n[0] == p+2
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}
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if p == 40 {
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// We'll never find (d/n) = -1 if n is a square.
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// If n is a non-square we expect to find a d in just a few attempts on average.
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// After 40 attempts, take a moment to check if n is indeed a square.
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t1 = t1.sqrt(n)
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t1 = t1.sqr(t1)
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if t1.cmp(n) == 0 {
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return false
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}
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}
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}
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// Grantham definition of "extra strong Lucas pseudoprime", after Thm 2.3 on p. 876
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// (D, P, Q above have become Δ, b, 1):
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//
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// Let U_n = U_n(b, 1), V_n = V_n(b, 1), and Δ = b²-4.
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// An extra strong Lucas pseudoprime to base b is a composite n = 2^r s + Jacobi(Δ, n),
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// 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,
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// or (ii) V_{2^t s} ≡ 0 mod n for some 0 ≤ t < r-1.
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//
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// We know gcd(n, Δ) = 1 or else we'd have found Jacobi(d, n) == 0 above.
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// We know gcd(n, 2) = 1 because n is odd.
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//
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// Arrange s = (n - Jacobi(Δ, n)) / 2^r = (n+1) / 2^r.
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s := nat(nil).add(n, natOne)
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r := int(s.trailingZeroBits())
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s = s.shr(s, uint(r))
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nm2 := nat(nil).sub(n, natTwo) // n-2
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// We apply the "almost extra strong" test, which checks the above conditions
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// except for U_s ≡ 0 mod n, which allows us to avoid computing any U_k values.
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// Jacobsen points out that maybe we should just do the full extra strong test:
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// "It is also possible to recover U_n using Crandall and Pomerance equation 3.13:
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// U_n = D^-1 (2V_{n+1} - PV_n) allowing us to run the full extra-strong test
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// at the cost of a single modular inversion. This computation is easy and fast in GMP,
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// so we can get the full extra-strong test at essentially the same performance as the
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// almost extra strong test."
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// Compute Lucas sequence V_s(b, 1), where:
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//
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// V(0) = 2
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// V(1) = P
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// V(k) = P V(k-1) - Q V(k-2).
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//
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// (Remember that due to method C above, P = b, Q = 1.)
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//
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// In general V(k) = α^k + β^k, where α and β are roots of x² - Px + Q.
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// Crandall and Pomerance (p.147) observe that for 0 ≤ j ≤ k,
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//
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// V(j+k) = V(j)V(k) - V(k-j).
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//
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// So in particular, to quickly double the subscript:
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//
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// V(2k) = V(k)² - 2
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// V(2k+1) = V(k) V(k+1) - P
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//
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// We can therefore start with k=0 and build up to k=s in log₂(s) steps.
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natP := nat(nil).setWord(p)
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vk := nat(nil).setWord(2)
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vk1 := nat(nil).setWord(p)
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t2 := nat(nil) // temp
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for i := int(s.bitLen()); i >= 0; i-- {
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if s.bit(uint(i)) != 0 {
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// k' = 2k+1
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// V(k') = V(2k+1) = V(k) V(k+1) - P.
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t1 = t1.mul(vk, vk1)
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t1 = t1.add(t1, n)
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t1 = t1.sub(t1, natP)
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t2, vk = t2.div(vk, t1, n)
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// V(k'+1) = V(2k+2) = V(k+1)² - 2.
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t1 = t1.sqr(vk1)
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t1 = t1.add(t1, nm2)
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t2, vk1 = t2.div(vk1, t1, n)
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} else {
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// k' = 2k
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// V(k'+1) = V(2k+1) = V(k) V(k+1) - P.
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t1 = t1.mul(vk, vk1)
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t1 = t1.add(t1, n)
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t1 = t1.sub(t1, natP)
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t2, vk1 = t2.div(vk1, t1, n)
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// V(k') = V(2k) = V(k)² - 2
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t1 = t1.sqr(vk)
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t1 = t1.add(t1, nm2)
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t2, vk = t2.div(vk, t1, n)
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}
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}
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// Now k=s, so vk = V(s). Check V(s) ≡ ±2 (mod n).
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if vk.cmp(natTwo) == 0 || vk.cmp(nm2) == 0 {
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// Check U(s) ≡ 0.
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// As suggested by Jacobsen, apply Crandall and Pomerance equation 3.13:
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//
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// U(k) = D⁻¹ (2 V(k+1) - P V(k))
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//
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// Since we are checking for U(k) == 0 it suffices to check 2 V(k+1) == P V(k) mod n,
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// or P V(k) - 2 V(k+1) == 0 mod n.
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t1 := t1.mul(vk, natP)
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t2 := t2.shl(vk1, 1)
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if t1.cmp(t2) < 0 {
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t1, t2 = t2, t1
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}
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t1 = t1.sub(t1, t2)
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t3 := vk1 // steal vk1, no longer needed below
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vk1 = nil
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_ = vk1
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t2, t3 = t2.div(t3, t1, n)
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if len(t3) == 0 {
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return true
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}
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}
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// Check V(2^t s) ≡ 0 mod n for some 0 ≤ t < r-1.
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for t := 0; t < r-1; t++ {
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if len(vk) == 0 { // vk == 0
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return true
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}
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// Optimization: V(k) = 2 is a fixed point for V(k') = V(k)² - 2,
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// so if V(k) = 2, we can stop: we will never find a future V(k) == 0.
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if len(vk) == 1 && vk[0] == 2 { // vk == 2
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return false
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}
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// k' = 2k
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// V(k') = V(2k) = V(k)² - 2
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t1 = t1.sqr(vk)
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t1 = t1.sub(t1, natTwo)
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t2, vk = t2.div(vk, t1, n)
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}
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return false
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}
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