prime.go

     1  // Copyright 2016 The Go Authors. All rights reserved.
     2  // Use of this source code is governed by a BSD-style
     3  // license that can be found in the LICENSE file.
     4  
     5  package big
     6  
     7  import "math/rand"
     8  
     9  // ProbablyPrime reports whether x is probably prime,
    10  // applying the Miller-Rabin test with n pseudorandomly chosen bases
    11  // as well as a Baillie-PSW test.
    12  //
    13  // If x is prime, ProbablyPrime returns true.
    14  // If x is chosen randomly and not prime, ProbablyPrime probably returns false.
    15  // The probability of returning true for a randomly chosen non-prime is at most ¼ⁿ.
    16  //
    17  // ProbablyPrime is 100% accurate for inputs less than 2⁶⁴.
    18  // See Menezes et al., Handbook of Applied Cryptography, 1997, pp. 145-149,
    19  // and FIPS 186-4 Appendix F for further discussion of the error probabilities.
    20  //
    21  // ProbablyPrime is not suitable for judging primes that an adversary may
    22  // have crafted to fool the test.
    23  //
    24  // As of Go 1.8, ProbablyPrime(0) is allowed and applies only a Baillie-PSW test.
    25  // Before Go 1.8, ProbablyPrime applied only the Miller-Rabin tests, and ProbablyPrime(0) panicked.
    26  func (x *Int) ProbablyPrime(n int) bool {
    27  	// Note regarding the doc comment above:
    28  	// It would be more precise to say that the Baillie-PSW test uses the
    29  	// extra strong Lucas test as its Lucas test, but since no one knows
    30  	// how to tell any of the Lucas tests apart inside a Baillie-PSW test
    31  	// (they all work equally well empirically), that detail need not be
    32  	// documented or implicitly guaranteed.
    33  	// The comment does avoid saying "the" Baillie-PSW test
    34  	// because of this general ambiguity.
    35  
    36  	if n < 0 {
    37  		panic("negative n for ProbablyPrime")
    38  	}
    39  	if x.neg || len(x.abs) == 0 {
    40  		return false
    41  	}
    42  
    43  	// primeBitMask records the primes < 64.
    44  	const primeBitMask uint64 = 1<<2 | 1<<3 | 1<<5 | 1<<7 |
    45  		1<<11 | 1<<13 | 1<<17 | 1<<19 | 1<<23 | 1<<29 | 1<<31 |
    46  		1<<37 | 1<<41 | 1<<43 | 1<<47 | 1<<53 | 1<<59 | 1<<61
    47  
    48  	w := x.abs[0]
    49  	if len(x.abs) == 1 && w < 64 {
    50  		return primeBitMask&(1<<w) != 0
    51  	}
    52  
    53  	if w&1 == 0 {
    54  		return false // x is even
    55  	}
    56  
    57  	const primesA = 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 37
    58  	const primesB = 29 * 31 * 41 * 43 * 47 * 53
    59  
    60  	var rA, rB uint32
    61  	switch _W {
    62  	case 32:
    63  		rA = uint32(x.abs.modW(primesA))
    64  		rB = uint32(x.abs.modW(primesB))
    65  	case 64:
    66  		r := x.abs.modW((primesA * primesB) & _M)
    67  		rA = uint32(r % primesA)
    68  		rB = uint32(r % primesB)
    69  	default:
    70  		panic("math/big: invalid word size")
    71  	}
    72  
    73  	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 ||
    74  		rB%29 == 0 || rB%31 == 0 || rB%41 == 0 || rB%43 == 0 || rB%47 == 0 || rB%53 == 0 {
    75  		return false
    76  	}
    77  
    78  	return x.abs.probablyPrimeMillerRabin(n+1, true) && x.abs.probablyPrimeLucas()
    79  }
    80  
    81  // probablyPrimeMillerRabin reports whether n passes reps rounds of the
    82  // Miller-Rabin primality test, using pseudo-randomly chosen bases.
    83  // If force2 is true, one of the rounds is forced to use base 2.
    84  // See Handbook of Applied Cryptography, p. 139, Algorithm 4.24.
    85  // The number n is known to be non-zero.
    86  func (n nat) probablyPrimeMillerRabin(reps int, force2 bool) bool {
    87  	nm1 := nat(nil).sub(n, natOne)
    88  	// determine q, k such that nm1 = q << k
    89  	k := nm1.trailingZeroBits()
    90  	q := nat(nil).shr(nm1, k)
    91  
    92  	nm3 := nat(nil).sub(nm1, natTwo)
    93  	rand := rand.New(rand.NewSource(int64(n[0])))
    94  
    95  	var x, y, quotient nat
    96  	nm3Len := nm3.bitLen()
    97  
    98  NextRandom:
    99  	for i := 0; i < reps; i++ {
   100  		if i == reps-1 && force2 {
   101  			x = x.set(natTwo)
   102  		} else {
   103  			x = x.random(rand, nm3, nm3Len)
   104  			x = x.add(x, natTwo)
   105  		}
   106  		y = y.expNN(x, q, n, false)
   107  		if y.cmp(natOne) == 0 || y.cmp(nm1) == 0 {
   108  			continue
   109  		}
   110  		for j := uint(1); j < k; j++ {
   111  			y = y.sqr(y)
   112  			quotient, y = quotient.div(y, y, n)
   113  			if y.cmp(nm1) == 0 {
   114  				continue NextRandom
   115  			}
   116  			if y.cmp(natOne) == 0 {
   117  				return false
   118  			}
   119  		}
   120  		return false
   121  	}
   122  
   123  	return true
   124  }
   125  
   126  // probablyPrimeLucas reports whether n passes the "almost extra strong" Lucas probable prime test,
   127  // using Baillie-OEIS parameter selection. This corresponds to "AESLPSP" on Jacobsen's tables (link below).
   128  // The combination of this test and a Miller-Rabin/Fermat test with base 2 gives a Baillie-PSW test.
   129  //
   130  // References:
   131  //
   132  // Baillie and Wagstaff, "Lucas Pseudoprimes", Mathematics of Computation 35(152),
   133  // October 1980, pp. 1391-1417, especially page 1401.
   134  // https://www.ams.org/journals/mcom/1980-35-152/S0025-5718-1980-0583518-6/S0025-5718-1980-0583518-6.pdf
   135  //
   136  // Grantham, "Frobenius Pseudoprimes", Mathematics of Computation 70(234),
   137  // March 2000, pp. 873-891.
   138  // https://www.ams.org/journals/mcom/2001-70-234/S0025-5718-00-01197-2/S0025-5718-00-01197-2.pdf
   139  //
   140  // Baillie, "Extra strong Lucas pseudoprimes", OEIS A217719, https://oeis.org/A217719.
   141  //
   142  // Jacobsen, "Pseudoprime Statistics, Tables, and Data", http://ntheory.org/pseudoprimes.html.
   143  //
   144  // Nicely, "The Baillie-PSW Primality Test", https://web.archive.org/web/20191121062007/http://www.trnicely.net/misc/bpsw.html.
   145  // (Note that Nicely's definition of the "extra strong" test gives the wrong Jacobi condition,
   146  // as pointed out by Jacobsen.)
   147  //
   148  // Crandall and Pomerance, Prime Numbers: A Computational Perspective, 2nd ed.
   149  // Springer, 2005.
   150  func (n nat) probablyPrimeLucas() bool {
   151  	// Discard 0, 1.
   152  	if len(n) == 0 || n.cmp(natOne) == 0 {
   153  		return false
   154  	}
   155  	// Two is the only even prime.
   156  	// Already checked by caller, but here to allow testing in isolation.
   157  	if n[0]&1 == 0 {
   158  		return n.cmp(natTwo) == 0
   159  	}
   160  
   161  	// Baillie-OEIS "method C" for choosing D, P, Q,
   162  	// as in https://oeis.org/A217719/a217719.txt:
   163  	// try increasing P ≥ 3 such that D = P² - 4 (so Q = 1)
   164  	// until Jacobi(D, n) = -1.
   165  	// The search is expected to succeed for non-square n after just a few trials.
   166  	// After more than expected failures, check whether n is square
   167  	// (which would cause Jacobi(D, n) = 1 for all D not dividing n).
   168  	p := Word(3)
   169  	d := nat{1}
   170  	t1 := nat(nil) // temp
   171  	intD := &Int{abs: d}
   172  	intN := &Int{abs: n}
   173  	for ; ; p++ {
   174  		if p > 10000 {
   175  			// This is widely believed to be impossible.
   176  			// If we get a report, we'll want the exact number n.
   177  			panic("math/big: internal error: cannot find (D/n) = -1 for " + intN.String())
   178  		}
   179  		d[0] = p*p - 4
   180  		j := Jacobi(intD, intN)
   181  		if j == -1 {
   182  			break
   183  		}
   184  		if j == 0 {
   185  			// d = p²-4 = (p-2)(p+2).
   186  			// If (d/n) == 0 then d shares a prime factor with n.
   187  			// Since the loop proceeds in increasing p and starts with p-2==1,
   188  			// the shared prime factor must be p+2.
   189  			// If p+2 == n, then n is prime; otherwise p+2 is a proper factor of n.
   190  			return len(n) == 1 && n[0] == p+2
   191  		}
   192  		if p == 40 {
   193  			// We'll never find (d/n) = -1 if n is a square.
   194  			// If n is a non-square we expect to find a d in just a few attempts on average.
   195  			// After 40 attempts, take a moment to check if n is indeed a square.
   196  			t1 = t1.sqrt(n)
   197  			t1 = t1.sqr(t1)
   198  			if t1.cmp(n) == 0 {
   199  				return false
   200  			}
   201  		}
   202  	}
   203  
   204  	// Grantham definition of "extra strong Lucas pseudoprime", after Thm 2.3 on p. 876
   205  	// (D, P, Q above have become Δ, b, 1):
   206  	//
   207  	// Let U_n = U_n(b, 1), V_n = V_n(b, 1), and Δ = b²-4.
   208  	// An extra strong Lucas pseudoprime to base b is a composite n = 2^r s + Jacobi(Δ, n),
   209  	// 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,
   210  	// or (ii) V_{2^t s} ≡ 0 mod n for some 0 ≤ t < r-1.
   211  	//
   212  	// We know gcd(n, Δ) = 1 or else we'd have found Jacobi(d, n) == 0 above.
   213  	// We know gcd(n, 2) = 1 because n is odd.
   214  	//
   215  	// Arrange s = (n - Jacobi(Δ, n)) / 2^r = (n+1) / 2^r.
   216  	s := nat(nil).add(n, natOne)
   217  	r := int(s.trailingZeroBits())
   218  	s = s.shr(s, uint(r))
   219  	nm2 := nat(nil).sub(n, natTwo) // n-2
   220  
   221  	// We apply the "almost extra strong" test, which checks the above conditions
   222  	// except for U_s ≡ 0 mod n, which allows us to avoid computing any U_k values.
   223  	// Jacobsen points out that maybe we should just do the full extra strong test:
   224  	// "It is also possible to recover U_n using Crandall and Pomerance equation 3.13:
   225  	// U_n = D^-1 (2V_{n+1} - PV_n) allowing us to run the full extra-strong test
   226  	// at the cost of a single modular inversion. This computation is easy and fast in GMP,
   227  	// so we can get the full extra-strong test at essentially the same performance as the
   228  	// almost extra strong test."
   229  
   230  	// Compute Lucas sequence V_s(b, 1), where:
   231  	//
   232  	//	V(0) = 2
   233  	//	V(1) = P
   234  	//	V(k) = P V(k-1) - Q V(k-2).
   235  	//
   236  	// (Remember that due to method C above, P = b, Q = 1.)
   237  	//
   238  	// In general V(k) = α^k + β^k, where α and β are roots of x² - Px + Q.
   239  	// Crandall and Pomerance (p.147) observe that for 0 ≤ j ≤ k,
   240  	//
   241  	//	V(j+k) = V(j)V(k) - V(k-j).
   242  	//
   243  	// So in particular, to quickly double the subscript:
   244  	//
   245  	//	V(2k) = V(k)² - 2
   246  	//	V(2k+1) = V(k) V(k+1) - P
   247  	//
   248  	// We can therefore start with k=0 and build up to k=s in log₂(s) steps.
   249  	natP := nat(nil).setWord(p)
   250  	vk := nat(nil).setWord(2)
   251  	vk1 := nat(nil).setWord(p)
   252  	t2 := nat(nil) // temp
   253  	for i := int(s.bitLen()); i >= 0; i-- {
   254  		if s.bit(uint(i)) != 0 {
   255  			// k' = 2k+1
   256  			// V(k') = V(2k+1) = V(k) V(k+1) - P.
   257  			t1 = t1.mul(vk, vk1)
   258  			t1 = t1.add(t1, n)
   259  			t1 = t1.sub(t1, natP)
   260  			t2, vk = t2.div(vk, t1, n)
   261  			// V(k'+1) = V(2k+2) = V(k+1)² - 2.
   262  			t1 = t1.sqr(vk1)
   263  			t1 = t1.add(t1, nm2)
   264  			t2, vk1 = t2.div(vk1, t1, n)
   265  		} else {
   266  			// k' = 2k
   267  			// V(k'+1) = V(2k+1) = V(k) V(k+1) - P.
   268  			t1 = t1.mul(vk, vk1)
   269  			t1 = t1.add(t1, n)
   270  			t1 = t1.sub(t1, natP)
   271  			t2, vk1 = t2.div(vk1, t1, n)
   272  			// V(k') = V(2k) = V(k)² - 2
   273  			t1 = t1.sqr(vk)
   274  			t1 = t1.add(t1, nm2)
   275  			t2, vk = t2.div(vk, t1, n)
   276  		}
   277  	}
   278  
   279  	// Now k=s, so vk = V(s). Check V(s) ≡ ±2 (mod n).
   280  	if vk.cmp(natTwo) == 0 || vk.cmp(nm2) == 0 {
   281  		// Check U(s) ≡ 0.
   282  		// As suggested by Jacobsen, apply Crandall and Pomerance equation 3.13:
   283  		//
   284  		//	U(k) = D⁻¹ (2 V(k+1) - P V(k))
   285  		//
   286  		// Since we are checking for U(k) == 0 it suffices to check 2 V(k+1) == P V(k) mod n,
   287  		// or P V(k) - 2 V(k+1) == 0 mod n.
   288  		t1 := t1.mul(vk, natP)
   289  		t2 := t2.shl(vk1, 1)
   290  		if t1.cmp(t2) < 0 {
   291  			t1, t2 = t2, t1
   292  		}
   293  		t1 = t1.sub(t1, t2)
   294  		t3 := vk1 // steal vk1, no longer needed below
   295  		vk1 = nil
   296  		_ = vk1
   297  		t2, t3 = t2.div(t3, t1, n)
   298  		if len(t3) == 0 {
   299  			return true
   300  		}
   301  	}
   302  
   303  	// Check V(2^t s) ≡ 0 mod n for some 0 ≤ t < r-1.
   304  	for t := 0; t < r-1; t++ {
   305  		if len(vk) == 0 { // vk == 0
   306  			return true
   307  		}
   308  		// Optimization: V(k) = 2 is a fixed point for V(k') = V(k)² - 2,
   309  		// so if V(k) = 2, we can stop: we will never find a future V(k) == 0.
   310  		if len(vk) == 1 && vk[0] == 2 { // vk == 2
   311  			return false
   312  		}
   313  		// k' = 2k
   314  		// V(k') = V(2k) = V(k)² - 2
   315  		t1 = t1.sqr(vk)
   316  		t1 = t1.sub(t1, natTwo)
   317  		t2, vk = t2.div(vk, t1, n)
   318  	}
   319  	return false
   320  }
   321
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