# Source file src/math/big/prime.go

## Documentation: math/big

```     1  // Copyright 2016 The Go Authors. All rights reserved.
2  // Use of this source code is governed by a BSD-style
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
49  	if len(x.abs) == 1 && w < 64 {
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)))
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)
105  		}
106  		y = y.expNN(x, q, n)
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", 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 {
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&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 = 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 == 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.
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)
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)
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)
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)
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 == 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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