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

```     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[0]
49		if len(x.abs) == 1 && w < 64 {
51		}
52
53		if w&1 == 0 {
54			return false // n 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)
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.mul(y, 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	// http://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	// http://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[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.mul(t1, 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.mul(vk1, 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.mul(vk, 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[0] == 2 { // vk == 2
311				return false
312			}
313			// k' = 2k
314			// V(k') = V(2k) = V(k)² - 2
315			t1 = t1.mul(vk, 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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