Update prebuilts to go1.8rc1 ab/3640477
Bug: 32982374
Test: m -j blueprint_tools
Change-Id: Ife36ed9cf5b2617ccc8fdf0fdd178d19378546cd
diff --git a/src/math/big/prime.go b/src/math/big/prime.go
new file mode 100644
index 0000000..3e9690e
--- /dev/null
+++ b/src/math/big/prime.go
@@ -0,0 +1,320 @@
+// 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 // n 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)
+ if y.cmp(natOne) == 0 || y.cmp(nm1) == 0 {
+ continue
+ }
+ for j := uint(1); j < k; j++ {
+ y = y.mul(y, 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.
+// http://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.
+// http://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", 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.mul(t1, 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.mul(vk1, 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.mul(vk, 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.mul(vk, vk)
+ t1 = t1.sub(t1, natTwo)
+ t2, vk = t2.div(vk, t1, n)
+ }
+ return false
+}
