| // Copyright 2009 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. |
| |
| // This file contains operations on unsigned multi-precision integers. |
| // These are the building blocks for the operations on signed integers |
| // and rationals. |
| |
| // This package implements multi-precision arithmetic (big numbers). |
| // The following numeric types are supported: |
| // |
| // - Int signed integers |
| // |
| // All methods on Int take the result as the receiver; if it is one |
| // of the operands it may be overwritten (and its memory reused). |
| // To enable chaining of operations, the result is also returned. |
| // |
| // If possible, one should use big over bignum as the latter is headed for |
| // deprecation. |
| // |
| package big |
| |
| import "rand" |
| |
| // An unsigned integer x of the form |
| // |
| // x = x[n-1]*_B^(n-1) + x[n-2]*_B^(n-2) + ... + x[1]*_B + x[0] |
| // |
| // with 0 <= x[i] < _B and 0 <= i < n is stored in a slice of length n, |
| // with the digits x[i] as the slice elements. |
| // |
| // A number is normalized if the slice contains no leading 0 digits. |
| // During arithmetic operations, denormalized values may occur but are |
| // always normalized before returning the final result. The normalized |
| // representation of 0 is the empty or nil slice (length = 0). |
| |
| // TODO(gri) - convert these routines into methods for type 'nat' |
| // - decide if type 'nat' should be exported |
| |
| func normN(z []Word) []Word { |
| i := len(z) |
| for i > 0 && z[i-1] == 0 { |
| i-- |
| } |
| z = z[0:i] |
| return z |
| } |
| |
| |
| func makeN(z []Word, m int, clear bool) []Word { |
| if len(z) > m { |
| z = z[0:m] // reuse z - has at least one extra word for a carry, if any |
| if clear { |
| for i := range z { |
| z[i] = 0 |
| } |
| } |
| return z |
| } |
| |
| c := 4 // minimum capacity |
| if m > c { |
| c = m |
| } |
| return make([]Word, m, c+1) // +1: extra word for a carry, if any |
| } |
| |
| |
| func newN(z []Word, x uint64) []Word { |
| if x == 0 { |
| return makeN(z, 0, false) |
| } |
| |
| // single-digit values |
| if x == uint64(Word(x)) { |
| z = makeN(z, 1, false) |
| z[0] = Word(x) |
| return z |
| } |
| |
| // compute number of words n required to represent x |
| n := 0 |
| for t := x; t > 0; t >>= _W { |
| n++ |
| } |
| |
| // split x into n words |
| z = makeN(z, n, false) |
| for i := 0; i < n; i++ { |
| z[i] = Word(x & _M) |
| x >>= _W |
| } |
| |
| return z |
| } |
| |
| |
| func setN(z, x []Word) []Word { |
| z = makeN(z, len(x), false) |
| for i, d := range x { |
| z[i] = d |
| } |
| return z |
| } |
| |
| |
| func addNN(z, x, y []Word) []Word { |
| m := len(x) |
| n := len(y) |
| |
| switch { |
| case m < n: |
| return addNN(z, y, x) |
| case m == 0: |
| // n == 0 because m >= n; result is 0 |
| return makeN(z, 0, false) |
| case n == 0: |
| // result is x |
| return setN(z, x) |
| } |
| // m > 0 |
| |
| z = makeN(z, m, false) |
| c := addVV(&z[0], &x[0], &y[0], n) |
| if m > n { |
| c = addVW(&z[n], &x[n], c, m-n) |
| } |
| if c > 0 { |
| z = z[0 : m+1] |
| z[m] = c |
| } |
| |
| return z |
| } |
| |
| |
| func subNN(z, x, y []Word) []Word { |
| m := len(x) |
| n := len(y) |
| |
| switch { |
| case m < n: |
| panic("underflow") |
| case m == 0: |
| // n == 0 because m >= n; result is 0 |
| return makeN(z, 0, false) |
| case n == 0: |
| // result is x |
| return setN(z, x) |
| } |
| // m > 0 |
| |
| z = makeN(z, m, false) |
| c := subVV(&z[0], &x[0], &y[0], n) |
| if m > n { |
| c = subVW(&z[n], &x[n], c, m-n) |
| } |
| if c != 0 { |
| panic("underflow") |
| } |
| z = normN(z) |
| |
| return z |
| } |
| |
| |
| func cmpNN(x, y []Word) (r int) { |
| m := len(x) |
| n := len(y) |
| if m != n || m == 0 { |
| switch { |
| case m < n: |
| r = -1 |
| case m > n: |
| r = 1 |
| } |
| return |
| } |
| |
| i := m - 1 |
| for i > 0 && x[i] == y[i] { |
| i-- |
| } |
| |
| switch { |
| case x[i] < y[i]: |
| r = -1 |
| case x[i] > y[i]: |
| r = 1 |
| } |
| return |
| } |
| |
| |
| func mulAddNWW(z, x []Word, y, r Word) []Word { |
| m := len(x) |
| if m == 0 || y == 0 { |
| return newN(z, uint64(r)) // result is r |
| } |
| // m > 0 |
| |
| z = makeN(z, m, false) |
| c := mulAddVWW(&z[0], &x[0], y, r, m) |
| if c > 0 { |
| z = z[0 : m+1] |
| z[m] = c |
| } |
| |
| return z |
| } |
| |
| |
| func mulNN(z, x, y []Word) []Word { |
| m := len(x) |
| n := len(y) |
| |
| switch { |
| case m < n: |
| return mulNN(z, y, x) |
| case m == 0 || n == 0: |
| return makeN(z, 0, false) |
| case n == 1: |
| return mulAddNWW(z, x, y[0], 0) |
| } |
| // m >= n && m > 1 && n > 1 |
| |
| z = makeN(z, m+n, true) |
| if &z[0] == &x[0] || &z[0] == &y[0] { |
| z = makeN(nil, m+n, true) // z is an alias for x or y - cannot reuse |
| } |
| for i := 0; i < n; i++ { |
| if f := y[i]; f != 0 { |
| z[m+i] = addMulVVW(&z[i], &x[0], f, m) |
| } |
| } |
| z = normN(z) |
| |
| return z |
| } |
| |
| |
| // q = (x-r)/y, with 0 <= r < y |
| func divNW(z, x []Word, y Word) (q []Word, r Word) { |
| m := len(x) |
| switch { |
| case y == 0: |
| panic("division by zero") |
| case y == 1: |
| q = setN(z, x) // result is x |
| return |
| case m == 0: |
| q = setN(z, nil) // result is 0 |
| return |
| } |
| // m > 0 |
| z = makeN(z, m, false) |
| r = divWVW(&z[0], 0, &x[0], y, m) |
| q = normN(z) |
| return |
| } |
| |
| |
| func divNN(z, z2, u, v []Word) (q, r []Word) { |
| if len(v) == 0 { |
| panic("Divide by zero undefined") |
| } |
| |
| if cmpNN(u, v) < 0 { |
| q = makeN(z, 0, false) |
| r = setN(z2, u) |
| return |
| } |
| |
| if len(v) == 1 { |
| var rprime Word |
| q, rprime = divNW(z, u, v[0]) |
| if rprime > 0 { |
| r = makeN(z2, 1, false) |
| r[0] = rprime |
| } else { |
| r = makeN(z2, 0, false) |
| } |
| return |
| } |
| |
| q, r = divLargeNN(z, z2, u, v) |
| return |
| } |
| |
| |
| // q = (uIn-r)/v, with 0 <= r < y |
| // See Knuth, Volume 2, section 4.3.1, Algorithm D. |
| // Preconditions: |
| // len(v) >= 2 |
| // len(uIn) >= len(v) |
| func divLargeNN(z, z2, uIn, v []Word) (q, r []Word) { |
| n := len(v) |
| m := len(uIn) - len(v) |
| |
| u := makeN(z2, len(uIn)+1, false) |
| qhatv := make([]Word, len(v)+1) |
| q = makeN(z, m+1, false) |
| |
| // D1. |
| shift := leadingZeroBits(v[n-1]) |
| shiftLeft(v, v, shift) |
| shiftLeft(u, uIn, shift) |
| u[len(uIn)] = uIn[len(uIn)-1] >> (_W - uint(shift)) |
| |
| // D2. |
| for j := m; j >= 0; j-- { |
| // D3. |
| var qhat Word |
| if u[j+n] == v[n-1] { |
| qhat = _B - 1 |
| } else { |
| var rhat Word |
| qhat, rhat = divWW_g(u[j+n], u[j+n-1], v[n-1]) |
| |
| // x1 | x2 = q̂v_{n-2} |
| x1, x2 := mulWW_g(qhat, v[n-2]) |
| // test if q̂v_{n-2} > br̂ + u_{j+n-2} |
| for greaterThan(x1, x2, rhat, u[j+n-2]) { |
| qhat-- |
| prevRhat := rhat |
| rhat += v[n-1] |
| // v[n-1] >= 0, so this tests for overflow. |
| if rhat < prevRhat { |
| break |
| } |
| x1, x2 = mulWW_g(qhat, v[n-2]) |
| } |
| } |
| |
| // D4. |
| qhatv[len(v)] = mulAddVWW(&qhatv[0], &v[0], qhat, 0, len(v)) |
| |
| c := subVV(&u[j], &u[j], &qhatv[0], len(qhatv)) |
| if c != 0 { |
| c := addVV(&u[j], &u[j], &v[0], len(v)) |
| u[j+len(v)] += c |
| qhat-- |
| } |
| |
| q[j] = qhat |
| } |
| |
| q = normN(q) |
| shiftRight(u, u, shift) |
| shiftRight(v, v, shift) |
| r = normN(u) |
| |
| return q, r |
| } |
| |
| |
| // log2 computes the integer binary logarithm of x. |
| // The result is the integer n for which 2^n <= x < 2^(n+1). |
| // If x == 0, the result is -1. |
| func log2(x Word) int { |
| n := 0 |
| for ; x > 0; x >>= 1 { |
| n++ |
| } |
| return n - 1 |
| } |
| |
| |
| // log2N computes the integer binary logarithm of x. |
| // The result is the integer n for which 2^n <= x < 2^(n+1). |
| // If x == 0, the result is -1. |
| func log2N(x []Word) int { |
| m := len(x) |
| if m > 0 { |
| return (m-1)*_W + log2(x[m-1]) |
| } |
| return -1 |
| } |
| |
| |
| func hexValue(ch byte) int { |
| var d byte |
| switch { |
| case '0' <= ch && ch <= '9': |
| d = ch - '0' |
| case 'a' <= ch && ch <= 'f': |
| d = ch - 'a' + 10 |
| case 'A' <= ch && ch <= 'F': |
| d = ch - 'A' + 10 |
| default: |
| return -1 |
| } |
| return int(d) |
| } |
| |
| |
| // scanN returns the natural number corresponding to the |
| // longest possible prefix of s representing a natural number in a |
| // given conversion base, the actual conversion base used, and the |
| // prefix length. The syntax of natural numbers follows the syntax |
| // of unsigned integer literals in Go. |
| // |
| // If the base argument is 0, the string prefix determines the actual |
| // conversion base. A prefix of ``0x'' or ``0X'' selects base 16; the |
| // ``0'' prefix selects base 8. Otherwise the selected base is 10. |
| // |
| func scanN(z []Word, s string, base int) ([]Word, int, int) { |
| // determine base if necessary |
| i, n := 0, len(s) |
| if base == 0 { |
| base = 10 |
| if n > 0 && s[0] == '0' { |
| if n > 1 && (s[1] == 'x' || s[1] == 'X') { |
| if n == 2 { |
| // Reject a string which is just '0x' as nonsense. |
| return nil, 0, 0 |
| } |
| base, i = 16, 2 |
| } else { |
| base, i = 8, 1 |
| } |
| } |
| } |
| if base < 2 || 16 < base { |
| panic("illegal base") |
| } |
| |
| // convert string |
| z = makeN(z, len(z), false) |
| for ; i < n; i++ { |
| d := hexValue(s[i]) |
| if 0 <= d && d < base { |
| z = mulAddNWW(z, z, Word(base), Word(d)) |
| } else { |
| break |
| } |
| } |
| |
| return z, base, i |
| } |
| |
| |
| // string converts x to a string for a given base, with 2 <= base <= 16. |
| // TODO(gri) in the style of the other routines, perhaps this should take |
| // a []byte buffer and return it |
| func stringN(x []Word, base int) string { |
| if base < 2 || 16 < base { |
| panic("illegal base") |
| } |
| |
| if len(x) == 0 { |
| return "0" |
| } |
| |
| // allocate buffer for conversion |
| i := (log2N(x)+1)/log2(Word(base)) + 1 // +1: round up |
| s := make([]byte, i) |
| |
| // don't destroy x |
| q := setN(nil, x) |
| |
| // convert |
| for len(q) > 0 { |
| i-- |
| var r Word |
| q, r = divNW(q, q, Word(base)) |
| s[i] = "0123456789abcdef"[r] |
| } |
| |
| return string(s[i:]) |
| } |
| |
| |
| // leadingZeroBits returns the number of leading zero bits in x. |
| func leadingZeroBits(x Word) int { |
| c := 0 |
| if x < 1<<(_W/2) { |
| x <<= _W / 2 |
| c = _W / 2 |
| } |
| |
| for i := 0; x != 0; i++ { |
| if x&(1<<(_W-1)) != 0 { |
| return i + c |
| } |
| x <<= 1 |
| } |
| |
| return _W |
| } |
| |
| const deBruijn32 = 0x077CB531 |
| |
| var deBruijn32Lookup = []byte{ |
| 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, |
| 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9, |
| } |
| |
| const deBruijn64 = 0x03f79d71b4ca8b09 |
| |
| var deBruijn64Lookup = []byte{ |
| 0, 1, 56, 2, 57, 49, 28, 3, 61, 58, 42, 50, 38, 29, 17, 4, |
| 62, 47, 59, 36, 45, 43, 51, 22, 53, 39, 33, 30, 24, 18, 12, 5, |
| 63, 55, 48, 27, 60, 41, 37, 16, 46, 35, 44, 21, 52, 32, 23, 11, |
| 54, 26, 40, 15, 34, 20, 31, 10, 25, 14, 19, 9, 13, 8, 7, 6, |
| } |
| |
| // trailingZeroBits returns the number of consecutive zero bits on the right |
| // side of the given Word. |
| // See Knuth, volume 4, section 7.3.1 |
| func trailingZeroBits(x Word) int { |
| // x & -x leaves only the right-most bit set in the word. Let k be the |
| // index of that bit. Since only a single bit is set, the value is two |
| // to the power of k. Multipling by a power of two is equivalent to |
| // left shifting, in this case by k bits. The de Bruijn constant is |
| // such that all six bit, consecutive substrings are distinct. |
| // Therefore, if we have a left shifted version of this constant we can |
| // find by how many bits it was shifted by looking at which six bit |
| // substring ended up at the top of the word. |
| switch _W { |
| case 32: |
| return int(deBruijn32Lookup[((x&-x)*deBruijn32)>>27]) |
| case 64: |
| return int(deBruijn64Lookup[((x&-x)*(deBruijn64&_M))>>58]) |
| default: |
| panic("Unknown word size") |
| } |
| |
| return 0 |
| } |
| |
| |
| func shiftLeft(dst, src []Word, n int) { |
| if len(src) == 0 { |
| return |
| } |
| |
| ñ := _W - uint(n) |
| for i := len(src) - 1; i >= 1; i-- { |
| dst[i] = src[i] << uint(n) |
| dst[i] |= src[i-1] >> ñ |
| } |
| dst[0] = src[0] << uint(n) |
| } |
| |
| |
| func shiftRight(dst, src []Word, n int) { |
| if len(src) == 0 { |
| return |
| } |
| |
| ñ := _W - uint(n) |
| for i := 0; i < len(src)-1; i++ { |
| dst[i] = src[i] >> uint(n) |
| dst[i] |= src[i+1] << ñ |
| } |
| dst[len(src)-1] = src[len(src)-1] >> uint(n) |
| } |
| |
| |
| // greaterThan returns true iff (x1<<_W + x2) > (y1<<_W + y2) |
| func greaterThan(x1, x2, y1, y2 Word) bool { return x1 > y1 || x1 == y1 && x2 > y2 } |
| |
| |
| // modNW returns x % d. |
| func modNW(x []Word, d Word) (r Word) { |
| // TODO(agl): we don't actually need to store the q value. |
| q := makeN(nil, len(x), false) |
| return divWVW(&q[0], 0, &x[0], d, len(x)) |
| } |
| |
| |
| // powersOfTwoDecompose finds q and k such that q * 1<<k = n and q is odd. |
| func powersOfTwoDecompose(n []Word) (q []Word, k Word) { |
| if len(n) == 0 { |
| return n, 0 |
| } |
| |
| zeroWords := 0 |
| for n[zeroWords] == 0 { |
| zeroWords++ |
| } |
| // One of the words must be non-zero by invariant, therefore |
| // zeroWords < len(n). |
| x := trailingZeroBits(n[zeroWords]) |
| |
| q = makeN(nil, len(n)-zeroWords, false) |
| shiftRight(q, n[zeroWords:], x) |
| |
| k = Word(_W*zeroWords + x) |
| return |
| } |
| |
| |
| // randomN creates a random integer in [0..limit), using the space in z if |
| // possible. n is the bit length of limit. |
| func randomN(z []Word, rand *rand.Rand, limit []Word, n int) []Word { |
| bitLengthOfMSW := uint(n % _W) |
| if bitLengthOfMSW == 0 { |
| bitLengthOfMSW = _W |
| } |
| mask := Word((1 << bitLengthOfMSW) - 1) |
| z = makeN(z, len(limit), false) |
| |
| for { |
| for i := range z { |
| switch _W { |
| case 32: |
| z[i] = Word(rand.Uint32()) |
| case 64: |
| z[i] = Word(rand.Uint32()) | Word(rand.Uint32())<<32 |
| } |
| } |
| |
| z[len(limit)-1] &= mask |
| |
| if cmpNN(z, limit) < 0 { |
| break |
| } |
| } |
| |
| return normN(z) |
| } |
| |
| |
| // If m != nil, expNNN calculates x**y mod m. Otherwise it calculates x**y. It |
| // reuses the storage of z if possible. |
| func expNNN(z, x, y, m []Word) []Word { |
| if len(y) == 0 { |
| z = makeN(z, 1, false) |
| z[0] = 1 |
| return z |
| } |
| |
| if m != nil { |
| // We likely end up being as long as the modulus. |
| z = makeN(z, len(m), false) |
| } |
| z = setN(z, x) |
| v := y[len(y)-1] |
| // It's invalid for the most significant word to be zero, therefore we |
| // will find a one bit. |
| shift := leadingZeros(v) + 1 |
| v <<= shift |
| var q []Word |
| |
| const mask = 1 << (_W - 1) |
| |
| // We walk through the bits of the exponent one by one. Each time we |
| // see a bit, we square, thus doubling the power. If the bit is a one, |
| // we also multiply by x, thus adding one to the power. |
| |
| w := _W - int(shift) |
| for j := 0; j < w; j++ { |
| z = mulNN(z, z, z) |
| |
| if v&mask != 0 { |
| z = mulNN(z, z, x) |
| } |
| |
| if m != nil { |
| q, z = divNN(q, z, z, m) |
| } |
| |
| v <<= 1 |
| } |
| |
| for i := len(y) - 2; i >= 0; i-- { |
| v = y[i] |
| |
| for j := 0; j < _W; j++ { |
| z = mulNN(z, z, z) |
| |
| if v&mask != 0 { |
| z = mulNN(z, z, x) |
| } |
| |
| if m != nil { |
| q, z = divNN(q, z, z, m) |
| } |
| |
| v <<= 1 |
| } |
| } |
| |
| return z |
| } |
| |
| |
| // lenN returns the bit length of z. |
| func lenN(z []Word) int { |
| if len(z) == 0 { |
| return 0 |
| } |
| |
| return (len(z)-1)*_W + (_W - leadingZeroBits(z[len(z)-1])) |
| } |
| |
| |
| const ( |
| primesProduct32 = 0xC0CFD797 // Π {p ∈ primes, 2 < p <= 29} |
| primesProduct64 = 0xE221F97C30E94E1D // Π {p ∈ primes, 2 < p <= 53} |
| ) |
| |
| var bigOne = []Word{1} |
| var bigTwo = []Word{2} |
| |
| // ProbablyPrime performs reps Miller-Rabin tests to check whether n is prime. |
| // If it returns true, n is prime with probability 1 - 1/4^reps. |
| // If it returns false, n is not prime. |
| func probablyPrime(n []Word, reps int) bool { |
| if len(n) == 0 { |
| return false |
| } |
| |
| if len(n) == 1 { |
| if n[0]%2 == 0 { |
| return n[0] == 2 |
| } |
| |
| // We have to exclude these cases because we reject all |
| // multiples of these numbers below. |
| if n[0] == 3 || n[0] == 5 || n[0] == 7 || n[0] == 11 || |
| n[0] == 13 || n[0] == 17 || n[0] == 19 || n[0] == 23 || |
| n[0] == 29 || n[0] == 31 || n[0] == 37 || n[0] == 41 || |
| n[0] == 43 || n[0] == 47 || n[0] == 53 { |
| return true |
| } |
| } |
| |
| var r Word |
| switch _W { |
| case 32: |
| r = modNW(n, primesProduct32) |
| case 64: |
| r = modNW(n, primesProduct64&_M) |
| default: |
| panic("Unknown word size") |
| } |
| |
| if r%3 == 0 || r%5 == 0 || r%7 == 0 || r%11 == 0 || |
| r%13 == 0 || r%17 == 0 || r%19 == 0 || r%23 == 0 || r%29 == 0 { |
| return false |
| } |
| |
| if _W == 64 && (r%31 == 0 || r%37 == 0 || r%41 == 0 || |
| r%43 == 0 || r%47 == 0 || r%53 == 0) { |
| return false |
| } |
| |
| nm1 := subNN(nil, n, bigOne) |
| // 1<<k * q = nm1; |
| q, k := powersOfTwoDecompose(nm1) |
| |
| nm3 := subNN(nil, nm1, bigTwo) |
| rand := rand.New(rand.NewSource(int64(n[0]))) |
| |
| var x, y, quotient []Word |
| nm3Len := lenN(nm3) |
| |
| NextRandom: |
| for i := 0; i < reps; i++ { |
| x = randomN(x, rand, nm3, nm3Len) |
| addNN(x, x, bigTwo) |
| y = expNNN(y, x, q, n) |
| if cmpNN(y, bigOne) == 0 || cmpNN(y, nm1) == 0 { |
| continue |
| } |
| for j := Word(1); j < k; j++ { |
| y = mulNN(y, y, y) |
| quotient, y = divNN(quotient, y, y, n) |
| if cmpNN(y, nm1) == 0 { |
| continue NextRandom |
| } |
| if cmpNN(y, bigOne) == 0 { |
| return false |
| } |
| } |
| return false |
| } |
| |
| return true |
| } |