// 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);
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 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 n Miller-Rabin tests to check whether n is prime.
// If it returns true, n is prime with probability 1 - 1/4^n.
// 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;
}
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