mirror of
https://github.com/autc04/Retro68.git
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1260 lines
27 KiB
Go
1260 lines
27 KiB
Go
// Copyright 2009 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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// This file implements unsigned multi-precision integers (natural
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// numbers). They are the building blocks for the implementation
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// of signed integers, rationals, and floating-point numbers.
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package big
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import (
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"math/rand"
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"sync"
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)
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// An unsigned integer x of the form
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//
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// x = x[n-1]*_B^(n-1) + x[n-2]*_B^(n-2) + ... + x[1]*_B + x[0]
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//
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// with 0 <= x[i] < _B and 0 <= i < n is stored in a slice of length n,
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// with the digits x[i] as the slice elements.
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//
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// A number is normalized if the slice contains no leading 0 digits.
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// During arithmetic operations, denormalized values may occur but are
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// always normalized before returning the final result. The normalized
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// representation of 0 is the empty or nil slice (length = 0).
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//
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type nat []Word
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var (
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natOne = nat{1}
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natTwo = nat{2}
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natTen = nat{10}
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)
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func (z nat) clear() {
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for i := range z {
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z[i] = 0
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}
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}
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func (z nat) norm() nat {
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i := len(z)
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for i > 0 && z[i-1] == 0 {
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i--
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}
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return z[0:i]
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}
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func (z nat) make(n int) nat {
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if n <= cap(z) {
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return z[:n] // reuse z
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}
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// Choosing a good value for e has significant performance impact
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// because it increases the chance that a value can be reused.
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const e = 4 // extra capacity
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return make(nat, n, n+e)
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}
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func (z nat) setWord(x Word) nat {
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if x == 0 {
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return z[:0]
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}
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z = z.make(1)
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z[0] = x
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return z
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}
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func (z nat) setUint64(x uint64) nat {
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// single-digit values
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if w := Word(x); uint64(w) == x {
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return z.setWord(w)
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}
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// compute number of words n required to represent x
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n := 0
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for t := x; t > 0; t >>= _W {
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n++
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}
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// split x into n words
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z = z.make(n)
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for i := range z {
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z[i] = Word(x & _M)
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x >>= _W
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}
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return z
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}
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func (z nat) set(x nat) nat {
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z = z.make(len(x))
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copy(z, x)
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return z
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}
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func (z nat) add(x, y nat) nat {
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m := len(x)
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n := len(y)
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switch {
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case m < n:
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return z.add(y, x)
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case m == 0:
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// n == 0 because m >= n; result is 0
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return z[:0]
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case n == 0:
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// result is x
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return z.set(x)
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}
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// m > 0
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z = z.make(m + 1)
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c := addVV(z[0:n], x, y)
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if m > n {
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c = addVW(z[n:m], x[n:], c)
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}
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z[m] = c
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return z.norm()
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}
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func (z nat) sub(x, y nat) nat {
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m := len(x)
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n := len(y)
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switch {
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case m < n:
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panic("underflow")
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case m == 0:
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// n == 0 because m >= n; result is 0
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return z[:0]
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case n == 0:
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// result is x
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return z.set(x)
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}
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// m > 0
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z = z.make(m)
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c := subVV(z[0:n], x, y)
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if m > n {
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c = subVW(z[n:], x[n:], c)
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}
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if c != 0 {
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panic("underflow")
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}
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return z.norm()
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}
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func (x nat) cmp(y nat) (r int) {
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m := len(x)
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n := len(y)
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if m != n || m == 0 {
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switch {
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case m < n:
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r = -1
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case m > n:
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r = 1
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}
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return
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}
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i := m - 1
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for i > 0 && x[i] == y[i] {
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i--
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}
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switch {
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case x[i] < y[i]:
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r = -1
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case x[i] > y[i]:
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r = 1
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}
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return
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}
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func (z nat) mulAddWW(x nat, y, r Word) nat {
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m := len(x)
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if m == 0 || y == 0 {
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return z.setWord(r) // result is r
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}
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// m > 0
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z = z.make(m + 1)
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z[m] = mulAddVWW(z[0:m], x, y, r)
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return z.norm()
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}
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// basicMul multiplies x and y and leaves the result in z.
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// The (non-normalized) result is placed in z[0 : len(x) + len(y)].
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func basicMul(z, x, y nat) {
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z[0 : len(x)+len(y)].clear() // initialize z
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for i, d := range y {
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if d != 0 {
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z[len(x)+i] = addMulVVW(z[i:i+len(x)], x, d)
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}
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}
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}
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// montgomery computes z mod m = x*y*2**(-n*_W) mod m,
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// assuming k = -1/m mod 2**_W.
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// z is used for storing the result which is returned;
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// z must not alias x, y or m.
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// See Gueron, "Efficient Software Implementations of Modular Exponentiation".
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// https://eprint.iacr.org/2011/239.pdf
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// In the terminology of that paper, this is an "Almost Montgomery Multiplication":
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// x and y are required to satisfy 0 <= z < 2**(n*_W) and then the result
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// z is guaranteed to satisfy 0 <= z < 2**(n*_W), but it may not be < m.
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func (z nat) montgomery(x, y, m nat, k Word, n int) nat {
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// This code assumes x, y, m are all the same length, n.
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// (required by addMulVVW and the for loop).
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// It also assumes that x, y are already reduced mod m,
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// or else the result will not be properly reduced.
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if len(x) != n || len(y) != n || len(m) != n {
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panic("math/big: mismatched montgomery number lengths")
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}
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z = z.make(n)
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z.clear()
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var c Word
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for i := 0; i < n; i++ {
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d := y[i]
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c2 := addMulVVW(z, x, d)
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t := z[0] * k
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c3 := addMulVVW(z, m, t)
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copy(z, z[1:])
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cx := c + c2
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cy := cx + c3
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z[n-1] = cy
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if cx < c2 || cy < c3 {
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c = 1
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} else {
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c = 0
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}
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}
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if c != 0 {
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subVV(z, z, m)
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}
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return z
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}
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// Fast version of z[0:n+n>>1].add(z[0:n+n>>1], x[0:n]) w/o bounds checks.
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// Factored out for readability - do not use outside karatsuba.
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func karatsubaAdd(z, x nat, n int) {
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if c := addVV(z[0:n], z, x); c != 0 {
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addVW(z[n:n+n>>1], z[n:], c)
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}
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}
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// Like karatsubaAdd, but does subtract.
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func karatsubaSub(z, x nat, n int) {
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if c := subVV(z[0:n], z, x); c != 0 {
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subVW(z[n:n+n>>1], z[n:], c)
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}
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}
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// Operands that are shorter than karatsubaThreshold are multiplied using
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// "grade school" multiplication; for longer operands the Karatsuba algorithm
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// is used.
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var karatsubaThreshold int = 40 // computed by calibrate.go
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// karatsuba multiplies x and y and leaves the result in z.
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// Both x and y must have the same length n and n must be a
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// power of 2. The result vector z must have len(z) >= 6*n.
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// The (non-normalized) result is placed in z[0 : 2*n].
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func karatsuba(z, x, y nat) {
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n := len(y)
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// Switch to basic multiplication if numbers are odd or small.
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// (n is always even if karatsubaThreshold is even, but be
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// conservative)
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if n&1 != 0 || n < karatsubaThreshold || n < 2 {
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basicMul(z, x, y)
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return
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}
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// n&1 == 0 && n >= karatsubaThreshold && n >= 2
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// Karatsuba multiplication is based on the observation that
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// for two numbers x and y with:
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//
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// x = x1*b + x0
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// y = y1*b + y0
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//
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// the product x*y can be obtained with 3 products z2, z1, z0
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// instead of 4:
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//
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// x*y = x1*y1*b*b + (x1*y0 + x0*y1)*b + x0*y0
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// = z2*b*b + z1*b + z0
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//
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// with:
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//
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// xd = x1 - x0
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// yd = y0 - y1
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//
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// z1 = xd*yd + z2 + z0
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// = (x1-x0)*(y0 - y1) + z2 + z0
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// = x1*y0 - x1*y1 - x0*y0 + x0*y1 + z2 + z0
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// = x1*y0 - z2 - z0 + x0*y1 + z2 + z0
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// = x1*y0 + x0*y1
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// split x, y into "digits"
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n2 := n >> 1 // n2 >= 1
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x1, x0 := x[n2:], x[0:n2] // x = x1*b + y0
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y1, y0 := y[n2:], y[0:n2] // y = y1*b + y0
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// z is used for the result and temporary storage:
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//
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// 6*n 5*n 4*n 3*n 2*n 1*n 0*n
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// z = [z2 copy|z0 copy| xd*yd | yd:xd | x1*y1 | x0*y0 ]
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//
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// For each recursive call of karatsuba, an unused slice of
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// z is passed in that has (at least) half the length of the
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// caller's z.
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// compute z0 and z2 with the result "in place" in z
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karatsuba(z, x0, y0) // z0 = x0*y0
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karatsuba(z[n:], x1, y1) // z2 = x1*y1
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// compute xd (or the negative value if underflow occurs)
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s := 1 // sign of product xd*yd
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xd := z[2*n : 2*n+n2]
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if subVV(xd, x1, x0) != 0 { // x1-x0
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s = -s
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subVV(xd, x0, x1) // x0-x1
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}
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// compute yd (or the negative value if underflow occurs)
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yd := z[2*n+n2 : 3*n]
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if subVV(yd, y0, y1) != 0 { // y0-y1
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s = -s
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subVV(yd, y1, y0) // y1-y0
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}
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// p = (x1-x0)*(y0-y1) == x1*y0 - x1*y1 - x0*y0 + x0*y1 for s > 0
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// p = (x0-x1)*(y0-y1) == x0*y0 - x0*y1 - x1*y0 + x1*y1 for s < 0
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p := z[n*3:]
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karatsuba(p, xd, yd)
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// save original z2:z0
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// (ok to use upper half of z since we're done recursing)
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r := z[n*4:]
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copy(r, z[:n*2])
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// add up all partial products
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//
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// 2*n n 0
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// z = [ z2 | z0 ]
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// + [ z0 ]
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// + [ z2 ]
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// + [ p ]
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//
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karatsubaAdd(z[n2:], r, n)
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karatsubaAdd(z[n2:], r[n:], n)
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if s > 0 {
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karatsubaAdd(z[n2:], p, n)
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} else {
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karatsubaSub(z[n2:], p, n)
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}
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}
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// alias reports whether x and y share the same base array.
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func alias(x, y nat) bool {
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return cap(x) > 0 && cap(y) > 0 && &x[0:cap(x)][cap(x)-1] == &y[0:cap(y)][cap(y)-1]
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}
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// addAt implements z += x<<(_W*i); z must be long enough.
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// (we don't use nat.add because we need z to stay the same
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// slice, and we don't need to normalize z after each addition)
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func addAt(z, x nat, i int) {
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if n := len(x); n > 0 {
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if c := addVV(z[i:i+n], z[i:], x); c != 0 {
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j := i + n
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if j < len(z) {
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addVW(z[j:], z[j:], c)
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}
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}
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}
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}
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func max(x, y int) int {
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if x > y {
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return x
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}
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return y
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}
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// karatsubaLen computes an approximation to the maximum k <= n such that
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// k = p<<i for a number p <= karatsubaThreshold and an i >= 0. Thus, the
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// result is the largest number that can be divided repeatedly by 2 before
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// becoming about the value of karatsubaThreshold.
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func karatsubaLen(n int) int {
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i := uint(0)
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for n > karatsubaThreshold {
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n >>= 1
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i++
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}
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return n << i
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}
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func (z nat) mul(x, y nat) nat {
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m := len(x)
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n := len(y)
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switch {
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case m < n:
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return z.mul(y, x)
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case m == 0 || n == 0:
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return z[:0]
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case n == 1:
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return z.mulAddWW(x, y[0], 0)
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}
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// m >= n > 1
|
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// determine if z can be reused
|
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if alias(z, x) || alias(z, y) {
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z = nil // z is an alias for x or y - cannot reuse
|
||
}
|
||
|
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// use basic multiplication if the numbers are small
|
||
if n < karatsubaThreshold {
|
||
z = z.make(m + n)
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basicMul(z, x, y)
|
||
return z.norm()
|
||
}
|
||
// m >= n && n >= karatsubaThreshold && n >= 2
|
||
|
||
// determine Karatsuba length k such that
|
||
//
|
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// x = xh*b + x0 (0 <= x0 < b)
|
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// y = yh*b + y0 (0 <= y0 < b)
|
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// b = 1<<(_W*k) ("base" of digits xi, yi)
|
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//
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k := karatsubaLen(n)
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// k <= n
|
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// multiply x0 and y0 via Karatsuba
|
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x0 := x[0:k] // x0 is not normalized
|
||
y0 := y[0:k] // y0 is not normalized
|
||
z = z.make(max(6*k, m+n)) // enough space for karatsuba of x0*y0 and full result of x*y
|
||
karatsuba(z, x0, y0)
|
||
z = z[0 : m+n] // z has final length but may be incomplete
|
||
z[2*k:].clear() // upper portion of z is garbage (and 2*k <= m+n since k <= n <= m)
|
||
|
||
// If xh != 0 or yh != 0, add the missing terms to z. For
|
||
//
|
||
// xh = xi*b^i + ... + x2*b^2 + x1*b (0 <= xi < b)
|
||
// yh = y1*b (0 <= y1 < b)
|
||
//
|
||
// the missing terms are
|
||
//
|
||
// x0*y1*b and xi*y0*b^i, xi*y1*b^(i+1) for i > 0
|
||
//
|
||
// since all the yi for i > 1 are 0 by choice of k: If any of them
|
||
// were > 0, then yh >= b^2 and thus y >= b^2. Then k' = k*2 would
|
||
// be a larger valid threshold contradicting the assumption about k.
|
||
//
|
||
if k < n || m != n {
|
||
var t nat
|
||
|
||
// add x0*y1*b
|
||
x0 := x0.norm()
|
||
y1 := y[k:] // y1 is normalized because y is
|
||
t = t.mul(x0, y1) // update t so we don't lose t's underlying array
|
||
addAt(z, t, k)
|
||
|
||
// add xi*y0<<i, xi*y1*b<<(i+k)
|
||
y0 := y0.norm()
|
||
for i := k; i < len(x); i += k {
|
||
xi := x[i:]
|
||
if len(xi) > k {
|
||
xi = xi[:k]
|
||
}
|
||
xi = xi.norm()
|
||
t = t.mul(xi, y0)
|
||
addAt(z, t, i)
|
||
t = t.mul(xi, y1)
|
||
addAt(z, t, i+k)
|
||
}
|
||
}
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// mulRange computes the product of all the unsigned integers in the
|
||
// range [a, b] inclusively. If a > b (empty range), the result is 1.
|
||
func (z nat) mulRange(a, b uint64) nat {
|
||
switch {
|
||
case a == 0:
|
||
// cut long ranges short (optimization)
|
||
return z.setUint64(0)
|
||
case a > b:
|
||
return z.setUint64(1)
|
||
case a == b:
|
||
return z.setUint64(a)
|
||
case a+1 == b:
|
||
return z.mul(nat(nil).setUint64(a), nat(nil).setUint64(b))
|
||
}
|
||
m := (a + b) / 2
|
||
return z.mul(nat(nil).mulRange(a, m), nat(nil).mulRange(m+1, b))
|
||
}
|
||
|
||
// q = (x-r)/y, with 0 <= r < y
|
||
func (z nat) divW(x nat, y Word) (q nat, r Word) {
|
||
m := len(x)
|
||
switch {
|
||
case y == 0:
|
||
panic("division by zero")
|
||
case y == 1:
|
||
q = z.set(x) // result is x
|
||
return
|
||
case m == 0:
|
||
q = z[:0] // result is 0
|
||
return
|
||
}
|
||
// m > 0
|
||
z = z.make(m)
|
||
r = divWVW(z, 0, x, y)
|
||
q = z.norm()
|
||
return
|
||
}
|
||
|
||
func (z nat) div(z2, u, v nat) (q, r nat) {
|
||
if len(v) == 0 {
|
||
panic("division by zero")
|
||
}
|
||
|
||
if u.cmp(v) < 0 {
|
||
q = z[:0]
|
||
r = z2.set(u)
|
||
return
|
||
}
|
||
|
||
if len(v) == 1 {
|
||
var r2 Word
|
||
q, r2 = z.divW(u, v[0])
|
||
r = z2.setWord(r2)
|
||
return
|
||
}
|
||
|
||
q, r = z.divLarge(z2, u, v)
|
||
return
|
||
}
|
||
|
||
// getNat returns a *nat of len n. The contents may not be zero.
|
||
// The pool holds *nat to avoid allocation when converting to interface{}.
|
||
func getNat(n int) *nat {
|
||
var z *nat
|
||
if v := natPool.Get(); v != nil {
|
||
z = v.(*nat)
|
||
}
|
||
if z == nil {
|
||
z = new(nat)
|
||
}
|
||
*z = z.make(n)
|
||
return z
|
||
}
|
||
|
||
func putNat(x *nat) {
|
||
natPool.Put(x)
|
||
}
|
||
|
||
var natPool sync.Pool
|
||
|
||
// q = (uIn-r)/v, with 0 <= r < y
|
||
// Uses z as storage for q, and u as storage for r if possible.
|
||
// See Knuth, Volume 2, section 4.3.1, Algorithm D.
|
||
// Preconditions:
|
||
// len(v) >= 2
|
||
// len(uIn) >= len(v)
|
||
func (z nat) divLarge(u, uIn, v nat) (q, r nat) {
|
||
n := len(v)
|
||
m := len(uIn) - n
|
||
|
||
// determine if z can be reused
|
||
// TODO(gri) should find a better solution - this if statement
|
||
// is very costly (see e.g. time pidigits -s -n 10000)
|
||
if alias(z, uIn) || alias(z, v) {
|
||
z = nil // z is an alias for uIn or v - cannot reuse
|
||
}
|
||
q = z.make(m + 1)
|
||
|
||
qhatvp := getNat(n + 1)
|
||
qhatv := *qhatvp
|
||
if alias(u, uIn) || alias(u, v) {
|
||
u = nil // u is an alias for uIn or v - cannot reuse
|
||
}
|
||
u = u.make(len(uIn) + 1)
|
||
u.clear() // TODO(gri) no need to clear if we allocated a new u
|
||
|
||
// D1.
|
||
var v1p *nat
|
||
shift := nlz(v[n-1])
|
||
if shift > 0 {
|
||
// do not modify v, it may be used by another goroutine simultaneously
|
||
v1p = getNat(n)
|
||
v1 := *v1p
|
||
shlVU(v1, v, shift)
|
||
v = v1
|
||
}
|
||
u[len(uIn)] = shlVU(u[0:len(uIn)], uIn, shift)
|
||
|
||
// D2.
|
||
vn1 := v[n-1]
|
||
for j := m; j >= 0; j-- {
|
||
// D3.
|
||
qhat := Word(_M)
|
||
if ujn := u[j+n]; ujn != vn1 {
|
||
var rhat Word
|
||
qhat, rhat = divWW(ujn, u[j+n-1], vn1)
|
||
|
||
// x1 | x2 = q̂v_{n-2}
|
||
vn2 := v[n-2]
|
||
x1, x2 := mulWW(qhat, vn2)
|
||
// test if q̂v_{n-2} > br̂ + u_{j+n-2}
|
||
ujn2 := u[j+n-2]
|
||
for greaterThan(x1, x2, rhat, ujn2) {
|
||
qhat--
|
||
prevRhat := rhat
|
||
rhat += vn1
|
||
// v[n-1] >= 0, so this tests for overflow.
|
||
if rhat < prevRhat {
|
||
break
|
||
}
|
||
x1, x2 = mulWW(qhat, vn2)
|
||
}
|
||
}
|
||
|
||
// D4.
|
||
qhatv[n] = mulAddVWW(qhatv[0:n], v, qhat, 0)
|
||
|
||
c := subVV(u[j:j+len(qhatv)], u[j:], qhatv)
|
||
if c != 0 {
|
||
c := addVV(u[j:j+n], u[j:], v)
|
||
u[j+n] += c
|
||
qhat--
|
||
}
|
||
|
||
q[j] = qhat
|
||
}
|
||
if v1p != nil {
|
||
putNat(v1p)
|
||
}
|
||
putNat(qhatvp)
|
||
|
||
q = q.norm()
|
||
shrVU(u, u, shift)
|
||
r = u.norm()
|
||
|
||
return q, r
|
||
}
|
||
|
||
// Length of x in bits. x must be normalized.
|
||
func (x nat) bitLen() int {
|
||
if i := len(x) - 1; i >= 0 {
|
||
return i*_W + bitLen(x[i])
|
||
}
|
||
return 0
|
||
}
|
||
|
||
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 least significant zero
|
||
// bits of x.
|
||
func trailingZeroBits(x Word) uint {
|
||
// 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. Multiplying 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.
|
||
// (Knuth, volume 4, section 7.3.1)
|
||
switch _W {
|
||
case 32:
|
||
return uint(deBruijn32Lookup[((x&-x)*deBruijn32)>>27])
|
||
case 64:
|
||
return uint(deBruijn64Lookup[((x&-x)*(deBruijn64&_M))>>58])
|
||
default:
|
||
panic("unknown word size")
|
||
}
|
||
}
|
||
|
||
// trailingZeroBits returns the number of consecutive least significant zero
|
||
// bits of x.
|
||
func (x nat) trailingZeroBits() uint {
|
||
if len(x) == 0 {
|
||
return 0
|
||
}
|
||
var i uint
|
||
for x[i] == 0 {
|
||
i++
|
||
}
|
||
// x[i] != 0
|
||
return i*_W + trailingZeroBits(x[i])
|
||
}
|
||
|
||
// z = x << s
|
||
func (z nat) shl(x nat, s uint) nat {
|
||
m := len(x)
|
||
if m == 0 {
|
||
return z[:0]
|
||
}
|
||
// m > 0
|
||
|
||
n := m + int(s/_W)
|
||
z = z.make(n + 1)
|
||
z[n] = shlVU(z[n-m:n], x, s%_W)
|
||
z[0 : n-m].clear()
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// z = x >> s
|
||
func (z nat) shr(x nat, s uint) nat {
|
||
m := len(x)
|
||
n := m - int(s/_W)
|
||
if n <= 0 {
|
||
return z[:0]
|
||
}
|
||
// n > 0
|
||
|
||
z = z.make(n)
|
||
shrVU(z, x[m-n:], s%_W)
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
func (z nat) setBit(x nat, i uint, b uint) nat {
|
||
j := int(i / _W)
|
||
m := Word(1) << (i % _W)
|
||
n := len(x)
|
||
switch b {
|
||
case 0:
|
||
z = z.make(n)
|
||
copy(z, x)
|
||
if j >= n {
|
||
// no need to grow
|
||
return z
|
||
}
|
||
z[j] &^= m
|
||
return z.norm()
|
||
case 1:
|
||
if j >= n {
|
||
z = z.make(j + 1)
|
||
z[n:].clear()
|
||
} else {
|
||
z = z.make(n)
|
||
}
|
||
copy(z, x)
|
||
z[j] |= m
|
||
// no need to normalize
|
||
return z
|
||
}
|
||
panic("set bit is not 0 or 1")
|
||
}
|
||
|
||
// bit returns the value of the i'th bit, with lsb == bit 0.
|
||
func (x nat) bit(i uint) uint {
|
||
j := i / _W
|
||
if j >= uint(len(x)) {
|
||
return 0
|
||
}
|
||
// 0 <= j < len(x)
|
||
return uint(x[j] >> (i % _W) & 1)
|
||
}
|
||
|
||
// sticky returns 1 if there's a 1 bit within the
|
||
// i least significant bits, otherwise it returns 0.
|
||
func (x nat) sticky(i uint) uint {
|
||
j := i / _W
|
||
if j >= uint(len(x)) {
|
||
if len(x) == 0 {
|
||
return 0
|
||
}
|
||
return 1
|
||
}
|
||
// 0 <= j < len(x)
|
||
for _, x := range x[:j] {
|
||
if x != 0 {
|
||
return 1
|
||
}
|
||
}
|
||
if x[j]<<(_W-i%_W) != 0 {
|
||
return 1
|
||
}
|
||
return 0
|
||
}
|
||
|
||
func (z nat) and(x, y nat) nat {
|
||
m := len(x)
|
||
n := len(y)
|
||
if m > n {
|
||
m = n
|
||
}
|
||
// m <= n
|
||
|
||
z = z.make(m)
|
||
for i := 0; i < m; i++ {
|
||
z[i] = x[i] & y[i]
|
||
}
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
func (z nat) andNot(x, y nat) nat {
|
||
m := len(x)
|
||
n := len(y)
|
||
if n > m {
|
||
n = m
|
||
}
|
||
// m >= n
|
||
|
||
z = z.make(m)
|
||
for i := 0; i < n; i++ {
|
||
z[i] = x[i] &^ y[i]
|
||
}
|
||
copy(z[n:m], x[n:m])
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
func (z nat) or(x, y nat) nat {
|
||
m := len(x)
|
||
n := len(y)
|
||
s := x
|
||
if m < n {
|
||
n, m = m, n
|
||
s = y
|
||
}
|
||
// m >= n
|
||
|
||
z = z.make(m)
|
||
for i := 0; i < n; i++ {
|
||
z[i] = x[i] | y[i]
|
||
}
|
||
copy(z[n:m], s[n:m])
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
func (z nat) xor(x, y nat) nat {
|
||
m := len(x)
|
||
n := len(y)
|
||
s := x
|
||
if m < n {
|
||
n, m = m, n
|
||
s = y
|
||
}
|
||
// m >= n
|
||
|
||
z = z.make(m)
|
||
for i := 0; i < n; i++ {
|
||
z[i] = x[i] ^ y[i]
|
||
}
|
||
copy(z[n:m], s[n:m])
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// greaterThan reports whether (x1<<_W + x2) > (y1<<_W + y2)
|
||
func greaterThan(x1, x2, y1, y2 Word) bool {
|
||
return x1 > y1 || x1 == y1 && x2 > y2
|
||
}
|
||
|
||
// modW returns x % d.
|
||
func (x nat) modW(d Word) (r Word) {
|
||
// TODO(agl): we don't actually need to store the q value.
|
||
var q nat
|
||
q = q.make(len(x))
|
||
return divWVW(q, 0, x, d)
|
||
}
|
||
|
||
// random creates a random integer in [0..limit), using the space in z if
|
||
// possible. n is the bit length of limit.
|
||
func (z nat) random(rand *rand.Rand, limit nat, n int) nat {
|
||
if alias(z, limit) {
|
||
z = nil // z is an alias for limit - cannot reuse
|
||
}
|
||
z = z.make(len(limit))
|
||
|
||
bitLengthOfMSW := uint(n % _W)
|
||
if bitLengthOfMSW == 0 {
|
||
bitLengthOfMSW = _W
|
||
}
|
||
mask := Word((1 << bitLengthOfMSW) - 1)
|
||
|
||
for {
|
||
switch _W {
|
||
case 32:
|
||
for i := range z {
|
||
z[i] = Word(rand.Uint32())
|
||
}
|
||
case 64:
|
||
for i := range z {
|
||
z[i] = Word(rand.Uint32()) | Word(rand.Uint32())<<32
|
||
}
|
||
default:
|
||
panic("unknown word size")
|
||
}
|
||
z[len(limit)-1] &= mask
|
||
if z.cmp(limit) < 0 {
|
||
break
|
||
}
|
||
}
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// If m != 0 (i.e., len(m) != 0), expNN sets z to x**y mod m;
|
||
// otherwise it sets z to x**y. The result is the value of z.
|
||
func (z nat) expNN(x, y, m nat) nat {
|
||
if alias(z, x) || alias(z, y) {
|
||
// We cannot allow in-place modification of x or y.
|
||
z = nil
|
||
}
|
||
|
||
// x**y mod 1 == 0
|
||
if len(m) == 1 && m[0] == 1 {
|
||
return z.setWord(0)
|
||
}
|
||
// m == 0 || m > 1
|
||
|
||
// x**0 == 1
|
||
if len(y) == 0 {
|
||
return z.setWord(1)
|
||
}
|
||
// y > 0
|
||
|
||
// x**1 mod m == x mod m
|
||
if len(y) == 1 && y[0] == 1 && len(m) != 0 {
|
||
_, z = z.div(z, x, m)
|
||
return z
|
||
}
|
||
// y > 1
|
||
|
||
if len(m) != 0 {
|
||
// We likely end up being as long as the modulus.
|
||
z = z.make(len(m))
|
||
}
|
||
z = z.set(x)
|
||
|
||
// If the base is non-trivial and the exponent is large, we use
|
||
// 4-bit, windowed exponentiation. This involves precomputing 14 values
|
||
// (x^2...x^15) but then reduces the number of multiply-reduces by a
|
||
// third. Even for a 32-bit exponent, this reduces the number of
|
||
// operations. Uses Montgomery method for odd moduli.
|
||
if x.cmp(natOne) > 0 && len(y) > 1 && len(m) > 0 {
|
||
if m[0]&1 == 1 {
|
||
return z.expNNMontgomery(x, y, m)
|
||
}
|
||
return z.expNNWindowed(x, y, m)
|
||
}
|
||
|
||
v := y[len(y)-1] // v > 0 because y is normalized and y > 0
|
||
shift := nlz(v) + 1
|
||
v <<= shift
|
||
var q nat
|
||
|
||
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)
|
||
// zz and r are used to avoid allocating in mul and div as
|
||
// otherwise the arguments would alias.
|
||
var zz, r nat
|
||
for j := 0; j < w; j++ {
|
||
zz = zz.mul(z, z)
|
||
zz, z = z, zz
|
||
|
||
if v&mask != 0 {
|
||
zz = zz.mul(z, x)
|
||
zz, z = z, zz
|
||
}
|
||
|
||
if len(m) != 0 {
|
||
zz, r = zz.div(r, z, m)
|
||
zz, r, q, z = q, z, zz, r
|
||
}
|
||
|
||
v <<= 1
|
||
}
|
||
|
||
for i := len(y) - 2; i >= 0; i-- {
|
||
v = y[i]
|
||
|
||
for j := 0; j < _W; j++ {
|
||
zz = zz.mul(z, z)
|
||
zz, z = z, zz
|
||
|
||
if v&mask != 0 {
|
||
zz = zz.mul(z, x)
|
||
zz, z = z, zz
|
||
}
|
||
|
||
if len(m) != 0 {
|
||
zz, r = zz.div(r, z, m)
|
||
zz, r, q, z = q, z, zz, r
|
||
}
|
||
|
||
v <<= 1
|
||
}
|
||
}
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// expNNWindowed calculates x**y mod m using a fixed, 4-bit window.
|
||
func (z nat) expNNWindowed(x, y, m nat) nat {
|
||
// zz and r are used to avoid allocating in mul and div as otherwise
|
||
// the arguments would alias.
|
||
var zz, r nat
|
||
|
||
const n = 4
|
||
// powers[i] contains x^i.
|
||
var powers [1 << n]nat
|
||
powers[0] = natOne
|
||
powers[1] = x
|
||
for i := 2; i < 1<<n; i += 2 {
|
||
p2, p, p1 := &powers[i/2], &powers[i], &powers[i+1]
|
||
*p = p.mul(*p2, *p2)
|
||
zz, r = zz.div(r, *p, m)
|
||
*p, r = r, *p
|
||
*p1 = p1.mul(*p, x)
|
||
zz, r = zz.div(r, *p1, m)
|
||
*p1, r = r, *p1
|
||
}
|
||
|
||
z = z.setWord(1)
|
||
|
||
for i := len(y) - 1; i >= 0; i-- {
|
||
yi := y[i]
|
||
for j := 0; j < _W; j += n {
|
||
if i != len(y)-1 || j != 0 {
|
||
// Unrolled loop for significant performance
|
||
// gain. Use go test -bench=".*" in crypto/rsa
|
||
// to check performance before making changes.
|
||
zz = zz.mul(z, z)
|
||
zz, z = z, zz
|
||
zz, r = zz.div(r, z, m)
|
||
z, r = r, z
|
||
|
||
zz = zz.mul(z, z)
|
||
zz, z = z, zz
|
||
zz, r = zz.div(r, z, m)
|
||
z, r = r, z
|
||
|
||
zz = zz.mul(z, z)
|
||
zz, z = z, zz
|
||
zz, r = zz.div(r, z, m)
|
||
z, r = r, z
|
||
|
||
zz = zz.mul(z, z)
|
||
zz, z = z, zz
|
||
zz, r = zz.div(r, z, m)
|
||
z, r = r, z
|
||
}
|
||
|
||
zz = zz.mul(z, powers[yi>>(_W-n)])
|
||
zz, z = z, zz
|
||
zz, r = zz.div(r, z, m)
|
||
z, r = r, z
|
||
|
||
yi <<= n
|
||
}
|
||
}
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// expNNMontgomery calculates x**y mod m using a fixed, 4-bit window.
|
||
// Uses Montgomery representation.
|
||
func (z nat) expNNMontgomery(x, y, m nat) nat {
|
||
numWords := len(m)
|
||
|
||
// We want the lengths of x and m to be equal.
|
||
// It is OK if x >= m as long as len(x) == len(m).
|
||
if len(x) > numWords {
|
||
_, x = nat(nil).div(nil, x, m)
|
||
// Note: now len(x) <= numWords, not guaranteed ==.
|
||
}
|
||
if len(x) < numWords {
|
||
rr := make(nat, numWords)
|
||
copy(rr, x)
|
||
x = rr
|
||
}
|
||
|
||
// Ideally the precomputations would be performed outside, and reused
|
||
// k0 = -m**-1 mod 2**_W. Algorithm from: Dumas, J.G. "On Newton–Raphson
|
||
// Iteration for Multiplicative Inverses Modulo Prime Powers".
|
||
k0 := 2 - m[0]
|
||
t := m[0] - 1
|
||
for i := 1; i < _W; i <<= 1 {
|
||
t *= t
|
||
k0 *= (t + 1)
|
||
}
|
||
k0 = -k0
|
||
|
||
// RR = 2**(2*_W*len(m)) mod m
|
||
RR := nat(nil).setWord(1)
|
||
zz := nat(nil).shl(RR, uint(2*numWords*_W))
|
||
_, RR = RR.div(RR, zz, m)
|
||
if len(RR) < numWords {
|
||
zz = zz.make(numWords)
|
||
copy(zz, RR)
|
||
RR = zz
|
||
}
|
||
// one = 1, with equal length to that of m
|
||
one := make(nat, numWords)
|
||
one[0] = 1
|
||
|
||
const n = 4
|
||
// powers[i] contains x^i
|
||
var powers [1 << n]nat
|
||
powers[0] = powers[0].montgomery(one, RR, m, k0, numWords)
|
||
powers[1] = powers[1].montgomery(x, RR, m, k0, numWords)
|
||
for i := 2; i < 1<<n; i++ {
|
||
powers[i] = powers[i].montgomery(powers[i-1], powers[1], m, k0, numWords)
|
||
}
|
||
|
||
// initialize z = 1 (Montgomery 1)
|
||
z = z.make(numWords)
|
||
copy(z, powers[0])
|
||
|
||
zz = zz.make(numWords)
|
||
|
||
// same windowed exponent, but with Montgomery multiplications
|
||
for i := len(y) - 1; i >= 0; i-- {
|
||
yi := y[i]
|
||
for j := 0; j < _W; j += n {
|
||
if i != len(y)-1 || j != 0 {
|
||
zz = zz.montgomery(z, z, m, k0, numWords)
|
||
z = z.montgomery(zz, zz, m, k0, numWords)
|
||
zz = zz.montgomery(z, z, m, k0, numWords)
|
||
z = z.montgomery(zz, zz, m, k0, numWords)
|
||
}
|
||
zz = zz.montgomery(z, powers[yi>>(_W-n)], m, k0, numWords)
|
||
z, zz = zz, z
|
||
yi <<= n
|
||
}
|
||
}
|
||
// convert to regular number
|
||
zz = zz.montgomery(z, one, m, k0, numWords)
|
||
|
||
// One last reduction, just in case.
|
||
// See golang.org/issue/13907.
|
||
if zz.cmp(m) >= 0 {
|
||
// Common case is m has high bit set; in that case,
|
||
// since zz is the same length as m, there can be just
|
||
// one multiple of m to remove. Just subtract.
|
||
// We think that the subtract should be sufficient in general,
|
||
// so do that unconditionally, but double-check,
|
||
// in case our beliefs are wrong.
|
||
// The div is not expected to be reached.
|
||
zz = zz.sub(zz, m)
|
||
if zz.cmp(m) >= 0 {
|
||
_, zz = nat(nil).div(nil, zz, m)
|
||
}
|
||
}
|
||
|
||
return zz.norm()
|
||
}
|
||
|
||
// bytes writes the value of z into buf using big-endian encoding.
|
||
// len(buf) must be >= len(z)*_S. The value of z is encoded in the
|
||
// slice buf[i:]. The number i of unused bytes at the beginning of
|
||
// buf is returned as result.
|
||
func (z nat) bytes(buf []byte) (i int) {
|
||
i = len(buf)
|
||
for _, d := range z {
|
||
for j := 0; j < _S; j++ {
|
||
i--
|
||
buf[i] = byte(d)
|
||
d >>= 8
|
||
}
|
||
}
|
||
|
||
for i < len(buf) && buf[i] == 0 {
|
||
i++
|
||
}
|
||
|
||
return
|
||
}
|
||
|
||
// setBytes interprets buf as the bytes of a big-endian unsigned
|
||
// integer, sets z to that value, and returns z.
|
||
func (z nat) setBytes(buf []byte) nat {
|
||
z = z.make((len(buf) + _S - 1) / _S)
|
||
|
||
k := 0
|
||
s := uint(0)
|
||
var d Word
|
||
for i := len(buf); i > 0; i-- {
|
||
d |= Word(buf[i-1]) << s
|
||
if s += 8; s == _S*8 {
|
||
z[k] = d
|
||
k++
|
||
s = 0
|
||
d = 0
|
||
}
|
||
}
|
||
if k < len(z) {
|
||
z[k] = d
|
||
}
|
||
|
||
return z.norm()
|
||
}
|
||
|
||
// sqrt sets z = ⌊√x⌋
|
||
func (z nat) sqrt(x nat) nat {
|
||
if x.cmp(natOne) <= 0 {
|
||
return z.set(x)
|
||
}
|
||
if alias(z, x) {
|
||
z = nil
|
||
}
|
||
|
||
// Start with value known to be too large and repeat "z = ⌊(z + ⌊x/z⌋)/2⌋" until it stops getting smaller.
|
||
// See Brent and Zimmermann, Modern Computer Arithmetic, Algorithm 1.13 (SqrtInt).
|
||
// https://members.loria.fr/PZimmermann/mca/pub226.html
|
||
// If x is one less than a perfect square, the sequence oscillates between the correct z and z+1;
|
||
// otherwise it converges to the correct z and stays there.
|
||
var z1, z2 nat
|
||
z1 = z
|
||
z1 = z1.setUint64(1)
|
||
z1 = z1.shl(z1, uint(x.bitLen()/2+1)) // must be ≥ √x
|
||
for n := 0; ; n++ {
|
||
z2, _ = z2.div(nil, x, z1)
|
||
z2 = z2.add(z2, z1)
|
||
z2 = z2.shr(z2, 1)
|
||
if z2.cmp(z1) >= 0 {
|
||
// z1 is answer.
|
||
// Figure out whether z1 or z2 is currently aliased to z by looking at loop count.
|
||
if n&1 == 0 {
|
||
return z1
|
||
}
|
||
return z.set(z1)
|
||
}
|
||
z1, z2 = z2, z1
|
||
}
|
||
}
|