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c007ce824d
Preparation was in CL 134570043. This CL contains only the effect of 'hg mv src/pkg/* src'. For more about the move, see golang.org/s/go14nopkg.
144 lines
4.6 KiB
Go
144 lines
4.6 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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package math
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// The original C code and the long comment below are
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// from FreeBSD's /usr/src/lib/msun/src/e_sqrt.c and
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// came with this notice. The go code is a simplified
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// version of the original C.
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//
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// ====================================================
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// Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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//
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// Developed at SunPro, a Sun Microsystems, Inc. business.
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// Permission to use, copy, modify, and distribute this
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// software is freely granted, provided that this notice
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// is preserved.
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// ====================================================
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//
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// __ieee754_sqrt(x)
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// Return correctly rounded sqrt.
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// -----------------------------------------
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// | Use the hardware sqrt if you have one |
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// -----------------------------------------
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// Method:
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// Bit by bit method using integer arithmetic. (Slow, but portable)
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// 1. Normalization
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// Scale x to y in [1,4) with even powers of 2:
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// find an integer k such that 1 <= (y=x*2**(2k)) < 4, then
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// sqrt(x) = 2**k * sqrt(y)
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// 2. Bit by bit computation
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// Let q = sqrt(y) truncated to i bit after binary point (q = 1),
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// i 0
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// i+1 2
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// s = 2*q , and y = 2 * ( y - q ). (1)
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// i i i i
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//
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// To compute q from q , one checks whether
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// i+1 i
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//
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// -(i+1) 2
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// (q + 2 ) <= y. (2)
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// i
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// -(i+1)
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// If (2) is false, then q = q ; otherwise q = q + 2 .
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// i+1 i i+1 i
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//
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// With some algebraic manipulation, it is not difficult to see
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// that (2) is equivalent to
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// -(i+1)
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// s + 2 <= y (3)
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// i i
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//
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// The advantage of (3) is that s and y can be computed by
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// i i
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// the following recurrence formula:
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// if (3) is false
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//
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// s = s , y = y ; (4)
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// i+1 i i+1 i
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//
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// otherwise,
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// -i -(i+1)
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// s = s + 2 , y = y - s - 2 (5)
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// i+1 i i+1 i i
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//
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// One may easily use induction to prove (4) and (5).
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// Note. Since the left hand side of (3) contain only i+2 bits,
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// it does not necessary to do a full (53-bit) comparison
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// in (3).
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// 3. Final rounding
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// After generating the 53 bits result, we compute one more bit.
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// Together with the remainder, we can decide whether the
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// result is exact, bigger than 1/2ulp, or less than 1/2ulp
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// (it will never equal to 1/2ulp).
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// The rounding mode can be detected by checking whether
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// huge + tiny is equal to huge, and whether huge - tiny is
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// equal to huge for some floating point number "huge" and "tiny".
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//
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//
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// Notes: Rounding mode detection omitted. The constants "mask", "shift",
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// and "bias" are found in src/math/bits.go
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// Sqrt returns the square root of x.
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//
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// Special cases are:
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// Sqrt(+Inf) = +Inf
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// Sqrt(±0) = ±0
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// Sqrt(x < 0) = NaN
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// Sqrt(NaN) = NaN
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func Sqrt(x float64) float64
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func sqrt(x float64) float64 {
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// special cases
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switch {
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case x == 0 || IsNaN(x) || IsInf(x, 1):
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return x
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case x < 0:
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return NaN()
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}
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ix := Float64bits(x)
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// normalize x
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exp := int((ix >> shift) & mask)
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if exp == 0 { // subnormal x
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for ix&1<<shift == 0 {
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ix <<= 1
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exp--
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}
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exp++
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}
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exp -= bias // unbias exponent
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ix &^= mask << shift
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ix |= 1 << shift
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if exp&1 == 1 { // odd exp, double x to make it even
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ix <<= 1
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}
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exp >>= 1 // exp = exp/2, exponent of square root
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// generate sqrt(x) bit by bit
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ix <<= 1
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var q, s uint64 // q = sqrt(x)
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r := uint64(1 << (shift + 1)) // r = moving bit from MSB to LSB
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for r != 0 {
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t := s + r
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if t <= ix {
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s = t + r
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ix -= t
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q += r
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}
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ix <<= 1
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r >>= 1
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}
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// final rounding
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if ix != 0 { // remainder, result not exact
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q += q & 1 // round according to extra bit
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}
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ix = q>>1 + uint64(exp-1+bias)<<shift // significand + biased exponent
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return Float64frombits(ix)
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}
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func sqrtC(f float64, r *float64) {
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*r = sqrt(f)
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}
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