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go/src/math/sqrt.go
Russ Cox c007ce824d build: move package sources from src/pkg to src
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.
2014-09-08 00:08:51 -04:00

144 lines
4.6 KiB
Go

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