skia / external / github.com / libsdl-org / SDL / 9fdc6cf307b50df62c074352b62892758f599a00 / . / src / libm / e_sqrt.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 algebric 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". | |

* | |

* Special cases: | |

* sqrt(+-0) = +-0 ... exact | |

* sqrt(inf) = inf | |

* sqrt(-ve) = NaN ... with invalid signal | |

* sqrt(NaN) = NaN ... with invalid signal for signaling NaN | |

* | |

* Other methods : see the appended file at the end of the program below. | |

*--------------- | |

*/ | |

#include "math_libm.h" | |

#include "math_private.h" | |

static const double one = 1.0, tiny = 1.0e-300; | |

double attribute_hidden __ieee754_sqrt(double x) | |

{ | |

double z; | |

int32_t sign = (int)0x80000000; | |

int32_t ix0,s0,q,m,t,i; | |

u_int32_t r,t1,s1,ix1,q1; | |

EXTRACT_WORDS(ix0,ix1,x); | |

/* take care of Inf and NaN */ | |

if((ix0&0x7ff00000)==0x7ff00000) { | |

return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf | |

sqrt(-inf)=sNaN */ | |

} | |

/* take care of zero */ | |

if(ix0<=0) { | |

if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ | |

else if(ix0<0) | |

return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ | |

} | |

/* normalize x */ | |

m = (ix0>>20); | |

if(m==0) { /* subnormal x */ | |

while(ix0==0) { | |

m -= 21; | |

ix0 |= (ix1>>11); ix1 <<= 21; | |

} | |

for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; | |

m -= i-1; | |

ix0 |= (ix1>>(32-i)); | |

ix1 <<= i; | |

} | |

m -= 1023; /* unbias exponent */ | |

ix0 = (ix0&0x000fffff)|0x00100000; | |

if(m&1){ /* odd m, double x to make it even */ | |

ix0 += ix0 + ((ix1&sign)>>31); | |

ix1 += ix1; | |

} | |

m >>= 1; /* m = [m/2] */ | |

/* generate sqrt(x) bit by bit */ | |

ix0 += ix0 + ((ix1&sign)>>31); | |

ix1 += ix1; | |

q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ | |

r = 0x00200000; /* r = moving bit from right to left */ | |

while(r!=0) { | |

t = s0+r; | |

if(t<=ix0) { | |

s0 = t+r; | |

ix0 -= t; | |

q += r; | |

} | |

ix0 += ix0 + ((ix1&sign)>>31); | |

ix1 += ix1; | |

r>>=1; | |

} | |

r = sign; | |

while(r!=0) { | |

t1 = s1+r; | |

t = s0; | |

if((t<ix0)||((t==ix0)&&(t1<=ix1))) { | |

s1 = t1+r; | |

if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1; | |

ix0 -= t; | |

if (ix1 < t1) ix0 -= 1; | |

ix1 -= t1; | |

q1 += r; | |

} | |

ix0 += ix0 + ((ix1&sign)>>31); | |

ix1 += ix1; | |

r>>=1; | |

} | |

/* use floating add to find out rounding direction */ | |

if((ix0|ix1)!=0) { | |

z = one-tiny; /* trigger inexact flag */ | |

if (z>=one) { | |

z = one+tiny; | |

if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;} | |

else if (z>one) { | |

if (q1==(u_int32_t)0xfffffffe) q+=1; | |

q1+=2; | |

} else | |

q1 += (q1&1); | |

} | |

} | |

ix0 = (q>>1)+0x3fe00000; | |

ix1 = q1>>1; | |

if ((q&1)==1) ix1 |= sign; | |

ix0 += (m <<20); | |

INSERT_WORDS(z,ix0,ix1); | |

return z; | |

} | |

/* | |

* wrapper sqrt(x) | |

*/ | |

#ifndef _IEEE_LIBM | |

double sqrt(double x) | |

{ | |

double z = __ieee754_sqrt(x); | |

if (_LIB_VERSION == _IEEE_ || isnan(x)) | |

return z; | |

if (x < 0.0) | |

return __kernel_standard(x, x, 26); /* sqrt(negative) */ | |

return z; | |

} | |

#else | |

strong_alias(__ieee754_sqrt, sqrt) | |

#endif | |

libm_hidden_def(sqrt) | |

/* | |

Other methods (use floating-point arithmetic) | |

------------- | |

(This is a copy of a drafted paper by Prof W. Kahan | |

and K.C. Ng, written in May, 1986) | |

Two algorithms are given here to implement sqrt(x) | |

(IEEE double precision arithmetic) in software. | |

Both supply sqrt(x) correctly rounded. The first algorithm (in | |

Section A) uses newton iterations and involves four divisions. | |

The second one uses reciproot iterations to avoid division, but | |

requires more multiplications. Both algorithms need the ability | |

to chop results of arithmetic operations instead of round them, | |

and the INEXACT flag to indicate when an arithmetic operation | |

is executed exactly with no roundoff error, all part of the | |

standard (IEEE 754-1985). The ability to perform shift, add, | |

subtract and logical AND operations upon 32-bit words is needed | |

too, though not part of the standard. | |

A. sqrt(x) by Newton Iteration | |

(1) Initial approximation | |

Let x0 and x1 be the leading and the trailing 32-bit words of | |

a floating point number x (in IEEE double format) respectively | |

1 11 52 ...widths | |

------------------------------------------------------ | |

x: |s| e | f | | |

------------------------------------------------------ | |

msb lsb msb lsb ...order | |

------------------------ ------------------------ | |

x0: |s| e | f1 | x1: | f2 | | |

------------------------ ------------------------ | |

By performing shifts and subtracts on x0 and x1 (both regarded | |

as integers), we obtain an 8-bit approximation of sqrt(x) as | |

follows. | |

k := (x0>>1) + 0x1ff80000; | |

y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits | |

Here k is a 32-bit integer and T1[] is an integer array containing | |

correction terms. Now magically the floating value of y (y's | |

leading 32-bit word is y0, the value of its trailing word is 0) | |

approximates sqrt(x) to almost 8-bit. | |

Value of T1: | |

static int T1[32]= { | |

0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, | |

29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, | |

83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, | |

16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; | |

(2) Iterative refinement | |

Apply Heron's rule three times to y, we have y approximates | |

sqrt(x) to within 1 ulp (Unit in the Last Place): | |

y := (y+x/y)/2 ... almost 17 sig. bits | |

y := (y+x/y)/2 ... almost 35 sig. bits | |

y := y-(y-x/y)/2 ... within 1 ulp | |

Remark 1. | |

Another way to improve y to within 1 ulp is: | |

y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) | |

y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) | |

2 | |

(x-y )*y | |

y := y + 2* ---------- ...within 1 ulp | |

2 | |

3y + x | |

This formula has one division fewer than the one above; however, | |

it requires more multiplications and additions. Also x must be | |

scaled in advance to avoid spurious overflow in evaluating the | |

expression 3y*y+x. Hence it is not recommended uless division | |

is slow. If division is very slow, then one should use the | |

reciproot algorithm given in section B. | |

(3) Final adjustment | |

By twiddling y's last bit it is possible to force y to be | |

correctly rounded according to the prevailing rounding mode | |

as follows. Let r and i be copies of the rounding mode and | |

inexact flag before entering the square root program. Also we | |

use the expression y+-ulp for the next representable floating | |

numbers (up and down) of y. Note that y+-ulp = either fixed | |

point y+-1, or multiply y by nextafter(1,+-inf) in chopped | |

mode. | |

I := FALSE; ... reset INEXACT flag I | |

R := RZ; ... set rounding mode to round-toward-zero | |

z := x/y; ... chopped quotient, possibly inexact | |

If(not I) then { ... if the quotient is exact | |

if(z=y) { | |

I := i; ... restore inexact flag | |

R := r; ... restore rounded mode | |

return sqrt(x):=y. | |

} else { | |

z := z - ulp; ... special rounding | |

} | |

} | |

i := TRUE; ... sqrt(x) is inexact | |

If (r=RN) then z=z+ulp ... rounded-to-nearest | |

If (r=RP) then { ... round-toward-+inf | |

y = y+ulp; z=z+ulp; | |

} | |

y := y+z; ... chopped sum | |

y0:=y0-0x00100000; ... y := y/2 is correctly rounded. | |

I := i; ... restore inexact flag | |

R := r; ... restore rounded mode | |

return sqrt(x):=y. | |

(4) Special cases | |

Square root of +inf, +-0, or NaN is itself; | |

Square root of a negative number is NaN with invalid signal. | |

B. sqrt(x) by Reciproot Iteration | |

(1) Initial approximation | |

Let x0 and x1 be the leading and the trailing 32-bit words of | |

a floating point number x (in IEEE double format) respectively | |

(see section A). By performing shifs and subtracts on x0 and y0, | |

we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. | |

k := 0x5fe80000 - (x0>>1); | |

y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits | |

Here k is a 32-bit integer and T2[] is an integer array | |

containing correction terms. Now magically the floating | |

value of y (y's leading 32-bit word is y0, the value of | |

its trailing word y1 is set to zero) approximates 1/sqrt(x) | |

to almost 7.8-bit. | |

Value of T2: | |

static int T2[64]= { | |

0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, | |

0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, | |

0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, | |

0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, | |

0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, | |

0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, | |

0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, | |

0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; | |

(2) Iterative refinement | |

Apply Reciproot iteration three times to y and multiply the | |

result by x to get an approximation z that matches sqrt(x) | |

to about 1 ulp. To be exact, we will have | |

-1ulp < sqrt(x)-z<1.0625ulp. | |

... set rounding mode to Round-to-nearest | |

y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) | |

y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) | |

... special arrangement for better accuracy | |

z := x*y ... 29 bits to sqrt(x), with z*y<1 | |

z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) | |

Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that | |

(a) the term z*y in the final iteration is always less than 1; | |

(b) the error in the final result is biased upward so that | |

-1 ulp < sqrt(x) - z < 1.0625 ulp | |

instead of |sqrt(x)-z|<1.03125ulp. | |

(3) Final adjustment | |

By twiddling y's last bit it is possible to force y to be | |

correctly rounded according to the prevailing rounding mode | |

as follows. Let r and i be copies of the rounding mode and | |

inexact flag before entering the square root program. Also we | |

use the expression y+-ulp for the next representable floating | |

numbers (up and down) of y. Note that y+-ulp = either fixed | |

point y+-1, or multiply y by nextafter(1,+-inf) in chopped | |

mode. | |

R := RZ; ... set rounding mode to round-toward-zero | |

switch(r) { | |

case RN: ... round-to-nearest | |

if(x<= z*(z-ulp)...chopped) z = z - ulp; else | |

if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; | |

break; | |

case RZ:case RM: ... round-to-zero or round-to--inf | |

R:=RP; ... reset rounding mod to round-to-+inf | |

if(x<z*z ... rounded up) z = z - ulp; else | |

if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; | |

break; | |

case RP: ... round-to-+inf | |

if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else | |

if(x>z*z ...chopped) z = z+ulp; | |

break; | |

} | |

Remark 3. The above comparisons can be done in fixed point. For | |

example, to compare x and w=z*z chopped, it suffices to compare | |

x1 and w1 (the trailing parts of x and w), regarding them as | |

two's complement integers. | |

...Is z an exact square root? | |

To determine whether z is an exact square root of x, let z1 be the | |

trailing part of z, and also let x0 and x1 be the leading and | |

trailing parts of x. | |

If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 | |

I := 1; ... Raise Inexact flag: z is not exact | |

else { | |

j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 | |

k := z1 >> 26; ... get z's 25-th and 26-th | |

fraction bits | |

I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); | |

} | |

R:= r ... restore rounded mode | |

return sqrt(x):=z. | |

If multiplication is cheaper then the foregoing red tape, the | |

Inexact flag can be evaluated by | |

I := i; | |

I := (z*z!=x) or I. | |

Note that z*z can overwrite I; this value must be sensed if it is | |

True. | |

Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be | |

zero. | |

-------------------- | |

z1: | f2 | | |

-------------------- | |

bit 31 bit 0 | |

Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd | |

or even of logb(x) have the following relations: | |

------------------------------------------------- | |

bit 27,26 of z1 bit 1,0 of x1 logb(x) | |

------------------------------------------------- | |

00 00 odd and even | |

01 01 even | |

10 10 odd | |

10 00 even | |

11 01 even | |

------------------------------------------------- | |

(4) Special cases (see (4) of Section A). | |

*/ |