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-
-/* @(#)e_sqrt.c 1.3 95/01/18 */
-/*
- * ====================================================
- * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
- *
- * Developed at SunSoft, a Sun Microsystems, Inc. business.
- * Permission to use, copy, modify, and distribute this
- * software is freely granted, provided that this notice
- * is preserved.
- * ====================================================
- */
-
-/* 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.h>
-#include "math_private.h"
-
-static const double one = 1.0, tiny=1.0e-300;
-
-double
-sqrt(double x)
-{
- double z;
- int32_t sign = (int)0x80000000;
- int32_t ix0,s0,q,m,t,i;
- uint32_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==(uint32_t)0xffffffff) { q1=0; q += 1;}
- else if (z>one) {
- if (q1==(uint32_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;
-}
-
-/*
-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).
-
- */
-
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