| /* @(#)e_sqrt.c 5.1 93/09/24 */ |
| /* |
| * ==================================================== |
| * 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. |
| * ==================================================== |
| */ |
| |
| #if defined(LIBM_SCCS) && !defined(lint) |
| static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $"; |
| #endif |
| |
| /* __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.h"*/ |
| #include "math_private.h" |
| |
| #ifdef __STDC__ |
| double SDL_NAME(copysign)(double x, double y) |
| #else |
| double SDL_NAME(copysign)(x,y) |
| double x,y; |
| #endif |
| { |
| u_int32_t hx,hy; |
| GET_HIGH_WORD(hx,x); |
| GET_HIGH_WORD(hy,y); |
| SET_HIGH_WORD(x,(hx&0x7fffffff)|(hy&0x80000000)); |
| return x; |
| } |
| |
| #ifdef __STDC__ |
| double SDL_NAME(scalbn) (double x, int n) |
| #else |
| double SDL_NAME(scalbn) (x,n) |
| double x; int n; |
| #endif |
| { |
| int32_t k,hx,lx; |
| EXTRACT_WORDS(hx,lx,x); |
| k = (hx&0x7ff00000)>>20; /* extract exponent */ |
| if (k==0) { /* 0 or subnormal x */ |
| if ((lx|(hx&0x7fffffff))==0) return x; /* +-0 */ |
| x *= two54; |
| GET_HIGH_WORD(hx,x); |
| k = ((hx&0x7ff00000)>>20) - 54; |
| if (n< -50000) return tiny*x; /*underflow*/ |
| } |
| if (k==0x7ff) return x+x; /* NaN or Inf */ |
| k = k+n; |
| if (k > 0x7fe) return huge*SDL_NAME(copysign)(huge,x); /* overflow */ |
| if (k > 0) /* normal result */ |
| {SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); return x;} |
| if (k <= -54) { |
| if (n > 50000) /* in case integer overflow in n+k */ |
| return huge*SDL_NAME(copysign)(huge,x); /*overflow*/ |
| else return tiny*SDL_NAME(copysign)(tiny,x); /*underflow*/ |
| } |
| k += 54; /* subnormal result */ |
| SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); |
| return x*twom54; |
| } |
| |
| #ifdef __STDC__ |
| double __ieee754_sqrt(double x) |
| #else |
| double __ieee754_sqrt(x) |
| double x; |
| #endif |
| { |
| 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(((int32_t)(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; |
| } |
| |
| /* |
| 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). |
| |
| */ |
| |
