/* @(#)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__ static const double one = 1.0, tiny=1.0e-300; #else static double one = 1.0, tiny=1.0e-300; #endif #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(((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). */

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