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Diffstat (limited to 'src/math/e_sqrt.c')

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diff --git a/src/math/e_sqrt.c b/src/math/e_sqrt.c new file mode 100644 index 00000000..2bc68747 --- /dev/null +++ b/src/math/e_sqrt.c @@ -0,0 +1,442 @@ + +/* @(#)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). + + */ + |