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Diffstat (limited to 'src/math/e_sqrt.c')
| -rw-r--r-- | src/math/e_sqrt.c | 442 |
1 files changed, 0 insertions, 442 deletions
diff --git a/src/math/e_sqrt.c b/src/math/e_sqrt.c deleted file mode 100644 index 2bc68747..00000000 --- a/src/math/e_sqrt.c +++ /dev/null @@ -1,442 +0,0 @@ - -/* @(#)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). - - */ - |