| /** @file | |
| Compute the logrithm of x. | |
| Copyright (c) 2010 - 2011, Intel Corporation. All rights reserved.<BR> | |
| This program and the accompanying materials are licensed and made available under | |
| the terms and conditions of the BSD License that accompanies this distribution. | |
| The full text of the license may be found at | |
| http://opensource.org/licenses/bsd-license. | |
| THE PROGRAM IS DISTRIBUTED UNDER THE BSD LICENSE ON AN "AS IS" BASIS, | |
| WITHOUT WARRANTIES OR REPRESENTATIONS OF ANY KIND, EITHER EXPRESS OR IMPLIED. | |
| * ==================================================== | |
| * 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. | |
| * ==================================================== | |
| e_sqrt.c 5.1 93/09/24 | |
| NetBSD: e_sqrt.c,v 1.12 2002/05/26 22:01:52 wiz Exp | |
| **/ | |
| #include <LibConfig.h> | |
| #include <sys/EfiCdefs.h> | |
| #include <errno.h> | |
| #include "math.h" | |
| #include "math_private.h" | |
| #if defined(_MSC_VER) /* Handle Microsoft VC++ compiler specifics. */ | |
| // potential divide by 0 -- near line 129, (x-x)/(x-x) is on purpose | |
| #pragma warning ( disable : 4723 ) | |
| #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. | |
| *--------------- | |
| */ | |
| static const double one = 1.0, tiny=1.0e-300; | |
| double | |
| __ieee754_sqrt(double x) | |
| { | |
| 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) { | |
| errno = EDOM; | |
| 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)==(u_int32_t)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 than 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). | |
| */ | |