1 | /* |
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2 | * ==================================================== |
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3 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. |
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4 | * |
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5 | * Developed at SunPro, a Sun Microsystems, Inc. business. |
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6 | * Permission to use, copy, modify, and distribute this |
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7 | * software is freely granted, provided that this notice |
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8 | * is preserved. |
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9 | * ==================================================== |
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10 | */ |
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11 | |
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12 | /* |
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13 | * Modified for ALMOS-MKH OS at UPMC, France, August 2018. (Alain Greiner) |
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14 | */ |
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15 | |
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16 | /* __ieee754_sqrt(x) |
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17 | * Return correctly rounded sqrt. |
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18 | * ------------------------------------------ |
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19 | * | Use the hardware sqrt if you have one | |
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20 | * ------------------------------------------ |
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21 | * Method: |
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22 | * Bit by bit method using integer arithmetic. (Slow, but portable) |
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23 | * 1. Normalization |
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24 | * Scale x to y in [1,4) with even powers of 2: |
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25 | * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then |
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26 | * sqrt(x) = 2^k * sqrt(y) |
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27 | * 2. Bit by bit computation |
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28 | * Let q = sqrt(y) truncated to i bit after binary point (q = 1), |
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29 | * i 0 |
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30 | * i+1 2 |
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31 | * s = 2*q , and y = 2 * ( y - q ). (1) |
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32 | * i i i i |
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33 | * |
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34 | * To compute q from q , one checks whether |
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35 | * i+1 i |
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36 | * |
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37 | * -(i+1) 2 |
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38 | * (q + 2 ) <= y. (2) |
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39 | * i |
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40 | * -(i+1) |
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41 | * If (2) is false, then q = q ; otherwise q = q + 2 . |
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42 | * i+1 i i+1 i |
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43 | * |
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44 | * With some algebric manipulation, it is not difficult to see |
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45 | * that (2) is equivalent to |
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46 | * -(i+1) |
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47 | * s + 2 <= y (3) |
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48 | * i i |
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49 | * |
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50 | * The advantage of (3) is that s and y can be computed by |
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51 | * i i |
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52 | * the following recurrence formula: |
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53 | * if (3) is false |
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54 | * |
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55 | * s = s , y = y ; (4) |
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56 | * i+1 i i+1 i |
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57 | * |
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58 | * otherwise, |
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59 | * -i -(i+1) |
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60 | * s = s + 2 , y = y - s - 2 (5) |
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61 | * i+1 i i+1 i i |
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62 | * |
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63 | * One may easily use induction to prove (4) and (5). |
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64 | * Note. Since the left hand side of (3) contain only i+2 bits, |
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65 | * it does not necessary to do a full (53-bit) comparison |
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66 | * in (3). |
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67 | * 3. Final rounding |
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68 | * After generating the 53 bits result, we compute one more bit. |
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69 | * Together with the remainder, we can decide whether the |
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70 | * result is exact, bigger than 1/2ulp, or less than 1/2ulp |
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71 | * (it will never equal to 1/2ulp). |
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72 | * The rounding mode can be detected by checking whether |
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73 | * huge + tiny is equal to huge, and whether huge - tiny is |
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74 | * equal to huge for some floating point number "huge" and "tiny". |
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75 | * |
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76 | * Special cases: |
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77 | * sqrt(+-0) = +-0 ... exact |
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78 | * sqrt(inf) = inf |
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79 | * sqrt(-ve) = NaN ... with invalid signal |
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80 | * sqrt(NaN) = NaN ... with invalid signal for signaling NaN |
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81 | * |
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82 | * Other methods : see the appended file at the end of the program below. |
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83 | *--------------- |
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84 | */ |
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85 | |
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86 | #include "math.h" |
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87 | #include "math_private.h" |
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88 | |
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89 | static const double one = 1.0, tiny = 1.0e-300; |
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90 | |
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91 | double __ieee754_sqrt(double x) |
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92 | { |
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93 | double z; |
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94 | int32_t sign = (int)0x80000000; |
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95 | int32_t ix0,s0,q,m,t,i; |
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96 | uint32_t r,t1,s1,ix1,q1; |
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97 | |
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98 | EXTRACT_WORDS(ix0,ix1,x); |
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99 | |
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100 | /* take care of Inf and NaN */ |
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101 | if((ix0&0x7ff00000)==0x7ff00000) { |
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102 | return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf |
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103 | sqrt(-inf)=sNaN */ |
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104 | } |
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105 | /* take care of zero */ |
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106 | if(ix0<=0) { |
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107 | if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ |
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108 | else if(ix0<0) |
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109 | return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ |
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110 | } |
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111 | /* normalize x */ |
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112 | m = (ix0>>20); |
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113 | if(m==0) { /* subnormal x */ |
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114 | while(ix0==0) { |
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115 | m -= 21; |
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116 | ix0 |= (ix1>>11); ix1 <<= 21; |
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117 | } |
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118 | for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; |
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119 | m -= i-1; |
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120 | ix0 |= (ix1>>(32-i)); |
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121 | ix1 <<= i; |
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122 | } |
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123 | m -= 1023; /* unbias exponent */ |
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124 | ix0 = (ix0&0x000fffff)|0x00100000; |
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125 | if(m&1){ /* odd m, double x to make it even */ |
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126 | ix0 += ix0 + ((ix1&sign)>>31); |
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127 | ix1 += ix1; |
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128 | } |
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129 | m >>= 1; /* m = [m/2] */ |
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130 | |
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131 | /* generate sqrt(x) bit by bit */ |
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132 | ix0 += ix0 + ((ix1&sign)>>31); |
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133 | ix1 += ix1; |
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134 | q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ |
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135 | r = 0x00200000; /* r = moving bit from right to left */ |
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136 | |
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137 | while(r!=0) { |
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138 | t = s0+r; |
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139 | if(t<=ix0) { |
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140 | s0 = t+r; |
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141 | ix0 -= t; |
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142 | q += r; |
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143 | } |
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144 | ix0 += ix0 + ((ix1&sign)>>31); |
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145 | ix1 += ix1; |
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146 | r>>=1; |
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147 | } |
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148 | |
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149 | r = sign; |
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150 | while(r!=0) { |
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151 | t1 = s1+r; |
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152 | t = s0; |
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153 | if((t<ix0)||((t==ix0)&&(t1<=ix1))) { |
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154 | s1 = t1+r; |
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155 | if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1; |
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156 | ix0 -= t; |
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157 | if (ix1 < t1) ix0 -= 1; |
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158 | ix1 -= t1; |
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159 | q1 += r; |
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160 | } |
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161 | ix0 += ix0 + ((ix1&sign)>>31); |
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162 | ix1 += ix1; |
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163 | r>>=1; |
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164 | } |
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165 | |
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166 | /* use floating add to find out rounding direction */ |
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167 | if((ix0|ix1)!=0) { |
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168 | z = one-tiny; /* trigger inexact flag */ |
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169 | if (z>=one) { |
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170 | z = one+tiny; |
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171 | if (q1==(uint32_t)0xffffffff) { q1=0; q += 1;} |
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172 | else if (z>one) { |
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173 | if (q1==(uint32_t)0xfffffffe) q+=1; |
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174 | q1+=2; |
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175 | } else |
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176 | q1 += (q1&1); |
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177 | } |
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178 | } |
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179 | ix0 = (q>>1)+0x3fe00000; |
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180 | ix1 = q1>>1; |
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181 | if ((q&1)==1) ix1 |= sign; |
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182 | ix0 += (m <<20); |
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183 | INSERT_WORDS(z,ix0,ix1); |
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184 | return z; |
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185 | } |
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186 | |
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187 | |
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188 | |
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189 | /* |
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190 | Other methods (use floating-point arithmetic) |
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191 | ------------- |
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192 | (This is a copy of a drafted paper by Prof W. Kahan |
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193 | and K.C. Ng, written in May, 1986) |
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194 | |
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195 | Two algorithms are given here to implement sqrt(x) |
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196 | (IEEE double precision arithmetic) in software. |
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197 | Both supply sqrt(x) correctly rounded. The first algorithm (in |
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198 | Section A) uses newton iterations and involves four divisions. |
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199 | The second one uses reciproot iterations to avoid division, but |
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200 | requires more multiplications. Both algorithms need the ability |
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201 | to chop results of arithmetic operations instead of round them, |
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202 | and the INEXACT flag to indicate when an arithmetic operation |
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203 | is executed exactly with no roundoff error, all part of the |
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204 | standard (IEEE 754-1985). The ability to perform shift, add, |
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205 | subtract and logical AND operations upon 32-bit words is needed |
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206 | too, though not part of the standard. |
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207 | |
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208 | A. sqrt(x) by Newton Iteration |
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209 | |
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210 | (1) Initial approximation |
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211 | |
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212 | Let x0 and x1 be the leading and the trailing 32-bit words of |
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213 | a floating point number x (in IEEE double format) respectively |
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214 | |
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215 | 1 11 52 ...widths |
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216 | ------------------------------------------------------ |
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217 | x: |s| e | f | |
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218 | ------------------------------------------------------ |
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219 | msb lsb msb lsb ...order |
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220 | |
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221 | |
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222 | ------------------------ ------------------------ |
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223 | x0: |s| e | f1 | x1: | f2 | |
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224 | ------------------------ ------------------------ |
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225 | |
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226 | By performing shifts and subtracts on x0 and x1 (both regarded |
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227 | as integers), we obtain an 8-bit approximation of sqrt(x) as |
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228 | follows. |
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229 | |
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230 | k := (x0>>1) + 0x1ff80000; |
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231 | y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits |
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232 | Here k is a 32-bit integer and T1[] is an integer array containing |
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233 | correction terms. Now magically the floating value of y (y's |
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234 | leading 32-bit word is y0, the value of its trailing word is 0) |
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235 | approximates sqrt(x) to almost 8-bit. |
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236 | |
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237 | Value of T1: |
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238 | static int T1[32]= { |
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239 | 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, |
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240 | 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, |
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241 | 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, |
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242 | 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; |
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243 | |
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244 | (2) Iterative refinement |
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245 | |
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246 | Apply Heron's rule three times to y, we have y approximates |
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247 | sqrt(x) to within 1 ulp (Unit in the Last Place): |
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248 | |
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249 | y := (y+x/y)/2 ... almost 17 sig. bits |
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250 | y := (y+x/y)/2 ... almost 35 sig. bits |
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251 | y := y-(y-x/y)/2 ... within 1 ulp |
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252 | |
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253 | |
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254 | Remark 1. |
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255 | Another way to improve y to within 1 ulp is: |
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256 | |
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257 | y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) |
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258 | y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) |
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259 | |
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260 | 2 |
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261 | (x-y )*y |
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262 | y := y + 2* ---------- ...within 1 ulp |
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263 | 2 |
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264 | 3y + x |
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265 | |
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266 | |
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267 | This formula has one division fewer than the one above; however, |
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268 | it requires more multiplications and additions. Also x must be |
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269 | scaled in advance to avoid spurious overflow in evaluating the |
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270 | expression 3y*y+x. Hence it is not recommended uless division |
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271 | is slow. If division is very slow, then one should use the |
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272 | reciproot algorithm given in section B. |
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273 | |
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274 | (3) Final adjustment |
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275 | |
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276 | By twiddling y's last bit it is possible to force y to be |
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277 | correctly rounded according to the prevailing rounding mode |
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278 | as follows. Let r and i be copies of the rounding mode and |
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279 | inexact flag before entering the square root program. Also we |
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280 | use the expression y+-ulp for the next representable floating |
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281 | numbers (up and down) of y. Note that y+-ulp = either fixed |
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282 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
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283 | mode. |
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284 | |
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285 | I := FALSE; ... reset INEXACT flag I |
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286 | R := RZ; ... set rounding mode to round-toward-zero |
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287 | z := x/y; ... chopped quotient, possibly inexact |
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288 | If(not I) then { ... if the quotient is exact |
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289 | if(z=y) { |
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290 | I := i; ... restore inexact flag |
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291 | R := r; ... restore rounded mode |
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292 | return sqrt(x):=y. |
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293 | } else { |
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294 | z := z - ulp; ... special rounding |
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295 | } |
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296 | } |
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297 | i := TRUE; ... sqrt(x) is inexact |
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298 | If (r=RN) then z=z+ulp ... rounded-to-nearest |
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299 | If (r=RP) then { ... round-toward-+inf |
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300 | y = y+ulp; z=z+ulp; |
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301 | } |
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302 | y := y+z; ... chopped sum |
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303 | y0:=y0-0x00100000; ... y := y/2 is correctly rounded. |
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304 | I := i; ... restore inexact flag |
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305 | R := r; ... restore rounded mode |
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306 | return sqrt(x):=y. |
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307 | |
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308 | (4) Special cases |
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309 | |
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310 | Square root of +inf, +-0, or NaN is itself; |
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311 | Square root of a negative number is NaN with invalid signal. |
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312 | |
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313 | |
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314 | B. sqrt(x) by Reciproot Iteration |
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315 | |
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316 | (1) Initial approximation |
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317 | |
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318 | Let x0 and x1 be the leading and the trailing 32-bit words of |
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319 | a floating point number x (in IEEE double format) respectively |
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320 | (see section A). By performing shifs and subtracts on x0 and y0, |
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321 | we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. |
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322 | |
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323 | k := 0x5fe80000 - (x0>>1); |
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324 | y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits |
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325 | |
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326 | Here k is a 32-bit integer and T2[] is an integer array |
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327 | containing correction terms. Now magically the floating |
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328 | value of y (y's leading 32-bit word is y0, the value of |
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329 | its trailing word y1 is set to zero) approximates 1/sqrt(x) |
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330 | to almost 7.8-bit. |
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331 | |
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332 | Value of T2: |
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333 | static int T2[64]= { |
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334 | 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, |
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335 | 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, |
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336 | 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, |
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337 | 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, |
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338 | 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, |
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339 | 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, |
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340 | 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, |
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341 | 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; |
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342 | |
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343 | (2) Iterative refinement |
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344 | |
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345 | Apply Reciproot iteration three times to y and multiply the |
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346 | result by x to get an approximation z that matches sqrt(x) |
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347 | to about 1 ulp. To be exact, we will have |
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348 | -1ulp < sqrt(x)-z<1.0625ulp. |
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349 | |
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350 | ... set rounding mode to Round-to-nearest |
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351 | y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) |
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352 | y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) |
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353 | ... special arrangement for better accuracy |
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354 | z := x*y ... 29 bits to sqrt(x), with z*y<1 |
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355 | z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) |
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356 | |
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357 | Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that |
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358 | (a) the term z*y in the final iteration is always less than 1; |
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359 | (b) the error in the final result is biased upward so that |
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360 | -1 ulp < sqrt(x) - z < 1.0625 ulp |
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361 | instead of |sqrt(x)-z|<1.03125ulp. |
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362 | |
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363 | (3) Final adjustment |
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364 | |
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365 | By twiddling y's last bit it is possible to force y to be |
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366 | correctly rounded according to the prevailing rounding mode |
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367 | as follows. Let r and i be copies of the rounding mode and |
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368 | inexact flag before entering the square root program. Also we |
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369 | use the expression y+-ulp for the next representable floating |
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370 | numbers (up and down) of y. Note that y+-ulp = either fixed |
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371 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
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372 | mode. |
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373 | |
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374 | R := RZ; ... set rounding mode to round-toward-zero |
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375 | switch(r) { |
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376 | case RN: ... round-to-nearest |
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377 | if(x<= z*(z-ulp)...chopped) z = z - ulp; else |
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378 | if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; |
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379 | break; |
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380 | case RZ:case RM: ... round-to-zero or round-to--inf |
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381 | R:=RP; ... reset rounding mod to round-to-+inf |
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382 | if(x<z*z ... rounded up) z = z - ulp; else |
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383 | if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; |
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384 | break; |
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385 | case RP: ... round-to-+inf |
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386 | if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else |
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387 | if(x>z*z ...chopped) z = z+ulp; |
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388 | break; |
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389 | } |
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390 | |
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391 | Remark 3. The above comparisons can be done in fixed point. For |
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392 | example, to compare x and w=z*z chopped, it suffices to compare |
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393 | x1 and w1 (the trailing parts of x and w), regarding them as |
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394 | two's complement integers. |
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395 | |
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396 | ...Is z an exact square root? |
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397 | To determine whether z is an exact square root of x, let z1 be the |
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398 | trailing part of z, and also let x0 and x1 be the leading and |
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399 | trailing parts of x. |
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400 | |
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401 | If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 |
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402 | I := 1; ... Raise Inexact flag: z is not exact |
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403 | else { |
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404 | j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 |
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405 | k := z1 >> 26; ... get z's 25-th and 26-th |
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406 | fraction bits |
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407 | I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); |
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408 | } |
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409 | R:= r ... restore rounded mode |
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410 | return sqrt(x):=z. |
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411 | |
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412 | If multiplication is cheaper then the foregoing red tape, the |
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413 | Inexact flag can be evaluated by |
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414 | |
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415 | I := i; |
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416 | I := (z*z!=x) or I. |
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417 | |
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418 | Note that z*z can overwrite I; this value must be sensed if it is |
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419 | True. |
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420 | |
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421 | Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be |
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422 | zero. |
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423 | |
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424 | -------------------- |
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425 | z1: | f2 | |
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426 | -------------------- |
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427 | bit 31 bit 0 |
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428 | |
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429 | Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd |
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430 | or even of logb(x) have the following relations: |
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431 | |
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432 | ------------------------------------------------- |
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433 | bit 27,26 of z1 bit 1,0 of x1 logb(x) |
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434 | ------------------------------------------------- |
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435 | 00 00 odd and even |
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436 | 01 01 even |
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437 | 10 10 odd |
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438 | 10 00 even |
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439 | 11 01 even |
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440 | ------------------------------------------------- |
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441 | |
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442 | (4) Special cases (see (4) of Section A). |
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443 | |
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444 | */ |
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