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1 /* |
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2 Copyright (C) 2003 David Bateman |
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3 |
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4 This file is part of Octave. |
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5 |
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6 Octave is free software; you can redistribute it and/or modify it |
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7 under the terms of the GNU General Public License as published by the |
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8 Free Software Foundation; either version 2, or (at your option) any |
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9 later version. |
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10 |
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11 Octave is distributed in the hope that it will be useful, but WITHOUT |
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12 ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or |
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13 FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License |
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14 for more details. |
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15 |
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16 You should have received a copy of the GNU General Public License |
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17 along with Octave; see the file COPYING. If not, write to the Free |
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18 Software Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA |
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19 02110-1301, USA. |
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20 |
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21 Code stolen in large part from Python's, listobject.c, which itself had |
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22 no license header. However, thanks to Tim Peters for the parts of the |
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23 code I ripped-off. |
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24 |
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25 As required in the Python license the short description of the changes |
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26 made are |
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27 |
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28 * convert the sorting code in listobject.cc into a generic class, |
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29 replacing PyObject* with the type of the class T. |
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30 |
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31 The Python license is |
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32 |
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33 PSF LICENSE AGREEMENT FOR PYTHON 2.3 |
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34 -------------------------------------- |
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35 |
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36 1. This LICENSE AGREEMENT is between the Python Software Foundation |
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37 ("PSF"), and the Individual or Organization ("Licensee") accessing and |
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38 otherwise using Python 2.3 software in source or binary form and its |
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39 associated documentation. |
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40 |
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41 2. Subject to the terms and conditions of this License Agreement, PSF |
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42 hereby grants Licensee a nonexclusive, royalty-free, world-wide |
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43 license to reproduce, analyze, test, perform and/or display publicly, |
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44 prepare derivative works, distribute, and otherwise use Python 2.3 |
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45 alone or in any derivative version, provided, however, that PSF's |
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46 License Agreement and PSF's notice of copyright, i.e., "Copyright (c) |
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47 2001, 2002, 2003 Python Software Foundation; All Rights Reserved" are |
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48 retained in Python 2.3 alone or in any derivative version prepared by |
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49 Licensee. |
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50 |
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51 3. In the event Licensee prepares a derivative work that is based on |
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52 or incorporates Python 2.3 or any part thereof, and wants to make |
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53 the derivative work available to others as provided herein, then |
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54 Licensee hereby agrees to include in any such work a brief summary of |
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55 the changes made to Python 2.3. |
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56 |
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57 4. PSF is making Python 2.3 available to Licensee on an "AS IS" |
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58 basis. PSF MAKES NO REPRESENTATIONS OR WARRANTIES, EXPRESS OR |
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59 IMPLIED. BY WAY OF EXAMPLE, BUT NOT LIMITATION, PSF MAKES NO AND |
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60 DISCLAIMS ANY REPRESENTATION OR WARRANTY OF MERCHANTABILITY OR FITNESS |
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61 FOR ANY PARTICULAR PURPOSE OR THAT THE USE OF PYTHON 2.3 WILL NOT |
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62 INFRINGE ANY THIRD PARTY RIGHTS. |
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63 |
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64 5. PSF SHALL NOT BE LIABLE TO LICENSEE OR ANY OTHER USERS OF PYTHON |
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65 2.3 FOR ANY INCIDENTAL, SPECIAL, OR CONSEQUENTIAL DAMAGES OR LOSS AS |
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66 A RESULT OF MODIFYING, DISTRIBUTING, OR OTHERWISE USING PYTHON 2.3, |
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67 OR ANY DERIVATIVE THEREOF, EVEN IF ADVISED OF THE POSSIBILITY THEREOF. |
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68 |
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69 6. This License Agreement will automatically terminate upon a material |
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70 breach of its terms and conditions. |
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71 |
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72 7. Nothing in this License Agreement shall be deemed to create any |
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73 relationship of agency, partnership, or joint venture between PSF and |
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74 Licensee. This License Agreement does not grant permission to use PSF |
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75 trademarks or trade name in a trademark sense to endorse or promote |
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76 products or services of Licensee, or any third party. |
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77 |
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78 8. By copying, installing or otherwise using Python 2.3, Licensee |
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79 agrees to be bound by the terms and conditions of this License |
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80 Agreement. |
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81 */ |
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82 |
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83 #ifdef HAVE_CONFIG_H |
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84 #include <config.h> |
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85 #endif |
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86 |
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87 #include "lo-mappers.h" |
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88 #include "quit.h" |
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89 #include "oct-sort.h" |
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90 |
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91 #define IFLT(a,b) if (compare == NULL ? ((a) < (b)) : compare ((a), (b))) |
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92 |
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93 template <class T> |
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94 octave_sort<T>::octave_sort (void) : compare (NULL) |
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95 { |
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96 merge_init (); |
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97 merge_getmem (1024); |
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98 } |
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99 |
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100 template <class T> |
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101 octave_sort<T>::octave_sort (bool (*comp) (T, T)) : compare (comp) |
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102 { |
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103 merge_init (); |
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104 merge_getmem (1024); |
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105 } |
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106 |
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107 /* Reverse a slice of a list in place, from lo up to (exclusive) hi. */ |
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108 template <class T> |
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109 void |
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110 octave_sort<T>::reverse_slice(T *lo, T *hi) |
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111 { |
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112 --hi; |
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113 while (lo < hi) |
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114 { |
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115 T t = *lo; |
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116 *lo = *hi; |
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117 *hi = t; |
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118 ++lo; |
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119 --hi; |
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120 } |
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121 } |
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122 |
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123 template <class T> |
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124 void |
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125 octave_sort<T>::binarysort (T *lo, T *hi, T *start) |
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126 { |
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127 register T *l, *p, *r; |
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128 register T pivot; |
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129 |
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130 if (lo == start) |
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131 ++start; |
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132 |
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133 for (; start < hi; ++start) |
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134 { |
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135 /* set l to where *start belongs */ |
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136 l = lo; |
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137 r = start; |
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138 pivot = *r; |
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139 /* Invariants: |
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140 * pivot >= all in [lo, l). |
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141 * pivot < all in [r, start). |
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142 * The second is vacuously true at the start. |
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143 */ |
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144 do |
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145 { |
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146 p = l + ((r - l) >> 1); |
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147 IFLT (pivot, *p) |
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148 r = p; |
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149 else |
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150 l = p+1; |
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151 } |
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152 while (l < r); |
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153 /* The invariants still hold, so pivot >= all in [lo, l) and |
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154 pivot < all in [l, start), so pivot belongs at l. Note |
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155 that if there are elements equal to pivot, l points to the |
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156 first slot after them -- that's why this sort is stable. |
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157 Slide over to make room. |
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158 Caution: using memmove is much slower under MSVC 5; |
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159 we're not usually moving many slots. */ |
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160 for (p = start; p > l; --p) |
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161 *p = *(p-1); |
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162 *l = pivot; |
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163 } |
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164 |
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165 return; |
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166 } |
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167 |
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168 /* |
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169 Return the length of the run beginning at lo, in the slice [lo, hi). lo < hi |
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170 is required on entry. "A run" is the longest ascending sequence, with |
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171 |
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172 lo[0] <= lo[1] <= lo[2] <= ... |
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173 |
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174 or the longest descending sequence, with |
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175 |
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176 lo[0] > lo[1] > lo[2] > ... |
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177 |
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178 Boolean *descending is set to 0 in the former case, or to 1 in the latter. |
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179 For its intended use in a stable mergesort, the strictness of the defn of |
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180 "descending" is needed so that the caller can safely reverse a descending |
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181 sequence without violating stability (strict > ensures there are no equal |
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182 elements to get out of order). |
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183 |
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184 Returns -1 in case of error. |
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185 */ |
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186 template <class T> |
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187 int |
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188 octave_sort<T>::count_run(T *lo, T *hi, int *descending) |
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189 { |
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190 int n; |
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191 |
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192 *descending = 0; |
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193 ++lo; |
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194 if (lo == hi) |
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195 return 1; |
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196 |
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197 n = 2; |
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198 |
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199 IFLT (*lo, *(lo-1)) |
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200 { |
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201 *descending = 1; |
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202 for (lo = lo+1; lo < hi; ++lo, ++n) |
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203 { |
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204 IFLT (*lo, *(lo-1)) |
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205 ; |
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206 else |
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207 break; |
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208 } |
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209 } |
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210 else |
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211 { |
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212 for (lo = lo+1; lo < hi; ++lo, ++n) |
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213 { |
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214 IFLT (*lo, *(lo-1)) |
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215 break; |
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216 } |
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217 } |
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218 |
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219 return n; |
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220 } |
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221 |
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222 /* |
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223 Locate the proper position of key in a sorted vector; if the vector contains |
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224 an element equal to key, return the position immediately to the left of |
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225 the leftmost equal element. [gallop_right() does the same except returns |
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226 the position to the right of the rightmost equal element (if any).] |
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227 |
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228 "a" is a sorted vector with n elements, starting at a[0]. n must be > 0. |
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229 |
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230 "hint" is an index at which to begin the search, 0 <= hint < n. The closer |
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231 hint is to the final result, the faster this runs. |
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232 |
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233 The return value is the int k in 0..n such that |
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234 |
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235 a[k-1] < key <= a[k] |
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236 |
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237 pretending that *(a-1) is minus infinity and a[n] is plus infinity. IOW, |
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238 key belongs at index k; or, IOW, the first k elements of a should precede |
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239 key, and the last n-k should follow key. |
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240 |
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241 Returns -1 on error. See listsort.txt for info on the method. |
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242 */ |
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243 template <class T> |
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244 int |
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245 octave_sort<T>::gallop_left(T key, T *a, int n, int hint) |
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246 { |
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247 int ofs; |
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248 int lastofs; |
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249 int k; |
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250 |
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251 a += hint; |
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252 lastofs = 0; |
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253 ofs = 1; |
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254 IFLT (*a, key) |
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255 { |
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256 /* a[hint] < key -- gallop right, until |
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257 * a[hint + lastofs] < key <= a[hint + ofs] |
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258 */ |
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259 const int maxofs = n - hint; /* &a[n-1] is highest */ |
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260 while (ofs < maxofs) |
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261 { |
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262 IFLT (a[ofs], key) |
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263 { |
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264 lastofs = ofs; |
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265 ofs = (ofs << 1) + 1; |
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266 if (ofs <= 0) /* int overflow */ |
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267 ofs = maxofs; |
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268 } |
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269 else /* key <= a[hint + ofs] */ |
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270 break; |
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271 } |
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272 if (ofs > maxofs) |
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273 ofs = maxofs; |
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274 /* Translate back to offsets relative to &a[0]. */ |
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275 lastofs += hint; |
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276 ofs += hint; |
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277 } |
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278 else |
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279 { |
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280 /* key <= a[hint] -- gallop left, until |
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281 * a[hint - ofs] < key <= a[hint - lastofs] |
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282 */ |
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283 const int maxofs = hint + 1; /* &a[0] is lowest */ |
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284 while (ofs < maxofs) |
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285 { |
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286 IFLT (*(a-ofs), key) |
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287 break; |
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288 /* key <= a[hint - ofs] */ |
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289 lastofs = ofs; |
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290 ofs = (ofs << 1) + 1; |
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291 if (ofs <= 0) /* int overflow */ |
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292 ofs = maxofs; |
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293 } |
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294 if (ofs > maxofs) |
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295 ofs = maxofs; |
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296 /* Translate back to positive offsets relative to &a[0]. */ |
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297 k = lastofs; |
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298 lastofs = hint - ofs; |
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299 ofs = hint - k; |
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300 } |
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301 a -= hint; |
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302 |
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303 /* Now a[lastofs] < key <= a[ofs], so key belongs somewhere to the |
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304 * right of lastofs but no farther right than ofs. Do a binary |
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305 * search, with invariant a[lastofs-1] < key <= a[ofs]. |
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306 */ |
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307 ++lastofs; |
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308 while (lastofs < ofs) |
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309 { |
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310 int m = lastofs + ((ofs - lastofs) >> 1); |
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311 |
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312 IFLT (a[m], key) |
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313 lastofs = m+1; /* a[m] < key */ |
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314 else |
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315 ofs = m; /* key <= a[m] */ |
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316 } |
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317 return ofs; |
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318 } |
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319 |
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320 /* |
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321 Exactly like gallop_left(), except that if key already exists in a[0:n], |
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322 finds the position immediately to the right of the rightmost equal value. |
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323 |
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324 The return value is the int k in 0..n such that |
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325 |
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326 a[k-1] <= key < a[k] |
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327 |
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328 or -1 if error. |
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329 |
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330 The code duplication is massive, but this is enough different given that |
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331 we're sticking to "<" comparisons that it's much harder to follow if |
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332 written as one routine with yet another "left or right?" flag. |
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333 */ |
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334 template <class T> |
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335 int |
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336 octave_sort<T>::gallop_right(T key, T *a, int n, int hint) |
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337 { |
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338 int ofs; |
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339 int lastofs; |
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340 int k; |
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341 |
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342 a += hint; |
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343 lastofs = 0; |
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344 ofs = 1; |
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345 IFLT (key, *a) |
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346 { |
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347 /* key < a[hint] -- gallop left, until |
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348 * a[hint - ofs] <= key < a[hint - lastofs] |
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349 */ |
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350 const int maxofs = hint + 1; /* &a[0] is lowest */ |
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351 while (ofs < maxofs) |
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352 { |
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353 IFLT (key, *(a-ofs)) |
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354 { |
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355 lastofs = ofs; |
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356 ofs = (ofs << 1) + 1; |
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357 if (ofs <= 0) /* int overflow */ |
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358 ofs = maxofs; |
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359 } |
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360 else /* a[hint - ofs] <= key */ |
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361 break; |
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362 } |
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363 if (ofs > maxofs) |
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364 ofs = maxofs; |
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365 /* Translate back to positive offsets relative to &a[0]. */ |
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366 k = lastofs; |
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367 lastofs = hint - ofs; |
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368 ofs = hint - k; |
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369 } |
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370 else |
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371 { |
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372 /* a[hint] <= key -- gallop right, until |
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373 * a[hint + lastofs] <= key < a[hint + ofs] |
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374 */ |
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375 const int maxofs = n - hint; /* &a[n-1] is highest */ |
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376 while (ofs < maxofs) |
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377 { |
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378 IFLT (key, a[ofs]) |
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379 break; |
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380 /* a[hint + ofs] <= key */ |
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381 lastofs = ofs; |
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382 ofs = (ofs << 1) + 1; |
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383 if (ofs <= 0) /* int overflow */ |
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384 ofs = maxofs; |
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385 } |
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386 if (ofs > maxofs) |
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387 ofs = maxofs; |
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388 /* Translate back to offsets relative to &a[0]. */ |
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389 lastofs += hint; |
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390 ofs += hint; |
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391 } |
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392 a -= hint; |
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393 |
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394 /* Now a[lastofs] <= key < a[ofs], so key belongs somewhere to the |
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395 * right of lastofs but no farther right than ofs. Do a binary |
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396 * search, with invariant a[lastofs-1] <= key < a[ofs]. |
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397 */ |
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398 ++lastofs; |
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399 while (lastofs < ofs) |
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400 { |
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401 int m = lastofs + ((ofs - lastofs) >> 1); |
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402 |
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403 IFLT (key, a[m]) |
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404 ofs = m; /* key < a[m] */ |
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405 else |
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406 lastofs = m+1; /* a[m] <= key */ |
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407 } |
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408 return ofs; |
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409 } |
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410 |
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411 /* Conceptually a MergeState's constructor. */ |
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412 template <class T> |
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413 void |
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414 octave_sort<T>::merge_init(void) |
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415 { |
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416 ms.a = NULL; |
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417 ms.alloced = 0; |
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418 ms.n = 0; |
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419 ms.min_gallop = MIN_GALLOP; |
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420 } |
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421 |
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422 /* Free all the temp memory owned by the MergeState. This must be called |
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423 * when you're done with a MergeState, and may be called before then if |
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424 * you want to free the temp memory early. |
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425 */ |
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426 template <class T> |
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427 void |
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428 octave_sort<T>::merge_freemem(void) |
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429 { |
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430 if (ms.a) |
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431 free (ms.a); |
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432 ms.alloced = 0; |
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433 ms.a = NULL; |
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434 } |
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435 |
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436 static inline int |
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437 roundupsize(int n) |
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438 { |
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439 unsigned int nbits = 3; |
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440 unsigned int n2 = (unsigned int)n >> 8; |
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441 |
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442 /* Round up: |
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443 * If n < 256, to a multiple of 8. |
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444 * If n < 2048, to a multiple of 64. |
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445 * If n < 16384, to a multiple of 512. |
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446 * If n < 131072, to a multiple of 4096. |
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447 * If n < 1048576, to a multiple of 32768. |
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448 * If n < 8388608, to a multiple of 262144. |
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449 * If n < 67108864, to a multiple of 2097152. |
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450 * If n < 536870912, to a multiple of 16777216. |
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451 * ... |
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452 * If n < 2**(5+3*i), to a multiple of 2**(3*i). |
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453 * |
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454 * This over-allocates proportional to the list size, making room |
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455 * for additional growth. The over-allocation is mild, but is |
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456 * enough to give linear-time amortized behavior over a long |
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457 * sequence of appends() in the presence of a poorly-performing |
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458 * system realloc() (which is a reality, e.g., across all flavors |
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459 * of Windows, with Win9x behavior being particularly bad -- and |
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460 * we've still got address space fragmentation problems on Win9x |
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461 * even with this scheme, although it requires much longer lists to |
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462 * provoke them than it used to). |
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463 */ |
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464 while (n2) { |
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465 n2 >>= 3; |
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466 nbits += 3; |
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467 } |
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|
468 return ((n >> nbits) + 1) << nbits; |
|
|
469 } |
|
|
470 |
|
|
471 /* Ensure enough temp memory for 'need' array slots is available. |
|
|
472 * Returns 0 on success and -1 if the memory can't be gotten. |
|
|
473 */ |
|
|
474 template <class T> |
|
|
475 int |
|
|
476 octave_sort<T>::merge_getmem(int need) |
|
|
477 { |
|
|
478 if (need <= ms.alloced) |
|
|
479 return 0; |
|
|
480 |
|
|
481 need = roundupsize(need); |
|
|
482 /* Don't realloc! That can cost cycles to copy the old data, but |
|
|
483 * we don't care what's in the block. |
|
|
484 */ |
|
|
485 merge_freemem( ); |
|
|
486 ms.a = (T *) malloc (need * sizeof (T)); |
|
|
487 if (ms.a) { |
|
|
488 ms.alloced = need; |
|
|
489 return 0; |
|
|
490 } |
|
|
491 merge_freemem( ); /* reset to sane state */ |
|
|
492 return -1; |
|
|
493 } |
|
|
494 |
|
|
495 #define MERGE_GETMEM(NEED) ((NEED) <= ms.alloced ? 0 : \ |
|
|
496 merge_getmem(NEED)) |
|
|
497 |
|
|
498 /* Merge the na elements starting at pa with the nb elements starting at pb |
|
|
499 * in a stable way, in-place. na and nb must be > 0, and pa + na == pb. |
|
|
500 * Must also have that *pb < *pa, that pa[na-1] belongs at the end of the |
|
|
501 * merge, and should have na <= nb. See listsort.txt for more info. |
|
|
502 * Return 0 if successful, -1 if error. |
|
|
503 */ |
|
|
504 template <class T> |
|
|
505 int |
|
|
506 octave_sort<T>::merge_lo(T *pa, int na, T *pb, int nb) |
|
|
507 { |
|
|
508 int k; |
|
|
509 T *dest; |
|
|
510 int result = -1; /* guilty until proved innocent */ |
|
|
511 int min_gallop = ms.min_gallop; |
|
|
512 |
|
|
513 if (MERGE_GETMEM(na) < 0) |
|
|
514 return -1; |
|
|
515 memcpy(ms.a, pa, na * sizeof(T)); |
|
|
516 dest = pa; |
|
|
517 pa = ms.a; |
|
|
518 |
|
|
519 *dest++ = *pb++; |
|
|
520 --nb; |
|
|
521 if (nb == 0) |
|
|
522 goto Succeed; |
|
|
523 if (na == 1) |
|
|
524 goto CopyB; |
|
|
525 |
|
|
526 for (;;) { |
|
|
527 int acount = 0; /* # of times A won in a row */ |
|
|
528 int bcount = 0; /* # of times B won in a row */ |
|
|
529 |
|
|
530 /* Do the straightforward thing until (if ever) one run |
|
|
531 * appears to win consistently. |
|
|
532 */ |
|
|
533 for (;;) { |
|
|
534 |
|
|
535 IFLT (*pb, *pa) |
|
|
536 { |
|
|
537 *dest++ = *pb++; |
|
|
538 ++bcount; |
|
|
539 acount = 0; |
|
|
540 --nb; |
|
|
541 if (nb == 0) |
|
|
542 goto Succeed; |
|
|
543 if (bcount >= min_gallop) |
|
|
544 break; |
|
|
545 } |
|
|
546 else |
|
|
547 { |
|
|
548 *dest++ = *pa++; |
|
|
549 ++acount; |
|
|
550 bcount = 0; |
|
|
551 --na; |
|
|
552 if (na == 1) |
|
|
553 goto CopyB; |
|
|
554 if (acount >= min_gallop) |
|
|
555 break; |
|
|
556 } |
|
|
557 } |
|
|
558 |
|
|
559 /* One run is winning so consistently that galloping may |
|
|
560 * be a huge win. So try that, and continue galloping until |
|
|
561 * (if ever) neither run appears to be winning consistently |
|
|
562 * anymore. |
|
|
563 */ |
|
|
564 ++min_gallop; |
|
|
565 do |
|
|
566 { |
|
|
567 min_gallop -= min_gallop > 1; |
|
|
568 ms.min_gallop = min_gallop; |
|
|
569 k = gallop_right(*pb, pa, na, 0); |
|
|
570 acount = k; |
|
|
571 if (k) |
|
|
572 { |
|
|
573 if (k < 0) |
|
|
574 goto Fail; |
|
|
575 memcpy(dest, pa, k * sizeof(T)); |
|
|
576 dest += k; |
|
|
577 pa += k; |
|
|
578 na -= k; |
|
|
579 if (na == 1) |
|
|
580 goto CopyB; |
|
|
581 /* na==0 is impossible now if the comparison |
|
|
582 * function is consistent, but we can't assume |
|
|
583 * that it is. |
|
|
584 */ |
|
|
585 if (na == 0) |
|
|
586 goto Succeed; |
|
|
587 } |
|
|
588 *dest++ = *pb++; |
|
|
589 --nb; |
|
|
590 if (nb == 0) |
|
|
591 goto Succeed; |
|
|
592 |
|
|
593 k = gallop_left(*pa, pb, nb, 0); |
|
|
594 bcount = k; |
|
|
595 if (k) { |
|
|
596 if (k < 0) |
|
|
597 goto Fail; |
|
|
598 memmove(dest, pb, k * sizeof(T)); |
|
|
599 dest += k; |
|
|
600 pb += k; |
|
|
601 nb -= k; |
|
|
602 if (nb == 0) |
|
|
603 goto Succeed; |
|
|
604 } |
|
|
605 *dest++ = *pa++; |
|
|
606 --na; |
|
|
607 if (na == 1) |
|
|
608 goto CopyB; |
|
|
609 } while (acount >= MIN_GALLOP || bcount >= MIN_GALLOP); |
|
|
610 ++min_gallop; /* penalize it for leaving galloping mode */ |
|
|
611 ms.min_gallop = min_gallop; |
|
|
612 } |
|
|
613 Succeed: |
|
|
614 result = 0; |
|
|
615 Fail: |
|
|
616 if (na) |
|
|
617 memcpy(dest, pa, na * sizeof(T)); |
|
|
618 return result; |
|
|
619 CopyB: |
|
|
620 /* The last element of pa belongs at the end of the merge. */ |
|
|
621 memmove(dest, pb, nb * sizeof(T)); |
|
|
622 dest[nb] = *pa; |
|
|
623 return 0; |
|
|
624 } |
|
|
625 |
|
|
626 /* Merge the na elements starting at pa with the nb elements starting at pb |
|
|
627 * in a stable way, in-place. na and nb must be > 0, and pa + na == pb. |
|
|
628 * Must also have that *pb < *pa, that pa[na-1] belongs at the end of the |
|
|
629 * merge, and should have na >= nb. See listsort.txt for more info. |
|
|
630 * Return 0 if successful, -1 if error. |
|
|
631 */ |
|
|
632 template <class T> |
|
|
633 int |
|
|
634 octave_sort<T>::merge_hi(T *pa, int na, T *pb, int nb) |
|
|
635 { |
|
|
636 int k; |
|
|
637 T *dest; |
|
|
638 int result = -1; /* guilty until proved innocent */ |
|
|
639 T *basea; |
|
|
640 T *baseb; |
|
|
641 int min_gallop = ms.min_gallop; |
|
|
642 |
|
|
643 if (MERGE_GETMEM(nb) < 0) |
|
|
644 return -1; |
|
|
645 dest = pb + nb - 1; |
|
|
646 memcpy(ms.a, pb, nb * sizeof(T)); |
|
|
647 basea = pa; |
|
|
648 baseb = ms.a; |
|
|
649 pb = ms.a + nb - 1; |
|
|
650 pa += na - 1; |
|
|
651 |
|
|
652 *dest-- = *pa--; |
|
|
653 --na; |
|
|
654 if (na == 0) |
|
|
655 goto Succeed; |
|
|
656 if (nb == 1) |
|
|
657 goto CopyA; |
|
|
658 |
|
|
659 for (;;) |
|
|
660 { |
|
|
661 int acount = 0; /* # of times A won in a row */ |
|
|
662 int bcount = 0; /* # of times B won in a row */ |
|
|
663 |
|
|
664 /* Do the straightforward thing until (if ever) one run |
|
|
665 * appears to win consistently. |
|
|
666 */ |
|
|
667 for (;;) |
|
|
668 { |
|
|
669 IFLT (*pb, *pa) |
|
|
670 { |
|
|
671 *dest-- = *pa--; |
|
|
672 ++acount; |
|
|
673 bcount = 0; |
|
|
674 --na; |
|
|
675 if (na == 0) |
|
|
676 goto Succeed; |
|
|
677 if (acount >= min_gallop) |
|
|
678 break; |
|
|
679 } |
|
|
680 else |
|
|
681 { |
|
|
682 *dest-- = *pb--; |
|
|
683 ++bcount; |
|
|
684 acount = 0; |
|
|
685 --nb; |
|
|
686 if (nb == 1) |
|
|
687 goto CopyA; |
|
|
688 if (bcount >= min_gallop) |
|
|
689 break; |
|
|
690 } |
|
|
691 } |
|
|
692 |
|
|
693 /* One run is winning so consistently that galloping may |
|
|
694 * be a huge win. So try that, and continue galloping until |
|
|
695 * (if ever) neither run appears to be winning consistently |
|
|
696 * anymore. |
|
|
697 */ |
|
|
698 ++min_gallop; |
|
|
699 do |
|
|
700 { |
|
|
701 min_gallop -= min_gallop > 1; |
|
|
702 ms.min_gallop = min_gallop; |
|
|
703 k = gallop_right(*pb, basea, na, na-1); |
|
|
704 if (k < 0) |
|
|
705 goto Fail; |
|
|
706 k = na - k; |
|
|
707 acount = k; |
|
|
708 if (k) |
|
|
709 { |
|
|
710 dest -= k; |
|
|
711 pa -= k; |
|
|
712 memmove(dest+1, pa+1, k * sizeof(T )); |
|
|
713 na -= k; |
|
|
714 if (na == 0) |
|
|
715 goto Succeed; |
|
|
716 } |
|
|
717 *dest-- = *pb--; |
|
|
718 --nb; |
|
|
719 if (nb == 1) |
|
|
720 goto CopyA; |
|
|
721 |
|
|
722 k = gallop_left(*pa, baseb, nb, nb-1); |
|
|
723 if (k < 0) |
|
|
724 goto Fail; |
|
|
725 k = nb - k; |
|
|
726 bcount = k; |
|
|
727 if (k) |
|
|
728 { |
|
|
729 dest -= k; |
|
|
730 pb -= k; |
|
|
731 memcpy(dest+1, pb+1, k * sizeof(T)); |
|
|
732 nb -= k; |
|
|
733 if (nb == 1) |
|
|
734 goto CopyA; |
|
|
735 /* nb==0 is impossible now if the comparison |
|
|
736 * function is consistent, but we can't assume |
|
|
737 * that it is. |
|
|
738 */ |
|
|
739 if (nb == 0) |
|
|
740 goto Succeed; |
|
|
741 } |
|
|
742 *dest-- = *pa--; |
|
|
743 --na; |
|
|
744 if (na == 0) |
|
|
745 goto Succeed; |
|
|
746 } while (acount >= MIN_GALLOP || bcount >= MIN_GALLOP); |
|
|
747 ++min_gallop; /* penalize it for leaving galloping mode */ |
|
|
748 ms.min_gallop = min_gallop; |
|
|
749 } |
|
|
750 Succeed: |
|
|
751 result = 0; |
|
|
752 Fail: |
|
|
753 if (nb) |
|
|
754 memcpy(dest-(nb-1), baseb, nb * sizeof(T)); |
|
|
755 return result; |
|
|
756 CopyA: |
|
|
757 /* The first element of pb belongs at the front of the merge. */ |
|
|
758 dest -= na; |
|
|
759 pa -= na; |
|
|
760 memmove(dest+1, pa+1, na * sizeof(T)); |
|
|
761 *dest = *pb; |
|
|
762 return 0; |
|
|
763 } |
|
|
764 |
|
|
765 /* Merge the two runs at stack indices i and i+1. |
|
|
766 * Returns 0 on success, -1 on error. |
|
|
767 */ |
|
|
768 template <class T> |
|
|
769 int |
|
|
770 octave_sort<T>::merge_at(int i) |
|
|
771 { |
|
|
772 T *pa, *pb; |
|
|
773 int na, nb; |
|
|
774 int k; |
|
|
775 |
|
|
776 pa = ms.pending[i].base; |
|
|
777 na = ms.pending[i].len; |
|
|
778 pb = ms.pending[i+1].base; |
|
|
779 nb = ms.pending[i+1].len; |
|
|
780 |
|
|
781 /* Record the length of the combined runs; if i is the 3rd-last |
|
|
782 * run now, also slide over the last run (which isn't involved |
|
|
783 * in this merge). The current run i+1 goes away in any case. |
|
|
784 */ |
|
|
785 ms.pending[i].len = na + nb; |
|
|
786 if (i == ms.n - 3) |
|
|
787 ms.pending[i+1] = ms.pending[i+2]; |
|
|
788 --ms.n; |
|
|
789 |
|
|
790 /* Where does b start in a? Elements in a before that can be |
|
|
791 * ignored (already in place). |
|
|
792 */ |
|
|
793 k = gallop_right(*pb, pa, na, 0); |
|
|
794 if (k < 0) |
|
|
795 return -1; |
|
|
796 pa += k; |
|
|
797 na -= k; |
|
|
798 if (na == 0) |
|
|
799 return 0; |
|
|
800 |
|
|
801 /* Where does a end in b? Elements in b after that can be |
|
|
802 * ignored (already in place). |
|
|
803 */ |
|
|
804 nb = gallop_left(pa[na-1], pb, nb, nb-1); |
|
|
805 if (nb <= 0) |
|
|
806 return nb; |
|
|
807 |
|
|
808 /* Merge what remains of the runs, using a temp array with |
|
|
809 * min(na, nb) elements. |
|
|
810 */ |
|
|
811 if (na <= nb) |
|
|
812 return merge_lo(pa, na, pb, nb); |
|
|
813 else |
|
|
814 return merge_hi(pa, na, pb, nb); |
|
|
815 } |
|
|
816 |
|
|
817 /* Examine the stack of runs waiting to be merged, merging adjacent runs |
|
|
818 * until the stack invariants are re-established: |
|
|
819 * |
|
|
820 * 1. len[-3] > len[-2] + len[-1] |
|
|
821 * 2. len[-2] > len[-1] |
|
|
822 * |
|
|
823 * See listsort.txt for more info. |
|
|
824 * |
|
|
825 * Returns 0 on success, -1 on error. |
|
|
826 */ |
|
|
827 template <class T> |
|
|
828 int |
|
|
829 octave_sort<T>::merge_collapse(void) |
|
|
830 { |
|
|
831 struct s_slice *p = ms.pending; |
|
|
832 |
|
|
833 while (ms.n > 1) |
|
|
834 { |
|
|
835 int n = ms.n - 2; |
|
|
836 if (n > 0 && p[n-1].len <= p[n].len + p[n+1].len) |
|
|
837 { |
|
|
838 if (p[n-1].len < p[n+1].len) |
|
|
839 --n; |
|
|
840 if (merge_at(n) < 0) |
|
|
841 return -1; |
|
|
842 } |
|
|
843 else if (p[n].len <= p[n+1].len) |
|
|
844 { |
|
|
845 if (merge_at(n) < 0) |
|
|
846 return -1; |
|
|
847 } |
|
|
848 else |
|
|
849 break; |
|
|
850 } |
|
|
851 return 0; |
|
|
852 } |
|
|
853 |
|
|
854 /* Regardless of invariants, merge all runs on the stack until only one |
|
|
855 * remains. This is used at the end of the mergesort. |
|
|
856 * |
|
|
857 * Returns 0 on success, -1 on error. |
|
|
858 */ |
|
|
859 template <class T> |
|
|
860 int |
|
|
861 octave_sort<T>::merge_force_collapse(void) |
|
|
862 { |
|
|
863 struct s_slice *p = ms.pending; |
|
|
864 |
|
|
865 while (ms.n > 1) |
|
|
866 { |
|
|
867 int n = ms.n - 2; |
|
|
868 if (n > 0 && p[n-1].len < p[n+1].len) |
|
|
869 --n; |
|
|
870 if (merge_at(n) < 0) |
|
|
871 return -1; |
|
|
872 } |
|
|
873 return 0; |
|
|
874 } |
|
|
875 |
|
|
876 /* Compute a good value for the minimum run length; natural runs shorter |
|
|
877 * than this are boosted artificially via binary insertion. |
|
|
878 * |
|
|
879 * If n < 64, return n (it's too small to bother with fancy stuff). |
|
|
880 * Else if n is an exact power of 2, return 32. |
|
|
881 * Else return an int k, 32 <= k <= 64, such that n/k is close to, but |
|
|
882 * strictly less than, an exact power of 2. |
|
|
883 * |
|
|
884 * See listsort.txt for more info. |
|
|
885 */ |
|
|
886 template <class T> |
|
|
887 int |
|
|
888 octave_sort<T>::merge_compute_minrun(int n) |
|
|
889 { |
|
|
890 int r = 0; /* becomes 1 if any 1 bits are shifted off */ |
|
|
891 |
|
|
892 while (n >= 64) { |
|
|
893 r |= n & 1; |
|
|
894 n >>= 1; |
|
|
895 } |
|
|
896 return n + r; |
|
|
897 } |
|
|
898 |
|
|
899 template <class T> |
|
|
900 void |
|
|
901 octave_sort<T>::sort (T *v, int elements) |
|
|
902 { |
|
|
903 /* Re-initialize the Mergestate as this might be the second time called */ |
|
|
904 ms.n = 0; |
|
|
905 ms.min_gallop = MIN_GALLOP; |
|
|
906 |
|
|
907 if (elements > 1) |
|
|
908 { |
|
|
909 int nremaining = elements; |
|
|
910 T *lo = v; |
|
|
911 T *hi = v + elements; |
|
|
912 |
|
|
913 /* March over the array once, left to right, finding natural runs, |
|
|
914 * and extending short natural runs to minrun elements. |
|
|
915 */ |
|
|
916 int minrun = merge_compute_minrun(nremaining); |
|
|
917 do |
|
|
918 { |
|
|
919 int descending; |
|
|
920 int n; |
|
|
921 |
|
|
922 /* Identify next run. */ |
|
|
923 n = count_run(lo, hi, &descending); |
|
|
924 if (n < 0) |
|
|
925 goto fail; |
|
|
926 if (descending) |
|
|
927 reverse_slice(lo, lo + n); |
|
|
928 /* If short, extend to min(minrun, nremaining). */ |
|
|
929 if (n < minrun) |
|
|
930 { |
|
|
931 const int force = nremaining <= minrun ? nremaining : minrun; |
|
|
932 binarysort(lo, lo + force, lo + n); |
|
|
933 n = force; |
|
|
934 } |
|
|
935 /* Push run onto pending-runs stack, and maybe merge. */ |
|
|
936 assert(ms.n < MAX_MERGE_PENDING); |
|
|
937 ms.pending[ms.n].base = lo; |
|
|
938 ms.pending[ms.n].len = n; |
|
|
939 ++ms.n; |
|
|
940 if (merge_collapse( ) < 0) |
|
|
941 goto fail; |
|
|
942 /* Advance to find next run. */ |
|
|
943 lo += n; |
|
|
944 nremaining -= n; |
|
|
945 } while (nremaining); |
|
|
946 |
|
|
947 merge_force_collapse( ); |
|
|
948 } |
|
|
949 |
|
|
950 fail: |
|
|
951 return; |
|
|
952 } |
|
|
953 |
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954 /* |
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955 ;;; Local Variables: *** |
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956 ;;; mode: C++ *** |
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957 ;;; End: *** |
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958 */ |
