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1 /* |
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2 * Copyright (c) 2007, Oracle and/or its affiliates. All rights reserved. |
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3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. |
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4 * |
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5 * This code is free software; you can redistribute it and/or modify it |
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6 * under the terms of the GNU General Public License version 2 only, as |
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7 * published by the Free Software Foundation. Oracle designates this |
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8 * particular file as subject to the "Classpath" exception as provided |
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9 * by Oracle in the LICENSE file that accompanied this code. |
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10 * |
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11 * This code 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 * version 2 for more details (a copy is included in the LICENSE file that |
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15 * accompanied this code). |
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16 * |
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17 * You should have received a copy of the GNU General Public License version |
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18 * 2 along with this work; if not, write to the Free Software Foundation, |
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19 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. |
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20 * |
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21 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA |
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22 * or visit www.oracle.com if you need additional information or have any |
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23 * questions. |
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24 */ |
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25 |
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26 package sun.java2d.pisces; |
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27 |
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28 import java.util.Iterator; |
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29 |
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30 class Curve { |
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31 |
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32 float ax, ay, bx, by, cx, cy, dx, dy; |
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33 float dax, day, dbx, dby; |
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34 |
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35 Curve() { |
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36 } |
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37 |
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38 void set(float[] points, int type) { |
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39 switch(type) { |
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40 case 8: |
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41 set(points[0], points[1], |
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42 points[2], points[3], |
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43 points[4], points[5], |
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44 points[6], points[7]); |
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45 break; |
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46 case 6: |
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47 set(points[0], points[1], |
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48 points[2], points[3], |
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49 points[4], points[5]); |
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50 break; |
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51 default: |
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52 throw new InternalError("Curves can only be cubic or quadratic"); |
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53 } |
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54 } |
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55 |
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56 void set(float x1, float y1, |
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57 float x2, float y2, |
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58 float x3, float y3, |
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59 float x4, float y4) |
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60 { |
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61 ax = 3 * (x2 - x3) + x4 - x1; |
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62 ay = 3 * (y2 - y3) + y4 - y1; |
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63 bx = 3 * (x1 - 2 * x2 + x3); |
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64 by = 3 * (y1 - 2 * y2 + y3); |
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65 cx = 3 * (x2 - x1); |
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66 cy = 3 * (y2 - y1); |
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67 dx = x1; |
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68 dy = y1; |
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69 dax = 3 * ax; day = 3 * ay; |
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70 dbx = 2 * bx; dby = 2 * by; |
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71 } |
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72 |
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73 void set(float x1, float y1, |
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74 float x2, float y2, |
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75 float x3, float y3) |
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76 { |
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77 ax = ay = 0f; |
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78 |
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79 bx = x1 - 2 * x2 + x3; |
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80 by = y1 - 2 * y2 + y3; |
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81 cx = 2 * (x2 - x1); |
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82 cy = 2 * (y2 - y1); |
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83 dx = x1; |
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84 dy = y1; |
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85 dax = 0; day = 0; |
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86 dbx = 2 * bx; dby = 2 * by; |
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87 } |
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88 |
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89 float xat(float t) { |
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90 return t * (t * (t * ax + bx) + cx) + dx; |
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91 } |
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92 float yat(float t) { |
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93 return t * (t * (t * ay + by) + cy) + dy; |
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94 } |
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95 |
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96 float dxat(float t) { |
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97 return t * (t * dax + dbx) + cx; |
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98 } |
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99 |
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100 float dyat(float t) { |
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101 return t * (t * day + dby) + cy; |
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102 } |
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103 |
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104 private float ddxat(float t) { |
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105 return 2 * dax * t + dbx; |
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106 } |
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107 |
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108 private float ddyat(float t) { |
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109 return 2 * day * t + dby; |
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110 } |
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111 |
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112 int dxRoots(float[] roots, int off) { |
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113 return Helpers.quadraticRoots(dax, dbx, cx, roots, off); |
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114 } |
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115 |
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116 int dyRoots(float[] roots, int off) { |
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117 return Helpers.quadraticRoots(day, dby, cy, roots, off); |
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118 } |
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119 |
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120 int infPoints(float[] pts, int off) { |
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121 // inflection point at t if -f'(t)x*f''(t)y + f'(t)y*f''(t)x == 0 |
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122 // Fortunately, this turns out to be quadratic, so there are at |
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123 // most 2 inflection points. |
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124 final float a = dax * dby - dbx * day; |
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125 final float b = 2 * (cy * dax - day * cx); |
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126 final float c = cy * dbx - cx * dby; |
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127 |
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128 return Helpers.quadraticRoots(a, b, c, pts, off); |
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129 } |
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130 |
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131 // finds points where the first and second derivative are |
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132 // perpendicular. This happens when g(t) = f'(t)*f''(t) == 0 (where |
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133 // * is a dot product). Unfortunately, we have to solve a cubic. |
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134 private int perpendiculardfddf(float[] pts, int off, final float err) { |
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135 assert pts.length >= off + 4; |
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136 |
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137 // these are the coefficients of g(t): |
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138 final float a = 2*(dax*dax + day*day); |
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139 final float b = 3*(dax*dbx + day*dby); |
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140 final float c = 2*(dax*cx + day*cy) + dbx*dbx + dby*dby; |
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141 final float d = dbx*cx + dby*cy; |
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142 // TODO: We might want to divide the polynomial by a to make the |
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143 // coefficients smaller. This won't change the roots. |
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144 return Helpers.cubicRootsInAB(a, b, c, d, pts, off, err, 0f, 1f); |
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145 } |
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146 |
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147 // Tries to find the roots of the function ROC(t)-w in [0, 1). It uses |
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148 // a variant of the false position algorithm to find the roots. False |
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149 // position requires that 2 initial values x0,x1 be given, and that the |
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150 // function must have opposite signs at those values. To find such |
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151 // values, we need the local extrema of the ROC function, for which we |
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152 // need the roots of its derivative; however, it's harder to find the |
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153 // roots of the derivative in this case than it is to find the roots |
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154 // of the original function. So, we find all points where this curve's |
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155 // first and second derivative are perpendicular, and we pretend these |
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156 // are our local extrema. There are at most 3 of these, so we will check |
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157 // at most 4 sub-intervals of (0,1). ROC has asymptotes at inflection |
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158 // points, so roc-w can have at least 6 roots. This shouldn't be a |
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159 // problem for what we're trying to do (draw a nice looking curve). |
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160 int rootsOfROCMinusW(float[] roots, int off, final float w, final float err) { |
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161 // no OOB exception, because by now off<=6, and roots.length >= 10 |
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162 assert off <= 6 && roots.length >= 10; |
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163 int ret = off; |
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164 int numPerpdfddf = perpendiculardfddf(roots, off, err); |
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165 float t0 = 0, ft0 = ROCsq(t0) - w*w; |
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166 roots[off + numPerpdfddf] = 1f; // always check interval end points |
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167 numPerpdfddf++; |
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168 for (int i = off; i < off + numPerpdfddf; i++) { |
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169 float t1 = roots[i], ft1 = ROCsq(t1) - w*w; |
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170 if (ft0 == 0f) { |
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171 roots[ret++] = t0; |
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172 } else if (ft1 * ft0 < 0f) { // have opposite signs |
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173 // (ROC(t)^2 == w^2) == (ROC(t) == w) is true because |
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174 // ROC(t) >= 0 for all t. |
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175 roots[ret++] = falsePositionROCsqMinusX(t0, t1, w*w, err); |
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176 } |
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177 t0 = t1; |
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178 ft0 = ft1; |
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179 } |
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180 |
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181 return ret - off; |
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182 } |
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183 |
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184 private static float eliminateInf(float x) { |
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185 return (x == Float.POSITIVE_INFINITY ? Float.MAX_VALUE : |
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186 (x == Float.NEGATIVE_INFINITY ? Float.MIN_VALUE : x)); |
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187 } |
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188 |
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189 // A slight modification of the false position algorithm on wikipedia. |
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190 // This only works for the ROCsq-x functions. It might be nice to have |
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191 // the function as an argument, but that would be awkward in java6. |
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192 // It is something to consider for java7, depending on how closures |
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193 // and function objects turn out. Same goes for the newton's method |
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194 // algorithm in Helpers.java |
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195 private float falsePositionROCsqMinusX(float x0, float x1, |
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196 final float x, final float err) |
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197 { |
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198 final int iterLimit = 100; |
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199 int side = 0; |
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200 float t = x1, ft = eliminateInf(ROCsq(t) - x); |
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201 float s = x0, fs = eliminateInf(ROCsq(s) - x); |
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202 float r = s, fr; |
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203 for (int i = 0; i < iterLimit && Math.abs(t - s) > err * Math.abs(t + s); i++) { |
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204 r = (fs * t - ft * s) / (fs - ft); |
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205 fr = ROCsq(r) - x; |
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206 if (fr * ft > 0) {// have the same sign |
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207 ft = fr; t = r; |
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208 if (side < 0) { |
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209 fs /= (1 << (-side)); |
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210 side--; |
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211 } else { |
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212 side = -1; |
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213 } |
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214 } else if (fr * fs > 0) { |
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215 fs = fr; s = r; |
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216 if (side > 0) { |
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217 ft /= (1 << side); |
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218 side++; |
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219 } else { |
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220 side = 1; |
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221 } |
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222 } else { |
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223 break; |
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224 } |
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225 } |
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226 return r; |
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227 } |
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228 |
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229 // returns the radius of curvature squared at t of this curve |
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230 // see http://en.wikipedia.org/wiki/Radius_of_curvature_(applications) |
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231 private float ROCsq(final float t) { |
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232 final float dx = dxat(t); |
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233 final float dy = dyat(t); |
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234 final float ddx = ddxat(t); |
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235 final float ddy = ddyat(t); |
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236 final float dx2dy2 = dx*dx + dy*dy; |
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237 final float ddx2ddy2 = ddx*ddx + ddy*ddy; |
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238 final float ddxdxddydy = ddx*dx + ddy*dy; |
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239 float ret = ((dx2dy2*dx2dy2) / (dx2dy2 * ddx2ddy2 - ddxdxddydy*ddxdxddydy))*dx2dy2; |
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240 return ret; |
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241 } |
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242 |
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243 // curve to be broken should be in pts[0] |
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244 // this will change the contents of both pts and Ts |
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245 // TODO: There's no reason for Ts to be an array. All we need is a sequence |
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246 // of t values at which to subdivide. An array statisfies this condition, |
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247 // but is unnecessarily restrictive. Ts should be an Iterator<Float> instead. |
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248 // Doing this will also make dashing easier, since we could easily make |
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249 // LengthIterator an Iterator<Float> and feed it to this function to simplify |
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250 // the loop in Dasher.somethingTo. |
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251 static Iterator<float[]> breakPtsAtTs(final float[][] pts, final int type, |
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252 final float[] Ts, final int numTs) |
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253 { |
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254 assert pts.length >= 2 && pts[0].length >= 8 && numTs <= Ts.length; |
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255 return new Iterator<float[]>() { |
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256 int nextIdx = 0; |
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257 int nextCurveIdx = 0; |
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258 float prevT = 0; |
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259 |
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260 @Override public boolean hasNext() { |
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261 return nextCurveIdx < numTs + 1; |
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262 } |
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263 |
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264 @Override public float[] next() { |
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265 float[] ret; |
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266 if (nextCurveIdx < numTs) { |
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267 float curT = Ts[nextCurveIdx]; |
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268 float splitT = (curT - prevT) / (1 - prevT); |
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269 Helpers.subdivideAt(splitT, |
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270 pts[nextIdx], 0, |
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271 pts[nextIdx], 0, |
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272 pts[1-nextIdx], 0, type); |
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273 updateTs(Ts, Ts[nextCurveIdx], nextCurveIdx + 1, numTs - nextCurveIdx - 1); |
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274 ret = pts[nextIdx]; |
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275 nextIdx = 1 - nextIdx; |
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276 } else { |
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277 ret = pts[nextIdx]; |
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278 } |
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279 nextCurveIdx++; |
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280 return ret; |
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281 } |
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282 |
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283 @Override public void remove() {} |
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284 }; |
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285 } |
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286 |
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287 // precondition: ts[off]...ts[off+len-1] must all be greater than t. |
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288 private static void updateTs(float[] ts, final float t, final int off, final int len) { |
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289 for (int i = off; i < off + len; i++) { |
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290 ts[i] = (ts[i] - t) / (1 - t); |
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291 } |
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292 } |
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293 } |
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294 |