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1 /* |
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2 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi |
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3 */ |
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4 |
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5 /* |
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6 Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. |
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7 Then for degree elevation, the equations are: |
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8 |
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9 Q0 = P0 |
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10 Q1 = 1/3 P0 + 2/3 P1 |
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11 Q2 = 2/3 P1 + 1/3 P2 |
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12 Q3 = P2 |
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13 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from |
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14 the equations above: |
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15 |
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16 P1 = 3/2 Q1 - 1/2 Q0 |
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17 P1 = 3/2 Q2 - 1/2 Q3 |
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18 If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since |
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19 it's likely not, your best bet is to average them. So, |
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20 |
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21 P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 |
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22 |
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23 |
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24 SkDCubic defined by: P1/2 - anchor points, C1/C2 control points |
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25 |x| is the euclidean norm of x |
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26 mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the |
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27 control point at C = (3·C2 - P2 + 3·C1 - P1)/4 |
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28 |
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29 Algorithm |
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30 |
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31 pick an absolute precision (prec) |
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32 Compute the Tdiv as the root of (cubic) equation |
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33 sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec |
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34 if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a |
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35 quadratic, with a defect less than prec, by the mid-point approximation. |
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36 Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv) |
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37 0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point |
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38 approximation |
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39 Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation |
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40 |
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41 confirmed by (maybe stolen from) |
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42 http://www.caffeineowl.com/graphics/2d/vectorial/cubic2quad01.html |
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43 // maybe in turn derived from http://www.cccg.ca/proceedings/2004/36.pdf |
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44 // also stored at http://www.cis.usouthal.edu/~hain/general/Publications/Bezier/bezier%20cccg04%20paper.pdf |
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45 |
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46 */ |
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47 |
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48 #include "SkPathOpsCubic.h" |
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49 #include "SkPathOpsLine.h" |
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50 #include "SkPathOpsQuad.h" |
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51 #include "SkReduceOrder.h" |
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52 #include "SkTArray.h" |
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53 #include "SkTSort.h" |
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54 |
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55 #define USE_CUBIC_END_POINTS 1 |
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56 |
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57 static double calc_t_div(const SkDCubic& cubic, double precision, double start) { |
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58 const double adjust = sqrt(3.) / 36; |
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59 SkDCubic sub; |
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60 const SkDCubic* cPtr; |
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61 if (start == 0) { |
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62 cPtr = &cubic; |
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63 } else { |
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64 // OPTIMIZE: special-case half-split ? |
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65 sub = cubic.subDivide(start, 1); |
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66 cPtr = ⊂ |
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67 } |
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68 const SkDCubic& c = *cPtr; |
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69 double dx = c[3].fX - 3 * (c[2].fX - c[1].fX) - c[0].fX; |
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70 double dy = c[3].fY - 3 * (c[2].fY - c[1].fY) - c[0].fY; |
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71 double dist = sqrt(dx * dx + dy * dy); |
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72 double tDiv3 = precision / (adjust * dist); |
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73 double t = SkDCubeRoot(tDiv3); |
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74 if (start > 0) { |
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75 t = start + (1 - start) * t; |
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76 } |
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77 return t; |
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78 } |
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79 |
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80 SkDQuad SkDCubic::toQuad() const { |
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81 SkDQuad quad; |
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82 quad[0] = fPts[0]; |
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83 const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2}; |
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84 const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2}; |
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85 quad[1].fX = (fromC1.fX + fromC2.fX) / 2; |
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86 quad[1].fY = (fromC1.fY + fromC2.fY) / 2; |
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87 quad[2] = fPts[3]; |
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88 return quad; |
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89 } |
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90 |
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91 static bool add_simple_ts(const SkDCubic& cubic, double precision, SkTArray<double, true>* ts) { |
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92 double tDiv = calc_t_div(cubic, precision, 0); |
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93 if (tDiv >= 1) { |
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94 return true; |
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95 } |
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96 if (tDiv >= 0.5) { |
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97 ts->push_back(0.5); |
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98 return true; |
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99 } |
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100 return false; |
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101 } |
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102 |
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103 static void addTs(const SkDCubic& cubic, double precision, double start, double end, |
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104 SkTArray<double, true>* ts) { |
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105 double tDiv = calc_t_div(cubic, precision, 0); |
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106 double parts = ceil(1.0 / tDiv); |
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107 for (double index = 0; index < parts; ++index) { |
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108 double newT = start + (index / parts) * (end - start); |
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109 if (newT > 0 && newT < 1) { |
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110 ts->push_back(newT); |
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111 } |
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112 } |
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113 } |
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114 |
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115 // flavor that returns T values only, deferring computing the quads until they are needed |
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116 // FIXME: when called from recursive intersect 2, this could take the original cubic |
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117 // and do a more precise job when calling chop at and sub divide by computing the fractional ts. |
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118 // it would still take the prechopped cubic for reduce order and find cubic inflections |
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119 void SkDCubic::toQuadraticTs(double precision, SkTArray<double, true>* ts) const { |
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120 SkReduceOrder reducer; |
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121 int order = reducer.reduce(*this, SkReduceOrder::kAllow_Quadratics); |
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122 if (order < 3) { |
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123 return; |
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124 } |
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125 double inflectT[5]; |
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126 int inflections = findInflections(inflectT); |
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127 SkASSERT(inflections <= 2); |
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128 if (!endsAreExtremaInXOrY()) { |
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129 inflections += findMaxCurvature(&inflectT[inflections]); |
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130 SkASSERT(inflections <= 5); |
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131 } |
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132 SkTQSort<double>(inflectT, &inflectT[inflections - 1]); |
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133 // OPTIMIZATION: is this filtering common enough that it needs to be pulled out into its |
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134 // own subroutine? |
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135 while (inflections && approximately_less_than_zero(inflectT[0])) { |
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136 memmove(inflectT, &inflectT[1], sizeof(inflectT[0]) * --inflections); |
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137 } |
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138 int start = 0; |
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139 int next = 1; |
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140 while (next < inflections) { |
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141 if (!approximately_equal(inflectT[start], inflectT[next])) { |
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142 ++start; |
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143 ++next; |
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144 continue; |
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145 } |
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146 memmove(&inflectT[start], &inflectT[next], sizeof(inflectT[0]) * (--inflections - start)); |
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147 } |
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148 |
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149 while (inflections && approximately_greater_than_one(inflectT[inflections - 1])) { |
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150 --inflections; |
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151 } |
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152 SkDCubicPair pair; |
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153 if (inflections == 1) { |
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154 pair = chopAt(inflectT[0]); |
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155 int orderP1 = reducer.reduce(pair.first(), SkReduceOrder::kNo_Quadratics); |
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156 if (orderP1 < 2) { |
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157 --inflections; |
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158 } else { |
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159 int orderP2 = reducer.reduce(pair.second(), SkReduceOrder::kNo_Quadratics); |
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160 if (orderP2 < 2) { |
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161 --inflections; |
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162 } |
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163 } |
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164 } |
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165 if (inflections == 0 && add_simple_ts(*this, precision, ts)) { |
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166 return; |
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167 } |
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168 if (inflections == 1) { |
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169 pair = chopAt(inflectT[0]); |
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170 addTs(pair.first(), precision, 0, inflectT[0], ts); |
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171 addTs(pair.second(), precision, inflectT[0], 1, ts); |
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172 return; |
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173 } |
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174 if (inflections > 1) { |
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175 SkDCubic part = subDivide(0, inflectT[0]); |
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176 addTs(part, precision, 0, inflectT[0], ts); |
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177 int last = inflections - 1; |
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178 for (int idx = 0; idx < last; ++idx) { |
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179 part = subDivide(inflectT[idx], inflectT[idx + 1]); |
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180 addTs(part, precision, inflectT[idx], inflectT[idx + 1], ts); |
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181 } |
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182 part = subDivide(inflectT[last], 1); |
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183 addTs(part, precision, inflectT[last], 1, ts); |
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184 return; |
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185 } |
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186 addTs(*this, precision, 0, 1, ts); |
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187 } |