1 /* 2 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi 3 */ 4 5 /* 6 Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. 7 Then for degree elevation, the equations are: 8 9 Q0 = P0 10 Q1 = 1/3 P0 + 2/3 P1 11 Q2 = 2/3 P1 + 1/3 P2 12 Q3 = P2 13 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from 14 the equations above: 15 16 P1 = 3/2 Q1 - 1/2 Q0 17 P1 = 3/2 Q2 - 1/2 Q3 18 If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since 19 it's likely not, your best bet is to average them. So, 20 21 P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3 22 23 24 SkDCubic defined by: P1/2 - anchor points, C1/C2 control points 25 |x| is the euclidean norm of x 26 mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the 27 control point at C = (3C2 - P2 + 3C1 - P1)/4 28 29 Algorithm 30 31 pick an absolute precision (prec) 32 Compute the Tdiv as the root of (cubic) equation 33 sqrt(3)/18 |P2 - 3C2 + 3C1 - P1|/2 Tdiv ^ 3 = prec 34 if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a 35 quadratic, with a defect less than prec, by the mid-point approximation. 36 Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv) 37 0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point 38 approximation 39 Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation 40 41 confirmed by (maybe stolen from) 42 http://www.caffeineowl.com/graphics/2d/vectorial/cubic2quad01.html 43 // maybe in turn derived from http://www.cccg.ca/proceedings/2004/36.pdf 44 // also stored at http://www.cis.usouthal.edu/~hain/general/Publications/Bezier/bezier%20cccg04%20paper.pdf 45 46 */ 47 48 #include "SkPathOpsCubic.h" 49 #include "SkPathOpsLine.h" 50 #include "SkPathOpsQuad.h" 51 #include "SkReduceOrder.h" 52 #include "SkTArray.h" 53 #include "SkTSort.h" 54 55 #define USE_CUBIC_END_POINTS 1 56 57 static double calc_t_div(const SkDCubic& cubic, double precision, double start) { 58 const double adjust = sqrt(3.) / 36; 59 SkDCubic sub; 60 const SkDCubic* cPtr; 61 if (start == 0) { 62 cPtr = &cubic; 63 } else { 64 // OPTIMIZE: special-case half-split ? 65 sub = cubic.subDivide(start, 1); 66 cPtr = ⊂ 67 } 68 const SkDCubic& c = *cPtr; 69 double dx = c[3].fX - 3 * (c[2].fX - c[1].fX) - c[0].fX; 70 double dy = c[3].fY - 3 * (c[2].fY - c[1].fY) - c[0].fY; 71 double dist = sqrt(dx * dx + dy * dy); 72 double tDiv3 = precision / (adjust * dist); 73 double t = SkDCubeRoot(tDiv3); 74 if (start > 0) { 75 t = start + (1 - start) * t; 76 } 77 return t; 78 } 79 80 SkDQuad SkDCubic::toQuad() const { 81 SkDQuad quad; 82 quad[0] = fPts[0]; 83 const SkDPoint fromC1 = {(3 * fPts[1].fX - fPts[0].fX) / 2, (3 * fPts[1].fY - fPts[0].fY) / 2}; 84 const SkDPoint fromC2 = {(3 * fPts[2].fX - fPts[3].fX) / 2, (3 * fPts[2].fY - fPts[3].fY) / 2}; 85 quad[1].fX = (fromC1.fX + fromC2.fX) / 2; 86 quad[1].fY = (fromC1.fY + fromC2.fY) / 2; 87 quad[2] = fPts[3]; 88 return quad; 89 } 90 91 static bool add_simple_ts(const SkDCubic& cubic, double precision, SkTArray<double, true>* ts) { 92 double tDiv = calc_t_div(cubic, precision, 0); 93 if (tDiv >= 1) { 94 return true; 95 } 96 if (tDiv >= 0.5) { 97 ts->push_back(0.5); 98 return true; 99 } 100 return false; 101 } 102 103 static void addTs(const SkDCubic& cubic, double precision, double start, double end, 104 SkTArray<double, true>* ts) { 105 double tDiv = calc_t_div(cubic, precision, 0); 106 double parts = ceil(1.0 / tDiv); 107 for (double index = 0; index < parts; ++index) { 108 double newT = start + (index / parts) * (end - start); 109 if (newT > 0 && newT < 1) { 110 ts->push_back(newT); 111 } 112 } 113 } 114 115 // flavor that returns T values only, deferring computing the quads until they are needed 116 // FIXME: when called from recursive intersect 2, this could take the original cubic 117 // and do a more precise job when calling chop at and sub divide by computing the fractional ts. 118 // it would still take the prechopped cubic for reduce order and find cubic inflections 119 void SkDCubic::toQuadraticTs(double precision, SkTArray<double, true>* ts) const { 120 SkReduceOrder reducer; 121 int order = reducer.reduce(*this, SkReduceOrder::kAllow_Quadratics); 122 if (order < 3) { 123 return; 124 } 125 double inflectT[5]; 126 int inflections = findInflections(inflectT); 127 SkASSERT(inflections <= 2); 128 if (!endsAreExtremaInXOrY()) { 129 inflections += findMaxCurvature(&inflectT[inflections]); 130 SkASSERT(inflections <= 5); 131 } 132 SkTQSort<double>(inflectT, &inflectT[inflections - 1]); 133 // OPTIMIZATION: is this filtering common enough that it needs to be pulled out into its 134 // own subroutine? 135 while (inflections && approximately_less_than_zero(inflectT[0])) { 136 memmove(inflectT, &inflectT[1], sizeof(inflectT[0]) * --inflections); 137 } 138 int start = 0; 139 do { 140 int next = start + 1; 141 if (next >= inflections) { 142 break; 143 } 144 if (!approximately_equal(inflectT[start], inflectT[next])) { 145 ++start; 146 continue; 147 } 148 memmove(&inflectT[start], &inflectT[next], sizeof(inflectT[0]) * (--inflections - start)); 149 } while (true); 150 while (inflections && approximately_greater_than_one(inflectT[inflections - 1])) { 151 --inflections; 152 } 153 SkDCubicPair pair; 154 if (inflections == 1) { 155 pair = chopAt(inflectT[0]); 156 int orderP1 = reducer.reduce(pair.first(), SkReduceOrder::kNo_Quadratics); 157 if (orderP1 < 2) { 158 --inflections; 159 } else { 160 int orderP2 = reducer.reduce(pair.second(), SkReduceOrder::kNo_Quadratics); 161 if (orderP2 < 2) { 162 --inflections; 163 } 164 } 165 } 166 if (inflections == 0 && add_simple_ts(*this, precision, ts)) { 167 return; 168 } 169 if (inflections == 1) { 170 pair = chopAt(inflectT[0]); 171 addTs(pair.first(), precision, 0, inflectT[0], ts); 172 addTs(pair.second(), precision, inflectT[0], 1, ts); 173 return; 174 } 175 if (inflections > 1) { 176 SkDCubic part = subDivide(0, inflectT[0]); 177 addTs(part, precision, 0, inflectT[0], ts); 178 int last = inflections - 1; 179 for (int idx = 0; idx < last; ++idx) { 180 part = subDivide(inflectT[idx], inflectT[idx + 1]); 181 addTs(part, precision, inflectT[idx], inflectT[idx + 1], ts); 182 } 183 part = subDivide(inflectT[last], 1); 184 addTs(part, precision, inflectT[last], 1, ts); 185 return; 186 } 187 addTs(*this, precision, 0, 1, ts); 188 } 189
