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| 1 /* | |
| 2 * Copyright 2012 Google Inc. | |
| 3 * | |
| 4 * Use of this source code is governed by a BSD-style license that can be | |
| 5 * found in the LICENSE file. | |
| 6 */ | |
| 7 #include "CurveIntersection.h" | |
| 8 #include "Intersections.h" | |
| 9 #include "LineUtilities.h" | |
| 10 #include "QuadraticUtilities.h" | |
| 11 | |
| 12 /* | |
| 13 Find the interection of a line and quadratic by solving for valid t values. | |
| 14 | |
| 15 From http://stackoverflow.com/questions/1853637/how-to-find-the-mathematical-fun
ction-defining-a-bezier-curve | |
| 16 | |
| 17 "A Bezier curve is a parametric function. A quadratic Bezier curve (i.e. three | |
| 18 control points) can be expressed as: F(t) = A(1 - t)^2 + B(1 - t)t + Ct^2 where | |
| 19 A, B and C are points and t goes from zero to one. | |
| 20 | |
| 21 This will give you two equations: | |
| 22 | |
| 23 x = a(1 - t)^2 + b(1 - t)t + ct^2 | |
| 24 y = d(1 - t)^2 + e(1 - t)t + ft^2 | |
| 25 | |
| 26 If you add for instance the line equation (y = kx + m) to that, you'll end up | |
| 27 with three equations and three unknowns (x, y and t)." | |
| 28 | |
| 29 Similar to above, the quadratic is represented as | |
| 30 x = a(1-t)^2 + 2b(1-t)t + ct^2 | |
| 31 y = d(1-t)^2 + 2e(1-t)t + ft^2 | |
| 32 and the line as | |
| 33 y = g*x + h | |
| 34 | |
| 35 Using Mathematica, solve for the values of t where the quadratic intersects the | |
| 36 line: | |
| 37 | |
| 38 (in) t1 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - x, | |
| 39 d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - g*x - h, x] | |
| 40 (out) -d + h + 2 d t - 2 e t - d t^2 + 2 e t^2 - f t^2 + | |
| 41 g (a - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2) | |
| 42 (in) Solve[t1 == 0, t] | |
| 43 (out) { | |
| 44 {t -> (-2 d + 2 e + 2 a g - 2 b g - | |
| 45 Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 - | |
| 46 4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) / | |
| 47 (2 (-d + 2 e - f + a g - 2 b g + c g)) | |
| 48 }, | |
| 49 {t -> (-2 d + 2 e + 2 a g - 2 b g + | |
| 50 Sqrt[(2 d - 2 e - 2 a g + 2 b g)^2 - | |
| 51 4 (-d + 2 e - f + a g - 2 b g + c g) (-d + a g + h)]) / | |
| 52 (2 (-d + 2 e - f + a g - 2 b g + c g)) | |
| 53 } | |
| 54 } | |
| 55 | |
| 56 Using the results above (when the line tends towards horizontal) | |
| 57 A = (-(d - 2*e + f) + g*(a - 2*b + c) ) | |
| 58 B = 2*( (d - e ) - g*(a - b ) ) | |
| 59 C = (-(d ) + g*(a ) + h ) | |
| 60 | |
| 61 If g goes to infinity, we can rewrite the line in terms of x. | |
| 62 x = g'*y + h' | |
| 63 | |
| 64 And solve accordingly in Mathematica: | |
| 65 | |
| 66 (in) t2 = Resultant[a*(1 - t)^2 + 2*b*(1 - t)*t + c*t^2 - g'*y - h', | |
| 67 d*(1 - t)^2 + 2*e*(1 - t)*t + f*t^2 - y, y] | |
| 68 (out) a - h' - 2 a t + 2 b t + a t^2 - 2 b t^2 + c t^2 - | |
| 69 g' (d - 2 d t + 2 e t + d t^2 - 2 e t^2 + f t^2) | |
| 70 (in) Solve[t2 == 0, t] | |
| 71 (out) { | |
| 72 {t -> (2 a - 2 b - 2 d g' + 2 e g' - | |
| 73 Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 - | |
| 74 4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')]) / | |
| 75 (2 (a - 2 b + c - d g' + 2 e g' - f g')) | |
| 76 }, | |
| 77 {t -> (2 a - 2 b - 2 d g' + 2 e g' + | |
| 78 Sqrt[(-2 a + 2 b + 2 d g' - 2 e g')^2 - | |
| 79 4 (a - 2 b + c - d g' + 2 e g' - f g') (a - d g' - h')])/ | |
| 80 (2 (a - 2 b + c - d g' + 2 e g' - f g')) | |
| 81 } | |
| 82 } | |
| 83 | |
| 84 Thus, if the slope of the line tends towards vertical, we use: | |
| 85 A = ( (a - 2*b + c) - g'*(d - 2*e + f) ) | |
| 86 B = 2*(-(a - b ) + g'*(d - e ) ) | |
| 87 C = ( (a ) - g'*(d ) - h' ) | |
| 88 */ | |
| 89 | |
| 90 | |
| 91 class LineQuadraticIntersections { | |
| 92 public: | |
| 93 | |
| 94 LineQuadraticIntersections(const Quadratic& q, const _Line& l, Intersections& i) | |
| 95 : quad(q) | |
| 96 , line(l) | |
| 97 , intersections(i) { | |
| 98 } | |
| 99 | |
| 100 int intersectRay(double roots[2]) { | |
| 101 /* | |
| 102 solve by rotating line+quad so line is horizontal, then finding the roots | |
| 103 set up matrix to rotate quad to x-axis | |
| 104 |cos(a) -sin(a)| | |
| 105 |sin(a) cos(a)| | |
| 106 note that cos(a) = A(djacent) / Hypoteneuse | |
| 107 sin(a) = O(pposite) / Hypoteneuse | |
| 108 since we are computing Ts, we can ignore hypoteneuse, the scale factor: | |
| 109 | A -O | | |
| 110 | O A | | |
| 111 A = line[1].x - line[0].x (adjacent side of the right triangle) | |
| 112 O = line[1].y - line[0].y (opposite side of the right triangle) | |
| 113 for each of the three points (e.g. n = 0 to 2) | |
| 114 quad[n].y' = (quad[n].y - line[0].y) * A - (quad[n].x - line[0].x) * O | |
| 115 */ | |
| 116 double adj = line[1].x - line[0].x; | |
| 117 double opp = line[1].y - line[0].y; | |
| 118 double r[3]; | |
| 119 for (int n = 0; n < 3; ++n) { | |
| 120 r[n] = (quad[n].y - line[0].y) * adj - (quad[n].x - line[0].x) * opp; | |
| 121 } | |
| 122 double A = r[2]; | |
| 123 double B = r[1]; | |
| 124 double C = r[0]; | |
| 125 A += C - 2 * B; // A = a - 2*b + c | |
| 126 B -= C; // B = -(b - c) | |
| 127 return quadraticRootsValidT(A, 2 * B, C, roots); | |
| 128 } | |
| 129 | |
| 130 int intersect() { | |
| 131 addEndPoints(); | |
| 132 double rootVals[2]; | |
| 133 int roots = intersectRay(rootVals); | |
| 134 for (int index = 0; index < roots; ++index) { | |
| 135 double quadT = rootVals[index]; | |
| 136 double lineT = findLineT(quadT); | |
| 137 if (pinTs(quadT, lineT)) { | |
| 138 _Point pt; | |
| 139 xy_at_t(line, lineT, pt.x, pt.y); | |
| 140 intersections.insert(quadT, lineT, pt); | |
| 141 } | |
| 142 } | |
| 143 return intersections.fUsed; | |
| 144 } | |
| 145 | |
| 146 int horizontalIntersect(double axisIntercept, double roots[2]) { | |
| 147 double D = quad[2].y; // f | |
| 148 double E = quad[1].y; // e | |
| 149 double F = quad[0].y; // d | |
| 150 D += F - 2 * E; // D = d - 2*e + f | |
| 151 E -= F; // E = -(d - e) | |
| 152 F -= axisIntercept; | |
| 153 return quadraticRootsValidT(D, 2 * E, F, roots); | |
| 154 } | |
| 155 | |
| 156 int horizontalIntersect(double axisIntercept, double left, double right, bool fl
ipped) { | |
| 157 addHorizontalEndPoints(left, right, axisIntercept); | |
| 158 double rootVals[2]; | |
| 159 int roots = horizontalIntersect(axisIntercept, rootVals); | |
| 160 for (int index = 0; index < roots; ++index) { | |
| 161 _Point pt; | |
| 162 double quadT = rootVals[index]; | |
| 163 xy_at_t(quad, quadT, pt.x, pt.y); | |
| 164 double lineT = (pt.x - left) / (right - left); | |
| 165 if (pinTs(quadT, lineT)) { | |
| 166 intersections.insert(quadT, lineT, pt); | |
| 167 } | |
| 168 } | |
| 169 if (flipped) { | |
| 170 flip(); | |
| 171 } | |
| 172 return intersections.fUsed; | |
| 173 } | |
| 174 | |
| 175 int verticalIntersect(double axisIntercept, double roots[2]) { | |
| 176 double D = quad[2].x; // f | |
| 177 double E = quad[1].x; // e | |
| 178 double F = quad[0].x; // d | |
| 179 D += F - 2 * E; // D = d - 2*e + f | |
| 180 E -= F; // E = -(d - e) | |
| 181 F -= axisIntercept; | |
| 182 return quadraticRootsValidT(D, 2 * E, F, roots); | |
| 183 } | |
| 184 | |
| 185 int verticalIntersect(double axisIntercept, double top, double bottom, bool flip
ped) { | |
| 186 addVerticalEndPoints(top, bottom, axisIntercept); | |
| 187 double rootVals[2]; | |
| 188 int roots = verticalIntersect(axisIntercept, rootVals); | |
| 189 for (int index = 0; index < roots; ++index) { | |
| 190 _Point pt; | |
| 191 double quadT = rootVals[index]; | |
| 192 xy_at_t(quad, quadT, pt.x, pt.y); | |
| 193 double lineT = (pt.y - top) / (bottom - top); | |
| 194 if (pinTs(quadT, lineT)) { | |
| 195 intersections.insert(quadT, lineT, pt); | |
| 196 } | |
| 197 } | |
| 198 if (flipped) { | |
| 199 flip(); | |
| 200 } | |
| 201 return intersections.fUsed; | |
| 202 } | |
| 203 | |
| 204 protected: | |
| 205 | |
| 206 // add endpoints first to get zero and one t values exactly | |
| 207 void addEndPoints() | |
| 208 { | |
| 209 for (int qIndex = 0; qIndex < 3; qIndex += 2) { | |
| 210 for (int lIndex = 0; lIndex < 2; lIndex++) { | |
| 211 if (quad[qIndex] == line[lIndex]) { | |
| 212 intersections.insert(qIndex >> 1, lIndex, line[lIndex]); | |
| 213 } | |
| 214 } | |
| 215 } | |
| 216 } | |
| 217 | |
| 218 void addHorizontalEndPoints(double left, double right, double y) | |
| 219 { | |
| 220 for (int qIndex = 0; qIndex < 3; qIndex += 2) { | |
| 221 if (quad[qIndex].y != y) { | |
| 222 continue; | |
| 223 } | |
| 224 if (quad[qIndex].x == left) { | |
| 225 intersections.insert(qIndex >> 1, 0, quad[qIndex]); | |
| 226 } | |
| 227 if (quad[qIndex].x == right) { | |
| 228 intersections.insert(qIndex >> 1, 1, quad[qIndex]); | |
| 229 } | |
| 230 } | |
| 231 } | |
| 232 | |
| 233 void addVerticalEndPoints(double top, double bottom, double x) | |
| 234 { | |
| 235 for (int qIndex = 0; qIndex < 3; qIndex += 2) { | |
| 236 if (quad[qIndex].x != x) { | |
| 237 continue; | |
| 238 } | |
| 239 if (quad[qIndex].y == top) { | |
| 240 intersections.insert(qIndex >> 1, 0, quad[qIndex]); | |
| 241 } | |
| 242 if (quad[qIndex].y == bottom) { | |
| 243 intersections.insert(qIndex >> 1, 1, quad[qIndex]); | |
| 244 } | |
| 245 } | |
| 246 } | |
| 247 | |
| 248 double findLineT(double t) { | |
| 249 double x, y; | |
| 250 xy_at_t(quad, t, x, y); | |
| 251 double dx = line[1].x - line[0].x; | |
| 252 double dy = line[1].y - line[0].y; | |
| 253 if (fabs(dx) > fabs(dy)) { | |
| 254 return (x - line[0].x) / dx; | |
| 255 } | |
| 256 return (y - line[0].y) / dy; | |
| 257 } | |
| 258 | |
| 259 void flip() { | |
| 260 // OPTIMIZATION: instead of swapping, pass original line, use [1].y - [0].y | |
| 261 int roots = intersections.fUsed; | |
| 262 for (int index = 0; index < roots; ++index) { | |
| 263 intersections.fT[1][index] = 1 - intersections.fT[1][index]; | |
| 264 } | |
| 265 } | |
| 266 | |
| 267 static bool pinTs(double& quadT, double& lineT) { | |
| 268 if (!approximately_one_or_less(lineT)) { | |
| 269 return false; | |
| 270 } | |
| 271 if (!approximately_zero_or_more(lineT)) { | |
| 272 return false; | |
| 273 } | |
| 274 if (precisely_less_than_zero(quadT)) { | |
| 275 quadT = 0; | |
| 276 } else if (precisely_greater_than_one(quadT)) { | |
| 277 quadT = 1; | |
| 278 } | |
| 279 if (precisely_less_than_zero(lineT)) { | |
| 280 lineT = 0; | |
| 281 } else if (precisely_greater_than_one(lineT)) { | |
| 282 lineT = 1; | |
| 283 } | |
| 284 return true; | |
| 285 } | |
| 286 | |
| 287 private: | |
| 288 | |
| 289 const Quadratic& quad; | |
| 290 const _Line& line; | |
| 291 Intersections& intersections; | |
| 292 }; | |
| 293 | |
| 294 // utility for pairs of coincident quads | |
| 295 static double horizontalIntersect(const Quadratic& quad, const _Point& pt) { | |
| 296 LineQuadraticIntersections q(quad, *((_Line*) 0), *((Intersections*) 0)); | |
| 297 double rootVals[2]; | |
| 298 int roots = q.horizontalIntersect(pt.y, rootVals); | |
| 299 for (int index = 0; index < roots; ++index) { | |
| 300 double x; | |
| 301 double t = rootVals[index]; | |
| 302 xy_at_t(quad, t, x, *(double*) 0); | |
| 303 if (AlmostEqualUlps(x, pt.x)) { | |
| 304 return t; | |
| 305 } | |
| 306 } | |
| 307 return -1; | |
| 308 } | |
| 309 | |
| 310 static double verticalIntersect(const Quadratic& quad, const _Point& pt) { | |
| 311 LineQuadraticIntersections q(quad, *((_Line*) 0), *((Intersections*) 0)); | |
| 312 double rootVals[2]; | |
| 313 int roots = q.verticalIntersect(pt.x, rootVals); | |
| 314 for (int index = 0; index < roots; ++index) { | |
| 315 double y; | |
| 316 double t = rootVals[index]; | |
| 317 xy_at_t(quad, t, *(double*) 0, y); | |
| 318 if (AlmostEqualUlps(y, pt.y)) { | |
| 319 return t; | |
| 320 } | |
| 321 } | |
| 322 return -1; | |
| 323 } | |
| 324 | |
| 325 double axialIntersect(const Quadratic& q1, const _Point& p, bool vertical) { | |
| 326 if (vertical) { | |
| 327 return verticalIntersect(q1, p); | |
| 328 } | |
| 329 return horizontalIntersect(q1, p); | |
| 330 } | |
| 331 | |
| 332 int horizontalIntersect(const Quadratic& quad, double left, double right, | |
| 333 double y, double tRange[2]) { | |
| 334 LineQuadraticIntersections q(quad, *((_Line*) 0), *((Intersections*) 0)); | |
| 335 double rootVals[2]; | |
| 336 int result = q.horizontalIntersect(y, rootVals); | |
| 337 int tCount = 0; | |
| 338 for (int index = 0; index < result; ++index) { | |
| 339 double x, y; | |
| 340 xy_at_t(quad, rootVals[index], x, y); | |
| 341 if (x < left || x > right) { | |
| 342 continue; | |
| 343 } | |
| 344 tRange[tCount++] = rootVals[index]; | |
| 345 } | |
| 346 return tCount; | |
| 347 } | |
| 348 | |
| 349 int horizontalIntersect(const Quadratic& quad, double left, double right, double
y, | |
| 350 bool flipped, Intersections& intersections) { | |
| 351 LineQuadraticIntersections q(quad, *((_Line*) 0), intersections); | |
| 352 return q.horizontalIntersect(y, left, right, flipped); | |
| 353 } | |
| 354 | |
| 355 int verticalIntersect(const Quadratic& quad, double top, double bottom, double x
, | |
| 356 bool flipped, Intersections& intersections) { | |
| 357 LineQuadraticIntersections q(quad, *((_Line*) 0), intersections); | |
| 358 return q.verticalIntersect(x, top, bottom, flipped); | |
| 359 } | |
| 360 | |
| 361 int intersect(const Quadratic& quad, const _Line& line, Intersections& i) { | |
| 362 LineQuadraticIntersections q(quad, line, i); | |
| 363 return q.intersect(); | |
| 364 } | |
| 365 | |
| 366 int intersectRay(const Quadratic& quad, const _Line& line, Intersections& i) { | |
| 367 LineQuadraticIntersections q(quad, line, i); | |
| 368 return q.intersectRay(i.fT[0]); | |
| 369 } | |
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Chromium Code Reviews
Issue 867213004:
remove prototype pathops code (Closed)
Base URL: https://skia.googlesource.com/skia.git@master