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-function-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 flipped) { 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 flipped) { 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 } 370