1 // geometry.cpp: Algebraic geometry helper functions
3 // Part of the Architektonas Project
4 // (C) 2011 Underground Software
5 // See the README and GPLv3 files for licensing and warranty information
7 // JLH = James Hammons <jlhamm@acm.org>
10 // --- ---------- ------------------------------------------------------------
11 // JLH 08/31/2011 Created this file
13 // NOTE: All methods in this class are static.
19 #include "dimension.h"
21 #include "mathconstants.h"
24 Point Geometry::IntersectionOfLineAndLine(Point p1, Point p2, Point p3, Point p4)
26 // Find the intersection of the lines by formula:
27 // px = (x1y2 - y1x2)(x3 - x4) - (x1 - x2)(x3y4 - y3x4)
28 // py = (x1y2 - y1x2)(y3 - y4) - (y1 - y2)(x3y4 - y3x4)
29 // d = (x1 - x2)(y3 - y4) - (y1 - y2)(x3 - x4) = 0 if lines are parallel
30 // Intersection is (px / d, py / d)
32 double d = ((p1.x - p2.x) * (p3.y - p4.y)) - ((p1.y - p2.y) * (p3.x - p4.x));
34 // Check for parallel lines, and return sentinel if so
36 return Point(0, 0, -1);
38 double px = (((p1.x * p2.y) - (p1.y * p2.x)) * (p3.x - p4.x))
39 - ((p1.x - p2.x) * ((p3.x * p4.y) - (p3.y * p4.x)));
40 double py = (((p1.x * p2.y) - (p1.y * p2.x)) * (p3.y - p4.y))
41 - ((p1.y - p2.y) * ((p3.x * p4.y) - (p3.y * p4.x)));
43 return Point(px / d, py / d, 0);
47 // Returns the parameter of a point in space to this vector. If the parameter
48 // is between 0 and 1, the normal of the vector to the point is on the vector.
49 // Note: lp1 is the tail, lp2 is the head of the line (vector).
50 double Geometry::ParameterOfLineAndPoint(Point tail, Point head, Point point)
52 // Geometric interpretation:
53 // The parameterized point on the vector lineSegment is where the normal of
54 // the lineSegment to the point intersects lineSegment. If the pp < 0, then
55 // the perpendicular lies beyond the 1st endpoint. If pp > 1, then the
56 // perpendicular lies beyond the 2nd endpoint.
58 Vector lineSegment = head - tail;
59 double magnitude = lineSegment.Magnitude();
60 Vector pointSegment = point - tail;
61 double t = lineSegment.Dot(pointSegment) / (magnitude * magnitude);
66 Point Geometry::MirrorPointAroundLine(Point point, Point tail, Point head)
68 // Get the vector of the intersection of the line and the normal on the
69 // line to the point in question.
70 double t = ParameterOfLineAndPoint(tail, head, point);
71 Vector v = Vector(tail, head) * t;
73 // Get the point normal to point to the line passed in
74 Point normalOnLine = tail + v;
76 // Make our mirrored vector (head - tail)
77 Vector mirror = -(point - normalOnLine);
79 // Find the mirrored point
80 Point mirroredPoint = normalOnLine + mirror;
86 Point Geometry::RotatePointAroundPoint(Point point, Point rotationPoint, double angle)
88 Vector v = Vector(point, rotationPoint);
89 // Vector v = Vector(rotationPoint, point);
90 double px = (v.x * cos(angle)) - (v.y * sin(angle));
91 double py = (v.x * sin(angle)) + (v.y * cos(angle));
93 return Vector(rotationPoint.x + px, rotationPoint.y + py, 0);
97 double Geometry::Determinant(Point p1, Point p2)
99 return (p1.x * p2.y) - (p2.x * p1.y);
104 Intersecting line segments:
106 Segment L1 has edges A=(a1,a2), A'=(a1',a2').
107 Segment L2 has edges B=(b1,b2), B'=(b1',b2').
108 Segment L1 is the set of points tA'+(1-t)A, where 0<=t<=1.
109 Segment L2 is the set of points sB'+(1-s)B, where 0<=s<=1.
110 Segment L1 meet segment L2 if and only if for some t and s we have
111 tA'+(1-t)A=sB'+(1-s)B
112 The solution of this with respect to t and s is
114 t=((-b?'a?+b?'b?+b?a?+a?b?'-a?b?-b?b?')/(b?'a?'-b?'a?-b?a?'+b?a?-a?'b?'+a?'b?+a?b?'-a?b?))
116 s=((-a?b?+a?'b?-a?a?'+b?a?+a?'a?-b?a?')/(b?'a?'-b?'a?-b?a?'+b?a?-a?'b??+a?'b?+a?b?'-a?b?))
118 So check if the above two numbers are both >=0 and <=1.
123 // Finds the intesection between two objects (if any)
124 bool Geometry::Intersects(Object * obj1, Object * obj2, double * t, double * s)
129 // Finds the intersection between two lines (if any)
130 int Geometry::Intersects(Line * l1, Line * l2, double * tp/*= 0*/, double * up/*= 0*/)
132 Vector r(l1->position, l1->endpoint);
133 Vector s(l2->position, l2->endpoint);
134 Vector v1 = l2->position - l1->position; // q - p
135 // Vector v2 = l1->position - l2->position; // p - q
136 //printf("l1: (%lf, %lf) (%lf, %lf), l2: (%lf, %lf) (%lf, %lf)\n", l1->position.x, l1->position.y, l1->endpoint.x, l1->endpoint.y, l2->position.x, l2->position.y, l2->endpoint.x, l2->endpoint.y);
137 double rxs = (r.x * s.y) - (s.x * r.y);
142 double qpxr = (v1.x * r.y) - (r.x * v1.y);
144 //printf(" --> R x S = 0! (q - p) x r = %lf\n", qpxr);
145 //printf(" -->(q - p) . r = %lf, r . r = %lf\n", v1.Dot(r), r.Dot(r));
146 //printf(" -->(p - q) . s = %lf, s . s = %lf\n", v2.Dot(s), s.Dot(s));
147 //printf(" -->(q - p) . s = %lf, (p - q) . r = %lf\n", v1.Dot(s), v2.Dot(r));
149 // Lines are parallel, so no intersection...
154 //this works IFF the vectors are pointing in the same direction. everything else
156 // If (q - p) . r == r . r, t = 1, u = 0
157 if (v1.Dot(r) == r.Dot(r))
159 // If (p - q) . s == s . s, t = 0, u = 1
160 else if (v2.Dot(s) == s.Dot(s))
165 // Check to see which endpoints are connected... Four possibilities:
166 if (l1->position == l2->position)
168 else if (l1->position == l2->endpoint)
170 else if (l1->endpoint == l2->position)
172 else if (l1->endpoint == l2->endpoint)
180 t = ((v1.x * s.y) - (s.x * v1.y)) / rxs;
181 u = ((v1.x * r.y) - (r.x * v1.y)) / rxs;
184 Now there are five cases (NOTE: only valid if vectors face the same way!):
186 1. If r × s = 0 and (q − p) × r = 0, then the two lines are collinear. If in addition, either 0 ≤ (q − p) · r ≤ r · r or 0 ≤ (p − q) · s ≤ s · s, then the two lines are overlapping.
188 2. If r × s = 0 and (q − p) × r = 0, but neither 0 ≤ (q − p) · r ≤ r · r nor 0 ≤ (p − q) · s ≤ s · s, then the two lines are collinear but disjoint.
190 3. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting.
192 4. If r × s ≠ 0 and 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1, the two line segments meet at the point p + t r = q + u s.
194 5. Otherwise, the two line segments are not parallel but do not intersect.
196 // Return parameter values, if user passed in valid pointers
203 // If the parameters are in range, we have overlap!
204 if ((t >= 0) && (t <= 1.0) && (u >= 0) && (u <= 1.0))
211 // Finds the intersection between two lines (if any)
212 int Geometry::Intersects(Line * l1, Dimension * d1, double * tp/*= 0*/, double * up/*= 0*/)
214 Line l2(d1->position, d1->endpoint);
215 return Intersects(l1, &l2, tp, up);
219 // Finds the intersection(s) between a line and a circle (if any)
220 int Geometry::Intersects(Line * l, Circle * c, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
223 Vector center = c->position;
224 Vector v1 = l->position - center;
225 Vector v2 = l->endpoint - center;
227 double dr = d.Magnitude();
228 double determinant = (v1.x * v2.y) - (v1.y * v2.x);
230 double discriminant = ((c->radius * c->radius) * (dr * dr)) - (determinant * determinant);
232 if (discriminant < 0)
240 I'm thinking a better approach to this might be as follows:
242 -- Get the distance of the circle's center from the line segment. If it's
243 > the radius, it doesn't intersect.
244 -- If the parameter is off the line segment, check distance to endpoints. (Not sure
245 how to proceed from here, it's different than the following.)
246 [Actually, you can use the following for all of it. You only know if you have
247 an intersection at the last step, which is OK.]
248 -- If the radius == distance, we have a tangent line.
249 -- If radius > distance, use Pythagorus to find the length on either side of the
250 normal to the spots where the hypotenuse (== radius' length) contact the line.
251 -- Use those points to find the parameter on the line segment; if they're not on
252 the line segment, no intersection.
254 double t = ParameterOfLineAndPoint(l->position, l->endpoint, c->position);
255 //small problem here: it clamps the result to the line segment. NOT what we want
256 //here! !!! FIX !!! [DONE]
257 Vector p = l->GetPointAtParameter(t);
258 double distance = Vector::Magnitude(c->position, p);
260 // If the center of the circle is farther from the line than the radius, fail.
261 if (distance > c->radius)
264 // Now we have to check for intersection points.
265 // Tangent case: (needs to return something)
266 if ((distance == c->radius) && (t >= 0.0) && (t <= 1.0))
268 // Need to set tp & up to something... !!! FIX !!!
273 *up = Vector(c->position, p).Angle();
278 // The line intersects the circle in two points (possibly). Use Pythagorus
279 // to find them for testing.
280 double offset = sqrt((c->radius * c->radius) - (distance * distance));
281 //need to convert distance to paramter value... :-/
282 //t = position on line / length of line segment, so if we divide the offset by length,
283 //that should give us what we want.
284 double length = Vector::Magnitude(l->position, l->endpoint);
285 double t1 = t + (offset / length);
286 double t2 = t - (offset / length);
288 //need to find angles for the circle...
289 Vector cp1 = l->position + (Vector(l->position, l->endpoint) * (length * t1));
290 Vector cp2 = l->position + (Vector(l->position, l->endpoint) * (length * t2));
291 double a1 = Vector(c->position, cp1).Angle();
292 double a2 = Vector(c->position, cp2).Angle();
294 //instead of this, return a # which is the # of intersections. [DONE]
295 int intersections = 0;
297 // Now check for if the parameters are in range
298 if ((t1 >= 0) && (t1 <= 1.0))
303 if ((t2 >= 0) && (t2 <= 1.0))
308 return intersections;
313 // Finds the intersection(s) between a circle and a circle (if any)
314 // There can be 0, 1, or 2 intersections.
315 // Returns the angles of the points of intersection in tp thru wp, with the
316 // angles returned as c1, c2, c1, c2 (if applicable--in the 1 intersection case,
317 // only the first two angles are returned: c1, c2).
318 int Geometry::Intersects(Circle * c1, Circle * c2, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/, Point * p1/*= 0*/, Point * p2/*= 0*/)
320 // Get the distance between centers. If the distance plus the radius of the
321 // smaller circle is less than the radius of the larger circle, there is no
322 // intersection. If the distance is greater than the sum of the radii,
323 // there is no intersection. If the distance is equal to the sum of the
324 // radii, they are tangent and intersect at one point. Otherwise, they
325 // intersect at two points.
326 Vector centerLine(c1->position, c2->position);
327 double d = centerLine.Magnitude();
328 //printf("Circle #1: pos=<%lf, %lf>, r=%lf\n", c1->position.x, c1->position.y, c1->radius);
329 //printf("Circle #2: pos=<%lf, %lf>, r=%lf\n", c2->position.x, c2->position.y, c2->radius);
330 //printf("Distance between #1 & #2: %lf\n", d);
332 // Check to see if we actually have an intersection, and return failure if not
333 if ((fabs(c1->radius - c2->radius) > d) || ((c1->radius + c2->radius) < d))
336 // There are *two* tangent cases!
337 if (((c1->radius + c2->radius) == d) || (fabs(c1->radius - c2->radius) == d))
339 // Need to return something in tp & up!! !!! FIX !!! [DONE]
341 *tp = centerLine.Angle();
344 *up = centerLine.Angle() + PI;
349 // Find the distance from the center of c1 to the perpendicular chord
350 // (which contains the points of intersection)
351 double x = ((d * d) - (c2->radius * c2->radius) + (c1->radius * c1->radius))
353 // Find the the length of the perpendicular chord
355 double a = sqrt((-d + c2->radius - c1->radius) * (-d - c2->radius + c1->radius) * (-d + c2->radius + c1->radius) * (d + c2->radius + c1->radius)) / d;
357 // Now, you can use pythagorus to find the length of the hypotenuse, but we
358 // already know that length, it's the radius! :-P
359 // What's needed is the angle of the center line and the radial line. Since
360 // there's two intersection points, there's also four angles (two for each
362 // We can use the arccos to find the angle using just the radius and the
363 // distance to the perpendicular chord...!
364 double angleC1 = acos(x / c1->radius);
365 double angleC2 = acos((d - x) / c2->radius);
368 *tp = centerLine.Angle() - angleC1;
371 *up = (centerLine.Angle() + PI) - angleC2;
374 *vp = centerLine.Angle() + angleC1;
377 *wp = (centerLine.Angle() + PI) + angleC2;
380 *p1 = c1->position + (centerLine.Unit() * x) + (Vector::Normal(Vector(), centerLine) * (a / 2.0));
383 *p2 = c1->position + (centerLine.Unit() * x) - (Vector::Normal(Vector(), centerLine) * (a / 2.0));
389 // should we just do common trig solves, like AAS, ASA, SAS, SSA?
391 // c^2 = a^2 + b^2 -2ab*cos(C)
393 // cos(C) = (c^2 - a^2 - b^2) / -2ab = (a^2 + b^2 - c^2) / 2ab
395 // a / sin A = b / sin B = c / sin C
397 // Solve the angles of the triangle given the sides. Angles returned are
398 // opposite of the given sides (so a1 consists of sides s2 & s3, and so on).
399 void Geometry::FindAnglesForSides(double s1, double s2, double s3, double * a1, double * a2, double * a3)
401 // Use law of cosines to find 1st angle
402 double cosine1 = ((s2 * s2) + (s3 * s3) - (s1 * s1)) / (2.0 * s2 * s3);
404 // Check for a valid triangle
405 if ((cosine1 < -1.0) || (cosine1 > 1.0))
408 double angle1 = acos(cosine1);
410 // Use law of sines to find 2nd & 3rd angles
411 // sin A / a = sin B / b
412 // sin B = (sin A / a) * b
413 double angle2 = asin(s2 * (sin(angle1) / s1));
414 double angle3 = asin(s3 * (sin(angle1) / s1));