[architektonas] / src / geometry.cpp

// geometry.cpp: Algebraic geometry helper functions
//
// Part of the Architektonas Project
// (C) 2011 Underground Software
// See the README and GPLv3 files for licensing and warranty information
//
// JLH = James Hammons <jlhamm@acm.org>
//
// WHO  WHEN        WHAT
// ---  ----------  ------------------------------------------------------------
// JLH  08/31/2011  Created this file
//
// NOTE: All methods in this class are static.
//

#include "geometry.h"
#include <math.h>
#include <stdio.h>
#include "mathconstants.h"


// This is unused
#if 0
Point Geometry::IntersectionOfLineAndLine(Point p1, Point p2, Point p3, Point p4)
{
        // Find the intersection of the lines by formula:
        // px = (x1y2 - y1x2)(x3 - x4) - (x1 - x2)(x3y4 - y3x4)
        // py = (x1y2 - y1x2)(y3 - y4) - (y1 - y2)(x3y4 - y3x4)
        // d = (x1 - x2)(y3 - y4) - (y1 - y2)(x3 - x4) = 0 if lines are parallel
        // Intersection is (px / d, py / d)

        double d = ((p1.x - p2.x) * (p3.y - p4.y)) - ((p1.y - p2.y) * (p3.x - p4.x));

        // Check for parallel lines, and return sentinel if so
        if (d == 0)
                return Point(0, 0, -1);

        double px = (((p1.x * p2.y) - (p1.y * p2.x)) * (p3.x - p4.x))
                - ((p1.x - p2.x) * ((p3.x * p4.y) - (p3.y * p4.x)));
        double py = (((p1.x * p2.y) - (p1.y * p2.x)) * (p3.y - p4.y))
                - ((p1.y - p2.y) * ((p3.x * p4.y) - (p3.y * p4.x)));

        return Point(px / d, py / d, 0);
}
#endif


// Returns the parameter of a point in space to this vector. If the parameter
// is between 0 and 1, the normal of the vector to the point is on the vector.
// Note: lp1 is the tail, lp2 is the head of the line (vector).
double Geometry::ParameterOfLineAndPoint(Point tail, Point head, Point point)
{
        // Geometric interpretation:
        // The parameterized point on the vector lineSegment is where the normal of
        // the lineSegment to the point intersects lineSegment. If the pp < 0, then
        // the perpendicular lies beyond the 1st endpoint. If pp > 1, then the
        // perpendicular lies beyond the 2nd endpoint.

        Vector lineSegment = head - tail;
        double magnitude = lineSegment.Magnitude();
        Vector pointSegment = point - tail;
        double t = lineSegment.Dot(pointSegment) / (magnitude * magnitude);
        return t;
}


Point Geometry::MirrorPointAroundLine(Point point, Point tail, Point head)
{
        // Get the vector of the intersection of the line and the normal on the
        // line to the point in question.
        double t = ParameterOfLineAndPoint(tail, head, point);
        Vector v = Vector(tail, head) * t;

        // Get the point normal to point to the line passed in
        Point normalOnLine = tail + v;

        // Make our mirrored vector (head - tail)
        Vector mirror = -(point - normalOnLine);

        // Find the mirrored point
        Point mirroredPoint = normalOnLine + mirror;

        return mirroredPoint;
}


//
// point: The point we're rotating
// rotationPoint: The point we're rotating around
//
Point Geometry::RotatePointAroundPoint(Point point, Point rotationPoint, double angle)
{
//      Vector v = Vector(point, rotationPoint);
        Vector v = Vector(rotationPoint, point);
        double px = (v.x * cos(angle)) - (v.y * sin(angle));
        double py = (v.x * sin(angle)) + (v.y * cos(angle));

        return Vector(rotationPoint.x + px, rotationPoint.y + py, 0);
}


double Geometry::Determinant(Point p1, Point p2)
{
        return (p1.x * p2.y) - (p2.x * p1.y);
}


int Geometry::Intersects(Object * obj1, Object * obj2, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
{
        if ((obj1->type == OTLine) && (obj2->type == OTLine))
                return CheckLineToLineIntersection(obj1, obj2, tp, up);

        return 0;
}


/*
Intersecting line segments:
An easier way:
Segment L1 has edges A=(a1,a2), A'=(a1',a2').
Segment L2 has edges B=(b1,b2), B'=(b1',b2').
Segment L1 is the set of points tA'+(1-t)A, where 0<=t<=1.
Segment L2 is the set of points sB'+(1-s)B, where 0<=s<=1.
Segment L1 meet segment L2 if and only if for some t and s we have
tA'+(1-t)A=sB'+(1-s)B
The solution of this with respect to t and s is

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?))

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?))

So check if the above two numbers are both >=0 and <=1.
*/


// Finds the intersection between two lines (if any)
int Geometry::CheckLineToLineIntersection(Object * l1, Object * l2, double * tp, double * up)
{
        Vector r(l1->p[0], l1->p[1]);
        Vector s(l2->p[0], l2->p[1]);
        Vector v1 = l2->p[0] - l1->p[0];        // q - p
#if 0
        Vector v2 = l1->p[0] - l2->p[0];        // p - q
printf("l1: (%lf, %lf) (%lf, %lf), l2: (%lf, %lf) (%lf, %lf)\n", l1->p[0].x, l1->p[0].y, l1->p[1].x, l1->p[1].y, l2->p[0].x, l2->p[0].y, l2->p[1].x, l2->p[1].y);
#endif
        double rxs = (r.x * s.y) - (s.x * r.y);
        double t, u;

        if (rxs == 0)
        {
                double qpxr = (v1.x * r.y) - (r.x * v1.y);

#if 0
printf("  --> R x S = 0! (q - p) x r = %lf\n", qpxr);
printf("  -->(q - p) . r = %lf, r . r = %lf\n", v1.Dot(r), r.Dot(r));
printf("  -->(p - q) . s = %lf, s . s = %lf\n", v2.Dot(s), s.Dot(s));
printf("  -->(q - p) . s = %lf, (p - q) . r = %lf\n", v1.Dot(s), v2.Dot(r));
#endif

                // Lines are parallel, so no intersection...
                if (qpxr != 0)
                        return 0;

#if 0
//this works IFF the vectors are pointing in the same direction. everything else
//is fucked!
                // If (q - p) . r == r . r, t = 1, u = 0
                if (v1.Dot(r) == r.Dot(r))
                        t = 1.0, u = 0;
                // If (p - q) . s == s . s, t = 0, u = 1
                else if (v2.Dot(s) == s.Dot(s))
                        t = 0, u = 1.0;
                else
                        return 0;
#else
                // Check to see which endpoints are connected... Four possibilities:
                if (l1->p[0] == l2->p[0])
                        t = 0, u = 0;
                else if (l1->p[0] == l2->p[1])
                        t = 0, u = 1.0;
                else if (l1->p[1] == l2->p[0])
                        t = 1.0, u = 0;
                else if (l1->p[1] == l2->p[1])
                        t = 1.0, u = 1.0;
                else
                        return 0;
#endif
        }
        else
        {
                t = ((v1.x * s.y) - (s.x * v1.y)) / rxs;
                u = ((v1.x * r.y) - (r.x * v1.y)) / rxs;
        }
/*
Now there are five cases (NOTE: only valid if vectors face the same way!):

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.

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.

3. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting.

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.

5. Otherwise, the two line segments are not parallel but do not intersect.
*/
        // Return parameter values, if user passed in valid pointers
        if (tp)
                *tp = t;

        if (up)
                *up = u;

        // If the parameters are in range, we have overlap!
        if ((t >= 0) && (t <= 1.0) && (u >= 0) && (u <= 1.0))
                return 1;

        return 0;
}


#if 0
// Finds the intersection between two lines (if any)
int Geometry::Intersects(Line * l1, Dimension * d1, double * tp/*= 0*/, double * up/*= 0*/)
{
        Line l2(d1->position, d1->endpoint);
        return Intersects(l1, &l2, tp, up);
}


// Finds the intersection(s) between a line and a circle (if any)
int Geometry::Intersects(Line * l, Circle * c, double * tp/*= 0*/, double * up/*= 0*/, double * vp/*= 0*/, double * wp/*= 0*/)
{
#if 0
        Vector center = c->position;
        Vector v1 = l->position - center;
        Vector v2 = l->endpoint - center;
        Vector d = v2 - v1;
        double dr = d.Magnitude();
        double determinant = (v1.x * v2.y) - (v1.y * v2.x);

        double discriminant = ((c->radius * c->radius) * (dr * dr)) - (determinant * determinant);

        if (discriminant < 0)
                return false;

        

        return true;
#else
/*
I'm thinking a better approach to this might be as follows:

-- Get the distance of the circle's center from the line segment. If it's
   > the radius, it doesn't intersect.
-- If the parameter is off the line segment, check distance to endpoints. (Not sure
   how to proceed from here, it's different than the following.)
   [Actually, you can use the following for all of it. You only know if you have
    an intersection at the last step, which is OK.]
-- If the radius == distance, we have a tangent line.
-- If radius > distance, use Pythagorus to find the length on either side of the
   normal to the spots where the hypotenuse (== radius' length) contact the line.
-- Use those points to find the parameter on the line segment; if they're not on
   the line segment, no intersection.
*/
        double t = ParameterOfLineAndPoint(l->position, l->endpoint, c->position);
//small problem here: it clamps the result to the line segment. NOT what we want
//here! !!! FIX !!! [DONE]
        Vector p = l->GetPointAtParameter(t);
        double distance = Vector::Magnitude(c->position, p);

        // If the center of the circle is farther from the line than the radius, fail.
        if (distance > c->radius)
                return 0;

        // Now we have to check for intersection points.
        // Tangent case: (needs to return something)
        if ((distance == c->radius) && (t >= 0.0) && (t <= 1.0))
        {
                // Need to set tp & up to something... !!! FIX !!!
                if (tp)
                        *tp = t;

                if (up)
                        *up = Vector(c->position, p).Angle();

                return 1;
        }

        // The line intersects the circle in two points (possibly). Use Pythagorus
        // to find them for testing.
        double offset = sqrt((c->radius * c->radius) - (distance * distance));
//need to convert distance to paramter value... :-/
//t = position on line / length of line segment, so if we divide the offset by length,
//that should give us what we want.
        double length = Vector::Magnitude(l->position, l->endpoint);
        double t1 = t + (offset / length);
        double t2 = t - (offset / length);

//need to find angles for the circle...
        Vector cp1 = l->position + (Vector(l->position, l->endpoint) * (length * t1));
        Vector cp2 = l->position + (Vector(l->position, l->endpoint) * (length * t2));
        double a1 = Vector(c->position, cp1).Angle();
        double a2 = Vector(c->position, cp2).Angle();

//instead of this, return a # which is the # of intersections. [DONE]
        int intersections = 0;

        // Now check for if the parameters are in range
        if ((t1 >= 0) && (t1 <= 1.0))
        {
                intersections++;
        }

        if ((t2 >= 0) && (t2 <= 1.0))
        {
                intersections++;
        }

        return intersections;
#endif
}


// Finds the intersection(s) between a circle and a circle (if any)
// There can be 0, 1, or 2 intersections.
// Returns the angles of the points of intersection in tp thru wp, with the
// angles returned as c1, c2, c1, c2 (if applicable--in the 1 intersection case,
// only the first two angles are returned: c1, c2).
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*/)
{
        // Get the distance between centers. If the distance plus the radius of the
        // smaller circle is less than the radius of the larger circle, there is no
        // intersection. If the distance is greater than the sum of the radii,
        // there is no intersection. If the distance is equal to the sum of the
        // radii, they are tangent and intersect at one point. Otherwise, they
        // intersect at two points.
        Vector centerLine(c1->position, c2->position);
        double d = centerLine.Magnitude();
//printf("Circle #1: pos=<%lf, %lf>, r=%lf\n", c1->position.x, c1->position.y, c1->radius);
//printf("Circle #2: pos=<%lf, %lf>, r=%lf\n", c2->position.x, c2->position.y, c2->radius);
//printf("Distance between #1 & #2: %lf\n", d);

        // Check to see if we actually have an intersection, and return failure if not
        if ((fabs(c1->radius - c2->radius) > d) || ((c1->radius + c2->radius) < d))
                return 0;

        // There are *two* tangent cases!
        if (((c1->radius + c2->radius) == d) || (fabs(c1->radius - c2->radius) == d))
        {
                // Need to return something in tp & up!! !!! FIX !!! [DONE]
                if (tp)
                        *tp = centerLine.Angle();

                if (up)
                        *up = centerLine.Angle() + PI;

                return 1;
        }

        // Find the distance from the center of c1 to the perpendicular chord
        // (which contains the points of intersection)
        double x = ((d * d) - (c2->radius * c2->radius) + (c1->radius * c1->radius))
                / (2.0 * d);
        // Find the the length of the perpendicular chord
// Not needed...!
        double a = sqrt((-d + c2->radius - c1->radius) * (-d - c2->radius + c1->radius) * (-d + c2->radius + c1->radius) * (d + c2->radius + c1->radius)) / d;

        // Now, you can use pythagorus to find the length of the hypotenuse, but we
        // already know that length, it's the radius! :-P
        // What's needed is the angle of the center line and the radial line. Since
        // there's two intersection points, there's also four angles (two for each
        // circle)!
        // We can use the arccos to find the angle using just the radius and the
        // distance to the perpendicular chord...!
        double angleC1 = acos(x / c1->radius);
        double angleC2 = acos((d - x) / c2->radius);

        if (tp)
                *tp = centerLine.Angle() - angleC1;

        if (up)
                *up = (centerLine.Angle() + PI) - angleC2;

        if (vp)
                *vp =  centerLine.Angle() + angleC1;

        if (wp)
                *wp = (centerLine.Angle() + PI) + angleC2;

        if (p1)
                *p1 = c1->position + (centerLine.Unit() * x) + (Vector::Normal(Vector(), centerLine) * (a / 2.0));

        if (p2)
                *p2 = c1->position + (centerLine.Unit() * x) - (Vector::Normal(Vector(), centerLine) * (a / 2.0));

        return 2;
}
#endif

// should we just do common trig solves, like AAS, ASA, SAS, SSA?
// Law of Cosines:
// c^2 = a^2 + b^2 -2ab*cos(C)
// Solving for C:
// cos(C) = (c^2 - a^2 - b^2) / -2ab = (a^2 + b^2 - c^2) / 2ab
// Law of Sines:
// a / sin A = b / sin B = c / sin C

// Solve the angles of the triangle given the sides. Angles returned are
// opposite of the given sides (so a1 consists of sides s2 & s3, and so on).
void Geometry::FindAnglesForSides(double s1, double s2, double s3, double * a1, double * a2, double * a3)
{
        // Use law of cosines to find 1st angle
        double cosine1 = ((s2 * s2) + (s3 * s3) - (s1 * s1)) / (2.0 * s2 * s3);

        // Check for a valid triangle
        if ((cosine1 < -1.0) || (cosine1 > 1.0))
                return;

        double angle1 = acos(cosine1);

        // Use law of sines to find 2nd & 3rd angles
// sin A / a = sin B / b
// sin B = (sin A / a) * b
        double angle2 = asin(s2 * (sin(angle1) / s1));
        double angle3 = asin(s3 * (sin(angle1) / s1));

        if (a1)
                *a1 = angle1;

        if (a2)
                *a2 = angle2;

        if (a3)
                *a3 = angle3;
}