79 lines
2 KiB
C
79 lines
2 KiB
C
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/*
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* This software is licensed under the terms of the MIT-License
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* See COPYING for further information.
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* ---
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* Copyright (c) 2011-2018, Lukas Weber <laochailan@web.de>.
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* Copyright (c) 2012-2018, Andrei Alexeyev <akari@alienslab.net>.
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*/
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#include "taisei.h"
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#include "geometry.h"
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bool point_in_ellipse(complex p, Ellipse e) {
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double Xp = creal(p);
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double Yp = cimag(p);
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double Xe = creal(e.origin);
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double Ye = cimag(e.origin);
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double a = e.angle;
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return (
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pow(cos(a) * (Xp - Xe) + sin(a) * (Yp - Ye), 2) / pow(creal(e.axes)/2, 2) +
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pow(sin(a) * (Xp - Xe) - cos(a) * (Yp - Ye), 2) / pow(cimag(e.axes)/2, 2)
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) <= 1;
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}
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// Is the point of shortest distance between the line through a and b
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// and a point c between a and b and closer than r?
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// if yes, return f so that a+f*(b-a) is that point.
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// otherwise return -1.
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double lineseg_circle_intersect(LineSegment seg, Circle c) {
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complex m, v;
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double projection, lv, lm, distance;
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m = seg.b - seg.a; // vector pointing along the line
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v = seg.a - c.origin; // vector from circle to point A
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lv = cabs(v);
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lm = cabs(m);
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if(lv < c.radius) {
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return 0;
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}
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if(lm == 0) {
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return -1;
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}
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projection = -creal(v*conj(m)) / lm; // project v onto the line
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// now the distance can be calculated by Pythagoras
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distance = sqrt(pow(lv, 2) - pow(projection, 2));
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if(distance <= c.radius) {
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double f = projection/lm;
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if(f >= 0 && f <= 1) { // it’s on the line!
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return f;
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}
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}
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return -1;
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}
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bool lineseg_ellipse_intersect(LineSegment seg, Ellipse e) {
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// Transform the coordinate system so that the ellipse becomes a circle
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// with origin at (0, 0) and diameter equal to its X axis. Then we can
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// calculate the segment-circle intersection.
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double ratio = creal(e.axes) / cimag(e.axes);
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complex rotation = cexp(I * -e.angle);
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seg.a *= rotation;
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seg.b *= rotation;
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seg.a = creal(seg.a) + I * ratio * cimag(seg.a);
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seg.b = creal(seg.b) + I * ratio * cimag(seg.b);
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Circle c = { .radius = creal(e.axes) / 2 };
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return lineseg_circle_intersect(seg, c) >= 0;
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}
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