Most common way to compute line line intersection C++? - c++

Question

Seems there is no way to compute line line intersection using boost::geometry, but I wonder what is the most common way to do it in C++?
I need intersection algorithms for two infinite lines in 2D, if it will be faster it can be two different functions like:
bool line_intersection(line,line);
point line_intersetion(line,line);
P.S. I really try to avoid a wheel invention, so incline to use some library.

Answer 1

The best algorithms that I've found for finding the intersection of lines are in: Real Time Collision Detection by Christer Ericson, a copy of the book can be found here.
Chapter 5 from page 146 onwards describes how to find the closest point of 3D lines which is also the crossing point of 2D lines... with example code in C.
Note: beware of parallel lines, they can cause divide by zero errors.

Answer 2

Express one of the lines in parametric form and the other in implicit form:
X = X0 + t (X1 - X0), Y= Y0 + t (Y1 - Y0)
S(X, Y) = (X - X2) (Y3 - Y2) - (Y - Y2) (X3 - X2) = 0
By linearity of the relations, you have
S(X, Y) = S(X0, Y0) + t (S(X1, Y1) - S(X0, Y0)) = S0 + t (S1 - S0) = 0
From this you get t, and from t the coordinates of the intersection.
It takes a total of 15 adds, 6 multiplies and a single divide.
Degeneracy is indicated by S1 == S0, meaning that the lines are parallel. In practice, the coordinates may not be exact because of truncation errors or others, so that test for equality to 0 can fail. A workaround is to consider the test
|S0 - S1| <= µ |S0|
for small µ.

Answer 3

You can try my code, I'm using boost::geometry and I put a small test case in the main function.
I define a class Line with two points as attributes.
Cross product is very a simple way to know if two lines intersect. In 2D, you can compute the perp dot product (see perp function) that is a projection of cross product on the normal vector of 2D plane. To compute it, you need to get a direction vector of each line (see getVector method).
In 2D, you can get the intersection point of two lines using perp dot product and parametric equation of line. I found an explanation here.
The intersect function returns a boolean to check if two lines intersect. If they intersect, it computes the intersection point by reference.
#include <cmath>
#include <iostream>
#include <boost/geometry/geometry.hpp>
#include <boost/geometry/geometries/point_xy.hpp>
namespace bg = boost::geometry;
// Define two types Point and Vector for a better understanding
// (even if they derive from the same class)
typedef bg::model::d2::point_xy<double> Point;
typedef bg::model::d2::point_xy<double> Vector;
// Class to define a line with two points
class Line
{
public:
Line(const Point& point1,const Point& point2): p1(point1), p2(point2) {}
~Line() {}
// Extract a direction vector
Vector getVector() const
{
Vector v(p2);
bg::subtract_point(v,p1);
return v;
}
Point p1;
Point p2;
};
// Compute the perp dot product of vectors v1 and v2
double perp(const Vector& v1, const Vector& v2)
{
return bg::get<0>(v1)*bg::get<1>(v2)-bg::get<1>(v1)*bg::get<0>(v2);
}
// Check if lines l1 and l2 intersect
// Provide intersection point by reference if true
bool intersect(const Line& l1, const Line& l2, Point& inter)
{
Vector v1 = l1.getVector();
Vector v2 = l2.getVector();
if(std::abs(perp(v1,v2))>0.)
{
// Use parametric equation of lines to find intersection point
Line l(l1.p1,l2.p1);
Vector v = l.getVector();
double t = perp(v,v2)/perp(v1,v2);
inter = v1;
bg::multiply_value(inter,t);
bg::add_point(inter,l.p1);
return true;
}
else return false;
}
int main(int argc, char** argv)
{
Point p1(0.,0.);
Point p2(1.,0.);
Point p3(0.,1.);
Point p4(0.,2.);
Line l1(p1,p2);
Line l2(p3,p4);
Point inter;
if( intersect(l1,l2,inter) )
{
std::cout<<"Coordinates of intersection: "<<inter.x()<<" "<<inter.y()<<std::endl;
}
return 0;
}
EDIT: more detail on cross product and perp dot product + delete tol argument (off topic)

Answer 4

This code should work for you. You may be able to optimize it a bit:
template <class Tpoint>
Tpoint line<Tpoint>::intersect(const line& other) const{
Tpoint x = other.first - first;
Tpoint d1 = second - first;
Tpoint d2 = other.second - other.first;
auto cross = d1.x*d2.y - d1.y*d2.x;
auto t1 = (x.x * d2.y - x.y * d2.x) / static_cast<float>(cross);
return first + d1 * t1;
}

Answer 5

Perhaps a common way is to approximate the infinity? From my library using boost::geometry:
// prev and next are segments and RAY_LENGTH is a very large constant
// create 'lines'
auto prev_extended = extendSeg(prev, -RAY_LENGTH, RAY_LENGTH);
auto next_extended = extendSeg(next, -RAY_LENGTH, RAY_LENGTH);
// intersect!
Points_t isection_howmany;
bg::intersection(prev_extended, next_extended, isection_howmany);
then you could test whether the 'lines' intersect like this:
if (isection_howmany.empty())
cout << "parallel";
else if (isection_howmany.size() == 2)
cout << "collinear";
extendSeg() simply extends the segment in both directions by the given amounts.
Also bear in mind - to support an infinite line arithmetic the point type should also support an infinite value. However here the assumption is that you are looking for a numerical solution!

Answer 6

To solve this problem, I pieced together the following function, but unexpectedly found that it cannot calculate the intersection of line segments, but the intersection of lines.
class Solution {
typedef complex<double> point;
#define x real()
#define y imag()
struct LinePara
{
double k;
double b;
};
LinePara getLinePara(float x1, float y1, float x2, float y2)
{
LinePara ret;
double m = x2 - x1;
if (m == 0)
{
ret.k = 1000.0;
ret.b = y1 - ret.k * x1;
}
else
{
ret.k = (y2 - y1) / (x2 - x1);
ret.b = y1 - ret.k * x1;
}
return ret;
}
struct line {
double a, b, c;
};
const double EPS = 1e-6;
double det(double a, double b, double c, double d) {
return a * d - b * c;
}
line convertLineParaToLine(LinePara s)
{
return line{ s.k,-1,s.b };
}
bool intersect(line m, line n, point& res) {
double zn = det(m.a, m.b, n.a, n.b);
if (abs(zn) < EPS)
return false;
res.real(-det(m.c, m.b, n.c, n.b) / zn);
res.imag(-det(m.a, m.c, n.a, n.c) / zn);
return true;
}
bool parallel(line m, line n) {
return abs(det(m.a, m.b, n.a, n.b)) < EPS;
}
bool equivalent(line m, line n) {
return abs(det(m.a, m.b, n.a, n.b)) < EPS
&& abs(det(m.a, m.c, n.a, n.c)) < EPS
&& abs(det(m.b, m.c, n.b, n.c)) < EPS;
}
vector<double> mian(vector<vector<double>> line1, vector<vector<double>> line2)
{
vector<point> points;
points.push_back(point(line1[0][0], line1[0][1]));
points.push_back(point(line1[1][0], line1[1][1]));
points.push_back(point(line2[0][0], line2[0][1]));
points.push_back(point(line2[1][0], line2[1][1]));
line li1 = convertLineParaToLine(getLinePara(line1[0][0], line1[0][1], line1[1][0], line1[1][1]));
line li2 = convertLineParaToLine(getLinePara(line2[0][0], line2[0][1], line2[1][0], line2[1][1]));
point pos;
if (intersect(li1, li2, pos))
{
return{ pos.x ,pos.y };
}
else
{
if (equivalent(li1, li2)) {
if (points[1].x < points[2].x)
{
return vector<double>{ points[1].x, points[1].y };
}
else if (points[1].x > points[2].x)
{
return vector<double>{ points[2].x, points[2].y };
}
else if (points[1].x == points[2].x)
{
if (points[1].y < points[2].y)
{
return vector<double>{ points[1].x, points[1].y };
}
else if (points[1].y > points[2].y)
{
return vector<double>{ points[2].x, points[2].y };
}
}
else
{
return vector<double>{ points[2].x, points[2].y };
}
}
else
{
return {}/* << "平行！"*/;
}
return {};
}
}
public:
vector<double> intersection(vector<int>& start1, vector<int>& end1, vector<int>& start2, vector<int>& end2) {
vector<vector<double>> line1{ {(double)start1[0],(double)start1[1]},{(double)end1[0],(double)end1[1] } };
vector<vector<double>> line2{ {(double)start2[0],(double)start2[1]},{(double)end2[0],(double)end2[1] } };
return mian(line1, line2);
}
};
From there