Given two line segments (p1, q1) and (p2, q2), find if the given line segments intersect with each other.
Before we discuss solution, let us define notion of orientation. Orientation of an ordered triplet of points in the plane can be
–counterclockwise
–clockwise
–colinear
The following diagram shows different possible orientations of (a, b, c)

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How is Orientation useful here?
Two segments (p1,q1) and (p2,q2) intersect if and only if one of the following two conditions is verified
1. General Case:
– (p1, q1, p2) and (p1, q1, q2) have different orientations and
– (p2, q2, p1) and (p2, q2, q1) have different orientations.


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2. Special Case
– (p1, q1, p2), (p1, q1, q2), (p2, q2, p1), and (p2, q2, q1) are all collinear and
– the x-projections of (p1, q1) and (p2, q2) intersect
– the y-projections of (p1, q1) and (p2, q2) intersect

Following is C++ implementation based on above idea.
#include <iostream>
using namespace std;
struct Point
{
int x;
int y;
};
bool onSegment(Point p, Point q, Point r)
{
if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
return true;
return false;
}
int orientation(Point p, Point q, Point r)
{
int val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if (val == 0) return 0;
return (val > 0)? 1: 2;
}
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
int o1 = orientation(p1, q1, p2);
int o2 = orientation(p1, q1, q2);
int o3 = orientation(p2, q2, p1);
int o4 = orientation(p2, q2, q1);
if (o1 != o2 && o3 != o4)
return true;
if (o1 == 0 && onSegment(p1, p2, q1)) return true;
if (o2 == 0 && onSegment(p1, q2, q1)) return true;
if (o3 == 0 && onSegment(p2, p1, q2)) return true;
if (o4 == 0 && onSegment(p2, q1, q2)) return true;
return false;
}
int main()
{
struct Point p1 = {1, 1}, q1 = {10, 1};
struct Point p2 = {1, 2}, q2 = {10, 2};
doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";
p1 = {10, 0}, q1 = {0, 10};
p2 = {0, 0}, q2 = {10, 10};
doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";
p1 = {-5, -5}, q1 = {0, 0};
p2 = {1, 1}, q2 = {10, 10};
doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";
return 0;
}
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Output:
No
Yes
No