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)

**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.

**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.

// A C++ program to check if two given line segments intersect #include <iostream> using namespace std; struct Point { int x; int y; }; // Given three colinear points p, q, r, the function checks if // point q lies on line segment 'pr' 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; } // To find orientation of ordered triplet (p, q, r). // The function returns following values // 0 --> p, q and r are colinear // 1 --> Clockwise // 2 --> Counterclockwise int orientation(Point p, Point q, Point r) { // See http://www.geeksforgeeks.org/orientation-3-ordered-points/ // for details of below formula. int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y); if (val == 0) return 0; // colinear return (val > 0)? 1: 2; // clock or counterclock wise } // The main function that returns true if line segment 'p1q1' // and 'p2q2' intersect. bool doIntersect(Point p1, Point q1, Point p2, Point q2) { // Find the four orientations needed for general and // special cases int o1 = orientation(p1, q1, p2); int o2 = orientation(p1, q1, q2); int o3 = orientation(p2, q2, p1); int o4 = orientation(p2, q2, q1); // General case if (o1 != o2 && o3 != o4) return true; // Special Cases // p1, q1 and p2 are colinear and p2 lies on segment p1q1 if (o1 == 0 && onSegment(p1, p2, q1)) return true; // p1, q1 and p2 are colinear and q2 lies on segment p1q1 if (o2 == 0 && onSegment(p1, q2, q1)) return true; // p2, q2 and p1 are colinear and p1 lies on segment p2q2 if (o3 == 0 && onSegment(p2, p1, q2)) return true; // p2, q2 and q1 are colinear and q1 lies on segment p2q2 if (o4 == 0 && onSegment(p2, q1, q2)) return true; return false; // Doesn't fall in any of the above cases } // Driver program to test above functions 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; }

Output:

No Yes No

**Sources:**

http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf

Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest

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