How to check if a given point lies inside or outside a polygon?

Given a polygon and a point ‘p’, find if ‘p’ lies inside the polygon or not. The points lying on the border are considered inside.


Following is a simple idea to check whether a point is inside or outside.

1)Draw a horizontal line to the right of each point and extend it to infinity.
2) Count the number of times a line intersects the polygon. We have:
    even number ---> point is outside
    odd number ----> point is inside

If point is same as one of the vertices of polygon, then it intersects with two lines.  We explicitly handle it by comparing the point with n vertices of the polygon.

Following is C++ implementation of the above idea.

// A C++ program to check if a given point lies inside a given polygon #include <iostream> #include <limits.h> 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) { 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 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 } // Returns true if the point p lies inside the polygon[] with n vertices bool isInside(Point polygon[], int n, Point p) { // There must be at least 3 vertices in polygon[] if (n < 3) return false; // Create a point for line segment from p to infinite Point extreme = {INT_MAX, p.y}; // Count intersections of the above line with sides of polygon int count = 0; for (int i = 0; i < n-1; i++) { // If p is same as one of the vertices if (p.x == polygon[i].x && p.y == polygon[i].y) return true; // Otherwise check for intersection if (doIntersect(polygon[i], polygon[i+1], p, extreme)) count++; } // If p is same as last vertex if (p.x == polygon[n-1].x && p.y == polygon[n-1].y) return true; // Last side (from last to first point) is missed in the // above loop, consider it now if (doIntersect(polygon[n-1], polygon[0], p, extreme)) count++; // Return true if count is odd, false otherwise return count&1; // Same as (count%2 == 1) } // Driver program to test above functions int main() { Point polygon[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}}; int n = sizeof(polygon)/sizeof(polygon[0]); Point p = {20, 20}; isInside(polygon, n, p)? cout << "Yes \n": cout << "No \n"; p = {5, 5}; isInside(polygon, n, p)? cout << "Yes \n": cout << "No \n"; return 0; }

Output:

No
Yes

Time Complexity: O(n) where n is the number of vertices in the given polygon.
Source:
http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf