Thursday, 11 July 2013

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.
polygon21

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
polygon3

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
 

2 comments:

  1. Hi,
    Thanks for the code. But hands down!!! This code shows me the wrong output.

    ReplyDelete
  2. Hi Himangi,

    Is this taking into account the co-ordinate system? The following did not work for me.

    Point polygon[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};
    Point p = {-1, 0};

    I was expecting Point (-1,0) to be reported as lying outside but the program did not.

    ReplyDelete