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

28 Jun, 2022
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.
We strongly recommend to see the following post first. How to check if two given line segments intersect?
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

1) Count the number of times the line intersects with polygon edges.

2) A point is inside the polygon if either count of intersections is odd or
   point lies on an edge of polygon.  If none of the conditions is true, then 
   point lies outside.
How to handle point ‘g’ in the above figure? Note that we should return true if the point lies on the line or same as one of the vertices of the given polygon. To handle this, after checking if the line from ‘p’ to extreme intersects, we check whether ‘p’ is collinear with vertices of current line of polygon. If it is collinear, then we check if the point ‘p’ lies on current side of polygon, if it lies, we return true, else false.
Following is the implementation of the above idea. 
C++
Java
Python3
C#
Javascript
// A C++ program to check if a given point lies inside a given polygon// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/// for explanation of functions onSegment(), orientation() and doIntersect()#include <iostream>using namespace std; // Define Infinite (Using INT_MAX caused overflow problems)#define INF 10000 struct Point{    int x;    int y;}; // Given three collinear 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 collinear// 1 --> Clockwise// 2 --> Counterclockwiseint 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; // collinear    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 collinear and p2 lies on segment p1q1    if (o1 == 0 && onSegment(p1, p2, q1)) return true;     // p1, q1 and p2 are collinear and q2 lies on segment p1q1    if (o2 == 0 && onSegment(p1, q2, q1)) return true;     // p2, q2 and p1 are collinear and p1 lies on segment p2q2    if (o3 == 0 && onSegment(p2, p1, q2)) return true;     // p2, q2 and q1 are collinear 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 verticesbool 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 = {INF, p.y};     // Count intersections of the above line with sides of polygon    int count = 0, i = 0;    do    {        int next = (i+1)%n;         // Check if the line segment from 'p' to 'extreme' intersects        // with the line segment from 'polygon[i]' to 'polygon[next]'        if (doIntersect(polygon[i], polygon[next], p, extreme))        {            // If the point 'p' is collinear with line segment 'i-next',            // then check if it lies on segment. If it lies, return true,            // otherwise false            if (orientation(polygon[i], p, polygon[next]) == 0)            return onSegment(polygon[i], p, polygon[next]);             count++;        }        i = next;    } while (i != 0);     // Return true if count is odd, false otherwise    return count&1; // Same as (count%2 == 1)} // Driver program to test above functionsint main(){    Point polygon1[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};    int n = sizeof(polygon1)/sizeof(polygon1[0]);    Point p = {20, 20};    isInside(polygon1, n, p)? cout << "Yes \n": cout << "No \n";     p = {5, 5};    isInside(polygon1, n, p)? cout << "Yes \n": cout << "No \n";     Point polygon2[] = {{0, 0}, {5, 5}, {5, 0}};    p = {3, 3};    n = sizeof(polygon2)/sizeof(polygon2[0]);    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n";     p = {5, 1};    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n";     p = {8, 1};    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n";     Point polygon3[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};    p = {-1,10};    n = sizeof(polygon3)/sizeof(polygon3[0]);    isInside(polygon3, n, p)? cout << "Yes \n": cout << "No \n";     return 0;}
// A Java program to check if a given point// lies inside a given polygon// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/// for explanation of functions onSegment(),// orientation() and doIntersect()class GFG{     // Define Infinite (Using INT_MAX    // caused overflow problems)    static int INF = 10000;     static class Point    {        int x;        int y;         public Point(int x, int y)        {            this.x = x;            this.y = y;        }    };     // Given three collinear points p, q, r,    // the function checks if point q lies    // on line segment 'pr'    static boolean onSegment(Point p, Point q, Point r)    {        if (q.x <= Math.max(p.x, r.x) &&            q.x >= Math.min(p.x, r.x) &&            q.y <= Math.max(p.y, r.y) &&            q.y >= Math.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 collinear    // 1 --> Clockwise    // 2 --> Counterclockwise    static 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; // collinear        }        return (val > 0) ? 1 : 2; // clock or counterclock wise    }     // The function that returns true if    // line segment 'p1q1' and 'p2q2' intersect.    static boolean 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 collinear and        // p2 lies on segment p1q1        if (o1 == 0 && onSegment(p1, p2, q1))        {            return true;        }         // p1, q1 and p2 are collinear and        // q2 lies on segment p1q1        if (o2 == 0 && onSegment(p1, q2, q1))        {            return true;        }         // p2, q2 and p1 are collinear and        // p1 lies on segment p2q2        if (o3 == 0 && onSegment(p2, p1, q2))        {            return true;        }         // p2, q2 and q1 are collinear and        // q1 lies on segment p2q2        if (o4 == 0 && onSegment(p2, q1, q2))        {            return true;        }         // Doesn't fall in any of the above cases        return false;    }     // Returns true if the point p lies    // inside the polygon[] with n vertices    static boolean 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 = new Point(INF, p.y);         // Count intersections of the above line        // with sides of polygon        int count = 0, i = 0;        do        {            int next = (i + 1) % n;             // Check if the line segment from 'p' to            // 'extreme' intersects with the line            // segment from 'polygon[i]' to 'polygon[next]'            if (doIntersect(polygon[i], polygon[next], p, extreme))            {                // If the point 'p' is collinear with line                // segment 'i-next', then check if it lies                // on segment. If it lies, return true, otherwise false                if (orientation(polygon[i], p, polygon[next]) == 0)                {                    return onSegment(polygon[i], p,                                    polygon[next]);                }                 count++;            }            i = next;        } while (i != 0);         // Return true if count is odd, false otherwise        return (count % 2 == 1); // Same as (count%2 == 1)    }     // Driver Code    public static void main(String[] args)    {        Point polygon1[] = {new Point(0, 0),                            new Point(10, 0),                            new Point(10, 10),                            new Point(0, 10)};        int n = polygon1.length;        Point p = new Point(20, 20);        if (isInside(polygon1, n, p))        {            System.out.println("Yes");        }        else        {            System.out.println("No");        }        p = new Point(5, 5);        if (isInside(polygon1, n, p))        {            System.out.println("Yes");        }        else        {            System.out.println("No");        }        Point polygon2[] = {new Point(0, 0),            new Point(5, 5), new Point(5, 0)};        p = new Point(3, 3);        n = polygon2.length;        if (isInside(polygon2, n, p))        {            System.out.println("Yes");        }        else        {            System.out.println("No");        }        p = new Point(5, 1);        if (isInside(polygon2, n, p))        {            System.out.println("Yes");        }        else        {            System.out.println("No");        }        p = new Point(8, 1);        if (isInside(polygon2, n, p))        {            System.out.println("Yes");        }        else        {            System.out.println("No");        }        Point polygon3[] = {new Point(0, 0),                            new Point(10, 0),                            new Point(10, 10),                            new Point(0, 10)};        p = new Point(-1, 10);        n = polygon3.length;        if (isInside(polygon3, n, p))        {            System.out.println("Yes");        }        else        {            System.out.println("No");        }    }} // This code is contributed by 29AjayKumar
# A Python3 program to check if a given point # lies inside a given polygon# Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/# for explanation of functions onSegment(),# orientation() and doIntersect()  # Define Infinite (Using INT_MAX # caused overflow problems)INT_MAX = 10000 # Given three collinear points p, q, r, # the function checks if point q lies# on line segment 'pr'def onSegment(p:tuple, q:tuple, r:tuple) -> bool:         if ((q[0] <= max(p[0], r[0])) &        (q[0] >= min(p[0], r[0])) &        (q[1] <= max(p[1], r[1])) &        (q[1] >= min(p[1], r[1]))):        return True             return False # To find orientation of ordered triplet (p, q, r).# The function returns following values# 0 --> p, q and r are collinear# 1 --> Clockwise# 2 --> Counterclockwisedef orientation(p:tuple, q:tuple, r:tuple) -> int:         val = (((q[1] - p[1]) *            (r[0] - q[0])) -           ((q[0] - p[0]) *            (r[1] - q[1])))                if val == 0:        return 0    if val > 0:        return 1 # Collinear    else:        return 2 # Clock or counterclock def doIntersect(p1, q1, p2, q2):         # Find the four orientations needed for     # general and special cases    o1 = orientation(p1, q1, p2)    o2 = orientation(p1, q1, q2)    o3 = orientation(p2, q2, p1)    o4 = orientation(p2, q2, q1)     # General case    if (o1 != o2) and (o3 != o4):        return True         # Special Cases    # p1, q1 and p2 are collinear and    # p2 lies on segment p1q1    if (o1 == 0) and (onSegment(p1, p2, q1)):        return True     # p1, q1 and p2 are collinear and    # q2 lies on segment p1q1    if (o2 == 0) and (onSegment(p1, q2, q1)):        return True     # p2, q2 and p1 are collinear and    # p1 lies on segment p2q2    if (o3 == 0) and (onSegment(p2, p1, q2)):        return True     # p2, q2 and q1 are collinear and    # q1 lies on segment p2q2    if (o4 == 0) and (onSegment(p2, q1, q2)):        return True     return False # Returns true if the point p lies # inside the polygon[] with n verticesdef is_inside_polygon(points:list, p:tuple) -> bool:         n = len(points)         # There must be at least 3 vertices    # in polygon    if n < 3:        return False             # Create a point for line segment    # from p to infinite    extreme = (INT_MAX, p[1])    count = i = 0         while True:        next = (i + 1) % n                 # Check if the line segment from 'p' to         # 'extreme' intersects with the line         # segment from 'polygon[i]' to 'polygon[next]'        if (doIntersect(points[i],                        points[next],                        p, extreme)):                                         # If the point 'p' is collinear with line             # segment 'i-next', then check if it lies             # on segment. If it lies, return true, otherwise false            if orientation(points[i], p,                           points[next]) == 0:                return onSegment(points[i], p,                                 points[next])                                              count += 1                     i = next                 if (i == 0):            break             # Return true if count is odd, false otherwise    return (count % 2 == 1) # Driver codeif __name__ == '__main__':         polygon1 = [ (0, 0), (10, 0), (10, 10), (0, 10) ]         p = (20, 20)    if (is_inside_polygon(points = polygon1, p = p)):      print ('Yes')    else:      print ('No')           p = (5, 5)    if (is_inside_polygon(points = polygon1, p = p)):      print ('Yes')    else:      print ('No')     polygon2 = [ (0, 0), (5, 0), (5, 5), (3, 3) ]         p = (3, 3)    if (is_inside_polygon(points = polygon2, p = p)):      print ('Yes')    else:      print ('No')           p = (5, 1)    if (is_inside_polygon(points = polygon2, p = p)):      print ('Yes')    else:      print ('No')           p = (8, 1)    if (is_inside_polygon(points = polygon2, p = p)):      print ('Yes')    else:      print ('No')         polygon3 = [ (0, 0), (10, 0), (10, 10), (0, 10) ]         p = (-1, 10)    if (is_inside_polygon(points = polygon3, p = p)):      print ('Yes')    else:      print ('No')       # This code is contributed by Vikas Chitturi
// A C# program to check if a given point// lies inside a given polygon// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/// for explanation of functions onSegment(),// orientation() and doIntersect()using System; class GFG{     // Define Infinite (Using INT_MAX    // caused overflow problems)    static int INF = 10000;     class Point    {        public int x;        public int y;         public Point(int x, int y)        {            this.x = x;            this.y = y;        }    };     // Given three collinear points p, q, r,    // the function checks if point q lies    // on line segment 'pr'    static bool onSegment(Point p, Point q, Point r)    {        if (q.x <= Math.Max(p.x, r.x) &&            q.x >= Math.Min(p.x, r.x) &&            q.y <= Math.Max(p.y, r.y) &&            q.y >= Math.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 collinear    // 1 --> Clockwise    // 2 --> Counterclockwise    static 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; // collinear        }        return (val > 0) ? 1 : 2; // clock or counterclock wise    }     // The function that returns true if    // line segment 'p1q1' and 'p2q2' intersect.    static 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 collinear and        // p2 lies on segment p1q1        if (o1 == 0 && onSegment(p1, p2, q1))        {            return true;        }         // p1, q1 and p2 are collinear and        // q2 lies on segment p1q1        if (o2 == 0 && onSegment(p1, q2, q1))        {            return true;        }         // p2, q2 and p1 are collinear and        // p1 lies on segment p2q2        if (o3 == 0 && onSegment(p2, p1, q2))        {            return true;        }         // p2, q2 and q1 are collinear and        // q1 lies on segment p2q2        if (o4 == 0 && onSegment(p2, q1, q2))        {            return true;        }         // Doesn't fall in any of the above cases        return false;    }     // Returns true if the point p lies    // inside the polygon[] with n vertices    static 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 = new Point(INF, p.y);         // Count intersections of the above line        // with sides of polygon        int count = 0, i = 0;        do        {            int next = (i + 1) % n;             // Check if the line segment from 'p' to            // 'extreme' intersects with the line            // segment from 'polygon[i]' to 'polygon[next]'            if (doIntersect(polygon[i],                            polygon[next], p, extreme))            {                // If the point 'p' is collinear with line                // segment 'i-next', then check if it lies                // on segment. If it lies, return true, otherwise false                if (orientation(polygon[i], p, polygon[next]) == 0)                {                    return onSegment(polygon[i], p,                                    polygon[next]);                }                count++;            }            i = next;        } while (i != 0);         // Return true if count is odd, false otherwise        return (count % 2 == 1); // Same as (count%2 == 1)    }     // Driver Code    public static void Main(String[] args)    {        Point []polygon1 = {new Point(0, 0),                            new Point(10, 0),                            new Point(10, 10),                            new Point(0, 10)};        int n = polygon1.Length;        Point p = new Point(20, 20);        if (isInside(polygon1, n, p))        {            Console.WriteLine("Yes");        }        else        {            Console.WriteLine("No");        }        p = new Point(5, 5);        if (isInside(polygon1, n, p))        {            Console.WriteLine("Yes");        }        else        {            Console.WriteLine("No");        }        Point []polygon2 = {new Point(0, 0),                            new Point(5, 5),                            new Point(5, 0)};        p = new Point(3, 3);        n = polygon2.Length;        if (isInside(polygon2, n, p))        {            Console.WriteLine("Yes");        }        else        {            Console.WriteLine("No");        }        p = new Point(5, 1);        if (isInside(polygon2, n, p))        {            Console.WriteLine("Yes");        }        else        {            Console.WriteLine("No");        }        p = new Point(8, 1);        if (isInside(polygon2, n, p))        {            Console.WriteLine("Yes");        }        else        {            Console.WriteLine("No");        }        Point []polygon3 = {new Point(0, 0),                            new Point(10, 0),                            new Point(10, 10),                            new Point(0, 10)};        p = new Point(-1, 10);        n = polygon3.Length;        if (isInside(polygon3, n, p))        {            Console.WriteLine("Yes");        }        else        {            Console.WriteLine("No");        }    }} // This code is contributed by 29AjayKumar
<script>// A Javascript program to check if a given point// lies inside a given polygon// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/// for explanation of functions onSegment(),// orientation() and doIntersect() // Define Infinite (Using INT_MAX    // caused overflow problems)let INF = 10000; class Point{    constructor(x,y)    {        this.x = x;        this.y = y;    }} // Given three collinear points p, q, r,    // the function checks if point q lies    // on line segment 'pr'function onSegment(p,q,r){     if (q.x <= Math.max(p.x, r.x) &&            q.x >= Math.min(p.x, r.x) &&            q.y <= Math.max(p.y, r.y) &&            q.y >= Math.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 collinear    // 1 --> Clockwise    // 2 --> Counterclockwisefunction orientation(p,q,r){    let val = (q.y - p.y) * (r.x - q.x)                - (q.x - p.x) * (r.y - q.y);          if (val == 0)        {            return 0; // collinear        }        return (val > 0) ? 1 : 2; // clock or counterclock wise} // The function that returns true if    // line segment 'p1q1' and 'p2q2' intersect.function  doIntersect(p1,q1,p2,q2){    // Find the four orientations needed for        // general and special cases        let o1 = orientation(p1, q1, p2);        let o2 = orientation(p1, q1, q2);        let o3 = orientation(p2, q2, p1);        let o4 = orientation(p2, q2, q1);          // General case        if (o1 != o2 && o3 != o4)        {            return true;        }          // Special Cases        // p1, q1 and p2 are collinear and        // p2 lies on segment p1q1        if (o1 == 0 && onSegment(p1, p2, q1))        {            return true;        }          // p1, q1 and p2 are collinear and        // q2 lies on segment p1q1        if (o2 == 0 && onSegment(p1, q2, q1))        {            return true;        }          // p2, q2 and p1 are collinear and        // p1 lies on segment p2q2        if (o3 == 0 && onSegment(p2, p1, q2))        {            return true;        }          // p2, q2 and q1 are collinear and        // q1 lies on segment p2q2        if (o4 == 0 && onSegment(p2, q1, q2))        {            return true;        }          // Doesn't fall in any of the above cases        return false;} // Returns true if the point p lies    // inside the polygon[] with n verticesfunction  isInside(polygon,n,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        let extreme = new Point(INF, p.y);          // Count intersections of the above line        // with sides of polygon        let count = 0, i = 0;        do        {            let next = (i + 1) % n;              // Check if the line segment from 'p' to            // 'extreme' intersects with the line            // segment from 'polygon[i]' to 'polygon[next]'            if (doIntersect(polygon[i], polygon[next], p, extreme))            {                // If the point 'p' is collinear with line                // segment 'i-next', then check if it lies                // on segment. If it lies, return true, otherwise false                if (orientation(polygon[i], p, polygon[next]) == 0)                {                    return onSegment(polygon[i], p,                                    polygon[next]);                }                  count++;            }            i = next;        } while (i != 0);          // Return true if count is odd, false otherwise        return (count % 2 == 1); // Same as (count%2 == 1)} // Driver Codepolygon1 = [new Point(0, 0),                            new Point(10, 0),                            new Point(10, 10),                            new Point(0, 10)];                             let n = polygon1.length;        let p = new Point(20, 20);        if (isInside(polygon1, n, p))        {            document.write("Yes<br>");        }        else        {            document.write("No<br>");        }        p = new Point(5, 5);        if (isInside(polygon1, n, p))        {            document.write("Yes<br>");        }        else        {            document.write("No<br>");        }        let polygon2 = [new Point(0, 0),            new Point(5, 5), new Point(5, 0)];        p = new Point(3, 3);        n = polygon2.length;        if (isInside(polygon2, n, p))        {            document.write("Yes<br>");        }        else        {            document.write("No<br>");        }        p = new Point(5, 1);        if (isInside(polygon2, n, p))        {            document.write("Yes<br>");        }        else        {            document.write("No<br>");        }        p = new Point(8, 1);        if (isInside(polygon2, n, p))        {            document.write("Yes<br>");        }        else        {            document.write("No<br>");        }        let polygon3 = [new Point(0, 0),                            new Point(10, 0),                            new Point(10, 10),                            new Point(0, 10)];        p = new Point(-1, 10);        n = polygon3.length;        if (isInside(polygon3, n, p))        {            document.write("Yes<br>");        }        else        {            document.write("No<br>");        }                            // This code is contributed by rag2127</script>
Output: 
No
Yes
Yes
Yes
No
No
Time Complexity: O(n) where n is the number of vertices in the given polygon.
Space Complexity: O(1)
Please suggest if someone has a better solution which is more efficient in terms of space and time.This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Source: http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
ImanSahebi
29AjayKumar
Vikas Chitturi
rag2127
sweetyty
sagartomar9927
_shinchancode
Geometric
Mathematical
Mathematical
Geometric
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