Check if a given point lies inside a Polygon

Data StructureAlgorithmsMisc Algorithms

In this problem, one polygon is given, and a point P is also given. We need to check whether the point is inside the polygon or outside the polygon.

For solving it we will draw a straight line from the point P. It extends to the infinity. The line is horizontal, or it is parallel to the x-axis.

From that line, we will count how many times the line intersects the sides of a polygon. When the point is inside the polygon, it will intersect the sides, an odd number of times, if P is placed on any side of the polygon, then it will cut an even number of times. If none of the condition is true, then it is outside polygon.

Input and Output

Input:
Points of a polygon {(0, 0), (10, 0), (10, 10), (0, 10)}. And point P (5, 3) to check.
Output:
Point is inside.

Algorithm

checkInside(Poly, n, p)

Input: The points of the polygon, the number of points of the polygon, the point p to check.

Output: True when p is inside the polygon, otherwise false.

Begin
   if n<3, then
      return false
   create a line named exLine from point p to infinity, Slope of the line is 0°.
   count :=0 and i := 0
   repeat
      create line called side, from point poly[i] to poly[(i+1) mod n]
      if the side and exLine intersects, then
         if side and exLine are collinear, then
            if point p on the side, then
               return true
            else return false
         count := count + 1
      i := (i + 1) mod n
   until i ≠ 0
   return true if count is odd
End

Example

#include<iostream>
using namespace std;

struct Point {
   int x, y;
};

struct line {
   Point p1, p2;
};

bool onLine(line l1, Point p) {        //check whether p is on the line or not
   if(p.x <= max(l1.p1.x, l1.p2.x) && p.x <= min(l1.p1.x, l1.p2.x) &&
      (p.y <= max(l1.p1.y, l1.p2.y) && p.y <= min(l1.p1.y, l1.p2.y)))
         return true;

   return false;
}

int direction(Point a, Point b, Point c) {
   int val = (b.y-a.y)*(c.x-b.x)-(b.x-a.x)*(c.y-b.y);
   if (val == 0)
      return 0;           //colinear
   else if(val < 0)
      return 2;          //anti-clockwise direction
      return 1;          //clockwise direction
}

bool isIntersect(line l1, line l2) {
   //four direction for two lines and points of other line
   int dir1 = direction(l1.p1, l1.p2, l2.p1);
   int dir2 = direction(l1.p1, l1.p2, l2.p2);
   int dir3 = direction(l2.p1, l2.p2, l1.p1);
   int dir4 = direction(l2.p1, l2.p2, l1.p2);

   if(dir1 != dir2 && dir3 != dir4)
      return true;           //they are intersecting
   if(dir1==0 && onLine(l1, l2.p1))        //when p2 of line2 are on the line1
      return true;
   if(dir2==0 && onLine(l1, l2.p2))         //when p1 of line2 are on the line1
      return true;
   if(dir3==0 && onLine(l2, l1.p1))       //when p2 of line1 are on the line2
      return true;
   if(dir4==0 && onLine(l2, l1.p2)) //when p1 of line1 are on the line2
      return true;
   return false;
}

bool checkInside(Point poly[], int n, Point p) {
   if(n < 3)
      return false;                  //when polygon has less than 3 edge, it is not polygon
   line exline = {p, {9999, p.y}};   //create a point at infinity, y is same as point p
   int count = 0;
   int i = 0;
   do {
      line side = {poly[i], poly[(i+1)%n]};     //forming a line from two consecutive points of poly
      if(isIntersect(side, exline)) {          //if side is intersects exline
         if(direction(side.p1, p, side.p2) == 0)
            return onLine(side, p);
         count++;
      }
      i = (i+1)%n;
   } while(i != 0);
      return count&1;             //when count is odd
}

int main() {
   // line polygon = {{{0,0},{10,0}},{{10,0},{10,10}},{{10,10},{0,10}},{{0,10},{0,0}}};
   Point polygon[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}};
   Point p = {5, 3};
   int n = 4;

   if(checkInside(polygon, n, p))
      cout << "Point is inside.";
   else
      cout << "Point is outside.";
}

Output

Point is inside.
raja
Published on 12-Jul-2018 07:33:05
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