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Check if a given point lies inside a Polygon
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.
Updated on: 17-Jun-2020
2K+ Views
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