Convex Hull Using Jarvis's Algorithm in C++

In this tutorial, we will be discussing a program to find the convex hull of a given set of points using Jarvis’s Algorithm.

Convex hull is the smallest polygon convex figure containing all the given points either on the boundary on inside the figure.

In Jarvis’s algorithm, we select the leftmost point and keep wrapping points moving in the clockwise direction.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
//structure of the point
struct Point{
   int x, y;
};
//calculating the position of the points
int cal_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 counterclockwise
}
//printing convex hull
void convexHull(Point points[], int n){
   if (n < 3) return;
   vector<Point> hull;
   //calculating the leftmost point
   int l = 0;
   for (int i = 1; i < n; i++)
   if (points[i].x < points[l].x)
   l = i;
   //moving in the clockwise direction
   int p = l, q;
   do{
      //adding current point to result
      hull.push_back(points[p]);
      q = (p+1)%n;
      for (int i = 0; i < n; i++){
         if (cal_orientation(points[p], points[i], points[q]) == 2)
         q = i;
      }
      p = q;
   } while (p != l); //if didn't reached the first point
   for (int i = 0; i < hull.size(); i++)
   cout << "(" << hull[i].x << ", "
   << hull[i].y << ")\n";
}
int main(){
   Point points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1},
   {3, 0}, {0, 0}, {3, 3}};
   int n = sizeof(points)/sizeof(points[0]);
   convexHull(points, n);
   return 0;
}

Output

(0, 3)
(0, 0)
(3, 0)
(3, 3)