# Convex Hull Jarvis’s Algorithm or Wrapping 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{
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)