Convex Hull using Jarvis’ Algorithm or Wrapping

Given a set of points in the plane. the convex hull of the set is the smallest convex polygon that contains all the points of it.

We strongly recommend to see the following post first. 
How to check if two given line segments intersect?
The idea of Jarvis’s Algorithm is simple, we start from the leftmost point (or point with minimum x coordinate value) and we keep wrapping points in counterclockwise direction. 

The big question is, given a point p as current point, how to find the next point in output? 

The idea is to use orientation() here. Next point is selected as the point that beats all other points at counterclockwise orientation, i.e., next point is q if for any other point r, we have “orientation(p, q, r) = counterclockwise”. 

 Algorithm:
Step 1) Initialize p as leftmost point. 
Step 2) Do following while we don’t come back to the first (or leftmost) point. 
            2.1) The next point q is the point, such that the triplet (p, q, r) is counter clockwise for any other point r. 

                    To find this, we simply initialize q as next point, then we traverse through all points. 

                    For any point i, if i is more counter clockwise, i.e., orientation(p, i, q) is counter clockwise, then we update q as i. 

                    Our final value of q is going to be the most counter clockwise point. 
           2.2) next[p] = q (Store q as next of p in the output convex hull). 
           2.3) p = q (Set p as q for next iteration).

Below is the implementation of above algorithm. 

C++

// A C++ program to find convex hull of a set of points. Refer
// https://www.geeksforgeeks.org/orientation-3-ordered-points/
// for explanation of orientation()
#include <bits/stdc++.h>
using namespace std;

struct Point
{
    int x, y;
};

// 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
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
}

// Prints convex hull of a set of n points.
void convexHull(Point points[], int n)
{
    // There must be at least 3 points
    if (n < 3) return;

    // Initialize Result
    vector<Point> hull;

    // Find the leftmost point
    int l = 0;
    for (int i = 1; i < n; i++)
        if (points[i].x < points[l].x)
            l = i;

    // Start from leftmost point, keep moving counterclockwise
    // until reach the start point again.  This loop runs O(h)
    // times where h is number of points in result or output.
    int p = l, q;
    do
    {
        // Add current point to result
        hull.push_back(points[p]);

        // Search for a point 'q' such that orientation(p, q,
        // x) is counterclockwise for all points 'x'. The idea
        // is to keep track of last visited most counterclock-
        // wise point in q. If any point 'i' is more counterclock-
        // wise than q, then update q.
        q = (p+1)%n;
        for (int i = 0; i < n; i++)
        {
           // If i is more counterclockwise than current q, then
           // update q
           if (orientation(points[p], points[i], points[q]) == 2)
               q = i;
        }

        // Now q is the most counterclockwise with respect to p
        // Set p as q for next iteration, so that q is added to
        // result 'hull'
        p = q;

    } while (p != l);  // While we don't come to first point

    // Print Result
    for (int i = 0; i < hull.size(); i++)
        cout << "(" << hull[i].x << ", "
              << hull[i].y << ")\n";
}

// Driver program to test above functions
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;
}

Java

// Java program to find convex hull of a set of points. Refer
// https://www.geeksforgeeks.org/orientation-3-ordered-points/
// for explanation of orientation()
import java.util.*;

class Point
{
    int x, y;
    Point(int x, int y){
        this.x=x;
        this.y=y;
    }
}

class GFG {

    // 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
    public 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
    }

    // Prints convex hull of a set of n points.
    public static void convexHull(Point points[], int n)
    {
        // There must be at least 3 points
        if (n < 3) return;

        // Initialize Result
        Vector<Point> hull = new Vector<Point>();

        // Find the leftmost point
        int l = 0;
        for (int i = 1; i < n; i++)
            if (points[i].x < points[l].x)
                l = i;

        // Start from leftmost point, keep moving
        // counterclockwise until reach the start point
        // again. This loop runs O(h) times where h is
        // number of points in result or output.
        int p = l, q;
        do
        {
            // Add current point to result
            hull.add(points[p]);

            // Search for a point 'q' such that
            // orientation(p, q, x) is counterclockwise
            // for all points 'x'. The idea is to keep
            // track of last visited most counterclock-
            // wise point in q. If any point 'i' is more
            // counterclock-wise than q, then update q.
            q = (p + 1) % n;

            for (int i = 0; i < n; i++)
            {
               // If i is more counterclockwise than
               // current q, then update q
               if (orientation(points[p], points[i], points[q])
                                                   == 2)
                   q = i;
            }

            // Now q is the most counterclockwise with
            // respect to p. Set p as q for next iteration,
            // so that q is added to result 'hull'
            p = q;

        } while (p != l);  // While we don't come to first
                           // point

        // Print Result
        for (Point temp : hull)
            System.out.println("(" + temp.x + ", " +
                                temp.y + ")");
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {

        Point points[] = new Point[7];
        points[0]=new Point(0, 3);
        points[1]=new Point(2, 3);
        points[2]=new Point(1, 1);
        points[3]=new Point(2, 1);
        points[4]=new Point(3, 0);
        points[5]=new Point(0, 0);
        points[6]=new Point(3, 3);

        int n = points.length;
        convexHull(points, n);

    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find convex hull of a set of points. Refer
# https://www.geeksforgeeks.org/orientation-3-ordered-points/
# for explanation of orientation()

# point class with x, y as point
class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y

def Left_index(points):

    '''
    Finding the left most point
    '''
    minn = 0
    for i in range(1,len(points)):
        if points[i].x < points[minn].x:
            minn = i
        elif points[i].x == points[minn].x:
            if points[i].y > points[minn].y:
                minn = i
    return minn

def orientation(p, q, r):
    '''
    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
    '''
    val = (q.y - p.y) * (r.x - q.x) - \
          (q.x - p.x) * (r.y - q.y)

    if val == 0:
        return 0
    elif val > 0:
        return 1
    else:
        return 2

def convexHull(points, n):

    # There must be at least 3 points
    if n < 3:
        return

    # Find the leftmost point
    l = Left_index(points)

    hull = []

    '''
    Start from leftmost point, keep moving counterclockwise
    until reach the start point again. This loop runs O(h)
    times where h is number of points in result or output.
    '''
    p = l
    q = 0
    while(True):

        # Add current point to result
        hull.append(p)

        '''
        Search for a point 'q' such that orientation(p, q,
        x) is counterclockwise for all points 'x'. The idea
        is to keep track of last visited most counterclock-
        wise point in q. If any point 'i' is more counterclock-
        wise than q, then update q.
        '''
        q = (p + 1) % n

        for i in range(n):

            # If i is more counterclockwise
            # than current q, then update q
            if(orientation(points[p],
                           points[i], points[q]) == 2):
                q = i

        '''
        Now q is the most counterclockwise with respect to p
        Set p as q for next iteration, so that q is added to
        result 'hull'
        '''
        p = q

        # While we don't come to first point
        if(p == l):
            break

    # Print Result
    for each in hull:
        print(points[each].x, points[each].y)

# Driver Code
points = []
points.append(Point(0, 3))
points.append(Point(2, 2))
points.append(Point(1, 1))
points.append(Point(2, 1))
points.append(Point(3, 0))
points.append(Point(0, 0))
points.append(Point(3, 3))

convexHull(points, len(points))

# This code is contributed by
# Akarsh Somani, IIIT Kalyani

C#

// C# program to find convex hull of a set of points. Refer
// https://www.geeksforgeeks.org/orientation-3-ordered-points/
// for explanation of orientation()
using System;
using System.Collections.Generic;

public class Point
{
    public int x, y;
    public Point(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}

public class GFG
{

    // 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
    public 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
    }

    // Prints convex hull of a set of n points.
    public static void convexHull(Point []points, int n)
    {
        // There must be at least 3 points
        if (n < 3) return;

        // Initialize Result
        List<Point> hull = new List<Point>();

        // Find the leftmost point
        int l = 0;
        for (int i = 1; i < n; i++)
            if (points[i].x < points[l].x)
                l = i;

        // Start from leftmost point, keep moving
        // counterclockwise until reach the start point
        // again. This loop runs O(h) times where h is
        // number of points in result or output.
        int p = l, q;
        do
        {
            // Add current point to result
            hull.Add(points[p]);

            // Search for a point 'q' such that
            // orientation(p, q, x) is counterclockwise
            // for all points 'x'. The idea is to keep
            // track of last visited most counterclock-
            // wise point in q. If any point 'i' is more
            // counterclock-wise than q, then update q.
            q = (p + 1) % n;

            for (int i = 0; i < n; i++)
            {
            // If i is more counterclockwise than
            // current q, then update q
            if (orientation(points[p], points[i], points[q])
                                                == 2)
                q = i;
            }

            // Now q is the most counterclockwise with
            // respect to p. Set p as q for next iteration,
            // so that q is added to result 'hull'
            p = q;

        } while (p != l); // While we don't come to first
                        // point

        // Print Result
        foreach (Point temp in hull)
            Console.WriteLine("(" + temp.x + ", " +
                                temp.y + ")");
    }

    /* Driver code */
    public static void Main(String[] args)
    {

        Point []points = new Point[7];
        points[0]=new Point(0, 3);
        points[1]=new Point(2, 3);
        points[2]=new Point(1, 1);
        points[3]=new Point(2, 1);
        points[4]=new Point(3, 0);
        points[5]=new Point(0, 0);
        points[6]=new Point(3, 3);

        int n = points.Length;
        convexHull(points, n);

    }
}

// This code is contributed by Princi Singh

Javascript

<script>
// Javascript program to find convex hull of a set of points. Refer
// https://www.geeksforgeeks.org/orientation-3-ordered-points/
// for explanation of orientation()

class Point
{
    constructor(x, y)
    {
        this.x = x;
        this.y = y;
    }
}

// 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
function 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
}

// Prints convex hull of a set of n points.
function convexHull(points, n)
{
    // There must be at least 3 points
        if (n < 3) return;

        // Initialize Result
        let hull = [];

        // Find the leftmost point
        let l = 0;
        for (let i = 1; i < n; i++)
            if (points[i].x < points[l].x)
                l = i;

        // Start from leftmost point, keep moving
        // counterclockwise until reach the start point
        // again. This loop runs O(h) times where h is
        // number of points in result or output.
        let p = l, q;
        do
        {

            // Add current point to result
            hull.push(points[p]);

            // Search for a point 'q' such that
            // orientation(p, q, x) is counterclockwise
            // for all points 'x'. The idea is to keep
            // track of last visited most counterclock-
            // wise point in q. If any point 'i' is more
            // counterclock-wise than q, then update q.
            q = (p + 1) % n;

            for (let i = 0; i < n; i++)
            {
               // If i is more counterclockwise than
               // current q, then update q
               if (orientation(points[p], points[i], points[q])
                                                   == 2)
                   q = i;
            }

            // Now q is the most counterclockwise with
            // respect to p. Set p as q for next iteration,
            // so that q is added to result 'hull'
            p = q;

        } while (p != l);  // While we don't come to first
                           // point

        // Print Result
        for (let temp of hull.values())
            document.write("(" + temp.x + ", " +
                                temp.y + ")<br>");
}

/* Driver program to test above function */
let points = new Array(7);
points[0] = new Point(0, 3);
points[1] = new Point(2, 3);
points[2] = new Point(1, 1);
points[3] = new Point(2, 1);
points[4] = new Point(3, 0);
points[5] = new Point(0, 0);
points[6] = new Point(3, 3);

let n = points.length;
convexHull(points, n);

// This code is contributed by avanitrachhadiya2155
</script>

Output: The output is points of the convex hull. 

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

Time Complexity: O(m * n), where n is number of input points and m is number of output or hull points (m <= n).  For every point on the hull we examine all the other points to determine the next point.

Worst case, Time complexity: O(n²).  The worst case occurs when all the points are on the hull (m = n).

Auxiliary Space: O(n), since n extra space has been taken.
Set 2- Convex Hull (Graham  Scan) 
Note : The above code may produce different results for different order of inputs, when there are collinear points in the convex hull. For example, it produces output as (0, 3) (0, 0) (3, 0) (3, 3) for input (0, 3), (0, 0), (0, 1), (3, 0), (3, 3) and output as (0, 3) (0, 1) (0, 0) (3, 0) (3, 3) for input as (0, 3), (0, 1), (0, 0), (3, 0), (3, 3). We generally need the farthest next point in case of collinear, we can get the desired result in case of collinear points by adding one more if condition. Please refer this modified code.
Sources:
http://www.cs.uiuc.edu/~jeffe/teaching/373/notes/x05-convexhull.pdf
http://www.dcs.gla.ac.uk/~pat/52233/slides/Hull1x1.pdf
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