Given a set of points, a Convex hull is the smallest convex polygon containing all the given points.

Input is an array of points specified by their x and y coordinates. Output is a convex hull of this set of points in ascending order of x coordinates.

Example :

Input : points[] = {{0, 3}, {1, 1}, {2, 2}, {4, 4}, {0, 0}, {1, 2}, {3, 1}, {3, 3}}; Output : The points in convex hull are: (0, 0) (0, 3) (3, 1) (4, 4) Input : points[] = {{0, 3}, {1, 1} Output : Not Possible There must be at least three points to form a hull. Input : points[] = {(0, 0), (0, 4), (-4, 0), (5, 0), (0, -6), (1, 0)}; Output : (-4, 0), (5, 0), (0, -6), (0, 4)

We have discussed following algorithms for Convex Hull problem.

Convex Hull | Set 1 (Jarvis’s Algorithm or Wrapping)

Convex Hull | Set 2 (Graham Scan)

The QuickHull algorithm is a Divide and Conquer algorithm similar to QuickSort. Let a[0…n-1] be the input array of points. Following are the steps for finding the convex hull of these points.

- Find the point with minimum x-coordinate lets say, min_x and similarly the point with maximum x-coordinate, max_x.
- Make a line joining these two points, say
**L**. This line will divide the whole set into two parts. Take both the parts one by one and proceed further. - For a part, find the point P with maximum distance from the line L. P forms a triangle with the points min_x, max_x. It is clear that the points residing inside this triangle can never be the part of convex hull.
- The above step divides the problem into two sub-problems (solved recursively). Now the line joining the points P and min_x and the line joining the points P and max_x are new lines and the points residing outside the triangle is the set of points. Repeat point no. 3 till there no point left with the line. Add the end points of this point to the convex hull.

Below is C++ implementation of above idea. The implementation uses set to store points so that points can be printed in sorted order. A point is represented as a pair.

`// C++ program to implement Quick Hull algorithm ` `// to find convex hull. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// iPair is integer pairs ` `#define iPair pair<int, int> ` ` ` `// Stores the result (points of convex hull) ` `set<iPair> hull; ` ` ` `// Returns the side of point p with respect to line ` `// joining points p1 and p2. ` `int` `findSide(iPair p1, iPair p2, iPair p) ` `{ ` ` ` `int` `val = (p.second - p1.second) * (p2.first - p1.first) - ` ` ` `(p2.second - p1.second) * (p.first - p1.first); ` ` ` ` ` `if` `(val > 0) ` ` ` `return` `1; ` ` ` `if` `(val < 0) ` ` ` `return` `-1; ` ` ` `return` `0; ` `} ` ` ` `// returns a value proportional to the distance ` `// between the point p and the line joining the ` `// points p1 and p2 ` `int` `lineDist(iPair p1, iPair p2, iPair p) ` `{ ` ` ` `return` `abs` `((p.second - p1.second) * (p2.first - p1.first) - ` ` ` `(p2.second - p1.second) * (p.first - p1.first)); ` `} ` ` ` `// End points of line L are p1 and p2. side can have value ` `// 1 or -1 specifying each of the parts made by the line L ` `void` `quickHull(iPair a[], ` `int` `n, iPair p1, iPair p2, ` `int` `side) ` `{ ` ` ` `int` `ind = -1; ` ` ` `int` `max_dist = 0; ` ` ` ` ` `// finding the point with maximum distance ` ` ` `// from L and also on the specified side of L. ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `{ ` ` ` `int` `temp = lineDist(p1, p2, a[i]); ` ` ` `if` `(findSide(p1, p2, a[i]) == side && temp > max_dist) ` ` ` `{ ` ` ` `ind = i; ` ` ` `max_dist = temp; ` ` ` `} ` ` ` `} ` ` ` ` ` `// If no point is found, add the end points ` ` ` `// of L to the convex hull. ` ` ` `if` `(ind == -1) ` ` ` `{ ` ` ` `hull.insert(p1); ` ` ` `hull.insert(p2); ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Recur for the two parts divided by a[ind] ` ` ` `quickHull(a, n, a[ind], p1, -findSide(a[ind], p1, p2)); ` ` ` `quickHull(a, n, a[ind], p2, -findSide(a[ind], p2, p1)); ` `} ` ` ` `void` `printHull(iPair a[], ` `int` `n) ` `{ ` ` ` `// a[i].second -> y-coordinate of the ith point ` ` ` `if` `(n < 3) ` ` ` `{ ` ` ` `cout << ` `"Convex hull not possible\n"` `; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `// Finding the point with minimum and ` ` ` `// maximum x-coordinate ` ` ` `int` `min_x = 0, max_x = 0; ` ` ` `for` `(` `int` `i=1; i<n; i++) ` ` ` `{ ` ` ` `if` `(a[i].first < a[min_x].first) ` ` ` `min_x = i; ` ` ` `if` `(a[i].first > a[max_x].first) ` ` ` `max_x = i; ` ` ` `} ` ` ` ` ` `// Recursively find convex hull points on ` ` ` `// one side of line joining a[min_x] and ` ` ` `// a[max_x] ` ` ` `quickHull(a, n, a[min_x], a[max_x], 1); ` ` ` ` ` `// Recursively find convex hull points on ` ` ` `// other side of line joining a[min_x] and ` ` ` `// a[max_x] ` ` ` `quickHull(a, n, a[min_x], a[max_x], -1); ` ` ` ` ` `cout << ` `"The points in Convex Hull are:\n"` `; ` ` ` `while` `(!hull.empty()) ` ` ` `{ ` ` ` `cout << ` `"("` `<<( *hull.begin()).first << ` `", "` ` ` `<< (*hull.begin()).second << ` `") "` `; ` ` ` `hull.erase(hull.begin()); ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `iPair a[] = {{0, 3}, {1, 1}, {2, 2}, {4, 4}, ` ` ` `{0, 0}, {1, 2}, {3, 1}, {3, 3}}; ` ` ` `int` `n = ` `sizeof` `(a)/` `sizeof` `(a[0]); ` ` ` `printHull(a, n); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Input :

The points in Convex Hull are: (0, 0) (0, 3) (3, 1) (4, 4)

**Time Complexity:** The analysis is similar to Quick Sort. On average, we get time complexity as O(n Log n), but in worst case, it can become O(n^{2})

This article is contributed by **Amritya Yagni**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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