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;` `}` |

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})

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