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 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 the square of distance between // p1 and p2. int dist(iPair p, iPair q) { return (p.second - q.second) * (p.second - q.second) + (p.first - q.first) * (p.first - q.first); } // 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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