# Convex Hull (Simple Divide and Conquer Algorithm)

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 the convex hull of this set of points.

Examples:

```Input : points[] = {(0, 0), (0, 4), (-4, 0), (5, 0),
(0, -6), (1, 0)};
Output : (-4, 0), (5, 0), (0, -6), (0, 4)
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Pre-requisite:
Tangents between two convex polygons

Algorithm:
Given the set of points for which we have to find the convex hull. Suppose we know the convex hull of the left half points and the right half points, then the problem now is to merge these two convex hulls and determine the convex hull for the complete set.
This can be done by finding the upper and lower tangent to the right and left convex hulls. This is illustrated here Tangents between two convex polygons

Let the left convex hull be a and the right convex hull be b. Then the lower and upper tangents are named as 1 and 2 respectively, as shown in the figure.
Then the red outline shows the final convex hull.

Now the problem remains, how to find the convex hull for the left and right half. Now recursion comes into picture, we divide the set of points until the number of points in the set is very small, say 5, and we can find the convex hull for these points by brute algorithm. The merging of these halves would result in the convex hull for the complete set of points.

Note:
We have used the brute algorithm to find the convex hull for small number of points and it has a time complexity of . But some people suggest the following, the convex hull for 3 or less points is the complete set of points. This is correct but the problem comes when we try to merge a left convex hull of 2 points and right convex hull of 3 points, then the program gets trapped in a infinite loop in some special cases. So, to get rid of this problem I directly found the convex hull for 5 or less points by algorithm, which is somewhat greater but does not effect the overall complexity of the algorithm.

```// A divide and conquer program to find convex
// hull of a given set of points.
#include<bits/stdc++.h>
using namespace std;

// stores the center of polygon (It is made
// global becuase it is used in comare function)
pair<int, int> mid;

// determines the quadrant of a point
// (used in compare())
{
if (p.first >= 0 && p.second >= 0)
return 1;
if (p.first <= 0 && p.second >= 0)
return 2;
if (p.first <= 0 && p.second <= 0)
return 3;
return 4;
}

// Checks whether the line is crossing the polygon
int orientation(pair<int, int> a, pair<int, int> b,
pair<int, int> c)
{
int res = (b.second-a.second)*(c.first-b.first) -
(c.second-b.second)*(b.first-a.first);

if (res == 0)
return 0;
if (res > 0)
return 1;
return -1;
}

// compare function for sorting
bool compare(pair<int, int> p1, pair<int, int> q1)
{
pair<int, int> p = make_pair(p1.first - mid.first,
p1.second - mid.second);
pair<int, int> q = make_pair(q1.first - mid.first,
q1.second - mid.second);

if (one != two)
return (one < two);
return (p.second*q.first < q.second*p.first);
}

// Finds upper tangent of two polygons 'a' and 'b'
// represented as two vectors.
vector<pair<int, int>> merger(vector<pair<int, int> > a,
vector<pair<int, int> > b)
{
// n1 -> number of points in polygon a
// n2 -> number of points in polygon b
int n1 = a.size(), n2 = b.size();

int ia = 0, ib = 0;
for (int i=1; i<n1; i++)
if (a[i].first > a[ia].first)
ia = i;

// ib -> leftmost point of b
for (int i=1; i<n2; i++)
if (b[i].first < b[ib].first)
ib=i;

// finding the upper tangent
int inda = ia, indb = ib;
bool done = 0;
while (!done)
{
done = 1;
while (orientation(b[indb], a[inda], a[(inda+1)%n1]) >=0)
inda = (inda + 1) % n1;

while (orientation(a[inda], b[indb], b[(n2+indb-1)%n2]) <=0)
{
indb = (n2+indb-1)%n2;
done = 0;
}
}

int uppera = inda, upperb = indb;
inda = ia, indb=ib;
done = 0;
int g = 0;
while (!done)//finding the lower tangent
{
done = 1;
while (orientation(a[inda], b[indb], b[(indb+1)%n2])>=0)
indb=(indb+1)%n2;

while (orientation(b[indb], a[inda], a[(n1+inda-1)%n1])<=0)
{
inda=(n1+inda-1)%n1;
done=0;
}
}

int lowera = inda, lowerb = indb;
vector<pair<int, int>> ret;

//ret contains the convex hull after merging the two convex hulls
//with the points sorted in anti-clockwise order
int ind = uppera;
ret.push_back(a[uppera]);
while (ind != lowera)
{
ind = (ind+1)%n1;
ret.push_back(a[ind]);
}

ind = lowerb;
ret.push_back(b[lowerb]);
while (ind != upperb)
{
ind = (ind+1)%n2;
ret.push_back(b[ind]);
}
return ret;

}

// Brute force algorithm to find convex hull for a set
// of less than 6 points
vector<pair<int, int>> bruteHull(vector<pair<int, int>> a)
{
// Take any pair of points from the set and check
// whether it is the edge of the convex hull or not.
// if all the remaining points are on the same side
// of the line then the line is the edge of convex
// hull otherwise not
set<pair<int, int> >s;

for (int i=0; i<a.size(); i++)
{
for (int j=i+1; j<a.size(); j++)
{
int x1 = a[i].first, x2 = a[j].first;
int y1 = a[i].second, y2 = a[j].second;

int a1 = y1-y2;
int b1 = x2-x1;
int c1 = x1*y2-y1*x2;
int pos = 0, neg = 0;
for (int k=0; k<a.size(); k++)
{
if (a1*a[k].first+b1*a[k].second+c1 <= 0)
neg++;
if (a1*a[k].first+b1*a[k].second+c1 >= 0)
pos++;
}
if (pos == a.size() || neg == a.size())
{
s.insert(a[i]);
s.insert(a[j]);
}
}
}

vector<pair<int, int>>ret;
for (auto e:s)
ret.push_back(e);

// Sorting the points in the anti-clockwise order
mid = {0, 0};
int n = ret.size();
for (int i=0; i<n; i++)
{
mid.first += ret[i].first;
mid.second += ret[i].second;
ret[i].first *= n;
ret[i].second *= n;
}
sort(ret.begin(), ret.end(), compare);
for (int i=0; i<n; i++)
ret[i] = make_pair(ret[i].first/n, ret[i].second/n);

return ret;
}

// Returns the convex hull for the given set of points
vector<pair<int, int>> divide(vector<pair<int, int>> a)
{
// If the number of points is less than 6 then the
// function uses the brute algorithm to find the
// convex hull
if (a.size() <= 5)
return bruteHull(a);

// left contains the left half points
// right contains the right half points
vector<pair<int, int>>left, right;
for (int i=0; i<a.size()/2; i++)
left.push_back(a[i]);
for (int i=a.size()/2; i<a.size(); i++)
right.push_back(a[i]);

// convex hull for the left and right sets
vector<pair<int, int>>left_hull = divide(left);
vector<pair<int, int>>right_hull = divide(right);

// merging the convex hulls
return merger(left_hull, right_hull);
}

// Driver code
int main()
{
vector<pair<int, int> > a;
a.push_back(make_pair(0, 0));
a.push_back(make_pair(1, -4));
a.push_back(make_pair(-1, -5));
a.push_back(make_pair(-5, -3));
a.push_back(make_pair(-3, -1));
a.push_back(make_pair(-1, -3));
a.push_back(make_pair(-2, -2));
a.push_back(make_pair(-1, -1));
a.push_back(make_pair(-2, -1));
a.push_back(make_pair(-1, 1));

int n = a.size();

// sorting the set of points according
// to the x-coordinate
sort(a.begin(), a.end());
vector<pair<int, int> >ans = divide(a);

cout << "convex hull:\n";
for (auto e:ans)
cout << e.first << " "
<< e.second << endl;

return 0;
}
```

Output:

```Convex Hull:
-5 -3
-1 -5
1 -4
0 0
-1 1
```

Time Complexity: The merging of the left and the right convex hulls take O(n) time and as we are dividing the points in two equal parts, so the time complexity of the above algorithm is O(n * log n).

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