# Dynamic Convex hull | Adding Points to an Existing Convex Hull

Given a convex hull, we need to add a given number of points to the convex hull and print the convex hull after every point addition. The points should be in anti-clockwise order after addition of every point.
Examples:

```Input :
Convex Hull : (0, 0), (3, -1), (4, 5), (-1, 4)
Point to add : (100, 100)

Output :
New convex hull : (-1, 4) (0, 0) (3, -1) (100, 100)
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We first check whether the point is inside the given convex hull or not. If it is, then nothing has to be done we directly return the given convex hull. If the point is outside the convex hull, we find the lower and upper tangents, and then merge the point with the given convex hull to find the new convex hull, as shown in the figure.

The red outline shows the new convex hull after merging the point and the given convex hull.

To find the upper tangent, we first choose a point on the hull that is nearest to the given point. Then while the line joining the point on the convex hull and the given point crosses the convex hull, we move anti-clockwise till we get the tangent line.

The figure shows the moving of the point on the convex hull for finding the upper tangent.

Note: It is assumed here that the input of the initial convex hull is in the anti-clockwise order, otherwise we have to first sort them in anti-clockwise order then apply the following code.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Code:

```// C++ program to add given a point p to a given
// convext hull. The program assumes that the
// point of given convext hull are in anti-clockwise
// order.
#include<bits/stdc++.h>
using namespace std;

// checks whether the point crosses the convex hull
// or not
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;
}

// Returns the square of distance between two input points
int sqDist(pair<int, int> p1, pair<int, int> p2)
{
return (p1.first-p2.first)*(p1.first-p2.first) +
(p1.second-p2.second)*(p1.second-p2.second);
}

// Checks whether the point is inside the convex hull or not
bool inside(vector<pair<int, int>> a, pair<int, int> p)
{
// Initialize the centroid of the convex hull
pair<int, int> mid = {0, 0};

int n = a.size();

// Multiplying with n to avoid floating point
// arithmetic.
p.first *= n;
p.second *= n;
for (int i=0; i<n; i++)
{
mid.first += a[i].first;
mid.second += a[i].second;
a[i].first *= n;
a[i].second *= n;
}

// if the mid and the given point lies always
// on the same side w.r.t every edge of the
// convex hull, then the point lies inside
// the convex hull
for (int i=0, j; i<n; i++)
{
j = (i+1)%n;
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 for_mid = a1*mid.first+b1*mid.second+c1;
int for_p = a1*p.first+b1*p.second+c1;
if (for_mid*for_p < 0)
return false;
}

return true;
}

// Adds a point p to given convex hull a[]
void addPoint(vector<pair<int, int>> &a, pair<int, int> p)
{
// If point is inside p
if (inside(a, p))
return;

// point having minimum distance from the point p
int ind = 0;
int n = a.size();
for (int i=1; i<n; i++)
if (sqDist(p, a[i]) < sqDist(p, a[ind]))
ind = i;

// Find the upper tangent
int up = ind;
while (orientation(p, a[up], a[(up+1)%n])>=0)
up = (up + 1) % n;

// Find the lower tangent
int low = ind;
while (orientation(p, a[low], a[(n+low-1)%n])<=0)
low = (n+low - 1) % n;

// Initialize result
vector<pair<int, int>>ret;

// making the final hull by traversing points
// from up to low of given convex hull.
int curr = up;
ret.push_back(a[curr]);
while (curr != low)
{
curr = (curr+1)%n;
ret.push_back(a[curr]);
}

// Modify the original vector
ret.push_back(p);
a.clear();
for (int i=0; i<ret.size(); i++)
a.push_back(ret[i]);
}

// Driver code
int main()
{
// the set of points in the convex hull
vector<pair<int, int> > a;
a.push_back({0, 0});
a.push_back({3, -1});
a.push_back({4, 5});
a.push_back({-1, 4});
int n = a.size();

pair<int, int> p = {100, 100};

// Print the modified Convex Hull
for (auto e : a)
cout << "(" << e.first << ", "
<< e.second << ") ";

return 0;
}
```

Output:

```(-1, 4) (0, 0) (3, -1) (100, 100)
```

Time Complexity:
The time complexity of the above algorithm is O(n*q), where q is the number of points to be added.

This article is contributed by Amritya Vagmi 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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