# Tangents between two Convex Polygons

Given two convex polygons, we need to find the lower and upper tangents to these polygons.
As shown in the figure below, and show upper and lower tangent respectively.

Image source: http://geomalgorithms.com/polypoly1.gif

Examples:

```Input : First Polygon  : {(2, 2), (3, 3), (5, 2), (4, 0), (3, 1)}
Second Polygon : {(-1, 0), (0, 1), (1, 0), (0, -2)}.
Output : Upper Tangent - line joining (0,1) and (3,3)
Lower Tangent - line joining (0,-2) and (4,0)
```

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

Overview:
Let’s have two convex polygons as shown,

For finding the upper tangent, we start by taking two points. The rightmost point of a and leftmost point of b. The line joining them is labeled as 1. As this line passes through the polygon b (is not above polygon b) so we take the clockwise next point on b, the line is labeled 2. Now the line is above the polygon b, fine! But the line is crossing the polygon a, so we move to the anti-clockwise next point, labeled as 3 in the picture. This again crossing the polygon a so we move to line 4. This line is crossing b so we move to line 5. Now this line is crossing neither of the points. So this is the upper tangent for the given polygons.
For finding the lower tangent we need to move inversely through the polygons i.e. if the line is crossing the polygon b we move to anti-clockwise next and to clockwise next if the line is crossing the polygon a.

Algorithm for upper tangent:

```L <- line joining the rightmost point of a
and leftmost point of b.
while (L crosses any of the polygons)
{
while(L crosses b)
L <- L' : the point on b moves up.
while(L crosses a)
L <- L' : the point on a moves up.
}
```

Algorithm for lower tangent:

```L <- line joining the rightmost point of a
and leftmost point of b.
while (L crosses any of the polygons)
{
while (L crosses b)
L <- L' : the point on b moves down.
while (L crosses a)
L <- L' : the point on a moves down.
}```
```// C++ program to find upper tangent of two polygons.
#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())
int quad(pair<int,int> p)
{
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);

int one = quad(p);
int two = quad(q);

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.
void findUpperTangent(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();

// To find a point inside the convex polygon(centroid),
// we sum up all the coordinates and then divide  by
// n(number of points). But this would be a floating-point
// value. So to get rid off this we multiply points
// initially with n1 and then find the center and
// then divided it by n1 again.
// Similarly we do divide and multiply for n2 (i.e.,
// elements of b)

// maxa and minb are used to check if polygon a
// is left of b.
int maxa = INT_MIN;
for (int i=0; i<n1; i++)
{
maxa = max(maxa, a[i].first);
mid.first  += a[i].first;
mid.second += a[i].second;
a[i].first *= n1;
a[i].second *= n1;
}

// sorting the points is counter clockwise order
// for polygon a
sort(a.begin(), a.end(), compare);

for (int i=0; i<n1; i++)
{
a[i].first /= n1;
a[i].second /= n1;
}

mid = {0, 0};

int minb = INT_MAX;
for (int i=0; i<n2; i++)
{
mid.first += b[i].first;
mid.second += b[i].second;
minb = min(minb, b[i].first);
b[i].first *= n2;
b[i].second *= n2;
}

// sorting the points in counter clockwise
// order for polygon b
sort(b.begin(), b.end(), compare);

for (int i=0; i<n2; i++)
{
b[i].first/=n2;
b[i].second/=n2;
}

// If a is to the right of b, swap a and b
// This makes sure a is left of b.
if (minb < maxa)
{
a.swap(b);
n1 = a.size();
n2 = b.size();
}

// ia -> rightmost point of a
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;
}
}

cout << "upper tangent (" << a[inda].first << ","
<< a[inda].second << ") (" << b[indb].first
<< "," << b[indb].second << ")\n";
}

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

vector<pair<int,int> > b;
b.push_back({0, 1});
b.push_back({1, 0});
b.push_back({0, -2});
b.push_back({-1, 0});

findUpperTangent(a, b);

return 0;
}
```

Output:

```Upper tangent (0,1) (3,3)
```

Note that the above code only finds upper tangent. We can similarly find lower tangent.

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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