# Polygon Clipping | Sutherland–Hodgman Algorithm

A convex polygon and a convex clipping area are given. The task is to clip polygon edges using the Sutherland–Hodgman Algorithm. Input is in the form of vertices of the polygon in clockwise order.

Examples:

```Input : Polygon : (100,150), (200,250), (300,200)
Clipping Area : (150,150), (150,200), (200,200),
(200,150) i.e. a Square
Output : (150, 162) (150, 200) (200, 200) (200, 174)

Example 2
Input : Polygon : (100,150), (200,250), (300,200)
Clipping Area : (100,300), (300,300), (200,100)
Output : (242, 185) (166, 166) (150, 200) (200, 250) (260, 220)

```

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

Overview of the algorithm:

```Consider each edge e of clipping Area  and do following:
a) Clip given polygon against e.
```

How to clip against an edge of clipping area?
The edge (of clipping area) is extended infinitely to create a boundary and all the vertices are clipped using this boundary. The new list of vertices generated is passed to the next edge of the clip polygon in clockwise fashion until all the edges have been used.

There are four possible cases for any given edge of given polygon against current clipping edge e.

1. Both vertices are inside : Only the second vertex is added to the output list
2. First vertex is outside while second one is inside : Both the point of intersection of the edge with the clip boundary and the second vertex are added to the output list
3. First vertex is inside while second one is outside : Only the point of intersection of the edge with the clip boundary is added to the output list
4. Both vertices are outside : No vertices are added to the output list

There are two sub-problems that need to be discussed before implementing the algorithm:-

To decide if a point is inside or outside the clipper polygon
If the vertices of the clipper polygon are given in clockwise order then all the points lying on the right side of the clipper edges are inside that polygon. This can be calculated using :

To find the point of intersection of an edge with the clip boundary
If two points of each line(1,2 & 3,4) are known, then their point of intersection can be calculated using the formula :-

```// C++ program for implementing Sutherland–Hodgman
// algorithm for polygon clipping
#include<iostream>
using namespace std;

const int MAX_POINTS = 20;

// Returns x-value of point of intersectipn of two
// lines
int x_intersect(int x1, int y1, int x2, int y2,
int x3, int y3, int x4, int y4)
{
int num = (x1*y2 - y1*x2) * (x3-x4) -
(x1-x2) * (x3*y4 - y3*x4);
int den = (x1-x2) * (y3-y4) - (y1-y2) * (x3-x4);
return num/den;
}

// Returns y-value of point of intersectipn of
// two lines
int y_intersect(int x1, int y1, int x2, int y2,
int x3, int y3, int x4, int y4)
{
int num = (x1*y2 - y1*x2) * (y3-y4) -
(y1-y2) * (x3*y4 - y3*x4);
int den = (x1-x2) * (y3-y4) - (y1-y2) * (x3-x4);
return num/den;
}

// This functions clips all the edges w.r.t one clip
// edge of clipping area
void clip(int poly_points[][2], int &poly_size,
int x1, int y1, int x2, int y2)
{
int new_points[MAX_POINTS][2], new_poly_size = 0;

// (ix,iy),(kx,ky) are the co-ordinate values of
// the points
for (int i = 0; i < poly_size; i++)
{
// i and k form a line in polygon
int k = (i+1) % poly_size;
int ix = poly_points[i][0], iy = poly_points[i][1];
int kx = poly_points[k][0], ky = poly_points[k][1];

// Calculating position of first point
// w.r.t. clipper line
int i_pos = (x2-x1) * (iy-y1) - (y2-y1) * (ix-x1);

// Calculating position of second point
// w.r.t. clipper line
int k_pos = (x2-x1) * (ky-y1) - (y2-y1) * (kx-x1);

// Case 1 : When both points are inside
if (i_pos < 0  && k_pos < 0)
{
new_points[new_poly_size][0] = kx;
new_points[new_poly_size][1] = ky;
new_poly_size++;
}

// Case 2: When only first point is outside
else if (i_pos >= 0  && k_pos < 0)
{
// Point of intersection with edge
// and the second point is added
new_points[new_poly_size][0] = x_intersect(x1,
y1, x2, y2, ix, iy, kx, ky);
new_points[new_poly_size][1] = y_intersect(x1,
y1, x2, y2, ix, iy, kx, ky);
new_poly_size++;

new_points[new_poly_size][0] = kx;
new_points[new_poly_size][1] = ky;
new_poly_size++;
}

// Case 3: When only second point is outside
else if (i_pos < 0  && k_pos >= 0)
{
//Only point of intersection with edge is added
new_points[new_poly_size][0] = x_intersect(x1,
y1, x2, y2, ix, iy, kx, ky);
new_points[new_poly_size][1] = y_intersect(x1,
y1, x2, y2, ix, iy, kx, ky);
new_poly_size++;
}

// Case 4: When both points are outside
else
{
}
}

// Copying new points into original array
// and changing the no. of vertices
poly_size = new_poly_size;
for (int i = 0; i < poly_size; i++)
{
poly_points[i][0] = new_points[i][0];
poly_points[i][1] = new_points[i][1];
}
}

// Implements Sutherland–Hodgman algorithm
void suthHodgClip(int poly_points[][2], int poly_size,
int clipper_points[][2], int clipper_size)
{
//i and k are two consecutive indexes
for (int i=0; i<clipper_size; i++)
{
int k = (i+1) % clipper_size;

// We pass the current array of vertices, it's size
// and the end points of the selected clipper line
clip(poly_points, poly_size, clipper_points[i][0],
clipper_points[i][1], clipper_points[k][0],
clipper_points[k][1]);
}

// Printing vertices of clipped polygon
for (int i=0; i < poly_size; i++)
cout << '(' << poly_points[i][0] <<
", " << poly_points[i][1] << ") ";
}

//Driver code
int main()
{
// Defining polygon vertices in clockwise order
int poly_size = 3;
int poly_points[20][2] = {{100,150}, {200,250},
{300,200}};

// Defining clipper polygon vertices in clockwise order
// 1st Example with square clipper
int clipper_size = 4;
int clipper_points[][2] = {{150,150}, {150,200},
{200,200}, {200,150} };

// 2nd Example with triangle clipper
/*int clipper_size = 3;
int clipper_points[][2] = {{100,300}, {300,300},
{200,100}};*/

//Calling the clipping function
suthHodgClip(poly_points, poly_size, clipper_points,
clipper_size);

return 0;
}
```

Output:

```(150, 162) (150, 200) (200, 200) (200, 174)
```

This article is contributed by Nabaneet Roy. If you like GeeksforGeeks 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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