Area of a polygon with given n ordered vertices

2.7

Given ordered coordinates of a polygon with n vertices. Find area of the polygon. Here ordered mean that the coordinates are given either in clockwise manner or anticlockwise from first vertex to last.

Examples:

Input :  X[] = {0, 4, 4, 0}, Y[] = {0, 0, 4, 4};
Output : 16

Input : X[] = {0, 4, 2}, Y[] = {0, 0, 4}
Output : 8

We can compute area of a polygon using Shoelace formula.

Area
formula
 = | 1/2 [ (x1y2 + x2y3 + ... + xn-1yn + xny1) -
           (x2y1 + x3y2 + ... + xnyn-1 + x1yn) ] |

Below is C++ implementation of above formula.

// C++ program to evaluate area of a polygon using
// shoelace formula
#include <bits/stdc++.h>
using namespace std;

// (X[i], Y[i]) are coordinates of i'th point.
double polygonArea(double X[], double Y[], int n)
{
    // Initialze area
    double area = 0.0;

    // Calculate value of shoelace formula
    int j = n - 1;
    for (int i = 0; i < n; i++)
    {
        area += (X[j] + X[i]) * (Y[j] - Y[i]);
        j = i;  // j is previous vertex to i
    }

    // Return absolute value
    return abs(area / 2.0);
}

// Driver program to test above function
int main()
{
    double X[] = {0, 2, 4};
    double Y[] = {1, 3, 7};

    int n = sizeof(X)/sizeof(X[0]);

    cout << polygonArea(X, Y, n);
}

Output :

2

Why is it called Shoelace Formula?
The formula is called so because of the way we evaluate it.

Example:

Let the input vertices be
 (0, 1), (2, 3), and (4, 7). 

Evaluation procedure matches with process of tying
shoelaces.

We write vertices as below
  0    1
  2    3
  4    7
  0    1  [written twice]

we evaluate positive terms as below
  0  \  1
  2  \  3
  4  \  7
  0     1  
i.e., 0*3 + 2*7 + 4*1 = 18 

we evaluate negative terms as below
  0     1
  2  /  3
  4  /  7
  0  /  1  
i.e., 0*7 + 4*3 + 2*1 = 14

Area = 1/2 (18 - 14) = 2 

See this for a clearer image.

How does this work?
We can always divide a polygon into triangles. The area formula is derived by taking each edge AB, and calculating the (signed) area of triangle ABO with a vertex at the origin O, by taking the cross-product (which gives the area of a parallelogram) and dividing by 2. As one wraps around the polygon, these triangles with positive and negative area will overlap, and the areas between the origin and the polygon will be cancelled out and sum to 0, while only the area inside the reference triangle remains. [Source : Wiki]

Related articles :
Minimum Cost Polygon Triangulation
Find Simple Closed Path for a given set of points

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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