Skip to content
Related Articles

Related Articles

Improve Article

Count of parallelograms in a plane

  • Difficulty Level : Medium
  • Last Updated : 07 Jan, 2017

Given some points on a plane, which are distinct and no three of them lie on the same line. We need to find number of Parallelograms with the vertices as the given points.
Examples:

Input : points[] = {(0, 0), (0, 2), (2, 2), (4, 2), 
                                   (1, 4), (3, 4)}
Output : 2
Two Parallelograms are possible by choosing above
given point as vertices, which are shown in below 
diagram.


We can solve this problem by using a special property of parallelograms that diagonals of a parallelogram intersect each other in the middle. So if we get such a middle point which is middle point of more than one line segment, then we can conclude that a parallelogram exists, more accurately if a middle point occurs x times, then diagonals of possible parallelograms can be chosen in xC2 ways, i.e. there will be x*(x-1)/2 parallelograms corresponding to this particular middle point with a frequency x. So we iterate over all pair of points and we calculate their middle point and increase frequency of middle point by 1. At the end, we count number of parallelograms according to the frequency of each distinct middle point as explained above. As we just need frequency of middle point, division by 2 is ignored while calculating middle point for simplicity.




// C++ program to get number of Parallelograms we
// can make by given points of the plane
#include <bits/stdc++.h>
using namespace std;
  
// Returns count of Parallelograms possible
// from given points
int countOfParallelograms(int x[], int y[], int N)
{
    // Map to store frequency of mid points
    map<pair<int, int>, int> cnt;
    for (int i=0; i<N; i++)
    {
        for (int j=i+1; j<N; j++)
        {
            // division by 2 is ignored, to get
            // rid of doubles
            int midX = x[i] + x[j];
            int midY = y[i] + y[j];
  
            // increase the frequency of mid point
            cnt[make_pair(midX, midY)]++;
        }
    }
  
    // Iterating through all mid points
    int res = 0;
    for (auto it = cnt.begin(); it != cnt.end(); it++)
    {
        int freq = it->second;
  
        // Increase the count of Parallelograms by
        // applying function on frequency of mid point
        res += freq*(freq - 1)/2
    }
  
    return res;
}
  
//  Driver code to test above methods
int main()
{
    int x[] = {0, 0, 2, 4, 1, 3};
    int y[] = {0, 2, 2, 2, 4, 4};
    int N = sizeof(x) / sizeof(int);
  
    cout << countOfParallelograms(x, y, N) << endl;
  
    return 0;
}

Output:

2

This article is contributed by Utkarsh Trivedi. 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.



Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :