Check if four segments form a rectangle

We are given four segments as a pair of coordinates of their end points. We need to tell whether those four line segments make a rectangle or not.

Input : segments[] =  [(4, 2), (7, 5),
                       (2, 4), (4, 2),
                       (2, 4), (5, 7),
                       (5, 7), (7, 5)]
Output : Yes
Given these segment make a rectangle of length 3X2.

Input : segment[] = [(7, 0), (10, 0),
                     (7, 0), (7, 3),
                     (7, 3), (10, 2),
                     (10, 2), (10, 0)]
Output : Not
These segments do not make a rectangle.

Above examples are shown in below diagram.

This problem is mainly an extension of How to check if given four points form a square

We can solve this problem by using properties of a rectangle. First, we check total unique end points of segments, if count of these points is not equal to 4 then the line segment can’t make a rectangle. Then we check distances between all pair of points, there should be at most 3 different distances, one for diagonal and two for sides and at the end we will check the relation among these three distances, for line segments to make a rectangle these distance should satisfy Pythagorean relation because sides and diagonal of rectangle makes a right angle triangle. If they satisfy mentioned conditions then we will flag polygon made by line segment as rectangle otherwise not.

//  C++ program to check whether it is possible
// to make a rectangle from 4 segments
#include <bits/stdc++.h>
using namespace std;
#define N 4

//  structure to represent a segment
struct Segment
    int ax, ay;
    int bx, by;

    Segment(int ax, int ay, int bx, int by) :
              ax(ax), ay(ay), bx(bx), by(by)

// Utility method to return square of distance
// between two points
int getDis(pair<int, int> a, pair<int, int> b)
    return (a.first - b.first)*(a.first - b.first) +
           (a.second - b.second)*(a.second - b.second);

// method returns true if line Segments make
// a rectangle
bool isPossibleRectangle(Segment segments[])
    set< pair<int, int> > st;

    // putiing all end points in a set to
    // count total unique points
    for (int i = 0; i < N; i++)
        st.insert(make_pair(segments[i].ax, segments[i].ay));
        st.insert(make_pair(segments[i].bx, segments[i].by));

    // If total unique points are not 4, then
    // they can't make a rectangle
    if (st.size() != 4)
        return false;

    //  dist will store unique 'square of distances'
    set<int> dist;

    //  calculating distance between all pair of
    // end points of line segments
    for (auto it1=st.begin(); it1!=st.end(); it1++)
        for (auto it2=st.begin(); it2!=st.end(); it2++)
            if (*it1 != *it2)
                dist.insert(getDis(*it1, *it2));

    // if total unique distance are more than 3,
    // then line segment can't make a rectangle
    if (dist.size() > 3)
        return false;

    // copying distance into array. Note that set maintains
    // sorted order.
    int distance[3];
    int i = 0;
    for (auto it = dist.begin(); it != dist.end(); it++)
        distance[i++] = *it;

    // If line seqments form a square
    if (dist.size() == 2)
      return (2*distance[0] == distance[1]);

    //  distance of sides should satisfy pythagorean
    // theorem
    return (distance[0] + distance[1] == distance[2]);

//  Driver code to test above methods
int main()
    Segment segments[] =
        {4, 2, 7, 5},
        {2, 4, 4, 2},
        {2, 4, 5, 7},
        {5, 7, 7, 5}

    if (isPossibleRectangle(segments))
        cout << "Yes";
        cout << "No";



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