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.
Examples:

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";
    else
        cout << "No";
}

Output:

Yes

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.

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