Check if four segments form a rectangle - GeeksforGeeks

09 Mar, 2022
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
 
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. 
 
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. 
CPP
// 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 segmentstruct Segment{    int ax, ay;    int bx, by;}; // Utility method to return square of distance// between two pointsint 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 rectanglebool isPossibleRectangle(Segment segments[]){    set< pair<int, int> > st;     // putting 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 methodsint main(){    Segment segments[] =    {        {4, 2, 7, 5},        {2, 4, 4, 2},        {2, 4, 5, 7},        {5, 7, 7, 5}    };     (isPossibleRectangle(segments))?cout << "Yes\n":cout << "No\n";}
Output:  
Yes
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