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

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