# 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.

## 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 segment` `struct` `Segment` `{` ` ` `int` `ax, ay;` ` ` `int` `bx, 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;` ` ` `// 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 methods` `int` `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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