# How to check if given four points form a square

Last Updated : 25 Oct, 2022

Given coordinates of four points in a plane, find if the four points form a square or not.

To check for square, we need to check for following.
a) All four sides formed by points are the same.
b) The angle between any two sides is 90 degree. (This condition is required as Rhombus also has same sides.
c) Check both the diagonals have the same distance

Examples:

Input: p1 = { 20, 10 }, p2 = { 10, 20 }, p3 = { 20, 20 }, p4 = { 10, 10 }
Output: Yes
Explanation:

Input: p1 = { 20, 20 }, p2 = { 10, 20 }, p3 = { 20, 20 }, p4 = { 10, 10 }
Output:  No

Approach: The idea is to pick any point and calculate its distance from the rest of the points. Let the picked point be ‘p’. To form a square, the distance of two points must be the same from ‘p’, let this distance be d. The distance from one point must be different from that d and must be equal to ?2 times d. Let this point with different distance be ‘q’.

The above condition is not good enough as the point with a different distance can be on the other side. We also need to check that q is at the same distance from 2 other points and this distance is the same as d.

Below are the implementations of the above idea.

## C++

 // A C++ program to check if four given points form a square or not. #include  using namespace std;   // Structure of a point in 2D space struct Point {     int x, y; };   // A utility function to find square of distance // from point 'p' to point 'q' int distSq(Point p, Point q) {     return (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y); }   // This function returns true if (p1, p2, p3, p4) form a // square, otherwise false bool isSquare(Point p1, Point p2, Point p3, Point p4) {     int d2 = distSq(p1, p2); // from p1 to p2     int d3 = distSq(p1, p3); // from p1 to p3     int d4 = distSq(p1, p4); // from p1 to p4       if (d2 == 0 || d3 == 0 || d4 == 0)             return false;       // If lengths if (p1, p2) and (p1, p3) are same, then     // following conditions must met to form a square.     // 1) Square of length of (p1, p4) is same as twice     // the square of (p1, p2)     // 2) Square of length of (p2, p3) is same     // as twice the square of (p2, p4)       if (d2 == d3 && 2 * d2 == d4         && 2 * distSq(p2, p4) == distSq(p2, p3)) {         return true;     }       // The below two cases are similar to above case     if (d3 == d4 && 2 * d3 == d2         && 2 * distSq(p3, p2) == distSq(p3, p4)) {         return true;     }     if (d2 == d4 && 2 * d2 == d3         && 2 * distSq(p2, p3) == distSq(p2, p4)) {         return true;     }       return false; }   // Driver program to test above function int main() {     Point p1 = { 20, 10 }, p2 = { 10, 20 },           p3 = { 20, 20 }, p4 = { 10, 10 };     isSquare(p1, p2, p3, p4) ? cout << "Yes" : cout << "No";     return 0; }

## Java

 // A Java program to check if four given points form a square or not.   class GFG {   // Structure of a point in 2D space static class Point  {     int x, y;           public Point(int x, int y)          {             this.x = x;             this.y = y;         }       };   // A utility function to find square of distance // from point 'p' to point 'q' static int distSq(Point p, Point q) {     return (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y); }   // This function returns true if (p1, p2, p3, p4) form a // square, otherwise false static boolean isSquare(Point p1, Point p2, Point p3, Point p4) {     int d2 = distSq(p1, p2); // from p1 to p2     int d3 = distSq(p1, p3); // from p1 to p3     int d4 = distSq(p1, p4); // from p1 to p4       if (d2 == 0 || d3 == 0 || d4 == 0)             return false;       // If lengths if (p1, p2) and (p1, p3) are same, then     // following conditions must met to form a square.     // 1) Square of length of (p1, p4) is same as twice     // the square of (p1, p2)     // 2) Square of length of (p2, p3) is same     // as twice the square of (p2, p4)       if (d2 == d3 && 2 * d2 == d4         && 2 * distSq(p2, p4) == distSq(p2, p3))     {         return true;     }       // The below two cases are similar to above case     if (d3 == d4 && 2 * d3 == d2         && 2 * distSq(p3, p2) == distSq(p3, p4))      {         return true;     }     if (d2 == d4 && 2 * d2 == d3         && 2 * distSq(p2, p3) == distSq(p2, p4))     {         return true;     }       return false; }   // Driver code public static void main(String[] args) {     Point p1 = new Point(20, 10), p2 = new Point( 10, 20 ),         p3 = new Point(20, 20 ), p4 = new Point( 10, 10 );     System.out.println(isSquare(p1, p2, p3, p4)==true ? "Yes" : "No"); } }   // This code is contributed by PrinciRaj1992

## Python3

 # A Python3 program to check if # four given points form a square or not. class Point:           # Structure of a point in 2D space     def __init__(self, x, y):         self.x = x         self.y = y   # A utility function to find square of  # distance from point 'p' to point 'q' def distSq(p, q):     return (p.x - q.x) * (p.x - q.x) +\            (p.y - q.y) * (p.y - q.y)   # This function returns true if (p1, p2, p3, p4)  # form a square, otherwise false def isSquare(p1, p2, p3, p4):       d2 = distSq(p1, p2) # from p1 to p2     d3 = distSq(p1, p3) # from p1 to p3     d4 = distSq(p1, p4) # from p1 to p4       if d2 == 0 or d3 == 0 or d4 == 0:             return False       # If lengths if (p1, p2) and (p1, p3) are same, then     # following conditions must be met to form a square.     # 1) Square of length of (p1, p4) is same as twice     # the square of (p1, p2)     # 2) Square of length of (p2, p3) is same     # as twice the square of (p2, p4)       if d2 == d3 and 2 * d2 == d4 and \                     2 * distSq(p2, p4) == distSq(p2, p3):         return True       # The below two cases are similar to above case     if d3 == d4 and 2 * d3 == d2 and \                     2 * distSq(p3, p2) == distSq(p3, p4):         return True       if d2 == d4 and 2 * d2 == d3 and \                     2 * distSq(p2, p3) == distSq(p2, p4):         return True       return False   # Driver Code if __name__=="__main__":     p1 = Point(20, 10)     p2 = Point(10, 20)     p3 = Point(20, 20)     p4 = Point(10, 10)           if isSquare(p1, p2, p3, p4):         print('Yes')      else:         print('No')   # This code is contributed by Mayank Chaudhary # aka chaudhary_19

## C#

 // A C# program to check if four given points form a square or not. using System;   class GFG {   // Structure of a point in 2D space class Point  {     public int x, y;       public Point(int x, int y)      {         this.x = x;         this.y = y;     }       };   // A utility function to find square of distance // from point 'p' to point 'q' static int distSq(Point p, Point q) {     return (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y); }   // This function returns true if (p1, p2, p3, p4) form a // square, otherwise false static bool isSquare(Point p1, Point p2, Point p3, Point p4) {     int d2 = distSq(p1, p2); // from p1 to p2     int d3 = distSq(p1, p3); // from p1 to p3     int d4 = distSq(p1, p4); // from p1 to p4       if (d2 == 0 || d3 == 0 || d4 == 0)             return false;       // If lengths if (p1, p2) and (p1, p3) are same, then     // following conditions must met to form a square.     // 1) Square of length of (p1, p4) is same as twice     // the square of (p1, p2)     // 2) Square of length of (p2, p3) is same     // as twice the square of (p2, p4)     if (d2 == d3 && 2 * d2 == d4         && 2 * distSq(p2, p4) == distSq(p2, p3))     {         return true;     }       // The below two cases are similar to above case     if (d3 == d4 && 2 * d3 == d2         && 2 * distSq(p3, p2) == distSq(p3, p4))      {         return true;     }     if (d2 == d4 && 2 * d2 == d3         && 2 * distSq(p2, p3) == distSq(p2, p4))     {         return true;     }     return false; }   // Driver code public static void Main(String[] args) {     Point p1 = new Point(20, 10), p2 = new Point(10, 20),         p3 = new Point(20, 20), p4 = new Point(10, 10);     Console.WriteLine(isSquare(p1, p2, p3, p4) == true ? "Yes" : "No"); } }   // This code is contributed by 29AjayKumar

## Javascript

 

Output

Yes

Time Complexity: O(1), all operations are being carried out in O(1) constant time.
Auxiliary Space: O(1), no extra space required

Extended Problem:

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