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Create an n x n square matrix, where all the sub-matrix have the sum of opposite corner elements as even

  • Last Updated : 04 Aug, 2021

Given an integer N. The task is to generate a square matrix of ( n x n ) having the elements ranging from 1 to n^2 with the following condition:

  • The elements of the matrix should be distinct i.e used only once
  • Numbers ranging from 1 to n^2
  • Every sub-matrix you choose should have the sum of opposite corner elements as even i.e sum of top left and bottom right should be even and the sum of top right and bottom left element should be even

This property should apply to all the submatrices of the matrix. You need to generate an Even Sub-Matrix

Examples:

Input:
Output: 1 2 
4 3 
Explanation: Here sum of 1+3=4 is even and 2+4=6 is even

Input:
Output: 1 2 3 
4 5 6 
7 8 9 
Explanation: The sub matrix [1 2 4 5], [2 3 5 6], [4 5 7 8], [5 6 8 9], [1 2 3 4 5 6 7 8 9] satisfies the condition of opposite corner 
elements having even sum 
 



Approach:

As we know for any two elements sum to be even it can be Sum of ODD and ODD or Sum of EVEN and EVEN. In either of the two cases for the corner elements sum to be even we need to ensure that the diagonal pattern arranged elements should be either odd or even. So we make the 2d array having diagonals as all odds or all evens to find any submatrix having corner elements sum even. The below approach can be followed for the same.

  • When n is odd the diagonals are already in all odd or even  elements, so we need not modify and generate a simple 2d array
  • When n is even the matrix generated does not satisfy the property having an even sum of opposite corner elements of the sub-matrices, so we reverse the alternate row elements so that diagonals of every submatrix are either all odd or all even.

Below is the implementation.

C++




// C++ program for
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
void sub_mat_even(int N)
{
  // Counter to initialize
  // the values in 2-D array
  int K = 1;
   
  // To create a 2-D array
  // from to 1 to N*2
  int A[N][N];
   
  for(int i = 0; i < N; i++)
  {
    for(int j = 0; j < N; j++)
    {
      A[i][j] = K;
      K++;
    }
  }
 
  // If found even we reverse
  // the alternate row elements
  // to get all diagonal elements
  // as all even or all odd
  if(N % 2 == 0)
  {
    for(int i = 0; i < N; i++)
    {
      if(i % 2 == 1)
      {
        int s = 0;
        int l = N - 1;
         
        // Reverse the row
        while(s < l)
        {
          swap(A[i][s],
               A[i][l]);
          s++;
          l--;
        }
      }
    }
  }
 
  // Print the formed array
  for(int i = 0; i < N; i++)
  {
    for(int j = 0; j < N; j++)
    {
      cout << A[i][j] << " ";
    }
    cout << endl;
  }
}
 
// Driver code
int main()
{
    int N = 4;
   
    // Function call
    sub_mat_even(N);
}
 
// This code is contributed by mishrapriyanshu557

Java




// Java program for
// the above approach
import java.io.*;
 
class GFG{
 
static void sub_mat_even(int N)
{
     
    // Counter to initialize
    // the values in 2-D array
    int K = 1;
 
    // To create a 2-D array
    // from to 1 to N*2
    int[][] A = new int[N][N];
 
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
            A[i][j] = K;
            K++;
        }
    }
 
    // If found even we reverse
    // the alternate row elements
    // to get all diagonal elements
    // as all even or all odd
    if (N % 2 == 0)
    {
        for(int i = 0; i < N; i++)
        {
            if (i % 2 == 1)
            {
                int s = 0;
                int l = N - 1;
 
                // Reverse the row
                while (s < l)
                {
                    swap(A[i], s, l);
                    s++;
                    l--;
                }
            }
        }
    }
 
    // Print the formed array
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
            System.out.print(A[i][j] + " ");
        }
        System.out.println();
    }
}
 
private static void swap(int[] A, int s, int l)
{
    int temp = A[s];
    A[s] = A[l];
    A[l] = temp;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4;
 
    // Function call
    sub_mat_even(N);
}
}
 
// This code is contributed by jithin

Python3




# Python3 program for
# the above approach
import itertools
 
 
def sub_mat_even(n):
     
    temp = itertools.count(1)
     
    # create a 2d array ranging
    # from 1 to n^2
    l = [[next(temp)for i in range(n)]for i in range(n)]
     
    # If found even we reverse the alternate
    # row elements to get all diagonal elements
    # as all even or all odd
    if n%2 == 0:
        for i in range(0,len(l)):
            if i%2 == 1:
                l[i][:] = l[i][::-1]
     
    # Printing the array formed
    for i in range(n):
        for j in range(n):
            print(l[i][j],end=" ")
        print()
 
n = 4
sub_mat_even(n)

C#




// C# program for
// the above approach
using System;
class GFG {
     
    static void sub_mat_even(int N)
    {
          
        // Counter to initialize
        // the values in 2-D array
        int K = 1;
      
        // To create a 2-D array
        // from to 1 to N*2
        int[,] A = new int[N, N];
      
        for(int i = 0; i < N; i++)
        {
            for(int j = 0; j < N; j++)
            {
                A[i, j] = K;
                K++;
            }
        }
      
        // If found even we reverse
        // the alternate row elements
        // to get all diagonal elements
        // as all even or all odd
        if (N % 2 == 0)
        {
            for(int i = 0; i < N; i++)
            {
                if (i % 2 == 1)
                {
                    int s = 0;
                    int l = N - 1;
      
                    // Reverse the row
                    while (s < l)
                    {
                        int temp = A[i, s];
                        A[i, s] = A[i, l];
                        A[i, l] = temp;
                        s++;
                        l--;
                    }
                }
            }
        }
      
        // Print the formed array
        for(int i = 0; i < N; i++)
        {
            for(int j = 0; j < N; j++)
            {
                Console.Write(A[i, j] + " ");
            }
            Console.WriteLine();
        }
    }
 
  static void Main() {
       
    int N = 4;
  
    // Function call
    sub_mat_even(N);
  }
}
 
// This code is contributed by divyeshrabadiya07

Output:

1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13

This approach takes O(n*2) time complexity.

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