Create matrix whose sum of diagonals in each sub matrix is even

Given a number N, the task is to create a square matrix of size N*N with values in range [1, N*N], such that the sum of each diagonal of an even sub-square matrix is even.

Examples:

Input: N = 3
Output:
1 2 3 
4 5 6 
7 8 9 
Explanation:For each even sub-square matrix the sum of each diagonal is a even number.
1 2 
4 5 
sum of each diagonal is 6 and 6 i.e even number.

Input: N = 4
Output:
1 2 3 4 
6 5 8 7 
9 10 11 12 
14 13 16 15 
Explanation:
For each even sub-square matrix the sum of each diagonal is a even number.
1 2 
6 5 
sum of each diagonal is 6 and 8 i.e even number.

Approach: The idea is to arrange elements from 1 to N*N in the below-given ways:

1. Initialize odd and even by 1 and 2 elements respectively.
2. Iterate two nested loop in the range [0, N].
3. If the sum of indices in the two nested loops is even the print the value of odd and increment odd by 2 and if the sum is odd then print the value of even, and increment even by 2.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to print N*N order matrix
// with all sub-matrix of even order
// is sum of its diagonal also even
void evenSubMatrix(int N)
{
    // Even index
    int even = 1;

    // Odd index
    int odd = 2;

    // Iterate two nested loop
    for (int i = 0; i < N; i++) {

        for (int j = 0; j < N; j++) {

            // For even index the element
            // should be consecutive odd
            if ((i + j) % 2 == 0) {
                cout << even << " ";
                even += 2;
            }

            // for odd index the element
            // should be consecutive even
            else {
                cout << odd << " ";
                odd += 2;
            }
        }
        cout << "\n";
    }
}

// Driver Code
int main()
{
    // Given order of matrix
    int N = 4;

    // Function call
    evenSubMatrix(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to print N*N order matrix
// with all sub-matrix of even order
// is sum of its diagonal also even
static void evenSubMatrix(int N)
{

    // Even index
    int even = 1;

    // Odd index
    int odd = 2;

    // Iterate two nested loop
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {

            // For even index the element
            // should be consecutive odd
            if ((i + j) % 2 == 0)
            {
                System.out.print(even + " ");
                even += 2;
            }

            // For odd index the element
            // should be consecutive even
            else
            {
                System.out.print(odd + " ");
                odd += 2;
            }
        }
        System.out.println();
    }
}

// Driver code
public static void main(String[] args)
{

    // Given order of matrix
    int N = 4;

    // Function call
    evenSubMatrix(N);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function to print N*N order matrix
# with all sub-matrix of even order
# is sum of its diagonal also even
def evenSubMatrix(N):

    # Even index
    even = 1

    # Odd index
    odd = 2

    # Iterate two nested loop
    for i in range(N):
        for j in range(N):

            # For even index the element
            # should be consecutive odd
            if ((i + j) % 2 == 0):
                print(even, end = " ")
                even += 2

            # For odd index the element
            # should be consecutive even
            else:
                print(odd, end = " ")
                odd += 2

        print()

# Driver Code

# Given order of matrix
N = 4

# Function call
evenSubMatrix(N)

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;

class GFG{

// Function to print N*N order matrix
// with all sub-matrix of even order
// is sum of its diagonal also even
static void evenSubMatrix(int N)
{

    // Even index
    int even = 1;

    // Odd index
    int odd = 2;

    // Iterate two nested loop
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {

            // For even index the element
            // should be consecutive odd
            if ((i + j) % 2 == 0)
            {
                Console.Write(even + " ");
                even += 2;
            }

            // For odd index the element
            // should be consecutive even
            else
            {
                Console.Write(odd + " ");
                odd += 2;
            }
        }
        Console.WriteLine();
    }
}

// Driver code
public static void Main(String[] args)
{

    // Given order of matrix
    int N = 4;

    // Function call
    evenSubMatrix(N);
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>
// Java script program for the above approach

// Function to print N*N order matrix
// with all sub-matrix of even order
// is sum of its diagonal also even
function evenSubMatrix( N)
{

    // Even index
    let even = 1;

    // Odd index
    let odd = 2;

    // Iterate two nested loop
    for(let i = 0; i < N; i++)
    {
        for(let j = 0; j < N; j++)
        {

            // For even index the element
            // should be consecutive odd
            if ((i + j) % 2 == 0)
            {
                document.write(even + " ");
                even += 2;
            }

            // For odd index the element
            // should be consecutive even
            else
            {
                document.write(odd + " ");
                odd += 2;
            }
        }
        document.write("<br>");
    }
}

// Driver code


    // Given order of matrix
    let N = 4;

    // Function call
    evenSubMatrix(N);

// This code is contributed by manoj
</script>

Output:

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

Time Complexity: O(N*N)
Auxiliary Space: O(1)