Find an N x N grid whose xor of every row and column is equal

Given an integer N which is a multiple of 4, the task is to find an N x N grid for which the bitwise xor of every row and column is same.

Examples:

Input: N = 4
Output:
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15



Input: N = 8
Output:
0 1 2 3 16 17 18 19
4 5 6 7 20 21 22 23
8 9 10 11 24 25 26 27
12 13 14 15 28 29 30 31
32 33 34 35 48 49 50 51
36 37 38 39 52 53 54 55
40 41 42 43 56 57 58 59
44 45 46 47 60 61 62 63

Approach: To solve this problem lets fix the xor of every row and column to 0 since xor of 4 consecutive numbers starting from 0 is 0. Here is an example of a 4 x 4 matrix:

0 ^ 1 ^ 2 ^ 3 = 0
4 ^ 5 ^ 6 ^ 7 = 0
8 ^ 9 ^ 10 ^ 11 = 0
12 ^ 13 ^ 14 ^ 15 = 0
and so on.

If you notice in the above example, the xor of every row and column is 0. Now we need to place the numbers in such a way that the xor of each row and column is 0.So we can divide our N x N matrix into smaller 4 x 4 matrices with N / 4 rows and columns and fill the cells in a way that the xor of every row and column is 0.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the n x n matrix
// that satisfies the given condition
void findGrid(int n)
{
    int arr[n][n];
  
    // Initialize x to 0
    int x = 0;
  
    // Divide the n x n matrix into n / 4 matrices
    // for each of the n / 4 rows where
    // each matrix is of size 4 x 4
    for (int i = 0; i < n / 4; i++) {
        for (int j = 0; j < n / 4; j++) {
            for (int k = 0; k < 4; k++) {
                for (int l = 0; l < 4; l++) {
                    arr[i * 4 + k][j * 4 + l] = x;
                    x++;
                }
            }
        }
    }
  
    // Print the generated matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << arr[i][j] << " ";
        }
        cout << "\n";
    }
}
  
// Driver code
int main()
{
    int n = 4;
  
    findGrid(n);
  
    return 0;
}

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Java














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// Java implementation of the approach 


import java.io.*; 


  


class GFG  


{ 


      


// Function to find the n x n matrix 


// that satisfies the given condition 


static void findGrid(int n) 


{ 


    int [][]arr = new int[n][n]; 


  


    // Initialize x to 0 


    int x = 0; 


  


    // Divide the n x n matrix into n / 4 matrices 


    // for each of the n / 4 rows where 


    // each matrix is of size 4 x 4 


    for (int i = 0; i < n / 4; i++) 


    { 


        for (int j = 0; j < n / 4; j++)  


        { 


            for (int k = 0; k < 4; k++)  


            { 


                for (int l = 0; l < 4; l++)  


                { 


                    arr[i * 4 + k][j * 4 + l] = x; 


                    x++; 


                } 


            } 


        } 


    } 


  


    // Print the generated matrix 


    for (int i = 0; i < n; i++)  


    { 


        for (int j = 0; j < n; j++)  


        { 


            System.out.print(arr[i][j] + " "); 


        } 


        System.out.println(" "); 


    } 


} 


  


// Driver code 


public static void main (String[] args) 


{ 


    int n = 4; 


      


    findGrid(n); 


} 


} 


  


// This code is contributed by ajit. 



















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Python3














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# Python3 implementation of the approach  


  


# Function to find the n x n matrix  


# that satisfies the given condition  


def findGrid(n):  


  


    arr = [[0 for k in range(n)]  


              for l in range(n)]  


  


    # Initialize x to 0  


    x = 0


  


    # Divide the n x n matrix into n / 4 matrices  


    # for each of the n / 4 rows where  


    # each matrix is of size 4 x 4  


    for i in range(n // 4):  


        for j in range(n // 4):  


            for k in range(4):  


                for l in range(4):  


                    arr[i * 4 + k][j * 4 + l] = x  


                    x += 1


  


    # Print the generated matrix  


    for i in range(n):  


        for j in range(n):  


            print(arr[i][j], end = " ") 


        print() 


  


# Driver code  


n = 4


findGrid(n)  


  


# This code is contributed by divyamohan123 



















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C#














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// C# implementation of the approach 


using System; 


      


class GFG  


{ 


      


// Function to find the n x n matrix 


// that satisfies the given condition 


static void findGrid(int n) 


{ 


    int [,]arr = new int[n, n]; 


  


    // Initialize x to 0 


    int x = 0; 


  


    // Divide the n x n matrix into n / 4 matrices 


    // for each of the n / 4 rows where 


    // each matrix is of size 4 x 4 


    for (int i = 0; i < n / 4; i++) 


    { 


        for (int j = 0; j < n / 4; j++)  


        { 


            for (int k = 0; k < 4; k++)  


            { 


                for (int l = 0; l < 4; l++)  


                { 


                    arr[i * 4 + k, j * 4 + l] = x; 


                    x++; 


                } 


            } 


        } 


    } 


  


    // Print the generated matrix 


    for (int i = 0; i < n; i++)  


    { 


        for (int j = 0; j < n; j++)  


        { 


            Console.Write(arr[i, j] + " "); 


        } 


        Console.WriteLine(" "); 


    } 


} 


  


// Driver code 


public static void Main (String[] args) 


{ 


    int n = 4; 


      


    findGrid(n); 


} 


} 


  


// This code is contributed by PrinciRaj1992 



















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Output:


0 1 2 3 
4 5 6 7 
8 9 10 11 
12 13 14 15





Time Complexity: O(N2)



          
          
          

          

    

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