XOR of XORs of all sub-matrices

Given a ‘N*N’ matrix, the task is to find the XOR of XORs of all possible sub-matrices.

Examples:

Input :arr =  {{3, 1},
               {1, 3}}
Output : 0
Explanation: All the elements lie in 4 submatrices each. 
4 being even, there total contribution towards 
final answer becomes 0. Thus, ans = 0.

Input : arr = {{6, 7, 13},
               {8, 3, 4},
               {9, 7, 6}};
Output : 4

A Simple Approach is to generate all the possible submatrices, find the XOR of each submatrice uniquely and then XOR them all up. The time complexity of this approach will be O(n6).



Better solution: For each index(R, C), we will try to find the number of sub-matrices in which that index lies. If the number of sub-matrices is odd, then the final answer will be updated as ans = (ans ^ arr[R][C]). In case of even, we don’t need to update answer. This works because a number XORed with itself gives zero and order of operation doesn’t effect final XOR value.

Assuming 0-based indexing, number of sub-matrices an index (R, C) lies in equals

(R + 1)*(C + 1)*(N - R)*(N - C)

Below is the implementation of the above approach:

C++

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// C++ program to find the XOR of XOR's of
// all submatrices
  
#include <iostream>
using namespace std;
  
#define n 3
  
// Function to find to required
// XOR value
int submatrixXor(int arr[][n])
{
    int ans = 0;
  
    // Nested loop to find the
    // number of sub-matrix each
    // index belongs to
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            // Number of ways to choose
            // from top-left elements
            int top_left = (i + 1) * (j + 1);
  
            // Number of ways to choose
            // from bottom-right elements
            int bottom_right = (n - i) * (n - j);
  
            if ((top_left % 2 == 1) && (bottom_right % 2 == 1))
                ans = (ans ^ arr[i][j]);
        }
    }
  
    return ans;
}
  
// Driver Code
int main()
{
    int arr[][n] = { { 6, 7, 13 },
                     { 8, 3, 4 },
                     { 9, 7, 6 } };
  
    cout << submatrixXor(arr);
  
    return 0;
}

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Java

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//Java program to find the XOR of XOR's 
// of all submatrices
class GFG
{
      
// Function to find to required
// XOR value
static int submatrixXor(int[][]arr)
{
    int n = 3;
    int ans = 0;
  
    // Nested loop to find the
    // number of sub-matrix each
    // index belongs to
    for (int i = 0; i < n; i++) 
    {
        for (int j = 0; j < n; j++) 
        {
            // Number of ways to choose
            // from top-left elements
            int top_left = (i + 1) * (j + 1);
  
            // Number of ways to choose
            // from bottom-right elements
            int bottom_right = (n - i) * (n - j);
  
            if ((top_left % 2 == 1) &&
                (bottom_right % 2 == 1))
                ans = (ans ^ arr[i][j]);
        }
    }
  
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    int[][] arr = {{ 6, 7, 13},
                   { 8, 3, 4 },
                   { 9, 7, 6 }};
  
    System.out.println(submatrixXor(arr));
}
}
  
// This code is contributed
// by Code_Mech.

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Python3

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# Python3 program to find the XOR of 
# XOR's of all submatrices
  
# Function to find to required
# XOR value
def submatrixXor(arr, n):
  
    ans = 0
  
    # Nested loop to find the
    # number of sub-matrix each
    # index belongs to
    for i in range(0, n):
        for j in range(0, n):
  
            # Number of ways to choose
            # from top-left elements
            top_left = (i + 1) * (j + 1)
  
            # Number of ways to choose
            # from bottom-right elements
            bottom_right = (n - i) * (n - j)
            if (top_left % 2 == 1 and 
                bottom_right % 2 == 1):
                ans = (ans ^ arr[i][j])
    return ans
  
# Driver code
n = 3
arr = [[6, 7, 13],
       [8, 3, 4],
       [9, 7, 6]]
print(submatrixXor(arr, n))
  
# This code is contributed by Shrikant13

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C#

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// C# program to find the XOR of XOR's 
// of all submatrices
using System;
  
class GFG
{
      
// Function to find to required
// XOR value
static int submatrixXor(int [,]arr)
{
    int n = 3;
    int ans = 0;
  
    // Nested loop to find the
    // number of sub-matrix each
    // index belongs to
    for (int i = 0; i < n; i++) 
    {
        for (int j = 0; j < n; j++) 
        {
            // Number of ways to choose
            // from top-left elements
            int top_left = (i + 1) * (j + 1);
  
            // Number of ways to choose
            // from bottom-right elements
            int bottom_right = (n - i) * (n - j);
  
            if ((top_left % 2 == 1) &&
                (bottom_right % 2 == 1))
                ans = (ans ^ arr[i, j]);
        }
    }
  
    return ans;
}
  
// Driver Code
public static void Main()
{
    int [, ]arr = {{ 6, 7, 13},
                   { 8, 3, 4 },
                   { 9, 7, 6 }};
  
    Console.Write(submatrixXor(arr));
}
}
  
// This code is contributed
// by Akanksha Rai

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PHP

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<?php
// PHP program to find the XOR of 
// XOR's of all submatrices 
  
// Function to find to required 
// XOR value 
function submatrixXor($arr
    $ans = 0; 
    $n = 3 ;
  
    // Nested loop to find the 
    // number of sub-matrix each 
    // index belongs to 
    for ($i = 0; $i < $n; $i++) 
    
        for ($j = 0; $j < $n; $j++) 
        
            // Number of ways to choose 
            // from top-left elements 
            $top_left = ($i + 1) * ($j + 1); 
  
            // Number of ways to choose 
            // from bottom-right elements 
            $bottom_right = ($n - $i) * ($n - $j); 
  
            if (($top_left % 2 == 1) &&
                ($bottom_right % 2 == 1)) 
                $ans = ($ans ^ $arr[$i][$j]); 
        
    
  
    return $ans
  
// Driver Code 
$arr = array(array( 6, 7, 13 ), 
             array( 8, 3, 4 ), 
             array( 9, 7, 6 )); 
  
echo submatrixXor($arr); 
  
# This code is contributed by Ryuga
?>

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Output:

4

Time Complexity: O(N2)



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