# 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++

 `// C++ program to find the XOR of XOR's of ` `// all submatrices ` ` `  `#include ` `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; ` `} `

## Java

 `//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. `

## Python3

 `# 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 `

## C#

 `// 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 `

## PHP

 ` `

Output:

```4
```

Time Complexity: O(N2)

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