Construct a Matrix whose sum of diagonals for each square submatrix is even

Given an integer N, the task is to construct a matrix M[][] of size N x N with numbers in range [1, N^2], such that the sum of diagonal elements for each square submatrix is even.
Examples:

Input: N = 2
Output:
1 2
4 3
Explanation:
This matrix has 5 square submatrix and 4 of them ([1], [2], [3], [4]) have a=0 so they satisfy the conditions.
The last square submatrix is the whole matrix M where r=c=a=1. We can see that M_{(1, 1)}+M_{(2, 2)}=1+3=4 and M_{(1, 2)}+M_{(2, 1)}=2+4=6 are both even.
Input: N = 4
Output:
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13

Approach: We know that the sum of two numbers is even when their parity is the same. Let us say the parity of M_{(i, j)} is odd that means the parity of M_{(i+1, j+1)}, M_{(i+1, j-1)}, M_{(i-1, j+1)}, M_{(i-1, j-1)} has to be odd.
Below is the illustration for N = 4 to generate a matrix of size 4×4:

So from the above illustration we have to fill the matrix in the Checkerboard Pattern. We can fill it in two ways:

All black cells have an odd integer and white cells have an even integer.
All black cells have an even integer and white cells have an odd integer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to print the desired matrix 
void UniqueMatrix(int N) 
{ 
    int element_value = 1; 
    int i = 0; 
  
    // element_value will start from 1 
    // and go up to N ^ 2 
  
    // i is row number and it starts 
    // from 0 and go up to N-1 
      
    // Iterate ove all [0, N] 
    while(i < N) 
    { 
          
        // If is even 
        if(i % 2 == 0) 
        { 
            for(int f = element_value; 
                    f < element_value + N; f++) 
            { 
                  
                // If row number is even print 
                // the row in forward order 
                cout << f << " "; 
            } 
            element_value += N; 
        } 
        else
        { 
            for(int k = element_value + N - 1; 
                    k > element_value - 1; k--) 
            { 
                  
                // If row number is odd print 
                // the row in reversed order 
                cout << k << " "; 
  
            } 
            element_value += N; 
        } 
        cout << endl; 
        i = i + 1; 
    } 
} 
  
// Driver Code 
int main() 
{ 
      
    // Given matrix size 
    int N = 4; 
  
    // Function call 
    UniqueMatrix(N); 
} 
  
// This code is contributed by chitranayal 

Python3

# Python3 program for the above approach 
  
# Function to print the desired matrix 
def UniqueMatrix(N): 
  
    element_value = 1
    i = 0
  
    # element_value will start from 1 
    # and go up to N ^ 2 
  
    # i is row number and it starts 
    # from 0 and go up to N-1 
      
    # Iterate ove all [0, N] 
    while(i < N): 
          
        # If is even 
        if(i % 2 == 0): 
  
            for f in range(element_value, element_value + N, 1): 
  
                # If row number is even print 
                # the row in forward order 
                print(f, end =' ') 
            element_value += N 
  
        else: 
  
            for k in range(element_value + N-1, element_value-1, -1): 
  
                # if row number is odd print 
                # the row in reversed order 
                print(k, end =' ') 
  
            element_value += N 
  
        print() 
        i = i + 1
  
  
# Driver Code 
  
# Given Matrix Size 
N = 4
  
# Function Call 
UniqueMatrix(N) 

Output:

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

Time Complexity: O(N^2)
Auxiliary Space: O(N^2)

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My Personal Notes

C++

Python3

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