Generate a matrix having sum of secondary diagonal equal to a perfect square

Given an integer N, the task is to generate a matrix of dimensions N x N using positive integers from the range [1, N] such that the sum of the secondary diagonal is a perfect square.

Examples:

Input: N = 3
Output:
1 2 3
2 3 1
3 2 1
Explanation:
The sum of secondary diagonal = 3 + 3 + 3 = 9(= 32).

Input: N = 7
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6
Explanation:
The sum of secondary diagonal = 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49(= 72).

Approach: Since the generated matrix needs to be of dimensions N x N, therefore, to make the sum of elements in the secondary diagonal a perfect square, the idea is to assign N at each index of the secondary diagonal. Therefore, the sum of all N elements in this diagonal is N2, which is a perfect square. Follow the steps below to solve the problem:



  1. Initialize a matrix mat[][] of dimension N x N.
  2. Initialize the first row of the matrix as {1 2 3 … N}.
  3. Now for the remaining rows of the matrix, fill each row by circular left shift of the arrangement of the previous row of the matrix by 1.
  4. Print the matrix after completing the above steps.

Below is the implementation of the above approach:

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
class GFG{
     
// Function to print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
static void diagonalSumPerfectSquare(int[] arr,
                                     int N)
{
     
    // Iterate for next N - 1 rows
    for(int i = 0; i < N; i++)
    {
         
        // Print the current row after
        // the left shift
        for(int j = 0; j < N; j++)
        {
            System.out.print(arr[(j + i) % 7] + " ");
        }
        System.out.println();
    }
}
 
// Driver Code
public static void main(String[] srgs)
{
     
    // Given N
    int N = 7;
 
    int[] arr = new int[N];
 
    // Fill the array with elements
    // ranging from 1 to N
    for(int i = 0; i < N; i++)
    {
        arr[i] = i + 1;
    }
 
    // Function Call
    diagonalSumPerfectSquare(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
 
# Function to print the matrix whose sum
# of element in secondary diagonal is a
# perfect square
def diagonalSumPerfectSquare(arr, N):
   
    # Print the current row
    print(*arr, sep =" ")
     
    # Iterate for next N - 1 rows
    for i in range(N-1):
        
        # Perform left shift by 1
        arr = arr[1::] + arr[:1:]
         
        # Print the current row after
        # the left shift
        print(*arr, sep =" ")
 
# Driver Code
 
# Given N
N = 7
 
arr = []
 
# Fill the array with elements
# ranging from 1 to N
for i in range(1, N + 1):
    arr.append(i)
 
# Function Call
diagonalSumPerfectSquare(arr, N)

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the
// above approach
using System;
class GFG{
     
// Function to print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
static void diagonalSumPerfectSquare(int[] arr,
                                     int N)
{   
  // Iterate for next N - 1 rows
  for(int i = 0; i < N; i++)
  {
    // Print the current row after
    // the left shift
    for(int j = 0; j < N; j++)
    {
      Console.Write(
      arr[(j + i) % 7] + " ");
    }
    Console.WriteLine();
  }
}
 
// Driver Code
public static void Main(String[] srgs)
{   
  // Given N
  int N = 7;
 
  int[] arr = new int[N];
 
  // Fill the array with elements
  // ranging from 1 to N
  for(int i = 0; i < N; i++)
  {
    arr[i] = i + 1;
  }
 
  // Function Call
  diagonalSumPerfectSquare(arr, N);
}
}
 
// This code is contributed by 29AjayKumar

chevron_right


Output: 

1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6






 

Time Complexity: O(N2)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up


If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.