Question:
Write a Program in Java to fill a square matrix of size ‘n*n” in a circular fashion (clockwise) with natural numbers from 1 to n*n, taking ‘n’ as input.
We will take a variable ‘k’ which will begin with 1 and will do the work of filling. i.e. for every cell, it will increase by 1. The below given processes will repeat till the value of ‘k’ becomes ‘n*n’
Write a Program in Java to fill a square matrix of size ‘n*n” in a circular fashion (clockwise) with natural numbers from 1 to n*n, taking ‘n’ as input.
For example: if n = 4, then n*n = 16, hence the array will be filled as given below
Note: This program is also known as Spiral Matrix
Solution:
import java.io.*;
class Circular_Matrix
{
public static void main(String args[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter the number of elements : ");
int n=Integer.parseInt(br.readLine());
int A[][]=new int[n][n];
int k=1, c1=0, c2=n-1, r1=0, r2=n-1;
while(k<=n*n)
{
for(int i=c1;i<=c2;i++)
{
A[r1][i]=k++;
}
for(int j=r1+1;j<=r2;j++)
{
A[j][c2]=k++;
}
for(int i=c2-1;i>=c1;i--)
{
A[r2][i]=k++;
}
for(int j=r2-1;j>=r1+1;j--)
{
A[j][c1]=k++;
}
c1++;
c2--;
r1++;
r2--;
}
/* Printing the Circular matrix */
System.out.println("The Circular Matrix is:");
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
System.out.print(A[i][j]+ "\t");
}
System.out.println();
}
}
}
Working:
We will take a variable ‘k’ which will begin with 1 and will do the work of filling. i.e. for every cell, it will increase by 1. The below given processes will repeat till the value of ‘k’ becomes ‘n*n’
C1 denotes the index of the column from where we have to begin. Hence its initial value will be 0.
C2 denotes the index of the column where we have to end. Hence its initial value will be ‘n-1′ (n is the size of the matrix).
R1 denotes the index of the row from where we have to begin. Hence its initial value will be 0.
R2 denotes the index of the row where we have to end. Hence its initial value will be ‘n-1′ (n is the size of the matrix).
The filling up of the matrix in circular fashion will consist of 4 different steps which will continue till the matrix is filled completely.
Step 1: We will fill the elements of Row 0 (R1), starting from Column 0 (C1) till ‘n-1′ (C2). The cells which will be filled are marked in the image above in yellow color.
The elements will be accessed as follows: A[R1][i], where ‘i’ will go from C1 to C2 (A[ ][ ] is the array)
Step 2: Now, we will fill the elements of Column ‘n-1′ (C2), starting from Row R1+1 till R2. The cells which will be filled are marked in the image above in grey color.
The elements will be accessed as follows: A[j][C2], where ‘j’ will go from R1+1 to R2 (A[ ][ ] is the array)
Step 3: Next we will fill the elements of Row ‘n-1′ (R2), starting from Column C2-1 till C1. The cells which will be filled are marked in the image above in green color.
The elements will be accessed as follows: A[R2][i], where ‘i’ will go from C2-1 to C1 (A[ ][ ] is the array)
Step 4: Now, we will fill the elements of Column C1, starting from Row R2-1 till R1+1. The cells which will be filled are marked in the image above in blue color.
The elements will be accessed as follows: A[j][C1], where ‘j’ will go from R2-1 to R1+1 (A[ ][ ] is the array)
The above 4 steps will now repeat with the inner matrix which is marked in white color in the above image. For the inner matrix,
C1 will increase by 1 i.e. it will be C1+1.
C2 will decrease by 1 i.e. it will be C2-1.
R1 will increase by 1 i.e. it will be R1+1.
R2 will decrease by 1 i.e. it will be R2-1.
The above processes will repeat till we have filled in ‘n*n’ values.
Output: