In this article, we have take a look at how to print a rectangular matrix of size 2 * 3 with integers in C Programming Language with numbers from 1 to 6.

The pattern is as follows:

246
135

Table of contents

• Problem statement
• Approach to solve
• Implementation
• Output

Problem statement

In this problem, we have to print the following rectangular matrix of size 2*3:

246
135

Approach to solve the problem

Following is the approach to solve this problem:

• We can solve this problem by using 2D array.

• First we declare a variable a[2][3] where 2 represents the number of rows and 3 represents the number of columns. We also declare a variable count and initialise it with a value 1.

• Then we run two loops-

• Outer loop for the rows and inner for the columns.

• The positions of the elements in the matrix are as follows-

2-(0,0)  4-(0,1)  6-(0,2)
1-(1,0)  3-(1,1)  5-(1,2)

• In each of the columns the second coordinate of the element is same and the first coordinate varies by 1 or 0. So we assume the second coordinate value to be in the outer loop and the first coordinate value to be in the inner loop.

• The second coordinate values vary from 0 to 2 so the outer loop runs from 0 till 2.

• The first coordinate values vary from 0 to 1. The first element lies in (1,0), second in (0,0), third in (1,1) and so on.
  So the inner loop runs from 1 to 0 while decrementing its value by 1.
  Inside this loop we take assign j as row and i as column and assign the value of count and increment count by 1 in the next line.

• Initially the value of i=0, we enter the loop. The value of j=1 so a[j][i]=a[1][0] i.e. 1, then j=0 so a[j][i]=a[0][0] i.e. 2 as we had incremented the value of count.
  Now j=-1 which doesn't satisfy the condition of j>=0 so it exits the loop and cursor goes to next line due to the command "printf("\n")". This way the loop continues and we get the desired output.

Implementation

Following is the implementation of the code in c programming language-

#include<stdio.h>
int main()
{
	int i,j, count=1,a[2][3];
	for(i=0;i<3;i++)
	{
		for(j=1;j>=0;j--)
		{
			a[j][i]=count;
			count++;
		}
	}
	for(i=0;i<2;i++)
	{
		for(j=0;j<3;j++)
		{
			printf("%d",a[i][j]);
		}
		printf("\n");
	}
	return 0;
}

Output

The command to run the program is:

gcc code.c
./a.out

The output will be:

246
135

With this article at OpenGenus, you must have the complete idea of printing the given pattern.