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Printing patterns with numbers using nested for-loops

Adithya Challa

Overview

We will look into some of the printing patterns given below.

  • Nested-Loops help to solve printing pattern problems.
  • We basically have an outer-loop nested within an inner-loop.

Syntax of nested-for loop

for(initialization;condition;updation)
{
for(initialization;condition;updation)
{
// inner-loop staments
}
// outer-loop statements
}

Explanation

  • To work with these patterns it is important to understand the looping variable condition applied to it.
  • We consider the outer loop for running it the necessary number of times, and the inner loop is used to print the pattern.

Pattern-1

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

#include <stdio.h>
int main() {
  for(int i=1;i<=5;i++)
  {
    for(int j=1;j<=i;j++)
    {
     printf("%d ",j);
    }
    printf("\n");
  }
}

Pattern-2

1
2 2
3 3 3
4 4 4 4
5 5 5 5 5

#include <stdio.h>
int main() {
  for(int i=1;i<=5;i++)
  {
    for(int j=1;j<=i;j++)
    {
     printf("%d ",i);
    }
    printf("\n");
  }
}

Pattern-3

1 2 3 4 5
1 2 3 4
1 2 3
1 2
1

#include <stdio.h>
int main() {
  for(int i=5;i>=1;i--)
  {
    for(int j=1;j<=i;j++)
    {
       printf("%d ",j);
    }
    printf("\n");
  }
}

Pattern-4

1 2 3 4 5
2 3 4 5
3 4 5
4 5
5

#include <stdio.h>
int main() {
  for(int i=1;i<=5;i++)
  {
    for(int j=i;j<=5;j++)
    {
       printf("%d ",j);
    }
    printf("\n");
  }
}

Pattern-5

5
5 4
5 4 3
5 4 3 2
5 4 3 2 1

#include <stdio.h>
int main() {
  for(int i=5;i>=1;i--)
  {
    for(int j=5;j>=i;j--)
    {
       printf("%d ",j);
    }
    printf("\n");
  }
}

Pattern-6

1
1 0
1 0 1
1 0 1 0
1 0 1 0 1

  • This pattern is simply an extension of Pattern-1.
  • This pattern is obtained by performing %2operation on each element.
#include <stdio.h>
int main() {
  for(int i=1;i<=5;i++)
  {
    for(int j=1;j<=i;j++)
    {
     printf("%d ",j%2);
    }
    printf("\n");
  }
}

Pattern-7

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

#include<stdio.h>
int main()
{
	int i,j,k=1;
	for(i=1;i<=5;i++)
	{
		for(j=1;j<=i;j++)
		{
			printf("%d ",k++);
		}
		printf("\n");
	}
}

Pattern-8

1
0 1
0 1 0
1 0 1 0
1 0 1 0 1

  • We can get this pattern by performing %2 operation on Pattern-7.
#include<stdio.h>
int main()
{
	int i,j,k=1;
	for(i=1;i<=5;i++)
	{
		for(j=1;j<=i;j++)
		{
			printf("%d ",(k++)%2);
		}
		printf("\n");
	}
}

Pattern-9

  • Now, we will add the required number of spaces in a given pattern.
  • We will be adding another loop for adding the required number of spaces.

        1
     1 2
        1 2 3
  1 2 3 4
  1 2 3 4 5

#include <stdio.h>

int main()
{
   int i,j,k;
   for(i=1;i<=5;i++)
   {
         for(k=5;k>=i;k--)
         {
       printf(" ");
         }
       for(j=1;j<=i;j++)
       {
       printf("%d",j); 
       }
       printf("\n");
   }
}

Pattern 10

        1
      1   2
       1   2   3
  1   2   3   4
  1   2   3   4   5

  • We can get this pattern by adding a space to Pattern-9.
#include <stdio.h>

int main()
{
   int i,j,k;
   for(i=1;i<=5;i++)
   {
         for(k=5;k>=i;k--)
         {
       printf(" ");
         }
       for(j=1;j<=i;j++)
       {
       printf("%d ",j); 
       }
       printf("\n");
   }
}

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