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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-loopnested within aninner-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
%2operation 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
spaceto 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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