Class 10 Computer Science Chapter 5 Nested Loops In C

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Class 10 Computer Science Chapter 5 Nested Loops In C

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NESTED LOOPS IN C

Chapter – 5

COMPUTER SCIENCE

1. Write C programs to display the following pattern using nested loop construct.

(a) 1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

Solution: Code:

``````#include <stdio.h>

int main() {

int rows, cols;

printf("Enter the number of rows: ");

scanf("%d", &rows);

printf("Enter the number of columns: ");

scanf("%d", &cols);

for (int i = 1; i <= rows; i++) {

for (int j = 1; j <= cols; j++) {

printf("%d ", j);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of rows: 5

Enter the number of columns: 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3``````

(b) 1 2 1

1 2 1

1 2 1

1 2 1

1 2 1

Solution:

Code:

``````#include <stdio.h>

int main() {

int rows, cols;

printf("Enter the number of rows: ");

scanf("%d", &rows);

printf("Enter the number of columns: ");

scanf("%d", &cols);

for (int i = 1; i <= rows; i++) {

for (int j = 1; j <= cols; j++) {

printf("%d ", j % 2 + 1);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of rows: 5

Enter the number of columns: 3

1 2 1

1 2 1

1 2 1

1 2 1

1 2 1``````

(c) 4 3 2 1

4 3 2 1

4 3 2 1

4 3 2 1

4 3 2 1

Solution: Code:

``````#include <stdio.h>

int main() {

int rows, cols;

printf("Enter the number of rows: ");

scanf("%d", &rows);

printf("Enter the number of columns: ");

scanf("%d", &cols);

for (int i = 1; i <= rows; i++) {

for (int j = cols; j >= 1; j--) {

printf("%d ", j);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of rows: 5

Enter the number of columns: 4

4 3 2 1

4 3 2 1

4 3 2 1

4 3 2 1

4 3 2 1``````

(d)                2

2  3  4

2 3 4 5 6

Solution: Code:

``````#include <stdio.h>

int main() {

int rows;

printf("Enter the number of rows: ");

scanf("%d", &rows);

for (int i = 1; i <= rows; i++) {

// Print spaces

for (int space = 1; space <= rows - i; space++) {

printf("  ");

}

// Print numbers

int num = i + 1;

for (int j = 1; j <= i; j++) {

printf("%d ", num);

num++;

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of rows: 3

2

2 3 4

2 3 4 5 6``````

(e)

``````1

1 2 1

1 2 3 2 1``````

Solution: Code:

``````#include <stdio.h>

int main() {

int rows;

printf("Enter the number of rows: ");

scanf("%d", &rows);

for (int i = 1; i <= rows; i++) {

// Print spaces

for (int space = 1; space <= rows - i; space++) {

printf("  ");

}

// Print increasing numbers

for (int j = 1; j <= i; j++) {

printf("%d ", j);

}

// Print decreasing numbers

for (int k = i - 1; k >= 1; k--) {

printf("%d ", k);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of rows: 3

1

1 2 1

1 2 3 2 1``````

(f)

``````      *

* * *

* * * * *

* * * * * * *

* * * * *

* * *

*``````

Solution:

Code:

``````#include <stdio.h>

int main() {

int i, j, k, p;

for (k = 1; k <= 4; k++) {

for (i = 1; i <= 4 - k; i++) {

printf("  ");  // Print spaces

}

for (j = 1; j <= k; j++) {

printf(" * ");  // Print asterisks

}

for (p = 1; p <= k - 1; p++) {

printf(" * ");  // Print additional asterisks

}

printf("\n");

}

for (k = 3; k >= 1; k--) {

for (i = 1; i <= 4 - k; i++) {

printf("  ");  // Print spaces

}

for (j = 1; j <= k; j++) {

printf(" * ");  // Print asterisks

}

for (p = 1; p <= k - 1; p++) {

printf(" * ");  // Print additional asterisks

}

printf("\n");

}

return 0;

}``````

Output:

``````      *

* * *

* * * * *

* * * * * * *

* * * * *

* * *

*``````

(g)

``````X             x

X X         x x

X X X     x x x

X X X X x x x x

X X X X X X X X``````

Solution: Code:

``````#include <stdio.h>

int main() {

int i, j, k, p = 4;

for (k = 1; k <= 5; k++) {

for (i = 1; i <= k; i++) {

printf(" X ");

}

for (j = 1; j <= p; j++) {

printf("   ");  // Print spaces

}

p--;

for (i = 1; i <= k; i++) {

printf(" x ");

}

printf("\n");

}

return 0;

}``````

Output:

`````` X             x

X X         x x

X X X     x x x

X X X X x x x x

X X X X X X X X``````

2. Modify the solution of question no. 1 to accept the number of lines as the input. The program should make the display pattern accordingly.

Solution:

(a) Code:

``````#include<stdio.h>

int main()

{

int i, j, n;

printf(“Enter the number of lines : “);

scanf(“%d”, &n);

for(i=1; i<=n ; i++)

{

for(j=1 ; j<=3; j++)

{

printf(“%d “,j);

}

printf(“\n”);

}

return(0);

}``````

Output:

``````Enter the number of lines : 5

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3``````

(b) Code:

``````#include <stdio.h>

int main()

{

int i, j, n;

printf("Enter the number of lines : ");

scanf("%d", &n);

for (i = 1; i <= n; i++)

{

for (j = 1; j <= 2; j++)

{

printf(" %d ", j);

}

for (j = 2; j >= 1; j--)

{

printf(" %d", j);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of lines : 3

1 2 2 1

1 2 2 1

1 2 2 1``````

(c) Code:

``````#include<stdio.h>

int main()

{

int i, j, n;

printf(“Enter the number of lines : “);

scanf(“%d “,j);

}

printf(“\n”);

}

return(0);

}``````

Output:

``````Enter the number of lines: 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5``````

(d) Code:

``````#include <stdio.h>

int main()

{

int i, j, k, m = 2, n;

printf("Enter the number of lines: ");

scanf("%d", &n);

for (i = n; i >= 1; i--)

{

for (j = 1; j < i; j++)

{

printf(" ");

}

for (k = 2; k <= m; k++)

{

printf(" %d", k);

}

m = m + 2;

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of lines: 8

2 4 6 8 10 12 14 16

2 4 6 8 10 12 14

2 4 6 8 10 12

2 4 6 8 10

2 4 6 8

2 4 6

2 4

2``````

(e) Code:

``````#include <stdio.h>

int main()

{

int i, j, k, p, n;

printf("Enter the number of lines: ");

scanf("%d", &n);

for (k = 1; k <= n; k++)

{

for (i = 1; i <= n - k; i++)

{

printf(" ");

}

for (j = 1; j <= k; j++)

{

printf(" %d ", j);

}

for (p = k - 1; p >= 1; p--)

{

printf(" %d ", p);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of lines: 8

1

1 2 1

1 2 3 2 1

1 2 3 4 3 2 1

1 2 3 4 5 4 3 2 1

1 2 3 4 5 6 5 4 3 2 1

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

1 2 3 4 5 6 7 8 7 6 5 4 3 2 1``````

(f) Code:

``````#include <stdio.h>

int main()

{

int i, j, k, p, a, n, b;

printf("Enter the odd number of lines: ");

scanf("%d", &n);

a = n / 2 + 1;

b = a - 1;

for (k = 1; k <= a; k++)

{

for (i = 1; i <= a - k; i++)

{

printf(" ");

}

for (j = 1; j <= k; j++)

{

printf(" * ");

}

for (p = 1; p <= k - 1; p++)

{

printf(" * ");

}

printf("\n");

}

for (k = b; k >= 1; k--)

{

for (i = 1; i <= a - k; i++)

{

printf(" ");

}

for (j = 1; j <= k; j++)

{

printf(" * ");

}

for (p = 1; p <= k - 1; p++)

{

printf(" * ");

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the odd number of lines: 7

*

* * *

* * * * *

* * * * * * *

* * * * * * * * *

* * * * * * * * * * *

* * * * * * * * * * * * *

* * * * * * * * * * *

* * * * * * * * * *

* * * * * * * *

* * * * * * *

* * * * * *

* * * * *

* * * *

* * *

* *

*``````

(g) Code:

``````#include<stdio.h>

int main()

{

int i, j, k, p, n;

printf(“Enter the number of lines : “);

scanf(“%d”, &n);

p = n-1;

for(k=1; k<=n; k++)

{

for(i=1; i<=k; i++)

{

printf(“ X “);

}

for(j=1; j<=p; j++)

{

printf(“ “);

}

p—;

for(i=1; i<=k; i++)

{

printf(“ X “);

}

printf(“\n”),

}

return 0;

}``````

Output:

``````Enter the number of lines: 4

X     X

X X   X X

X X X X X X

X X   X X``````

3. Extend the programs of Example 5.6 and Example 5.7 to make it dynamic by accepting the number of lines as input from the keyboard.

Solution: Example 5.6: Code:

``````#include <stdio.h>

int main()

{

int i, j, k, n;

printf("Enter the number of lines: ");

scanf("%d", &n);

for (k = 1; k <= n; k++)

{

for (i = 1; i <= (n - k); i++)

{

printf(" ");

}

for (j = 1; j <= (2 * k - 1); j++)

{

printf(" X ");

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of lines: 10

X

X X

X X X

X X X X

X X X X X

X X X X X X

X X X X X X X

X X X X X X X X

X X X X X X X X X

X X X X X X X X X X``````

Example: 5.7: Code:

``````#include <stdio.h>

int main()

{

int i, j, k, n;

printf("Enter the number of lines: ");

scanf("%d", &n);

for (k = 1; k <= n; k++)

{

for (i = 1; i <= (n - k); i++)

{

printf(" ");

}

for (j = 1; j <= (2 * k - 1); j++)

{

printf(" %d ", j);

}

printf("\n");

}

return 0;

}``````

Output:

``````Enter the number of lines: 4

1

1 2 3

4 5 6 7 8

9 10 11 12 13 14 15``````

4. What is nested loop? Why do one use nested loops in our programs?

Ans. A nested loop is a loop inside another loop. Although all kinds of loops can be nested, the most common nested loop involves loops.

These loops are particularly useful when displaying multidimensional data. When using these loops, the first iteration of the first loop will initialize, followed by the second loop.

Example: Any number of loops can be defined inside another loop, i, e., There is no restriction for defining any number of loops. The nesting level can be defined at n times.

5. Do we need to use same type of loops as outer and inner loops? Justify your answer with some code segments.

Ans. It is not mandatory to nest the same type of loop. We can put a for loop inside a while loop or a do-while loop inside a for loop.

Example: Code:

``````#include <stdio.h>

int main() {

int rows = 5;

for (int i = 1; i <= rows; i++) {

int j = 1;

while (j <= i) {

printf("%d ", j);

j++;

}

printf("\n");

}

return 0;

}``````

Output:

``````1

1 2

1 2 3

1 2 3 4

1 2 3 4 5``````

6. Can we put third loop inside the inner loop of a nested loop construct? Write a C program to justify your answer.

Ans. Yes, you can place a third loop inside the inner loop of a nested loop construct.

Example:

Code:

``````#include <stdio.h>

int main() {

int i, j, k;

for (i = 1; i <= 3; i++) {

for (j = 1; j <= 3; j++) {

for (k = 1; k <= 3; k++) {

printf("%d %d %d\n", i, j, k);

}

}

}

return 0;

}``````

Output:

``````1 1 1

1 1 2

1 1 3

1 2 1

1 2 2

1 2 3

1 3 1

1 3 2

1 3 3

2 1 1

2 1 2

2 1 3

2 2 1

2 2 2

2 2 3

2 3 1

2 3 2

2 3 3

3 1 1

3 1 2

3 1 3

3 2 1

3 2 2

3 2 3

3 3 1

3 3 2

3 3 3``````
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