1. Write a C program to print the following pattern:
  2. *
  3. * *
  4. * * *
  5. * * * *
  1. Write a C program to print the following pattern:
  2. * *
  3. * * * * * *
  4. * * * * * * * * * *
  5. * * * * * * * * * * *
  1. Write a C program to print the following pattern:
  2. * *
  3. * *
  4. * * * *
  5. * * * *
  6. * * * * *
  7. * * * *
  8. * * * *
  9. * *
  10. * *
  1. Write a C program to print the following pattern:
  2. * * * * *
  3. * * * *
  4. * * *
  5. * *
  6. *
  7. * *
  8. * * *
  9. * * * *
  10. * * * * *
  1. Write a C program to print the following pattern:
  2. *
  3. * * *
  4. * * * * *
  5. * * * * * * *
  6. * * * * * * * * *
  7. * * * * * * *
  8. * * * * *
  9. * * *
  10. *
  11. * * *
  12. * * * * *
  1. Write a C program to print the following pattern:
  2. *
  3. * *
  4. * * *
  5. * * * *
  6. * * *
  7. * *
  8. *
  1. Write a C program to print the following pattern:
  2. * * * * * * * * *
  3. * * * * * * * *
  4. * * * * * *
  5. * * * *
  6. * *
  7. * * * *
  8. * * * * * *
  9. * * * * * * * *
  10. * * * * * * * * *
  1. Write a C program to print the following pattern:
  2. * * * * * * * * * * * * * * * * *
  3. * * * * * * * * * * * * * *
  4. * * * * * * * * * *
  5. * * * * * *
  6. * * * * * * * * *
  7. * * * * * * *
  8. * * * * *
  9. * * *
  10. *
  1. Write a C program to print the following pattern:
  2. *
  3. * * *
  4. * * * * *
  5. * * * * * * *
  6. * *
  7. * * * *
  8. * * * * * *
  9. * * * * * * *
  10. * * * * * *
  11. * * * *
  12. * *
  13. * * * * * * *
  14. * * * * *
  15. * * *
  16. *
  1. Write a C program to print the following pattern:
  2. * * * * * * * * * * * * * * * * * * * * * * * * *
  3. * * * * * *
  4. * * * * * *
  5. * * * * * *
  6. * * *
  7. * * * * * *
  8. * * * * * *
  9. * * * * * *
  10. * * * * * * * * * * * * * * * * * * * * * * * * *
  1. Write a C program to print the following pattern:

*

* *

* * *

* * * *

Program:

?

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18 / /* This is a simple mirror-image of a right angle triangle */
#include <stdio.h>
intmain() {
charprnt = '*';
inti, j, nos = 4, s;
for(i = 1; i <= 5; i++) {
for(s = nos; s >= 1; s--) { // Spacing factor
printf(" ");
}
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
printf("\n");
--nos; // Controls the spacing factor
}
return0;
}

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  1. Write C program to print the following pattern:
  2. * *
  3. * * * * * *
  4. * * * * * * * * * *

* * * * * * * * * * *

Program:

?

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29 / #include<stdio.h>
intmain() {
charprnt = '*';
inti, j, k, s, c = 1, nos = 9;
for(i = 1; c <= 4; i++) {
// As we want to print the columns in odd sequence viz. 1,3,5,.etc
if((i % 2) != 0) {
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for(s = nos; s >= 1; s--) { //The spacing factor
if(c == 4 & s == 1) {
break;
}
printf(" ");
}
for(k = 1; k <= i; k++) {
if(c == 4 & k == 5) {
break;
}
printf("%2c", prnt);
}
printf("\n");
nos = nos - 4; // controls the spacing factor
++c;
}
}
return0;
}

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  1. Write C program to print the following pattern:
  2. * *
  3. * *
  4. * * * *
  5. * * * *
  6. * * * * *
  7. * * * *
  8. * * * *
  9. * *

* *

Program:

?

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62 / #include<stdio.h>
intmain() {
charprnt = '*';
inti, j, k, s, p, r, nos = 7;
for(i = 1; i <= 5; i++) {
for(j = 1; j <= i; j++) {
if((i % 2) != 0 & (j % 2) != 0) {
printf("%3c", prnt);
}
elseif((i % 2) == 0 & (j % 2) == 0) {
printf("%3c", prnt);
}
else{
printf(" ");
}
}
for(s = nos; s >= 1; s--) { // for the spacing factor
printf(" ");
}
for(k = 1; k <= i; k++) { //Joining seperate figures
if(i == 5 & k == 1) {
continue;
}
if((k % 2) != 0) {
printf("%3c", prnt);
}
else{
printf(" ");
}
}
printf("\n");
nos = nos - 2; // space control
} nos = 1; // remaining half..
for(p = 4; p >= 1; p--) {
for(r = 1; r <= p; r++) {
if((p % 2) != 0 & (r % 2) != 0) {
printf("%3c", prnt);
}
elseif((p % 2) == 0 & (r % 2) == 0) {
printf("%3c", prnt);
}
else{
printf(" ");
}
}
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(k = 1; k <= p; k++) {
if((k % 2) != 0) {
printf("%3c", prnt);
}
else{
printf(" ");
}
}
nos = nos + 2; // space control
printf("\n");
}
return0;
}

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Explanation:
This can be seen as an inverted diamond composed of stars. It can be noted that the composition of this figure follows sequential pattern of consecutive stars and spaces.

In case of odd row number, the odd column positions will be filled up with ‘*’, else a space will be spaced and vice-versa in case of even numbered row.

In order to achieve this we will construct four different right angle triangles

aligned as per the requirement.
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  1. Write a C program to print the following pattern:
  2. * * * * *
  3. * * * *
  4. * * *
  5. * *
  6. *
  7. * *
  8. * * *
  9. * * * *

* * * * *

Program:

?

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36 / #include<stdio.h>
intmain() {
charprnt = '*';
inti, j, s, nos = 0;
for(i = 9; i >= 1; (i = i - 2)) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
if((i % 2) != 0 & (j % 2) != 0) {
printf("%2c", prnt);
} else{
printf(" ");
}
}
printf("\n");
nos++;
}
nos = 3;
for(i = 3; i <= 9; (i = i + 2)) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
if((i % 2) != 0 & (j % 2) != 0) {
printf("%2c", prnt);
} else{
printf(" ");
}
}
nos--;
printf("\n");
}
return0;
}

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  1. Write a C program to print the following pattern:
  2. *
  3. * * *
  4. * * * * *
  5. * * * * * * *
  6. * * * * * * * * *
  7. * * * * * * *
  8. * * * * *
  9. * * *
  10. *
  11. * * *

* * * * *

Program:

?

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49 / #include<stdio.h>
intmain() {
charprnt = '*';
inti, j, k, s, nos = 4;
for(i = 1; i <= 5; i++) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for(k = 1; k <= (i - 1); k++) {
if(i == 1) { continue;
}
printf("%2c", prnt);
}
printf("\n"); nos--;
}
nos = 1;
for(i = 4; i >= 1; i--) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for(k = 1; k <= (i - 1); k++) {
printf("%2c", prnt);
}
nos++;
printf("\n");
}
nos = 3;
for(i = 2; i <= 5; i++) {
if((i % 2) != 0) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
}
if((i % 2) != 0) {
printf("\n");
nos--;
}
}
return0;
}

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  1. Write a C program to print the following pattern:
  2. *
  3. * *
  4. * * *
  5. * * * *
  6. * * *
  7. * *

*

Program:

?

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35 / /*
This can be seen as two right angle triangles sharing the same base
which is modified by adding few extra shifting spaces
*/
#include <stdio.h>
// This function controls the inner loop and the spacing
// factor guided by the outer loop index and the spacing index.
inttriangle(intnos, inti) {
charprnt = '*';
ints, j;
for(s = nos; s >= 1; s--) { // Spacing factor
printf(" ");
}
for(j = 1; j <= i; j++) { //The inner loop
printf("%2c", prnt);
}
return0;
}
intmain() {
inti, nos = 5;
//draws the upper triangle
for(i = 1; i <= 4; i++) {
triangle(nos, i); //Inner loop construction
nos++; // Increments the spacing factor
printf("\n"); }
nos = 7; //Draws the lower triangle skipping its base.
for(i = 3; i >= 1; i--) {
intj = 1;
triangle(nos, i); // Inner loop construction
nos = nos - j; // Spacing factor
printf("\n");
}
return0;
}

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  1. Write a C program to print the following pattern:
  2. * * * * * * * * *
  3. * * * * * * * *
  4. * * * * * *
  5. * * * *
  6. * *
  7. * * * *
  8. * * * * * *
  9. * * * * * * * *

* * * * * * * * *

Program:

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40 / #include <stdio.h>
intmain() {
charprnt = '*';
inti, j, k, s, nos = -1;
for(i = 5; i >= 1; i--) {
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(k = 1; k <= i; k++) {
if(i == 5 & k == 5) {
continue;
}
printf("%2c", prnt);
}
nos = nos + 2;
printf("\n");
}
nos = 5;
for(i = 2; i <= 5; i++) {
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(k = 1; k <= i; k++) {
if(i == 5 & k == 5) {
break;
}
printf("%2c", prnt);
}
nos = nos - 2;
printf("\n");
}
return0;
}

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  1. Write a C program to print the following pattern:
  2. * * * * * * * * * * * * * * * * *
  3. * * * * * * * * * * * * * *
  4. * * * * * * * * * *
  5. * * * * * *
  6. * * * * * * * * *
  7. * * * * * * *
  8. * * * * *
  9. * * *

*

Program:

?

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38 / #include <stdio.h>
intmain() {
charprnt = '*';
inti, j, k, s, sp, nos = 0, nosp = -1;
for(i = 9; i >= 3; (i = i - 2)) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
for(sp = nosp; sp >= 1; sp--) {
printf(" ");
}
for(k = 1; k <= i; k++) {
if(i == 9 & k == 1) {
continue;
}
printf("%2c", prnt);
}
nos++;
nosp = nosp + 2;
printf("\n");
}
nos = 4;
for(i = 9; i >= 1; (i = i - 2)) {
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
printf("%2c", prnt);
}
nos++;
printf("\n");
}
return0;
}

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  1. Write a C program to print the following pattern:
  2. *
  3. * * *
  4. * * * * *
  5. * * * * * * *
  6. * *
  7. * * * *
  8. * * * * * *
  9. * * * * * * *
  10. * * * * * *
  11. * * * *
  12. * *
  13. * * * * * * *
  14. * * * * *
  15. * * *

*

Program:

?

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54 / #include <stdio.h>
/*
* nos = Num. of spaces required in the triangle.
* i = Counter for the num. of charcters to print in each row
* skip= A flag for checking whether to
* skip a character in a row.
*
*/
inttriangle(intnos, inti, intskip) {
charprnt = '*';
ints, j;
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
if(skip != 0) {
if(i == 4 & j == 1) {
continue;
}
}
printf("%2c", prnt);
}
return0;
}
intmain() {
inti, nos = 4;
for(i = 1; i <= 7; (i = i + 2)) {
triangle(nos, i, 0);
nos--;
printf("\n");
}
nos = 5;
for(i = 1; i <= 4; i++) {
triangle(1, i, 0); //one space needed in each case of the formation
triangle(nos, i, 1); //skip printing one star in the last row.
nos = nos - 2;
printf("\n");
}
nos = 1;
for(i = 3; i >= 1; i--) {
triangle(1, i, 0);
triangle(nos, i, 0);
nos = nos + 2;
printf("\n");
}
nos = 1;
for(i = 7; i >= 1; (i = i - 2)) {
triangle(nos, i, 0);
nos++;
printf("\n");
}
return0;
}

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  1. Write a C program to print the following pattern:
  2. * * * * * * * * * * * * * * * * * * * * * * * * *
  3. * * * * * *
  4. * * * * * *
  5. * * * * * *
  6. * * *
  7. * * * * * *
  8. * * * * * *
  9. * * * * * *

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

Program:

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54 / #include <stdio.h>
/*
* nos = Num. of spaces required in the triangle.
* i = Counter for the num. of characters to print in each row
* skip= A flag for check whether to
* skip a character in a row.
*
*/
inttriangle(intnos, inti, intskip) {
charprnt = '*';
ints, j;
for(s = nos; s >= 1; s--) {
printf(" ");
}
for(j = 1; j <= i; j++) {
if(skip != 0) {
if(i == 9 & j == 1) {
continue;
}
}
if(i == 1 || i == 9) {
printf("%2c", prnt);
}
elseif(j == 1 || j == i) {
printf("%2c", prnt);
} else{
printf(" ");
} }
return0; }
intmain() {
inti, nos = 0, nosp = -1, nbsp = -1;
for(i = 9; i >= 1; (i = i - 2)) {
triangle(nos, i, 0);
triangle(nosp, i, 1);
triangle(nbsp, i, 1);
printf("\n");
nos++;
nosp = nosp + 2;
nbsp = nbsp + 2;
}
nos = 3, nosp = 5, nbsp = 5;
for(i = 3; i <= 9; (i = i + 2)) {
triangle(nos, i, 0);
triangle(nosp, i, 1);
triangle(nbsp, i, 1);
printf("\n");
nos--;
nosp = nosp - 2;
nbsp = nbsp - 2;
}
return0;
}