10 Challenging star pattern programs in C
Here are few challenging C program questions for you. Let’s see how many of them you would be able to write without seeing the program solution given below. These questions are to print patterns using asterisk(star) character. Comment below if you have tougher pattern questions.
Also read Number Pattern Programs
- Write a C program to print the following pattern:
* * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
- Write a C program to print the following pattern:
- Write C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program: - Write a C program to print the following pattern:
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Program:
*
* *
* * *
* * * *
Program:/* This is a simple mirror-image of a right angle triangle */
#include
int main() {
char prnt = '*';
int i, 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
}
return 0;
}
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#include
int main() {
char prnt = '*';
int i, 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;
}
}
return 0;
}
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#include
int main() {
char prnt = '*';
int i, 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); } else if ((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); } else if ((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");
}
return 0;
}
Download CodeExplanation:
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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#include
int main() {
char prnt = '*';
int i, 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");
}
return 0;
}
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#include
int main() {
char prnt = '*';
int i, 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--;
}
}
return 0;
}
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/*
This can be seen as two right angle triangles sharing the same base
which is modified by adding few extra shifting spaces
*/
#include
// This function controls the inner loop and the spacing
// factor guided by the outer loop index and the spacing index.
int triangle(int nos, int i) {
char prnt = '*';
int s, j;
for (s = nos; s >= 1; s--) { // Spacing factor
printf(" ");
}
for (j = 1; j <= i; j++) { //The inner loop
printf("%2c", prnt);
}
return 0;
}
int main() {
int i, 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--) {
int j = 1;
triangle(nos, i); // Inner loop construction
nos = nos - j; // Spacing factor
printf("\n");
}
return 0;
}
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#include
int main() {
char prnt = '*';
int i, 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");
}
return 0;
}
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#include
int main() {
char prnt = '*';
int i, 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");
}
return 0;
}
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#include
/*
* 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.
*
*/
int triangle(int nos, int i, int skip) {
char prnt = '*';
int s, 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);
}
return 0;
}
int main() {
int i, 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");
}
return 0;
}
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#include
/*
* 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.
*
*/
int triangle(int nos, int i, int skip) {
char prnt = '*';
int s, 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); } else if (j == 1 || j == i) { printf("%2c", prnt); } else { printf(" "); } } return 0; } int main() { int i, 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;
}
return 0;
}
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