How to print diamond shape pattern in C? - c

Question

how to print diamond shape pattern like this?
the pattern isn't commonly using "*" only but it't modified with "."
The input:
5
The output:
....0.....
...000....
..0.0.0...
.0..0..0..
000000000.
.0..0..0..
..0.0.0...
...000....
....0.....
Here my code:
#include <stdio.h>
int main()
{
int number, i, k, count = 1;
scanf("%d", &number);
count = number - 1;
for (k = 1; k <= number; k++)
{
for (i = 1; i <= count; i++)
printf(".");
count--;
for (i = 1; i <= 2 * k - 1; i++)
printf("0");
printf("\n");
}
count = 1;
for (k = 1; k <= number - 1; k++)
{
for (i = 1; i <= count; i++)
printf(".");
count++;
for (i = 1 ; i <= 2 *(number - k)- 1; i++)
printf("0");
printf("\n");
}
return 0;
}

Answer 1

Think of a Cartesian coordinate system whose origin is the center of symmetry of the diamond pattern to be drawn. The 0s reside in the lines y = x + (n - 1), y = -x + (n - 1), y = x - (n - 1), y = -x - (n - 1), and the axes y = 0, x = 0. Then we can easily codify this information to draw the diamond pattern:
#include <stdio.h>
void print_diamond (int n)
{
--n;
for (int y = n; y >= -n; --y) {
for (int x = -n; x <= n; ++x)
putchar(x == 0 || y == 0
|| y == x + n || y == -x + n || y == x - n || y == -x - n
? '0' : '.');
putchar('\n');
}
}
int main (void)
{
print_diamond(5);
}
The line
|| y == x + n || y == -x + n || y == x - n || y == -x - n
can be reduced to
|| abs(x - y) == n || abs(x + y) == n
after stdlib.h is #included.

Answer 2

I rapidly wrote this for fun keeping your approach, however this is not the most efficient nor the most elegant way of doing this
int number, i, k, count = 1;
scanf("%d", &number);
for (k = 0; k < number-1; k++){
i = 1;
while(i < number-k){
printf(".");
i++;
}
printf("0");
i++;
while(i< number){
printf(".");
i++;
}
if(i==number){
printf("0");
i++;
}
while(i<number+k){
printf(".");
i++;
}
if(k>0){
printf("0");
i++;
}
while(i<number*2){
printf(".");
i++;
}
printf("\n");
}
You will probably find better answers if you search a bit deeper
Also, you can just
void printpir(int number, int k){
int i = 1;
while(i < number-k){
printf(".");
i++;
}
printf("0");
i++;
while(i< number){
printf(".");
i++;
}
if(i==number){
printf("0");
i++;
}
while(i<number+k){
printf(".");
i++;
}
if(k>0){
printf("0");
i++;
}
while(i<number*2){
printf(".");
i++;
}
printf("\n");
}
int main(int argc, char** argv) {
int number, i, k, count = 1;
scanf("%d", &number);
for (k = 0; k < number-1; k++){
printpir(number, k);
}
for(i= 1; i<number*2; i++){
printf("0");
}
printf("\n");
k--;
for (; k >= 0; k--)
{
printpir(number, k);
}
return 0;
}

Answer 3

Here is a solution with only 2 for-loops. I can explain my answer if you need it.
Hope you can understand...
#include <stdio.h>
#include <stdlib.h>
int main() {
int n, _i, i, _j, j, k;
printf("Number: ");
scanf("%d", &n);
for (_i = 1; _i < 2*n; _i++) {
i = abs(_i - n);
for (_j = 1; _j <= 2*n; _j++) {
j = abs(_j - n);
if (j < n-i) {
if (i==0 || j == 0 || (i+j==n-1)) {
printf("0");
} else {
printf(".");
}
} else {
printf(".");
}
}
printf("\n");
}
return 0;
}