Star pattern practicing on alignment control

Question

I am practicing to generate the star in the following result, but I was fail.

1.

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

2.

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

For my logic on this case, I am thinking that could be generated by using width alignment control on the loop, but the result shown that it is wrong...

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

With the code:

    int i,j;
    char ch='*';//created for using of alignment
    for(i=1;i<=5;i++)
    {
        for(j=5;j>=i;j--)
        {
            printf("%5c", ch);
        }
        printf("\n");
    }

For my logic, I assume that could be generated by using the method of alignment (%5c). However, the result is not as same as what I expected.

I have already succeeded to generate the star as shown as below:

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

With Code:

    int i,j;
    for(i=1;i<=5;i++)
    {
        for(j=5;j>=i;j--)
        {
            printf("*");
        }
        printf("\n");
    }

Do I have the wrong thinking on the method to make pattern 1 and 2? Or the right thinking with the wrong coding?

Answer 1

To align, you might be actually drawing a rectangle. In each row, you output a series of asterisks (possibly empty, i.e., zero asterisks) and a series of spaces (possibly empty, i.e., no spaces).

It is an ideal application of the (condition) ? (value_if_true) : (value_if_false) construct. You always draw the full rectangle, and pattern only changes due to condition: "is this a space or an asterisk?"

int i, j;
for(i=1;i<=5;i++)
{
    for(j=1;j<=5;j++)
    {
        // This generates a full rectangle
        // printf("%c", '*');
        // This generates an empty rectangle ;-)
        // printf("%c", ' ');
        // This generates a triangle
        printf("%c", (i >= j) ? '*' : ' ');
    }
    printf("\n");
}

There are two triangles oriented top-down governed by the condition (i >= j) or (i <= j); to flip the triangles horizontally you replace the increasing sequence with a decreasing sequence, i.e., (6-j) instead of (j):

   when j is       6-j is
       1              5
       2              4
      ...            ...
       5               1

so that the test is ((6-j) <= i).