Simple C question *shape* (Diamond with an empty box inside)

[SM-Count]

Since you're not allowed to have jump conditions, the easiest way to do it with the code you already have is to break it into 4 different for loops rather than the 2 that you have.

First thing you'd have to do is have a variable that tells you how many holes you need Max(N-2,0) or just n-2 if you can assume N has min 3.

Then you can do a bit of trickery to know when you need holes. Implement a counter variable that starts at zero and increments every time you print a *. Wrap your first loop in a while statement that runs until the counter variable is equal to N (i.e. the number of starts you printed is equal to N which starts the empty box).

The second/third part of the code is simple, it's basically what you have but with empty boxes, since you have a variable with the box height/width (n-2) modify your code so that you draw n-2 spaces after stars starting with 2 stars until you get to one above the center. Then draw the bottom part of the with the n-2 spaces in the middle.

Last part of the code is like the first part, wrap it in a while loop with a counter, exit the while loop when your counter = 1.

[NuclearSilo]

first analyse:
- n = half size of the box
- center index = [2;2] <=> [(n+1)/2;(n+1)/2]
- this box is symmetric, you can create small task then just do the "mirror"
- it's hard if you print directly the "*"

algorithm (not in C):
- create an array of arrays size n*2-1 x n*2-1 (n=5 => size = 9)
- fill all the arrays with "*"
- fill the top-left corner a blank triangle

for (i=0; i<n-1; i++)
{ --rows
for (i=0; j<n-i-1; i++) --column
fill blank
}

- repeat for all corner
- fill the middle with blank where
index range = 0 -> n-1
start position = [(n+1)/2 ; (n+1)/2
end position = [(n-1) - (n+1)/2 ; (n-1) - (n+1)/2]

[NuclearSilo]

nevermind about what I said.

check if the current point is inside the zone [(n+1)/2 ; 2(n-1) - (n+1)/2], if yes then do not print it

[SM-Count]

I'm at school, I'll edit this post when I get home in 1.5 hrs with some concrete code, hopefully that'll help.

[SM-Count]

http://codepad.org/e0L2crAN

Here you go, it's a shell for what you should be thinking of. Obviously it's more verbose than you need so you can change the var names to i/j/z/whatever but it should be pretty clear, if it's not you can im me or something, just shoot me a pm. I've obviously commented out the part you should be trying to do, but just follow my first code block example. You'll have to draw the bottom of the diamond as well.

[NuclearSilo]

http://codepad.org/vUSSC2TY

Spoiler!

Code: Select all

#include <stdio.h>

void main()
{
    int N = 5;
    int c = N-1;   //center
    int end = 2*(N-1);
    int half = (N+1)/2;
    int i, j;

    for (i=0; i< end+1 ; i++) 
    {
      //draw blank
      for (j=0; j<fabs(c-i) ; j++) printf("   ");

    //draw *
    for (j=fabs(c-i); j <= (end-fabs(c-i)) ; j++)
    {
       //blank box condition
       if (i >= half
        && j >= half
        && i <= (end-half)
        && j <= (end-half))
          printf("   "); 
       else
          printf(" * ");
     }


      //line break
      printf("\r\n");
     }
}

Replace fabs for optimization, but it's less understandable than the first
http://codepad.org/N4I72K7C

[SM-Count]

He can't use if statements nuke :/ That's why my code looks retarded lol.

[SM-Count]

Oh, if you can use if statements use nuke's code, my code is downright retarded because I had to do a workaround for no ifs. fabs is just absolute value, just say you googled it.