A fractal in c - Sierpinsky triangle

Question

I am trying to code a project in c, that displays a fractal called Sierpinski fractal, (where the nodes are represented by '#'). So a 1-sierpinski triangle looks like :

##
#

a 2-sierpinski triangle

####
# #
##
# 

and so on... Here's a link to find what it looks like : http://fr.wikipedia.org/wiki/Triangle_de_Sierpiński

I was told it could be done without any loop, just by recursive method. So I tried something like :

//extracting the power of two's index
int puiss_2(int N){
    int i=0,j=1;
    for(i=0;i<N;i++){
        j=j*2;
        i++;
    }
    return j;
}

//the recursive method
void fractal(int N)
{
    int M;
    M= puiss_2(N);


    if(M==0){
        printf("##\n");
        printf("# ");
    }
    else{
        fractal(N-1);
        fractal(N-1);
        printf("\n");
        fractal(N-1);
        printf(" ");
    }
}

int main()
{
    int N;
    scanf("%d",&N);
    fractal(N);
}

Of course it didn't work because, when I jump to a line, I can't reverse it. So when I call it two times :

fractal(N-1); fractal(N-1);

two contiguous motives are not gathered one aside the other... Does anyone has an idea on how to make that ? Or perhaps I went completely wrong in my algo's design?

Answer 1

Here's some code that is perhaps complicated but recursive !

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void sierpinsky(int N, char c[1000]){
    int i=0,j,k,l,born;

    for(i=0;i<N;i++){printf("%c",c[i]);}
    printf("\n");

    if(N==1){}
    else{
        if((c[0]=='#')&&(c[1]=='#')&&(c[2]=='#')){
            for (j=0;2*j<N;j++){
                if(c[2*j]=='#'){
                    c[2*j]='#';c[2*j+1]=' ';
                }
                else{
                    c[2*j]=' ';c[2*j+1]=' ';
                }
            }
        }
        else if ((c[0]=='#')&&(c[1]!='#')&&(c[2]=='#')){
                for (j=0;4*j<N;j++){
                    if(c[4*j]=='#'){
                        c[4*j]='#';c[4*j+1]='#';c[4*j+2]=' ';c[4*j+3]=' ';
                    }
                    else{
                        c[4*j]=' ';c[4*j+1]=' ';c[4*j+2]=' ';c[4*j+3]=' ';
                    }
                }
            }
        else if ((c[0]=='#')&&(c[1]!='#')&&(c[2] !='#')){
                k=0;
            while(c[k+1] !='#'){k++;}
            born = k+1;
            j=0;

            while(j<N){
                if((c[j]=='#')&&(c[j+born]=='#')){
                for(l=0;l<born;l++){
                    c[j+l]='#';
                    }
                    j=j+born+1;
                }

                else if ((c[j]!='#')&&(c[j-1+born]=='#')&&(c[j-1+2*born] !='#'))
                {
                    c[j-1]='#';
                    for(l=0;l<born;l++){
                        c[j+l]='#';
                    }
                    j=j+born+1;
                }
                else{
                    c[j-1]= ' ';
                    c[j]=' ';
                    j++;
                }
            }
        }
        else if ((c[0] =='#')&&(c[1] =='#')&&(c[2] !='#')){
            for (j=0;4*j<N;j++){
                if(c[4*j]=='#'){
                    c[4*j]='#';c[4*j+1]=' ';c[4*j+2]=' ';c[4*j+3]=' ';
                }
                else{
                    c[4*j]=' ';c[4*j+1]=' ';c[4*j+2]=' ';c[4*j+3]=' ';
                }
            }

        }
            else{}

            sierpinsky(N-1, c);
        }
    }

int main()
{   int i,size;
    scanf("%d",&size);
    char c[1000];
    for(i=0;i<size;i++){c[i]='#';}
    for(i=size;i<1000;i++){c[i]='a';}
    sierpinsky(size, c);
}