Fractal via bit twiddling

Posted on by John

The Sierpinski triangle has come up several times on this blog. Here’s yet another way to produce a Sierpinski triangle, this time by taking the bitwise-and of two integers. The ith bit of x&y is 1 if and only if the ith bit of x and the ith bit of y are both 1.

The following C program prints an asterisk when the bitwise-and of i and j is zero.

    #include <stdio.h>

    int main() {
        int N = 32;
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++)
                printf("%c", (i&j) ? ' ' : '*');
            printf("\n");
        }
    }

Here’s a screenshot of the output.

screen shot that looks like Sierpinski triangle

More posts on Sierpinski triangle

2 thoughts on “Fractal via bit twiddling”

  1. John Mocambique

    Thanks John. That’s an interesting pattern.
    Below is a c# implementation:

     static void Main(string[] args)
        {
            int N = 32;
            for (int i = 0; i < N; i++)
            {
                for (int j = 0; j < N; j++)
                {
                    Console.Write((i&j) == 0 ? '*' : ' ');
                }
                Console.WriteLine();
            }
            Console.ReadLine();
        }

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