TechplexEngineer
10/27/2013 - 9:10 PM

FizzBuzz solved using only bit twiddling. It essentially uses two deterministic finite automata for divisibility testing.

FizzBuzz solved using only bit twiddling. It essentially uses two deterministic finite automata for divisibility testing.

fizzbuzz.c
content_copyfile_download
#include <stdio.h>
int f0(unsigned int x) { return x? (x&(1<<31)? f1(x<<1) : f0(x<<1)) : 1; }
int f1(unsigned int x) { return x? (x&(1<<31)? f3(x<<1) : f2(x<<1)) : 0; }
int f2(unsigned int x) { return x? (x&(1<<31)? f0(x<<1) : f4(x<<1)) : 0; }
int f3(unsigned int x) { return x? (x&(1<<31)? f2(x<<1) : f1(x<<1)) : 0; }
int f4(unsigned int x) { return x? (x&(1<<31)? f4(x<<1) : f3(x<<1)) : 0; }

int t0(unsigned int x) { return x? (x&(1<<31)? t1(x<<1) : t0(x<<1)) : 1; }
int t1(unsigned int x) { return x? (x&(1<<31)? t0(x<<1) : t2(x<<1)) : 0; }
int t2(unsigned int x) { return x? (x&(1<<31)? t2(x<<1) : t1(x<<1)) : 0; }

int main(void) {
    unsigned int i;
    for(i=1; i <= 100; i++) {
        if(t0(i)) printf("Fizz");
        if(f0(i)) printf("Buzz");
        if(!(t0(i)|f0(i))) printf("%u",i);
        printf("\n");
    }
}