FizzBuzz solved using only bit twiddling

Re:FizzBuzz solved using only bit twiddling by waps on Monday October 28, 2013 @03:39PM (#6629277)

I love this answer. It's like "yes I know and hate fizzbuzz, also you don't know crap", but in code.

Poetry.

In python:

  from itertools import *
  fizz = cycle(["", "", "fizz"])
  buzz = cycle([""] * 4 + ["buzz"])
  fizzbuzz = izip(fizz, buzz, xrange(1,100))
  print "\n".join(map(lambda (x,y,z): x+y if x or y else str(z), fizzbuzz))

(bugs found by marekmroz fixed)

Can you do [1,2,3] * 4 in haskell as well ?

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Re:FizzBuzz solved using only bit twiddling by marekmroz on Monday October 28, 2013 @04:00PM (#6629372)

Funny thing is, as soon as I saw OP's solution I fired up IPython to do the same thing in Python. :)

Admittedly mine was not as elegant as I haven't thought of using izip with 3 iterables. Still, I don't think that lambda and islice is really needed... xrange(100) guarantees a finite number of iterations already.

   from itertools import *
   fizzer = cycle(['Fizz','',''])
   buzzer = cycle(['Buzz','','','',''])
   fizzbuzzer = izip(xrange(100), fizzer, buzzer)
   for f in fizzbuzzer:
       print f[1] + f[2] if f[1] or f[2] else f[0]

EDIT: the output from your version does not look right. "Fizz" and "Buzz" are on the wrong index positions. 0 1 fizz 3 buzz fizz 6 7 fizz buzz ...

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Re:FizzBuzz solved using only bit twiddling by waps on Monday October 28, 2013 @05:03PM (#6629770)

heh you're right. Now I feel so dumb. Off by one error. I needed to use xrange(1, 100) instead of just xrange.

You're also correct islice is not needed.

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Re:FizzBuzz solved using only bit twiddling by gamegoblin on Monday October 28, 2013 @04:00PM (#6629374)

I more gray-bearded haskell wizard than I might have a better solution, but to me the direct haskell equivalent of python's [1,2,3]*4 is

  concat $ replicate 4 [1,2,3]

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