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Writing FizzBuzz with flip-flops

View fizzbuzz.rb
1 2 3 4 5 6
a=b=c=(1..100).each do |num|
print num, ?\r,
("Fizz" unless (a = !a) .. (a = !a)),
("Buzz" unless (b = !b) ... !((c = !c) .. (c = !c))),
?\n
end

Ok I'll say it. WTF?

Please explain.

Edit: Please explain https://gist.github.com/3230984/a13fc25d6b2f78524c865bbb622e4f3cd9e97b6a

p "TT" if (true..true)
p "TF" if (true..false)
p "FT" if (false..true)
p "FF" if (false..false)

I was actually a little bit scared to run it in irb.

Thank you.

Owner

There's more to it though. Every flip-flop has an internal state (that starts out as false) that is persisted when there's a loop around the flip-flop.

For double dots it works like this:

if state == false
  return_value = start.call
else
  return_value = true
end
state = !stop.call

For triple dots:

if state == false
  return_value = start.call
else
  return_value = true
  state = !stop.call
end

(Where start/stop is the first and second part of the flip-flop).

When you use assignments and negation in the flip-flop you essentially have two state variables (the flip-flop's internal state + your variable) which triggers the return value in a certain pattern.

@judofyr: I was going to ask you to explain the rest of it, because I was clearly ducking that :-P

Thank you all.

Owner

I don't have a proof, but it seems to me that it's impossible to make it return false for every 6th iteration. I've solved it for 1, 2, 3, 4, 5, 7, 8, 9, 10, but not 6 :-(

@judofyr: Try building a truth table for them. That might help.

Owner

The problem is that it branches. In the 2-flip-flop if the state is false and we want the next values to be [true, false], then there are two possible inputs. There's also several ways to achieve the same input. One solution to the "dividable by 3" is the input: {start:true, stop:false}, {stop:true}, {start:false}. This can be solved either by having the same 2-cycle (true, false, true, false) in both start and stop ((s = !s) .. (s = !s)), but you can also use two different variables having a separate 2-cycle ((a = !a) .. (b = !b)).

Owner

Yay, after some improvements in my verifier-program, I now know there are solutions for everything from 1 to 17 \o/

Using flip-flops has to be benchmarked. I think that there is nothing wrong with using faster, more common methods. It doesn't make sense to iterate every value of the 1 to 100 range and then still have to eval each condition of a flip-flop.
Have you done a ven test ever? Just a thought. Map the state at every eval.

I'm confused about the !
Is this a not or a bang?

Owner

And here's a version of FizzBuzzBazz:

a=b=c=d=(e=1..100).each do |num|
  print num, ?\r,
    ("Fizz" unless (a = !a) .. (a = !a)),
    ("Buzz" unless (b = !b) ... !((c = !c) .. (c = !c))),
    ("Bazz" unless ((d = !d) .. (d = !d)) ... (e = !e)),
    ?\n
end
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