Writing FizzBuzz with flip-flops

fizzbuzz.rb

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

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)).

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?

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