{{ message }}

Instantly share code, notes, and snippets.

judofyr/fizzbuzz.rb

Forked from JEG2/fizzbuzz.rb
Created Aug 1, 2012
Writing FizzBuzz with flip-flops
 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

practicingruby commented Aug 2, 2012

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

23inhouse commented Aug 2, 2012

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

judofyr commented Aug 2, 2012

 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.

practicingruby commented Aug 2, 2012

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

23inhouse commented Aug 2, 2012

 Thank you all.

judofyr commented Aug 2, 2012

 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 :-(

practicingruby commented Aug 2, 2012

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

judofyr commented Aug 2, 2012

 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)`).

judofyr commented Aug 2, 2012

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

DouglasAllen commented Aug 11, 2012

 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.

DouglasAllen commented Aug 11, 2012

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

judofyr commented Oct 5, 2012

 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```