Thanks for your response Erik. It appears I have not articulated my question well enough.
When p is odd, why is the return value
(f*(f+2*g), f^2 + g^2)
as opposed to the return value of

(f^2+g^2, g*(2*f-g))

What is it about the boolean value that requires two entirely seperate things to be done?

On Fri, May 6, 2016 at 7:38 PM, Erik Rantapaa <erantapaa@gmail.com> wrote:


On Friday, May 6, 2016 at 6:46:26 PM UTC-5, Michael Litchard wrote:
I've been working on a project that needs a good fibonacci generator, and I'm to the point where can now improve upon this one:
https://wiki.haskell.org/The_Fibonacci_sequence#Fastest_Fib_in_the_West

thanks to this guy:
https://groups.google.com/forum/#!topic/haskell-cafe/HUgbAUCvCp4

He suggested breaking up a guard into two diffeent functions, which I can do, but I don't know what to call them because I don't know why the operations are different. I'm referring to this section:

fib' (f, g) p
            | p         = (f*(f+2*g), f^2 + g^2)
            | otherwise = (f^2+g^2,   g*(2*f-g))

I'd like to know the reason why each guard does two entirely different things, so I know what to call the functions when I seperate them out.

Clearly `p` is a Bool, and it comes from the expression:

     map (toEnum . fromIntegral) $ unfoldl divs n

What's going on in `toEnum . fromIntegral` is that a remainder (either 0 or 1 - it blows up for anything else) is being converted to a Bool, with 0 mapping to False and 1 mapping to True. So `isOdd` would be a more descriptive name for `p`.