[Haskell-cafe] The maths behind the Fastest Fibb In The West.
michael at schmong.org
Sat May 7 05:58:58 UTC 2016
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
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 at 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:
>> thanks to this guy:
>> 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`.
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