code review of birthday paradox, "no instance for" help

[jjared...@gmail.com]

Commenting out the definitions of birthdayparadox as well as test suppress all errors. I've tried to do add type annotations (Floating a) => a -> a, but truthfully I just want someone to point me in the right direction, because I know I'm just not able to read the error properly. I'm on day 2 Haskell learning so I just haven't absorbed enough of how Haskell infers types, or even the machinery of what types do. Thanks for any help.

https://gist.github.com/1223492

birthdayparadox.hs:3:17:
No instance for (Floating (a0 -> c0))
arising from a use of `logBase'
Possible fix: add an instance declaration for (Floating (a0 -> c0))
In the expression: logBase (1 - pShare) (1 - p)
In an equation for `uniquePairs':
uniquePairs p = logBase (1 - pShare) (1 - p)

birthdayparadox.hs:26:26:
No instance for (Num (a0 -> c0))
arising from the literal `1'
Possible fix: add an instance declaration for (Num (a0 -> c0))
In the second argument of `(-)', namely `1'
In the expression: 2 * x - 1
In the expression: \ x -> 2 * x - 1

birthdayparadox.hs:27:6:
No instance for (Ord (a0 -> c0))
arising from a use of `newtonRaphson'
Possible fix: add an instance declaration for (Ord (a0 -> c0))
In the expression: newtonRaphson f f' a0 1.0e-3
In the expression:
let
a0 = sqrt . uniquePairs p
f n = arcp n p
f' = \ x -> ...
in newtonRaphson f f' a0 1.0e-3
In an equation for `birthdayparadox':
birthdayparadox p
= let
a0 = sqrt . uniquePairs p
f n = arcp n p
....
in newtonRaphson f f' a0 1.0e-3

birthdayparadox.hs:29:24:
No instance for (Fractional (a0 -> c0))
arising from the literal `0.5'
Possible fix:
add an instance declaration for (Fractional (a0 -> c0))
In the first argument of `birthdayparadox', namely `0.5'
In the expression: birthdayparadox 0.5
In an equation for `test': test = birthdayparadox 0.5

[Dirk Thierbach]

jjared...@gmail.com wrote:

> Commenting out the definitions of birthdayparadox as well as test suppress
> all errors. I've tried to do add type annotations (Floating a) => a -> a,
> but truthfully I just want someone to point me in the right direction,
> because I know I'm just not able to read the error properly.

"No instance for Floating (x -> y)" means that the compiler tries to
use a function type as some (floating point) number. That usually
means you've forgotten to apply an argument to a function somewhere,
you need to add parenthesis, or you made a typo.

> https://gist.github.com/1223492

The program is shorter than the error message, so why don't post it?

> pShare = 1 / 365

>
> uniquePairs p = logBase (1 - pShare) (1 - p)

> arcp n p = n^2 - n - uniquePairs p
>
> newtonRaphsonStep f f' ai = ai - f ai / f' ai
>
> newtonRaphson f f' ai epsilon =
> let ai' = newtonRaphsonStep f f' ai
> in if abs (ai' - ai) < epsilon then ai' else newtonRaphson f f' ai' epsilon

>
> birthdayparadox p =
> let a0 = sqrt . uniquePairs p
> f n = arcp n p

> f' = \x -> 2 * x - 1
> in newtonRaphson f f' a0 0.001
>
> test = birthdayparadox 0.5

I debugged this by commenting out the offending code, and then I used :t
in the toplevel to give my the types inferred by the compiler. Then I added
those in the source (which also means they serve as documentation for future
readers).

I got:

pShare :: Double
uniquePairs :: Double -> Double
arcp :: Double -> Double -> Double
newtonRaphson :: (Ord a, Fractional a) => (a -> a) -> (a -> a) -> a -> a -> a

So I can see that in birthdayparadox, a0 is supposed to be a Double (you
could specialize newtonRaphson to Double if you want, or keep it as it is
if you want to use it in other contexts, too).

However, (.) has a different binding priority than you think:

*Main> :t sqrt . uniquePairs 0.5
<interactive>:1:8:
Couldn't match expected type `a0 -> c0' with actual type `Double'
In the return type of a call of `uniquePairs'

Either add parenthesis:

*Main> :t (sqrt . uniquePairs) 0.5
(sqrt . uniquePairs) 0.5 :: Double

or use ($) instead:

*Main> :t sqrt $ uniquePairs 0.5
sqrt $ uniquePairs 0.5 :: Double

After that change, it compiles successfully.

HTH,

- Dirk

[jjared...@gmail.com]

Oooooooooooooooh!

Thanks so much. I kept seeing a0 -> c0, and I couldn't understand why it wasn't something like a0 -> b0. But it is because uniquePairs is a->b then sqrt is b->c.

Excellent, very helpful.