composing floor and sqrt in haskell

问题

I'm just learning haskell (on my own, for fun) and I've come up against a wall.

My Question:

How can I define a function

flrt = (floor . sqrt)

When I try it in a file and compile, GCHi complains with the following:

AKS.hs:11:9:
    No instance for (RealFrac Integer)
      arising from a use of `floor'
    Possible fix: add an instance declaration for (RealFrac Integer)
    In the first argument of `(.)', namely `floor'
    In the expression: (floor . sqrt)
    In an equation for `flrt': flrt = (floor . sqrt)

AKS.hs:11:17:
    No instance for (Floating Integer)
      arising from a use of `sqrt'
    Possible fix: add an instance declaration for (Floating Integer)
    In the second argument of `(.)', namely `sqrt'
    In the expression: (floor . sqrt)
    In an equation for `flrt': flrt = (floor . sqrt)

I don't understand why the resulting function isn't just Int -> Int.

I've just finished my second year of CS and done a basic PL course. I've heard of, but don't quite get types yet. I tried reading through a few haskell tutorials but it's all going above my head.

P.S. - I also don't understand what a monad is. (a lot of the other questions that my search turned up talked about these)

P.P.S. - My full source

bar = \a b -> if (2^a) > b
                then (a-1)
                else bar (a+1) b
foo = bar 1

flrt :: Integer -> Integer
flrt = (floor . sqrt)

aks target = if (target < 2)
                then putStr "Not a Prime.\n\n"
                else if elem (mod target 10) [0,2,4,5,6,8]
                        then putStr "Composite\n\n"
                        else if (elem target) [a^b | a <- [3,5..(flrt target)], b <- [1.. (foo target)]]

                                then putStr "Composite\n\n"--}
                            else 
                            putStr "filler"

回答1:

As copumpkin remarked, it might actually be a bad idea to convert to floating point here, because this comes with loss of precision and therefore might, even with rounding, yield incorrect results for sufficiently large integer inputs.

I assume all numbers you're dealing with will at least be small enough that there is some floating-point representation for them, e.g. all are < 10³⁰⁰. But, for instance

Prelude> round(sqrt.fromInteger$10^60 :: Double) ^ 2
1000000000000000039769249677312000395398304974095154031886336
Prelude>  {-   and not   -}     10^60    {-  == (10^30)^2 == (sqrt$10^60) ^ 2  -}
1000000000000000000000000000000000000000000000000000000000000

Which is way off, in terms of absolute difference. Still it's certainly a rather good approximation relative to the numbers themselves, so you can use it as a quickly determined starting point for an algorithm to find the exact result. You can implement Newton/Raphson (in this case AKA Heron) with Integers:

flrt :: Integer -> Integer  -- flrt x ≈ √x,  with  flrt x^2 ≤ x < flrt(x+1)^2
flrt x = approx (round . (sqrt::Double->Double) . fromInteger $ x)
   where approx r
            | ctrl <= x, (r+1)^2 > x  = r
            | otherwise               = approx $ r - diff
          where ctrl = r^2
                diff = (ctrl - x) // (2*r)    -- ∂/∂x x² = 2x

         a//b = a`div`b + if (a>0)==(b>0) then 1 else 0   -- always away from 0

This now works as desired:

*IntegerSqrt> (flrt $ 10^60) ^ 2
1000000000000000000000000000000000000000000000000000000000000

The division always away from 0 in the Newton-Raphson correction is here necessary to prevent getting stuck in an infinite recursion.

回答2:

The problem is that you're trying to use Integer as the input. Haskell is strongly typed, which means there are no implicit coercions or conversions of any kind. Look at signatures of functions you're trying to compose:

sqrt  :: Floating a => a -> a
floor :: (RealFrac a, Integral b) => a -> b

And at the signature of your function inferred by GHC:

> :t floor . sqrt
floor . sqrt :: (RealFrac b, Integral c, Floating b) => b -> c

So, to have a function that goes from Integer (which doesn't have a Floating instance) to Integer, you have to first convert your argument to Floating, which can be done by using fromIntegral:

> :t floor . sqrt . fromIntegral
floor . sqrt . fromIntegral :: (Integral a, Integral c) => a -> c

来源：https://stackoverflow.com/questions/10863132/composing-floor-and-sqrt-in-haskell

标签

haskell

sqrt

floor