composing floor and sqrt in haskell

Question

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 . s         


        

Answer 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.