[Haskell-beginners] Square root algorithm
mike h
mike_k_houghton at yahoo.co.uk
Fri Sep 15 15:49:10 UTC 2017
On the original ‘using pairs’ algorithm I’ve now coded something slightly cleaner than the original State based one.
There’s still a lot of code for a seemingly simple problem but I s’pose a lot of the functions are helpers for various things and the core of the solution is contained in the fold.
An easier to read copy of the code is at http://gitcommit.co.uk/2017/09/15/the-root-of-the-problem-part-2/ <http://gitcommit.co.uk/2017/09/15/the-root-of-the-problem-part-2/>
and the raw code is below. Thanks
Mike
———————
import Data.List.Split
import Data.Char
import Numeric
-- The entry point.
root :: Float -> Float
root n = f where
((f, _):_) = readFloat (lhs ++ "." ++ rhs)
(_, rootDigits) = rootFold n
(lhs, rhs) = splitAt (dpLocation n) rootDigits
--
-- fold with the initial value based on intSquareRoot function
-- and subsequent calculations based on doubling and the biggestN function.
rootFold :: Float -> (Integer, String)
rootFold n = foldr calculate (makeStartVal p1 p2) pairs where
(p1:p2:pairs) = digitList n
makeStartVal :: Integer -> Integer -> (Integer, String)
makeStartVal p1 p2 = res where
rt = intSquareRoot p1
res = (p2 + (p1 - rt * rt) * 100 , show rt)
calculate :: Integer -> (Integer, String) -> (Integer, String)
calculate p (n, res) = next where
(toAppend, remain) = biggestN (2 * read res) n
-- bring down the next pair and accumulate the result
next = (remain * 100 + p, res ++ show toAppend)
-- Where should decimal point be?
dpLocation :: Float -> Int
dpLocation n = if (even len) then len `div` 2 else (len + 1) `div` 2 where
[left, _] = splitOn "." . show $ n
len = length left
-- helper for formatting
formatFloatN numOfDecimals floatNum = showFFloat (Just numOfDecimals) floatNum ""
showFlt = formatFloatN 16
-- Takes float and makes list of 'paired' integers
digitList :: Float -> [Integer]
digitList n = res where
[l, r] = splitOn "." . showFlt $ n
res = map read $ (pairs . pad $ l) ++ (pairs . pad $ r) where
pairs [] = []
pairs xs =
let (ys, zs) = splitAt 2 xs
in ys : pairs zs
pad xs
| odd . length $ xs = "0" ++ xs
| otherwise = xs
-- eg largest number N such that 4N x N <= 161
-- and biggestN 4 161 = (3, 32)
--
biggestN :: Integer -> Integer -> (Integer, Integer)
biggestN = get 0 where
get n x y
| (x*10 + n) * n > y = (n-1, y - (x*10 + n - 1)*(n - 1))
| (x*10 + n) * n == y = (n , y - (x*10 + n) * n )
| otherwise = get (n + 1) x y
-- gives the largest int whose square is <= n
intSquareRoot :: Integer -> Integer
intSquareRoot n = root 0 where
root i
| i*i <= n = root (i + 1)
| otherwise = i - 1
———————
> On 13 Sep 2017, at 02:37, mukesh tiwari <mukeshtiwari.iiitm at gmail.com> wrote:
>
> Hi Patrick,
>
> On Tue, Sep 12, 2017 at 7:21 PM, PATRICK BROWNE <patrick.browne at dit.ie <mailto:patrick.browne at dit.ie>> wrote:
> Mukesh,
> Thanks for your reply.
> Does the mean that my original sqrt0 makes (2^25=33554432) function calls?
>
> Technically Yes, and more accurately for any given number n, you have 2 ^ (n - 1) call because your base case is 1.
>
> How many function calls does the let/where versions make?
>
> With let/where version you store the value of function call in in variable, so no more second call and total functional in this scenario is linear in number n. For you case with n = 25 the number of function calls is just 24.
>
> Best,
> Mukesh Tiwari
>
> Thanks,
> Pat
>
> On 12 September 2017 at 01:36, mukesh tiwari <mukeshtiwari.iiitm at gmail.com <mailto:mukeshtiwari.iiitm at gmail.com>> wrote:
> Hi Patrick,
>
> On Mon, Sep 11, 2017 at 10:44 PM, PATRICK BROWNE <patrick.browne at dit.ie <mailto:patrick.browne at dit.ie>> wrote:
> Why is it the sqrt0 function is so much slower than sqrt1. Does the where clause allow intermediate values to be stored?
> Regards,
> Pat
> sqrt0 :: Int -> Int
> sqrt0 0 = 0
> sqrt0 1 = 1
> sqrt0 n = ((sqrt0 (n - 1)) + (n `quot` sqrt0 (n-1))) `quot` 2
> -- sqrt0 25 several minutes
>
> In sqrt0, each function call with n > 1 creates two more function call, and this creates exponential blow up (factor of 2). You can make your code it faster by storing the intermediate result
>
> sqrt0 :: Int -> Int
> sqrt0 0 = 0
> sqrt0 1 = 1
> sqrt0 n = let k = sqrt0 (n - 1) in (k + (n `quot` k)) `quot` 2
>
> This code is not blowing exponentially because of you storing intermediate result leading to faster computation.
>
> sqrt1 :: Int -> Int
> sqrt1 n
> | n == 0 = 0
> | n == 1 = 1
> | otherwise = div (k + ( div n k)) 2
> where k = sqrt1(n-1)
> -- sqrt1 25 instant
>
>
> On 9 September 2017 at 05:49, KC <kc1956 at gmail.com <mailto:kc1956 at gmail.com>> wrote:
> One approach
>
> One function to compute the next iterate
>
> Another function to call the computation function until results are within some tolerance
>
> It's usually presented as separation of control and computation 😎
>
> --
> Sent from an expensive device which will be obsolete in a few months
> Casey
>
> On Sep 3, 2017 1:23 AM, "mike h" <mike_k_houghton at yahoo.co.uk <mailto:mike_k_houghton at yahoo.co.uk>> wrote:
> Hi,
>
> To help me in learning Haskell I started blogging about some of the things I’ve looked at.
> One such topic was calculating square roots ‘by hand’ and then deriving a Haskell algorithm.
> I wrote about the well known technique here
> http://gitcommit.co.uk/2017/08/25/the-root-of-the-problem-part-1/ <http://gitcommit.co.uk/2017/08/25/the-root-of-the-problem-part-1/>
>
> and it it is really quite a simple method.
>
> The second part of the post will be an implementation in Haskell.
>
> I then tried implementing it and got something that works but really its not very pleasant to look at! And its something I don’t want to post! Some parts are fine but I think I locked myself into the notion that it had to be using State and really the end result is pretty poor.
>
> I know this i perhaps a ‘big ask’ but I’d really appreciate any suggestions, solutions, hints etc. I will of course give full attribution.
>
> I’ve created a gist of the code here
> https://gist.github.com/banditpig <https://gist.github.com/banditpig>
>
> Many Thanks
>
> Mike
>
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