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 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 Many Thanks Mike _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
I don't know why this email was marked as spam, but it could be worth discussing. On Sun, Sep 3, 2017 at 4:22 PM, mike h <[hidden email]> wrote:
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In reply to this post by Mike Houghton
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" <[hidden email]> wrote:
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Thanks I’ll look into that.
To recap I have an unattractive but working implementation here https://gist.github.com/banditpig and the algorithm is described here http://gitcommit.co.uk/2017/08/25/the-root-of-the-problem-part-1/ Thanks Mike
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In reply to this post by KC
I’m sort of seeing this as a foldr over a list of digit pairs with the seed value of the fold being the first step of the algorithm (which is different from the other steps).
The list of pairs will be of indeterminate length as ’00’ is used until the required number of digits in the result is achieved. A minor wrinkle with this is if the input number is a perfect square then a lot of ‘0’s would be in the result. Then, after the fold, apply some simple function to determine where the decimal point should go. M
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In reply to this post by KC
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 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 <[hidden email]> wrote:
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Hi Patrick,
On Mon, Sep 11, 2017 at 10:44 PM, PATRICK BROWNE <[hidden email]> wrote:
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.
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Mukesh, Thanks for your reply. Does the mean that my original sqrt0 makes (2^25=33554432) function calls? How many function calls does the let/where versions make? Thanks, Pat On 12 September 2017 at 01:36, mukesh tiwari <[hidden email]> wrote:
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Hi Patrick,
On Tue, Sep 12, 2017 at 7:21 PM, PATRICK BROWNE <[hidden email]> wrote:
Technically Yes, and more accurately for any given number n, you have 2 ^ (n - 1) call because your base case is 1.
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
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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/ 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 ———————
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