Re: [Haskell-beginners] Square root algorithm

Hi Patrick, On Tue, Sep 12, 2017 at 7:21 PM, PATRICK BROWNE 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

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wrote:

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Hi Patrick,

On Mon, Sep 11, 2017 at 10:44 PM, PATRICK BROWNE wrote:

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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

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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 wrote:

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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" wrote:

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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/

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

Many Thanks

Mike

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