[Haskell-beginners] Square root algorithm

[PATRICK BROWNE]

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 <mukeshtiwari.iiitm at gmail.com>
wrote:

> Hi Patrick,
>
> On Mon, Sep 11, 2017 at 10:44 PM, PATRICK BROWNE <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> 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> 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/
>>>>
>>>> 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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>>>>
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