[Haskell-beginners] Square root algorithm

[mukesh tiwari]

Hi Patrick,

On Tue, Sep 12, 2017 at 7:21 PM, PATRICK BROWNE <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
> > 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
>>>>>
>>>>> _______________________________________________
>>>>> Beginners mailing list
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>>>>>
>>>>>
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