Beginners Digest, Vol 111, Issue 7

Tue, 12 Sep 2017 01:09:56 -0700 

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Today's Topics:

   1. Re:  Square root algorithm (PATRICK BROWNE)
   2. Re:  Square root algorithm (mukesh tiwari)
   3.  HSpec output option (Baa)


----------------------------------------------------------------------

Message: 1
Date: Mon, 11 Sep 2017 13:44:51 +0100
From: PATRICK BROWNE <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
        beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Square root algorithm
Message-ID:
        <cagflrkewcx53a6ur5jobem1+d7g6rqoq0lirigeqk4_lsnt...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

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 <[email protected]> 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" <[email protected]> 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
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>

-- 


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Message: 2
Date: Tue, 12 Sep 2017 10:36:58 +1000
From: mukesh tiwari <[email protected]>
To: [email protected],  The Haskell-Beginners Mailing List -
        Discussion of primarily beginner-level topics related to Haskell
        <[email protected]>
Subject: Re: [Haskell-beginners] Square root algorithm
Message-ID:
        <cafhzve_shodkqri8o0tgk4rsqkbqk9yd2---qwaoygy9vvf...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

Hi Patrick,

On Mon, Sep 11, 2017 at 10:44 PM, PATRICK BROWNE <[email protected]>
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 <[email protected]> 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" <[email protected]> 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
>>> [email protected]
>>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>>
>>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
>
> This email originated from DIT. If you received this email in error,
> please delete it from your system. Please note that if you are not the
> named addressee, disclosing, copying, distributing or taking any action
> based on the contents of this email or attachments is prohibited.
> www.dit.ie
>
> Is Ăł ITBÁC a thĂĄinig an rĂ­omhphost seo. MĂĄ fuair tĂș an rĂ­omhphost seo trĂ­
> earrĂĄid, scrios de do chĂłras Ă© le do thoil. Tabhair ar aird, mura tĂș an
> seolaĂ­ ainmnithe, go bhfuil dianchosc ar aon nochtadh, aon chĂłipeĂĄil, aon
> dåileadh nó ar aon ghníomh a dhéanfar bunaithe ar an åbhar atå sa
> rĂ­omhphost nĂł sna hiatĂĄin seo. www.dit.ie
>
> TĂĄ ITBÁC ag aistriĂș go GrĂĄinseach GhormĂĄin – DIT is on the move to
> Grangegorman <http://www.dit.ie/grangegorman>
>
> _______________________________________________
> Beginners mailing list
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> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
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------------------------------

Message: 3
Date: Tue, 12 Sep 2017 11:06:55 +0300
From: Baa <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
        beginner-level  topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] HSpec output option
Message-ID: <20170912110655.1cea1a86@Pavel>
Content-Type: text/plain; charset=US-ASCII

Hello, Dear List!

I have tests: I'm using HSpec (and QuickCheck too). And I have tests
like this:

  describe "Something" $ do
    it "something is correct" $ do
      ...blah-blah...
    it "any string is correct" $ property $
      \s -> all (=='*') (Something s) -- it's for example only!!!

so something like unit-test and property checks in one SomethingSpec.hs.

I'm running them with this Makefile:

  .PHONY: test fast-test

  fast-test:
        stack exec runhaskell -- -isrc -itest test/Spec.hs

  test:
        stack test

and in Spec.hs I have:

  {-# OPTIONS_GHC -F -pgmF hspec-discover #-}

That's all. So, when I find failed test, I get a trace like this:

  ...
  Failures:

    test/SomethingSpec.hs:172: 
     1) BlahBlah.superFunc any string is correct:
        result Gave up after 48 tests
  ...etc...

So, my question is: when QuichCheck runs my property test, it passes
argument to property's lambda. And on 48th test attempt with some
concreate argument value my check fails. How can I get detailed output
from such test environment, to see what concreate arguments lead to
failure? To see something (or similar/or more detailed even):

  Failed with arguments: s = ""

Is it possible (I run them with stack and with runhaskell too) ?

===
Best regards, Paul


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