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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 <patrick.bro...@dit.ie>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
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 <kc1...@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 đ
>
> --
> 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_hough...@yahoo.co.uk> wrote:
>
>> Hi,
>>
>> To help me in learning Haskell I started blogging about some of the
>> things Iâve looked at.
>> One such topic was calculating square roots âby handâ 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 donât 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 âbig askâ but Iâd really appreciate any
>> suggestions, solutions, hints etc. I will of course give full attribution.
>>
>> Iâve created a gist of the code here
>> https://gist.github.com/banditpig
>>
>> Many Thanks
>>
>> Mike
>>
>> _______________________________________________
>> Beginners mailing list
>> Beginners@haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> 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 <mukeshtiwari.ii...@gmail.com>
To: patrick.bro...@dit.ie, The Haskell-Beginners Mailing List -
Discussion of primarily beginner-level topics related to Haskell
<beginners@haskell.org>
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 <patrick.bro...@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 <kc1...@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 đ
>>
>> --
>> 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_hough...@yahoo.co.uk> wrote:
>>
>>> Hi,
>>>
>>> To help me in learning Haskell I started blogging about some of the
>>> things Iâve looked at.
>>> One such topic was calculating square roots âby handâ 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 donât 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 âbig askâ but Iâd really appreciate any
>>> suggestions, solutions, hints etc. I will of course give full attribution.
>>>
>>> Iâve created a gist of the code here
>>> https://gist.github.com/banditpig
>>>
>>> Many Thanks
>>>
>>> Mike
>>>
>>> _______________________________________________
>>> Beginners mailing list
>>> Beginners@haskell.org
>>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>>
>>>
>> _______________________________________________
>> Beginners mailing list
>> Beginners@haskell.org
>> 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 ITBĂC 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
> dĂĄileadh nĂł ar aon ghnĂomh a dhĂ©anfar bunaithe ar an ĂĄbhar atĂĄ sa
> rĂomhphost nĂł sna hiatĂĄin seo. www.dit.ie
>
> TĂĄ ITBĂC ag aistriĂș go GrĂĄinseach GhormĂĄin â DIT is on the move to
> Grangegorman <http://www.dit.ie/grangegorman>
>
> _______________________________________________
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>
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------------------------------
Message: 3
Date: Tue, 12 Sep 2017 11:06:55 +0300
From: Baa <aqua...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
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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