Hi Guys,
Thanks all for the suggestions, I have certainly improved my knowledge.
I made a blog post to show all the possible solution a problem can
have. you can check it out at katacoder.blogspot.com
Giba
On Sun, Apr 10, 2011 at 3:35 AM, Johan Tibell wrote:
> Hi Gilberto,
>
> On Wed, Mar 30, 2
Hi Gilberto,
On Wed, Mar 30, 2011 at 4:39 PM, Gilberto Garcia wrote:
> fkSum :: Int -> [Int] -> Int
> fkSum a [] = 0
> fkSum a (b) = foldl (+) 0 (filter (\x -> isMultiple x b) [1..a])
>
> isMultiple :: Int -> [Int] -> Bool
> isMultiple a [] = False
> isMultiple a (x:xs) = if (mod a x == 0) then T
Gilberto Garcia schrieb:
> isMultiple :: Int -> [Int] -> Bool
> isMultiple a [] = False
> isMultiple a (x:xs) = if (mod a x == 0) then True else isMultiple a xs
I think this one can be written in terms of 'List.any'.
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On Thu, Mar 31, 2011 at 7:29 AM, Daniel Fischer <
daniel.is.fisc...@googlemail.com> wrote:
> Err, terminology problem here.
> Strictly speaking, a function is strict iff
>
> f _|_ = _|_
>
> while we are talking here about evaluation strategies, so we should better
> have spoken of eager vs. deferr
On Thursday 31 March 2011 14:27:59, Yves Parès wrote:
> Just to be sure, because I am not quite familiar with the dark hairy
>
> internals of GHC:
> > Of course, given a type signature that allows strictness to be
> > inferred.
>
> You mean a signature with no type variables and types that are kn
Just to be sure, because I am not quite familiar with the dark hairy
internals of GHC:
> Of course, given a type signature that allows strictness to be inferred.
You mean a signature with no type variables and types that are know to GHC
as being strict?
(Like Int -> Int -> Int instead of (Num a)
On Thursday 31 March 2011 11:45:00, Christian Maeder wrote:
> Since we don't have a function sum' in the Prelude (should we have it?)
I think we should.
> I wonder what happens if you just use "sum". Will the "sum" (based on
> sum' so without -DUSE_REPORT_PRELUDE) be strict enough?
I don't kno
Am 31.03.2011 05:59, schrieb Felipe Almeida Lessa:
On Wed, Mar 30, 2011 at 2:39 PM, Gilberto Garcia wrote:
fkSum :: Int -> [Int] -> Int
fkSum a [] = 0
fkSum a (b) = foldl (+) 0 (filter (\x -> isMultiple x b) [1..a])
Daniel Fischer and Yves Parès gave you good suggestions about
implementing
Thank you very much for the suggestions.
giba
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On Wed, Mar 30, 2011 at 2:39 PM, Gilberto Garcia wrote:
> fkSum :: Int -> [Int] -> Int
> fkSum a [] = 0
> fkSum a (b) = foldl (+) 0 (filter (\x -> isMultiple x b) [1..a])
Daniel Fischer and Yves Parès gave you good suggestions about
implementing a different, better algorithm for you problem. How
If I'm not wrong :
sum [1..n] = (n² + n)/2
2011/3/30 Daniel Fischer
> On Wednesday 30 March 2011 16:39:49, Gilberto Garcia wrote:
> > Hi Haskellers,
> >
> > I was solving this problem from project euler to study haskell.
> > I came up whit the following solution and I was wondering if there is
On Wednesday 30 March 2011 16:39:49, Gilberto Garcia wrote:
> Hi Haskellers,
>
> I was solving this problem from project euler to study haskell.
> I came up whit the following solution and I was wondering if there is
> a more optimized and concise solution.
Yes. There's a constant-time formula fo
Hi Haskellers,
I was solving this problem from project euler to study haskell.
I came up whit the following solution and I was wondering if there is
a more optimized and concise solution.
fkSum :: Int -> [Int] -> Int
fkSum a [] = 0
fkSum a (b) = foldl (+) 0 (filter (\x -> isMultiple x b) [1..a])
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