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Today's Topics:
1. Re: Haskell triangular loop (correct use of (++)) (atomly)
----------------------------------------------------------------------
Message: 1
Date: Thu, 14 Apr 2016 17:31:18 -0500
From: atomly <ato...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Haskell triangular loop (correct use
of (++))
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<caa_y42yqipcuo6m0g-hu2nuhlkwvked59q67bztvkunsolq...@mail.gmail.com>
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https://wiki.haskell.org/99_questions/Solutions/26
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On Sun, Apr 10, 2016 at 11:59 AM, rohan sumant <r.s.sum...@gmail.com> wrote:
> Suppose I have a list of distinct integers and I wish to generate all
> possible unordered pairs (a,b) where a/=b.
>
> Ex: [1,2,3,0] --> [(1,2),(1,3),(1,0),(2,3),(2,0),(3,0)]
>
> The approach I am following is this :-
>
> mkpairs [] = []
> mkpairs (x:xs) = (map (fn x) xs) ++ (mkpairs xs)
>
> fn x y = (x,y)
>
> It is generating the desired output but I am a bit unsure about the time
> complexity of the function mkpairs. In an imperative language a nested
> triangular for loop would do the trick in O(n^2) or more precisely
> (n*(n-1)/2) operations. Does my code follow the same strategy? I am
> particularly worried about the (++) operator. I think that (++) wouldn't
> add to the time complexity since the initial code fragment (map (fn x) xs)
> is to be computed anyway. Am I wrong here? Is this implementation running
> O(n^2)? If not, could you please show me how to write a nested triangular
> loop in Haskell?
>
> Rohan Sumant
>
>
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