https://wiki.haskell.org/99_questions/Solutions/26
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On Sun, Apr 10, 2016 at 11:59 AM, rohan sumant
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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