[Haskell-cafe] How the following works

[Lennart Augustsson]

Assuming you already think you know what cinits does, you can convince
yourself using induction.


On Tue, Apr 14, 2009 at 11:16 AM, Tsunkiet Man <temp.tsun at gmail.com> wrote:
> Let's see, if I execute it by hand:
>
> cinits :: [a] -> [[a]]
> cinits [] = [[]]
> cinits (x:xs) = [] : map (x:) (cinits xs)
>
> cinits [1,2,3] = [] : map (1:) ( [] : map (2:) ( [] : map (3:) ( [[]]) ) )
>                    = [] : map (1:) ( [] : map (2:) ( [] : map (3:) [[]] ) )
>                    = [] : map (1:) ( [] : map (2:) ( [] : [[3]] )
>                    = [] : map (1:) ( [] : map (2:) ( [[], [3]] )
>                    = [] : map (1:) ( [] : [[2], [2,3]])
>                    = [] : map (1:) ( [[], [2], [2,3]])
>                    = [[],[1], [1,2], [1,2,3]]
>
> Well, I understand this part. But isn't there an easier way to "see" the
> answer?
>
> Thank you for your help!
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