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
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! _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe