Jose Luis Reyes F. wrote:
Thanks for your help, I tried to solve the problem this weekend but I have some dudes.
My reasoning is:
The sequence is [1,a1,a2,a3,..] where a1 = h 1 1, a2 = h 1 a1, and so on
So, I want to construct the list of lists [[1,a1,a2,a3], [h 1 1, h 1 a1, h 1 a2,..], [h a1 1, h a1 a1, h a1 a2,...] ... [h an 1, h an a2,h an a3,...] ... ]
The function is
hammingx = 1 : foldl1 merge [ map (h x) hammingx | x <- hammingx]
However this is not a solution because the list is a infinite list of infinite lists.
Oh, but it's pretty close working, actually. We only need two small modifications: 1) use the right fold. foldl on an infinite list *never* produces a result. 'merge' is associative, so we can just use foldr instead: hammingx = 1 : foldr1 merge [map (h x) hammingx | x <- hammingx ] 2) we still haven't exploited the property that h x 1 > x. This property ensures that of the two arguments that 'merge' is applied to by 'foldr', the first one starts with the smaller element. So we can write merge' (x:xs) ys = x : merge xs ys hammingx = 1 : foldr1 merge' [map (h x) hammingx | x <- hammingx] which actually works. I'd actually define 'hamming' as a higher order function: hamming :: Ord a => (a -> a -> a) -> a -> [a] hamming f a = result where result = a : foldr1 merge' [map (f x) result | x <- result] hammingx = hamming h 1 HTH, Bertram