Hi, after the factorial function I tried to write my second Haskell program, tackling the longest common subsequence algorithm: lcsList :: [a] -> [a] -> [a] lcsList [] _ = [] lcsList _ [] = [] lcsList (x:xs) (y:ys) = if x == y then lcsList xs ys else let lcs1 = lcsList x:xs ys lcs2 = lcsList xs y:ys in if (length lcs1) > (length lcs2) then lcs1 else lcs2 main = do putStrLn("Result: " show lcsList [1,2,3] [1,3]) GHC has several problems with it: lcs.hs:4:27: Could not deduce (Eq a) from the context () arising from a use of `==' at lcs.hs:4:27-32 Possible fix: add (Eq a) to the context of the type signature for `lcsList' I understand that I need to specify that type 'a' needs to be a derived type of something that can be compared, but how do I specify this in the code? On a related note, how can I make the function more flexible, to discern between case-sensitive and case-insensitive string comparison, for example? In Lisp I would just write this lambda list: (defun lcs-list (list1 list2 &key (test #'eql)) ...) and it would allow me to specify the test but use some sensible default if I don't. And two other errors which I don't quite get: lcs.hs:7:50: Couldn't match expected type `[a] -> [[a1] -> [a1]]' against inferred type `[a]' In the second argument of `(:)', namely `xs ys' In the expression: lcsList x : xs ys In the definition of `lcs1': lcs1 = lcsList x : xs ys lcs.hs:8:33: Occurs check: cannot construct the infinite type: a = [a] When generalising the type(s) for `lcs2' In the expression: let lcs1 = lcsList x : xs ys lcs2 = lcsList xs y : ys in if (length lcs1) > (length lcs2) then lcs1 else lcs2 in if (length lcs1) > (length lcs2) then lcs1 else lcs2 Can you help me with that? Thanks, Leslie -- LinkedIn Profile: http://www.linkedin.com/in/polzer Xing Profile: https://www.xing.com/profile/LeslieP_Polzer Blog: http://blog.viridian-project.de/