Daniel Fischer wrote:
isPalindrome xs = isPalindrome' 0 (length xs) °°°where °°°°°°isPalindrome' i j °°°°°°°°°| j <= i = True °°°°°°°°°| xs !! i /= xs !! (j-1) = False °°°°°°°°°| otherwise = isPalindrome (i+1) (j-1)
I only want to point out, that it is possible to put "where" ("do" or "let") at the end of the previous line, but a separate line for "where" is a good idea, too.
I have replaced your nested ifs by guards (increases readability, IMO) and corrected the stopping condition so that it also works on words of odd length.
Instead of course I prefer boolean expressions in some cases: isPalindrome xs = isPalindrome' 0 (length xs) where isPalindrome' i j = j <= i || xs !! i == xs !! (j-1) && isPalindrome' (i+1) (j-1)
However, note that Haskell lists aren't arrays, but singly linked lists, so to find xs !! k, all the first (k+1) cells of the list must be visited, making your algorithm less efficient than the naive one, since you must visit O((length xs)^2) cells.
Right, and only computing the length takes n steps, which is hard to avoid when one wants to avoid double equality tests. isPalindrome xs = and $ zipWith (==) xs $ reverse $ drop (div (length xs + 1) 2) xs Cheers Christian