[Haskell-beginners] Palindromic solution??

Date: Mon, 16 Feb 2009 16:32:23 +0100 From: Miguel Pignatelli Subject: [Haskell-beginners] Indentation of local functions To: beginners@haskell.org Message-ID: <90C28A9E-A67D-47CA-8416-5F1C5C27A1F9@uv.es> Content-Type: text/plain; charset="us-ascii"

Hi all,

This is my first post in this forum, I'm pretty new to Haskell (although I have some previous experience in functional programming with OCaml)

I'm trying to write the typical function that determines if a list is a palindrome. the typical answer would be something like:

isPalindrome xs = xs == (reverse xs)

But I find this pretty inefficient (duplication of the list and double of needed comparisons). So I tried my own version using just indexes:

isPalindrome xs = isPalindrome' 0 (length xs) where isPalindrome' i j = if i == j -- line 43 then True else if (xs !! i) == (xs !! (j-1)) then isPalindrome' (i+1) (j-1) else False

But, when trying to load this in ghci it throws the following error:

xxx.hs:43:12: parse error (possibly incorrect indentation) Failed, modules loaded: none. (Line 43 is marked in the code)

I seems that the definition of isPalindrome' must be in one line. So, this works as expected:

isPalindrome xs = isPalindrome' 0 (length xs) where isPalindrome' i j = if i == j then True else if (xs !! i) == (xs !! (j-1)) then isPalindrome' (i+1) (j-1) else False

Is there any way to make the local definition of isPalindrome' more readable?

Any help in understanding this would be appreciated

Thanks in advance,

I have found one solution to your problem isPalindrome xs = isPalindrome' 0 (length xs) where isPalindrome' i j = if i == j -- line 43 then True else if (xs !! i) == (xs !! (j-1)) then isPalindrome' (i+1) (j-1) else False This loads without error but poses a second problem it generates an index too large exception. Alan Cameron