Re: [Haskell-beginners] Palindromic solution??

On 16 Feb 2009, at 17:26, Alan Cameron wrote:

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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,

Of note by the way – the new solution is less efficient than the old one, for each letter that it compares it must traverse the list to find the (j-1)'th element. We can get a nice efficient solution by simply using the take function: isPalindrome xs = take l xs == (take l $ reverse xs) where l = length xs `div` 2 Bob