# 99 questions/Solutions/6

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(Added solution using Control.Arrows fan out operator.) |
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isPalindrome''' :: (Eq a) => [a] -> Bool | isPalindrome''' :: (Eq a) => [a] -> Bool | ||

isPalindrome''' = Control.Monad.liftM2 (==) id reverse | isPalindrome''' = Control.Monad.liftM2 (==) id reverse | ||

+ | </haskell> | ||

+ | |||

+ | Or even: | ||

+ | |||

+ | <haskell> | ||

+ | isPalindrome'''' :: (Eq a) => [a] -> Bool | ||

+ | isPalindrome'''' = (==) Control.Applicative.<*> reverse | ||

</haskell> | </haskell> | ||

Line 42: | Line 49: | ||

<haskell> | <haskell> | ||

palindrome :: (Eq a) => [a] -> Bool | palindrome :: (Eq a) => [a] -> Bool | ||

− | + | palindrome xs = foldr (&&) True $ zipWith (==) xs (reverse xs) | |

+ | palindrome' xs = and $ zipWith (==) xs (reverse xs) -- same, but easier | ||

+ | </haskell> | ||

+ | |||

+ | |||

+ | <haskell> | ||

+ | isPalindrome list = take half_len list == reverse (drop (half_len + (len `mod` 2)) list) | ||

+ | where | ||

+ | len = length list | ||

+ | half_len = len `div` 2 | ||

+ | |||

+ | isPalindrome' list = f_part == reverse s_part | ||

+ | where | ||

+ | len = length list | ||

+ | half_len = len `div` 2 | ||

+ | (f_part, s_part') = splitAt half_len list | ||

+ | s_part = drop (len `mod` 2) s_part' | ||

+ | </haskell> | ||

+ | |||

+ | |||

+ | Using Control.Arrows (&&&) fan out operator. | ||

+ | |||

+ | With monomorphism restriction: | ||

+ | |||

+ | <haskell> | ||

+ | isPalindrome1 xs = (uncurry (==) . (id &&& reverse)) xs | ||

+ | </haskell> | ||

+ | |||

+ | Point free with no monomorphism restriction: | ||

+ | |||

+ | <haskell> | ||

+ | isPalindrome1 = (uncurry (==) . (id &&& reverse)) | ||

</haskell> | </haskell> |

## Revision as of 20:49, 30 November 2012

(*) Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).

isPalindrome :: (Eq a) => [a] -> Bool isPalindrome xs = xs == (reverse xs)

isPalindrome' [] = True isPalindrome' [_] = True isPalindrome' xs = (head xs) == (last xs) && (isPalindrome' $ init $ tail xs)

Here's one to show it done in a fold just for the fun of it. Do note that it is less efficient then the previous 2 though.

isPalindrome'' :: (Eq a) => [a] -> Bool isPalindrome'' xs = foldl (\acc (a,b) -> if a == b then acc else False) True input where input = zip xs (reverse xs)

Another one just for fun:

isPalindrome''' :: (Eq a) => [a] -> Bool isPalindrome''' = Control.Monad.liftM2 (==) id reverse

Or even:

isPalindrome'''' :: (Eq a) => [a] -> Bool isPalindrome'''' = (==) Control.Applicative.<*> reverse

Here's one that does half as many compares:

palindrome :: (Eq a) => [a] -> Bool palindrome xs = p [] xs xs where p rev (x:xs) (_:_:ys) = p (x:rev) xs ys p rev (x:xs) [_] = rev == xs p rev xs [] = rev == xs

Here's one using foldr and zipWith.

palindrome :: (Eq a) => [a] -> Bool palindrome xs = foldr (&&) True $ zipWith (==) xs (reverse xs) palindrome' xs = and $ zipWith (==) xs (reverse xs) -- same, but easier

isPalindrome list = take half_len list == reverse (drop (half_len + (len `mod` 2)) list) where len = length list half_len = len `div` 2 isPalindrome' list = f_part == reverse s_part where len = length list half_len = len `div` 2 (f_part, s_part') = splitAt half_len list s_part = drop (len `mod` 2) s_part'

Using Control.Arrows (&&&) fan out operator.

With monomorphism restriction:

isPalindrome1 xs = (uncurry (==) . (id &&& reverse)) xs

Point free with no monomorphism restriction:

isPalindrome1 = (uncurry (==) . (id &&& reverse))