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Problem Let us suppose that we have a list xs (possibly a very big one), and we want to check that all its elements are the same. I came up with various ideas: Solution 0 checking that all elements in tail xs are equal to head xs : allTheSame :: (Eq a) => [a] -> Bool allTheSame xs = and $ map (== head xs) (tail xs) Solution 1 checking that length xs is equal to the length of the list obtained by taking elements from xs while they're equal to head xs allTheSame' :: (Eq a) => [a] -> Bool allTheSame' xs = (length xs) == (length $ takeWhile (== head xs) xs) Solution 2 recursive solution: allTheSame returns True if the first two elements of xs are equal and allTheSame returns True on the rest of xs allTheSame'' :: (Eq a) => [a] -> Bool allTheSame'' xs | n == 0 = False | n == 1 = True | n == 2 = xs !! 0 == xs !! 1 | otherwise = (xs !! 0 == xs !! 1) && (allTheSame'' $ snd $ splitAt 2 xs) where n = length xs Solution 3 divide and conquer: allTheSame''' :: (Eq a) => [a] -> Bool allTheSame''' xs | n == 0 = False | n == 1 = True | n == 2 = xs !! 0 == xs !! 1 | n == 3 = xs !! 0 == xs !! 1 && xs !! 1 == xs !! 2 | otherwise = allTheSame''' (fst split) && allTheSame''' (snd split) where n = length xs split = splitAt (n `div` 2) xs Solution 4 I just thought about this while writing this question: allTheSame'''' :: (Eq a) => [a] -> Bool allTheSame'''' xs = all (== head xs) (tail xs) Questions I think Solution 0 is not very efficient, at least in terms of memory, because ma

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