Why does application of `sequence` on List of Lists lead to computation of its Cartesian Product?

Question

My question is about the sequence function in Prelude, the signature of which is as follows:

sequence :: Monad m => [m a] -> m [a]

I understand how this function works for List of Maybes. For example, applying sequence on [Just 3, Just 9] gives Just [3, 9].

I noticed that applying sequence on List of Lists gives its Cartesian Product. Can someone please help me understand how/why this happens?

Solution 1

This works because using lists as monads in Haskell makes them model indeterminism. Consider:

sequence [[1,2],[3,4]]

By definition this is the same as:

do x <- [1,2]
   y <- [3,4]
   return [x,y]

Just read it as "First a choice between 1 and 2, then a choice between 3 and 4". The list monad will now accumulate all possible outcomes - hence the answer [[1,3],[1,4],[2,3],[2,4]].

(for an even more obfuscated example, see here)

Solution 2

sequence acts as if it were defined like this.

sequence [] = return []
sequence (m:ms) = do
    x <- m
    xs <- sequence ms
    return (x:xs)

(Or sequence = foldr (liftM2 (:)) (return []) but anyhow…)

Just think about what happens when applied to a list of lists.

sequence [] = [[]]
sequence (list : lists) =
    [ x : xs
    | x <- list
    , xs <- sequence lists
    ]

Solution 3

Just to explain, why the application of sequence to a list of lists is so different from the application of sequence to a list of Maybe-values:

When you apply sequence to a list of lists, then the type of sequence is specialized from

sequence :: Monad m => [m a] -> m [a]

to (with the type constructor m set to [])

sequence :: [[] a] -> [] [a] 

(which is the same as sequence :: [[a]] -> [[a]])

internally, sequence uses (>>=) -- i.e. the monadic bind function. For lists this bind function is implemented completly different than for m set to Maybe!

Comments

[phynfo]

BTW, the last term is exactly the same as the respective list-comprehension-expression [[x,y] | x <- [1,2], y <-[3,4]] -- this perhaps makes it clearer that it yield the Cartesian Product.

[Peaker]

"is so different from" --> "is not so different from" ?