用于列表列表中元组序列的Haskell算法

Question

I was playing around a bit in Haskell to get familiar with it, but got stuck at the following problem:

I want to define a function that, given a list containing some amount of other lists, each containing 0 or more tuples, creates a new list as following:

*Main> foo 
    [
      [ (1,2), (3,4)  ],
      [ (5,6)         ],
      [ (7,8), (9,10) ]
    ]


  = [
      [ (1,2), (5,6), (7,8)  ],
      [ (1,2), (5,6), (9,10) ],
      [ (3,4), (5,6), (7,8)  ],
      [ (3,4), (5,6), (9,10) ]
    ]

So, in other words, the function should compose a list with every tuple from the first list combined with in each case one of the other tuples in the N remaining lists.

I was trying to write a recursive algorithm for this, but can't wrap my head around dealing with the N amount of other lists to combine tuples with. For just two lists of tuples, I would write something like:

composeList [] _        = []
composeList (x:xs) list = composeTuples x list ++ composeList xs list

composeTuples _ []     = []
composeTuples t (x:xs) = [t,x] : composeTuples t xs

This gives me:

*Main Data.List> composeList [(1,2),(3,4)] [(5,6),(7,8)]

    [
      [ (1,2), (5,6) ],
      [ (1,2), (7,8) ],
      [ (3,4), (5,6) ],
      [ (3,4), (7,8) ]
    ]

Though I can't seem to put the pieces together and make it work for any number of lists, each with any (>=0) number of tuples.

I'm both interested in solving this issue with some of Haskell's predefined functions (if possible), as well as with a somewhat similar approach as the one I was going for in the example above.

Thanks in advance!

Answer 1

This is simply the list monad, selecting an element from each list non-deterministically.

The function you're looking for is sequence :: Monad m => [ma] -> m [a] from Control.Monad

λ. let as = [(1,2),(3,4)]
λ. let bs = [(5,6)]
λ. let cs = [(7,8),(9,10)]
λ. let xss = [as, bs, cs]
λ. sequence xss
  [[(1,2),(5,6),(7,8)]
  ,[(1,2),(5,6),(9,10)]
  ,[(3,4),(5,6),(7,8)]
  ,[(3,4),(5,6),(9,10)]
  ]

Answer 2

Here's a recursive solution

solution :: [[a]] -> [[a]]
solution (x: xs) = [y: ys | y <- x, ys <- solution xs]
solution []      = [[]]

The idea behind the solution is the following: prepend each element of the head of list to every list you get from recursively computing the result for the tail of the input list.