Removing syntactic sugar: List comprehension in Haskell

Question

Can I unsugar list comprehension in this expression:

[(i,j) | i <- [1..4], j <- [i+1..4]]

This is the output:

[(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)]

How can I, with map, filter and so on, write that piece of code?

edit

Here an other:

[(i,j,k) | i <- [1..6], j <- [i+1..6],k <- [j+1..6]]

This is the output:

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

Answer 1

List comprehensions (in fact, Monad comprehensions) can be desugared into do notation.

do i <- [1..4]
   j <- [i+1..4]
   return (i,j)

Which can be desugared as usual:

[1..4]   >>= \i ->
[i+1..4] >>= \j ->
return (i,j)

It is well known that a >>= \x -> return b is the same as fmap (\x -> b) a. So an intermediate desugaring step:

[1..4] >>= \i -> 
fmap (\j -> (i,j)) [i+1..4]

For lists, (>>=) = flip concatMap, and fmap = map

(flip concatMap) [1..4] (\i -> map (\j -> (i,j) [i+1..4])

flip simply switches the order of the inputs.

concatMap (\i -> map (\j -> (i,j)) [i+1..4]) [1..4]

And this is how you wind up with Tsuyoshi's answer.

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The second can similarly be desugared into:

concatMap (\i ->
  concatMap (\j ->
    map       (\k ->
      (i,j,k))
    [j+1..6])
  [i+1..6])
[1..6]