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)]
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.
The second can similarly be desugared into:
concatMap (\i ->
concatMap (\j ->
map (\k ->
(i,j,k))
[j+1..6])
[i+1..6])
[1..6]
The desugared code is:
concatMap (\i -> concatMap (\j -> (i, j) : []) [i+1..4]) [1..4]
Which can be refactored to Tsuyoshi Ito's answer.
There is yet another translation scheme, it is due to Wadler, as far as I know.
This would give:
let
lc_outer (x:xs) = let lc_inner (y:ys) = (x,y) : lc_inner ys
lc_inner [] = lc_outer xs
in lc_inner [x+1.. 4]
lc_outer [] = []
in lc_outer [1..4]
This translation avoids needless construction of singleton lists in the innermost level that would need to get flattened with concatMap later.
