Your first definition, kart xs ys = [(x,y) | x <- xs, y <- ys]
, is equivalent to
kart xs ys = xs >>= (x ->
ys >>= (y -> [(x,y)]))
where
(x:xs) >>= g = g x ++ (xs >>= g)
(x:xs) ++ ys = x : (xs ++ ys)
are sequential operations. Redefine them as alternating operations,
(x:xs) >>/ g = g x +/ (xs >>/ g)
(x:xs) +/ ys = x : (ys +/ xs)
[] +/ ys = ys
and your definition should be good to go for infinite lists as well:
kart_i xs ys = xs >>/ (x ->
ys >>/ (y -> [(x,y)]))
testing,
Prelude> take 20 $ kart_i [1..] [100..]
[(1,100),(2,100),(1,101),(3,100),(1,102),(2,101),(1,103),(4,100),(1,104),(2,102)
,(1,105),(3,101),(1,106),(2,103),(1,107),(5,100),(1,108),(2,104),(1,109),(3,102)]
courtesy of "The Reasoned Schemer". (see also conda, condi, conde, condu).
another way, more explicit, is to create separate sub-streams and combine them:
kart_i2 xs ys = foldr g [] [map (x,) ys | x <- xs]
where
g a b = head a : head b : g (tail a) (tail b)
this actually produces exactly the same results. But now we have more control over how we combine the sub-streams. We can be more diagonal:
kart_i3 xs ys = g [] [map (x,) ys | x <- xs]
where -- works both for finite
g [] [] = [] -- and infinite lists
g a b = concatMap (take 1) a
++ g (filter (not.null) (take 1 b ++ map (drop 1) a))
(drop 1 b)
so that now we get
Prelude> take 20 $ kart_i3 [1..] [100..]
[(1,100),(2,100),(1,101),(3,100),(2,101),(1,102),(4,100),(3,101),(2,102),(1,103)
,(5,100),(4,101),(3,102),(2,103),(1,104),(6,100),(5,101),(4,102),(3,103),(2,104)]
With some searching on SO I've also found an answer by Norman Ramsey with seemingly yet another way to generate the sequence, splitting these sub-streams into four areas - top-left tip, top row, left column, and recursively the rest. His merge
there is the same as our +/
here.
Your second definition,
genFromPair (e1, e2) = [x*e1 + y*e2 | x <- [0..], y <- [0..]]
is equivalent to just
genFromPair (e1, e2) = [0*e1 + y*e2 | y <- [0..]]
Because the list [0..]
is infinite there's no chance for any other value of x
to come into play. This is the problem that the above definitions all try to avoid.