[Haskell-cafe] Re: Need some help with an infinite list - Ouch
Lee Duhem
lee.duhem at gmail.com
Thu Jun 18 11:57:30 EDT 2009
On Wed, Jun 17, 2009 at 7:30 PM, GüŸnther Schmidt<gue.schmidt at web.de> wrote:
> Hi all,
>
> you have come up with so many solutions it's embarrassing to admit that I
> didn't come up with even one.
I have the similarly difficulties, but I found to understand some of
these answers,
equational reasoning is a very useful tool, I have prepared a blog post for how
I worked out some of these answers, here is the draft of it, I hope it
can help you
too.
Oh, if it doesn't help you at all, please let know why :-)
lee
====
Understanding Functions Which Use 'instance Monad []' by Equational Reasoning
GüŸnther Schmidt asked in Haskell-Cafe how to get a stream like this:
["a", ... , "z", "aa", ... , "az", "ba", ... , "bz", ... ]
and people in Haskell-Cafe offer some interesting answer for this question.
On the one hand, these answers show the power of Haskell and GHC base libraries,
but on the other hand, understanding them is a challenge for Haskell
newbie like me.
But I found to understand these answers, equational reasoning is very helpful,
here is why I think so.
Answer 1 (by Matthew Brecknell):
concat $ tail $ iterate (map (:) ['a' .. 'z'] <*>) [[]]
Well, how does this expression do what we want? concat, tail, iterate,
map, are easy,
looks like the magic is in (<*>).
What's this operator mean? (<*>) comes from class Applicative of
Control.Applicative,
class Functor f => Applicative f where
-- | Lift a value.
pure :: a -> f a
-- | Sequential application.
(<*>) :: f (a -> b) -> f a -> f b
and 'instance Applicative []' is
instance Applicative [] where
pure = return
(<*>) = ap
ap comes from Control.Monad
ap :: (Monad m) => m (a -> b) -> m a -> m b
ap = liftM2 id
liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
liftM2 f m1 m2 = do { x1 <- m1; x2 <- m2; return (f x1 x2) }
so the key to understand (<*>) is understanding the meaning of liftM2.
liftM2 uses, hum, do-notation, so by Haskell 98 report, this can be
translated to
liftM2 f m1 m2
(1.0) = m1 >>= \x1 ->
m2 >>= \x2 ->
return (f x1 x2)
When it is applied to list (you can convince yourself of this by type
inference),
wee need 'instance Monad []'
instance Monad [] where
m >>= k = foldr ((++) . k) [] m
m >> k = foldr ((++) . (\ _ -> k)) [] m
return x = [x]
fail _ = []
so
liftM2 f m1 m2
= m1 >>= \x1 ->
m2 >>= \x2 ->
return (f x1 x2)
let
f1
= \x1 ->
m2 >>= \x2 ->
return (f x1 x2)
f2
= \x2 -> return (f x1 x2)
we can write
m1 >>= f1
= foldr ((++) . f1) [] m1
m2 >>= f2
= foldr ((++) . f2) [] m2
Now we can see for list m1, m2, how does 'liftM2 f m1 m2' work
z1 = []
foreach x1 in (reverse m1); do -- foldr ((++) . f1) [] m1
z2 = []
foreach x2 in (reverse m2); do -- foldr ((++) . f2) [] m2
z2 = [f x1 x2] ++ z2
done
z1 = z2 ++ z1
done
Now we are ready to see how to apply (<*>):
map (:) ['a' .. 'z'] <*> [[]]
= (map (:) ['a' .. 'z']) <*> [[]]
= [('a':), ..., ('z':)] <*> [[]] -- misuse of [...] notation
= ap [('a':), ..., ('z':)] [[]]
= liftM2 id [('a':), ..., ('z':)] [[]]
= [('a':), ..., ('z':)] >>= \x1 ->
[[]] >>= \x2 ->
return (id x1 x2)
Here x1 bind to ('z':), ..., ('a':) in turn, x2 always bind to [], and
noticed that
return (id ('z':) []) -- f = id; x1 = ('a':); x2 = []
= return (('z':) [])
= return ((:) 'z' [])
= return "z"
= ["z"]
we have
map (:) ['a', .., 'z'] <*> [[]]
= liftM2 id [('a':), ..., ('z':)] [[]]
= ["a", ..., "z"]
(If you can't follow the this, work through the definition of foldr
step by step will be very helpful.)
map (:) ['a', .., 'z'] <*> (map (:) ['a', .., 'z'] <*> [[]])
= map (:) ['a', .., 'z'] <*> ["a", .., "z"]
= liftM2 id [('a':), ..., ('z':)] ["a", ..., "z"]
= ["aa", ..., "az", "ba", ..., "bz", ..., "za", ..., "zz"]
Now it's easy to know what we get from
iterate (map (:) ['a' .. 'z'] <*>) [[]]
= [[], f [[]], f (f [[]]), ...] -- f = map (:) ['a' .. 'z'] <*>
so
concat $ tail $ iterate (map (:) ['a' .. 'z'] <*>) [[]]
is exactly what we want.
Understanding Haskell codes by equational reasoning could be a very
tedious process, but it's also
a very helpful and instructive process for the beginners, because it
make you think slowly, check
the computation process step by step, just like the compiler does. And
in my opinion, this is exactly
what a debugger does.
Answer 2 (by Reid Barton):
concatMap (\n -> replicateM n ['a'..'z']) [1..]
In this solution, the hardest part is replicatM, which come from Control.Monad
replicateM :: (Monad m) => Int -> m a -> m [a]
replicateM n x = sequence (replicate n x)
sequence :: Monad m => [m a] -> m [a]
sequence ms = foldr k (return []) ms
where
k m m' = do { x <- m; xs <- m'; return (x:xs) }
recall the defintion of liftM2:
liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
liftM2 f m1 m2 = do { x1 <- m1; x2 <- m2; return (f x1 x2) }
so k in definition of sequence is an application of liftM2, and
sequence itself is a normal foldr.
Exercise 1:
Prove that for n >= 1
replicateM n ['a' .. 'z'] = (iterate (map (:) ['a' .. 'z'] <*>) [[]]) !! n
or more generally
replicateM = \n xs -> (iterate (map (:) xs <*>) [[]]) !! n
Answer:
replicateM 1 ['a' .. 'z']
= sequence [ ['a' .. 'z'] ]
= foldr k (return []) [['a' .. 'z']]
= k ['a' .. 'z'] [[]] -- return [] = [[]]
= liftM2 (:) ['a' .. 'z'] [[]]
= map (:) ['a' .. 'z'] <*> [[]]
= ["a", ..., "z"]
replicateM 2 ['a' .. 'z']
= sequence [['a' .. 'z'], ['a' .. 'z']]
= foldr k [[]] [['a' .. 'z'], ['a' .. 'z']]
= k ['a' .. 'z'] (k ['a' .. 'z'] [[]])
= k ['a' .. 'z'] (f [[]]) -- f = map (:) ['a' .. 'z'] <*>
= f (f [[]])
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