I've been learning Haskell in my spare time working through LYAH. Would like to improve upon my Haskell (/ Functional programming) skills by solving some problems from the imperative world. One of the problems from EPI is to print an "almost sorted array", in a sorted fashion where it is guaranteed that no element in the array is more than k away from its correct position. The input is a stream of elements and the requirement is to do this in O(n log k) time complexity and O(k) space complexity.
I've attempted to re-implement the imperative solution in Haskell as follows:
import qualified Data.Heap as Heap
-- print the k-sorted list in a sorted fashion
ksorted :: (Ord a, Show a) => [a] -> Int -> IO ()
ksorted [] _ = return ()
ksorted xs k = do
heap <- ksorted' xs Heap.empty
mapM_ print $ (Heap.toAscList heap) -- print the remaining elements in the heap.
where
ksorted' :: (Ord a, Show a) => [a] -> Heap.MinHeap a -> IO (Heap.MinHeap a)
ksorted' [] h = return h
ksorted' (x:xs) h = do let (m, h') = getMinAndBuildHeap h x in
(printMin m >> ksorted' xs h')
printMin :: (Show a) => Maybe a -> IO ()
printMin m = case m of
Nothing -> return ()
(Just item) -> print item
getMinAndBuildHeap :: (Ord a, Show a) => Heap.MinHeap a -> a -> (Maybe a, Heap.MinHeap a)
getMinAndBuildHeap h item= if (Heap.size h) > k
then ((Heap.viewHead h), (Heap.insert item (Heap.drop 1 h)))
else (Nothing, (Heap.insert item h))
I would like to know a better way of solving this in Haskell. Any inputs would be appreciated.
[Edit 1]: The input is stream, but for now I assumed a list instead (with only a forward iterator/ input iterator in some sense.)
[Edit 2]: added Data.Heap import to the code.
Thanks.
I think the main improvement is to separate the production of the sorted list from the printing of the sorted list. So:
import Data.Heap (MinHeap)
import qualified Data.Heap as Heap
ksort :: Ord a => Int -> [a] -> [a]
ksort k xs = go (Heap.fromList b) e where
(b, e) = splitAt (k-1) xs
go :: Ord a => MinHeap a -> [a] -> [a]
go heap [] = Heap.toAscList heap
go heap (x:xs) = x' : go heap' xs where
Just (x', heap') = Heap.view (Heap.insert x heap)
printKSorted :: (Ord a, Show a) => Int -> [a] -> IO ()
printKSorted k xs = mapM_ print (ksort k xs)
If I were feeling extra-special-fancy, I might try to turn go into a foldr or perhaps a mapAccumR, but in this case I think the explicit recursion is relatively readable, too.
Related
Specifically I'm searching for a function 'maximumWith',
maximumWith :: (Foldable f, Ord b) => (a -> b) -> f a -> a
Which behaves in the following way:
maximumWith length [[1, 2], [0, 1, 3]] == [0, 1, 3]
maximumWith null [[(+), (*)], []] == []
maximumWith (const True) x == head x
My use case is picking the longest word in a list.
For this I'd like something akin to maximumWith length.
I'd thought such a thing existed, since sortWith etc. exist.
Let me collect all the notes in the comments together...
Let's look at sort. There are 4 functions in the family:
sortBy is the actual implementation.
sort = sortBy compare uses Ord overloading.
sortWith = sortBy . comparing is the analogue of your desired maximumWith. However, this function has an issue. The ranking of an element is given by applying the given mapping function to it. However, the ranking is not memoized, so if an element needs to compared multiple times, the ranking will be recomputed. You can only use it guilt-free if the ranking function is very cheap. Such functions include selectors (e.g. fst), and newtype constructors. YMMV on simple arithmetic and data constructors. Between this inefficiency, the simplicity of the definition, and its location in GHC.Exts, it's easy to deduce that it's not used that often.
sortOn fixes the inefficiency by decorating each element with its image under the ranking function in a pair, sorting by the ranks, and then erasing them.
The first two have analogues in maximum: maximumBy and maximum. sortWith has no analogy; you may as well write out maximumBy (comparing _) every time. There is also no maximumOn, even though such a thing would be more efficient. The easiest way to define a maximumOn is probably just to copy sortOn:
maximumOn :: (Functor f, Foldable f, Ord r) => (a -> r) -> f a -> a
maximumOn rank = snd . maximumBy (comparing fst) . fmap annotate
where annotate e = let r = rank e in r `seq` (r, e)
There's a bit of interesting code in maximumBy that keeps this from optimizing properly on lists. It also works to use
maximumOn :: (Foldable f, Ord r) => (a -> r) -> f a -> a
maximumOn rank = snd . fromJust . foldl' max' Nothing
where max' Nothing x = let r = rank x in r `seq` Just (r, x)
max' old#(Just (ro, xo)) xn = let rn = rank xn
in case ro `compare` rn of
LT -> Just (rn, xo)
_ -> old
These pragmas may be useful:
{-# SPECIALIZE maximumOn :: Ord r => (a -> r) -> [a] -> a #-}
{-# SPECIALIZE maximumOn :: (a -> Int) -> [a] -> a #-}
HTNW has explained how to do what you asked, but I figured I should mention that for the specific application you mentioned, there's a way that's more efficient in certain cases (assuming the words are represented by Strings). Suppose you want
longest :: [[a]] -> [a]
If you ask for maximumOn length [replicate (10^9) (), []], then you'll end up calculating the length of a very long list unnecessarily. There are several ways to work around this problem, but here's how I'd do it:
data MS a = MS
{ _longest :: [a]
, _longest_suffix :: [a]
, _longest_bound :: !Int }
We will ensure that longest is the first of the longest strings seen thus far, and that longest_bound + length longest_suffix = length longest.
step :: MS a -> [a] -> MS a
step (MS longest longest_suffix longest_bound) xs =
go longest_bound longest_suffix xs'
where
-- the new list is not longer
go n suffo [] = MS longest suffo n
-- the new list is longer
go n [] suffn = MS xs suffn n
-- don't know yet
go !n (_ : suffo) (_ : suffn) =
go (n + 1) suffo suffn
xs' = drop longest_bound xs
longest :: [[a]] -> [a]
longest = _longest . foldl' step (MS [] [] 0)
Now if the second to longest list has q elements, we'll walk at most q conses into each list. This is the best possible complexity. Of course, it's only significantly better than the maximumOn solution when the longest list is much longer than the second to longest.
I've been solving a few combinatoric problems on Haskell, so I wrote down those 2 functions:
permutations :: (Eq a) => [a] -> [[a]]
permutations [] = [[]]
permutations list = do
x <- list
xs <- permutations (filter (/= x) list)
return (x : xs)
combinations :: (Eq a, Ord a) => Int -> [a] -> [[a]]
combinations 0 _ = [[]]
combinations n list = do
x <- list
xs <- combinations (n-1) (filter (> x) list)
return (x : xs)
Which works as follows:
*Main> permutations [1,2,3]
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
*Main> combinations 2 [1,2,3,4]
[[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Those were uncomfortably similar, so I had to abstract it. I wrote the following abstraction:
combinatoric next [] = [[]]
combinatoric next list = do
x <- list
xs <- combinatoric next (next x list)
return (x : xs)
Which receives a function that controls how to filter the elements of the list. It can be used to easily define permutations:
permutations :: (Eq a) => [a] -> [[a]]
permutations = combinatoric (\ x ls -> filter (/= x) ls)
But I couldn't define combinations this way since it carries an state (n). I could extend the combinatoric with an additional state argument, but that'd become too clunky and I remember such approach was not necessary in a somewhat similar situation. Thus, I wonder: is it possible to define combinations using combinatorics? If not, what is a better abstraction of combinatorics which successfully subsumes both functions?
This isn't a direct answer to your question (sorry), but I don't think your code is correct. The Eq and Ord constraints tipped me off - they shouldn't be necessary - so I wrote a couple of QuickCheck properties.
prop_numberOfPermutations xs = length (permutations xs) === factorial (length xs)
where _ = (xs :: [Int]) -- force xs to be instantiated to [Int]
prop_numberOfCombinations (Positive n) (NonEmpty xs) = n <= length xs ==>
length (combinations n xs) === choose (length xs) n
where _ = (xs :: [Int])
factorial :: Int -> Int
factorial x = foldr (*) 1 [1..x]
choose :: Int -> Int -> Int
choose n 0 = 1
choose 0 r = 0
choose n r = choose (n-1) (r-1) * n `div` r
The first property checks that the number of permutations of a list of length n is n!. The second checks that the number of r-combinations of a list of length n is C(n, r). Both of these properties fail when I run them against your definitions:
ghci> quickCheck prop_numberOfPermutations
*** Failed! Falsifiable (after 5 tests and 4 shrinks):
[0,0,0]
3 /= 6
ghci> quickCheck prop_numberOfCombinations
*** Failed! Falsifiable (after 4 tests and 1 shrink):
Positive {getPositive = 2}
NonEmpty {getNonEmpty = [3,3]}
0 /= 1
It looks like your functions fail when the input list contains duplicate elements. Writing an abstraction for an incorrect implementation isn't a good idea - don't try and run before you can walk! You might find it helpful to read the source code for the standard library's definition of permutations, which does not have an Eq constraint.
First let's improve the original functions. You assume that all elements are distinct wrt their equality for permutations, and that they're distinct and have an ordering for combinations. These constraints aren't necessary and as described in the other answer, the code can produce wrong results. Following the robustness principle, let's accept just unconstrained lists. For this we'll need a helper function that produces all possible splits of a list:
split :: [a] -> [([a], a, [a])]
split = loop []
where
loop _ [] = []
loop rs (x:xs) = (rs, x, xs) : loop (x:rs) xs
Note that the implementation causes prefixes returned by this function to be reversed, but it's nothing we require.
This allows us to write generic permutations and combinations.
permutations :: [a] -> [[a]]
permutations [] = [[]]
permutations list = do
(pre, x, post) <- split list
-- reversing 'pre' isn't really necessary, but makes the output
-- order natural
xs <- permutations (reverse pre ++ post)
return (x : xs)
combinations :: Int -> [a] -> [[a]]
combinations 0 _ = [[]]
combinations n list = do
(_, x, post) <- split list
xs <- combinations (n-1) post
return (x : xs)
Now what they have in common:
At each step they pick an element to output,
update the list of elements to pick from and
stop after some condition is met.
The last point is a bit problematic, as for permutations we end once the list to choose from is empty, while for combinations we have a counter. This is probably the reason why it was difficult to generalize. We can work around this by realizing that for permutations the number of steps is equal to the length of the input list, so we can express the condition in the number of repetitions.
For such problems it's often very convenient to express them using StateT s [] monad, where s is the state we're working with. In our case it'll be the list of elements to choose from. The core of our combinatorial functions can be then expressed with StateT [a] [] a: pick an element from the state and update the state for the next step. Since the stateful computations all happen in the [] monad, we automatically branch all possibilities. With that, we can define a generic function:
import Control.Monad.State
combinatoric :: Int -> StateT [a] [] b -> [a] -> [[b]]
combinatoric n k = evalStateT $ replicateM n k
And then define permutations and combinations by specifying the appropriate number of repetitions and what's the core StateT [a] [] a function:
permutations' :: [a] -> [[a]]
permutations' xs = combinatoric (length xs) f xs
where
f = StateT $ map (\(pre, x, post) -> (x, reverse pre ++ post)) . split
combinations' :: Int -> [a] -> [[a]]
combinations' n xs = combinatoric n f xs
where
f = StateT $ map (\(_, x, post) -> (x, post)) . split
I am relatively new to Haskell, but I am trying to learn both by reading and trying to solve problems on Project Euler. I am currently trying to implement a function that takes an infinite list of integers and returns the ordered list of pairwise sums of elements in said list. I am really looking for solutions to the specific issue I am facing, rather than advice on different strategies or approaches, but those are welcome as well, as being a coder doesn't mean knowing how to implement a strategy, but also choosing the best strategy available.
My approach relies on traversing an infinite list of infinite generators and retrieving elements in order, with several mathematical properties that are useful in implementing my solution.
If I were trying to obtain the sequence of pairwise sums of the natural numbers, for example, this would be my code:
myList :: [Integer]
myList = [1..]
myGens :: [[Integer]]
myGens = gens myList
where
gens = \xs -> map (\x -> [x+y|y<-(dropWhile (<x) xs)]) xs
Regardless of the number set used, provided that it is sorted, the following conditions hold:
∀ i ≥ 0, head (gens xs !! i) == 2*(myList !! i)
∀ i,j,k ≥ 0, l > 0, (((gens xs) !! i) !! j) < (((gens xs) !! i+k) !! j+l)
Special cases for the second condition are:
∀ i,j ≥ 0, (((gens xs) !! i) !! j) < (((gens xs) !! i+1) !! j)
∀ i,j ≥ 0, k > 0, (((gens xs) !! i) !! j) < (((gens xs) !! i+k) !! j)
Here is the particular code I am trying to modify:
stride :: [Integer] -> [Int] -> [[Integer]] -> [Integer]
stride xs cs xss = x : stride xs counts streams
where
(x,i) = step xs cs xss
counts = inc i cs
streams = chop i xss
step :: [Integer] -> [Int] -> [[Integer]] -> (Integer,Int)
step xs cs xss = pace xs (defer cs xss)
pace :: [Integer] -> [(Integer,Int)] -> (Integer,Int)
pace hs xs#((x,i):xt) = minim (x,i) hs xt
where
minim :: (Integer,Int) -> [Integer] -> [(Integer,Int)] -> (Integer,Int)
minim m _ [] = m
minim m#(g,i) hs (y#(h,n):ynt) | g > h && 2*(hs !! n) > h = y
| g > h = minim y hs ynt
| 2*(hs !! n) > g = m
| otherwise = minim m hs ynt
defer :: [Int] -> [[a]] -> [(a,Int)]
defer cs xss = (infer (zip cs (zip (map head xss) [0..])))
infer :: [(Int,(a,Int))] -> [(a,Int)]
infer [] = []
infer ((c,xi):xis) | c == 0 = xi:[]
| otherwise = xi:(infer (dropWhile (\(p,(q,r)) -> p>=c) xis))
The set in question I am using has the property that multiple distinct pairs produce an identical sum. I want an efficient method of handling all duplicate elements at once, in order to avoid an increased cost of computing all the pairwise sums up to N, as it requires M more tests if M is the number of duplicates.
Does anyone have any suggestions?
EDIT:
I made some changes to the code, independently of what was suggested, and would appreciate feedback on the relative efficiencies of my original code, my revised code, and the proposals so far.
stride :: [Integer] -> [Int] -> [[Integer]] -> [Integer]
stride xs cs xss = x : stride xs counts streams
where
(x,is) = step xs cs xss
counts = foldr (\i -> inc i) cs is
streams = foldr (\i -> chop i) xss is
step :: [Integer] -> [Int] -> [[Integer]] -> (Integer,[Int])
step xs cs xss = pace xs (defer cs xss)
pace :: [Integer] -> [(Integer,Int)] -> (Integer,[Int])
pace hs xs#((x,i):xt) = minim (x,(i:[])) hs xt
where
minim :: (Integer,[Int]) -> [Integer] -> [(Integer,Int)] -> (Integer,[Int])
minim m _ [] = m
minim m#(g,is#(i:_)) hs (y#(h,n):ynt) | g > h && 2*(hs !! n) > h = (h,[n])
| g > h = minim (h,[n]) hs ynt
| g == h && 2*(hs !! n) > h = (g,n:is)
| g == h = minim (g,n:is) hs ynt
| g < h && 2*(hs !! n) > g = m
| g < h = minim m hs ynt
Also, I left out the code for inc and chop:
alter :: (a->a) -> Int -> [a] -> [a]
alter = \f -> \n -> \xs -> (take (n) xs) ++ [f (xs !! n)] ++ (drop (n+1) xs)
inc :: Int -> [Int] -> [Int]
inc = alter (1+)
chop :: Int -> [[a]] -> [[a]]
chop = alter (tail)
I'm going to present a solution that uses an infinite pairing heap. We'll have logarithmic overhead per element constructed, but no one knows how to do better (in a model with comparison-based methods and real numbers).
The first bit of code is just the standard pairing heap.
module Queue where
import Data.Maybe (fromMaybe)
data Queue k = E
| T k [Queue k]
deriving Show
fromOrderedList :: (Ord k) => [k] -> Queue k
fromOrderedList [] = E
fromOrderedList [k] = T k []
fromOrderedList (k1 : ks'#(k2 : _ks''))
| k1 <= k2 = T k1 [fromOrderedList ks']
mergePairs :: (Ord k) => [Queue k] -> Queue k
mergePairs [] = E
mergePairs [q] = q
mergePairs (q1 : q2 : qs'') = merge (merge q1 q2) (mergePairs qs'')
merge :: (Ord k) => Queue k -> Queue k -> Queue k
merge (E) q2 = q2
merge q1 (E) = q1
merge q1#(T k1 q1's) q2#(T k2 q2's)
= if k1 <= k2 then T k1 (q2 : q1's) else T k2 (q1 : q2's)
deleteMin :: (Ord k) => Queue k -> Maybe (k, Queue k)
deleteMin (E) = Nothing
deleteMin (T k q's) = Just (k, mergePairs q's)
toOrderedList :: (Ord k) => Queue k -> [k]
toOrderedList q
= fromMaybe [] $
do (k, q') <- deleteMin q
return (k : toOrderedList q')
Note that fromOrderedList accepts infinite lists. I think that this can be justified theoretically by pretending as though the infinite list of descendants effectively are merged "just in time". This feels like the kind of thing that should be in the literature on purely functional data structures already, but I'm going to be lazy and not look right now.
The function mergeOrderedByMin takes this one step further and merges a potentially infinite list of queues, where the min element in each queue is nondecreasing. I don't think that we can reuse merge, since merge appears to be insufficiently lazy.
mergeOrderedByMin :: (Ord k) => [Queue k] -> Queue k
mergeOrderedByMin [] = E
mergeOrderedByMin (E : qs') = mergeOrderedByMin qs'
mergeOrderedByMin (T k q's : qs')
= T k (mergeOrderedByMin qs' : q's)
The next function removes duplicates from a sorted list. It's in the library that m09 suggested, but for the sake of completeness, I'll define it here.
nubOrderedList :: (Ord k) => [k] -> [k]
nubOrderedList [] = []
nubOrderedList [k] = [k]
nubOrderedList (k1 : ks'#(k2 : _ks''))
| k1 < k2 = k1 : nubOrderedList ks'
| k1 == k2 = nubOrderedList ks'
Finally, we put it all together. I'll use the squares as an example.
squares :: [Integer]
squares = map (^ 2) [0 ..]
sumsOfTwoSquares :: [Integer]
sumsOfTwoSquares
= nubOrderedList $ toOrderedList $
mergeOrderedByMin
[fromOrderedList (map (s +) squares) | s <- squares]
If you don't want to modify your code that much, you can use the nub function of Data.List.Ordered (installable by cabal install data-ordlist) to filter duplicates out.
It runs in linear time, ie complexity wise your algorithm won't change.
for your example [1..] the result is just [2..]. A "very smart compiler" could deduce this from the general solution with implicit heap, that follows.
gens xs is better expressed as
gens xs = map (\t#(x:_) -> map (x+) t) $ tails xs -- or should it be
-- map (\(x:ys) -> map (x+) ys) $ tails xs -- ?
Its resulting list of lists is easily merged without duplicates by tree-like folding1 (pictured here), with
pairsums xs = foldi (\(x:l) r-> x : union l r) $ gens xs
This assumes the input list is ordered in increasing order. If it's merely in non-decreasing order (with only finite runs of equals in it, of course), you'll need to slap an orderedNub on top of that (as m09 mentions),
pairsums' = orderedNub . pairsums
Just by using foldi where foldr would work, we often get an algorithmic improvement in complexity from a factor of n to log n, a pretty significant speedup. I use it as a general tool all the time.
1The code, adjusted for infinite lists only:
foldi f (x:xs) = f x (foldi f (pairs f xs))
pairs f (x:y:t) = f x y : pairs f t
union (x:xs) (y:ys) = case compare x y of
LT -> x : union xs (y:ys)
EQ -> x : union xs ys
GT -> y : union (x:xs) ys
See also:
mergesort as foldtree (by Heinrich Apfelmus)
infinite tree folding (by Dave Bayer)
Implicit Heap (by apfelmus)
I propose to build the pairs above the diagonal, that way a lot of duplicates are not even generated:
sums xs = zipWith (map . (+)) hs ts where
(hs:ts) = tails xs
Now you have a list of lists, each containing sorted sums. Because they are sorted, it is possible to determine the next element of the sequence in a finite number of steps:
filtermerge :: (Ord a) => [[a]]->[a]
filtermerge ((h:t):ts) = h : filtermerge (insert t ts) where
insert [] ts = ts
insert xs [] = [xs]
insert h ([]:t) = insert h t
insert (h:t) ts#((h1:t1):t2)
| h < h1 = (h:t):ts
| h == h1 = insert (h:t) $ insert t1 t2
| otherwise = insert (h1:t1) $ insert (h:t) t2
filtermerge _ = []
I'm beginning to try and get my head round haskell performance, and what makes things fast and slow, and I'm a little confused by this.
I have two implementations of a function that generates a list of primes up to a certain value. The first is straight off the Haskell wiki:
primesTo :: (Ord a, Num a, Enum a) => a -> [a]
primesTo m = eratos [2..m] where
eratos [] = []
eratos (p:xs) = p : eratos (xs `minus` [p*p, p*p+p..m])
The second is the same, but using an infinite list internally:
primes2 :: (Ord a, Num a, Enum a) => a -> [a]
primes2 m = takeWhile (<= m) (eratos [2..]) where
eratos [] = []
eratos (p:xs) = p : eratos (xs `minus` [p*p, p*p+p..])
In both cases, the minus function is:
minus :: (Ord a) => [a] -> [a] -> [a]
minus (x:xs) (y:ys) = case (compare x y) of
LT -> x : minus xs (y:ys)
EQ -> minus xs ys
GT -> minus (x:xs) ys
minus xs _ = xs
The latter implementation is significantly (~100x) slower than the former, and I don't get why. I would have thought that haskell's lazy evalutation would make them fairly equivalent under the hood.
This is obviously a reduced test case for the purposes of the question - in real life the optimisation would be no problem (although I don't understand why it is needed), but to me a function that just generates an infinite list of primes is more generically useful than a finite list, but appears slower to work with.
Looks like to me that there's a big difference between
(xs `minus` [p*p, p*p+p..m]) -- primesTo
(xs `minus` [p*p, p*p+p..]) -- primes2
The function minus steps through lists pairwise and terminates when one list reaches the end. In the first minus expression above, this occurs in no more than (m-p*p)/p steps when the latter list is exhausted. In the second one, it will always take steps on the order of length xs.
So your infinite lists have disabled at least one meaningful optimization.
One difference is that in the second case you need to generate one extra prime. You need to generate the first prime greater than m before takeWhile knows its time to stop.
Additionally, the [..m] bounds on both the list to filter and the lists of multiples help reduce the number of calculations. Whenever one of these lists gets empty minus immediately returns via its secons clause while in the infinite case the minus gets stuck in the first case. You can explore this a bit better if you also test the cases where only one of the lists is infinite:
--this is also slow
primes3 :: (Ord a, Num a, Enum a) => a -> [a]
primes3 m = takeWhile (<= m) (eratos [2..m]) where
eratos [] = []
eratos (p:xs) = p : eratos (xs `minus` [p*p, p*p+p..])
--this fast
primes4 :: (Ord a, Num a, Enum a) => a -> [a]
primes4 m = takeWhile (<= m) (eratos [2..]) where
eratos [] = []
eratos (p:xs) = p : eratos (xs `minus` [p*p, p*p+p..m])
Given the following simple BST definition:
data Tree x = Empty | Leaf x | Node x (Tree x) (Tree x)
deriving (Show, Eq)
inOrder :: Tree x -> [x]
inOrder Empty = []
inOrder (Leaf x) = [x]
inOrder (Node root left right) = inOrder left ++ [root] ++ inOrder right
I'd like to write an in-order function that can have side effects. I achieved that with:
inOrderM :: (Show x, Monad m) => (x -> m a) -> Tree x -> m ()
inOrderM f (Empty) = return ()
inOrderM f (Leaf y) = f y >> return ()
inOrderM f (Node root left right) = inOrderM f left >> f root >> inOrderM f right
-- print tree in order to stdout
inOrderM print tree
This works fine, but it seems repetitive - the same logic is already present in inOrder and my experience with Haskell leads me to believe that I'm probably doing something wrong if I'm writing a similar thing twice.
Is there any way that I can write a single function inOrder that can take either pure or monadic functions?
In inOrder you are mapping a Tree x to a [x], i. e. you sequentialize your tree. Why not just use mapM or mapM_ on the resulting list?
mapM_ print $ inOrder tree
Just to remind the types of the functions I've mentioned:
mapM :: (Monad m) => (a -> m b) -> [a] -> m [b]
mapM_ :: (Monad m) => (a -> m b) -> [a] -> m ()
You might want to look at implementing the Data.Traversable class or Data.Foldable class for your tree structure. Each only requires the definition of a single method.
In particular, if you implement the Data.Foldable class, you get the following two functions for free:
mapM_ :: (Foldable t, Monad m) => (a -> m b) -> t a -> m ()
toList :: Foldable t => t a -> [a]
It will also give you the rich set of functions (foldr, concatMap, any, ...) that you are used to using with the list type.
You only have to implement one of the following functions to create an instance of Data.Foldable:
foldMap :: Monoid m => (a -> m) -> t a -> m
foldr :: (a -> b -> b) -> b -> t a -> b