This is from the nicta course (hence List = [], Optional = Maybe, ...), so I'm not looking for a full solution, but I am stuck on a State Transformer question. The aim is to filter duplicates from a List and completely fail if passed any number > 100.
-- filtering :: Applicative f => (a -> f Bool) -> List a -> f (List a)
distinctF :: (Ord a, Num a) => List a -> Optional (List a)
distinctF lst = case runStateT (filtering go lst) S.empty of
Full (val, _) -> Full val
Empty -> Empty
where
--go :: a -> StateT (S.Set a) Optional Bool
go x = do
s <- getT
if x > 100 then do
return *?*Empty / False*?*
This typechecks while go = undefined, but I'm struggling to put Empty into as return wraps e.g. False in a Full/Just. fail gets me someway forward but I don't think that is the solution.
In practice I am probably missing a more important issue and would welcome enlightenment.
If the goal is to write function making both: unique filtering and failing on large input at the same time, you got the skeleton quite right:
distinctF :: (Ord a, Num a) => List a -> Optional (List a)
distinctF lst = evalStateT (go lst) S.empty -- evalStateT is your case runStateT part
where -- on empty input we just return empty list
go [] = return []
-- otherwise
go (x:xs)
-- we check whether we should 'fail'
-- for that we *lift* the value from underlying monad (Optional) into our StateT Optional
| x > 100 = lift $ Empty
| otherwise = do
-- the stuff to do
-- get the state, do nothing if x is in there
-- otherwise add x to the state and recurse
So for your question, you need to lift Empty, not return it.
OK, so I finally found a way, by realising that I could construct the precisely correct return type, rather than trying to rely on return
go x = do
if x > 100 then
StateT (\_ -> Empty) -- `return` a fail
else do
st <- getT
However, I am still not quite sure how <- unwraps both the StateT and the inner monadic container
Related
I'm trying to write an evaluation function for a language that I am working on in which non-determinism can be permitted within an if-block, called a selection block. What I'm trying to achieve is the ability to pick an if/selection statement from the block whose guard is true and evaluate it but it doesn't matter which one I pick.
From searching, I found an example that performs in a similar way to what I would like to achieve through modelling coinflips. Below is my adapation of it but I'm having issue in applying this logic to my problem.
import Control.Monad
data BranchType = Valid | Invalid deriving (Show)
data Branch = If (Bool, Integer) deriving (Show, Eq)
f Valid = [If (True, 1)]
f Invalid = [If (False, 0)]
pick = [Invalid, Invalid, Valid, Invalid, Valid]
experiment = do
b <- pick
r <- f b
guard $ fstB r
return r
s = take 1 experiment
fstB :: Branch -> Bool
fstB (If (cond, int)) = cond
main :: IO ()
main = putStrLn $ show $ s -- shows first branch which could be taken.
Below is my ADT and what I have been trying to make work:
data HStatement
= Eval HVal
| Print HVal
| Skip String
| Do HVal [HStatement]
| If (HVal, [HStatement])
| IfBlock [HStatement] -- made up of many If
| Select [HStatement] -- made up of many If
deriving (Eq, Read)
fstIf :: HStatement -> Bool
fstIf (If (cond, body)) = if hval2bool cond == True
then True
else False
h :: Env -> HStatement -> IOThrowsError ()
h env sb = do
x <- g env sb
guard $ fstIf x -- Couldn't match expected type ‘HStatement’ with actual type ‘[HStatement]’
-- after guard, take 1 x then evaluate
g :: Env -> HStatement -> IOThrowsError [HStatement]
g env (Select sb) = mapM (\x -> f env x) sb
f :: Env -> HStatement -> IOThrowsError HStatement
f env (If (cond, body)) = evalHVal env cond >>= \x -> case x of
Bool True -> return $ If (Bool True, body)
Bool False -> return $ If (Bool False, body)
The error I receive is the following : Couldn't match expected type ‘HStatement’ with actual type ‘[HStatement]’ at the guard line. I believe the reason as to why the first section of code was successful was because the values were being drawn from List but in the second case although they're being drawn from a list, they're being drawn from a [HStatement], not something that just represents a list...if that makes any sort of sense, I feel like I'm missing the vocabulary.
In essence then what should occur is given a selection block of n statement, a subset of these are produced whose guards are true and only one statement is taken from it.
The error message is pretty clear now that you have some types written down. g returns IOThrowsError [HStatement], so when you bind its result to x in h, you have an [HStatement]. You then call fstIf, which expects a single HStatement, not a list. You need to decide how to handle the multiple results from g.
I have random number generator
rand :: Int -> Int -> IO Int
rand low high = getStdRandom (randomR (low,high))
and a helper function to remove an element from a list
removeItem _ [] = []
removeItem x (y:ys) | x == y = removeItem x ys
| otherwise = y : removeItem x ys
I want to shuffle a given list by randomly picking an item from the list, removing it and adding it to the front of the list. I tried
shuffleList :: [a] -> IO [a]
shuffleList [] = []
shuffleList l = do
y <- rand 0 (length l)
return( y:(shuffleList (removeItem y l) ) )
But can't get it to work. I get
hw05.hs:25:33: error:
* Couldn't match expected type `[Int]' with actual type `IO [Int]'
* In the second argument of `(:)', namely
....
Any idea ?
Thanks!
Since shuffleList :: [a] -> IO [a], we have shuffleList (xs :: [a]) :: IO [a].
Obviously, we can't cons (:) :: a -> [a] -> [a] an a element onto an IO [a] value, but instead we want to cons it onto the list [a], the computation of which that IO [a] value describes:
do
y <- rand 0 (length l)
-- return ( y : (shuffleList (removeItem y l) ) )
shuffled <- shuffleList (removeItem y l)
return y : shuffled
In do notation, values to the right of <- have types M a, M b, etc., for some monad M (here, IO), and values to the left of <- have the corresponding types a, b, etc..
The x :: a in x <- mx gets bound to the pure value of type a produced / computed by the M-type computation which the value mx :: M a denotes, when that computation is actually performed, as a part of the combined computation represented by the whole do block, when that combined computation is performed as a whole.
And if e.g. the next line in that do block is y <- foo x, it means that a pure function foo :: a -> M b is applied to x and the result is calculated which is a value of type M b, denoting an M-type computation which then runs and produces / computes a pure value of type b to which the name y is then bound.
The essence of Monad is thus this slicing of the pure inside / between the (potentially) impure, it is these two timelines going on of the pure calculations and the potentially impure computations, with the pure world safely separated and isolated from the impurities of the real world. Or seen from the other side, the pure code being run by the real impure code interacting with the real world (in case M is IO). Which is what computer programs must do, after all.
Your removeItem is wrong. You should pick and remove items positionally, i.e. by index, not by value; and in any case not remove more than one item after having picked one item from the list.
The y in y <- rand 0 (length l) is indeed an index. Treat it as such. Rename it to i, too, as a simple mnemonic.
Generally, with Haskell it works better to maximize the amount of functional code at the expense of non-functional (IO or randomness-related) code.
In your situation, your “maximum” functional component is not removeItem but rather a version of shuffleList that takes the input list and (as mentioned by Will Ness) a deterministic integer position. List function splitAt :: Int -> [a] -> ([a], [a]) can come handy here. Like this:
funcShuffleList :: Int -> [a] -> [a]
funcShuffleList _ [] = []
funcShuffleList pos ls =
if (pos <=0) || (length(take (pos+1) ls) < (pos+1))
then ls -- pos is zero or out of bounds, so leave list unchanged
else let (left,right) = splitAt pos ls
in (head right) : (left ++ (tail right))
Testing:
λ>
λ> funcShuffleList 4 [0,1,2,3,4,5,6,7,8,9]
[4,0,1,2,3,5,6,7,8,9]
λ>
λ> funcShuffleList 5 "#ABCDEFGH"
"E#ABCDFGH"
λ>
Once you've got this, you can introduce randomness concerns in simpler fashion. And you do not need to involve IO explicitely, as any randomness-friendly monad will do:
shuffleList :: MonadRandom mr => [a] -> mr [a]
shuffleList [] = return []
shuffleList ls =
do
let maxPos = (length ls) - 1
pos <- getRandomR (0, maxPos)
return (funcShuffleList pos ls)
... IO being just one instance of MonadRandom.
You can run the code using the default IO-hosted random number generator:
main = do
let inpList = [0,1,2,3,4,5,6,7,8]::[Integer]
putStrLn $ "inpList = " ++ (show inpList)
-- mr automatically instantiated to IO:
outList1 <- shuffleList inpList
putStrLn $ "outList1 = " ++ (show outList1)
outList2 <- shuffleList outList1
putStrLn $ "outList2 = " ++ (show outList2)
Program output:
$ pickShuffle
inpList = [0,1,2,3,4,5,6,7,8]
outList1 = [6,0,1,2,3,4,5,7,8]
outList2 = [8,6,0,1,2,3,4,5,7]
$
$ pickShuffle
inpList = [0,1,2,3,4,5,6,7,8]
outList1 = [4,0,1,2,3,5,6,7,8]
outList2 = [2,4,0,1,3,5,6,7,8]
$
The output is not reproducible here, because the default generator is seeded by its launch time in nanoseconds.
If what you need is a full random permutation, you could have a look here and there - Knuth a.k.a. Fisher-Yates algorithm.
I'm writing a very simple two-pass assembler in Haskell and I've come across a scenario that I don't yet have the experience to solve. I think the solution is likely to involve monad transformers, which I don't really understand.
The assembler parses the assembly code into a list of Statements, which are either instructions or labels. Some Statements may refer to labels. The assembler needs to convert the Statements into Instructions, which involves eliminating the labels and substituting the label references with an appropriate value.
I have written the first pass of the assembler, which produces a [(String, Int)] representing a map from labels to addresses. I have also written the following function for translating a Statement into an Instruction:
stmtToInstruction :: Int -> [(String, Int)] -> Statement -> Either String [I.Instruction]
stmtToInstruction addr labels stmt = case stmt of
ADD d s1 s2 -> Right [I.ADD d s1 s2]
BEQL s1 s2 l -> case do label <- find (\e -> fst e == l) labels
let labelAddr = snd label
let relativeAddr = I.ImmS $ fromIntegral (labelAddr - addr)
return (I.BEQ s1 s2 relativeAddr) of
Just i -> Right [i]
Nothing -> Left $ "Label " ++ l ++ " not defined"
LABEL _ -> Right []
I've omitted several cases for brevity, but you can see all the possible results here:
ADD always succeeds and produces an instruction
BEQL can either succeed or fail, depending on whether a label is found
LABEL always succeeds, even though it produces no actual instructions
This works as expected. The problem I now have is writing this function:
replaceLabels :: [Statement] -> Either String [I.Instruction]
replaceLabels takes a list of statements, and runs stmtToInstruction on each one. The addr argument to stmtToInstruction must be the length of the [Instruction] accumulated so far. The output may either be a Left String, if one of the label references was invalid, or a Right [I.Instruction], if there were no errors.
mapM :: Monad m => (a -> m b) -> [a] -> m [b] gets us some of the way there, but provides no way to inject the current address into the (a -> m b) function. How do I make this work?
You're right: the StateT monad transformer will do the trick:
imapM :: (Traversable t, Monad m)
=> (Int -> a -> m b) -> t a -> m (t b)
imapM f = flip runStateT 0 .
mapM (\a ->
do
count <- get
put $! count + 1
f count a)
But writing the specialized version for lists might be better:
itraverse :: Applicative f
=> (Int -> a -> f b) -> [a] -> f [b]
itraverse f = go 0 where
go !_ [] = pure []
go !count (x:xs) = (:) <$> f count x <*> go (count + 1) xs
I've implemented a recursive solution that I'm sure is very inefficient. I'd still be interested to see the 'proper' way of doing this.
replaceLabels :: [Statement] -> Either String [I.Instruction]
replaceLabels [] = Right []
replaceLabels stmts#(s:ss) = replaceLabels' labels stmts 0
where labels = process stmts
replaceLabels' :: [(String, Int)] -> [Statement] -> Int -> Either String [I.Instruction]
replaceLabels' _ [] _ = Right []
replaceLabels' labels (s:ss) addr = do
instructions <- stmtToInstruction addr labels s
restInstructions <- replaceLabels' labels ss (addr + length instructions)
return (instructions ++ restInstructions)
I would start by changing
stmtToInstruction :: Int -> [(String, Int)] -> Statement -> Either String [I.Instruction]
into
stmtToInstruction :: [(String, Int)] -> Statement -> Either String (Int -> [I.Instruction])
That is, moving the function that takes the address into the Right branch of the Either. The reason is that label reference errors seem to be independent of addresses, so it's better to handle reference errors first and then worry about the address stuff in isolation.
This function resolves the references:
resolveRefs :: [(String,Int)] -> [Statement] -> Either String [Int -> [Instruction]]
resolveRefs environment = traverse (stmtToInstruction environment)
(traverse is equivalent to mapM but it only requires an Applicative constraint. They are different functions merely for historical reasons.)
Ok, after having handled the errors, lets now focus on the [Int -> [Instruction]] list. It seems that we have to map over it from the left while carrying an accumulated address that we must supply to each function. The mapAccumL function is perfect for this:
resolveAddrs :: [Int -> [Instruction]] -> [Instruction]
resolveAddrs funcs = mconcat . snd $ accumulate funcs
where
accumulate :: [Int -> [Instruction]] -> (Int,[[Instruction]])
accumulate = mapAccumL step 0
step address func = let is = func address in (address + length is,is)
So I'm trying to make a little program that can take in data captured during an experiment, and for the most part I think I've figured out how to recursively take in data until the user signals there is no more, however upon termination of data taking haskell throws Exception: <<loop>> and I can't really figure out why. Here's the code:
readData :: (Num a, Read a) => [Point a] -> IO [Point a]
readData l = do putStr "Enter Point (x,y,<e>) or (d)one: "
entered <- getLine
if (entered == "d" || entered == "done")
then return l
else do let l = addPoint l entered
nl <- readData l
return nl
addPoint :: (Num a, Read a) => [Point a] -> String -> [Point a]
addPoint l s = l ++ [Point (dataList !! 0) (dataList !! 1) (dataList !! 2)]
where dataList = (map read $ checkInputData . splitOn "," $ s) :: (Read a) => [a]
checkInputData :: [String] -> [String]
checkInputData xs
| length xs < 2 = ["0","0","0"]
| length xs < 3 = (xs ++ ["0"])
| length xs == 3 = xs
| length xs > 3 = ["0","0","0"]
As far as I can tell, the exception is indication that there is an infinite loop somewhere, but I can't figure out why this is occurring. As far as I can tell when "done" is entered the current level should simply return l, the list it's given, which should then cascade up the previous iterations of the function.
Thanks for any help. (And yes, checkInputData will have proper error handling once I figure out how to do that.)
<<loop>> basically means GHC has detected an infinite loop caused by a value which depends immediately on itself (cf. this question, or this one for further technical details if you are curious). In this case, that is triggered by:
else do let l = addPoint l entered
This definition, which shadows the l you passed as an argument, defines l in terms of itself. You meant to write something like...
else do let l' = addPoint l entered
... which defines a new value, l', in terms of the original l.
As Carl points out, turning on -Wall (e.g. by passing it to GHC at the command line, or with :set -Wall in GHCi) would make GHC warn you about the shadowing:
<interactive>:171:33: warning: [-Wname-shadowing]
This binding for ‘l’ shadows the existing binding
bound at <interactive>:167:10
Also, as hightlighted by dfeuer, the whole do-block in the else branch can be replaced by:
readData (addPoint l entered)
As an unrelated suggestion, in this case it is a good idea to replace your uses of length and (!!) with pattern matching. For instance, checkInputData can be written as:
checkInputData :: [String] -> [String]
checkInputData xs = case xs of
[_,_] -> xs ++ ["0"]
[_,_,_] -> xs
_ -> ["0","0","0"]
addPoint, in its turn, might become:
addPoint :: (Num a, Read a) => [Point a] -> String -> [Point a]
addPoint l s = l ++ [Point x y z]
where [x,y,z] = (map read $ checkInputData . splitOn "," $ s) :: (Read a) => [a]
That becomes even neater if you change checkInputData so that it returns a (String, String, String) triple, which would better express the invariant that you are reading exactly three values.
I'm writing a program to allocate pizzas to people; each person will get one pizza, ideally of their favorite type, unless stock has run out, in which case they are given their next favorite type recursively.
My approach is to compute a ((User, Pizza), Int) for the amount a person would like said pizza, sort those, and recurse through using a state monad to keep inventory counts.
The program is written and type checks:
allocatePizzasImpl :: [((User, Pizza), Int)]
-> State [(Pizza, Int)] [(User, Pizza)]
allocatePizzasImpl [] = return []
allocatePizzasImpl ((user, (flavor, _)):ranks) =
do inventory <- get
-- this line is never hit
put $ updateWith inventory (\i -> if i <= 0
then Nothing
else Just $ i - 1) flavor
next <- allocatePizzasImpl $ filter ((/= user) . fst) ranks
return $ (user, flavor) : next
and I have a helper function to extract the result:
allocatePizzas :: [Pizza]
-> [((User, Pizza), Int)]
-> [(User, Pizza)]
allocatePizzas pizzas rank = fst
. runState (allocatePizzasImpl rank)
$ buildQuotas pizzas
but the line indicated by -- this line is never hit is... never hit by any GHCI breakpoints; furthermore, if I break on the return call, GHCI says inventory isn't in scope.
When run, the result is assigning the same pizza (with one inventory count) to all users. Something is going wrong, but I have absolutely no idea how to proceed. I'm new to Haskell, so any comments on style would be appreciated as well =)
Thanks!
PS: For completeness, updateWith is defined as:
updateWith :: (Eq a, Eq b)
=> [(a, b)] -- inventory
-> (b -> Maybe b) -- update function; Nothing removes it
-> a -- key to update
-> [(a, b)]
updateWith set update key =
case lookup key set of
Just b -> replace set
(unwrapPair (key, update b))
(fromMaybe 0 $ elemIndex (key, b) set)
Nothing -> set
where replace :: [a] -> Maybe a -> Int -> [a]
replace [] _ _ = []
replace (_:xs) (Just val) 0 = val:xs
replace (_:xs) Nothing 0 = xs
replace (x:xs) val i = x : (replace xs val $ i - 1)
unwrapPair :: Monad m => (a, m b) -> m (a, b)
unwrapPair (a, mb) = do b <- mb
return (a, b)
I think your function replace is broken:
replace (_:xs) (Just val) 0 = val:xs
This doesn't pay any attention to the value it's replacing. Wasn't your intention to replace just the pair corresponding to key?
I think you want
updateWith [] e k = []
updateWith ((k', v):kvs) e k
| k' == k = case e v of
Just v' -> (k, v'):kvs
Nothing -> kvs
| otherwise = (k', v) : updateWith kvs e k
The issue (ignoring other conceptual things mentioned by the commenters) turned out to be using fst to extract the result from the State would for some reason not cause the State to actually be computed. Running the result through seq fixed it.
I'd be interested in knowing why this is the case, though!
Edit: As Daniel Wagner pointed out in the comments, I wasn't actually using inventory, which turned out to be the real bug. Marking this as accepted.