The problem sounds like this: write a program that reads a number n and then n persons, for each persons, read their name and age and then return the oldest persons/persons.
Example input:
3
Ion Ionel Ionescu
70
Gica Petrescu
99
Mustafa ben Muhamad
7
Example output
Oldest is Gica Petrescu (99 years).
My code so far:
readPers :: IO(String, Int)
readPers = do
name <- getLine
age <- readLn :: IO Int
return (name, age)
readPerss :: (Ord t, Num t) => t -> [IO (String, Int)]
readPerss n
| n > 0 = readPers : readPerss(n-1)
| otherwise = []
pFunc = do
print "Numer of persons:"
n <- readLn :: IO Int
let persons = readPerss n
return persons
I first read n, then I try to make a list of persons using readPers and readPerss, but I am stuck, I don't know how to tackle it from that point forward and I guess that my implementation thus far is not quite right.
How should I solve the problem?
You are very close! What you are doing in readPerss :: (Ord t, Num t) => t -> [IO (String, Int)] is returning a list of IO actions; each action returns a pair of String and Int when it is executed. Currently in pFunc you are only building this list of actions, storing it in a variable with let, and returning it from pFunc; you are never executing them with a <- “bind” statement.
There are a few simple ways to do what you want. The smallest change to your code that does what you want is to add sequence, which takes a container of actions and produces an action that returns a container:
sequence :: (Traversable t, Monad m) => t (m a) -> m (t a)
Here t is [], m is IO, and a is (String, Int):
sequence :: [IO (String, Int)] -> IO [(String, Int)]
Another way is to rewrite readPerss so that it executes the actions directly, accumulating the (String, Int) results in a list instead of accumulating the IO actions:
readPerss :: (Ord t, Num t) => t -> IO [(String, Int)]
-- Change [IO …] to IO […]: ~~~~~~~~~~~~~~~~~~
readPerss n
| n > 0 = do
pers <- readPers
perss <- readPerss (n - 1)
return (pers : perss)
| otherwise = return []
I know you may not be supposed to use library functions if this is a homework assignment or exercise, but in typical code “repeat x action n times and accumulate the results” is often represented with replicateM n x:
replicateM :: Applicative m => Int -> m a -> m [a]
This is how I always do this (it is from a code challenge isn’t it). I always seperate IO and logic as soon as possible. Works perfect (unless N is very big).
import Data.List.Split (chunksOf)
type Person = (String, Int)
main = do
x <- getContents
putStrLn $ program x
program :: String -> String
program s = “Oldest is “ ++ x ++ “ (“ ++ (show y) ++ “ years old).”
where
(x, y) = solve persons
persons = [(name, read age :: Int) | [name, age] <- chunksOf 2 . tail . lines $ s]
solve :: [Person] -> Person
solve ls = undefined
I leave the undefined to you.
Related
I wrote my first program in Haskell today. It compiles and runs successfully. And since it is not a typical "Hello World" program, it in fact does much more than that, so please congrats me :D
Anyway, I've few doubts regarding my code, and the syntax in Haskell.
Problem:
My program reads an integer N from the standard input and then, for each integer i in the range [1,N], it prints whether i is a prime number or not. Currently it doesn't check for input error. :-)
Solution: (also doubts/questions)
To solve the problem, I wrote this function to test primality of an integer:
is_prime :: Integer -> Bool
is_prime n = helper n 2
where
helper :: Integer -> Integer -> Bool
helper n i
| n < 2 * i = True
| mod n i > 0 = helper n (i+1)
| otherwise = False
It works great. But my doubt is that the first line is a result of many hit-and-trials, as what I read in this tutorial didn't work, and gave this error (I suppose this is an error, though it doesn't say so):
prime.hs:9:13:
Type constructor `Integer' used as a class
In the type signature for `is_prime':
is_prime :: Integer a => a -> Bool
According to the tutorial (which is a nicely-written tutorial, by the way), the first line should be: (the tutorial says (Integral a) => a -> String, so I thought (Integer a) => a -> Bool should work as well.)
is_prime :: (Integer a) => a -> Bool
which doesn't work, and gives the above posted error (?).
And why does it not work? What is the difference between this line (which doesn't work) and the line (which works)?
Also, what is the idiomatic way to loop through 1 to N? I'm not completely satisfied with the loop in my code. Please suggest improvements. Here is my code:
--read_int function
read_int :: IO Integer
read_int = do
line <- getLine
readIO line
--is_prime function
is_prime :: Integer -> Bool
is_prime n = helper n 2
where
helper :: Integer -> Integer -> Bool
helper n i
| n < 2 * i = True
| mod n i > 0 = helper n (i+1)
| otherwise = False
main = do
n <- read_int
dump 1 n
where
dump i x = do
putStrLn ( show (i) ++ " is a prime? " ++ show (is_prime i) )
if i >= x
then putStrLn ("")
else do
dump (i+1) x
You are misreading the tutorial. It would say the type signature should be
is_prime :: (Integral a) => a -> Bool
-- NOT Integer a
These are different types:
Integer -> Bool
This is a function that takes a value of type Integer and gives back a value of type Bool.
Integral a => a -> Bool
This is a function that takes a value of type a and gives back a value of type Bool.
What is a? It can be any type of the caller's choice that implements the Integral type class, such as Integer or Int.
(And the difference between Int and Integer? The latter can represent an integer of any magnitude, the former wraps around eventually, similar to ints in C/Java/etc.)
The idiomatic way to loop depends on what your loop does: it will either be a map, a fold, or a filter.
Your loop in main is a map, and because you're doing i/o in your loop, you need to use mapM_.
let dump i = putStrLn ( show (i) ++ " is a prime? " ++ show (is_prime i) )
in mapM_ dump [1..n]
Meanwhile, your loop in is_prime is a fold (specifically all in this case):
is_prime :: Integer -> Bool
is_prime n = all nondivisor [2 .. n `div` 2]
where
nondivisor :: Integer -> Bool
nondivisor i = mod n i > 0
(And on a minor point of style, it's conventional in Haskell to use names like isPrime instead of names like is_prime.)
Part 1: If you look at the tutorial again, you'll notice that it actually gives type signatures in the following forms:
isPrime :: Integer -> Bool
-- or
isPrime :: Integral a => a -> Bool
isPrime :: (Integral a) => a -> Bool -- equivalent
Here, Integer is the name of a concrete type (has an actual representation) and Integral is the name of a class of types. The Integer type is a member of the Integral class.
The constraint Integral a means that whatever type a happens to be, a has to be a member of the Integral class.
Part 2: There are plenty of ways to write such a function. Your recursive definition looks fine (although you might want to use n < i * i instead of n < 2 * i, since it's faster).
If you're learning Haskell, you'll probably want to try writing it using higher-order functions or list comprehensions. Something like:
module Main (main) where
import Control.Monad (forM_)
isPrime :: Integer -> Bool
isPrime n = all (\i -> (n `rem` i) /= 0) $ takeWhile (\i -> i^2 <= n) [2..]
main :: IO ()
main = do n <- readLn
forM_ [1..n] $ \i ->
putStrLn (show (i) ++ " is a prime? " ++ show (isPrime i))
It is Integral a, not Integer a. See http://www.haskell.org/haskellwiki/Converting_numbers.
map and friends is how you loop in Haskell. This is how I would re-write the loop:
main :: IO ()
main = do
n <- read_int
mapM_ tell_prime [1..n]
where tell_prime i = putStrLn (show i ++ " is a prime? " ++ show (is_prime i))
My code aims to create a word search puzzle. There is a data called Orientation representing the direction of each word in the puzzle.
data Orientation =
Forward | Back | Up | Down | UpForward | UpBack | DownForward | DownBack
deriving (Eq, Ord, Show, Read)
Now given a input of strings which is [String], I want to randomly assign each string an orientation like [(Orientation, String)]
assignWordDir :: [String] -> [(Orientation, String)]
assignWordDir [] = []
assignWordDir (s:strs) = (ori, s) : assignWordDir
where ori = pickOri [Forward, Back, Up, Down, UpForward, UpBack, DownForward, DownBack]
pickOri :: [a] -> IO a
pickOri xs = do
i <- randomRIO (0, len)
pure $ xs !! i
where len = length xs - 1
I cannot compile because the output of pickOri is IO Orientation, is there any suggestions on how to modify my code? Thanks a lot
Couldn't match expected type ‘[(IO Orientation, String)]’
with actual type ‘[String] -> [(Orientation, String)]’
You might consider modifying the functions so that they stay pure by taking a RandomGen parameter. The pickOri function, for example, might be modified thusly:
pickOri :: RandomGen g => g -> [a] -> (a, g)
pickOri rnd xs =
let len = length xs - 1
(i, g) = randomR (0, len) rnd
in (xs !! i, g)
It's necessary to return the new RandomGen value g together with the selected list element, so that it'll generate another pseudo-random number the next time around.
Likewise, you can modify assignWordDir like this:
assignWordDir :: RandomGen g => g -> [b] -> [(Orientation, b)]
assignWordDir _ [] = []
assignWordDir rnd (s:strs) = (ori, s) : assignWordDir g strs
where (ori, g) =
pickOri rnd [Forward, Back, Up, Down, UpForward, UpBack, DownForward, DownBack]
Notice that when recursing into to assignWordDir, the recursive function call uses the g it receives from pickOri.
You can use mkStdGen or newStdGen to produce RandomGen values. Here's an example using newStdGen:
*Q65132918> rnd <- newStdGen
*Q65132918> assignWordDir rnd ["foo", "bar", "baz"]
[(UpBack,"foo"),(Up,"bar"),(UpBack,"baz")]
*Q65132918> assignWordDir rnd ["foo", "bar", "baz"]
[(UpBack,"foo"),(Up,"bar"),(UpBack,"baz")]
Notice that when you use the same RandomGen value, you get the same sequence. That's because assignWordDir is a pure function, so that's expected.
You can, however, produce a new random sequence by creating or getting a new StdGen value:
*Q65132918> rnd <- newStdGen
*Q65132918> assignWordDir rnd ["foo", "bar", "baz"]
[(Up,"foo"),(Up,"bar"),(Forward,"baz")]
If you want to play with this in a compiled module, you can keep these functions as presented here, and then compose them with a newStdGen-generated StdGen in the main entry point.
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.