Fibonacci sequence in Haskell - haskell

I am trying to build a fibonacci wrapper function that takes in multiple command line arguments and computes the associated Fibonacci number for each argument

You should use the read or readMaybe functions to get an Int from the String:
fibonacci_wrapper :: [String] -> Maybe Int
fibonacci_wrapper (x:_) =
case readMaybe x of
Just n -> Just (fibonacci n)
Nothing -> Nothing -- read failed
-- Not enough args
fibonacci_wrapper _ = Nothing
However, if you want to take multiple arguments, you should return the rest of the argument list along with the Int result so that you can continue consuming arguments.

Wrapper solution could be:
wrapperStringToInt :: [String] -> [Int]
wrapperStringToInt [] = []
wrapperStringToInt (x:xs) = (read x :: Int) : wrapperStringToInt xs
You could use tuples to return fibonacci 4 is 3. The function printFibonacci prints that for each tuple.
printFibonacci :: [(Int, Int)] -> IO ()
printFibonacci [] = return ()
printFibonacci ((x, z):xs) =
do putStrLn $ "fibonacci " ++ show x ++ " is " ++ show z
printFibonacci xs
The command
map f (wrapperStringToInt arguments)
where
f = (\ x -> (x, fibonacci x))
returns an array with tuples ( Input, Output ) (e.g. With [3,4,5] this command returns [(3,2),(4,3),(5,5)]).
main = do
arguments <- getArgs
printFibonacci $ map f (wrapperStringToInt arguments)
where f = (\ x -> (x, fibonacci x))
After that, run in the terminal:
runhaskell Fibonacci.hs <INPUT>
(e.g. runhaskell Fibonacci.hs 3 4 5)
Finally, to execute this program like ./program 4 5 6, you could write an executable bash program, for example:
#!/bin/bash
runhaskell Fibonacci.hs "$#"
If error control is important you should use Maybe as other users have commented. I hope you find this information useful.

Related

Command line arguments as Ints in Haskell

I have a Haskell program which accepts 2 or 3 Ints from the command line:
-- test.hs
main :: IO ()
main = do
args <- fmap (read . head) getArgs
case args of
[x,y,a] -> doGeneration x y a
[x,y] -> doGeneration x y 10
_ -> usage
However, when I run it with arguments:
$ ./test 100 200
divide: Prelude.read: no parse
Why?
getArgs :: IO [String] returns a list of Strings, by taking the head and then args and it will then read that item.
You however never specified to what it should read, since you use args in a case … of … clause with [x,y,a] and [x, y], it will try to read it as a list of numbers (the type of the number is specified by the doGeneration signature. This thus means that you should write it as:
$ ./test [100,200]
But I think it makes not much sense to do that, you can rewrite the parsing part to:
main :: IO ()
main = do
args <- fmap (map read) getArgs
case args of
[x,y,a] -> doGeneration x y a
[x,y] -> doGeneration x y 10
_ -> usage
This means that it will read every parameter individually, and construct a list with the parsed items, and then we can pattern match on the parsed parts of the program parameters. In that case we thus can still use:
$ ./test 100 200
Your running code is equivalent to
....
case (read "100") of
[x,y,a :: Int] -> doGeneration x y a
[x,y :: Int] -> doGeneration x y 10
....
but reading a string "100" as a list of Ints is impossible, a.k.a. "there's no parse". That's why.
The Int comes from doGeneration's signature which you haven't included. But it must be a Num since you use a and 10 interchangeably.
It's better to use more variables in your do block instead of fmap, when you learn Haskell. It lessens the cognitive load and lets you see clearer what's going on:
main :: IO ()
main = do
args <- getArgs -- getArgs :: IO [String]
-- args :: [String]
let arg1 = head args -- arg1 :: String
val1 = read arg1
case val1 of
[x,y,a] -> doGeneration x y a
[x,y] -> doGeneration x y 10
_ -> usage

Inconsistent do notation in functions

Why is this function allowed:
-- function 1
myfunc :: String
myfunc = do
x <- (return True)
show x
and this is not:
-- function 2
myfunc :: String
myfunc = do
x <- getLine
show x
The compile error:
Couldn't match type `[]' with `IO'
Expected type: IO Char
Actual type: String
I get why function 2 shouldn't work, but why then thus function 1 work?
and why does this then work:
-- function 3
myfunc = do
x <- getLine
return (show x)
I get that it returns IO String then, but why is function 1 also not forced to do this?
In function1 the do block in myfunc is working in the list monad, because String is really just [Char]. In there, return True just creates [True]. When you do x <- return True that "extracts" True out of [True] and binds it to x. The next line show x converts True into a String "True". which being the return value the compiler value expects to see, ends up working fine.
Meanwhile in function2, the do block in myfunc is also working on the list monad (for the same reason, String being really [Char]) but calls on getLine which is only available in the IO monad. So unsurprisingly, this fails.
-- EDIT 1
OP has added a function3
-- function 3
myfunc :: String
myfunc = do
x <- getLine
return (show x)
No this should not work for the same reason function2 fails.
-- EDIT 2
OP has updated function3 to fix a copy paste error.
-- function 3
myfunc = do
x <- getLine
return (show x)
This is mentioned in the comments, but for clarity sake, this works because, when the type information is unspecified, GHC makes it best inference and after seeing getLine, it figures it’s IO String which does provide getLine.
Note - I wrote this answer with as casual a tone as I could manage without being wrong with the intention of making it approachable to a beginner level.
do blocks work in the context of an arbitrary Monad. The Monad, in this case, is []. The Monad instance for lists is based on list comprehensions:
instance Monad [] where
return x = [x]
xs >>= f = [y | x <- xs, y <- f x]
You can desugar the do notation thus:
myfunc :: String
myfunc = do
x <- (return True)
show x
-- ==>
myfunc = [y | x <- return True, y <- show x]
-- ==>
myfunc = [y | x <- [True], y <- show x]
In a list comprehension, x <- [True] is really just the same as let x = True, because you're only drawing one element from the list. So
myfunc = [y | y <- show True]
Of course, "the list of all y such that y is in show True" is just show True.

How to randomly shuffle a list

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.

Technique for reading in multiple lines for Haskell IO

Basically I would like to find a way so that a user can enter the number of test cases and then input their test cases. The program can then run those test cases and print out the results in the order that the test cases appear.
So basically I have main which reads in the number of test cases and inputs it into a function that will read from IO that many times. It looks like this:
main = getLine >>= \tst -> w (read :: String -> Int) tst [[]]
This is the method signature of w: w :: Int -> [[Int]]-> IO ()
So my plan is to read in the number of test cases and have w run a function which takes in each test case and store the result into the [[]] variable. So each list in the list will be an output. w will just run recursively until it reaches 0 and print out each list on a separate line. I'd like to know if there is a better way of doing this since I have to pass in an empty list into w, which seems extraneous.
As #bheklilr mentioned you can't update a value like [[]]. The standard functional approach is to pass an accumulator through a a set of recursive calls. In the following example the acc parameter to the loop function is this accumulator - it consists of all of the output collected so far. At the end of the loop we return it.
myTest :: Int -> [String]
myTest n = [ "output line " ++ show k ++ " for n = " ++ show n | k <- [1..n] ]
main = do
putStr "Enter number of test cases: "
ntests <- fmap read getLine :: IO Int
let loop k acc | k > ntests = return $ reverse acc
loop k acc = do
-- we're on the kth-iteration
putStr $ "Enter parameter for test case " ++ show k ++ ": "
a <- fmap read getLine :: IO Int
let output = myTest a -- run the test
loop (k+1) (output:acc)
allOutput <- loop 1 []
print allOutput
As you get more comfortable with this kind of pattern you'll recognize it as a fold (indeed a monadic fold since we're doing IO) and you can implement it with foldM.
Update: To help explain how fmap works, here are equivalent expressions written without using fmap:
With fmap: Without fmap:
n <- fmap read getLine :: IO [Int] line <- getLine
let n = read line :: Int
vals <- fmap (map read . words) getLine line <- getLine
:: IO [Int] let vals = (map read . words) line :: [Int]
Using fmap allows us to eliminate the intermediate variable line which we never reference again anyway. We still need to provide a type signature so read knows what to do.
The idiomatic way is to use replicateM:
runAllTests :: [[Int]] -> IO ()
runAllTests = {- ... -}
main = do
numTests <- readLn
tests <- replicateM numTests readLn
runAllTests tests
-- or:
-- main = readLn >>= flip replicateM readLn >>= runAllTests

Non-exhaustive patterns in lambda

I am getting Non-exhaustive patterns in lambda. I am not sure of the cause yet. Please anyone how to fix it. The code is below:
import Control.Monad
import Data.List
time_spent h1 h2 = max (abs (fst h1 - fst h2)) (abs (snd h1 - snd h2))
meeting_point xs = foldl' (find_min_time) maxBound xs
where
time_to_point p = foldl' (\tacc p' -> tacc + (time_spent p p')) 0 xs
find_min_time min_time p = let x = time_to_point p in if x < min_time then x else min_time
main = do
n <- readLn :: IO Int
points <- fmap (map (\[x,y] -> (x,y)) . map (map (read :: String->Int)) . map words . lines) getContents
putStrLn $ show $ meeting_point points
This is the lambda with the non-exhaustive patterns: \[x,y] -> (x,y).
The non-exhaustive pattern is because the argument you've specified, [x,y] doesn't match any possible list - it only matches lists with precisely two elements.
I would suggest replacing it with a separate function with an error case to print out the unexpected data in an error message so you can debug further, e.g.:
f [x,y] = (x, y)
f l = error $ "Unexpected list: " ++ show l
...
points <- fmap (map f . map ...)
As an addition to #GaneshSittampalam's answer, you could also do this with more graceful error handling using the Maybe monad, the mapM function from Control.Monad, and readMaybe from Text.Read. I would also recommend refactoring your code so that the parsing is its own function, it makes your main function much cleaner and easier to debug.
import Control.Monad (mapM)
import Text.Read (readMaybe)
toPoint :: [a] -> Maybe (a, a)
toPoint [x, y] = Just (x, y)
toPoint _ = Nothing
This is just a simple pattern matching function that returns Nothing if it gets a list with length not 2. Otherwise it turns it into a 2-tuple and wraps it in Just.
parseData :: String -> Maybe [(Int, Int)]
parseData text = do
-- returns Nothing if a non-Int is encountered
values <- mapM (mapM readMaybe . words) . lines $ text
-- returns Nothing if a line doesn't have exactly 2 values
mapM toPoint values
Your parsing can actually be simplified significantly by using mapM and readMaybe. The type of readMaybe is Read a => String -> Maybe a, and in this case since we've specified the type of parseData to return Maybe [(Int, Int)], the compiler can infer that readMaybe should have the local type of String -> Maybe Int. We still use lines and words in the same way, but now since we use mapM the type of the right hand side of the <- is Maybe [[Int]], so the type of values is [[Int]]. What mapM also does for us is if any of those actions fails, the overall computation exits early with Nothing. Then we simply use mapM toPoint to convert values into a list of points, but also with the failure mechanism built in. We actually could use the more general signature of parseData :: Read a => String -> Maybe [(a, a)], but it isn't necessary.
main = do
n <- readLn :: IO Int
points <- fmap parseData getContents
case points of
Just ps -> print $ meeting_point ps
Nothing -> putStrLn "Invalid data!"
Now we just use fmap parseData on getContents, making points have the type Maybe [(Int, Int)]. Finally, we pattern match on points to print out the result of the meeting_point computation or print a helpful message if something went wrong.
If you wanted even better error handling, you could leverage the Either monad in a similar fashion:
toPoint :: [a] -> Either String (a, a)
toPoint [x, y] = Right (x, y)
toPoint _ = Left "Invalid number of points"
readEither :: Read a => String -> Either String a
readEither text = maybe (Left $ "Invalid parse: " ++ text) Right $ readMaybe text
-- default value ^ Wraps output on success ^
-- Same definition with different type signature and `readEither`
parseData :: String -> Either String [(Int, Int)]
parseData text = do
values <- mapM (mapM readEither . words) . lines $ text
mapM toPoint values
main = do
points <- fmap parseData getContents
case points of
Right ps -> print $ meeting_point ps
Left err -> putStrLn $ "Error: " ++ err

Resources