haskell-chat/code-challenges

A place to share coding challenges in Haskell.

Baldur

@Iceland_jack_twitter

Double every other element of a list

Deepak Kapiswe

@deepakkapiswe_twitter

doubleElement::[a]->[a]

doubleElement = concatMap (\x->x:x:[])

Baldur

@Iceland_jack_twitter

Actually meant (*2) every number

But that

Deepak Kapiswe

@deepakkapiswe_twitter

then its very simple

Baldur

@Iceland_jack_twitter

But that solution is good (bonus points for using infinite lists of functions)

Deepak Kapiswe

@deepakkapiswe_twitter

dE::(Num a)=> [a]->[a]
dE = ((*2) <$>)

Baldur

@Iceland_jack_twitter

How about doubling every OTHER element @deepakkapiswe_twitter :)

(challenge: use lens)

(challenge: make the offset (2) an argument)

f 3 [1,11,111,1111,...] gives [2,11,111,2222,...]

Olli Helenius

@liff

f n = zipWith ($) (cycle ((*2) : replicate (n - 1) id))

Baldur

@Iceland_jack_twitter

Great @liff !!

Deepak Kapiswe

@deepakkapiswe_twitter

but how to use lenses ... I have used lens only till (^. , %~ , .~ . , & )

Baldur

@Iceland_jack_twitter

@liff Here is a lens solution in an old comment of mine

I create an (indexed) traversal

that focuses on every 2nd, 3rd, ... element

>>> toListOf (every 1) [0..4]
[0,1,2,3,4]
>>> toListOf (every 2) [0..4]
[0,2,4]
>>> toListOf (every 3) [0..4]
[0,3]

>>> itoListOf (every 2) "Hello"
[(0,'H'),(2,'l'),(4,'o')]

This is very powerful

Baldur

@Iceland_jack_twitter

Where every has the definition

every :: Int -> IndexedTraversal' Int [a] a
every n = traversed.indices (n `divides`)

Using operators

>>> [0..4] ^.. every 1
[0,1,2,3,4]
>>> [0..4] ^.. every 2
[0,2,4]
>>> [0..4] ^.. every 3
[0,3]

Deepak Kapiswe

@deepakkapiswe_twitter

nice one ...

but can you explain what is every

Baldur

@Iceland_jack_twitter

Are you familiar with traversals?

Deepak Kapiswe

@deepakkapiswe_twitter

no not now ...

Baldur

@Iceland_jack_twitter

a traversal can focus on many elements of a structure

so if you use a regular traverse (or traversed) you are focusing on every element

Baldur

@Iceland_jack_twitter

Continued in PM

Senenmut

@Senenmut

Hi. i want to send a email with Haskell and have here a copy of all code.

Hi there ,

i want to run a simple email sender with smtp from the Module :
Network.HaskellNet.SMTP.
The Modul is already installed. cabal list --installed tells this !!!
However, there is a compilation error that is asking for the Module :
Network.HaskellNet.Auth.
But cabal cannot FIND this module ! ANY IDEA WOULD HELP !

Here is the example code of the session :

import Network.HaskellNet.SMTP
import Network.HaskellNet.Auth
import qualified Data.Text.Lazy as T

main = doSMTP "your.smtp.server.com" $ \conn ->
authSucceed <- authenticate PLAIN "username" "password" conn
if authSucceed
then sendPlainTextMail "receiver@server.com" "sender@server.com" "subject" (T.pack "Hello! This is the mail body!") conn
else print "Authentication failed."

Here is the result of compilation with the right smtp data.

$ ghc -o sendmail_4 sendmail_4.hs

sendmail_4.hs:2:8:
Could not find module `Network.HaskellNet.Auth'
Use -v to see a list of the files searched for.
hans@mx1:~/Documents/Haskell
$

cabal has listed the following Network Module :

network installed version 2.4.1.2 , 2.6.3.3

cabal install Network.HaskellNet.Auth does NOT work. cabal cannot FIND the Package , but normally AUTH should be IN the main Network Package that is already installed.

juliopc

@juliopc

hello guys!! im tryin to write a trade assistant plugin. anyone can send a transaction from A type client to B type client... can someone advise me a open source for that

Jeyabalaji Subramanian

@jeyabalajis

Hello everyone

I am new to Haskell. was looking for a simple yet elegant way to find whether a word contains alternate vowels.

ram, rama, arav as inputs will return True. ramy would return False

have broken this down into a series of checks on each letter which checks if the letter is a vowel

Jeyabalaji Subramanian

@jeyabalajis

isVowel :: Char -> Bool
isVowel x = x `elem` ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']

if you are evaluating "arjun", this will translate to: isVowel 'a' && not (isVowel 'r') && isVowel 'j' && not (isVowel 'u') && isVowel 'n'

but not sure how to codify this. any help / hint is appreciated. thanks!

Jeyabalaji Subramanian

@jeyabalajis

isVowel :: Char -> Bool
isVowel x = x `elem` ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']

isNotVowel :: Char -> Bool
isNotVowel x = not (isVowel x)

recursiveAlternateVowels :: String -> Bool -> Bool
recursiveAlternateVowels xs v = go xs v where
    go [] _ = True
    go (x:xs) True = isVowel x && go xs False
    go (x:xs) False = isNotVowel x && go xs True

wordContainsAlternateVowels :: String -> Bool
wordContainsAlternateVowels [] = True
wordContainsAlternateVowels (x:xs)
    | isVowel x = recursiveAlternateVowels ([x]++xs) True
    | isNotVowel x = recursiveAlternateVowels ([x]++xs) False

solved this using "inner" recursive functions, which was pointed out to me in the beginners channel.

Ana

@anabeatrizzz

Hello, my teacher challenged me to show him the first 12 and last 12 digits of the result of the expression 2 ^ 17179869184. I already got the last 12 digits, I would like help to get the first 12 digits.

Enzo Crance

@cl0udyhug_twitter

This might be a shorter solution for the problem from July 27th. Sorry for being late I just installed the app and I am new to Haskell.

isVowel :: Char -> Bool
isVowel = flip elem "aeiouAEIOU"

checkAlternate :: (a -> Bool) -> [a] -> Bool
checkAlternate p0 l = all id $ reverse bools
  where bools = snd $ foldl (\(p, acc) c -> (not . p, p c : acc)) (p0, []) l

checkAlternateVowels :: String -> Bool
checkAlternateVowels = checkAlternate isVowel

Rita123-j

@Rita123-j

Hello everyone ,

I have a exercise for you guys, but don't know how to upload it here

can you tell me how

Microtribute

@Microtribute

Can you guys convert this Java code into Haskell?

int firstDuplicate(int[] a) {
    int n = a.length;

    boolean[] b = new boolean[n];

    for (int i = 0; i < n; i++) {
        if (b[a[i] - 1]) return a[i];
        b[a[i] - 1] = true;
    }

    return -1;
}

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