How to write a password strength testing program in Haskell? - haskell

I am very new to Haskell, and I found some exercises on the internet, and one of them asks to write a program that takes a String and checks if the word has at least 8 characters, 1 uppercase letter, and 1 digit. In case it does, it returns true.
However I have absolutely no idea what to do.
I have tried to work with a Lambda expressions but I think it's not the way to go.
Would appreciate some help,
Thanks

I would suggest something like this.
import Data.Char
checkPassword :: String -> Bool
checkPassword xs = all (\f -> f xs) [checkOneDigit, checkOneUppercase, checkLength]
where checkOneDigit xs = (== 1) $ length $ filter isDigit xs
checkOneUppercase xs = (== 1) $ length $ filter isUpper xs
checkLength xs = (>= 8) $ length xs

Related

Haskell - Exclude lists based on a test in a nested list comprehension

I want to create a series of possible equations based on a general specification:
test = ["12", "34=", "56=", "78"]
Each string (e.g. "12") represents a possible character at that location, in this case '1' or '2'.)
So possible equations from test would be "13=7" or "1=68".
I know the examples I give are not balanced but that's because I'm deliberately giving a simplified short string.
(I also know that I could use 'sequence' to search all possibilities but I want to be more intelligent so I need a different approach explained below.)
What I want is to try fixing each of the equals in turn and then removing all other equals in the equation. So I want:
[["12","=","56","78"],["12","34","=","78”]]
I've written this nested list comprehension:
(it needs: {-# LANGUAGE ParallelListComp #-} )
fixEquals :: [String] -> [[String]]
fixEquals re
= [
[
if index == outerIndex then equals else remain
| equals <- map (filter (== '=')) re
| remain <- map (filter (/= '=')) re
| index <- [1..]
]
| outerIndex <- [1..length re]
]
This produces:
[["","34","56","78"],["12","=","56","78"],["12","34","=","78"],["12","34","56","”]]
but I want to filter out any with empty lists within them. i.e. in this case, the first and last.
I can do:
countOfEmpty :: (Eq a) => [[a]] -> Int
countOfEmpty = length . filter (== [])
fixEqualsFiltered :: [String] -> [[String]]
fixEqualsFiltered re = filter (\x -> countOfEmpty x == 0) (fixEquals re)
so that "fixEqualsFiltered test" gives:
[["12","=","56","78"],["12","34","=","78”]]
which is what I want but it doesn’t seem elegant.
I can’t help thinking there’s another way to filter these out.
After all, it’s whenever "equals" is used in the if statement and is empty that we want to drop the equals so it seems a waste to build the list (e.g. ["","34","56","78”] and then ditch it.)
Any thoughts appreciated.
I don't know if this is any cleaner than your code, but it might be a bit more clear and maybe more efficient using a recursion:
fixEquals = init . f
f :: [String] -> [[String]]
f [] = [[]]
f (x:xs) | '=' `elem` x = ("=":removeEq xs) : map (removeEq [x] ++) (f xs)
| otherwise = map (x:) (f xs)
removeEq :: [String] -> [String]
removeEq = map (filter (/= '='))
The way it works is that, if there's an '=' in the current string, then it splits the return into two, if not just calls recursively. The init is needed as in the last element returned there's no equal in any string.
Finally, I believe you can probably find a better data structure to do what you need to achieve instead of using list of strings
Let
xs = [["","34","56","78"],["12","=","56","78"],["12","34","=","78"],["12","34","56",""]]
in
filter (not . any null) xs
will give
[["12","=","56","78"],["12","34","=","78"]]
If you want list comprehension then do
[x | x <- xs, and [not $ null y | y <- x]]
I think I'd probably do it this way. First, a preliminary that I've written so many times it's practically burned into my fingers by now:
zippers :: [a] -> [([a], a, [a])]
zippers = go [] where
go _ [] = []
go b (h:e) = (b,h,e):go (h:b) e
Probably running it once or twice in ghci will be a more clear explanation of what this does than any English writing I could do:
> zippers "abcd"
[("",'a',"bcd"),("a",'b',"cd"),("ba",'c',"d"),("cba",'d',"")]
In other words, it gives a way of selecting each element of a list in turn, giving the "leftovers" of what was before and after the selection point. Given that tool, here's our plan: we'll nondeterministically choose a String to serve as our equals sign, double-check that we've got an equals sign in the first place, and then clear out the equals from the others. So:
fixEquals ss = do
(prefix, s, suffix) <- zippers ss
guard ('=' `elem` s)
return (reverse (deleteEquals prefix) ++ ["="] ++ deleteEquals suffix)
deleteEquals = map (filter ('='/=))
Let's try it:
> fixEquals ["12", "34=", "56=", "78"]
[["12","=","56","78"],["12","34","=","78"]]
Perfect! But this is just a stepping-stone to actually generating the equations, right? It turns out to be not that hard to go all the way in one step, skipping this intermediate. Let's do that:
equations ss = do
(prefixes, s, suffixes) <- zippers ss
guard ('=' `elem` s)
prefix <- mapM (filter ('='/=)) (reverse prefixes)
suffix <- mapM (filter ('='/=)) suffixes
return (prefix ++ "=" ++ suffix)
And we can try it in ghci:
> equations ["12", "34=", "56=", "78"]
["1=57","1=58","1=67","1=68","2=57","2=58","2=67","2=68","13=7","13=8","14=7","14=8","23=7","23=8","24=7","24=8"]
The easiest waty to achieve what you want is to create all the combinations and to filter the ones that have a meaning:
Prelude> test = ["12", "34=", "56=", "78"]
Prelude> sequence test
["1357","1358","1367","1368","13=7","13=8","1457","1458","1467","1468","14=7","14=8","1=57","1=58","1=67","1=68","1==7","1==8","2357","2358","2367","2368","23=7","23=8","2457","2458","2467","2468","24=7","24=8"
Prelude> filter ((1==).length.filter('='==)) $ sequence test
["13=7","13=8","14=7","14=8","1=57","1=58","1=67","1=68","23=7","23=8","24=7","24=8","2=57","2=58","2=67","2=68"]
You pointed the drawback: imagine we have the followig list of strings: ["=", "=", "0123456789", "0123456789"]. We will generate 100 combinations and drop them all.
You can look at the combinations as a tree. For the ["12", "34"], you have:
/ \
1 2
/ \ / \
3 4 3 4
You can prune the tree: just ignore the subtrees when you have two = on the path.
Let's try to do it. First, a simple combinations function:
Prelude> :set +m
Prelude> let combinations :: [String] -> [String]
Prelude| combinations [] = [""]
Prelude| combinations (cs:ts) = [c:t | c<-cs, t<-combinations ts]
Prelude|
Prelude> combinations test
["1357","1358","1367","1368","13=7","13=8","1457","1458","1467","1468","14=7","14=8","1=57","1=58","1=67","1=68","1==7","1==8","2357","2358","2367","2368","23=7","23=8","2457","2458","2467","2468","24=7","24=8", ...]
Second, we need a variable to store the current number of = signs met:
if we find a second = sign, just drop the subtree
if we reach the end of a combination with no =, drop the combination
That is:
Prelude> let combinations' :: [String] -> Int -> [String]
Prelude| combinations' [] n= if n==1 then [""] else []
Prelude| combinations' (cs:ts) n = [c:t | c<-cs, let p = n+(fromEnum $ c=='='), p <= 1, t<-combinations' ts p]
Prelude|
Prelude> combinations' test 0
["13=7","13=8","14=7","14=8","1=57","1=58","1=67","1=68","23=7","23=8","24=7","24=8","2=57","2=58","2=67","2=68"]
We use p as the new number of = sign on the path: if p>1, drop the subtree.
If n is zero, we don't have any = sign in the path, drop the combination.
You may use the variable n to store more information, eg type of the last char (to avoid +* sequences).

Haskell - Removing non-letter characters but ignoring white spaces?

I am very new to Haskell. I am trying to return a list of strings from a given string (which could contain non-letter characters) but I get a single string in the list.
The below code shows What I have tried so far:
toLowerStr xs = map toLower xs
--drop non-letters characters
dropNonLetters xs = words $ (filter (\x -> x `elem` ['a'..'z'])) $ toLowerStr xs
lowercase all the characters by using toLower function
remove non-letter characters by using filter function
return a list of strings by using words function
I think the filter function is removing the white spaces and therefore it becomes a single string. I tried using isSpace function but I don't know exactly how to implement it in this case.
What is it that I am doing wrong? I get this output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orangeappleapple"]
But I want to achieve the below output:
λ> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I think the filter function is removing the white spaces and therefore it becomes a single string.
That is correct. As filter predicate you write \x -> x `elem` ['a'..'z']. ['a'..'z'] is a list that contains lowercase letters, so for whitespace, the predicate will fail, and thus you should allow spaces as well.
We can for instance add the space character to the list:
dropNonLetters xs = words $ (filter (\x -> x `elem` (' ':['a'..'z'])))) $ toLowerStr xs
But this is inelegant and does not really explain itself. The Data.Char module however ships with two functions that are interesting here: isLower :: Char -> Bool, and isSpace :: Char -> Bool. We can use this like:
dropNonLetters xs = words $ (filter (\x -> isLower x || isSpace x)) $ toLowerStr xs
isLower and isSpace are not only more "descriptive" and elegant. Usually these functions will be faster than a membership check (which will usually be done in O(n)), and furthermore it will also take into account tabs, new lines, etc.
We can also perform an eta-reduction on the function:
dropNonLetters = words . (filter (\x -> isLower x || isSpace x)) . toLowerStr
This then produces:
Prelude Data.Char> dropNonLetters "ORANGE, apple! APPLE!!"
["orange","apple","apple"]
I advise you to rename the function dropNonLetters, since now it does not fully explain that it will generate a list of words. Based on the name, I would think that it only drops non-letters, not that it converts the string to lowercase nor that it constructs words.
here's an example of separating characters into separate string lists:
sortNumbers :: [Char] -> [String]
sortNumbers args = filter (\strings ->strings/= "") $ zipWith (\x numbers -> filter (\char -> char == numbers) x) (repeat args)
['1'..'9']

rules of operator's $ usage in haskell

I tried the $ operator and came to a weird point where I don't understand what is happening.
f = double $ 3+4
but this does not
f xs = xs !! $ length xs - 1
Can the operator $ only appear after function names and not after operators?
If so is there a way to write the following function without the parenthesis
f xs = xs !! (length xs - 1)
You can't have two operators next to each other, it's invalid syntax. This holds true for all operators. So xs !! $ length xs - 1 just will not work. There isn't really a good way to get rid of the parens in that statement without resorting to some tricks that aren't worth it and make the code less readable.
The $ operator does not have to appear after function names, though, you can do
> let add x y = x + y
> add 1 $ add 2 $ add 3 4
Haskell infix notation is problematic, avoid it like this
f xs = (!!) xs $ length xs - 1
The brackets turn an infix operator into a prefix function.

Haskell function that returns number of elements that overlap

How would you go about defining a function that takes two strings, say string x and string y and return the number of elements at the end of the first string (string x) which overlap with the beginning of the second string (second y).
I'm thinking that it would be a good idea to use isPrefixOf from the prelude to do this since it checks if a string is in another string and returns True if it does or False otherwise. But I'm a little confused at how to return the count of how many elements overlap. I wrote some pseudocode based off how I think you would approach the problem.
countElements :: Eq a => (Str a, Str a) -> Int
countElements (x,y) =
if x `isPrefixOf` y == True
then return the count of how many elements overlap
otherwise 0
A sample output would be:
countElements (board, directors) = 1
countElements (bend, ending) = 3
Any help here? I'm not very good at writing Haskell code.
You have exactly the right idea, but your pseudocode misses the fact that you'll have to iterate on all of the possible tails of the first string passed to the function.
countElements :: String -> String -> Int
countElements s t = length $ head $ filter (`isPrefixOf` t) (tails s)
> countElements "board" "directors"
1
> countElements "bend" "endings"
3
Version without isPrefixOf:
import Data.List
countElements xs ys = length . fst . last . filter (uncurry (==)) $ zip (reverse $ tails xs) (inits ys)

Haskell Assignment - direction needed to split a String into words

we started a paper on Haskell a few weeks ago and just received our first assignment. I'm aware that SO doesn't like homework questions, so I'm not going to ask how to do it. Instead, it would be very much appreciated if anyone could push me in the right direction with this. Seeing as it might not be a specific question, would it be more appropriate in a discussion / community wiki?
Question: Tokenize a String, that is: "Hello, World!" -> ["Hello", "World"]
Coming from a Java background, I have to forget everything about the usual way to go about this. The problem is that I am still very clueless with Haskell. This is what I've come up with:
module Main where
main :: IO()
main = do putStrLn "Type in a string:\n"
x <- getLine
putStrLn "The string entered was:"
putStrLn x
putStrLn "\n"
print (tokenize x)
tokenize :: String -> [String]
tokenize [] = []
tokenize l = token l ++ tokenize l
token :: String -> String
token [] = []
token l = takeWhile (isAlphaNum) l
What would be the first glaring mistake?
Thank you.
The first glaring mistake is
tokenize l = token l ++ tokenize l
(++) :: [a] -> [a] -> [a] appends two lists of the same type. Since token :: String -> String (and type String = [Char]), the type of tokenize that is inferred from that line is tokenize :: String -> String.
You should use (:) :: a -> [a] -> [a] here.
The next mistake in that line is that in the recursive call, you pass the same input l once again, so you have an infinite recursion, always doing the same without change. You have to remove the first token (and a bit more) from the input for the argument to the recursive call.
Another problem is that your token supposes that the input begins with alphanumeric characters.
You also need a function that ensures that condition for what you pass to token.
This line results in an infinite list (which is OK, since Haskell is lazy, so the list only gets constructed "on demand"), because it is recurring with no change in the arguments:
tokenize l = token l ++ tokenize l
We can visualise what is happening when tokenize is called as:
tokenize l = token l ++ tokenize l
= token l ++ (token l ++ tokenize l)
= token l ++ (token l ++ (token l ++ tokenize l))
= ...
To stop this happening, you need to change what the argument to tokenize so that it recurs sensibly:
tokenize l = token l ++ tokenize <something goes here>
As others already pointed out your mistake, just a little hint: While you found already the very useful takeWhile function, you should have a look at span, as this could be even more helpful here.
This has something in it that feels similar to a parser monad. However, as you're a newcomer to Haskell, it's unlikely that you're in a position to understand how parsing monads work (or use them in your code) quite yet. To give you the basics, consider what you want:
tokenize :: String -> [String]
This takes a String, chomps it up into more pieces, and generates a list of strings corresponding to the words in the input string. How might we represent this? What we want to do is find a function that processes a single string, and at the first sign of whitespace, adds that string on to the sequence of words. But then you have to process what's left over. (I.e., the rest of the string.) For example, let's say you want to tokenize:
The brown fox jumped
You first pull out "The" and then continue processing " brown fox jumped" (note the space at the beginning of the second string). You will do this recursively, so naturally you will need a recursive function.
The natural solution that sticks out is to take something where you accumulate a set of strings you've tokenized so far, keep munching on the current input until you hit whitespace, then also accumulate what you've seen in the current string (this leads to an implementation where you're mostly consing stuff, and then occasionally reversing stuff).
Your exercise seemed a bit challenging to me so I decided to solve it just for self-training. Here's what I came up with:
import Data.List
import Data.Maybe
splitByAnyOf yss xs =
foldr (\ys acc -> concat $ map (splitBy ys) acc) [xs] yss
splitBy ys xs =
case (precedingElements ys xs, succeedingElements ys xs) of
(Just "", Just s) -> splitBy ys s
(Just p, Just "") -> [p]
(Just p, Just s) -> p : splitBy ys s
otherwise -> [xs]
succeedingElements ys xs =
fromMaybe Nothing . find isJust $ map (stripPrefix ys) $ tails xs
precedingElements ys xs =
fromMaybe Nothing . find isJust $ map (stripSuffix ys) $ inits xs
where
stripSuffix ys xs =
if ys `isSuffixOf` xs then Just $ take (length xs - length ys) xs
else Nothing
main = do
print $ splitBy "!" "Hello, World!"
print $ splitBy ", " "Hello, World!"
print $ splitByAnyOf [", ", "!"] "Hello, World!"
outputs:
["Hello, World"]
["Hello","World!"]
["Hello","World"]

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