I'm new to haskell and I got stuck in a little program I tried to make. I want to count number of times my guard statement goes through in all recursion rounds and then return that as Int. For example if c1 is 'a', c2 is 'b', g is 2 and s is "aaabbb" then returned int would be 2, because my guard statement is true in 2 cases.
I tried to make variable x and then add x + 1 to it every time guard statement happens. That didn't work because I learnt that in Haskell variable you set is always static so for example setting x = 0 at start would set that x to 0 every recursion round.
Here's my code:
gaps :: (Char, Char) -> Int -> String -> Int
gaps (c1,c2) g (s:xs)
| c1 == s && c2 == (s:xs) !! g = --Count how many times this statement happens --
| otherwise = gaps (c1,c2) g xs
Just add 1 and call the function recursively
gaps :: (Char, Char) -> Int -> String -> Int
gaps _ _ [] = 0 -- base case
gaps (c1,c2) g (s:xs)
| c1 == s && c2 == (s:xs) !! g = 1 + gaps (c1,c2) g xs -- add one to final result
| otherwise = gaps (c1,c2) g xs
> gaps ('a','b') 2 "aaabbb"
2
> gaps ('a','b') 3 "aaaabbbb"
3
Be carefull when using !!. It isn't total and might fail if your input string has c1's values less than g positions before the end of the string
> gaps ('a','b') 3 "aaaababbb" -- doesn't fail
3
> gaps ('a','b') 3 "aaaabbabb" -- does fail!!!!
Off the back of questions on how to make this thing safer, I have made the following code snippet, borrowing from Ismor's answer.
-- a way to safely get the nth item from a list
get' :: [a] -> Int -> Maybe a
get' [] _ = Nothing
get' (x:xs) 0 = Just x
get' (x:xs) n
| n > 0 = get' xs (n-1)
| otherwise = Nothing
-- takes a Maybe value. if it's Nothing, return 0. if it's Just a value, compare
-- the value and a given param, if equal return 1, else 0
seeEqual:: (Num b, Eq a) => Maybe a -> a -> b
seeEqual Nothing _ = 0
seeEqual (Just a) b
| a==b = 1
| otherwise = 0
-- I have edited the first guard so that it checks c1 and s, then tries to calculate
-- whether c2 and the specific list item are equal, and then recurses as before
gaps :: (Char, Char) -> Int -> String -> Int
gaps _ _ [] = 0 -- base case
gaps (c1,c2) g (s:xs)
| c1 == s = (seeEqual (get' (s:xs) g) c2) + gaps (c1,c2) g xs -- add one to final result
| otherwise = gaps (c1,c2) g xs
I do not claim that this is perfect, but I do think this is safe and shouldn't throw any exceptions or raise any errors.
Prelude> gaps ('a','b') 3 "aaaababbb"
3
Prelude> gaps ('a','b') 3 "aaaabbabb"
2
Related
I am writing some code to work with arbitrary radix numbers in haskell. They will be stored as lists of integers representing the digits.
I almost managed to get it working, but I have run into the problem of converting a list of tuples [(a_1,b_1),...,(a_n,b_n)] into a single list which is defined as follows:
for all i, L(a_i) = b_i.
if there is no i such that a_i = k, a(k)=0
In other words, this is a list of (position,value) pairs for values in an array. If a position does not have a corresponding value, it should be set to zero.
I have read this (https://wiki.haskell.org/How_to_work_on_lists) but I don't think any of these methods are suitable for this task.
baseN :: Integer -> Integer -> [Integer]
baseN n b = convert_digits (baseN_digits n b)
chunk :: (Integer, Integer) -> [Integer]
chunk (e,m) = m : (take (fromIntegral e) (repeat 0))
-- This is broken because the exponents don't count for each other's zeroes
convert_digits :: [(Integer,Integer)] -> [Integer]
convert_digits ((e,m):rest) = m : (take (fromIntegral (e)) (repeat 0))
convert_digits [] = []
-- Converts n to base b array form, where a tuple represents (exponent,digit).
-- This works, except it ignores digits which are zero. thus, I converted it to return (exponent, digit) pairs.
baseN_digits :: Integer -> Integer -> [(Integer,Integer)]
baseN_digits n b | n <= 0 = [] -- we're done.
| b <= 0 = [] -- garbage input.
| True = (e,m) : (baseN_digits (n-((b^e)*m)) b)
where e = (greedy n b 0) -- Exponent of highest digit
m = (get_coef n b e 1) -- the highest digit
-- Returns the exponent of the highest digit.
greedy :: Integer -> Integer -> Integer -> Integer
greedy n b e | n-(b^e) < 0 = (e-1) -- We have overshot so decrement.
| n-(b^e) == 0 = e -- We nailed it. No need to decrement.
| n-(b^e) > 0 = (greedy n b (e+1)) -- Not there yet.
-- Finds the multiplicity of the highest digit
get_coef :: Integer -> Integer -> Integer -> Integer -> Integer
get_coef n b e m | n - ((b^e)*m) < 0 = (m-1) -- We overshot so decrement.
| n - ((b^e)*m) == 0 = m -- Nailed it, no need to decrement.
| n - ((b^e)*m) > 0 = get_coef n b e (m+1) -- Not there yet.
You can call "baseN_digits n base" and it will give you the corresponding array of tuples which needs to be converted to the correct output
Here's something I threw together.
f = snd . foldr (\(e,n) (i,l') -> ( e , (n : replicate (e-i-1) 0) ++ l')) (-1,[])
f . map (fromIntegral *** fromIntegral) $ baseN_digits 50301020 10 = [5,0,3,0,1,0,2,0]
I think I understood your requirements (?)
EDIT:
Perhaps more naturally,
f xs = foldr (\(e,n) fl' i -> (replicate (i-e) 0) ++ (n : fl' (e-1))) (\i -> replicate (i+1) 0) xs 0
I have this code that will return the index of a char in a char array but I want my function to return something like -1 if the value isn't in the array. As it stands the function returns the size of the array if the element isn't in the array. Any ideas on how to change my code in order to apply this feature?
I am trying not to use any fancy functions to do this. I just want simple code without built-in functions.
isPartOf :: [(Char)] -> (Char) -> Int
isPartOf [] a = 0
isPartOf (a:b) c
| a == c = 0
| otherwise = 1 + isPartOf b c
For example:
*Main> isPartOf [('a'),('b'),('c')] ('z')
3
But I want:
*Main> isPartOf [('a'),('b'),('c')] ('z')
-1
Let's try to define such a function, but instead of returning -1 in case of element being not a part of the list, we can return Nothing:
isPartOf :: Eq a => [a] -> a -> Maybe Int
isPartOf [] _ = Nothing
isPartOf (x : xs) a | x == a = Just 0
| otherwise = fmap ((+) 1) (isPartOf xs a)
So, it works like that:
>> isPartOf [('a'),('b'),('c')] ('z')
Nothing
it :: Maybe Int
>> isPartOf [('a'),('b'),('c')] ('c')
Just 2
it :: Maybe Int
After that we can use built-in function fromMaybe to convert the Nothing case to -1:
>> fromMaybe (-1) $ isPartOf [('a'),('b'),('c')] ('c')
2
it :: Int
>> fromMaybe (-1) $ isPartOf [('a'),('b'),('c')] ('z')
-1
it :: Int
In case you're curios if such a function already exist, you can use Hoogle for that, searching the [a] -> a -> Maybe Int function: https://www.haskell.org/hoogle/?hoogle=%5Ba%5D+-%3E+a+-%3E+Maybe+Int
And the first answer will be elemIndex:
>> elemIndex 'c' [('a'),('b'),('c')]
Just 2
it :: Maybe Int
>> elemIndex 'z' [('a'),('b'),('c')]
Nothing
it :: Maybe Int
Hope this helps.
The smallest change to achieve this is
isPartOf :: [Char] -> Char -> Int
isPartOf [] a = (-1) -- was: 0
isPartOf (a:b) c
| a == c = 0
| otherwise = 1 + -- was: isPartOf b c
if (isPartOf b c) < 0 then (-2) else (isPartOf b c)
This is terrible computationally though. It recalculates the same value twice; what's worse is that the calculation is done with the recursive call and so the recursive call will be done twice and the time complexity overall will change from linear to exponential!
Let's not do that. But also, what's so special about Char? There's lots of stuff special about the Char but none are used here, except the comparison, (==).
The types the values of which can be compared by equality are known as those belonging to the Eq (for "equality") type class: Eq a => a. a is a type variable capable of assuming any type whatsoever; but here it is constrained to be such that ... yes, belongs to the Eq type class.
And so we write
isPartOf :: Eq a => [a] -> a -> Int
isPartOf [] a = (-1)
isPartOf (a:b) c
| a == c = 0
| otherwise = let d = isPartOf b c in
1 + if d < 0 then (-2) else d
That (-2) looks terribly ad-hoc! A more compact and idiomatic version using guards will also allow us to address this:
isPartOf :: Eq a => [a] -> a -> Int
isPartOf [] a = (-1)
isPartOf (a:b) c
| a == c = 0
| d < 0 = d
| otherwise = 1 + d
where
d = isPartOf b c
Yes, we can define d in the where clause, and use it in our guards, as well as in the body of each clause. Thanks to laziness it won't even be calculated once if its value wasn't needed, like in the first clause.
Now this code is passable.
The conditional passing and transformation is captured by the Maybe data type's Functor interface / instance:
fmap f Nothing = Nothing -- is not changed
fmap f (Just x) = Just (f x) -- is changed
which is what the other answer here is using. But it could be seen as "fancy" when we only start learning Haskell.
When you've written more functions like that, and become "fed up" with repeating the same pattern manually over and over, you'll come to appreciate it and will want to use it. But only then.
Yet another concern is that our code calculates its result on the way back from the recursion's base case.
But it could instead calculate it on the way forward, towards it, so it can return it immediately when the matching character is found. And if the end of list is found, discard the result calculated so far, and return (-1) instead. This is the approach taken by the second answer.
Though creating an additional function litters the global name space. It is usual to do this by defining it internally, in the so called "worker/wrapper" transformation:
isPartOf :: Eq a => [a] -> a -> Int
isPartOf xs c = go xs 0
where
go [] i = (-1)
go (a:b) i
| a == c = i
| otherwise = -- go b (1 + i)
go b $! (1 + i)
Additional boon is that we don't need to pass around the unchanged value c -- it is available in the outer scope, from the point of view of the internal "worker" function go, "wrapped" by and accessible only to our function, isPartOf.
$! is a special call operator which ensures that its argument value is calculated right away, and not delayed. This eliminates an unwanted (in this case) laziness and improves the code efficiency even more.
But from the point of view of overall cleanliness of the design it is better to return the index i wrapped in a Maybe (i.e. Just i or Nothing) instead of using a "special" value which is not so special after all -- it is still an Int.
It is good to have types reflect our intentions, and Maybe Int expresses it clearly and cleanly, so we don't have to remember which of the values are special and which regular, so that that knowledge is not external to our program text, but inherent to it.
It is a small and easy change, combining the best parts from the two previous variants:
isPartOf :: Eq a => [a] -> a -> Maybe Int
isPartOf .....
.......
....... Nothing .....
.......
....... Just i .....
.......
(none of the code was tested. if there are errors, you're invited to find them and correct them, and validate it by testing).
You can achieve it easily if you just pass current element idx to the next recursion:
isPartOf :: [Char] -> Char -> Int
isPartOf lst c = isPartOf' lst c 0
isPartOf' :: [Char] -> Char -> Int -> Int
isPartOf' [] a _ = -1
isPartOf' (a:b) c idx
| a == c = idx
| otherwise = isPartOf' b c (idx + 1)
You are using your function as an accumulator. This is cool except the additions with negative one. An accumulator cannot switch from accumulating to providing a negative 1. You want two different things from your function accumulator. You can use a counter for one thing then if the count becomes unnecessary because no match is found and a negative 1 is issued and nothing is lost. The count would be yet another parameter. ugh. You can use Maybe but that complicates. Two functions, like above is simpler. Here are two functions. The first is yours but the accumulator is not additive it's concatenative.
cIn (x:xs) c | x == c = [1]
| null xs = [-1]
| otherwise = 1:cIn xs c
Cin ['a','b','c'] 'c'
[1,1,1]
cIn ['a','b','c'] 'x'
[1,1,-1]
So the second function is
f ls = if last ls == 1 then sum ls else -1
It will
f $ Cin ['a','b','c'] 'c'
3
and
f $ Cin ['a','b','c'] 'x'
-1
You can zero the index base by changing [1] to [0]
I am trying to build a function that converts a Decimal(Int) into a Binary number.
Unfortunately other than in java it is not possible to divide an int by two in haskell.
I am very new to functional programming so the problem could be something trivial.
So far I could not find another solution to this problem but
here is my first try :
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = if (mod n 2 == 0) then
do
0:fromDecimal(n/2)
else
do
1:fromDecimal(n/2)
I got an java implementation here which I did before :
public void fromDecimal(int decimal){
for (int i=0;i<values.length;i++){
if(decimal % 2 = 0)
values[i]=true ;
decimal = decimal/ 2;
else {values[i]= false;
} }
}
Hopefully this is going to help to find a solution!
There are some problems with your solution. First of all, I advise not to use do at all, until you understand what do does. Here we do not need do at all.
Unfortunately other than in java it is not possible to divide an int by two in haskell.
It actually is, but the / operator (which is in fact the (/) function), has type (/) :: Fractional a => a -> a -> a. An Int is not Fractional. You can perform integer division with div :: Integral a => a -> a -> a.
So then the code looks like:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = if (mod n 2 == 0) then 0:fromDecimal (div n 2) else 1:fromDecimal (div n 2)
But we can definitely make this more elegant. mod n 2 can only result in two outcomes: 0 and 1, and these are exactly the ones that we use at the left side of the (:) operator.
So we do not need to use an if-then-else at all:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = mod n 2 : fromDecimal (div n 2)
Likely this is still not exactly what you want: here we write the binary value such that the last element, is the most significant one. This function will add a tailing zero, which does not make a semantical difference (due to that order), but it is not elegant either.
We can define an function go that omits this zero, if the given value is not zero, like:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = go n
where go 0 = []
go k = mod k 2 : go (div k 2)
If we however want to write the most significant bit first (so in the same order as we write decimal numbers), then we have to reverse the outcome. We can do this by making use of an accumulator:
fromDecimal :: Int -> [Int]
fromDecimal 0 = [0]
fromDecimal n = go n []
where go 0 r = r
go k rs = go (div k 2) (mod k 2:rs)
You cannot / integers in Haskell – division is not defined in terms of integral numbers! For integral division use div function, but in your case more suitable would be divMod that comes with mod gratis.
Also, you are going to get reversed output, so you can reverse manually it after that, or use more memory-efficient version with accumulator:
decToBin :: Int -> [Int]
decToBin = go [] where
go acc 0 = acc
go acc n = let (d, m) = n `divMod` 2 in go (m : acc) d
go will give you an empty list for 0. You may add it manually if the list is empty:
decToBin = (\l -> if null l then [0] else l) . go [] where ...
Think through how your algorithm will work. It starts from 2⁰, so it will generate bits backward from how we ordinarily think of them, i.e., least-significant bit first. Your algorithm can represent non-negative binary integers only.
fromDecimal :: Int -> [Int]
fromDecimal d | d < 0 = error "Must be non-negative"
| d == 0 = [0]
| otherwise = reverse (go d)
where go 0 = []
go d = d `rem` 2 : go (d `div` 2)
In Haskell, when we generate a list in reverse, go ahead and do so but then reverse the result at the end. The reason for this is consing up a list (gluing new items at the head with :) has a constant cost and the reverse at the end has a linear cost — but appending with ++ has a quadratic cost.
Common Haskell style is to have a private inner loop named go that the outer function applies when it’s happy with its arguments. The base case is to terminate with the empty list when d reaches zero. Otherwise, we take the current remainder modulo 2 and then proceed with d halved and truncated.
Without the special case for zero, fromDecimal 0 would be the empty list rather than [0].
The binary numbers are usually strings and not really used in calculations.
Strings are also less complicated.
The pattern of binary numbers is like any other. It repeats but at a faster clip.
Only a small set is necessary to generate up to 256 (0-255) binary numbers.
The pattern can systematically be expanded for more.
The starting pattern is 4, 0-3
bd = ["00","01","10","11"]
The function to combine them into larger numbers is
d2b n = head.drop n $ [ d++e++f++g | d <- bd, e <- bd, f <- bd, g <- bd]
d2b 125
"01111101"
If it's not obvious how to expand, then
bd = ["000","001","010","011","100","101","110","111"]
Will give you up to 4096 binary digits (0-4095). All else stays the same.
If it's not obvious, the db2 function uses 4 pairs of binary numbers so 4 of the set. (2^8) - 1 or (2^12) - 1 is how many you get.
By the way, list comprehension are sugar coated do structures.
Generate the above patterns with
[ a++b | a <- ["0","1"], b <- ["0","1"] ]
["00","01","10","11"]
and
[ a++b++c | a <- ["0","1"], b <- ["0","1"], c <- ["0","1"] ]
["000","001","010","011","100","101","110","111"]
More generally, one pattern and one function may serve the purpose
b2 = ["0","1"]
b4 = [ a++b++c++d | a <- b2, b <- b2, c <- b2, d <- b2]
b4
["0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"]
bb n = head.drop n $ [ a++b++c++d | a <- b4, b <- b4, c <- b4, d <- b4]
bb 32768
"1000000000000000"
bb 65535
"1111111111111111"
To calculate binary from decimal directly in Haskell using subtraction
cvtd n (x:xs) | x>n = 0:(cvtd n xs)
| n>x = 1:(cvtd (n-x) xs)
| True = 1:[0|f<-xs]
Use any number of bits you want, for example 10 bits.
cvtd 639 [2^e|e<-[9,8..0]]
[1,0,0,1,1,1,1,1,1,1]
import Data.List
dec2bin x =
reverse $ binstr $ unfoldr ndiv x
where
binstr = map (\x -> "01" !! x)
exch (a,b) = (b,a)
ndiv n =
case n of
0 -> Nothing
_ -> Just $ exch $ divMod n 2
I'm trying the solve the first question in Advent of Code 2017, and come up with the following solution to calculate the needed value:
checkRepetition :: [Int] -> Bool
checkRepetition [] = False
checkRepetition (x:xs)
| x == ( head xs ) = True
| otherwise = False
test :: [Int] -> Int
test [] = 0
test [x] = 0
test xs
| checkRepetition xs == True = ((head xs)*a) + (test (drop a xs))
| otherwise = test (tail xs)
where
a = (go (tail xs)) + 1
go :: [Int] -> Int
go [] = 0
go xs
| checkRepetition xs == True = 1 + ( go (tail xs) )
| otherwise = 0
However, when I give an input that contains repetitive numbers such as [1,3,3], it gives the error
*** Exception: Prelude.head: empty list
However, for 1.5 hours, I couldn't figure out exactly where this error is generated. I mean any function that is used in test function have a definition for [], but still it throws this error, so what is the problem ?
Note that, I have checked out this question, and in the given answer, it is advised not to use head and tail functions, but I have tested those function for various inputs, and they do not throw any error, so what exactly is the problem ?
I would appreciate any help or hint.
As was pointed out in the comments, the issue is here:
checkRepetition (x:xs)
| x == ( head xs ) = True
xs is not guaranteed to be a non-empty list (a one-element list is written as x:[], so that (x:xs) pattern matches that xs = []) and calling head on an empty list is a runtime error.
You can deal with this by changing your pattern to only match on a 2+ element list.
checkRepetition [] = False
checkRepetition [_] = False
checkRepetition (x1:x2:_) = x1 == x2
-- No need for the alternations on this function, by the way.
That said, your algorithm seems needlessly complex. All you have to do is check if the next value is equal, and if so then add the current value to the total. Assuming you can get your String -> [Int] on your own, consider something like:
filteredSum :: [Int] -> Int
filteredSum [] = 0 -- by definition, zero- and one-element lists
filteredSum [_] = 0 -- cannot produce a sum, so special case them here
filteredSum xss#(first:_) = go xss
where
-- handle all recursive cases
go (x1:xs#(x2:_)) | x1 == x2 = x1 + go xs
| otherwise = go xs
-- base case
go [x] | x == first = x -- handles last character wrapping
| otherwise = 0 -- and if it doesn't wrap
-- this should be unreachable
go [] = 0
For what it's worth, I think it's better to work in the Maybe monad and operate over Maybe [Int] -> Maybe Int, but luckily that's easy since Maybe is a functor.
digitToMaybeInt :: Char -> Maybe Int
digitToMaybeInt '0' = Just 0
digitToMaybeInt '1' = Just 1
digitToMaybeInt '2' = Just 2
digitToMaybeInt '3' = Just 3
digitToMaybeInt '4' = Just 4
digitToMaybeInt '5' = Just 5
digitToMaybeInt '6' = Just 6
digitToMaybeInt '7' = Just 7
digitToMaybeInt '8' = Just 8
digitToMaybeInt '9' = Just 9
digitToMaybeInt _ = Nothing
maybeResult :: Maybe Int
maybeResult = fmap filteredSum . traverse digitToMaybeInt $ input
result :: Int
result = case maybeResult of
Just x -> x
Nothing -> 0
-- this is equivalent to `maybe 0 id maybeResult`
Thank you for the link. I went there first to glean the purpose.
I assume the input will be a string. The helper function below constructs a numeric list to be used to sum if predicate is True, that is, the zipped values are equal, that is, each number compared to each successive number (the pair).
The helper function 'nl' invokes the primary function 'invcap' Inverse Captcha with a list of numbers.
The nl function is a list comprehension. The invcap function is a list comprehension. Perhaps the logic in this question is at fault. Overly complicated logic is more likely to introduce errors. Proofs are very much easier when logic is not cumbersome.
The primary function "invcap"
invcap l = sum [ x | (x,y) <- zip l $ (tail l) ++ [head l], x == y]
The helper function that converts a string to a list of digits and invokes invcap with a list of numeric digits.
nl cs = invcap [ read [t] :: Int | t <- cs]
Invocation examples
Prelude> nl "91212129" ......
9 ' ' ' ' ' ' ' ' ' ' ' ' '
Prelude> nl "1122" ......
3
Right now I'm working on a problem in Haskell in which I'm trying to check a list for a particular pair of values and return True/False depending on whether they are present in said list. The question goes as follows:
Define a function called after which takes a list of integers and two integers as parameters. after numbers num1 num2 should return true if num1 occurs in the list and num2 occurs after num1. If not it must return false.
My plan is to check the head of the list for num1 and drop it, then recursively go through until I 'hit' it. Then, I'll take the head of the tail and check that against num2 until I hit or reach the end of the list.
I've gotten stuck pretty early, as this is what I have so far:
after :: [Int] -> Int -> Int -> Bool
after x y z
| y /= head x = after (drop 1 x) y z
However when I try to run something such as after [1,4,2,6,5] 4 5 I get a format error. I'm really not sure how to properly word the line such that haskell will understand what I'm telling it to do.
Any help is greatly appreciated! Thanks :)
Edit 1: This is the error in question:
Program error: pattern match failure: after [3,Num_fromInt instNum_v30 4] 3 (Num_fromInt instNum_v30 2)
Try something like this:
after :: [Int] -> Int -> Int -> Bool
after (n:ns) a b | n == a = ns `elem` b
| otherwise = after ns a b
after _ _ _ = False
Basically, the function steps through the list, element by element. If at any point it encounters a (the first number), then it checks to see if b is in the remainder of the list. If it is, it returns True, otherwise it returns False. Also, if it hits the end of the list without ever seeing a, it returns False.
after :: Eq a => [a] -> a -> a -> Bool
after ns a b =
case dropWhile (/= a) ns of
[] -> False
_:xs -> b `elem` xs
http://hackage.haskell.org/package/base-4.8.2.0/docs/src/GHC.List.html#dropWhile
after xs p1 p2 = [p1, p2] `isSubsequenceOf` xs
So how can we define that? Fill in the blanks below!
isSubsequenceOf :: Eq a => [a] -> [a] -> Bool
[] `isSubsequenceOf` _ = ?
(_ : _) `isSubsequenceOf` [] = ?
xss#(x : xs) `isSubsequenceOf` (y:ys)
| x == y = ?
| otherwise = ?
after :: [Int] -> Int -> Int -> Bool
Prelude> let after xs a b = elem b . tail $ dropWhile (/=a) xs
Examples:
Prelude> after [1,2,3,4,3] 88 7
*** Exception: Prelude.tail: empty list
It raises an exception because of tail. It's easy to write tail' such that it won't raise that exception. Otherwise it works pretty well.
Prelude> after [1,2,3,4,3] 2 7
False
Prelude> after [1,2,3,4,3] 2 4
True