I'm trying to solve the 8 queens problem in Haskell without the use of any advanced functions, only with basic knowledge. I have come this far only but I'm getting an error that I can't understand.
The code:
queens = [[x1,x2,x3,x4,x5,x6,x7,x8] | x1<-[1..8], x2<-[1..8],
x3<-[1..8], x4<-[1..8], x5<-[1..8],
x6<-[1..8], x7<-[1..8], x8<-[1..8],
safeH [x2,x3,x4,x5,x6,x7,x8] x1]
safeH xs e = if length xs == 1 then head xs
else e /= safeH (tail xs) (head xs)
and the error message is:
y.hs:1:42:
No instance for (Num Bool) arising from the literal `1'
Possible fix: add an instance declaration for (Num Bool)
In the expression: 1
In the expression: [1 .. 8]
In a stmt of a list comprehension: x1 <- [1 .. 8]
[1 of 1] Compiling Main ( y.hs, interpreted )
Failed, modules loaded: none.
The culprit is
.........
safeH [x2,x3,x4,x5,x6,x7,x8] x1]
safeH xs e = if length xs == 1 then head xs
else e /= safeH (tail xs) (head xs)
specifically,
else e /= safeH (tail xs) (head xs)
because e == x1. So on the one hand safeH returns Bool, being used as a test in the list comprehension. OTOH you compare its result with x1. Which is 1, among other things (x1<-[1..8]). I.e. Num1. Which must also be a Bool. Hence the error.
1 A numeric literal such as 1 is parsed as a value of a polymorphic type Num a => a. I.e. its concrete type must belong to the Num type class. Since the concrete type is also determined to be Bool here, this means that Bool must belong to the Num type class, for this code to typecheck. Hence the instance for (Num Bool) is sought.
The if ... then ... else ... expression in safeH is not well-typed:
safeH l e = if length l == 1 then head l
else e /= safeH(tail l)(head l)
The then branch is incorrectly returning a numeric type, while the else branch is returning a boolean type Bool, as I think you intended.
You should add type signatures to all your top-level functions as a way of documenting what your code does, organizing your thoughts, and making errors easy to understand; the error message here is needlessly confusing because GHC infers that your code is returning some Num type thing from the first branch, and so when the second branch returns Bool GHC complains about the wrong thing: there being no instance of Num for the Bool type).
You should also read about pattern matching on lists, and take a look at the implementation of length and think about why it's not the best way to implement your function here.
So instead of using length and head, start with this framework:
safeH :: [Int] -> Int -> Bool
safeH [n] e = -- the case for a 1-length list
safeH (n:ns) e = -- ???
When you get something working then try redefining it where the base case is the empty list [].
Related
insert_at insert an element e into a list xs in specific location n
testgetting Left if n less than 0
test2 getting Left if n larger than xs's length or The type e isn't match the element's type in listxs. Otherwise passing Right xs to the next.
import Data.Typeable
insert_at :: a -> [a] -> Int -> [a]
insert_at e xs n = a++(e:b) where
t = splitAt n xs
a = fst t
b = snd t
test :: (Ord a, Num a) => b -> a -> Either [Char] b
test xs n = if n < 0 then Left "n<0" else Right xs
test2 :: (Typeable a1, Typeable a2) =>
a1 -> [a2] -> Int -> Either [Char] [a2]
test2 e xs n
| n> ( length xs )= Left "n> $ length xs "
| (typeOf e) /= (typeOf (head xs) ) = Left "(typeOf e) /= (typeOf (head xs) ) "
|otherwise = Right xs
sf :: Typeable a => a -> [a] -> Int -> Either [Char] [a]
sf e xs n = test xs n >>test2 e xs n >> Right (insert_at e xs n)
All the other error got properly handled, expect this.
* No instance for (Num Char) arising from the literal `1'
* In the expression: 1
In the second argument of `sf', namely `[1, 2, 3, 4, ....]'
In the expression: sf 'a' [1, 2, 3, 4, ....] 3
The error message states that you're trying to evaluate the expression sf 'a' [1, 2, 3, 4, ....] 3. Since this expression is not shown in your question, I'm assuming you're using it in GHCi to test out your code, right?
The type signature of sf says that the first parameter has type a, and the second parameter has type [a], which is a list whose elements are of the same type as the first parameter.
So the compiler sees that the first parameter is 'a'. That's a character, type Char.
"Got it," - thinks the compiler, - "now I know that a is Char. And now I know that the second parameter must have type [Char] - that is, a list of Char".
And yes, the second parameter is indeed a list, but wait! The first element of the list is not a character, but number 1! That does not compute!
Fortunately, number literals are special in Haskell. Number literals are not merely of type Int or even of type Integer, no! Numbers can be of any type as long as that type has an instance of class Num.
So since the compiler already knows that the elements of that list must be of type Char, but it sees a number literal, it concludes that it must now find an instance of class Num for type Char.
But there is no such instance! And so the compiler rightly complains: "no instance Num Char"
To fix the problem, I need to better understand what you were actually trying to do.
Did you intend the whole function to work on numbers? Then the first parameter must be a number, not a character.
Or did you intend it to work on characters? Then the second parameter must be a list of characters, not numbers.
Or did you intend the first two parameters not to be the same type at all? Then you must change the type signature of sf to indicate that.
sf expects a value and a list as its first two arguments. The elements of the list must have the same type as the first argument. This is the meaning of
sf :: a -> [a] -> ...
When you write sf 'a' [1], this means 1 and 'a' must have the same type. So the type checker looks for a way to interpret 1 as a Char; it fails, because this is not possible. Some fixes might include:
sf 'a' "1234" 3
sf 'a' [toEnum 1, toEnum 2, toEnum 3, toEnum 4] 3
sf (fromEnum 'a') [1, 2, 3, 4] 3
I am writing a small function in Haskell to check if a list is a palindrome by comparing it with it's reverse.
checkPalindrome :: [Eq a] -> Bool
checkPalindrome l = (l == reverse l)
where
reverse :: [a] -> [a]
reverse xs
| null xs = []
| otherwise = (last xs) : reverse newxs
where
before = (length xs) - 1
newxs = take before xs
I understand that I should use [Eq a] in the function definition because I use the equality operator later on, but I get this error when I compile:
Expected kind ‘*’, but ‘Eq a’ has kind ‘GHC.Prim.Constraint’
In the type signature for ‘checkPalindrome’:
checkPalindrome :: [Eq a] -> Bool
P.s Feel free to correct me if I am doing something wrong with my indentation, I'm very new to the language.
Unless Haskell adopted a new syntax, your type signature should be:
checkPalindrome :: Eq a => [a] -> Bool
Declare the constraint on the left hand side of a fat-arrow, then use it on the right hand side.
Unlike OO languages, Haskell makes a quite fundamental distinction between
Constraints – typeclasses like Eq.
Types – concrete types like Bool or lists of some type.
In OO languages, both of these would be represented by classes†, but a Haskell type class is completely different. You never have “values of class C”, only “types of class C”. (These concrete types may then contain values, but the classes don't.)
This distinction may seem pedantic, but it's actually very useful. What you wrote, [Eq a] -> Bool, would supposedly mean: each element of the list must be comparable... but comparable to what? You could have elements of different type in the list, how do you know that these elements are comparable to each other? In Haskell, that's no issue, because whenever the function is used you first settle on one type a. This type must be in the Eq class. The list then must have all elements from the same type a. This way you ensure that each element of the list is comparable to all of the others, not just, like, comparable to itself. Hence the signature
checkPalindrome :: Eq a => [a] -> Bool
This is the usual distinction on the syntax level: constraints must always‡ be written on the left of an => (implication arrow).
The constraints before the => are “implicit arguments”: you don't explicitly “pass Eq a to the function” when you call it, instead you just pass the stuff after the =>, i.e. in your example a list of some concrete type. The compiler will then look at the type and automatically look up its Eq typeclass instance (or raise a compile-time error if the type does not have such an instance). Hence,
GHCi, version 7.10.2: http://www.haskell.org/ghc/ :? for help
Prelude> let palin :: Eq a => [a] -> Bool; palin l = l==reverse l
Prelude> palin [1,2,3,2,1]
True
Prelude> palin [1,2,3,4,5]
False
Prelude> palin [sin, cos, tan]
<interactive>:5:1:
No instance for (Eq (a0 -> a0))
(maybe you haven't applied enough arguments to a function?)
arising from a use of ‘palin’
In the expression: palin [sin, cos, tan]
In an equation for ‘it’: it = palin [sin, cos, tan]
...because functions can't be equality-compared.
†Constraints may in OO also be interfaces / abstract base classes, which aren't “quite proper classes” but are still in many ways treated the same way as OO value-classes. Most modern OO languages now also support Haskell-style parametric polymorphism in addition to “element-wise”/covariant/existential polymorphism, but they require somewhat awkward extends trait-mechanisms because this was only implemented as an afterthought.
‡There are also functions which have “constraints in the arguments”, but that's a more advanced concept called rank-n polymorphism.
This is really an extended comment. Aside from your little type error, your function has another problem: it's extremely inefficient. The main problem is your definition of reverse.
reverse :: [a] -> [a]
reverse xs
| null xs = []
| otherwise = (last xs) : reverse newxs
where
before = (length xs) - 1
newxs = take before xs
last is O(n), where n is the length of the list. length is also O(n), where n is the length of the list. And take is O(k), where k is the length of the result. So your reverse will end up taking O(n^2) time. One fix is to just use the standard reverse function instead of writing your own. Another is to build up the result recursively, accumulating the result as you go:
reverse :: [a] -> [a]
reverse xs0 = go [] xs0
go acc [] = acc
go acc (x : xs) = go (x : acc) xs
This version is O(n).
There's another source of inefficiency in your implementation:
checkPalindrome l = (l == reverse l)
This isn't nearly as bad, but let's look at what it does. Suppose we have the string "abcdefedcba". Then we test whether "abcdefedcba" == "abcdefedcba". By the time we've checked half the list, we already know the answer. So we'd like to stop there! There are several ways to accomplish this. The simplest efficient one is probably to calculate the length of the list as part of the process of reversing it so we know how much we'll need to check:
reverseCount :: [a] -> (Int, [a])
reverseCount xs0 = go 0 [] xs0 where
go len acc [] = (len, acc)
go len acc (x : xs) = len `seq`
go (len + 1) (x : acc) xs
Don't worry about the len `seq` bit too much; that's just a bit of defensive programming to make sure laziness doesn't make things inefficient; it's probably not even necessary if optimizations are enabled. Now you can write a version of == that only looks at the first n elements of the lists:
eqTo :: Eq a => Int -> [a] -> [a] -> Bool
eqTo 0 _ _ = True
eqTo _ [] [] = True
eqTo n (x : xs) (y : ys) =
x == y && eqTo (n - 1) xs ys
eqTo _ _ _ = False
So now
isPalindrome xs = eqTo ((len + 1) `quot` 2) xs rev_xs
where
(len, rev_xs) = reverseCount xs
Here's another way, that's more efficient and arguably more elegant, but a bit tricky. We don't actually need to reverse the whole list; we only need to reverse half of it. This saves memory allocation. We can use a tortoise and hare trick:
splitReverse ::
[a] ->
( [a] -- the first half, reversed
, Maybe a -- the middle element
, [a] ) -- the second half, in order
splitReverse xs0 = go [] xs0 xs0 where
go front rear [] = (front, Nothing, rear)
go front (r : rs) [_] = (front, Just r, rs)
go front (r : rs) (_ : _ : xs) =
go (r : front) rs xs
Now
isPalindrome xs = front == rear
where
(front, _, rear) = splitReverse xs
Now for some numbers, using the test case
somePalindrome :: [Int]
somePalindrome = [1..10000] ++ [10000,9999..1]
Your original implementation takes 7.523s (2.316 mutator; 5.204 GC) and allocates 11 gigabytes to build the test list and check if it's a palindrome. My counting implementation takes less than 0.01s and allocates 2.3 megabytes. My tortoise and hare implementation takes less than 0.01s and allocates 1.7 megabytes.
Here is my code:
test :: (Num a) => [a] -> a
test [] = 0
test [x:xs] = x + test xs
Yet when I run it through ghci as :l test, I get this error:
[1 of 1] Compiling Main ( test.hs, interpreted )
test.hs:3:7:
Couldn't match type `a' with `[a]'
`a' is a rigid type variable bound by
the type signature for spew :: Num a => [a] -> a at test.hs:2:1
In the pattern: x : xs
In the pattern: [x : xs]
In an equation for `spew': spew [x : xs] = x + spew xs
Failed, modules loaded: none.
Try not to laugh :) it's my first attempt at haskell. Any help or explanations would be awesome.
PS: I know this could be easily done with a fold, but I'm trying to practice writing my own type signatures. Thanks in advance!!
You mean
test :: (Num a) => [a] -> a
test [] = 0
test (x:xs) = x + test xs -- note round brackets
with round brackets.
[x:xs] is a list with one element, itself a list, whereas (x:xs) is a list with a first element x and tail xs.
If you type length (1:[1,1,1]) you'll get 4, but if you type length [1:[1,1,1]] you'll get 1 - the only element is a list.
You probably meant to match the list as a whole, not the first element of the list:
test (x:xs) = ...
If you do otherwise, the pattern has the inferred type [[b]], so a == [b] according to tests signature, so xs must have type [b], so test xs must have type b but also type a according to the signature of test, which would mean that a == [a], which is a contradiction and leads to the unification error :)
I'm trying to write a program in Haskell
that gets a list (of integer) and prints out the number of elements that are bigger than the list's average
So far I tried
getAVG::[Integer]->Double
getAVG x = (fromIntegral (sum x)) / (fromIntegral (length x))
smallerThanAVG:: [Integer]->Integer
smallerThanAVG x = (map (\y -> (if (getAVG x > y) then 1 else 0)) x)
For some reason I'm getting this error
Couldn't match expected type `Double'
against inferred type `Integer'
Expected type: [Double]
Inferred type: [Integer]
In the second argument of `map', namely `x'
It could be that I haven't written the logic correctly, although I think I did..
Ideas?
These errors are the best kind, because they pinpoint where you have made a type error.
So let's do some manual type inference. Let's consider the expression:
map (\y -> (if (getAvg x > y) then 1 else 0)) x
There are a few constraints we know off the bat:
map :: (a -> b) -> [a] -> [b] -- from definition
(>) :: Num a => a -> a -> Bool -- from definition
getAvg :: [Integer] -> Double -- from type declaration
1, 0 :: Num a => a -- that's how Haskell works
x :: [Integer] -- from type declaration of smallerThanAVG
Now let's look at the larger expressions.
expr1 = getAvg x
expr2 = (expr1 > y)
expr3 = (if expr2 then 1 else 0)
expr4 = (\y -> expr3)
expr5 = map expr4 x
Now let's work backwards. expr5 is the same as the RHS of smallerThanAVG, so that means it has the same result type as what you've declared.
expr5 :: Integer -- wrong
However, this doesn't match our other constraint: the result of map must be [b] for some b. Integer is definitely not a list (although if you get facetious, it could be coerced into a list of bits). You probably meant to sum that list up.
expr6 = sum expr5
sum :: Num a => [a] -> a
Now let's work forwards.
expr1 :: Double -- result type of getAvg
y :: Double -- (>) in expr2 requires both inputs to have the same type
expr4 :: (Integer -> [a]) -- because for `map foo xs` (expr5)
-- where xs :: [a], foo must accept input a
y :: Integer -- y must have the input type of expr4
Herein lies the conflict: y cannot be both a Double and an Integer. I could equivalently restate this as: x cannot be both a [Double] and [Integer], which is what the compiler is saying. So tl;dr, the kicker is that (>) doesn't compare different types of Nums. The meme for this sort of problem is: "needs more fromIntegral".
(getAvg x > fromIntegral y)
Your code has two errors.
Although the type signature in the code declares that smallerThanAVG x evaluates to an Integer, its code is map ... x, which clearly evaluates to a list instead of a single Integer.
In the code getAVG x > y, you are comparing a Double to an Integer. In Haskell, you can only compare two values of the same type. Therefore, you have to use fromIntegral (or fromInteger) to convert an Integer to a Double. (This is essentially what caused the error message in the question, but you have to get used to it to figure it out.)
There are several ways to fix item 1 above, and I will not write them (because doing so would take away all the fun). However, if I am not mistaken, the code you are aiming at seems like counting the number of elements that are smaller than the average, in spite of what you write before the code.
Styling tips:
You have many superfluous parentheses in your code. For example, you do not have to parenthesize the condition in an if expression in Haskell (unlike if statement in C or Java). If you want to enjoy Haskell, you should learn Haskell in a good style instead of the Haskell which just works.
You call the variable which represents a list “x” in your code. It is conventional to use a variable name such as xs to represent a list.
Others have explained the errors beautifully.
I'm still learning Haskell but I'd like to provide an alternative version of this function:
greaterThanAvg :: [Int] -> [Int]
greaterThanAvg xs = filter (>avg) xs
where avg = sum xs `div` length xs
cnt = length $ greaterThanAvg [1,2,3,4,5]
I recently started learning Haskell and I'm trying to rewrite something I did for an interview in python in Haskell. I'm trying to convert a string from camel case to underscore separated ("myVariableName" -> "my_variable_name"), and also throw an error if the first character is upper case.
Here's what I have:
import qualified Data.Char as Char
translate_java :: String -> String
translate_java xs = translate_helper $ enumerate xs
where
translate_helper [] = []
translate_helper ((a, num):xs)
| num == 1 and Char.isUpper a = error "cannot start with upper"
| Char.isUpper a = '_' : Char.toLower a : translate_helper xs
| otherwise = a : translate_helper xs
enumerate :: (Num b, Enum b) => [a] -> [(a,b)]
enumerate xs = zip xs [1..]
I realize It's pretty likely I'm going about this in a weird way, and I'd love advice about better ways to implement this, but I'd like to get this to compile as well. Here's the error I'm getting now:
Prelude> :r
[1 of 1] Compiling Main ( translate.hs, interpreted )
translate.hs:4:20:
No instance for (Num
(([Bool] -> Bool) -> (Char -> Bool) -> Char -> t))
arising from a use of `translate_helper' at translate.hs:4:20-35
Possible fix:
add an instance declaration for
(Num (([Bool] -> Bool) -> (Char -> Bool) -> Char -> t))
In the first argument of `($)', namely `translate_helper'
In the expression: translate_helper $ enumerate xs
In the definition of `translate_java':
translate_java xs
= translate_helper $ enumerate xs
where
translate_helper [] = []
translate_helper ((a, num) : xs)
| num == 1 and Char.isUpper a
= error "cannot start with upper
"
| Char.isUpper a
= '_' : Char.toLower a : transla
te_helper xs
| otherwise = a : translate_help
er xs
Failed, modules loaded: none.
Any explanation of what's going on here would be great. I really don't understand where "(Num (([Bool] -> Bool) -> (Char -> Bool) -> Char -> t))" is coming from. I'd think the type declaration for translate_helper would be something like [(a,b)] -> [a]?
You have to replace and by &&. The first one is a function (prefix) that receives a list of boolean values and calculates an and of them all. The second one is a true logical and. The error message is a little bit confusing though. Whenever I get such a strange error message, I usually start to annotate my code with type signatures. Then the compiler is able to give you a more detailed description of what went wrong.
Others have mentioned that you should use (&&) instead of and, so I'll answer your other question: no, I don't think you're going about this in a weird way.
But... I do think it can be even more elegant!
translate_java (x:xs) | isUpper x = error "cannot start with an upper"
translate_java xs = concatMap translate xs where
translate x = ['_' | isUpper x] ++ [toLower x]
There's a few interesting things going on here:
The special case is checked straight away. Don't wait until you're recursing to do this!
The concatMap function is really handy in a lot of cases. It's just a map followed by a concat. If I were writing this myself, I'd probably use xs >>= translate instead.
That ['_' | isUpper x] is a list comprehension; this is a cute idiom for making a list with either 0 or 1 elements in it, depending on whether a predicate holds.
Other than that, the code should be fairly self-explanatory.
The problem is this:
| num == 1 and Char.isUpper a = ...
and is not an infix operator; rather it is a function:
and :: [Bool] -> Bool
So it is interpreting 1 and Char.isUpper a as applying three arguments to the "function" 1. Use && instead.
The error message comes from the way numerals are interpreted. A numeral, say, 1 is actually polymorphic; the specific type it gets depends on the type that is needed. That's why you can say x+1 and it will work whether x is an integer or a double or whatever. So the compiler inferred that the type of 1 needs to be a three-argument function, and then tried to find a numeric type matching that so it could convert 1 into that type (and, naturally, failed).
Here's my solution. It's not as masterful as the answer Daniel Wagner gave using concatMap and the list comprehension, but it's perhaps easier to understand for the beginner.
conv :: String -> String
conv [] = []
conv s#(x:xs) = if Char.isUpper x
then error "First character cannot be uppercase"
else change s
change :: String -> String
change [] = []
change (x:xs) = if Char.isUpper x
then '_' : Char.toLower x : change xs
else x : change xs
The function conv really just checks your criterion that the first character must not be uppercase, and if it isn't it hands over the string to the function change, which does the work. It goes through all the characters one by one, building a list, and if the character is uppercase, it adds an underscore followed by the lowercase version of the character, otherwise if the character is already lowercase it just adds it as it is.