My code:
isOrdered :: (a -> a -> Bool) -> [a] -> Bool
isOrdered mark xs =(head xs) `mark`(head(tail xs))
Compiles perfectly but when I try to call it with
isOrdered < [1,2,3]
I get an error:
Couldn't match expected type `(a0 -> a0 -> Bool) -> [a0] -> Bool'
with actual type `[t0]'
In the second argument of `(<)', namely `[1, 2, 3]'
In the expression: isOrdered < [1, 2, 3]
In an equation for `it': it = isOrdered < [1, 2, 3]
What am I missing here?
Since < is infix, you have to wrap it in parens. This converts it to be prefixed.
1 < 2 ==> (<) 1 2
1 + 5 ==> (+) 1 5
Then you're code becomes
isOrdered (<) [1, 2, 3]
This is actually part of a more general concept of sectioning. You can completely convert an infix operator to prefix with parens, or partially apply it like this
\x -> x + 1 ===> (+1)
\x -> 2 ^ x ===> (2^)
The only place where this goes a bit pear-shaped is with -. Since - is a super special prefix operator defined by the Haskell language, you can't do (-2), since it's not clear whether this is a section or a number. Haskell chooses a number, but if you want a section, there is a function subtract.
\x -> x - 2 ==> subtract 2
This works:
isOrdered (<) [1,2,3]
Not sure why, though. It’s a general thing:
Prelude> :type <
<interactive>:1:1: parse error on input `<'
Prelude> :type (<)
(<) :: Ord a => a -> a -> Bool
I’m sure others will explain.
When passing an operator to a higher level function, you have to surround it with parentheses:
isOrdered (<) [1, 2, 3]
You also have to do this for partially applied operators
map (== 2) [1, 2, 3]
This is not an answer to your question, but just a stylistic tip: idiomatic Haskell code should use pattern matching instead of head/tail. Pattern matching statically ensures that you don't accidentally try to access inexistent elements.
The way you would write your function using pattern matching is:
isOrdered :: (a -> a -> Bool) -> [a] -> Maybe Bool
isOrdered mark (x:y:_) = Just (x `mark` y)
isOrdered _ _ = Nothing
Note that I've included a case for when the list does not contain two elements and wrapped the result in a Maybe. This prevents run-time errors by forcing you to handle that case.
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
Given a list like:
[1, 2, 2, 6, 7, 8, 10, 11, 12, 15]
Split it into blandly increasing ranges (maybe equal):
[[1, 2, 2], [6, 7, 8], [10, 11, 12], [15]]
I tried using a recursive approach:
splitRanges [] = [[]]
splitRanges (x:y:xs)
| x `elem` [y, y + 1] = [x, y] : splitRanges xs
| otherwise = xs
So if the item is one less or equal to the item after I fuse them. But it says I am trying to build an infinite type:
Occurs check: cannot construct the infinite type: a0 = [a0]
Expected type: [[a0]]
Actual type: [a0]
But what does [the fact that it is monotone] have to do with how the list is split?
That being strictly increasing would give different results.
Or are you really trying to say something else?
I hope I am not.
Will the list always be monotone?
No, splitting a monotone list means making it into just one sub-list.
If not, how should that affect the results?
If it is not monotone, you will have many sublists.
Is it always brown into groups of three?
No, the groups may contain n elements.
More examples would be good
splitRanges [1, 3] == [[1], [3]]
splitRanges [1, 2, 5] == [[1, 2], [3]]
splitRanges [0, 0, 1] == [[0, 0, 1]]
splitRanges [1, 5, 7, 9] == [[1], [5], [7], [9]]
I appreciate hints rather than full answers, as I would like to improve myself, copy-pasting is not improvement.
Try breaking the problem into more manageable parts.
First, how would you split just one blandly increasing range from the start of a list? Lets guess that should be splitOne :: [Integer] -> ([Integer], [Integer]).
Second, how can you repeatedly apply splitOne to the left-over list? Try implementing splitMany :: [Integer] -> [[Integer]] by using splitOne.
For splitOne, what should you be trying to find? The first position to split at. What are "split positions"? Lets make that up.
split 0 1 2 3 4 …
list [ | x1, | x2, | x3, | x4, | x5, …]
So a split at 0 is ([], [x1,x2,x3,x4,x5,…]), and a split at 3 is ([x1,x2,x3],[x4,x5,…]). What relationship can you see between the split position and the split list?
How do you determine the first split position of the list? Lets say that is implemented as firstSplitPos :: [Integer] -> Integer. What is the first split position of an empty list?
Can you now implement splitOne using firstSplitPos?
One Possible Answer
-- What are the adjacencies for:
-- 1) empty lists?
-- 2) lists with one element?
-- 3) lists with more than one element?
--
-- Bonus: rewrite in point-free form using <*>
--
adjacencies :: [a] -> [(a,a)]
adjacencies xxs = zip xxs (drop 1 xxs)
-- Bonus: rewrite in point-free form
--
withIndices :: [a] -> [(Int,a)]
withIndices xxs = zip [0..] xxs
-- This is the most involved part of the answer. Pay close
-- attention to:
-- 1) empty lists
-- 2) lists with one element
-- 3) lists which are a blandly increasing sequence
--
firstSplitPos :: (Eq a, Num a) => [a] -> Int
firstSplitPos xxs = maybe (length xxs) pos (find q searchList)
where q (_,(a,b)) = a /= b && a + 1 /= b
searchList = withIndices (adjacencies xxs)
-- Why is the split position one more than the index?
pos (i,_) = i + 1
--
-- Bonus: rewrite in point-free form using <*>
--
splitOne :: (Eq a, Num a) => [a] -> ([a],[a])
splitOne xxs = splitAt (firstSplitPos xxs) xxs
splitMany :: (Eq a, Num a) => [a] -> [[a]]
-- What happens if we remove the case for []?
splitMany [] = []
splitMany xxs = let (l, r) = splitOne xxs in l : splitMany r
Another Approach
This is my explanation of Carsten's solution. It is already succinct but I have elected for a variation which does not use a 2-tuple.
We know that Haskell lists are defined inductively. To demonstrate this, we can define an equivalent data type.
data List a = Cons a (List a) -- Cons = (:)
| Nil -- Nil = []
Then ask the question: can we use induction on lists for the solution? If so, we only have to solve two cases: Cons and Nil. The type signature of foldr shows us exactly that:
foldr :: (a -> b -> b) -- Cons case
-> b -- Nil case
-> [a] -- The list
-> b -- The result
What if the list is Nil? Then the only blandly increasing sequence is the empty sequence. Therefore:
nilCase = [[]]
We might want nilCase = [] instead, as that also seems reasonable — i.e. there are no blandly increasing sequences.
Now you need some imagination. In the Cons case we only get to look at one new element at a time. With this new element, we could decide whether it belongs to the right-adjacent sequence or if it begins a new sequence.
What do I mean by right-adjacent? In [5,4,1,2,2,7], 1 belongs to the right-adjacent sequence [2,2].
How might this look?
-- The rest of the list is empty
consCase new [] = [new] : []
-- The right-adjacent sequence is empty
consCase new ([]:ss) = [new] : ss
-- The right-adjacent sequence is non-empty
-- Why `new + 1 == x` and not `new == x + 1`?
consCase new sss#(xxs#(x:_):ss)
| new == x || new + 1 == x = (new:xxs):ss
| otherwise = [new]:sss
Now that we solved the Nil case and the Cons case, we are done!
splitRanges = foldr consCase nilCase
It would be useful and idiomatic to write your function to take a predicate, instead of writing your split condition into the function itself:
splitBy2 :: (a -> a -> Bool) -> [a] -> [[a]]
splitBy2 ok xs = snd $ f xs [] []
where f (a:b:xs) acc_list acc_out_lists | ok a b = ...
I hope you don't mind spoiling part of it, but as the comments are discussing what you want (and I hope I've got it) maybe you are interested in another possible solution?
I don't want to spoil it all but I think you can easily work this out:
blandly :: (Ord a, Num a) => [a] -> [[a]]
blandly = g . foldr f ([],[])
where f x ([],xss) = ([x],xss)
f x (y:ys,xss)
| abs (x-y) <= 1 = undefined
| otherwise = undefined
g (ys,xss) = undefined
you just have to fill in the undefined holes
The idea is just to fold the list from the right, accumulating your inner lists in the first item of the tuple, s long as the elements are not to far away; and if they are: to push it to the second item.
If done correctly it will yield:
λ> blandly [1,3]
[[1],[3]]
λ> blandly [1,2,5]
[[1,2],[5]]
λ> blandly [0,0,1]
[[0,0,1]]
λ> blandly [1,5,7,9]
[[1],[5],[7],[9]]
which seems to be what you want
1 hour later - I think I can post my solution - just stop reading if you don't want to get spoiled
blandly :: (Ord a, Num a) => [a] -> [[a]]
blandly = uncurry (:) . foldr f ([],[])
where f x ([],xs) = ([x],xs)
f x (y:ys,xs)
| abs (x-y) <= 1 = (x:y:ys,xs)
| otherwise = ([x],(y:ys):xs)
maybe I have a slight misunderstanding here (the examples did not specify it) - but if you want on only monotonic increasing inner lists you just have to change the abs part:
blandly :: (Ord a, Num a) => [a] -> [[a]]
blandly = uncurry (:) . foldr f ([],[])
where f x ([],xss) = ([x],xss)
f x (y:ys,xss)
| 0 <= y-x
&& y-x <= 1 = (x:y:ys,xss)
| otherwise = ([x],(y:ys):xss)
i write a little code in Haskell to learn haskell. It looks like:
first :: [Int] -> Int -> Int
first [] x = x
first xs y = y
I wanted to express that i get a list([] or xs) and an argument(x or y) and no matter how the lists look like, the argument should presented.
So, when I write the following: first [1,2,3] 4, then ghci says:
Couldnt match expected type ´a0 -> t´ wth actual type ´[t0]´. Relevant bindings
include it :: t(boudn at <>:6:1)but its type ´[t 0]´ has none
In the expression : [1,2,3] 4
In an equation for ´it´: it = [1,2,3] 4
How can i fix it? What mistakes i´ve made? Can someone help me?
It looks like you use [1,2,3,4] 4 instead of first [1,2,3,4] 4.
> [1,2,3,4] 4
<interactive>:1:0:
Couldn't match expected type `t1 -> t' against inferred type `[a]'
In the expression: [1, 2, 3, 4] 4
In the definition of `it': it = [1, 2, 3, 4] 4
By the way, your definition of first is the same as the following:
first :: [Int] -> Int -> Int
first _ x = x
which means no matter what the first argument is, always returns the second argument.
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 [].
I've come to the realization that when I have nested data structures, I've been manually writing code to delve into them. Like this:
--one level
Prelude> map (*2) [1,2,3]
[2,4,6]
--nested two levels
Prelude> let t2 = map $ map (*2)
Prelude> t2 [[1,2,3],[4,5,6]]
[[2,4,6],[8,10,12]]
--nested three levels
Prelude> let t3 = map $ map $ map (*2)
Prelude> t3 [[ [1,2,3],[4,5,6] ],[ [1,2,3],[4,5,6] ]]
[[[2,4,6],[8,10,12]],[[2,4,6],[8,10,12]]]
so it occurs to me that I should be able to automatically construct a function for delving into my nested data structures using a higher order function:
Prelude> let t f n = (iterate map f) !! n
<interactive>:35:22:
Occurs check: cannot construct the infinite type: b0 = [b0]
Expected type: (a0 -> b0) -> a0 -> b0
Actual type: (a0 -> b0) -> [a0] -> [b0]
In the first argument of `iterate', namely `map'
In the first argument of `(!!)', namely `(iterate map f)'
Its strikes me that
I understand its finding a list where it expected...something else
I don't know how to fix this - should I write code to do repeated application even though thats what I thought iterate was for?
This seems similar to the concept of "lifting" - but I don't know how to apply that intuition.
The problem is that these "iterations" have different types. For each iteration, you get an extra level of nesting, so you'd want
t f 0 :: a -> b
t f 1 :: [a] -> [b]
t f 2 :: [[a]] -> [[b]]
But iterate :: (a -> a) -> a -> [a] requires that the iterations all have the same type. In fact, a direct implementation of the above would require some form of dependent types since the return type depends on the value of n.
Unless you have a good reason not to, I suggest keeping it simple and just writing out the required number of map calls. It's possible to use Template Haskell to generate them, but this will likely be more trouble than it's worth.
However, if you have complicated nested data structures, you may want to look into SYB which can automatically take care of the boilerplate of applying such transformations in nested structures.
Here's a quick example:
> import Data.Generics
> let t x = everywhere (mkT (*2)) x
> :t t
t :: Data a => a -> a
> t [2,4,6]
[4,8,12]
> t [[2,4,6],[8,10,12]]
[[4,8,12],[16,20,24]]
> t (Just [(1, 2, 3), (4, 5, 6)])
Just [(2,4,6),(8,10,12)]
Think about the type of (iterate map f) !! n. You want it to be a -> a for n = 0, [a] -> [a] for n = 1, [[a]] -> [[a]] for n = 2 - in general, you want the type of this expression to depend on the value of n. But that is impossible to do in Haskell, since it's not a dependently-typed language.