I want to create a data type that's either New [Int] or Empty; then I want to use an Int and check if the data type contains the Int. This is probably an extremely simple question but I'm new to creating data types and using them in functions, specifically getting a handle on the syntax; I don't have a grounded understanding of it yet.
data New = New [Int]
| Empty
deriving Show
check :: Int -> New -> Bool
check x ys
| x `elem` New ys = True
| otherwise = False
You'll want check to pattern-match on its New argument to access the underlying list, if any.
check :: Int -> New -> Bool
check x Empty = False
check x (New ys) = x `elem` ys
One way to collapse the definition a bit is to realize that as far as check is concerned, the values Empty and New [] are equivalent.
check :: Int -> New -> Bool
check x ys = x `elem` (case ys of
New ys' -> ys'
Empty -> [])
Even though this is an exercise in using custom data types, I feel like expanding on chepner's question, "What does Empty provide that New [] does not?" since it was not met with any response: Lists can already be empty, so you could make your check without a custom data type definition:
check :: Int -> [Int] -> Bool
check = elem
With the type that Robin said was isomorphic to your New type, Maybe, it would look like:
type New = Maybe [Int]
check :: Int -> New -> Bool
check _ Nothing = False
check x (Just xs) = x `elem` xs
Generally when the standard library offers a data type isomorphic to your current one, you should ask yourself if you want to create your own one. Sometimes it is warranted, but often you will want to use either type or newtype and not data.
With this type, Maybe [Int], you must ask yourself why you want to differentiate between Nothing and Just []. For example, Nothing could indicate that the function failed, while Just [] could indicate that it succeeded, but that the (correct) result was empty.
An example of a function where this makes sense is the Change exercise on Exercism:
Correctly determine the fewest number of coins to be given to a customer such that the sum of the coins' value would equal the correct amount of change.
Here Nothing means that it's not possible to give exact change, and Just [] means that the customer paid the exact amount and needs no change.
If, instead, you'd like some exercises that deal with writing functions for custom data types, see the "W3" exercises on https://github.com/opqdonut/haskell-exercises.
Related
I want to make a function that checks to see if each row of the board for the Bert Bos puzzle is red one row at a time, but conceptually I'm having a hard time with this. Initially I make the board with all blue squares, but once the squares have been flipped with a flip function, the allRed function should be able to tell if the row is all red or not. Each row is represented by a list of colors, either Blue or Red
I know I should be using the all function, but I'm having some problems actually writing it out for my situation
Here is what I have so far:
generateboard :: Int -> [[Color]]
generateboard n = replicate n (replicate n Blue)
allRed :: [[Color]] -> Bool
let board = generateboard
allRed board = []
allRed board = all ([x:_ | x <- board, x == Red])
allRed board
There are many mistakes and misunderstandings here. I recommend reading any of the introductory Haskell materials to strengthen your basic understanding of the language. I will answer the question directly nonetheless.
generateboard looks great.
You are right to think all :: Foldable t => (a -> Bool) -> t a -> Bool will help us define allRed. If the type is confusing you can instead think of it as (a -> Bool) -> [a] -> Bool. The documentation says:
Determines whether all elements of the [list] satisfy the predicate.
To use all we need a predicate (a function) with type a -> Bool and a list of type [a]. We know what the predicate needs to be:
\x -> x == Red
Another way to write this is:
(==) Red
The predicate has type Color -> Bool and so our list must then have type [Color]. However, we have a list of type [[Color]]. There are two ways I can see to go about this.
The simpler idea is to observe that the board structure is irrelevant if all we care about is the cells. Therefore, we can flatten the structure with concat :: [[a]] -> [a]. Then our solution is thus:
allRed xs = all ((==) Red) (concat xs)
Which is also written:
allRed = all ((==) Red) . concat
Another solution is to observe that if all rows are red then the whole board must be red. This solution is:
allRed xs = all (all ((==) Red)) xs
Which is also written:
allRed = all (all ((==) Red))
First, the all function:
all :: (a -> Bool) -> [a] -> Bool
all p xs = ...
takes a function p representing a property and a list xs and tests if p x is true (i.e., if x has property p) for every element x of xs. (For example, all even [2,4,7] checks if all elements of the given list are even, and it returns False because even 7 equals False.) So, to use all, you need two arguments -- a list of items to check, and a function that checks one item.
Second, when faced with the problem of processing a data structure in Haskell (in this case [[Color]]), an excellent rule of thumb is to the deconstruct the structure from the outside in, using one function for each level of structure. You have an (outer) list of (inner) lists of colors, so start with the outer list, the list of rows.
How would you write a function that checks if all the rows in the outer list satisfy the property that they "contain only red colors"? Or, to put it more simply, how would you write this function using all if you already had a helper function redRow that expressed the property of a row having only red colors?
redRow :: [Color] -> Bool
redRow row = ...
If you can write allRed board using all, board, and redRow, you'll have reduced the problem to writing the definition of redRow, which operates on a simpler data structure, an (inner) list of colors.
To write redRow, you should likewise be able to use all again with a function expressing the property of a color being red:
isRed :: Color -> Bool
isRed col = ...
(or using an equivalent lambda or "section" directly).
In this case, another approach is possible, too -- you could use concat to "flatten" the outer and inner list together and then tackle the easier problem of checking if all colors in a big long list are red.
I have function in haskell (lets call it 'dumb') which calls 3 different functions. These three different functions return different types, for example, a boolean or a list of booleans. How can I define function 'dumb' to either return a boolean or a list of booleans?
data Sumtype = givelist Integer | getprod Integer
prod :: Int -> Int
prod x = x*3
listnums :: Int -> [Int]
listnums x = [1...x]
dumb :: Sumtype -> (what comes here..?)
dumb (givelist x) -> listnums x
dum (getprod x) -> prod x
You make it return Either Boolean [Boolean]. But I'm suspicious about your motives. It sounds like an X/Y problem.
You're probably looking for the the Either type, although with it your function will return Either values. It's defined like this:
data Either a b = Left a | Right b
When you want to define a function that can return either a Bool or a list of Bools its type should look something like this:
dumb :: Either Bool [Bool]
In this case 'dumb' will be a function that doesn't take any arguments and return either a Bool or a list of Bools. In the function's body you can return a Bool like this:
Left bool
Or a list of bools like this:
Right [bool]
You can see a concrete example here: http://en.wikibooks.org/wiki/Haskell/More_on_datatypes#More_than_one_type_parameter
All that said though, the reason Sebastian asked you for more details is that Either is rarely used outside of error handling (AFAIK I know anyway). It's possible that in your case you don't really need it at all, but we can't be sure unless you tell us more about the other functions you use in 'dumb' and about your goals.
Unrelated Probems
It appears you are a beginner - welcome to Haskell! I strongly suggest you read and work through one of the many tutorials as that is more efficient and complete than asking individual questions.
Syntax
Let's start with correcting the syntax errors. Constructors, such as Givelist and Getprod must start with a capital letter. The function dumb was typo'ed once. Function definitions use = and not ->.
Types
Now we have type errors to address. The Sumtype uses Integer and you then switch to using Int. Lets just stick with Integer for simplicity.
With these fixes we get:
data Sumtype = Givelist Integer | Getprod Integer
prod :: Integer -> Integer
prod x = x*3
listnums :: Integer -> [Integer]
listnums x = [1...x]
dumb :: Sumtype -> (what comes here..?)
dumb (Givelist x) = listnums x
dumb (Getprod x) = prod x
The Question
You want to know "what comes here" where 'here' is the result type. As written, the function is actually invalid. One definition yields a list of integers, [Integer], while the other yields a single integer Integer. One solution is to use a sum type such as Either Integer [Integer] - this is very much like your pre-existing Sumtype:
dumb :: Sumtype -> Either Integer [Integer]
So now we need to return a constructor of Either in our function definitions. You can lookup the documentation or use :info Either in GHCi to learn the constructors if you don't have them memorized.
dumb (Givelist x) = Right (listnums x)
dumb (Getprod x) = Left (prod x)
Notice we had to use Left for the second case which returns an Integer, because the first type we wrote after Either (the left type) is Integer.
I am doing a haskell exercise, regarding define a function accumulate :: [IO a] -> IO [a]
which performs a sequence of interactions and accumulates their result in a list.
What makes me confused is how to express a list of IO a ? (action:actions)??
how to write recursive codes using IO??
This is my code, but these exists some problem...
accumulate :: [IO a] -> IO [a]
accumulate (action:actions) = do
value <- action
list <- accumulate (action:actions)
return (convert_to_list value list)
convert_to_list:: Num a =>a -> [a]-> [a]
convert_to_list a [] = a:[]
convert_to_list x xs = x:xs
What you are trying to implement is sequence from Control.Monad.
Just to let you find the answer instead of giving it, try searching for [IO a] -> IO [a] on hoogle (there's a Source link on the right hand side of the page when you've chosen a function).
Try to see in your code what happens when list of actions is empty list and see what does sequence do to take care of that.
There is already such function in Control.Monad and it called sequence (no you shouldn't look at it). You should denote the important decision taken during naming of it. Technically [IO a] says nothing about in which order those Monads should be attached to each other, but name sequence puts a meaning of sequential attaching.
As for the solving you problem. I'd suggest to look more at types and took advice of #sacundim. In GHCi (interpreter from Glasgow Haskell Compiler) there is pretty nice way to check type and thus understand expression (:t (:) will return (:) :: a -> [a] -> [a] which should remind you one of you own function but with less restrictive types).
First of all I'd try to see at what you have showed with more simple example.
data MyWrap a = MyWrap a
accumulate :: [MyWrap a] -> MyWrap [a]
accumulate (action:actions) = MyWrap (convert_to_list value values) where
MyWrap value = action -- use the pattern matching to unwrap value from action
-- other variant is:
-- value = case action of
-- MyWrap x -> x
MyWrap values = accumulate (action:actions)
I've made the same mistake that you did on purpose but with small difference (values is a hint). As you probably already have been told you could try to interpret any of you program by trying to inline appropriate functions definitions. I.e. match definitions on the left side of equality sign (=) and replace it with its right side. In your case you have infinite cycle. Try to solve it on this sample or your and I think you'll understand (btw your problem might be just a typo).
Update: Don't be scary when your program will fall in runtime with message about pattern match. Just think of case when you call your function as accumulate []
Possibly you looking for sequence function that maps [m a] -> m [a]?
So the short version of the answer to your question is, there's (almost) nothing wrong with your code.
First of all, it typechecks:
Prelude> let accumulate (action:actions) = do { value <- action ;
list <- accumulate (action:actions) ; return (value:list) }
Prelude> :t accumulate
accumulate :: (Monad m) => [m t] -> m [t]
Why did I use return (value:list) there? Look at your second function, it's just (:). Calling g
g a [] = a:[]
g a xs = a:xs
is the same as calling (:) with the same arguments. This is what's known as "eta reduction": (\x-> g x) === g (read === as "is equivalent").
So now just one problem remains with your code. You've already taken a value value <- action out of the action, so why do you reuse that action in list <- accumulate (action:actions)? Do you really have to? Right now you have, e.g.,
accumulate [a,b,c] ===
do { v1<-a; ls<-accumulate [a,b,c]; return (v1:ls) } ===
do { v1<-a; v2<-a; ls<-accumulate [a,b,c]; return (v1:v2:ls) } ===
do { v1<-a; v2<-a; v3<-a; ls<-accumulate [a,b,c]; return (v1:v2:v3:ls) } ===
.....
One simple fix and you're there.
For a silly challenge I am trying to implement a list type using as little of the prelude as possible and without using any custom types (the data keyword).
I can construct an modify a list using tuples like so:
import Prelude (Int(..), Num(..), Eq(..))
cons x = (x, ())
prepend x xs = (x, xs)
head (x, _) = x
tail (_, x) = x
at xs n = if n == 0 then xs else at (tail xs) (n-1)
I cannot think of how to write an at (!!) function. Is this even possible in a static language?
If it is possible could you try to nudge me in the right direction without telling me the answer.
There is a standard trick known as Church encoding that makes this easy. Here's a generic example to get you started:
data Foo = A Int Bool | B String
fooValue1 = A 3 False
fooValue2 = B "hello!"
Now, a function that wants to use this piece of data must know what to do with each of the constructors. So, assuming it wants to produce some result of type r, it must at the very least have two functions, one of type Int -> Bool -> r (to handle the A constructor), and the other of type String -> r (to handle the B constructor). In fact, we could write the type that way instead:
type Foo r = (Int -> Bool -> r) -> (String -> r) -> r
You should read the type Foo r here as saying "a function that consumes a Foo and produces an r". The type itself "stores" a Foo inside a closure -- so that it will effectively apply one or the other of its arguments to the value it closed over. Using this idea, we can rewrite fooValue1 and fooValue2:
fooValue1 = \consumeA consumeB -> consumeA 3 False
fooValue2 = \consumeA consumeB -> consumeB "hello!"
Now, let's try applying this trick to real lists (though not using Haskell's fancy syntax sugar).
data List a = Nil | Cons a (List a)
Following the same format as before, consuming a list like this involves either giving a value of type r (in case the constructor was Nil) or telling what to do with an a and another List a, so. At first, this seems problematic, since:
type List a r = (r) -> (a -> List a -> r) -> r
isn't really a good type (it's recursive!). But we can instead demand that we first reduce all the recursive arguments to r first... then we can adjust this type to make something more reasonable.
type List a r = (r) -> (a -> r -> r) -> r
(Again, we should read the type List a r as being "a thing that consumes a list of as and produces an r".)
There's one final trick that's necessary. What we would like to do is to enforce the requirement that the r that our List a r returns is actually constructed from the arguments we pass. That's a little abstract, so let's give an example of a bad value that happens to have type List a r, but which we'd like to rule out.
badList = \consumeNil consumeCons -> False
Now, badList has type List a Bool, but it's not really a function that consumes a list and produces a Bool, since in some sense there's no list being consumed. We can rule this out by demanding that the type work for any r, no matter what the user wants r to be:
type List a = forall r. (r) -> (a -> r -> r) -> r
This enforces the idea that the only way to get an r that gets us off the ground is to use the (user-supplied) consumeNil function. Can you see how to make this same refinement for our original Foo type?
If it is possible could you try and nudge me in the right direction without telling me the answer.
It's possible, in more than one way. But your main problem here is that you've not implemented lists. You've implemented fixed-size vectors whose length is encoded in the type.
Compare the types from adding an element to the head of a list vs. your implementation:
(:) :: a -> [a] -> [a]
prepend :: a -> b -> (a, b)
To construct an equivalent of the built-in list type, you'd need a function like prepend with a type resembling a -> b -> b. And if you want your lists to be parameterized by element type in a straightforward way, you need the type to further resemble a -> f a -> f a.
Is this even possible in a static language?
You're also on to something here, in that the encoding you're using works fine in something like Scheme. Languages with "dynamic" systems can be regarded as having a single static type with implicit conversions and metadata attached, which obviously solves the type mismatch problem in a very extreme way!
I cannot think of how to write an at (!!) function.
Recalling that your "lists" actually encode their length in their type, it should be easy to see why it's difficult to write functions that do anything other than increment/decrement the length. You can actually do this, but it requires elaborate encoding and more advanced type system features. A hint in this direction is that you'll need to use type-level numbers as well. You'd probably enjoy doing this as an exercise as well, but it's much more advanced than encoding lists.
Solution A - nested tuples:
Your lists are really nested tuples - for example, they can hold items of different types, and their type reveals their length.
It is possible to write indexing-like function for nested tuples, but it is ugly, and it won't correspond to Prelude's lists. Something like this:
class List a b where ...
instance List () b where ...
instance List a b => List (b,a) b where ...
Solution B - use data
I recommend using data construct. Tuples are internally something like this:
data (,) a b = Pair a b
so you aren't avoiding data. The division between "custom types" and "primitive types" is rather artificial in Haskell, as opposed to C.
Solution C - use newtype:
If you are fine with newtype but not data:
newtype List a = List (Maybe (a, List a))
Solution D - rank-2-types:
Use rank-2-types:
type List a = forall b. b -> (a -> b -> b) -> b
list :: List Int
list = \n c -> c 1 (c 2 n) -- [1,2]
and write functions for them. I think this is closest to your goal. Google for "Church encoding" if you need more hints.
Let's set aside at, and just think about your first four functions for the moment. You haven't given them type signatures, so let's look at those; they'll make things much clearer. The types are
cons :: a -> (a, ())
prepend :: a -> b -> (a, b)
head :: (a, b) -> a
tail :: (a, b) -> b
Hmmm. Compare these to the types of the corresponding Prelude functions1:
return :: a -> [a]
(:) :: a -> [a] -> [a]
head :: [a] -> a
tail :: [a] -> [a]
The big difference is that, in your code, there's nothing that corresponds to the list type, []. What would such a type be? Well, let's compare, function by function.
cons/return: here, (a,()) corresponds to [a]
prepend/(:): here, both b and (a,b) correspond to [a]
head: here, (a,b) corresponds to [a]
tail: here, (a,b) corresponds to [a]
It's clear, then, that what you're trying to say is that a list is a pair. And prepend indicates that you then expect the tail of the list to be another list. So what would that make the list type? You'd want to write type List a = (a,List a) (although this would leave out (), your empty list, but I'll get to that later), but you can't do this—type synonyms can't be recursive. After all, think about what the type of at/!! would be. In the prelude, you have (!!) :: [a] -> Int -> a. Here, you might try at :: (a,b) -> Int -> a, but this won't work; you have no way to convert a b into an a. So you really ought to have at :: (a,(a,b)) -> Int -> a, but of course this won't work either. You'll never be able to work with the structure of the list (neatly), because you'd need an infinite type. Now, you might argue that your type does stop, because () will finish a list. But then you run into a related problem: now, a length-zero list has type (), a length-one list has type (a,()), a length-two list has type (a,(a,())), etc. This is the problem: there is no single "list type" in your implementation, and so at can't have a well-typed first parameter.
You have hit on something, though; consider the definition of lists:
data List a = []
| a : [a]
Here, [] :: [a], and (:) :: a -> [a] -> [a]. In other words, a list is isomorphic to something which is either a singleton value, or a pair of a value and a list:
newtype List' a = List' (Either () (a,List' a))
You were trying to use the same trick without creating a type, but it's this creation of a new type which allows you to get the recursion. And it's exactly your missing recursion which allows lists to have a single type.
1: On a related note, cons should be called something like singleton, and prepend should be cons, but that's not important right now.
You can implement the datatype List a as a pair (f, n) where f :: Nat -> a and n :: Nat, where n is the length of the list:
type List a = (Int -> a, Int)
Implementing the empty list, the list operations cons, head, tail, and null, and a function convert :: List a -> [a] is left as an easy exercise.
(Disclaimer: stole this from Bird's Introduction to Functional Programming in Haskell.)
Of course, you could represent tuples via functions as well. And then True and False and the natural numbers ...
Why does this throw an error?
myTest :: Int -> [Int]
myTest a
| a == 0 = []
| otherwise = x ++ map(myTest) x
where x = [a-1]
I would expect that it would make a list going from a to 1. Instead I get the error:
couldn't match the expected type 'Int' against inferred type '[Int]'
in the first argument of 'map', namely '(myTest)'
in the second argument of '(++)', namely 'map (myTest) x'
This obviously isn't the best way to make this list, but it is a simplified version of a more complicated problem I am having.
I basically have a function foo :: a -> [a], and in the resulting list I need to call foo on every element expanding it into another list. In the end I want one big list where every element is a base case.
I am fairly new at Haskell so I am probably missing something fairly basic.
The signature of myTest is Int -> [Int].
The signature of map is (a -> b) -> [a] -> [b], and since myTest is the first argument, that makes it (Int -> [Int]) -> [Int] -> [[Int]].
But your function expects it to produce an [Int], not an [[Int]].
Edit to add: I think what you want is simply
myTest a
| a == 0 = []
| otherwise = [a] ++ myTest (a-1)
although this is probably not what you should want (it's much more heavyweight than an idiomatic Haskell solution), but without seeing your actual problem that this is a simplification of, it's hard to tell.
myTest :: Int -> [Int]
myTest 0 = []
myTest a = a : myTest (a-1)
Else's answer is better because is uses (:) instead of (++). In your case you can use either, because (:) adds a single element to the front of the list, and that's what you're trying to do.
(++) is a relatively expensive operation because it copies the spine of the left hand list. (:) is cheap because it never copies anything, it just creates a new cons cell that contains your new head element.
So always use (:) instead of (++) if you can.
EDIT: Just trying to explain since Else didn't post an explanation for his code.