Scala's Partial Functions in Haskell - scala

Scala has a very nice support of partial functions, mainly because in Scala when you define a partial function it also defines an isDefinedAt function for it. And also Scala has orElse and andThen functions to work with partial functions.
Haskell does support partial functions by simply non-exhaustively defining a function (though they are strongly discouraged in Haskell community). But to define isDefinedAt function in general you have to use some sort of exception handling, which I'm not being able to figure out. Once isDefinedAt function is defined then it can be used to define orElse and andThen function is already there as (.).
In short, I want to define a function,
isDefinedAt :: (a -> b) -> a -> Bool
isDefinedAt f x = -- returns True if f is defined at x else False
Can anyone please tell me how such a function can be written.
Note, I can define a function with signature
isDefinedAt :: (a -> b) -> a -> IO Bool
for generic b. But I want a function without IO in co-domain.
A nice article on Scala's Partial Functions is - How to create and use partial functions in Scala By Alvin Alexander

I recommend that, like in Scala, you use a separate type for partial functions.
import Control.Arrow
import Data.Maybe
type Partial = Kleisli Maybe
isDefinedAt :: Partial a b -> a -> Bool
isDefinedAt f x = isJust $ runKleisli f x
-- laziness should save some of the work, if possible
orElse :: Partial a b -> Partial a b -> Partial a b
orElse = (<+>)
andThen :: Partial a b -> Partial b c -> Partial a c
andThen = (>>>)

Your versions of isDefinedAt are not what Scala does (even in signature); it's very possible (though discouraged) for a PartialFunction to throw an exception when isDefinedAt is true. Or, when you define one explicitly (not using a literal), vice versa: apply doesn't have to throw when isDefinedAt is false, it's user responsibility not to call it then. So the direct equivalent would just be
data PartialFunction a b = PartialFunction { apply :: a -> b, isDefinedAt :: a -> Boolean }
which isn't particularly useful.
apply and isDefinedAt are only really linked in Scala for PartialFunction literals which requires compiler support:
A PartialFunction's value receives an additional isDefinedAt member, which is derived from the pattern match in the function literal, with each case's body being replaced by true, and an added default (if none was given) that evaluates to false.
You can emulate this by using Template Haskell, I believe, but I honestly think using the more Haskell-like approach as described in Daniel Wagner's answer is better. As he mentions, laziness helps.
Though it works even better if you make sure runKleisli f x is executed only once; optimizing cases where you have both isDefinedAt and runKleisli requires Common Subexpression Elimination, and the compiler is cautious about doing that, see Under what circumstances could Common Subexpression Elimination affect the laziness of a Haskell program?

You could do something like this (DISCLAIMER: I have not checked the laws of the relevant typeclasses, and the presence of a string in the constructor for the exception in Alternative makes me wonder if it is lawful). Scala's heterogeneous andThen is covered by fmap; its homogeneous andThen / compose are covered by the >>> / <<< from Category; orElse is covered by <|>; lift is runToMaybe.
However, without a deep compiler integration such as exists in Scala, the pattern incompleteness warnings will interact poorly with this. Haskell only has module-level pragmas for these things, and you won't want to just indiscriminately turn them off in any module where you declare inexhaustive functions, or you may get nasty surprises. Depending on your usecase, you may find optics more idiomatic and less problematic; you can have the boilerplate generated for you through Template Haskell.
(Note: I called it Inexhaustive because PartialFunction is something of a misnomer, in that it implies that Function is total. But Scala has no termination or positivity checkers, so the compiler is not actually able to talk about totality; so you get this weird situation where a function that is not a partial function is just a "regular" Function, whereas you should be able to call it a "total Function". The question here is not partially or totality, which is a broader idea, but inexhaustivity of pattern matches.)
{-# LANGUAGE TypeApplications #-}
module Inexhaustive
( Inexhaustive, inexhaustive
, runToMaybe, isDefinedAt
) where
import Prelude hiding ((.), id)
import Control.Applicative
import Control.Exception
import Control.Category
import Data.Maybe
import System.IO.Unsafe (unsafePerformIO)
newtype Inexhaustive a b = Inexhaustive (a -> b)
inexhaustive :: (a -> b) -> Inexhaustive a b
inexhaustive = Inexhaustive
runToMaybe :: Inexhaustive a b -> a -> Maybe b
runToMaybe (Inexhaustive f) x =
let io = fmap Just $ evaluate $ f x
in unsafePerformIO $ catch #PatternMatchFail io (\_ -> return Nothing)
isDefinedAt :: Inexhaustive a b -> a -> Bool
isDefinedAt f = isJust . runToMaybe f
instance Functor (Inexhaustive z) where
fmap f (Inexhaustive g) = inexhaustive (f . g)
instance Applicative (Inexhaustive z) where
pure x = inexhaustive (const x)
(Inexhaustive zab) <*> (Inexhaustive za) = Inexhaustive (\z -> zab z $ za z)
instance Alternative (Inexhaustive z) where
empty = inexhaustive (\_ -> throw $ PatternMatchFail "inexhaustive empty")
f <|> g =
inexhaustive $ \x ->
case runToMaybe f x <|> runToMaybe g x of
Just y -> y
instance Category Inexhaustive where
id = inexhaustive id
(Inexhaustive f) . (Inexhaustive g) = Inexhaustive (f . g)

Related

Questions about a la carte data types

I was reading the original paper about data types a la carte and decided to try to implement the idea in Scala (I know it's already implemented in many functional libraries). Unfortunately I found the original paper is hard to comprehend and I stuck somewhere in the beginning. Then I found another paper that was easier to understand and I managed to rewrite Haskell code from the paper into Scala, you can find it here. However I still struggling to understand a few moments:
A quote from the second paper
Orignal Expr data type
data Expr = Val Int | Add Expr Expr
New type signature:
data Arith e = Val Int | Add e e
For any functor f, its induced recursive datatype, Fix f, is defined as the least fixpoint of f, implemented as follows:
data Fix f = In (f (Fix f))
Now that we have tied the recursive knot of a signature,
Fix Arith is a language equivalent to the original Expr datatype
which allowed integer values and addition.
What does it mean exactly "we have tied the recursive knot of a signature" and what does it mean Fix Arith is a language equivalent to the original Expr ?
The actual type of In is In :: f (Fix f) -> Fix f
If we try to construct a value using In construct and Val 1 variable we'll get the following result:
> :t In(Val 1)
> In(Val 1) :: Fix Arith
Scala encoding of the same data types:
sealed trait Arith[A]
case class Val[A](x: Int) extends Arith[A]
case class Add[A](a: A, b: A) extends Arith[A]
trait Fix[F[_]]
case class In[F[_]](exp: F[Fix[F]]) extends Fix[F]
fold function
The fold function has the following signature and implementation
Haskell:
fold :: Functor f => (f a -> a) -> Fix f -> a
fold f (In t) = f (fmap (fold f) t)
Scala variant I came up with
def fold[F[_] : Functor, A](f: F[A] => A): Fix[F] => A = {
case In(t) =>
val g: F[Fix[F]] => F[A] = implicitly[Functor[F]].lift(fold(f))
f(g(t))
}
The thing that I'm curious about is that in my Scala version function g has the following type F[Fix[F]] => F[A] but the type of variable t after pattern matching is LaCarte$Add with value Add(In(Val(1)),In(Val(2))), how it happens that it's valid to apply function g to LaCarte$Add ? Also, I'd very appreciate if you can help me to understand fold function ?
Quote from the paper:
The first argument of fold is an f-algebra, which provides
the behavior of each constructor associated with a given signature f.
What does it mean exactly “we have tied the ‘recursive knot’ of a signature”?
The original Expr datatype is recursive, referring to itself in its own definition:
data Expr = Val Int | Add Expr Expr
The Arith type “factors out” the recursion by replacing recursive calls with a parameter:
data Arith e = Val Int | Add e e
The original Expr type can have any depth of nesting, which we want to support with Arith as well, but the maximum depth depends on what type we choose for e:
Arith Void can’t be nested: it can only be a literal value (Val n) because we can’t construct an Add, because we can’t obtain a value of type Void (it has no constructors)
Arith (Arith Void) can have one level of nesting: the outer constructor can be an Add, but the inner constructors can only be Lit.
Arith (Arith (Arith Void)) can have two levels
And so on
What Fix Arith gives us is a way to talk about the fixed point Arith (Arith (Arith …)) with no limit on the depth.
This is just like how we can replace a recursive function with a non-recursive function and recover the recursion with the fixed-point combinator:
factorial' :: (Integer -> Integer) -> Integer -> Integer
factorial' recur n = if n <= 1 then 1 else n * recur (n - 1)
factorial :: Integer -> Integer
factorial = fix factorial'
factorial 5 == 120
What does it mean Fix Arith is a language equivalent to the original Expr?
The language (grammar) that Fix Arith represents is equivalent to the language that Expr represents; that is, they’re isomorphic: you can write a pair of total functions Fix Arith -> Expr and Expr -> Fix Arith.
How it happens that it’s valid to apply function g to LaCarte$Add?
I’m not very familiar with Scala, but it looks like Add is a subtype of Arith, so the parameter of g of type F[Fix[F]] can be filled with a value of type Arith[Fix[Arith]] which you get by matching on the In constructor to “unfold” one level of recursion.

What kind of morphism is `filter` in category theory?

In category theory, is the filter operation considered a morphism? If yes, what kind of morphism is it? Example (in Scala)
val myNums: Seq[Int] = Seq(-1, 3, -4, 2)
myNums.filter(_ > 0)
// Seq[Int] = List(3, 2) // result = subset, same type
myNums.filter(_ > -99)
// Seq[Int] = List(-1, 3, -4, 2) // result = identical than original
myNums.filter(_ > 99)
// Seq[Int] = List() // result = empty, same type
One interesting way of looking at this matter involves not picking filter as a primitive notion. There is a Haskell type class called Filterable which is aptly described as:
Like Functor, but it [includes] Maybe effects.
Formally, the class Filterable represents a functor from Kleisli Maybe to Hask.
The morphism mapping of the "functor from Kleisli Maybe to Hask" is captured by the mapMaybe method of the class, which is indeed a generalisation of the homonymous Data.Maybe function:
mapMaybe :: Filterable f => (a -> Maybe b) -> f a -> f b
The class laws are simply the appropriate functor laws (note that Just and (<=<) are, respectively, identity and composition in Kleisli Maybe):
mapMaybe Just = id
mapMaybe (g <=< f) = mapMaybe g . mapMaybe f
The class can also be expressed in terms of catMaybes...
catMaybes :: Filterable f => f (Maybe a) -> f a
... which is interdefinable with mapMaybe (cf. the analogous relationship between sequenceA and traverse)...
catMaybes = mapMaybe id
mapMaybe g = catMaybes . fmap g
... and amounts to a natural transformation between the Hask endofunctors Compose f Maybe and f.
What does all of that have to do with your question? Firstly, a functor is a morphism between categories, and a natural transformation is a morphism between functors. That being so, it is possible to talk of morphisms here in a sense that is less boring than the "morphisms in Hask" one. You won't necessarily want to do so, but in any case it is an existing vantage point.
Secondly, filter is, unsurprisingly, also a method of Filterable, its default definition being:
filter :: Filterable f => (a -> Bool) -> f a -> f a
filter p = mapMaybe $ \a -> if p a then Just a else Nothing
Or, to spell it using another cute combinator:
filter p = mapMaybe (ensure p)
That indirectly gives filter a place in this particular constellation of categorical notions.
To answer are question like this, I'd like to first understand what is the essence of filtering.
For instance, does it matter that the input is a list? Could you filter a tree? I don't see why not! You'd apply a predicate to each node of the tree and discard the ones that fail the test.
But what would be the shape of the result? Node deletion is not always defined or it's ambiguous. You could return a list. But why a list? Any data structure that supports appending would work. You also need an empty member of your data structure to start the appending process. So any unital magma would do. If you insist on associativity, you get a monoid. Looking back at the definition of filter, the result is a list, which is indeed a monoid. So we are on the right track.
So filter is just a special case of what's called Foldable: a data structure over which you can fold while accumulating the results in a monoid. In particular, you could use the predicate to either output a singleton list, if it's true; or an empty list (identity element), if it's false.
If you want a categorical answer, then a fold is an example of a catamorphism, an example of a morphism in the category of algebras. The (recursive) data structure you're folding over (a list, in the case of filter) is an initial algebra for some functor (the list functor, in this case), and your predicate is used to define an algebra for this functor.
In this answer, I will assume that you are talking about filter on Set (the situation seems messier for other datatypes).
Let's first fix what we are talking about. I will talk specifically about the following function (in Scala):
def filter[A](p: A => Boolean): Set[A] => Set[A] =
s => s filter p
When we write it down this way, we see clearly that it's a polymorphic function with type parameter A that maps predicates A => Boolean to functions that map Set[A] to other Set[A]. To make it a "morphism", we would have to find some categories first, in which this thing could be a "morphism". One might hope that it's natural transformation, and therefore a morphism in the category of endofunctors on the "default ambient category-esque structure" usually referred to as "Hask" (or "Scal"? "Scala"?). To show that it's natural, we would have to check that the following diagram commutes for every f: B => A:
- o f
Hom[A, Boolean] ---------------------> Hom[B, Boolean]
| |
| |
| |
| filter[A] | filter[B]
| |
V ??? V
Hom[Set[A], Set[A]] ---------------> Hom[Set[B], Set[B]]
however, here we fail immediately, because it's not clear what to even put on the horizontal arrow at the bottom, since the assignment A -> Hom[Set[A], Set[A]] doesn't even seem functorial (for the same reasons why A -> End[A] is not functorial, see here and also here).
The only "categorical" structure that I see here for a fixed type A is the following:
Predicates on A can be considered to be a partially ordered set with implication, that is p LEQ q if p implies q (i.e. either p(x) must be false, or q(x) must be true for all x: A).
Analogously, on functions Set[A] => Set[A], we can define a partial order with f LEQ g whenever for each set s: Set[A] it holds that f(s) is subset of g(s).
Then filter[A] would be monotonic, and therefore a functor between poset-categories. But that's somewhat boring.
Of course, for each fixed A, it (or rather its eta-expansion) is also just a function from A => Boolean to Set[A] => Set[A], so it's automatically a "morphism" in the "Hask-category". But that's even more boring.
filter can be written in terms of foldRight as:
filter p ys = foldRight(nil)( (x, xs) => if (p(x)) x::xs else xs ) ys
foldRight on lists is a map of T-algebras (where here T is the List datatype functor), so filter is a map of T-algebras.
The two algebras in question here are the initial list algebra
[nil, cons]: 1 + A x List(A) ----> List(A)
and, let's say the "filter" algebra,
[nil, f]: 1 + A x List(A) ----> List(A)
where f(x, xs) = if p(x) x::xs else xs.
Let's call filter(p, _) the unique map from the initial algebra to the filter algebra in this case (it is called fold in the general case). The fact that it is a map of algebras means that the following equations are satisfied:
filter(p, nil) = nil
filter(p, x::xs) = f(x, filter(p, xs))

Universally Quantified Types in Haskell and Scala?

Existentially quantified types explains:
any use of a lowercase type implicitly begins with a forall keyword, so the two type declarations for map are equivalent, as are the declarations below:
id :: a -> a
id :: forall a . a -> a
Given Scala's scala.Predef#identity, is there a forall equivalent, i.e. per Haskell's above second function?
There is no explicit forall in Scala, but it has equivalents to the different ways explicit foralls are used in Haskell:
To enable ScopedTypeVariables: not necessary, as Scala type variables are scoped by default.
For existential types: Scala supports them directly.
For higher-rank types, i.e. ones which have a forall nested under function arrows: consider this example:
f2 :: (forall a. a->a) -> Int -> Int
f2 f x = f x
In Scala [A](A => A) can't be used directly, but
trait Poly1 {
def apply[A](x: A): A
}
is equivalent to it, so you can implement f2:
def f2(p: Poly1, x: Int) = p(x)
I am not a Scala expert, but my understanding is that all Scala type parameters are universally quantified (i.e. declared with a forall) unless they are explicitly quantified with forSome. See, for instance:
What is the forSome keyword in Scala for?
In Haskell we use the forall keyword to declare existential types. It may seem counter-intuitive to use forall when we mean for some, but it can be done using the logical equivalence of the following statements:
(forSome x. P(x)) implies Q
(forall x. P(x) implies Q)
Here Q is a statement not containing x.

What does >>= mean in purescript?

I was reading the purescript wiki and found following section which explains do in terms of >>=.
What does >>= mean?
Do notation
The do keyword introduces simple syntactic sugar for monadic
expressions.
Here is an example, using the monad for the Maybe type:
maybeSum :: Maybe Number -> Maybe Number -> Maybe Number
maybeSum a b = do
n <- a
m <- b
let result = n + m
return result
maybeSum takes two
values of type Maybe Number and returns their sum if neither number is
Nothing.
When using do notation, there must be a corresponding
instance of the Monad type class for the return type. Statements can
have the following form:
a <- x which desugars to x >>= \a -> ...
x which desugars to x >>= \_ -> ... or just x if this is the last statement.
A let binding let a = x. Note the lack of the in keyword.
The example maybeSum desugars to ::
maybeSum a b =
a >>= \n ->
b >>= \m ->
let result = n + m
in return result
>>= is a function, nothing more. It resides in the Prelude module and has type (>>=) :: forall m a b. (Bind m) => m a -> (a -> m b) -> m b, being an alias for the bind function of the Bind type class. You can find the definitions of the Prelude module in this link, found in the Pursuit package index.
This is closely related to the Monad type class in Haskell, which is a bit easier to find resources. There's a famous question on SO about this concept, which is a good starting point if you're looking to improve your knowledge on the bind function (if you're starting on functional programming now, you can skip it for a while).

Implementing the map function using only foldRight, foldLeft and unfold in Scala

I have to implement the Map function using only the foldRight, foldLeft and unfold. This means that I have to loop through every element in the list and apply a function f to it.
I have declared my own list as follow:
abstract class IntList
case class Nil() extends IntList
case class Cons(h: Int, t: IntList) extends IntList
And I've implemented the foldRight, foldLeft and unfold functions.
and the implementation of the new map function:
def map(ls: IntList, f: Int => Int): IntList = // ??
I've been thinking for a while now, but I don't have a clue where to begin. I may not use recursion in the map function. I'm pretty sure that I have to combine the power of fold and unfold together. Unfold returns a IntList, which is the return type of map. But I'm not sure what I have to give with this function.
Anyone has a clue? :)
Match the types, fill in the arguments to match.
For instance, if you are going to use foldRight, then B must be IntList, because that's the type returned by map. Now fill in the arguments to foldRight with whatever values you have that match the types.
[In reply to previous comments.]
I don't know which exact variant of unfold you are given. Assuming it's something like this (in Ocaml, sorry, don't have Scala installed right now):
(* unfold : ('a -> ('b * 'a) option) -> 'a -> 'b list *)
let rec unfold f x =
match f x with
| None -> []
| Some (y, x') -> y :: unfold f x'
Then a solution for map is the following:
let map f = unfold (function [] -> None | x::xs -> Some (f x, xs))
Hope that helps.