在Scala中组成免费monad

Question

I think I understand what Free monad is. I hope I understand also that functors compose but monads do not , ie if M1 and M2 are monads then M1[M2] is not necessarily a monad.

My questions are:

• Do Free monads compose ?
• Suppose we have functors F1 and F2 and their composition F1[F2] . Suppose also we have Free1 and Free2 -- Free monads for F1 and F2 . Can we define a Free monad for F1[F2] with just Free1 and Free2 ?

Answer 1

Hopefully I can answer your question:

Do Free monads compose?

No . For the same reasons as "normal" monads don't. To write monadic bind we need to know something about the underlying monad, or about the underlying functor in a free monad case.

Hopefully the Haskell syntax doesn't scare you too much:

type X f g a = Free f (Free g a)

bind :: X f g a -> (a -> X f g b) -> X f g b
bind (Pure (Pure x)) k = k x
bind (Pure (Free f)) k = error "not implemented"
bind _ = error "we don't even consider other cases"

In the second case we have f :: g (Free ga) and k :: a -> Free f (Free gb) . We could fmap , as it's the only thing we can do:

bind (Pure (Free f)) k = let kf = fmap (fmap k) f -- fmapping inside g ∘ Free g
                         in = error "not implement"

The type of kf will be: g (Free g (Free f (Free gb))) , when we'd need Free f (Free gb) . You'll arrive at the same problem as when writing a monad instance for any Compose m1 m2 , we'd need to reorder "bind layers" from ggfg to fggg , and to do that commutation we need to know more about f and g .

Please comment, if you want to see the Scala version of above. It will be much more obscure though :(

Can we define a Free monad for F1[F2] with just Free1 and Free2

In other words given:

type Free1[A] = Free[F1, A]
type Free2[A] = Free[F2, B]

type FreeDist[A] = Free1[Free2[A]] = Free[F1, Free[F2, A]]
type FreeComp[A] = Free[F1[F2[_]], A]

Could we write a monad homomorphism (a good mapping) from FreeDist[A] to FreeComp[A] ? We can't, for the same reasons as in a previous part.

---

Scala version

Scalaz has private definitions of subclasses of Free , so I have to implement Free myself to have an "runnable" example. Some of the code scrapped from http://eed3si9n.com/learning-scalaz/Free+Monad.html

First simplest definition of Free in Scala:

import scala.language.higherKinds

trait Functor[F[_]] {
  def map[A, B](x: F[A])(f: A => B): F[B]
}

sealed trait Free[F[_], A] {
  def map[B](f: A => B)(implicit functor: Functor[F]): Free[F, B]
  def flatMap[B](f: A => Free[F, B])(implicit functor: Functor[F]): Free[F, B]
}
case class Pure[F[_], A](x: A) extends Free[F, A] {
  def map[B](f: A => B)(implicit functor: Functor[F]): Free[F, B] = Pure[F, B](f(x))
  def flatMap[B](f: A => Free[F, B])(implicit functor: Functor[F]): Free[F, B] = f(x)
}
case class Bind[F[_], A](x: F[Free[F, A]]) extends Free[F, A]  {
  def map[B](f: A => B)(implicit functor: Functor[F]): Free[F, B] = Bind {
    functor.map[Free[F,A], Free[F,B]](x) { y => y.map(f) }
  }
  // omitted 
  def flatMap[B](f: A => Free[F, B])(implicit functor: Functor[F]): Free[F, B] = ???
}

Using that we can translate the Haskell example into Scala:

type X[F[_], G[_], A] = Free[F, Free[G, A]]

// bind :: X f g a -> (a -> X f g b) -> X f g b
def xFlatMap[F[_], G[_], A, B](x: X[F, G, A], k: A => X[F, G, B])(implicit functorG: Functor[G]): X[F, G, B] =
  x match {
    case Pure(Pure(y)) => k(y)
    case Pure(Bind(f)) => {
      // kf :: g (Free g (Free f (Free g b)))
      val kf: G[Free[G, Free[F, Free[G, B]]]] = functorG.map(f) { y => y.map(k) }
      // But we need Free[F, Free[G, B]]
      ???
    }
    // we don't consider other cases
    case _ => ???
  }

The result is the same, we can't make types match, We'd need transform Free[G, Free[F, A]] into Free[F, Free[G, A]] somehow.