Tuesday, September 6, 2011

Fix from Fold

My two previous posts, Fold Right from Fold Left and Folding Stream with Scala are actually my interpretation to (Hutton, 1999) paper. Now, I would like to continue with another paper (Pope, 2010 ?) that also talks about fold. Note that, Pope's work cites of  Hutton's work.

This post ends my fold trilogy. It has been an exciting and fun to play with.  I would love to continue with scalaz Fold (Foldr, Foldl, FoldMap Foldable are interesting), but I think I have to stop having fun :-)

Yet Another dropWhile Implementation

In Folding Stream with Scala , I implemented dropWhile using fold using (Hutton, 1999) paper. Pope proposes two more implementations, both work very well with infinite stream.

First, a reminder of fold implementation:
def foldr[A, B](combine:(A, =>B) => B, base:B)(xs:Stream[A]): B = { 
    if (xs.isEmpty) base 
    else combine(xs.head, foldr(combine, base)(xs.tail)) 

And here is the solution:

  def dwHo[A](pred:A=>Boolean, xs:Stream[A]):Stream[A]=>Stream[A] = { 
    val id =(s:Stream[A])=>s 
    val tail= (s:Stream[A])=>s.tail 
    def combine(next:A, rec: =>Stream[A]=>Stream[A]) = { 
     if (pred(next)) (rec compose tail) 
     else id 
    foldr(combine, id)(xs) 

  scala> val xs = Stream.range(20, 120) 
  xs: scala.collection.immutable.Stream[Int] = Stream(20, ?) 
  scala> dwHo( (_:Int)<100, xs)(xs).toList 
  res33: List[Int] = List(100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119) 
It also works with infinite Stream:

scala> val ys = Stream.from(1) 
ys: scala.collection.immutable.Stream[Int] = Stream(1, ?) 
scala> dwHo( (_:Int)<5, ys)(ys).take(10).toList 
res38: List[Int] = List(5, 6, 7, 8, 9, 10, 11, 12, 13, 14)
Under the hood, here is what happen:

foldr combine id [1..] 
 =combine 1 (foldr combine id [2..]) = 
 =(foldr combine id [2..]) . tail 
 =(combine 2 (foldr combine id [3..]) . tail 
 =(foldr combine id [3..]) .tail . tail 
 =(combine 3 (foldr combine id [4..]) . tail . tail 
 =(foldr combine id [4..]) .tail . tail . tail 
 =(combine 4 (foldr combine id [5..]) . tail . tail . tail 
 =(foldr combine id [5..]) . tail . tail . tail . tail 
 =(combine 5 (foldr combine id [6..]) . tail . tail . tail 
 =id . tail . tail . tail . tail
And (id.tail.tail.tail.tail)[1..] = [5..].

Fix from Fold
The most interesting part of the (Pope, 2010) is not on dropWhile though, but on an implementation of a function using fold. The function is called fix, also known as Y combinator. Check this page to have an idea of what fix function is.

Basically, using fix, we can encode recursion function. The following is an example of factorial function using fix in haskell:

Prelude> :m Control.Monad.Fix 
Prelude Control.Monad.Fix> fix (\rec n -> if n == 0 then 1 else n * rec (n-1)) 5  
In Scala, fix function is implemented in a much trickier way. The difficulties come from the Scala strictness. Here is the implementation of fix I found after googling a little bit:

def fix[A](f: (A=>A)=>(A=>A)): A=>A = f(fix(f))(_)

scala> fix[Long](f=>x=> if (x == 0) 1 else x * f(x - 1))(5)
res5: Long = 120

One question that may arise is whether fold can be implemented using fix. Well, apparently, yes, after all, it's a recursion, but I haven't tried it yet. A more interesting question would be if fix can be implemented using fold. The answer is yes, and that's the essence of (Pope, 2010) article. Here is the implementation of fix using foldr in Scala:

def fix[A](f: (A=>A)=>(A=>A)): A=>A = { 
  def combine( a:(A=>A)=>(A=>A), b: =>A=>A):A=>A = f(b)(_) 
  foldr(combine, null)(Stream.continually(null)) 

Give a try:

scala> fix[Long](f=>x=> if (x == 0) 1 else x * f(x - 1))(5) 
res6: Long = 120

Youpi..., isn't it awesome? This shows how expressive fold is, since it can now be used to implement (any?) recursion functions.


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