Can you always structure a recursive function for tail-call elimination? If not, what are other strategies to limit the stack size?
For example:
(inspired by Break or shortcircuit a fold in Scala)
// Depth-first search of labyrinth, with large depth > stacklimit
def search ( labyrinth: SearchDomain,
path: List[SolutionNode],
goal: DesiredResult ) = {
if ( path.head == goal ) return path
candidates: List[SolutionNode] = labyrinth.childNodes(path)
candidates.find { c =>
Nil != search( labyrinth, c :: path, goal ) // potential boom!
} match {
case Some(c) => c :: path
case None => Nil
}
}
The goal is not to nit-pick this particular function, but to use it as a prop to learn techniques to limit stack-size.
UPDATE
My take-away from this is:
If the problem domain is such that recursion may hit limitation of stack-size:
Rewrite the code to be the scala-compiler-version-of-tailcall-optimizable. This can be aided/verified by the new (2.8) #scala.annotation.tailrec annotation.
If that is not feasible, rewrite it to use iterative looping constructs.
I'm also getting the sense that this rewriting (either case) is something that takes a certain amount of skill/talent/smarts/practice.
All recursive functions can be expressed as an iterative process, and all iterative processes can be expressed as tail-recursion, so yes, strictly speaking, you can transform any recursive algorithm into a tail-recursive one. However, don't don't assume that this will actually save you space. In many cases the record-keeping done by the stack is necessary, and you'll end up needing to essentially emulate the stack in your iterative or tail-recursive version. This can still be useful, say when you've got a small stack but a large heap.
You should accept Laurence Gonsalves answer, but, as for the code:
// Depth-first search of labyrinth, with large depth > stacklimit
def search ( labyrinth: SearchDomain,
path: List[SolutionNode],
goal: DesiredResult ) = {
if ( path.head == goal ) return path
candidates: List[SolutionNode] = labyrinth.childNodes(path)
candidates.find { c =>
Nil != search( labyrinth, c :: path, goal ) // potential boom!
} match {
case Some(c) => c :: path
case None => Nil
}
}
becomes
// Depth-first search of labyrinth
def search ( labyrinth: SearchDomain,
path: List[SolutionNode],
goal: DesiredResult ) = {
def recurse( candidates: List[List[SolutionNode]],
path: List[SolutionNode] ) = candidates match {
case List(Nil) => Nil
case Nil :: tail => recurse(tail, path.tail)
case (nextCandidate :: otherCandidates) :: tail =>
if (nextCandidate == goal)
nextCandidate :: path
else
recurse(labyrinth.childNodes(nextCandidate :: path) :: otherCandidates,
nextCandidate :: path)
}
if ( path.head == goal )
path
else
recurse(labyrinth.childNodes(path), path)
}
All recursive functions can't expressed as a tail recursions, I think.
However you can express all recursive functions as iterative processes.
There are two cases to consider here. In the general case, are there some recursive functions that can't be expressed as tail calls? [UPDATE] As pointed out in another answer, there are not.
However, in the specific case of scala, there are some tail recursive functions that cannot be optimized to run in a tail recursive manner (meaning that they reuse stack frames.) This is mostly due to the limitations of the JVM I believe.
see previous question for more details about how this works.
Related
Scenario:
val col: IndexedSeq[Array[Char]] = for (i <- 1 to n) yield {
val x = for (j <- 1 to m) yield 'x'
x.toArray
}
This is a fairly simple char matrix. toArray used to allow updating.
var west = last.x - 1
while (west >= 0 && arr(last.y)(west) == '.') {
arr(last.y)(west) = ch;
west -= 1;
}
This is updating all . to ch until a non-dot char is found.
Generically, update until stop condition is met, unknown number of steps.
What is the idiomatic equivalent of it?
Conclusion
It's doable, but the trade-off isn't worth it, a lot of performance is lost to expressive syntax when the collection allows updating.
Your wish for a "cleaner, more idiomatic" solution is of course a little fuzzy, because it leaves a lot of room for subjectivity. In general, I'd consider a tail-recursive updating routine more idiomatic, but it might not be "cleaner" if you're more familiar with a non-functional programming style. I came up with this:
#tailrec
def update(arr:List[Char], replace:Char, replacement:Char, result:List[Char] = Nil):List[Char] = arr match {
case `replace` :: tail =>
update(tail, replace, replacement, replacement :: result)
case _ => result.reverse ::: arr
}
This takes one of the inner sequences (assuming a List for easier pattern matching, since Arrays are trivially convertible to lists), and replaces the replace char with the replacement recursively.
You can then use map to update the outer sequence, like so:
col.map { x => update(x, '.', ch) }
Another more reusable alternative is writing your own mapUntil, or using one which is implemented in a supplemental library (Scalaz probably has something like it). The one I came up with looks like this:
def mapUntil[T](input:List[T])(f:(T => Option[T])) = {
#tailrec
def inner(xs:List[T], result:List[T]):List[T] = xs match {
case Nil => Nil
case head :: tail => f(head) match {
case None => (head :: result).reverse ::: tail
case Some(x) => inner(tail, x :: result)
}
}
inner(input, Nil)
}
It does the same as a regular map invocation, except that it stops as soon as the passed function returns None, e.g.
mapUntil(List(1,2,3,4)) {
case x if x >= 3 => None
case x => Some(x-1)
}
Will result in
List[Int] = List(0, 1, 3, 4)
If you want to look at Scalaz, this answer might be a good place to start.
x3ro's answer is the right answer, esp. if you care about performance or are going to be using this operation in multiple places. I would like to add simple solution using only what you find in the collections API:
col.map { a =>
val (l, r) = a.span(_ == '.')
l.map {
case '.' => ch
case x => x
} ++ r
}
I'd like to be able to use a single variable multiple times within one pattern, so that it will only match if the same value is present in all places, such as
list match {
case x :: x :: xs => // recurse
}
which would match List(1,1,2) but would not match List(1,2,1). But this does not compile with error: x is already defined as value x.
In researching this question, I found out that I can also include a guard in the case clause, so I can do
list match {
case x1 :: x2 :: xs if x1==x2 => // recurse
}
which seems to work the same way (it does, right?). This is good, but it wouldn't look as clean if I wanted the same value in many places, like
list match {
case x1::x2::x3::x4::xs if x1==x2 && x2==x3 && x3==x4 => // recurse
}
Is there any more elegant way I can do this?
A few notes: Yes, I am just learning scala, if that wasn't clear, so I'm not sure this is something I'd ever really want to do, but I'm just interested in what's possible. In that regard, I'm not really looking for a completely different solution, like takeWhile or filter or something, but more so am specifically interested in pattern matching.
Scala doesn't provide quite that much flexibility with its matches (which may be a good thing, as one has to be aware of errors arising from unintentional variable re-use).
If you have a large number of identical items, you might want to consider a nested match (but note that you won't fail out of the inner match to be completed later down the outer match, so you have to handle everything locally):
list match {
case x :: rest => rest match {
case `x` :: `x` :: `x` :: xs => println("Four of the same")
case _ => println("Well, nonempty at least")
}
case _ => println("Boring, there's nothing here!")
}
Note the backticks which mean "we've already got this variable, check against it, don't set it!".
Alternatively, if you have specialized functionality that you use repeatedly, you can create a custom matcher:
object FourOf {
def unapplySeq(xs: List[Int]): Option[(Int, List[Int])] = xs match {
case x :: y :: z :: a :: rest if x==y && y==z && z==a => Some((x,rest))
case _ => None
}
}
and then use it whenever you need that complicated pattern:
list match {
case FourOf(x,rest) => println("four of the same")
case x :: more => println("Nonempty")
case _ => println("Yawn")
}
Neither of these are quite as tidy and flexible as what you were apparently hoping for, but then again, I'm not sure flipping between assigning and testing the same variable in a match statement is a good way to write clear code anyway.
For many repeats you might use stable identifiers to do a comparison (instead of catching a value):
val x = list.head
list match {
case `x`::`x`::`x`::`x`::xs => ....
}
But note that this won't work on empty list (you just cannot get head of it).
I think Rex's answer rocks. I am a fan of unapplySeq. But here's a not-so-clever-and-maybe-wasteful alternative, if your main bother is just with the sequence of =='s in each guard.
So in the TMTOWTDI spirit:
def same[A](xs: A*) = xs forall (xs.head==)
// Then in your pattern match,
list match {
// case x1::x2::x3::x4::xs if x1==x2 && x2==x3 && x3==x4 => // recurse
case x1::x2::x3::x4::xs if same(x1,x2,x3,x4) => // recurse
}
I like Om's answer as well, so here's an adaptation:
list.headOption map (x => list match {
case `x`::`x`::`x`::`x`::xs => //...;
case _ => // ...
}) getOrElse {
// do what you'd have done for an empty list...
}
I have a question about writing recursive algorithms in a functional style. I will use Scala for my example here, but the question applies to any functional language.
I am doing a depth-first enumeration of an n-ary tree where each node has a label and a variable number of children. Here is a simple implementation that prints the labels of the leaf nodes.
case class Node[T](label:T, ns:Node[T]*)
def dfs[T](r:Node[T]):Seq[T] = {
if (r.ns.isEmpty) Seq(r.label) else for (n<-r.ns;c<-dfs(n)) yield c
}
val r = Node('a, Node('b, Node('d), Node('e, Node('f))), Node('c))
dfs(r) // returns Seq[Symbol] = ArrayBuffer('d, 'f, 'c)
Now say that sometimes I want to be able to give up on parsing oversize trees by throwing an exception. Is this possible in a functional language? Specifically is this possible without using mutable state? That seems to depend on what you mean by "oversize". Here is a purely functional version of the algorithm that throws an exception when it tries to handle a tree with a depth of 3 or greater.
def dfs[T](r:Node[T], d:Int = 0):Seq[T] = {
require(d < 3)
if (r.ns.isEmpty) Seq(r.label) else for (n<-r.ns;c<-dfs(n, d+1)) yield c
}
But what if a tree is oversized because it is too broad rather than too deep? Specifically what if I want to throw an exception the n-th time the dfs() function is called recursively regardless of how deep the recursion goes? The only way I can see how to do this is to have a mutable counter that is incremented with each call. I can't see how to do it without a mutable variable.
I'm new to functional programming and have been working under the assumption that anything you can do with mutable state can be done without, but I don't see the answer here. The only thing I can think to do is write a version of dfs() that returns a view over all the nodes in the tree in depth-first order.
dfs[T](r:Node[T]):TraversableView[T, Traversable[_]] = ...
Then I could impose my limit by saying dfs(r).take(n), but I don't see how to write this function. In Python I'd just create a generator by yielding nodes as I visited them, but I don't see how to achieve the same effect in Scala. (Scala's equivalent to a Python-style yield statement appears to be a visitor function passed in as a parameter, but I can't figure out how to write one of these that will generate a sequence view.)
EDIT Getting close to the answer.
Here is an function that returns a Stream of nodes in depth-first order.
def dfs[T](r: Node[T]): Stream[Node[T]] = {
(r #:: Stream.empty /: r.ns)(_ ++ dfs(_))
}
That is almost it. The only problem is that Stream memoizes all results, which is a waste of memory. I want a traversable view. The following is the idea, but does not compile.
def dfs[T](r: Node[T]): TraversableView[Node[T], Traversable[Node[T]]] = {
(Traversable(r).view /: r.ns)(_ ++ dfs(_))
}
It gives a "found TraversableView[Node[T], Traversable[Node[T]]], required TraversableView[Node[T], Traversable[_]] error for the ++ operator. If I change the return type to TraversableView[Node[T], Traversable[_]], I get the same problem with the "found" and "required" clauses switched. So there's some magic type variance incantation I haven't lit upon yet, but this is close.
It can be done: you just have to write some code to actually iterate through the children in the way you want (as opposed to relying on for).
More explicitly, you'll have to write code to iterate through a list of children and check if the "depth" crossed your threshold. Here's some Haskell code (I'm really sorry, I'm not fluent in Scala, but this can probably be easily transliterated):
http://ideone.com/O5gvhM
In this code, I've basically replaced the for loop for an explicit recursive version. This allows me to stop the recursion if the number of visited nodes is already too deep (i.e., limit is not positive). When I recurse to examine the next child, I subtract the number of nodes the dfs of the previous child visited and set this as the limit for the next child.
Functional languages are fun, but they're a huge leap from imperative programming. It really makes you pay attention to the concept of state, because all of it is excruciatingly explicit in the arguments when you go functional.
EDIT: Explaining this a bit more.
I ended up converting from "print just the leaf nodes" (which was the original algorithm from the OP) to "print all nodes". This enabled me to have access to the number of nodes the subcall visited through the length of the resulting list. If you want to stick to the leaf nodes, you'll have to carry around how many nodes you have already visited:
http://ideone.com/cIQrna
EDIT again To clear up this answer, I'm putting all the Haskell code on ideone, and I've transliterated my Haskell code to Scala, so this can stay here as the definite answer to the question:
case class Node[T](label:T, children:Seq[Node[T]])
case class TraversalResult[T](num_visited:Int, labels:Seq[T])
def dfs[T](node:Node[T], limit:Int):TraversalResult[T] =
limit match {
case 0 => TraversalResult(0, Nil)
case limit =>
node.children match {
case Nil => TraversalResult(1, List(node.label))
case children => {
val result = traverse(node.children, limit - 1)
TraversalResult(result.num_visited + 1, result.labels)
}
}
}
def traverse[T](children:Seq[Node[T]], limit:Int):TraversalResult[T] =
limit match {
case 0 => TraversalResult(0, Nil)
case limit =>
children match {
case Nil => TraversalResult(0, Nil)
case first :: rest => {
val trav_first = dfs(first, limit)
val trav_rest =
traverse(rest, limit - trav_first.num_visited)
TraversalResult(
trav_first.num_visited + trav_rest.num_visited,
trav_first.labels ++ trav_rest.labels
)
}
}
}
val n = Node(0, List(
Node(1, List(Node(2, Nil), Node(3, Nil))),
Node(4, List(Node(5, List(Node(6, Nil))))),
Node(7, Nil)
))
for (i <- 1 to 8)
println(dfs(n, i))
Output:
TraversalResult(1,List())
TraversalResult(2,List())
TraversalResult(3,List(2))
TraversalResult(4,List(2, 3))
TraversalResult(5,List(2, 3))
TraversalResult(6,List(2, 3))
TraversalResult(7,List(2, 3, 6))
TraversalResult(8,List(2, 3, 6, 7))
P.S. this is my first attempt at Scala, so the above probably contains some horrid non-idiomatic code. I'm sorry.
You can convert breadth into depth by passing along an index or taking the tail:
def suml(xs: List[Int], total: Int = 0) = xs match {
case Nil => total
case x :: rest => suml(rest, total+x)
}
def suma(xs: Array[Int], from: Int = 0, total: Int = 0) = {
if (from >= xs.length) total
else suma(xs, from+1, total + xs(from))
}
In the latter case, you already have something to limit your breadth if you want; in the former, just add a width or somesuch.
The following implements a lazy depth-first search over nodes in a tree.
import collection.TraversableView
case class Node[T](label: T, ns: Node[T]*)
def dfs[T](r: Node[T]): TraversableView[Node[T], Traversable[Node[T]]] =
(Traversable[Node[T]](r).view /: r.ns) {
(a, b) => (a ++ dfs(b)).asInstanceOf[TraversableView[Node[T], Traversable[Node[T]]]]
}
This prints the labels of all the nodes in depth-first order.
val r = Node('a, Node('b, Node('d), Node('e, Node('f))), Node('c))
dfs(r).map(_.label).force
// returns Traversable[Symbol] = List('a, 'b, 'd, 'e, 'f, 'c)
This does the same thing, quitting after 3 nodes have been visited.
dfs(r).take(3).map(_.label).force
// returns Traversable[Symbol] = List('a, 'b, 'd)
If you want only leaf nodes you can use filter, and so forth.
Note that the fold clause of the dfs function requires an explicit asInstanceOf cast. See "Type variance error in Scala when doing a foldLeft over Traversable views" for a discussion of the Scala typing issues that necessitate this.
I have written a simple depth-first search in Scala with a recursive function like that:
search(labyrinth, path, goal)
where labyrinth is a specification of the problem (as graph or whatever), path is a list that holds the path taken so far and goal is a specification of the goal state. The function returns a path to the goal as a List and Nil if no path can be found.
The function expands, e.g. finds all suitable next nodes (candidates) and then has to recursively call itself.
I do this by
candidates.foldLeft(Nil){
(solution, next) =>
if( solution == Nil )
search( labyrinth, next :: path, goal )
else
solution
}
Please note that I have omitted some unescessary details. Everything is working fine so far. But once a solution is found inside the foldLeft call, this solution gets simply copied by the else part of the if-statement. Is there a way to avoid this by breaking the foldLeft or maybe by using a different function instead of foldLeft? Actually I could probably write a version of foldLeft which breaks once "not Nil" is returned myself. But is there one inside the API?
I'm not sure I understand the desire to short-circuit the loop. Is it expensive to iterate through the candidates? Is the candidates list potentially large?
Maybe you could use the "find" method:
candidates.find { c =>
Nil != search( labyrinth, c :: path, goal )
} match {
case Some(c) => c :: path
case None => Nil
}
If the potential depth of the search space is large, you could overflow your stack (fitting, given this site name). But, that is a topic for another post.
For giggles, here is an actual runnable implementation. I had to introduce a local mutable variable (fullPath) mostly out of laziness, but I'm sure you could take those out.
object App extends Application {
// This impl searches for a specific factor in a large int
type SolutionNode = Int
case class SearchDomain(number: Int) {
def childNodes(l: List[Int]): List[Int] = {
val num = if (l.isEmpty) number else l.head
if (num > 2) {
(2 to (num - 1)) find {
n => (num % n)==0
} match {
case Some(i) => List(i, num / i)
case None => List()
}
}
else List()
}
}
type DesiredResult = Int
def search ( labyrinth: SearchDomain, path: List[SolutionNode], goal: DesiredResult ): List[SolutionNode] = {
if ( !path.isEmpty && path.head == goal ) return path
if ( path.isEmpty ) return search(labyrinth, List(labyrinth.number), goal)
val candidates: List[SolutionNode] = labyrinth.childNodes(path)
var fullPath: List[SolutionNode] = List()
candidates.find { c =>
fullPath = search( labyrinth, c :: path, goal )
!fullPath.isEmpty
} match {
case Some(c) => fullPath
case None => Nil
}
}
// Is 5 a factor of 800000000?
val res = search(SearchDomain(800000000), Nil, 5)
println(res)
}
I like Mitch Blevins solution, as it is a perfect match for your algorithm. You may be interested in my own solution to another maze problem.
Ok so there are a lot of interpretations here, your question lacks some degrees of specificity.
I will try to point out all assumptions as I go along.
Existing Function: (First assumptions: Type signatures)
type Labyrinth // Not particularly important to logic
type Position // The grid "position" of the path
def search(labyrinth: Labyrinth, path: List[Position], goal: Position): List[Position]
The function expands, e.g. finds all suitable next nodes (candidates) and then has to recursively call itself.
Assumption: the following code is within the search function
def search(labyrinth: Labyrinth, path: List[Position], goal: Position): List[Position] = {
...
return candidates.foldLeft(Nil){
(solution, next) =>
if( solution == Nil )
search( labyrinth, next :: path, goal )
else
solution
}
}
Please note that I have omitted some unescessary details.
If my assumptions have been correct, then perhaps, but I still had to deduce some things.
Never assume the necessity of implementation details when asking for assistance on implementation, specificity makes answering questions far simpler.
Is there a way to avoid this by breaking the foldLeft or maybe by using a different function instead of foldLeft?
Why not use recursion?
def search(labyrinth: Labyrinth, path: List[Position], goal: Position): List[Position] = {
...
val candidates: List[Position] = ... // However you calculate those
def resultPathOpt(candidates: List[position]): Option[List[Position]] = {
candidates match {
// No candidates
case Nil => None
// At least one candidate
case firstCandidate :: restOfCandidates => {
// Search
search(labyrinth, firstCandidate :: path, goal) match {
// Search didn't pan out, no more candidates, done
case Nil if restOfCandidates.isEmpty => None
// Search didn't pan out, try other candidates
case Nil => resultPathOpt(restOfCandidates)
// Search panned out, done
case validPath => Some(validPath)
}
}
}
val maybeValidResult: Option[List[Position]] = resultPathOpt(candidates)
maybeValidResult.getOrElse(List.empty)
}
Now as soon as we find a "valid" path, note there is no check for this in the logic I have provided, so I am not quite sure if this would ever end as you have implemented it and as I have implemented this, but the concepts are still valid.
I have a recursive function that takes a Map as single parameter. It then adds new entries to that Map and calls itself with this larger Map. Please ignore the return values for now. The function isn't finished yet. Here's the code:
def breadthFirstHelper( found: Map[AIS_State,(Option[AIS_State], Int)] ): List[AIS_State] = {
val extension =
for(
(s, v) <- found;
next <- this.expand(s) if (! (found contains next) )
) yield (next -> (Some(s), 0))
if ( extension.exists( (s -> (p,c)) => this.isGoal( s ) ) )
List(this.getStart)
else
breadthFirstHelper( found ++ extension )
}
In extension are the new entries that shall get added to the map. Note that the for-statement generates an iterable, not a map. But those entries shall later get added to the original map for the recursive call. In the break condition, I need to test whether a certain value has been generated inside extension. I try to do this by using the exists method on extension. But the syntax for extracting values from the map entries (the stuff following the yield) doesn't work.
Questions:
How do I get my break condition (the boolean statement to the if) to work?
Is it a good idea to do recursive work on a immutable Map like this? Is this good functional style?
When using a pattern-match (e.g. against a Tuple2) in a function, you need to use braces {} and the case statement.
if (extension.exists { case (s,_) => isGoal(s) } )
The above also uses the fact that it is more clear when matching to use the wildcard _ for any allowable value (which you subsequently do not care about). The case xyz gets compiled into a PartialFunction which in turn extends from Function1 and hence can be used as an argument to the exists method.
As for the style, I am not functional programming expert but this seems like it will be compiled into a iterative form (i.e. it's tail-recursive) by scalac. There's nothing which says "recursion with Maps is bad" so why not?
Note that -> is a method on Any (via implicit conversion) which creates a Tuple2 - it is not a case class like :: or ! and hence cannot be used in a case pattern match statement. This is because:
val l: List[String] = Nil
l match {
case x :: xs =>
}
Is really shorthand/sugar for
case ::(x, xs) =>
Similarly a ! b is equivalent to !(a, b). Of course, you may have written your own case class ->...
Note2: as Daniel says below, you cannot in any case use a pattern-match in a function definition; so while the above partial function is valid, the following function is not:
(x :: xs) =>
This is a bit convoluted for me to follow, whatever Oxbow Lakes might think.
I'd like first to clarify one point: there is no break condition in for-comprehensions. They are not loops like C's (or Java's) for.
What an if in a for-comprehension means is a guard. For instance, let's say I do this:
for {i <- 1 to 10
j <- 1 to 10
if i != j
} yield (i, j)
The loop isn't "stopped" when the condition is false. It simply skips the iterations for which that condition is false, and proceed with the true ones. Here is another example:
for {i <- 1 to 10
j <- 1 to 10
if i % 2 != 0
} yield (i, j)
You said you don't have side-effects, so I can skip a whole chapter about side effects and guards on for-comprehensions. On the other hand, reading a blog post I made recently on Strict Ranges is not a bad idea.
So... give up on break conditions. They can be made to work, but they are not functional. Try to rephrase the problem in a more functional way, and the need for a break condition will be replaced by something else.
Next, Oxbow is correct in that (s -> (p,c) => isn't allowed because there is no extractor defined on an object called ->, but, alas, even (a :: b) => would not be allowed, because there is no pattern matching going on in functional literal parameter declaration. You must simply state the parameters on the left side of =>, without doing any kind of decomposition. You may, however, do this:
if ( extension.exists( t => val (s, (p,c)) = t; this.isGoal( s ) ) )
Note that I replaced -> with ,. This works because a -> b is a syntactic sugar for (a, b), which is, itself, a syntactic sugar for Tuple2(a, b). As you don't use neither p nor c, this works too:
if ( extension.exists( t => val (s, _) = t; this.isGoal( s ) ) )
Finally, your recursive code is perfectly fine, though probably not optimized for tail-recursion. For that, you either make your method final, or you make the recursive function private to the method. Like this:
final def breadthFirstHelper
or
def breadthFirstHelper(...) {
def myRecursiveBreadthFirstHelper(...) { ... }
myRecursiveBreadthFirstHelper(...)
}
On Scala 2.8 there is an annotation called #TailRec which will tell you if the function can be made tail recursive or not. And, in fact, it seems there will be a flag to display warnings about functions that could be made tail-recursive if slightly changed, such as above.
EDIT
Regarding Oxbow's solution using case, that's a function or partial function literal. It's type will depend on what the inference requires. In that case, because that's that exists takes, a function. However, one must be careful to ensure that there will always be a match, otherwise you get an exception. For example:
scala> List(1, 'c') exists { case _: Int => true }
res0: Boolean = true
scala> List(1, 'c') exists { case _: String => true }
scala.MatchError: 1
at $anonfun$1.apply(<console>:5)
... (stack trace elided)
scala> List(1, 'c') exists { case _: String => true; case _ => false }
res3: Boolean = false
scala> ({ case _: Int => true } : PartialFunction[AnyRef,Boolean])
res5: PartialFunction[AnyRef,Boolean] = <function1>
scala> ({ case _: Int => true } : Function1[Int, Boolean])
res6: (Int) => Boolean = <function1>
EDIT 2
The solution Oxbow proposes does use pattern matching, because it is based on function literals using case statements, which do use pattern matching. When I said it was not possible, I was speaking of the syntax x => s.