How to stop backtracking in Scala? - scala

Question

Suppose I am solving a problem (e.g. N-Queen) with backtracking. What if I want to find the only one (the 1-st) solution rather than all of them.
I guess I can do it imperatively (e.g. with a mutable boolean flag). I wonder how I can do it functionally.

Answer 1

While Scala's Option will work here, as two other answers have pointed out, a more idiomatic functional approach would be to use a "lazy list"—or a Stream, in Scala—to represent the set of solutions.
I've found myself writing code like this, for example:
trait Node[A] {
def children: Stream[A with Node[A]]
def dfs(f: A => Boolean): Stream[A] = this.children.flatMap {
child => if (f(child)) Stream(child) else child.dfs(f)
}
}
Now suppose I have a Board class that extends Node[Board] and that has an implementation of the children method that returns all valid boards with one additional piece. Suppose it also has some other useful stuff, like a size method, a companion object with an empty, etc.
I can then write the following to get a Stream of solutions:
val solutions = Board.empty.dfs(_.size == 8)
A Stream is lazy and only evaluates its head, so right now we've only searched the tree far enough to find the first solution. We can get this solution using head:
scala> solutions.head
res1: Board =
o . . . . . . .
. . . . o . . .
. . . . . . . o
. . . . . o . .
. . o . . . . .
. . . . . . o .
. o . . . . . .
. . . o . . . .
Or whatever. But I can also get other results if I want them:
scala> solutions(10)
res2: Board =
. o . . . . . .
. . . . . . o .
. . . . o . . .
. . . . . . . o
o . . . . . . .
. . . o . . . .
. . . . . o . .
. . o . . . . .
This searches just enough more of the tree to find the tenth solution, and then stops.
The big advantage of Stream over the Option approach is that I can get additional results if I need them, without paying more for the first.

Answer 2

Here's a simple case with a depth-first search that stops when it finds what it's looking for. It makes use of Option, as mentioned in Chris K's answer.
case class Tree[A](v: A, subtrees: Tree[A]*) {
def dfs(s: A): Option[A] = {
println("visiting " + v)
subtrees.foldLeft(if(v == s) Some(v) else None)((r, t) =>
if(r.isDefined)
r
else
t.dfs(s)
)
}
override def toString() = "Tree(%s%s%s)".format(v, if(subtrees.nonEmpty) ", " else "", subtrees.mkString(", "))
}
Usage:
scala> val t = Tree(1, Tree(2, Tree(3), Tree(4)), Tree(5, Tree(6), Tree(7)))
t: Tree[Int] = Tree(1, Tree(2, Tree(3), Tree(4)), Tree(5, Tree(6), Tree(7)))
The tree t looks like
1
/ \
2 5
/ \ / \
3 4 6 7
So we can search for an element and trace the nodes it visits:
scala> t.dfs(6)
visiting 1
visiting 2
visiting 3
visiting 4
visiting 5
visiting 6
res42: Option[Int] = Some(6)
Notice that we don't visit any more nodes after we find what we're looking for.

Answer 3

Assuming you're using a recursive search function, your function should either return a result (i.e. positioning of the queens) or an indication that no result could be found for that branch. Scala probably has an option/maybe type, you can use that. This advice is equally applicable to any functional language.