Mapping sub-sets of parentheses to chars - string

Question

I am attempting to create a Scala method that will take one parent group of parentheses, represented as a String, and then map each subgroup of parentheses to a different letter. It should then put these in a map which it returns, so basically I call the following method like this:
val s = "((2((x+3)+6)))"
val map = mapParentheses(s)
Where s could contain any number of sets of parentheses, and the Map returned should contain:
"(x+3)" -> 'a'
"(a+6)" -> 'b'
"(2b)" -> 'c'
"(c)" -> 'd'
So that elsewhere in my program I can recall 'd' and get "(c)" which will become "((2b))" then ((2(a+6))) and finally ((2((x+3)+6))). The string sent to the method mapParentheses will never have unmatched parentheses, or extra chars outside of the main parent parentheses, so the following items will never be sent:
"(fsf)a" because the a is outside the parent parentheses
"(a(aa))(a)" because the (a) is outside the parent parentheses
"((a)" because the parentheses are unmatched
")a(" because the parentheses are unmatched
So I was wondering if anyone knew of an easy (or not easy) way of creating this mapParentheses method.

Answer 1

You can do this pretty easily with Scala's parser combinators. First for the import and some simple data structures:
import scala.collection.mutable.Queue
import scala.util.parsing.combinator._
sealed trait Block {
def text: String
}
case class Stuff(text: String) extends Block
case class Paren(m: List[(String, Char)]) extends Block {
val text = m.head._2.toString
def toMap = m.map { case (k, v) => "(" + k + ")" -> v }.toMap
}
I.e., a block represents a substring of the input that is either some non-parenthetical stuff or a parenthetical.
Now for the parser itself:
class ParenParser(fresh: Queue[Char]) extends RegexParsers {
val stuff: Parser[Stuff] = "[^\\(\\)]+".r ^^ (Stuff(_))
def paren: Parser[Paren] = ("(" ~> insides <~ ")") ^^ {
case (s, m) => Paren((s -> fresh.dequeue) :: m)
}
def insides: Parser[(String, List[(String, Char)])] =
rep1(paren | stuff) ^^ { blocks =>
val s = blocks.flatMap(_.text)(collection.breakOut)
val m = blocks.collect {
case Paren(n) => n
}.foldLeft(List.empty[(String, Char)])(_ ++ _)
(s, m)
}
def parse(input: String) = this.parseAll(paren, input).get.toMap
}
Using get in the last line is very much not ideal, but is justified by your assertion that we can expect well-formed input.
Now we can create a new parser and pass in a mutable queue with some fresh variables:
val parser = new ParenParser(Queue('a', 'b', 'c', 'd', 'e', 'f'))
And now try out your test string:
scala> println(parser parse "((2((x+3)+6)))")
Map((c) -> d, (2b) -> c, (a+6) -> b, (x+3) -> a)
As desired. A more interesting exercise (left to the reader) would be to thread some state through the parser to avoid the mutable queue.

Answer 2

Classic recursive parsing problem. It can be handy to hold the different bits. We'll add a few utility methods to help us out later.
trait Part {
def text: String
override def toString = text
}
class Text(val text: String) extends Part {}
class Parens(val contents: Seq[Part]) extends Part {
val text = "(" + contents.mkString + ")"
def mapText(m: Map[Parens, Char]) = {
val inside = contents.collect{
case p: Parens => m(p).toString
case x => x.toString
}
"(" + inside.mkString + ")"
}
override def equals(a: Any) = a match {
case p: Parens => text == p.text
case _ => false
}
override def hashCode = text.hashCode
}
Now you need to parse into these things:
def str2parens(s: String): (Parens, String) = {
def fail = throw new Exception("Wait, you told me the input would be perfect.")
if (s(0) != '(') fail
def parts(s: String, found: Seq[Part] = Vector.empty): (Seq[Part], String) = {
if (s(0)==')') (found,s)
else if (s(0)=='(') {
val (p,s2) = str2parens(s)
parts(s2, found :+ p)
}
else {
val (tx,s2) = s.span(c => c != '(' && c != ')')
parts(s2, found :+ new Text(tx))
}
}
val (inside, more) = parts(s.tail)
if (more(0)!=')') fail
(new Parens(inside), more.tail)
}
Now we've got the whole thing parsed. So let's find all the bits.
def findParens(p: Parens): Set[Parens] = {
val inside = p.contents.collect{ case q: Parens => findParens(q) }
inside.foldLeft(Set(p)){_ | _}
}
Now we can build the map you want.
def mapParentheses(s: String) = {
val (p,_) = str2parens(s)
val pmap = findParens(p).toSeq.sortBy(_.text.length).zipWithIndex.toMap
val p2c = pmap.mapValues(i => ('a'+i).toChar)
p2c.map{ case(p,c) => (p.mapText(p2c), c) }.toMap
}
Evidence that it works:
scala> val s = "((2((x+3)+6)))"
s: java.lang.String = ((2((x+3)+6)))
scala> val map = mapParentheses(s)
map: scala.collection.immutable.Map[java.lang.String,Char] =
Map((x+3) -> a, (a+6) -> b, (2b) -> c, (c) -> d)
I will leave it as an exercise to the reader to figure out how it works, with the hint that recursion is a really powerful way to parse recursive structures.

Answer 3

def parse(s: String,
c: Char = 'a', out: Map[Char, String] = Map() ): Option[Map[Char, String]] =
"""\([^\(\)]*\)""".r.findFirstIn(s) match {
case Some(m) => parse(s.replace(m, c.toString), (c + 1).toChar , out + (c -> m))
case None if s.length == 1 => Some(out)
case _ => None
}
This outputs an Option containing a Map if it parses, which is better than throwing an exception if it doesn't. I suspect you really wanted a map from Char to the String, so that's what this outputs. c and out are default parameters so you don't need to input them yourself. The regex just means "any number of characters that aren't parens, eclosed in parens" (the paren characters need to be escaped with "\"). findFirstIn finds the first match and returns an Option[String], which we can pattern match on, replacing that string with the relevant character.
val s = "((2((x+3)+6)))"
parse(s) //Some(Map(a -> (x+3), b -> (a+6), c -> (2b), d -> (c)))
parse("(a(aa))(a)") //None