Removing unnecessary parentheses in a regular expression

Question

Suppose I have (in a javascript regular expression)

((((A)B)C)D)

Of course that really reads

ABCD

Is there an algorithm to eliminate unnecessary parentheses in a string like that?

Answer 1

This function will remove all groups that is not followed by a quantifier, and is not a look-around. It assumes ECMAScript flavor regex, and that capture-groups (( ... )) are unimportant.

function removeUnnecessaryParenthesis(s) {
   // Tokenize the pattern
   var pieces = s.split(/(\\.|\[(?:\\.|[^\]\\])+]|\((?:\?[:!=])?|\)(?:[*?+]\??|\{\d+,?\d*}\??)?)/g);
   var stack = [];
   for (var i = 0; i < pieces.length; i++) {
      if (pieces[i].substr(0,1) == "(") {
         // Opening parenthesis
         stack.push(i);
      } else if (pieces[i].substr(0,1) == ")") {
         // Closing parenthesis
         if (stack.length == 0) {
            // Unbalanced; Just skip the next one.
            continue;
         }
         var j = stack.pop();
         if ((pieces[j] == "(" || pieces[j] == "(?:") && pieces[i] == ")") {
             // If it is a capturing group, or a non-capturing group, and is
             // not followed by a quantifier;
             // Clear both the opening and closing pieces.
             pieces[i] = "";
             pieces[j] = "";
         }
      }
   }
   return pieces.join("");
}

Examples:

removeUnnecessaryParenthesis("((((A)B)C)D)")  --> "ABCD"
removeUnnecessaryParenthesis("((((A)?B)C)D)") --> "(A)?BCD"
removeUnnecessaryParenthesis("((((A)B)?C)D)") --> "(AB)?CD"

It does not try to determine if the parenthesis contains only a single token ((A)?). That would require a longer tokenizing pattern.

Answer 2

1) Use a parser that understands parenthesis

2) Use a Perl recursive regex that can match parenthesis (discouraged in this case IMHO) I don't think Boost regex's support the type of recursion needed.

3) Perhaps they are needed? Leave them alone.