The Universe of Discourse

Generating strings of balanced parentheses
About a year and a half ago, the Perl Quiz of the Week question was to write a program which, given an argument n, printed out all the strings containing n balanced pairs of parentheses. For example, if the argument was 3, the output was to contain the lines:

        ()()()
        (())()
        (()())
        ()(())
        ((()))

Enumerating parentheses is important because the parentheses obviously correspond to all the forms that an infix expression can take, so any attempt to enumerate all possible expressions of some form can be built atop a parenthesis-enumerator. But also, there's a straightforward correspondence between parentheses and tree structures, so by enumerating parentheses you are also enumerating all possible tree structures.

There were quite a few correct solutions posted to the list, and also some wrong ones. The commonest mistake to make was to assume that any string of n+1 pairs of balanced parentheses must have the form S(), ()S, or (S), where S is a string of n pairs of balanced parentheses. But this isn't so; (())(()) doesn't have this form. The commonest strategy that worked correctly was to generate all the strings of the right length of the form (S)S, where S is a shorter balanced string. This can be done recursively as follows:

        sub parens {
          my ($N) = @_;
          return ("") if $N == 0;
          my @result;
          for my $i (0 .. $N-1) {
            for $a (parens($i)) {
              for $b (parens($N-$i-1)) {
                push @result, "($a)$b";
              }
            }
          }
          return @result;
        }

        print join "\n", parens(shift()), "";

        sub parens {
          my $N = shift;
          $pattern = @_ ? shift : "S";
          if ($N == 0) {
            $pattern =~ tr/S//d;
            return $pattern;
          }
          return unless $pattern =~ /S/;
          my $new_pattern_a = my $new_pattern_b = $pattern;
          $new_pattern_a =~ s/S/S(S)/;
          $new_pattern_b =~ s/S//;
          return parens($N-1, $new_pattern_a), parens($N, $new_pattern_b);
        }

        print join "\n", parens(shift()), "";

        sub parens {
            my ($N, $unclosed, $s) = @_;
            $unclosed ||= 0;
            $s ||= "";
            return $s if $N == 0;
            my @result;
            push @result, parens($N,   $unclosed+1, "$s(")  if $unclosed < $N;
            push @result, parens($N-1, $unclosed-1, "$s)")  if $unclosed > 0 ;
            return @result;
        }

        print join "\n", parens(shift()), "";

        #!/usr/bin/perl -l

        print $_ = "()" x shift();
        print while s{^  ( \(+ )  ( \)+ ) \(  }
                     {"()" x (length($2) - 1)
                      . "("  x (length($1) - length($2) + 2) 
                      . ")" 
                     }xe;

I can explain how I thought it up, although I'm not sure whether the explanation will be helpful or whether it will sound like "I just counted the legs and divided by four."

One of the other list members suggested an alternative representation for the strings that seemed promising, so I tried out some examples. Close examination of those examples reminded me of a technique I had used to solve some other problems, so I tried to apply it in this case with the necesary variations. It worked well enough to continue the investigation, but the alternative representation was getting in the way. So I took what I had learned from applying the technique to the alternative representation and tried to write an algorithm to do the same thing directly to the strings of parentheses. Then I optimized the result for speed and code compactness.

I had originally planned to write a recursive solution, but before I started I thought it would be smart to investigate alternative representations. (Someday I'm going to write a long essay about this. I mentioned to Kurt Starsinic last week that the one sure sign of a program written by a novice programmer is that it will represent the data internally in the same format in which it needs to be output, typically as strings.)

One of the representations I looked at was one that had earlier been mentioned by Roger Burton West on the list. In this representation, a string of one "(" followed by n ")"'s becomes the number n, thus:

        ()()()()        1111
        ()()(())        1102
        ()(())()        1021
        ()(()())        1012
        ()((()))        1003
        (())()()        0211
        (())(())        0202
        (()())()        0121
        (()()())        0112
        (()(()))        0103
        ((()))()        0031
        ((())())        0022
        ((()()))        0013
        (((())))        0004

The essence of a counting process is that you find the rightmost column that has a certain property, and then you do a little transformation on it so that it gets a little less of that property and the columns to the right get reset to have more of it.

I realize that if you've never thought of it that way that is going to sound totally bizarre, so here's an example. Let's count base-10 numerals. The magic property in this case is the property of being less than 9. 0 has the largest possible amount of this property, since it is as much less than 9 as any digit can be. 8 has very little of this property, since it is just a little less than 9. 9 itself does not have any of this property, since it is not less than 9.

When you count, you find the rightmost column that is less than 9 and then you do a little transformation on it so that it has a little less of that property, so that it a little closer to 9:

        387
        388
        389

        390

        391
        392
        ...
        399

        400

Probably the next simplest example is when the property of interest is that the n'th column contains a number less than or equal to n:

        0000
        0010
        0100
        0110
        0200
        0210
        1000
        1010
        1100
        1110
        1200
        1210
        2000
        2010
        ...
        3210

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Another example I was already familiar with is like the base-2 counting example, but with the added restriction that you are not allowed to have two adjacent 1's:

        0000000
        0000001
        0000010
        0000100
        0000101
        0001000
        0001001
        0001010
        0010000
        0010001
        0010010
        0010100
        0010101
        0100000
        ...

        s/00((10)*1?)$/"01" . "0" x length $1/e;

So anyway, I was already familiar with this idea, and when I saw the parenthesis numbers, it reminded me strongly of one of these counting processes. Here they are again:

        1111
        1102        
        1021
        1012
        1003
        0211
        0202
        0121
        0112
        0103
        0031
        0022
        0013
        0004

Since a number n here represents the string ()))...), with n close parentheses, we have the constraints that the leftmost column may not exceed 1; the sum of the two leftmost columns may not exceed 2, and so on. Under these constraints, 1111 is clearly an extreme case, and no string has the 1's any farther to the left. To go from 1111 to 1102, the rightmost moveable 1, which is in the third column, moves to the right, into the fourth column, which I now imagine contains a heap of two 1's.

Now the rightmost 1 is in the second column. So I reset the columns to the right of that (getting back 1111) and then move the 1 from the second to the third column, yielding 1021. Again, I imagine that the 2 here represents a heap of two 1's. These 1's can move to the right, yielding 1012 and then 1003.

Now I'll need to move the 1 from the first to the second column, so I reset the columns to the right of that (getting back 1111 again) and move the 1 from the first to the second column, yielding 0211. Then the 1 in the third column moves, yielding 0202, and then I reset the third and fourth columns back to 0211 so that I can move a 1 from the second to the third column, yielding 0121.

My first cut at implementing this actually manipulated these digit strings directly, like this:

        output while s/([1-9])(0*)([1-9])$/($1-1).($3-length($2)+1).1x length($2)/e;

Translated back into parentheses, the algorithm might even be simpler to understand. The property we are trying to reduce is appearances of )(, in which ( appears to the right of ). In ()()()...(), the open parentheses are as far to the right as they can be, so this represents the maximal configuration. At each step, we're going to find the rightmost moveable open parenthesis, and we're going to move it one step to the left. That is, we're going to find the rightmost )( and change it to (). If there are other parentheses to the right of the ones that moved, we'll reset them into their minimal configuration. In what follows, the )( that just changed to () is in red, and the parentheses to its right that were reset are in blue. The )( that is about to change to () in the next step is in boldface.

In the initial configuration, the rightmost )( is in positions 6-7:

        ()()()()                        1111

        ()()(())                        1102

        ()(())()                       1021

        ()(()())                        1012

        ()((()))                        1003

        (())()()                        0211

        (())(())                        0202
        (()())()                        0121
        (()()())                        0112
        (()(()))                        0103
        ((()))()                        0031
        ((())())                        0022
        ((()()))                        0013
        (((())))                        0004

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