Exercise 11.2.5 (Partial) Solution
This solution differs a bit from the one provided earlier; that solution inducted over the #
of productions applied, rather than the height of the expression tree. They are very
similar, however.
We are to show that the following grammar generates the balanced profile strings, as defined in
section 2.6:
• <BalancedE> → ε
• <BalancedE> → (<BalancedE>)
• <BalancedE> → <BalancedE><BalancedE>
Part I: Show that all strings in L(<BalancedE>) are profile-balanced
We will use induction, on the height of the expression tree that produces strings s Є
L(<BalancedE>).
Our base case is simple. We have only one possible tree of height one:
Trivially, ε (the empty, or null string) is balanced. So, by the inductive hypothesis,
we may assume that any such tree of height <= n produces a profile-balanced string.
Section 2.6 defines a profile-balanced string s so: Set ctr=0, scan s from left to
right. For each character ‘(‘ that is encountered, increment ctr. For each ‘)’ that is encountered,
decrement ctr. If, and only if, after the entire string s is scanned, ctr is 0, and it at no time was
negative, then s is profile-balanced.
We now need to show that any string, s’, represented by an expression tree of height n+1 is also
balanced. There are 2 productions in the grammar that allow us to build on existing expressions.
Lets consider:
<BalancedE> → <BalancedE><BalancedE>
Represented:
If the tree rooted at the red node has height n+1, then the 2 (blue and green) have height <= n, so,
by our hypothesis, s1 and s2 are profile balanced. We need to show that the string s’ =s1 s2 is
profile balanced. Start on the left, ctr = 0. After we scan s1 ctr is 0, and was never negative.
<BalancedE>
<BalancedE> <BalancedE>
s1 s2
<BalancedE>
ε
