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In computer science, a string is a finite sequence of characters. For strings $A$ and $B$, we express $AB$ as $A$ followed by $B$. A balanced string of parentheses is a string of open and closed parentheses that is $()$, $ST$ or $(S)T$ where $S$ and $T$ are both balanced strings of parentheses. For example,

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

are balanced strings of parentheses where as

)( ())( (()

are not.

I am trying to prove by induction that a balanced string of parentheses has the same number of open parentheses as closed parentheses. I started with the base case N(1) being true, since any of $()$, $ST$, $S(T)$ or $(S)T$ has an equal number of left and right parentheses. But I got stumped when thinking of the assumption for $N(k)$ since there are so many possible arrangements. Can anyone help please? Much appreciated!

Marc van Leeuwen
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Bryan
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    Write down the actual grammar, and use structural induction on the righthand sides of rules. Your base case should only deal with $()$. – BrianO Jan 30 '16 at 07:19
  • Your rules for forming strings with balanced parentheses aren't quite right: they don't allow things like $(()$). The following grammar $$B \mathrel{::=} () ;|; (B) ;|; BB$$ works: it says $()$ is a $B$, if $S$ is a $B$ then so is $(S)$, and if $S$ and $T$ are $B$s then so is $ST$. – Rob Arthan Feb 02 '16 at 01:01

1 Answers1

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Use strong induction on the length of the string: while trying to prove this for balanced strings of some length $n>0$, you may assume the result for balanced string for any length shorter than $n$.

Now a balanced string is either empty, or of the form $(S)T$ where $S$ and $T$ are balanced strings (to see this, the first character of a non-empty balanced string must be '(', find its matching ')' which must exist, and then you've got your form $(S)T$, since both in between and after the ')' there must be balanced strings). Since both $S$ and $T$ are strictly shorter than $(S)T$, your induction hypothesis applies to them. It easily follows that the claimed result follows for $(S)T$ as well.

Marc van Leeuwen
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