Recursive Definitions
and Structural Induction
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Recursive Definitions
• Common way of defining mathematical
objects
– Base case(s)
– Constructor rule(s)
– “Nothing else” (generally implicit)
• Have already seen recursive definition of
the Fibonacci numbers
– F0=0, F1=1
– Fn+1 = Fn+Fn-1 (n ≥ 1)
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The set of all strings of balanced
parentheses
• (), (()()), ()() but not )( or (()
• Familiar “Counting rule”—an algorithm:
– Start count at 0
– When you see a “(“, add 1
– When you see a “)”, subtract 1
– Balanced if count never goes negative and
ends at 0
• But we want a structural definition
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A Structural Definition
• Base case:
– The empty string εis balanced
• Constructor rules:
– C1: If x is balanced then so is (x), that is, the result of
writing a “(“, then x, then “)”
– C2: If x and y are balanced then so is xy
• (Implicit “that’s-all” clause)
– (No string is balanced unless it can be constructed
using the base and constructor rules)
• NB: x and y are variables whose values are strings
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Some Balanced Strings
ε: Base case
( ) : C1 rule, (x), where x =ε
(( )): C1 rule, (x), where x = ( )
( )( ): C2 rule, xy, where x = y = ( )
(( ))( )( ): C2 rule, xy, where x = (( )), y = ( )( )
…
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Showing a String is Balanced
• Is (( )( )) balanced? Yes, because
– ε is balanced (base rule)
– (ε) is balanced by first constructor rule, but
that is just another way of writing ()
– ( )( ) is balanced by second constructor rule
– (( )( )) is balanced by first constructor rule
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Proof that every balanced string
has equal numbers of “(“ and “)”
• Suppose x is balanced. Then it is balanced
because of either the base rule or a
constructor rule
• If by the base rule, then x is ε and has 0 left
and 0 right parentheses. ✓
• If by the first constructor rule then x=(y) for
some previously constructed balanced string
y. Then y has the same number of “(“ and “)”,
say n. Then x = (y) has one more “(“ and one
more “)”, so n+1 of each. ✓
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Every balanced string has equal
numbers of “(“ and “)”
• If by the second constructor rule then x=yz
for some previously constructed balanced
strings y and z. Then y has the same
number of “(” and “)”, say n, and z has the
same number of “(” and “)”, say m. Then
x=yz has m+n “(” and m+n “)”. ✓
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Structural Induction:
The General Schema
• To prove P(x) holds for all x in a
recursively defined set S, prove
– P(b) for each base case b∈S
– P(c(x1,…,xk)) for each constructor c, assuming
as the induction hypothesis that P(x1), …, and
P(xk) all hold.
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