Unformatted text preview: hod directly exploits the inductive definition
of the set.
The method is more powerful than strong induction in
the sense that one can prove statements that are
difficult (or impossible) to prove with strong induction.
Typically, though, it is simply used because it is more
convenient than (strong) induction.
22 Structural Induction
In structural induction, the proof of the assertion that
every element of an inductively defined set S has a
certain property P proceeds by showing that
Basis: Every element in the basis of the definition of S
satisfies the property P.
Induction: Assuming that every argument of a
constructor has property P, show that the constructed
element has the property P. 23 Example 1: Binary Trees
Recall that the set B of binary trees over an alphabet
A is defined as follows:
Basis: ⟨⟩∈B.
Induction: If L, R∈B and x∈A, then ⟨L,x,R⟩∈B.
We can now prove that every binary tree has a
property P by arguing that
Basis: P(⟨⟩) is true.
Induction: For all binary trees L and R and x in A,
if P(L) and P(R), then P(⟨L,x,R⟩).
24 Binary Trees (Cont.)
Let f: B > N be the function defined by f ()
f (L, x, R) =
= 0
1
f ( L) + f ( R ) if L = R =
otherwise Theorem: Let T in B be a binary tree. Then f(T) yields
the number of leaves of T. 25 Binary Trees (Cont.)
Theorem: Let T in B be a binary tree. Then f(T) yields
the number of leaves of T.
Proof: By structural induction on T.
Basis: The empty tree has no leaves, so f(⟨⟩) = 0 is correct.
Induction: Let L,R be trees in B, x in A.
Suppose that f(L) and f(R) denotes the number of leaves of L
and R, respectively. If L=R=⟨⟩, then ⟨L,x,R⟩ = ⟨⟨⟩,x,⟨⟩⟩ has one leaf, namely x, so f
(⟨⟨⟩,x,⟨⟩⟩)=1 is correct.
26 Binary Trees (Cont.)
If L and R are not both empty, then the number of
leaves of the tree ⟨L,x,R⟩ is equal to the number of
leaves of L plus the number of leaves of R. Hence, by
induction hypothesis, we get
f(⟨L,x,R⟩) = f(L)+f(R)
as claimed. This completes the proo...
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This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.
 Fall '11
 math
 Recursion

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