Maggie Johnson
Handout #12
CS103B
Key Topics
* Structural Induction
* Other Methods of Proof for Trees
* Node Induction
_____________________________________________________________________
In 103A, we saw a number of inductive proofs concerning integers and sequences.
We would
assume some statement for n, or for all integers less than or equal to n (strong induction), and use
this "inductive hypothesis" to prove the same statement for n+1.
A similar form of proof is useful
for concluding facts about trees.
Suppose we want to prove that a statement S(T) is true for all trees.
For a basis, we show that S(T) is
true whenever T consists of one node.
For the induction, we suppose that T is a tree with root r and
children c1, c2, .
.., ck.
Let T1, T2, .
.., Tk be the subtrees of T rooted at c1, c2, .
.., ck respectively as
shown below:
The inductive step is to assume S(T1), S(T2), .
.., S(Tk) are all true, and prove S(T).
If we can do
this, we have proven S(T).
Why does structural induction work?
Example 1:
S(T) denotes that there are at most m
h
leaves in an mary tree of height h.
(Note: the
root is at level 1 with a height of 0)
basis:
Consider an mary tree of height 1.
These trees consist of a root with no more than m
children, each of which is a leaf.
Therefore, there are no more than m
1
= m leaves in an m
ary tree of height 1.
inductive hypothesis
: We assume the above statement is true for all mary trees of height less
than h, and show it is true for all mary trees of height h.
PROOF:
Let T be an mary tree of height h.
The leaves of T are the leaves of the subtrees of T
obtained by deleting the edges from the root to each of the vertices connected to it, as shown
below:
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View Full DocumentEach of these subtrees has height less than or equal to h1.
So, by the inductive hypothesis,
these trees have at most m
(h1)
leaves.
Since there are at most m such subtrees, each with a
minimum of m
(h1)
leaves, there are at most m * m
(h1)
= m
h
leaves in the rooted tree.
Thus, by the principle of mathematical induction applied to the structure of trees, S(T) is true
for all mary trees.
Example 2:
S(T) denotes a
strictly
binary tree with n leaves has 2n2 edges.
(Note: strictly binary
means every node has either 0 or 2 children.)
basis:
A strictly binary tree with two leaves consists of the root connected to its only two
children: 2 * 2  2 = 2.
inductive hypothesis
: We assume S(T) is true for all strictly binary trees with i leaves where i
>= 2 and i < n leaves; Show that S(T) is true for a tree with n leaves.
PROOF:
Consider a tree T with n leaves where n > 1.
By the definition of a binary tree, T must
consist of two subtrees T
L
and T
R
(since n > 1).
If T
L
has n
L
leaves and T
R
has n
R
leaves,
then n
L
+ n
R
= n.
By the inductive hypothesis, T
L
has 2n
L
 2 edges and T
R
has 2n
R
 2
edges.
Since T has two additional edges that are not part of T
L
or T
R
, the number of edges
in T is (2n
L
 2) + (2n
R
 2) + 2 = 2(n
L
+ n
R
)  2 = 2n  2.
Thus, by the principle of mathematical induction applied to the structure of trees, S(T) is true
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 Winter '08
 SAHAMI,M
 Mathematical Induction, Inductive Reasoning, Mathematical proof, inductive hypothesis

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