CS 473: Algorithms, Fall 2010
HBS 0
1. The following is an inductive proof of the statement that in every tree
T
= (
V
(
T
)
,E
(
T
)),

E
(
T
)

=

V
(
T
)
 
1, i.e a tree with
n
vertices has
n

1 edges.
Proof:
The proof is by induction on

V
(
T
)

.
Base case:
Base case is when

V
(
T
)

= 1. A tree with a single vertex has no edge, so

E
(
T
)

= 0. Therefore in this case the formula is true since 0 = 1

1.
Inductive step:
Assume that the formula is true for all trees
T
where

V
(
T
)

=
k
. We will
prove that the formula is true for trees with
k
+ 1 nodes. A tree
T
with
k
+ 1 nodes can be
obtained from a tree
T
0
with
k
nodes by attaching a new vertex to a leaf of
T
0
. This way we
add exactly one vertex and one edge to
T
0
, so

V
(
T
)

=

V
(
T
0
)

+1 and

E
(
T
)

=

E
(
T
0
)

+1.
Since

V
(
T
0
)

=
k
by induction hypothesis we have

E
(
T
0
)

=

V
(
T
0
)
 
1.
Combining the last three relations we have

E
(
T
)

=

E
(
T
0
)

+1 =

V
(
T
0
)

1+1 =

V
(
T
)

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 Fall '08
 Chekuri,C
 Algorithms, Graph Theory, Mathematical Induction, Recursion, Inductive Reasoning, Structural induction

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