Question

e:= 3. (i) Let T = (V, E) be a graph. Prove that the following are equivalent: (a) T is maximally acyclic: T does not contain a cycle but, for any u tv in V with {u, v} not in E, the graph (V, E U{e}) does contain a cycle. (b) T is minimally connected: T is connected but, for any e E E, the graph (V, E {e}) is not connected. (ii) Suppose that I (V, E) is a connected acyclic graph. Using (i), or otherwise, prove that [V] = |E| + 1. What is the significance of this equation for Euler's polyhedral formula?

Answer 1

a) Suppose
T= (V, E)
is maximally acyclic. Let
u, υεΙ
; we will show that there is a path in connecting .

If then of course this itself is a path in . Suppose that
{u, w} E E
. Then
T = (V, EU{u, v})
contains a cycle, and this cycle must contain the edge
{u, }
because otherwise it will be a cycle in
T= (V, E)
which is acyclic. But then, this cycle must be of the form

,

U1

showing that
T= (V, E)
contains the path
Ta 'n
connecting .

This proves that
T= (V, E)
is connected. To show that it is minimally connected, consider any edge
e EE
, say
e = {ຍ, º}
. If
T = (V, E {e})
is connected, then there is a path

01

in , showing that

T = (V, E {e})

V1 n

is a cycle in
T= (V, E)
. Again, this is impossible since
T= (V, E)
is acyclic. Thus,
T = (V, E {e})
is not connected. This shows that
T= (V, E)
is minimally connected.

Conversely, suppose is minimally connected. We will show that it is acyclic.

T= (V, E)

Assume, if possible, that there is a cycle

V1 Un

in
T= (V, E)
. Then
T = (V, E {v, v1})
is connected because every vertex in is connected to via a path in
T= (V, E)
and if this path contains $\{v,v_1\}$ we can replace it with

V1 Un

This contradicts that
T= (V, E)
is minimally connected. Thus,
T= (V, E)
must be acyclic.

Now consider vertices
V, ULEV
such that
{v, 01} & E
. Since
T= (V, E)
connected, there is a path

U1 n

in
T= (V, E)
, showing that
T = (V, EU{v, V1})
contains the cycle

V1 Un

This proves that is maximally acyclic.

T= (V, E)

This proves the equivalence.

b) Suppose that
T= (V, E)
is connected acyclic graph. We will use induction on
E
to prove that
V = E +1
.

If
E = 1
then
E = {{u, v}}
, and
V = {u, v}
, showing
V = E +1
if
E = 1
.

Suppose that $T_0=(V_0,E_0)$ is connected acyclic, such that
Eo =k+1
for some $k\geq 1$ . Suppose that
V = E +1
holds for all
T= (V, E)
such that . Let $\{u,v\}\in E_0$ . Since $T_0=(V_0,E_0)$ is connected acyclic, $T_0=(V_0,E_0\setminus\{u,v\})$ can not have any path between . Thus, $T_0=(V_0,E_0\setminus\{u,v\})$ is not connected, and since this is obtained by removing just one vertex from a connected graph, it has two connected components, say

$(V_1,E_1),(V_2,E_2)$

Being connected components, these are connected; being subgraph of acyclic graph, these are acyclic. Thus, by induction hypothesis, we know

$|V_1|=1+|E_1|,~~|V_2|=1+|E_2|$

This shows

$\begin{align*}|V|&=|V_1|+|V_2|\\ &=(1+|E_1|)+(1+|E_2|)\\ &=2+|E\setminus \{u,v\}|\\ &=2+|E|-1\\ &=|E|+1\end{align*}$

By induction, we have
V = E +1
for all
T= (V, E)
connected acyclic.

By Euler's formula, we have $|V|-|E|+|F|=2$ . If is connected acyclic, then

T= (V, E)

$2=|V|-|E|+|F|=1+|F|$

so that $|F|=1$