Proof. Clearly
G has
and
c as the
cut vertices. Observe that, for any
, a graph
has the path
from
to
c where
. To prove all cases of
, we may relabel the path
P to be
. Therefore,
and
. We see that
where
are vertices in
defined in
Section 3. Therefore,
.
Let D be a -set of G. If , then, to dominate , we have . If , then as is connected. In both cases, .
Suppose that . Because , to dominate B, . By the connectedness of , we have . Thus, implying that . Hence, suppose that . As and is connected, it follows that where . Thus, implying that .
Now, we will establish the criticality. Let u and v be a pair of non-adjacent vertices of G and let . We first assume that . Therefore, . Thus, and for some . Without loss of generality let . Clearly . We see that
.
Therefore, .
Thus, we assume that . If for some , then and, clearly, . Thus, . Therefore . Let . As , there exists such that . Suppose that . Thus, . Therefore, . If , then implying that .
Finally, we assume that . Thus, for some and . Clearly and . Thus, G is a k--critical graph and this completes the proof. □
Lemma 14. Let G be a graph in the class . Then, G is a k--critical graph with cut vertices.
Proof. Let D be a -set of G. As are cut vertices, by Lemma 2(3), . Recall that . Because , we can let . Thus . By Lemma 2(2), . So, is a connected dominating set of B. By , there exists a -set of B of size 3. By the minimality of , . Therefore, . Further, because and , it follows that . Therefore, and this implies that .
We will prove the criticality. Let u and v be non-adjacent vertices of G. Suppose first that , say. If , then . If , then, by Property of the class , there exists a -set of size 3 of B such that and . So . These imply that .
We then suppose that . If , then there exists i and j such that and . Without loss of generality let . Clearly, . Therefore, . For the case when , we have and thus . Finally, if , then, by Property of the class , there exists a -set such that , and . Thus, . This implies that . Clearly, are the cut vertices of G. This completes the proof. □
In the following, we let having a -set D. In view of Lemma 9, G has only two end blocks and all other blocks contain two cut vertices. Thus, we let and be the two end blocks and all other blocks contain two cut vertices. Without loss of generality let and for . Moreover, let for all . If is a -set of G such that , by the minimality of k, we have for all i. Thus, we can let
the class of a graph such that
for .
Lemma 15. For a -set of G, either or .
Proof. Suppose to the contrary that and . Lemmas 5 and 6 imply that . This contradicts Lemma 8. Thus, either or and this completes the proof. □
By Lemma 15, we may suppose without loss of generality that .
Lemma 16. where for some and for all .
Proof. By the definition, .
Conversely, let . Thus, by Lemma 9, G has only two end blocks and and all other blocks contain two cut vertices. Moreover, and for . In view of Lemma 2(3), . So . Thus, for all . Recall that . Lemma 15 implies that either or . That is, . This completes the proof. □
By Lemma 16, to characterize a graph G in the class , it suffices to consider when G is either in or . We first consider the case when . Let and be vertices and a copy of a complete graph.
Lemma 17. Let with a block containing two cut vertices and and . Then, where .
Proof. As
and
for some
, we must have
for all
. Because
contains two cut vertices,
. Therefore,
. Lemma 5 then implies that
Let
. We first show that
is complete. Let
x and
y be non-adjacent vertices of
. Consider
. Lemma 1(2) implies that
. As
, we must have
contradicting Corollary 1. Therefore,
is complete.
We will show that . Thus, we may assume to the contrary that . We let
and
.
As , it follows that . Because is complete, is complete. In fact, and X satisfy and of bad subgraphs. We then let
and
.
Thus, G has and as a bad subgraph. This contradicts Lemma 3. Therefore, .
We finally show that . We may assume to the contrary that there exists a vertex u of which is not adjacent to . Consider . Corollary 1 gives that . Lemma 4 gives further that . As , it follows that is not connected, a contradiction. Therefore, and, similarly, . As is a block, and this completes the proof. □
We will prove the following two theorems, both of which give main contribution to the proof of our first main theorem, Theorem 2.
Theorem 10. For , where for all .
Proof. Let for all . In views of Lemma 13, we have that . Thus, it suffices to show that .
We will show that
. Clearly,
is either 0 or 1. If
, then Lemma 5 implies that
. However, if
, then Lemma 6 implies that
. Thus, we let
and
As
, by Lemma 7,
. If
, then
where
is given at the definition of
. We then let
and
Clearly, G has a bad subgraph, contradicting Lemma 3. Thus, .
We now consider the case when . Thus, . By Lemma 6, implying that . Further, Lemma 10 implies also that for . Thus, .
We finally consider the case when
. Thus,
for some
and
for
. Similarly,
for all
by Lemma 10. Moreover, Lemma 17 yields that
. Let
. We will show that
. We let
a be a vertex in
if
. Then, we let
Consider
. As
is a cut vertex of
,
by Lemma 2(3). That is
when
. When
, by Lemma 17,
. As
, by Lemma 1(3),
. In both cases,
. If
, then
contradicting
. Therefore,
by Lemma 1(1). Thus,
implying that
. Therefore,
. This completes the proof. □
Theorem 11. For , .
Proof. By Lemma 14, . Thus, it is sufficient to show that .
As for all , by Lemma 10, . By Lemma 5 and similar arguments in Theorem 10, we have that .
We will show that
satisfies
of the class
. Recall that
. Let
be a
-set of
. Suppose that
. To dominate
, we have
. Thus,
. However,
contradicting the minimality of
k. Therefore,
. Thus, to prove that
satisfies
, it suffices to give a connected dominating set of size 3 of
containing a chosen vertex from
. For a vertex
v of
, consider
. Lemma 1(2) implies that
. Lemma 1(1) implies also that
. We first show that
. Suppose to the contrary that
. As
is connected, there exists
which is adjacent to
v. Because
,
w is adjacent to a vertex of
. Thus,
This contradicts the minimality of
k. Thus,
. Therefore,
or
.
Let
We first consider the case when
. Thus
. As
and
, we must have
Therefore,
and
, otherwise
where
, contradicting the minimality of
k. Let
such that
u is adjacent to a vertex
a in
A. Thus,
and so
satisfies
.
We now consider the case when . Let . Clearly, . As and , it follows that . If , then implying that , in particular, . Thus, where contradicting the minimality of D. Therefore, we let . As , . Thus, for a vertex a in A, . Therefore, satisfies .
We now consider the case when . Thus . Therefore, and, to dominate . So, satisfies .
We finally consider the case when . To dominate , we have that . By the connectedness of , . Thus, implying that
.
Therefore, . To dominate , . Thus satisfies and this completes the proof of Case 1.
To dominate , we have that . By the connectedness of , and . As , we must have . So satisfies . This completes the proof of Case 2.
We finally show that satisfies of the class . Let x and y be non-adjacent vertices of . Lemma 1(2) implies that . Lemma 1(1) implies also that . To dominate , we have that . Let . By the connectedness of , and . As , we must have . Thus, satisfies . Therefore, . This completes the proof. □
Now, we are ready to prove our first main result, Theorem 2. For completeness, we restate the theorem.
Theorem 2. For an integer , .
Proof. In view of Lemma 16, where for some and for all . Moreover, Theorems 10 and 11 imply that . This completes the proof. □