1 Introduction
In this paper, all graphs under our consideration are simple, finite and undirected. We follow the notation and terminology of [1]. For a graph
G
, we denote by
V
(
G
)
,
E
(
G
)
and
diam
(
G
)
the vertex set, edge set and diameter of
G
, respectively. The distance between two vertices
u
and
v
in a connected graph
G
, denoted by
dist
(
u
,
v
)
, is the length of a shortest path between them in
G
. The eccentricity of a vertex
v
in
G
is defined as
ecc
G
(
v
)
=
max
x
∈
V
(
G
)
dist
(
v
,
x
)
. For convenience, a set of internally pairwise vertex disjoint paths will be called disjoint.
In recent years, colored notions of connectivity in graphs becomes a new and active subject in graph theory. Stating from rainbow connection [2], rainbow vertex connection [3] and total rainbow connection [4,5] appeared later. Many researchers are working in this field, and a lot of papers have been published in journals, see [6,7,8, 9,10,11, 12,13,14, 15,16] for details. The reader can also see [17] for a survey, [18] for a dynamic survey and [19] for a new monograph on this topic.
In 2012, Borozan et al. [20] introduced the concept of proper
k
-connection number. A path in an edge-colored graph is a proper path if any two adjacent edges on the path differ in color. An edge-colored graph is proper k-connected if any two distinct vertices of the graph are connected by
k
disjoint proper paths. The proper k-connection number of a
k
-connected graph
G
, denoted by
p
c
k
(
G
)
, is defined as the smallest number of colors that are needed in order to make
G
proper
k
-connected. For more results, the reader can see [21,22, 23,24] for details.
As a natural generalization, Jiang et al. [25] presented the concept of proper vertex
k
-connection number. A path in a vertex-colored graph is a vertex proper path if any two internal adjacent vertices of the path differ in color. A vertex-colored graph is proper vertex k-connected if any two distinct vertices of the graph are connected by
k
disjoint vertex proper paths. For a
k
-connected graph
G
, the proper vertex k-connection number of
G
, denoted by
p
v
c
k
(
G
)
, is defined as the smallest number of colors required to make
G
proper vertex
k
-connected.
Motivated by the concept of total chromatic number of graph, now for proper connection and proper vertex connection, the concept of total proper connection was introduced by Jiang et al. [26]. A total coloring of a graph
G
is a mapping from the set
V
(
G
)
∪
E
(
G
)
to some finite set of colors. A path in a total-colored graph is a total proper path if the coloring of the edges and internal vertices is proper, that is, any two adjacent or incident elements of edges and internal vertices on the path differ in color. A total-colored graph is total proper k-connected if any two distinct vertices of the graph are connected by
k
disjoint total proper paths. For a connected graph
G
, the total proper k-connection number of a
k
-connected graph
G
, denoted by
tpc
k
(
G
)
, is defined as the smallest number of colors that are needed in order to make
G
total proper
k
-connected. For convenience, we write
tpc
(
G
)
for
tpc
1
(
G
)
. Obviously,
tpc
(
G
)
≤
tpc
2
(
G
)
≤
tpc
3
(
G
)
. By [26], if
G
is complete, then
tpc
(
G
)
=
1
; if
G
has a Hamiltonian path that is not complete, then
tpc
(
G
)
=
3
. Note that if
G
is a nontrivial connected graph and
H
is a connected spanning subgraph of
G
, then
tpc
(
G
)
≤
tpc
(
H
)
.
In this paper, we investigate the total proper connection number of a graph
G
under some constraints on its complement graph
G
¯
.
Theorem 1.1
Let
G
be a connected graph of order
n
≥
3
, if
diam
(
G
¯
)
does not belong to the following two cases: (i)
diam
(
G
¯
)
=
2
,
3
, (ii)
G
¯
contain exactly two components and one of them is trivial, then
tpc
(
G
)
≤
4
.
For the remaining cases,
tpc
(
G
)
can be very large as discussed in Section 2. Then we add a constraint, i.e., we let
G
¯
be triangle-free. Hence,
G
is claw-free, and we can derive our next main result:
Theorem 1.2
For a connected graph
G
, if
G
¯
is triangle-free, then
tpc
(
G
)
=
3
.
Recall that a clique of a graph is a set of mutually adjacent vertices, and that a maximum clique is a clique of the largest possible size in a given graph. The clique number
ω
(
G
)
of a graph
G
is the number of vertices in a maximum clique in
G
. Let
G
be a connected graph, and let
X
be a maximum clique of
G
. We say that
N
X
(
u
)
is the set of neighbors of
u
in
G
[
X
]
and
d
X
(
u
)
=
∣
N
X
(
u
)
∣
. Let
F
=
G
[
V
(
G
)
\
X
]
. Kemnitz and Schiermeyer [9] considered graphs with
r
c
(
G
)
=
2
and large clique number. In this paper, we characterize graphs with small total proper connection number with respect to their large clique number. If
ω
(
G
)
=
n
, then
G
is a complete graph, which implies
tpc
(
G
)
=
1
. If
G
is connected and
ω
(
G
)
=
n
−
1
, then
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
. For the cases
ω
(
G
)
=
n
−
2
,
n
−
3
, we obtain the following three main results.
Theorem 1.3
Let
G
be a connected graph of order
n
. If
ω
(
G
)
=
n
−
2
and
X
is a maximum clique of
G
with
V
(
G
)
\
X
=
{
u
1
,
u
2
}
, then
tpc
(
G
)
=
3
.
Theorem 1.4
Let
G
be a connected graph of order
n
,
diam
(
G
)
=
2
. If
ω
(
G
)
=
n
−
3
and
X
is a maximum clique of
G
with
V
(
G
)
\
X
=
{
u
1
,
u
2
,
u
3
}
, then
tpc
(
G
)
=
3
or
tpc
(
G
)
=
4
for the following case
F
≅
3
K
3
,
∣
N
X
(
u
1
)
∩
N
X
(
u
2
)
∩
N
X
(
u
3
)
∣
=
1
and
d
X
(
u
1
)
=
d
X
(
u
2
)
=
d
X
(
u
3
)
=
1
.
Theorem 1.5
Let
G
be a connected graph of order
n
,
diam
(
G
)
≥
3
. If
ω
(
G
)
=
n
−
3
and
X
is a maximum clique of
G
with
V
(
G
)
\
X
=
{
u
1
,
u
2
,
u
3
}
, then
tpc
(
G
)
=
3
, or
tpc
(
G
)
=
4
and one of the following holds.
F
≅
3
K
3
,
N
X
(
u
)
∩
N
X
(
v
)
≠
∅
and
d
X
(
u
)
=
d
X
(
v
)
=
1
, where
u
and
v
are any two distinct vertices in
V
(
G
)
\
X
.
F
≅
3
K
3
,
d
X
(
u
1
)
=
d
X
(
u
2
)
=
d
X
(
u
3
)
=
1
and for any two vertices in
V
(
G
)
\
X
, there is no common neighbor in
G
[
X
]
.
For an integer
n
≥
3
, the circular ladder
CL
2
n
of order
2
n
is a cubic graph constructed by taking two copies of the cycle
C
n
on disjoint vertex sets
(
u
1
,
u
2
,
…
,
u
n
)
and
(
v
1
,
v
2
,
…
,
v
n
)
, then joining the corresponding vertices
u
i
v
i
for
1
≤
i
≤
n
. The Möbius ladder
M
2
n
of order
2
n
is obtained from the ladder by deleting the edges
u
1
u
n
and
v
1
v
n
, and then inserting two edges
u
1
v
n
and
u
n
v
1
. Subscripts are considered modulo
n
, and we can derive our next main result:
Theorem 1.6
Let
n
be an integer with
n
≥
3
. Then
tpc
(
CL
2
n
)
=
tpc
2
(
CL
2
n
)
=
3
,
tpc
3
(
CL
2
n
)
=
4
.
tpc
(
M
2
n
)
=
tpc
2
(
M
2
n
)
=
3
,
tpc
3
(
M
2
n
)
=
4
.
In [7], Fujie-Okamoto et al. investigated the rainbow
k
-connection numbers of all small cubic graphs of order 8 or less. In this paper, we determine the total proper
k
-connection numbers of all small cubic graphs of order 8 or less. We can easily verify that all such cubic graphs have orders 4, 6, or 8, and those with orders 4 or 6 are
K
4
,
K
3
,
3
, and
K
3
□
K
2
(where
□
denotes Cartesian product). In [27], it was shown that all connected cubic graphs of order 8 are
Q
3
,
M
8
,
F
1
,
F
2
, and
F
3
, and these graphs are depicted in Figure 1. Our last main result is stated as follows:
Theorem 1.7
tpc
(
K
4
)
=
1
,
tpc
2
(
K
4
)
=
3
,
tpc
3
(
K
4
)
=
4
.
tpc
(
K
3
,
3
)
=
tpc
2
(
K
3
,
3
)
=
3
,
tpc
3
(
K
3
,
3
)
=
4
.
tpc
(
K
3
□
K
2
)
=
tpc
2
(
K
3
□
K
2
)
=
3
,
tpc
3
(
K
3
□
K
2
)
=
4
.
tpc
(
Q
3
)
=
tpc
2
(
Q
3
)
=
3
,
tpc
3
(
Q
3
)
=
4
.
tpc
(
M
8
)
=
tpc
2
(
M
8
)
=
3
,
tpc
3
(
M
8
)
=
4
.
tpc
(
F
1
)
=
tpc
2
(
F
1
)
=
3
,
tpc
3
(
F
1
)
=
4
.
tpc
(
F
2
)
=
tpc
2
(
F
2
)
=
3
.
tpc
(
F
3
)
=
tpc
2
(
F
3
)
=
3
,
tpc
3
(
F
3
)
=
4
.
2 Proof of Theorem 1.1
In order to prove Theorem 1.1, we need the following lemmas.
Lemma 2.1
[26] For
2
≤
m
≤
n
, we have
tpc
(
K
m
,
n
)
=
3
.
Lemma 2.2
[26] If
G
is a complete multipartite graph that is neither a complete graph nor a tree, then
tpc
(
G
)
=
3
.
Let
N
G
¯
i
(
x
)
=
{
v
:
dist
G
¯
(
x
,
v
)
=
i
}
, where
0
≤
i
≤
3
, and
N
G
¯
4
(
x
)
=
{
v
:
dist
(
x
,
v
)
≥
4
}
. In this paper, we use
N
G
¯
i
instead of
N
G
¯
i
(
x
)
for convenience. Then
N
G
¯
0
=
{
x
}
and
N
G
¯
1
=
N
G
¯
(
x
)
.
Lemma 2.3
For a connected graph
G
, if
G
¯
is connected and
diam
(
G
¯
)
≥
4
, then
tpc
(
G
)
≤
4
.
Proof
Choose a vertex
x
with
ecc
G
¯
(
x
)
=
diam
(
G
¯
)
. By the definition of
N
G
¯
i
, we know
u
v
∈
E
(
G
)
for any
u
∈
N
G
¯
i
,
v
∈
N
G
¯
j
with
∣
i
−
j
∣
≥
2
. Now we define a total coloring of
G
as follows: assign color 1 to the edges
x
u
for
u
∈
N
G
¯
2
, all edges between
N
G
¯
1
and
N
G
¯
3
, and all vertices and edges in
N
G
¯
4
; assign color 2 to the edges
x
u
for
u
∈
N
G
¯
3
, all edges between
N
G
¯
2
and
N
G
¯
4
, and all vertices and edges in
N
G
¯
1
; assign color 3 to the edges
x
u
for
u
∈
N
G
¯
4
, all edges between
N
G
¯
1
and
N
G
¯
4
, and all vertices and edges in
N
G
¯
2
,
N
G
¯
3
; assign color 4 to the vertex
x
.
We prove that there is a total proper path between any two vertices
u
and
v
of
G
. It is trivial when
u
v
∈
E
(
G
)
. Thus, we only need to consider the pairs
u
,
v
∈
N
G
¯
i
or
u
∈
N
G
¯
i
,
v
∈
N
G
¯
i
+
1
. Note that
P
=
x
x
3
x
1
x
4
x
2
is a total proper path, where
x
i
∈
N
G
¯
i
. By means of the path
P
, we can find that
u
and
v
are connected by some total proper path for any
u
∈
N
G
¯
i
,
v
∈
N
G
¯
i
+
1
. If
i
=
1
, then
P
=
u
x
3
x
x
2
x
4
v
is a total proper path, where
x
i
∈
N
G
¯
i
. If
i
=
2
, then
P
=
u
x
x
4
v
is a total proper path, where
x
4
∈
N
G
¯
4
. If
i
=
3
, then
P
=
u
x
1
x
4
x
2
x
v
is a total proper path, where
x
i
∈
N
G
¯
i
. If
i
=
4
, then
P
=
u
x
x
2
v
is a total proper path, where
x
2
∈
N
G
¯
2
. Hence,
tpc
(
G
)
≤
4
.□
Proof of Theorem 1.1
Assume that
G
¯
is connected. Since
diam
(
G
¯
)
≥
4
, we have
tpc
(
G
)
≤
4
by Lemma 2.3. Assume that
G
¯
is disconnected. By the assumption, we know that there exist either at least three connected components or exactly two nontrivial components. Let
G
i
¯
be the components of
G
¯
with
t
i
=
∣
V
(
G
i
¯
)
∣
, where
1
≤
i
≤
h
. Then
G
contains a connected spanning subgraph
K
t
1
,
t
2
,
…
,
t
h
, and we have
tpc
(
G
)
≤
tpc
(
K
t
1
,
t
2
,
…
,
t
h
)
=
3
by Lemma 2.2. Note that
G
is not complete. Thus,
tpc
(
G
)
≥
3
, and so
tpc
(
G
)
=
3
.□
Next, we will give three examples to show that
tpc
(
G
)
can be arbitrarily large if one of the following three conditions holds:
diam
(
G
¯
)
=
2
,
diam
(
G
¯
)
=
3
,
G
¯
contains exactly two connected components and one of them is trivial.
Example 2.4
For the graph
G
¯
shown in Figure 2, we choose a vertex
x
with
ecc
G
¯
(
x
)
=
diam
(
G
¯
)
. Let
N
G
¯
1
(
x
)
=
{
u
i
∣
1
≤
i
≤
k
}
,
N
G
¯
2
(
x
)
=
{
v
j
∣
1
≤
j
≤
k
}
, and let
E
(
G
¯
)
=
{
x
u
i
∣
1
≤
i
≤
k
}
∪
{
u
i
1
u
i
2
∣
1
≤
i
1
,
i
2
≤
k
}
∪
{
v
j
1
v
j
2
∣
1
≤
j
1
,
j
2
≤
k
}
∪
{
u
i
v
j
∣
1
≤
i
,
j
≤
k
}
\
{
u
i
v
i
∣
1
≤
i
≤
k
}
, where
k
≥
3
. Obviously,
diam
(
G
¯
)
=
2
and
G
is a tree. Then
tpc
(
G
)
=
Δ
(
G
)
+
1
=
k
+
1
by [26, Theorem 1]. Observe that
tpc
(
G
)
will be arbitrarily large based on the increase of
k
.
Example 2.5
For the graph
G
¯
shown in Figure 3, we choose a vertex
x
with
ecc
G
¯
(
x
)
=
diam
(
G
¯
)
. Let
N
G
¯
1
(
x
)
=
{
u
i
∣
1
≤
i
≤
k
}
,
N
G
¯
2
(
x
)
=
{
v
j
∣
1
≤
j
≤
k
}
, and
N
G
¯
3
(
x
)
=
{
w
s
∣
1
≤
s
≤
k
}
, where
k
≥
3
. Furthermore, let
E
(
G
¯
)
=
{
x
u
i
∣
1
≤
i
≤
k
}
∪
{
u
i
v
j
∣
1
≤
i
,
j
≤
k
}
∪
{
v
j
w
s
∣
1
≤
j
,
s
≤
k
}
∪
{
v
j
1
v
j
2
∣
1
≤
j
1
,
j
2
≤
k
}
. Obviously,
diam
(
G
¯
)
=
3
and
G
is a connected graph. Note that
N
G
¯
2
(
x
)
is a stable set in
G
¯
, and each edge between
x
and
N
G
¯
2
(
x
)
is a cut edge in
G
. Therefore,
tpc
(
G
)
≥
k
+
1
by [26, Proposition 2], and so
tpc
(
G
)
will be arbitrarily large based on the increase of
k
.
Example 2.6
Let
G
¯
contains exactly two components
G
1
¯
and
G
2
¯
, where
G
1
¯
is trivial and
G
2
¯
is a clique of
G
¯
. Clearly,
G
is a star, and
tpc
(
G
)
=
∣
V
(
G
2
¯
)
∣
+
1
. Thus,
tpc
(
G
)
can be made arbitrarily large by increasing
∣
V
(
G
2
¯
)
∣
.
3 Proof of Theorem 1.2
Lemma 3.1
For a connected graph
G
, if
G
¯
is connected with
diam
(
G
¯
)
≥
4
, and
G
¯
is triangle-free, then
tpc
(
G
)
=
3
.
Proof
Choose a vertex
x
with
ecc
G
¯
(
x
)
=
diam
(
G
¯
)
. Since
G
¯
is triangle-free, we know that
N
G
¯
1
is a clique in
G
. Now we define a total coloring of
G
as follows: assign color 1 to the edges
x
u
for
u
∈
N
G
¯
2
, all edges between
N
G
¯
1
and
N
G
¯
3
, and all vertices and edges in
N
G
¯
4
; assign color 2 to the edges between
N
G
¯
2
and
N
G
¯
4
, all vertices and edges in
N
G
¯
1
, and the vertex
x
; assign color 3 to the edges
x
u
for
u
∈
N
G
¯
3
,
N
G
¯
4
, all edges between
N
G
¯
1
and
N
G
¯
4
, and all vertices and edges in
N
G
¯
2
,
N
G
¯
3
.
We prove that there is a total proper path between any two distinct vertices
u
and
v
in
G
. Note that
P
=
x
x
2
x
4
x
1
x
3
is a total proper path, where
x
i
∈
N
G
¯
i
. By means of the path
P
, we can find that
u
and
v
are connected by some total proper path for any
u
∈
N
G
¯
i
,
v
∈
N
G
¯
i
+
1
. Thus, we only need to consider the pairs
u
,
v
∈
N
G
¯
i
. For
i
=
2
,
P
=
u
x
x
4
v
is a total proper path, where
x
4
∈
N
G
¯
4
. For
i
=
4
,
P
=
u
x
x
2
v
is a total proper path, where
x
2
∈
N
G
¯
2
. For
i
=
3
,
P
=
u
x
1
x
4
x
2
x
v
is a total proper path, where
x
i
∈
N
G
¯
i
. Thus,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
.□
Lemma 3.2
Let
G
be a connected graph. If
diam
(
G
¯
)
=
3
and
G
¯
is triangle-free, then
tpc
(
G
)
=
3
.
Proof
For a vertex
x
of
G
¯
satisfying
ecc
G
¯
(
x
)
=
diam
(
G
¯
)
=
3
, let
n
i
represent the number of vertices with distance
i
from
x
. If
n
1
=
n
2
=
n
3
=
1
, then
G
≅
P
4
, and so
tpc
(
G
)
=
3
.
Case 1. Two of
n
1
,
n
2
,
n
3
are equal to 1. Without loss of generality, we may assume
n
1
=
n
2
=
1
. Since
G
¯
is triangle-free, we have that
N
G
¯
3
is a stable set in
G
¯
, and so a clique in
G
. We can find that
G
has a Hamiltonian path. Thus,
tpc
(
G
)
=
3
.
Case 2. One of
n
1
,
n
2
,
n
3
is equal to 1. Suppose
n
2
=
1
. Since
G
¯
is triangle-free, we know that
N
G
¯
1
and
N
G
¯
3
is a stable set in
G
¯
, and so a clique in
G
. Note that
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
.
Subcase 2.1.
n
1
=
1
. Since
G
¯
is triangle-free, we obtain that
N
G
¯
2
is a clique in
G
. Define a total coloring of
G
as follows: assign color 3 to the vertex
x
, all edges between
N
G
¯
2
and
N
G
¯
3
, and all edges between
N
G
¯
1
and
N
G
¯
3
; assign color 2 to the edges
x
u
for
u
∈
N
G
¯
2
, and all vertices and edges in
N
G
¯
3
; assign color 1 to the edges
x
u
for
u
∈
N
G
¯
3
, and all vertices and edges in
N
G
¯
1
,
N
G
¯
2
. We prove that there is a total proper path between any two distinct vertices
u
and
v
in
G
. Note that
P
=
x
1
x
3
x
x
2
is a total proper path, where
x
i
∈
N
G
¯
i
. By means of the path
P
, we know that
u
and
v
are connected by some total proper path for any
u
∈
N
G
¯
i
,
v
∈
N
G
¯
i
+
1
. For any two vertices
u
,
v
∈
N
G
¯
3
, it is trivial if
u
v
∈
E
(
G
)
. If
u
v
∉
E
(
G
)
, since
u
,
v
∈
N
G
¯
3
, there exist two vertices
u
′
,
v
′
∈
N
G
¯
3
such that
u
u
′
,
v
v
′
∈
E
(
G
¯
)
. Since
G
¯
is triangle-free, we can see that
u
′
≠
v
′
and
v
u
′
,
u
v
′
∈
E
(
G
)
. Then
P
=
u
x
u
′
v
ia a total proper path. Hence,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
.
Subcase 2.2.
n
3
=
1
. Since
G
¯
is triangle-free, we know that
N
G
¯
1
is a stable set in
G
¯
, and so a clique in
G
. Define a total coloring of
G
as follows: assign color 3 to the vertex
x
, and all edges between
N
G
¯
1
and
N
G
¯
3
; assign color 2 to the edges
x
u
for
u
∈
N
G
¯
2
, and all vertices and edges in
N
G
¯
3
; assign color 1 to the edges
x
u
for
u
∈
N
G
¯
3
, and all vertices and edges in
N
G
¯
1
,
N
G
¯
2
. We prove that there is a total proper path between any two distinct vertices
u
and
v
in
G
. Note that
P
=
x
1
x
3
x
x
2
is a total proper path, where
x
i
∈
N
G
¯
i
. By means of the path
P
, we obtain that
u
and
v
are connected by some total proper path for any
u
∈
N
G
¯
i
,
v
∈
N
G
¯
i
+
1
. Let
u
,
v
be any two distinct vertices of
N
G
¯
2
, and
N
G
¯
3
=
{
y
}
. If
y
is adjacent to any vertex of
N
G
¯
2
in
G
¯
, then
N
G
¯
2
is a clique in
G
, and so
G
has a Hamiltonian path. Otherwise, let
V
y
denote the set of neighbors of
y
in
N
G
¯
2
in
G
. We can check that
P
=
u
y
x
v
is a total proper path, where
u
,
v
∈
V
y
. If
∣
N
G
¯
2
\
V
y
∣
=
1
, then
P
=
u
y
x
v
is a total proper path, where
u
∈
V
y
,
v
∈
N
G
¯
2
\
V
y
. If
∣
N
G
¯
2
\
V
y
∣
≥
2
, then
G
is claw-free since
G
¯
is triangle-free, and
G
[
x
∪
N
G
¯
2
\
V
y
]
is a complete graph. Note that
P
=
u
y
x
v
is a total proper path, where
u
∈
V
y
,
v
∈
N
G
¯
2
. Thus,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
.
Case 3.
n
1
,
n
2
,
n
3
≥
2
. Since
G
¯
is triangle-free, we have that
N
G
¯
1
is a stable set in
G
¯
, and so a clique in
G
. If any vertex in
N
G
¯
3
is adjacent to all vertices of
N
G
¯
2
in
G
¯
, then both
N
G
¯
2
and
N
G
¯
3
are stable sets in
G
¯
, and so cliques in
G
. Thus,
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
.
Otherwise, we choose a vertex
u
∈
N
G
¯
3
, let
V
u
denote the set of neighbors of
u
in
N
G
¯
2
in
G
, we have
V
u
≠
∅
,
N
G
¯
2
. Define a total coloring of
G
: assign color 2 to the vertex
x
, all vertices and edges in
N
G
¯
1
, and all edges between
N
G
¯
2
,
N
G
¯
3
; assign color 3 to the vertex
u
, all edges between
N
G
¯
1
and
N
G
¯
3
\
{
u
}
, and all edges between
x
and
V
u
; assign color 2 to the remaining vertices and edges. Note that
P
=
x
v
u
x
1
is a total proper path, where
v
∈
V
u
,
x
1
∈
N
G
¯
1
. For any two vertices
w
,
z
∈
N
G
¯
3
,
P
=
w
x
1
u
v
x
z
is a total proper path, where
x
1
∈
N
G
¯
1
,
v
∈
V
u
. For any two vertices
w
,
z
∈
N
G
¯
2
,
P
=
w
u
x
v
is a total proper path, where
u
,
v
∈
V
y
. If
∣
N
G
¯
2
\
V
y
∣
=
1
, then
P
=
u
x
v
is a total proper path, where
u
∈
V
y
,
v
∈
N
G
¯
2
\
V
y
. If
∣
N
G
¯
2
\
V
y
∣
≥
2
, since
G
¯
is triangle-free, we know that
G
is claw-free, and the subgraph
G
[
x
∪
N
G
¯
2
\
V
y
]
is a complete graph. Note that
P
=
u
x
v
is a total proper path, where
u
∈
V
y
,
v
∈
N
G
¯
2
\
V
y
. For any
w
∈
N
G
¯
2
,
z
∈
N
G
¯
3
,
P
=
w
x
v
u
x
1
z
is a total proper path, where
v
∈
V
y
,
x
1
∈
N
G
¯
1
. Similarly, there is a total proper path connecting any two vertices
w
∈
N
G
¯
2
,
z
∈
N
G
¯
1
. Hence,
G
is total proper connected, and so
tpc
(
G
)
=
3
.□
Lemma 3.3
For a connected graph
G
, if
G
¯
is triangle-free and
diam
(
G
¯
)
=
2
, then
tpc
(
G
)
=
3
.
Proof
Choose a vertex
x
with
ecc
G
¯
(
x
)
=
diam
(
G
¯
)
=
2
. Since
G
is connected, we have
n
1
≥
2
,
n
2
=
1
or
n
1
,
n
2
≥
2
, and there exist two vertices
u
∈
N
G
¯
1
,
v
∈
N
G
¯
2
such that
u
v
∈
E
(
G
)
. Assume
n
1
≥
2
and
n
2
=
1
. Since
G
¯
is triangle-free, we know that
N
G
¯
1
is a stable set in
G
¯
, and so a clique in
G
. Note that
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
.
Assume
n
1
,
n
2
≥
2
. Observe that
N
G
¯
1
is a stable set in
G
¯
since
G
¯
is triangle-free, and so a clique in
G
. We show a total coloring of
G
as follows: assign color 1 to the vertex
x
, the edge
u
v
and all vertices in
N
G
¯
1
\
u
; assign color 3 to the vertex
v
and all edges in
N
G
¯
1
; assign color 2 to the remaining vertices and edges. If there exist some vertices
w
∈
N
G
¯
2
with
d
G
¯
(
w
)
=
n
−
2
, then
w
is adjacent to the remaining vertices except
x
in
G
¯
. Since
diam
(
G
¯
)
=
2
, there exists an edge
w
1
w
2
∈
E
(
G
¯
)
with
w
1
∈
N
G
¯
1
,
w
2
∈
N
G
¯
2
. Thus,
w
,
w
1
,
w
2
is a triangle in
G
¯
, a contradiction. Hence,
d
G
¯
(
w
)
<
n
−
2
for all
w
∈
N
G
¯
2
, and so
d
G
(
w
)
≥
2
. For any
z
∈
N
G
¯
1
, we know that
P
=
x
v
u
z
is a total proper path. For any
y
∈
N
G
¯
2
\
{
v
}
and
z
∈
N
G
¯
1
, if
N
G
(
y
)
∩
N
G
¯
1
≠
∅
, let
w
∈
N
G
(
y
)
∩
N
G
¯
1
. Then
y
w
z
is a total proper path. Otherwise, let
N
G
(
y
)
∩
N
G
¯
1
=
∅
. We claim that
y
is adjacent to all the other vertices of
N
G
¯
2
in
G
. In fact, for any vertex
w
∈
N
G
¯
2
\
{
y
}
, there exists a vertex
w
′
∈
N
G
¯
1
such that
w
w
′
∈
E
(
G
¯
)
. Since
y
w
′
∈
E
(
G
¯
)
, we know that
y
w
∈
E
(
G
)
. Then
y
v
u
z
is a total proper path. Next we consider
w
,
z
∈
N
G
¯
2
such that
w
z
∉
E
(
G
)
. Since
G
¯
is triangle-free, we have that
G
is claw-free, and at least one of
w
and
z
is adjacent to the
v
, without loss of generality, assume that
w
v
∈
E
(
G
)
. Since
w
,
z
∈
N
G
¯
2
, there exist two vertices
w
′
,
z
′
∈
N
G
¯
2
such that
w
w
′
,
z
z
′
∈
E
(
G
¯
)
, and
w
′
≠
z
′
. Then
z
w
′
,
w
z
′
∈
E
(
G
)
and
P
=
w
v
u
w
′
z
is a total proper path. Thus,
G
is total proper connected with the above coloring. Hence,
tpc
(
G
)
=
3
.□
Lemma 3.4
Let
G
be a connected graph of order
n
≥
3
. If
G
¯
is disconnected and triangle-free, then
tpc
(
G
)
=
3
.
Proof
Suppose
G
¯
is triangle-free and contains two connected components one of which is trivial. Let
G
1
¯
and
G
2
¯
be the two components of
G
¯
, where
V
(
G
1
¯
)
=
{
u
}
. Then
u
is adjacent to any other vertex in
G
. We will consider two cases according to the value of
δ
, where
δ
is the minimum degree of
G
. If
δ
=
1
, let
d
(
v
)
=
δ
. Since
G
¯
is triangle-free, we know that
G
is claw-free, and the subgraph
G
[
V
(
G
)
\
{
v
}
]
is a complete graph. Thus,
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
. If
δ
≥
2
, let
d
(
v
)
=
δ
,
D
=
V
(
G
)
\
{
u
,
v
}
, and
V
v
be the set of neighbors of
v
in
G
. Now we define a total coloring of
G
as follows: assign color 1 to the vertex
v
and all the edges between
u
and
V
v
; assign color 3 to the vertex
u
and all the edges between
v
and
V
v
; assign color 2 to the remaining vertices and edges. Since
G
is claw-free, we can find that the subgraph
G
[
V
(
G
)
\
{
v
}
∪
V
v
]
is a complete graph, and
P
=
v
1
u
v
v
2
is a total proper path, where
v
1
,
v
2
∈
V
v
. For any
w
∈
V
v
,
z
∈
D
\
V
v
, we obtain that
P
=
w
u
z
is a total proper path. Thus,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
. Suppose
G
¯
contains at least three connected components or exactly two nontrivial components. Then we have
tpc
(
G
)
=
3
by the similar proof of Theorem 1.1.□
Proof of Theorem 1.2
If
G
¯
is connected, the result holds for the case
diam
(
G
¯
)
≥
4
by Lemma 3.1, the case
diam
(
G
¯
)
=
3
by Lemma 3.2, and the case
diam
(
G
¯
)
=
2
by Lemma 3.3. If
G
¯
is disconnected, the result holds by Lemma 3.4.□
4 Proof of Theorem 1.3
Suppose
F
≅
K
2
. Note that
G
has a Hamiltonian path, and thus
tpc
(
G
)
=
3
. Next, we compute the total proper connection number of
G
by proving the following claim.
Claim 1. Let
G
be a graph obtained by adding two pendant vertices
{
u
1
,
u
2
}
to a vertex
v
1
of a complete graph
K
t
. Then
tpc
(
G
)
=
3
.
Proof
Since
G
is not a complete graph, we have
tpc
(
G
)
≥
3
. Now we only need to prove
tpc
(
G
)
≤
3
by the following cases.
Case 1.
t
≡
0
(
mod
3
)
. Assign a total coloring
c
to
G
as follows: Let
c
(
u
1
v
1
)
=
1
,
c
(
u
2
v
1
)
=
3
;
c
(
v
3
i
+
1
)
=
c
(
v
3
i
+
2
v
3
i
+
3
)
=
2
,
c
(
v
3
i
+
2
)
=
c
(
v
3
i
+
3
v
3
i
+
4
)
=
1
, and
c
(
v
3
i
+
3
)
=
c
(
v
3
i
+
1
u
3
i
+
2
)
=
3
, where
0
≤
i
≤
t
3
−
1
. Observe that
P
1
=
u
1
v
1
v
2
⋯
v
t
−
1
v
t
and
P
2
=
u
2
v
1
v
t
v
t
−
1
⋯
v
3
v
2
are two total proper paths.
Case 2.
t
≡
1
(
mod
3
)
. Assign a total coloring
c
to
G
as follows: Let
c
(
u
1
v
1
)
=
c
(
v
t
−
1
v
1
)
=
1
,
c
(
v
t
)
=
2
,
c
(
u
2
v
1
)
=
c
(
v
t
v
1
)
=
3
. Let
i
be an integer with
0
≤
i
≤
t
3
−
1
,
c
(
v
3
i
+
1
)
=
c
(
v
3
i
+
2
v
3
i
+
3
)
=
2
,
c
(
v
3
i
+
2
)
=
c
(
v
3
i
+
3
v
3
i
+
4
)
=
1
, and
c
(
v
3
i
+
3
)
=
c
(
v
3
i
+
1
u
3
i
+
2
)
=
3
. We can find that
P
1
=
u
1
v
1
v
2
⋯
v
t
−
1
v
t
and
P
2
=
u
2
v
1
v
t
−
1
⋯
v
3
v
2
v
t
are two total proper paths.
Case 3.
t
≡
2
(
mod
3
)
. Assign a total coloring
c
to
G
as follows: Let
c
(
u
1
v
1
)
=
c
(
v
t
v
1
)
=
c
(
v
t
)
=
c
(
v
t
−
2
v
1
)
=
1
,
c
(
v
t
−
1
)
=
2
,
c
(
u
2
v
1
)
=
c
(
v
t
v
t
−
1
)
=
c
(
v
t
−
1
v
2
)
=
3
. Let
i
be an integer with
0
≤
i
≤
t
3
−
1
,
c
(
v
3
i
+
1
)
=
c
(
v
3
i
+
2
v
3
i
+
3
)
=
2
,
c
(
v
3
i
+
2
)
=
c
(
v
3
i
+
3
v
3
i
+
4
)
=
1
, and
c
(
v
3
i
+
3
)
=
c
(
v
3
i
+
1
u
3
i
+
2
)
=
3
. We can easily verify that
P
1
=
u
1
v
1
v
2
⋯
v
t
−
1
v
t
,
P
2
=
u
2
v
1
v
t
−
2
v
t
−
3
⋯
v
3
v
2
v
t
−
1
, and
P
3
=
v
t
v
1
u
2
are three total proper paths.
Thus,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
. This completes the proof of Claim 1.□
Suppose
F
≅
2
K
1
. Assume that
N
X
(
u
1
)
∩
N
X
(
u
2
)
=
∅
. Then
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
. Otherwise, if
∣
N
X
(
u
1
)
∩
N
X
(
u
2
)
∣
=
d
X
(
u
1
)
=
d
X
(
u
2
)
=
1
, then we know that
tpc
(
G
)
=
3
from Claim 1. If
∣
N
X
(
u
1
)
∩
N
X
(
u
2
)
∣
≥
2
, or
∣
N
X
(
u
1
)
∩
N
X
(
u
2
)
∣
=
1
, and
max
{
d
X
(
u
1
)
,
d
X
(
u
2
)
}
≥
2
, then we can find that
G
has a Hamiltonian path. Thus,
tpc
(
G
)
=
3
.
5 Proof of Theorem 1.4
Suppose
F
≅
K
3
or
P
3
. Note that
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
. The following three claims will be used later.
Claim 2. Let
G
be a graph obtained by adding a pendant vertex
u
3
adjacent to vertex
u
1
or
u
2
of graph in Claim 1. Then
tpc
(
G
)
=
3
.
Proof
Without loss of generality, assume that
u
3
is adjacent to
u
1
. Let
c
(
u
1
)
=
{
1
,
2
,
3
}
\
{
c
(
u
1
v
1
)
,
c
(
v
1
)
}
,
c
(
u
1
u
3
)
=
c
(
v
1
)
, and the remaining vertices and edges are assigned the same color as Claim 1. We can verify that
G
is total proper connected with the above coloring. Then
tpc
(
G
)
=
3
. This completes the proof of Claim 2.□
Claim 3. Let
G
be a graph obtained by adding three vertices
{
u
1
,
u
2
,
u
3
}
to a complete graph
K
t
such that
d
(
u
1
)
=
d
(
u
3
)
=
1
,
d
(
u
2
)
=
2
,
N
(
u
1
)
∩
N
(
u
3
)
=
∅
, and
∣
N
(
u
2
)
∩
N
(
u
i
)
∣
=
1
, where
1
≤
i
≤
2
. Then
tpc
(
G
)
=
3
.
Proof
Without loss of generality, assume that
N
(
u
2
)
∩
N
(
u
1
)
=
v
1
,
N
(
u
2
)
∩
N
(
u
3
)
=
v
i
. Let
c
(
u
3
v
i
)
=
c
(
v
i
v
i
+
1
)
,
c
(
v
1
u
2
)
=
c
(
v
i
v
i
−
1
)
, and the remaining vertices and edges are assigned the same color as Claim 1. Suppose
t
≡
0
(
mod
3
)
. Note that
P
1
=
u
1
v
1
v
2
⋯
v
t
−
1
v
t
,
P
2
=
u
2
v
1
v
t
v
t
−
1
⋯
v
3
v
2
,
P
3
=
u
3
v
i
u
2
,
P
4
=
u
1
v
1
v
2
⋯
v
i
u
3
, and
P
5
=
u
3
v
i
v
i
−
1
⋯
v
1
v
t
v
t
−
1
⋯
v
i
+
1
are five total proper paths. Suppose
t
≡
1
(
mod
3
)
. Note that
P
1
=
u
1
v
1
v
2
⋯
v
t
−
1
v
t
,
P
2
=
u
2
v
1
v
t
−
1
⋯
v
3
v
2
v
t
,
P
3
=
u
3
v
i
u
2
,
P
4
=
u
1
v
1
v
2
⋯
v
i
u
3
and
P
5
=
u
3
v
i
v
i
−
1
⋯
v
2
v
t
v
t
−
1
⋯
v
i
+
1
are five total proper paths. Suppose
t
≡
2
(
mod
3
)
. We can find that
P
1
=
u
1
v
1
v
2
⋯
v
t
−
1
v
t
,
P
2
=
u
2
v
1
v
t
−
2
⋯
v
3
v
2
v
t
−
1
,
P
3
=
u
3
v
i
u
2
,
P
4
=
u
1
v
1
v
2
⋯
v
i
u
3
,
P
5
=
u
3
v
i
v
i
−
1
⋯
v
2
v
t
−
1
v
t
−
2
⋯
v
i
+
1
, and
P
6
=
u
3
v
i
v
i
−
1
v
t
are six total proper paths. Therefore,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
. This completes the proof of Claim 3.□
Claim 4. Let
G
be a graph obtained by adding a vertex
u
to the graph in Claim 1 such that
d
(
u
)
=
2
and
v
1
is adjacent to
u
. Then
tpc
(
G
)
=
3
.
Proof
Since
d
(
u
)
=
2
, without loss of generality, we assume that
v
i
is adjacent to
u
. Let
c
(
u
v
1
)
=
1
,
c
(
u
v
i
)
=
{
1
,
2
,
3
}
\
{
c
(
v
i
)
,
c
(
v
i
v
i
−
1
)
}
, and the remaining vertices and edges are assigned the same color as Claim 1. Suppose
t
≡
0
(
mod
3
)
. Note that
P
1
=
u
v
i
v
i
−
1
⋯
v
1
v
t
v
t
−
1
⋯
v
i
+
1
,
P
2
=
u
v
i
v
i
−
1
⋯
v
1
u
1
and
P
3
=
u
v
1
u
2
are three total proper paths. Suppose
t
≡
1
(
mod
3
)
. Note that
P
1
=
u
v
i
v
i
−
1
⋯
v
1
u
1
,
P
2
=
u
v
i
v
i
−
1
⋯
v
2
v
t
v
t
−
1
⋯
v
i
+
1
, and
P
3
=
u
v
1
u
2
are three total proper paths. Suppose
t
≡
2
(
mod
3
)
. We can find that
P
1
=
u
v
i
v
i
−
1
⋯
v
1
u
1
,
P
2
=
u
v
i
v
i
−
1
⋯
v
2
v
t
−
1
v
t
−
2
⋯
v
i
+
1
,
P
3
=
u
v
1
u
2
, and
P
4
=
u
v
i
v
i
−
1
⋯
v
1
v
t
are four total proper paths. Hence,
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
3
. This completes the proof of Claim 4.□
Suppose
F
≅
K
2
+
K
1
. Let
V
(
K
2
)
=
{
u
1
,
u
2
}
and
V
(
K
1
)
=
{
u
3
}
. Since
diam
(
G
)
=
2
, we have
N
X
(
u
1
)
∩
N
X
(
u
2
)
∩
N
X
(
u
3
)
=
{
v
}
, and so
tpc
(
G
)
=
3
by Claim 2. Suppose
F
≅
3
K
1
. Assume
N
X
(
u
1
)
∩
N
X
(
u
2
)
∩
N
X
(
u
3
)
=
∅
. Then
tpc
(
G
)
=
3
by Claim 3. Assume
N
X
(
u
1
)
∩
N
X
(
u
2
)
∩
N
X
(
u
3
)
≠
∅
. If
d
X
(
u
1
)
=
d
X
(
u
2
)
=
d
X
(
u
3
)
=
1
, then
tpc
(
G
)
≥
4
by [26, Proposition 2]. Define a total coloring of
G
as follows:
c
(
u
3
v
)
=
4
with
v
∈
N
X
(
u
3
)
, and the remaining vertices and edges are assigned the same color as Claim 1. We check that any two vertices have a total proper path, and so
tpc
(
G
)
=
4
. Otherwise, we have
d
X
(
u
1
)
+
d
X
(
u
2
)
+
d
X
(
u
3
)
≥
4
. Without loss of generality, let
d
X
(
u
1
)
≥
2
, and
u
∈
X
\
{
v
}
where
v
∈
N
X
(
u
1
)
∩
N
X
(
u
2
)
∩
N
X
(
u
3
)
. Thus,
tpc
(
G
)
=
3
by Claim 4.
6 Proof of Theorem 1.5
Case 1.
diam
(
G
)
=
3
. We prove Case 1 by analyzing the structure of
F
.
Subcase 1.1.
F
≅
K
3
or
P
3
. Note that
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
.
Subcase 1.2.
F
≅
K
2
+
K
1
. Denote
V
(
K
2
)
=
{
u
1
,
u
2
}
and
V
(
K
1
)
=
{
u
3
}
. Suppose
N
X
(
u
1
)
∩
N
X
(
u
3
)
≠
∅
or
N
X
(
u
2
)
∩
N
X
(
u
3
)
≠
∅
. Without of loss generality, we may assume that
N
X
(
u
1
)
∩
N
X
(
u
3
)
≠
∅
. Then
tpc
(
G
)
=
3
by Claim 2. Suppose
N
X
(
u
1
)
∩
N
X
(
u
3
)
=
∅
and
N
X
(
u
2
)
∩
N
X
(
u
3
)
=
∅
. Since
diam
(
G
)
=
3
, we have
N
X
(
u
1
)
≠
∅
and
N
X
(
u
3
)
≠
∅
. Then
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
.
Subcase 1.3.
F
≅
3
K
1
. Let
V
(
F
)
=
{
u
1
,
u
2
,
u
3
}
. Since
diam
(
G
)
=
3
, we have
N
X
(
u
1
)
∩
N
X
(
u
2
)
∩
N
X
(
u
3
)
=
∅
. Suppose there exists two vertices
u
i
,
u
j
∈
V
(
F
)
satisfy
N
X
(
u
i
)
∩
N
X
(
u
j
)
≠
∅
. Without loss of generality, let
u
1
and
u
2
satisfy
N
X
(
u
1
)
∩
N
X
(
u
2
)
≠
∅
and
v
1
∈
N
X
(
u
1
)
∩
N
X
(
u
2
)
. Assume
d
X
(
u
1
)
=
d
X
(
u
2
)
=
1
and
v
i
∈
N
(
u
3
)
. Since
G
is not complete, we have
tpc
(
G
)
≥
3
. To the contrary, suppose there exists a total coloring
c
of
G
using three colors such that
G
is total proper connected. Since any two vertices of
G
are connected by a total proper path, we have
c
(
u
1
v
1
)
≠
c
(
v
1
)
≠
c
(
u
2
v
1
)
. Without loss of generality, let
c
(
u
1
v
1
)
=
1
,
c
(
v
1
)
=
2
and
c
(
u
2
v
1
)
=
3
. Consider the total proper path
P
between
u
1
and
u
, then the color of vertices and edges in
P
follows the sequence
1
,
2
,
3
,
…
,
1
,
2
,
3
,
…
. Thus, the value of
(
c
(
v
i
)
,
c
(
v
i
u
)
)
is
(
1
,
2
)
,
(
2
,
3
)
, or
(
3
,
1
)
. Consider the total proper path
Q
between
u
2
and
u
, then the color of vertices and edges in
Q
follows the sequence
3
,
2
,
1
,
…
,
3
,
2
,
1
,
…
. But the value of
(
c
(
v
i
)
,
c
(
v
i
u
)
)
is
(
3
,
2
)
,
(
2
,
1
)
, or
(
1
,
3
)
, a contradiction. Assign a total coloring
c
to
G
as follows:
c
(
u
1
v
1
)
=
1
,
c
(
v
1
)
=
c
(
v
i
u
)
=
2
,
c
(
u
2
v
1
)
=
3
, assign 4 to the remaining edges, and assign 1 to the remaining vertices. We can check that
G
is total proper connected with the above coloring, and so
tpc
(
G
)
=
4
. Assume
d
X
(
u
1
)
+
d
X
(
u
2
)
≥
3
, without loss of generality, let
d
X
(
u
1
)
≥
2
. If
d
X
(
u
3
)
=
1
and
N
X
(
u
1
)
∩
N
X
(
u
3
)
≠
∅
, then we have
tpc
(
G
)
=
3
by Claim 3; if
N
X
(
u
1
)
∩
N
X
(
u
3
)
=
∅
, or
N
X
(
u
1
)
∩
N
X
(
u
3
)
≠
∅
and
d
X
(
u
3
)
≥
2
, then
G
has a Hamiltonian path, and so
tpc
(
G
)
=
3
.
Now, we may suppose
N
X
(
u
1
)
∩
N
X
(
u
2
)
=
∅
,
N
X
(
u
1
)
∩
N
X
(
u
3
)
=
∅
, and
N
X
(
u
2
)
∩
N
X
(
u
3
)
=
∅
. Since
G
is not complete, we have
tpc
(
G
)
≥
3
. To the contrary, assume that
v
i
∈
N
(
u
i
)
, and there exists a total coloring
c
of
G
using three colors such that
G
is total proper connected. Since any two vertices of
G
are connected by a total proper path, we have
c
(
u
1
v
1
)
≠
c
(
v
1
)
. Without loss of generality, let
c
(
u
1
v
1
)
=
1
and
c
(
v
1
)
=
2
. Consider the total proper path
P
between
u
1
and
u
2
, then the color of vertices and edges in
P
follows the sequence
1
,
2
,
3
,
…
,
1
,
2
,
3
,
…
. Thus, the value of
(
c
(
v
2
)
,
c
(
v
2
u
2
)
)
is
(
1
,
2
)
,
(
2
,
3
)
, or
(
3
,
1
)
. Consider the total proper path
Q
between
u
1
and
