Total Global Dominator Coloring of Trees and Unicyclic Graphs

: A total global dominator coloring of a gra ph 𝐺 is a proper vertex coloring of 𝐺 with respect to which every vertex 𝑣 in 𝑉 dominates a color class, not containing 𝑣 and does not dominate another color class. The minimum number of colors required in such a coloring of 𝐺 is called the total global dominator chromatic number, denoted by 𝜒 𝑡𝑔𝑑 (𝐺) . In this paper, the total global dominator chromatic number of trees and unicyclic graphs are explored.

A set is called a dominating set of if [ ] = ( ). If ∈ ( ), then is called a dominator of or is dominated by and vice versa. If ( ) = ( ), then is called a total dominating set of . The minimum cardinality of a dominating set of is known as its domination number, denoted by ( ) and the minimum cardinality of a total dominating set of is its total domination number, denoted by ( ). If is a dominating set (total dominating set) of both and complement of , then is called a global dominating set (total global dominating set) of and cardinality of a minimum global dominating set (resp. total global dominating set) is called the global domination number (resp. total global domination number) of , denoted by ( ) ( . ( )) 1 . Many variations of domination have been widely studied 2,3 .
The distance between two vertices and of a graph , denoted by ( , ), is the number of edges in a minimal path between and . A tree is a connected acyclic graph. If two non-adjacent vertices in a tree are connected by an edge, then the new graph will consist of one (unique) cycle. Such graphs are known to be unicyclic graphs. In other words, unicyclic graphs are the graphs containing exactly one cycle in it.
A proper vertex coloring (or a proper coloring) of a graph is a mapping from the vertex set of to a non-empty set of colors such that adjacent vertices receive distinct colors. The chromatic number of , denoted by ( ) is the minimum number of colors in its proper coloring. Unless stated otherwise, all the vertex coloring discussed in this paper are proper.

Dominator Coloring of Graphs
Coloring and domination in graphs are two major areas of research in graph theory. Recently the concept dominator coloring, which is a combination of domination and coloring, has emerged as a promising area for further research.
A dominator coloring of is a coloring such that every vertex in ( ) dominates some color class and the minimum number of colors used to color  5 . Further studies on ( ) and its variations can be found in 6,7 .
Corresponding to a coloring of a graph , a color class which is dominated by a vertex is known as a dom-color class whereas a color class such that no vertex of is adjacent to any vertex in that color class is known as an anti-dom-color class of . A coloring is called a global dominator coloring if each vertex ∈ ( ) has a dom color class and an antidom color class. Global dominator chromatic number of is the minimum number of colors required for a global dominator coloring of and is denoted by ( ) 8 . A color class is called a proper dom-color class of if the ( ) contains that color class. Using this definition, the notion of total global dominator coloring (tgd-coloring) has been introduced as a coloring for which every vertex has proper dom color class and an antidom color class. The minimum number of colors used in a tgdcoloring of is called the total global dominator chromatic number of and is denoted by ( ) see 9 . Where the relationship between ( ) with different graph parameters have been studied and of graphs such as paths, cycles, complete multipartite graphs, complement of paths and complement of cycles the For any graph admitting tgd-coloring, ( ) ≥ 4 and 1 ≤ ( ) ≤ ( ) ≤ − 2.
This paper is an extension of the study of the parameter ( ) by examining the tgd-coloring for the families of trees and unicyclic graphs. For the terminology and results of graph theory refer 10 , for more about domination in graphs refer to 11 and for the terminology of graph coloring, see 12 . Some of the results which are significant and relevant in this study are listed below: Theorem 1 9 For any graph which admits tgdcoloring, ( ) ≤ ( ) + ( ). Observation 1 6 If is a tree of order ≥ 2, then ( ) ≤ ( ) ≤ ( ) + 2.

Main Results:
Let's begin with a general result that links the parameters ( ) and ( ) if ( ) = 1.

Theorem 2 Let
Proof. If admits a tgd-coloring, then ( ) ≤ ( ) 2 . It remains to prove that ( ) ≤ ( ) + 2. Consider a -coloring of such that ( ) = . From the very definition of TD-coloring it follows that every vertex of has a proper dom-color class. Since ( ) = 1, there exists a support vertex, say in . Let be a pendant vertex adjacent to and be a non-neighbour of in . Now define a new coloring ′ such that ′ ( ) = ( ) for all ∈ ( ), except for the vertices and and assign two new colors to the vertices and . Then it is easy to verify tht ′ is a tgd-coloring of . Every vertex of has a proper dom-color class with respect to the coloring ′ as is a -coloring. Further being a pendant vertex having a unique color, { } acts as an anti dom-color class for all vertices of other than the vertices , . Again { } acts as an anti-dom-color class of vertices and . Hence ′ is a tgd-coloring of , proving that ( ) ≤ ( ) + 2. □ The bounds in the inequality of the above theorem are sharp. The examples for both cases are in the coming sections. Let be the number of support vertices of . Then, the following theorem establishes a lower bound for ( ) in terms of .

Theorem 3 If admits tgd-coloring, then
Proof. The only possible member of a proper domcolor class of a pendent vertex is its support vertex. Therefore, the cardinality of the color class of a support vertex is 1. Since there are support vertices and cardinality of each of the support vertices is 1, + 1 colors are needed to color the all the support vertices and at least one more color is need to color the remaining vertices of . That is, □ The bound of the above theorem is sharp. It can be verified with the help of an example. Consider the graph given in Fig. 1, which has four support vertices. That is, = 4. If we color the four support vertices with 4 different colors and all the remaining vertices with a color different from the four colors used, then the defined coloring is a tgd-coloring. Therefore, ( ) ≤ 5. By Theorem 3, ( ) ≥ + 1 = 5. From the above two inequalities, ( ) = 5 = + 1.

Theorem 4
If is a graph having at least two support vertices such that the distance between them is at least 3, then ( ) = ( ). Proof. Assume that admits a tgd-coloring such that contains at least two support vertices which are at a distance of more than two. Obviously admits a coloring. Moreover, under any TD-coloring of , the colors received by the support vertices would be unique. Since there exist at least two support vertices which are at a distance more than 2, the closed neighbourhood of these two vertices will not have a common vertex. Therefore, one of these support vertices will constitute the anti-dom-color class for any vertex of . Hence any coloring of will also be a tgd coloring. □ The converse of the above theorem is not true. For example, consider the path 5 . Here, ( 5 ) = ( 5 ) = 4, even though there are no two support vertices which are at a distance at least 3.

Trees
This section discusses the condition for a tree to admit tgd-coloring and determine the values of of trees in terms of the graph parameters such as , , the diameter ( ), and the number of support vertices . The following result presents a necessary and sufficient condition for any tree to admit a tgdcoloring.

Theorem 5 A tree admits a tgd-coloring if and only
Proof. Let admits a tgd-coloring. If possible, assume that ( ) < 3. In this case the central vertex of is adjacent to all other vertices. That means the central vertex does not have an anti-domcolor class, contradicting the assumption that admits a tgd-coloring. Therefore, ( ) ≥ 3. Conversely assume that is a tree such that ( ) ≥ 3. Assign distinct colors to all vertices of , which obviously gives rise to a proper domcolor class to each vertex of . Since the ( ) ≥ 3, corresponding to any vertex of there exist at least one non-adjacent vertex serving as the antidom-color class. □ In view of Theorem 5, all trees considered for further discussion would be trees of diameter at least 3. The following theorem gives the exact value of for trees with diameter 3.

Theorem 6
If ( ) = 3, then ( ) = 4. Proof. Consider a tree of diameter 3. Then by Theorem 5, admits a tgd-coloring and hence ( ) ≥ 4. Now it remains to show the existance of tgd-coloring that requires only four colors. T being a tree with diameter 3 contains exactly two support vertices. Use two distinct colors to color these support vertices. Color the pendant vertices adjacent to one of the support vertices with color 3 and the remaining vertices with color 4, is a tgdcoloring, thus completing the proof. □ Next theorem provides of a tree with diameter 4 in terms of its number of support vertices.
Note that trees of diameter 4 can be classified into two types as given in Fig. 2.

Type-1
Type -2 The first being trees for which the central vertex is not a support vertex. This class of trees may be referred to as Type 1. The second type are those trees whose central vertex is a support vertex as well. The theorem can be proved by considering these two cases separately.
Case 1: Let T be a tree of Type-1.
In this case the center of , say , is not a support vertex. Then by Theorem 2, ( ) ≥ + 1. It can be claimed that ( ) ≠ + 1. On the contrary, assume that ( ) = + 1. Then, the colors assigned to the support vertices should be distinct. Hence there exist only one color that remains to color the pendent vertices and the central vertex which is not possible as this coloring does not yield an anti-dom-color class for . Therefore ( ) ≥ + 2. Now it remains to show that admits a tgd-coloring with + 2 colors. Define a coloring scheme such that the support vertices are colored with different colors and the central vertex with a color different from the colors used and the remaining vertices with yet another new color. This yields a tgd-coloring and hence ( ) = + 2.
Case 2: Let the tree is of Type-2.
In this case, the center of , say , is a support vertex. By Theorem 2, ( ) ≥ + 1, first show that ( ) ≠ + 1. If possible, assume that is a tree with ( ) = + 1. Since colors required to color the support vertices, the remaing all vertices should be color with the remaining one color. But it is not possible since in this case would not have an anti-dom-color class, implies ( ) ≥ + 2. To prove ( ) = + 2 it is enough to show that there exists a tgd-coloring of with + 2 colors. Define a coloring of as follows. Color all the support vertices with different colors, the support vertices at with + 1-th color and the remaining vertices with + 2-th color, yields a tgd-coloring. □

Remark 1
The result of Theorem 7 is true for trees of diameter 3 as well.
The following theorem establishes the relationship between and for trees with diameter 4. Proof. This follows from Theorem 7 and the result that, ( ) = + 2 for trees of Type-1 and, ( ) = + 1 for trees of Type-2 5 . □ In view of the above theorem note that, if, the set of support vertices of a tree of diameter 4 form an independent set, then ( ) = ( ).

Theorem 9
For a tree with diameter more than 4, ( ) = ( ). Proof. Let be a tree with diameter more than 4, contains at least two support vertices such that the distance between them is 3. Therefore, the proof follows from Theorem 3. □ In view of the Theorem 9 note that, if, the set of support vertices of a tree of diameter 4 form an independent set, then ( ) = ( ). And from Theorem 2 it obvious that ( ) ≤ ( ) ≤ ( ) + 2. □ Next theorem is a characterization of trees.

Theorem 10 For a tree ( ) = ( ) + 1 if and only if is of diameter 3 or of diameter 4 and the center vertex is a support vertex.
Proof. By Theorem 9, all trees of diameter 5 or above ( ) = ( ). Hence it remains to consider only trees of diameter 3 or 4. For a tree of diameter 3, one may get ( ) = 4 = ( ) + 1. Also, if is of diameter 4 then by Theorem 7, ( ) = ( ) if center of is not a support vertex. Therefore, ( ) = ( ) + 1 if and only if is of diameter 3 or of diameter 4 and the center vertex is a support vertex. □ In view of the results mentioned above, the following inequality is immediate. All trees which admit tgdcoloring satisfies Theorem 2. From the above results it is obvious that the maximum value of ( ) is ( ) + 1. Therefore, one has the following observation.

Observation 2 For a tree with diameter at least 3,
( ) ≤ ( ) ≤ ( ) + 1. For trees which admit tgd-coloring, the next theorem is a characterization with respect to its support vertices.

Theorem 11 For a tree , ( ) = + 1 if and only if the following conditions hold.
i) ( ) ≥ 5; ii) If is a support vertex of then there exist another support vertex of such that ( , ) = 1; and iii) The minimum distance between any two nonsupport non-leaf vertices is 2.
Proof. Let be a tree with the given three conditions. Consider a proper coloring such that all the support vertices in receive distinct colors, say 1 , 2 , … , and all the remaining vertices receive same color, say +1 , which is different from all ; 1 ≤ ≤ . This coloring is possible, as the distance between two non-support vertices is at least 2. Since ( ) ≥ + 1, it is enough to prove that is a tgd-coloring. Let be an arbitrary vertex of . Then, there exists a support vertex ′ in that is adjacent to . That is, { ′} is a proper dom-color class of the vertex . since ( ) ≥ 5, there exist two support vertices and ′ in such that ( , ′) ≥ 3. Therefore, either { } or { ′} is an anti-dom color class of . Therefore, is a tgd-coloring.
Conversely, assume that ( ) = + 1. The first condition is immediate from Theorem 6 and Theorem 7. Let be a support vertex of which is not adjacent to any other support vertex. Therefore, the only possible proper dom-color class of is the color class of the leaf vertices. Since ( ) ≥ 5, there is at least one leaf which is not adjacent to . This leads to a contradiction that does not have a proper dom-color class. This proves the second condition. If there exist two adjacent non-leaf nonsupport vertices, then ( ) = + 1 as these two vertices cannot have the same color. Thus, the third condition also holds. □ A caterpillar graph is a tree where the removal of its pendent vertices gives a path of order greater than or equal to 2. A complete caterpillar graph is a caterpillar where ( ) contains only support vertices and pendant vertices. The following result follows from Theorem 11.

Unicyclic Graphs
In this section, let denote the cycle in the unicyclic graph . If is a unicyclic graph, a vertex lies on is called an extreme vertex if it is adjacent to a vertex of degree at least 2 lying outside . It is known that the minimum value of and for a graph are 2 and 4 respectively. In this section, the total dominator chromatic number and total global dominator chromatic number of unicyclic graphs are studied.

Theorem 14
If is a unicyclic graph other than 4 , then ( ) ≥ 3. Further, ( ) = 3 if and only if is isomorphic to one of the graphs in Fig. 3. Proof. Note that the total dominator chromatic number of any cycle other than 4 is more than 2. Therefore, the proof of the first part is trivial. Next, it is to be proved the necessary and sufficient condition for ( ) = 3. It is easy to verify that of each of the graphs in Fig. 3 is 3. So, it remains to prove that ( ) ≠ 3, for any unicyclic graph which is not present in Fig. 3. It is known that of any cycle of length other than 3 or 4 is greater than 3. Therefore, one must consider unicyclic graphs with cycle of length 3 and 4.
Case 1: Let be a unicyclic graph which has a cycle of length 3. In this case, there exist two types of unicyclic graphs, one without extreme vertices and the other is with extreme vertices. First, let us assume that has an extreme vertex. Therefore, there exists a support vertex outside . Since the color of a support vertex cannot be assigned to any other vertex in TD-coloring, another three colors are required to color the vertices of the cycle. In this case, ( ) ≥ 4. If does not have an extreme vertex, then either is a cycle , or the graph obtained by adding pendent vertices at the vertices of . Here, the number of vertices of at which pendent vertices can be added such that ( ) = 3 is to be determined. It can be observed that if pendent vertices are added at one or two vertices of , then ( ) = 3. But, if pendent vertices are added at all three vertices of , then ( ) = 5. Therefore, the unicyclic graphs which have a cycle of length 3 with ( ) = 3 are those present in Fig. 3.

Case 2:
Let be a unicyclic graph which has a cycle of length 4. Also, let ( ) = 3. Therefore, should not have an extreme vertex. Because if contains an extreme vertex there exists a support vertex outside the cycle say , whose color class is solitary. If the remaining vertices of are colored with another two colors, then will not have a proper dom-color class. Next, it is required to find which of can be a support vertex. If pendent vertices are added only at a vertex of , then ( ) = 3. Since ( ) = 3, it is not possible to add pendent vertices at any three vertices of . Suppose that support vertices are added at two adjacent vertices, at least two more colors are required to color , which leads to a contradiction. Let the non-adjacent vertices of be support vertices. If the two support vertices are colored using two distinct colors and the remaining vertices with a third color, then the support vertices will not have a proper dom color class. That is, ( ) ≥ 3, which is a contradiction. Therefore, a unicyclic graphs with a cycle of length 4 with ( ) = 3 is as shown in Fig. 3. □ Proof. Let be a unicyclic graph that admits tgdcoloring. If is a unicyclic graph with ( ) = 1 then by Theorem 2, ( ) ≤ ( ) ≤ ( ) + 2. It remains to prove that cycles also admit this bound. It is proved in 9 that ( ) = ( ), where is a cycle of length at least 5 and for cycle of length 4, ( ) = ( ) + 2. □

Theorem 17
If is a unicyclic graph with a cycle of length at least 5, then, ( ) ≥ 4. Further, ( ) = 4, then the length of the cycle is 5 or 6.
Proof. As contains a cycle of length more than 4, the first part of the theorem is clear. Assume that ( ) = 4. It remains to prove that the length of the cycle is 5 or 6. Let the length of the cycle be at least 7. Note that ( ) ≥ 5 for cycle of length at least 7. Therefore, ( ) > 4 for unicyclic graphs with cycle of length more than 6. That is, the length of the cycle is either 5 or 6. □

Theorem 18
Let be a unicyclic graph with two or more extreme vertices. Then, ( ) = ( ). Proof. Let be a unicyclic graph with at least two extreme vertices. Then, there exist at least two support vertices such that the distance between them is at least three. Therefore, by Theorem 4, ( ) = ( ). □

Theorem 19
If is a unicyclic graph without extreme vertices and ( ) ≥ 5, then ( ) = ( ). Proof. Let be a unicyclic graph with no extreme vertex. Also, let ( ) ≥ 5 with respect to the coloring . To prove is a tgd-coloring, it is required to show that all vertices of has an anti dom-color class. One may note that each support is solitary. Let be an arbitrary vertex of . If is not a support vertex, then ( ) ≤ 2. As ( ) ≥ 5, is not adjacent to any of the vertices in at least 2 color classes. That is, has an anti dom-color class. If is a support vertex of , then lies on and can be adjacent to vertices in at most three color classes. Therefore, there exist at least one color class which contains no vertex adjacent to . That is, has an anti