#### Abstract

Let be a graph. The vertex (edge) arboricity of denoted by is the minimum number of subsets into which the vertex (edge) set of can be partitioned so that each subset induces an acyclic subgraph. Let be a graphical sequence and let be the class of realizations of . We prove that if , then there exist integers and such that has a realization with if and only if is an integer satisfying . Thus, for an arbitrary graphical sequence and , the two invariants and naturally arise and hence We write for the degree sequence of an -regular graph of order . We prove that . We consider the corresponding extremal problem on vertex arboricity and obtain in all situations and for all .

#### 1. Introduction

We limit our discussion to graphs that are simple and finite. For the most part, our notation and terminology follows that of Chartrand and Lesniak [1]. A typical problem in graph theory deals with decomposition of a graph into various subgraphs possessing some prescribed property. There are ordinarily two problems of this type, one dealing with a decomposition of the vertex set and the other with a decomposition of the edge set. The vertex coloring problem is an example of vertex decomposition while the edge coloring problem is an example of edge decomposition with some additional property. For a graph , it is always possible to partition into subsets , such that each induced subgraph contains no cycle. The *vertex arboricity * of a graph is the minimum number of subsets into which can be partitioned so that each subset induces an acyclic subgraph. Thus, if and only if is a forest. The vertex arboricity was first defined in [2] by Chartrand et al. For a few classes of graphs, the vertex arboricity is easily determined. For example, . . Also if or and otherwise. It is clear that, for every graph of order , . We now turn to the second decomposition problem. The * edge arboricity* or simply the * arboricity * of a nonempty graph is the minimum number of subsets into which can be partitioned so that each subset induces an acyclic subgraph. The arboricity was first defined by Nash-Williams in [3]. As with vertex arboricity, a nonempty graph has arboricity 1 if and only if it is a forest. Also .

A * linear forest* of a graph is a forest of in which each component is a path. The * linear arboricity* is the minimum number of subsets into which can be partitioned so that each subset induces a linear forest. It was first introduced by Harary in [4] and is denoted by . Note that the Greek letter, capital, , looks like three paths! It was proved in [5] that for all cubic graphs and for all 4-regular graphs . It was conjectured that for all -regular graphs .

A sequence of nonnegative integers is called a * degree sequence* of a graph if the vertices of can be labeled so that for all . If a sequence of nonnegative integers is a degree sequence of some graph , then is called * graphical sequence*. In this case, is called a * realization* of . A graph is * regular* of degree if for each vertex of . Such graphs are called *-regular*. We write for the sequence of length , where is a nonnegative integer and is a positive integer. It is well known that an -regular graph of order exists if and only if and . Furthermore, there exists a disconnected -regular graph of order if and only if .

Let be a graph and be independent where . Put

The operation is called a * switching operation*. It is easy to see that the graph obtained from by a switching has the same degree sequence as .

Havel [6] and Hakimi [7] independently obtained that if and are any two realizations of , then can be obtained from by a finite sequence of switchings. Let denote the set of all realizations of degree sequence . As a consequence of this result, Eggleton and Holton [8] defined, in 1978, * the graph * of realizations of whose vertex set is the set , two vertices being adjacent in the graph if one can be obtained from the other by a switching. As a consequence of Havel and Hakimi, we have the following theorem.

Theorem 1.1. *Let be a graphical sequence. Then, the graph is connected.*

Let and be a graphical sequence. Put

A graph parameter is said to satisfy an *intermediate value theorem over a class of graphs * if with , then, for every integer with , there is a graph such that . If a graph parameter satisfies an intermediate value theorem over , then we write IVT. The main purpose of this section is to prove that if , then IVT.

Theorem 1.2. *Let be a graphical sequence and . Then, there exist integers and such that there exists a graph with if and only if is an integer satisfying . That is, .*

The proof of Theorem 1.2 follows from Theorem 1.1 and the following results.

Lemma 1.3. *Let and be a switching on a graph . Then, .*

*Proof. *Let be a graph and be a switching on . Then, and hence . Since , we have that and hence .

Let . By Lemma 1.3, if , then we have that . If , then, by Lemma 1.3, we see that , where .

We have the following corollary.

Corollary 1.4. *Let be a graph and let be a switching on . If , then .*

#### 2. Extremal Results

Let . By Theorem 1.2, is uniquely determined by and , thus it is reasonable to denote and . We first state a famous result on edge arboricity of Nash-Williams [3].

Theorem 2.1. *For every nonempty graph , , where the maximum is taken over all nontrivial induced subgraphs of .*

As a consequence of the theorem, it follows that and .

If and is a subgraph of , then . It is well known that if is a graph with a degree sequence , then can be embedded as an induced subgraph of a -regular graph and . Thus, it is reasonable to determine and in the case where . It should be noted that if , then for all . Therefore, we may assume from now on that .

Theorem 2.2. *. *

*Proof. *Let be an -regular graph of order . By Theorem 2.1, we see that for any -regular graph of order . Since , it follows that . Let be an induced subgraph of of order . If , then . It follows that . If , then . It is clear that . Thus, . Therefore, .

In order to obtain the corresponding extremal results on vertex arboricity, we need to introduce some notation on graph construction.

Let and be finite nonempty sets. We denote by the complete bipartite graph with partite sets and . Let and be any two graphs. Then, is the graph with and . It is clear that the binary operation “” is associative. If and are disjoint (i.e., ), then . We also use for the union of disjoint copies of .

It should be noted once again that an -regular graph of order have if and only if . Thus, if and only if . Moreover, .

Let and be integers with and . Then, (1)if is even and , then there exists an -regular bipartite graph of order . Therefore, ; (2)if is odd and , then is even and there exists an -regular bipartite graph of order . Let be a set of independent edges of . Then, is a bipartite graph of order having vertices of degree and vertices of degree . Let be a graph obtained from by joining the vertices of to a new vertex. Therefore, is an -regular graph of order having ; (3)if is even and , then is the complete -regular bipartite graph of order . Let and be the corresponding partite sets of . Let be defined by . Then is an -regular graph of order having ; (4)if , then is even. Let and be as described in the previous case and be a new vertex. Then, , where , is a graph of order having vertices of degree and vertices of degree . Put and . Thus, .

*Case 1. *If is even, then we can construct an -regular graph from by adding independent edges to each of and . This construction yields .

*Case 2. *If is odd, then is a graph having vertices of degree and vertices of degree . Thus, it has vertices of degree in and also vertices in . Since is even, we can easily construct an -regular graph from by adding appropriate independent edges so that .

With these observations, we easily obtain the following result.

Lemma 2.3. *. *

In order to obtain in all other cases, we first state some known result concerning the forest number of regular graphs. Let be a graph and . is called an * induced forest* of if the subgraph of contains no cycle. The maximum cardinality of an induced forest of a graph is called the * forest number* of and is denoted by . That is,

The second author proved in [9] the following result: Let and be integers with . Then, (1) for all , (2)there exists a graph such that .

With these observations, we have the following result.

Lemma 2.4. *If , then .*

It is well known that . Thus, . It is also well known that , where is a 1-factor of , is the unique -regular graph of order . Since , it follows that a subgraph of induced by any four vertices of contains at most two edges. Equivalently, a subgraph of induced by any four vertices must contain a cycle. Thus, . It is easy to show that . Thus, .

Let be the Turán graph of order containing no clique of order . Thus, there exists a unique pair of integers and such that . Note that is a complete -partite graph of order consisting of partite sets of cardinality and partite sets of cardinality . Therefore, is an -regular graph if and only if . If , then contains vertices of degree and vertices of degree . A little modification of yields a regular graph with prescribed arboricity number as in the following theorem.

Theorem 2.5. *Let , and be positive integers satisfying . Then,*

(1)*there exists an -regular graph of order such that if is odd, *(2)*there exists an -regular graph of order such that if and are even, *(3)*there exists an -regular graph of order such that if is even and is odd.*

*Proof. *Note that is a complete -partite graph of order consisting of partite sets of cardinality and partite sets of cardinality . Let be partite sets of of cardinality and be partite sets of of cardinality . It is clear that if and if . For each , let ; similarly, for each , let .

(1) Suppose that is odd. Since for each , an -regular graph can be obtained from by adding independent edges to each . It is clear that .

(2) Suppose that and are even. Let where is a path with as its vertex set and as its edge set. Let be defined by and put . Finally, we obtain a graph which is an -regular graph of order and .

(3) Suppose that is even and is odd. Then, is odd. If , then, by Dirac [10], a subgraph of induced by contains a Hamiltonian cycle . Since is even and is of order , it follows that is of even order. Let be a set of independent edges of . Then, is an -regular graph of order and .

Suppose that . Let and . Then, is an -regular graph of order and .

Note that if is an -regular graph on vertices where , then . It follows, by Dirac [10], that is Hamiltonian.

Theorem 2.6. *Let and be integers with and . Then,
*

*Proof. *By Lemma 2.4, we have that if . We have already shown that if . Thus, in order to prove this theorem, it suffices to construct an -regular graph of order such that for and . Suppose that . Let and put . Then, and if and only if and is a multiple of .

Note that is a complete -partite graph of order consisting of partite sets of cardinality . Suppose that is odd. Let be a graph obtained from by adding independent edges to each partite set. Then, is an -regular graph of order having .

If is even, then is even. By the same argument with 2 in Theorem 2.5, there exists an -regular graph obtained from with .

Now suppose that and therefore . Thus, contains partite sets of cardinality and partite sets of cardinality . By Theorem 2.5, there exists an -regular graph of order with if is odd or both and are even and there exists an -regular graph of order with if is even and is odd. Since , it follows that . We will consider two cases according to the parity of and as in Theorem 2.5.*Case 1. *If is odd or both and are even, then, by Theorem 2.5, there exists an -regular graph of order with . Since , it follows that . If , then with , as required. Suppose that is even and . By Dirac [10], contains enough edge-disjoint Hamiltonian cycles whose removal produces an -regular graph of order and , as required.

Suppose is odd. Then, either or is odd and therefore is even. Also by Dirac [10], contains a 1-factor where is an -regular graph and . Since is even, the result follows by the same argument as above.*Case 2. *If is even and is odd, then, by Theorem 2.5, there exists an -regular graph of order and . Since is odd, it follows that both and are even. Since , it follows that . Therefore, it follows by the parity that . The result follows easily by the same argument as in Case 1.

This completes the proof.

Chartrand and Kronk [11] obtained a good upper bound for the vertex arboricity. In particular, they proved the following theorem.

Theorem 2.7. *For each graph , , where the maximum is taking over all induced subgraphs of . In particular, if is an -regular graph.*

In general, the bound given in Theorem 2.7 is not sharp but it is sharp in the class of -regular graphs of order as we will prove in the next theorem.

Theorem 2.8. *For .*

*Proof. *The result follows easily if . Suppose that . We write . Thus, . Note that an -regular graph of order exists implying must be even. Thus, if is odd, then must be even and hence must be even. Put , where is an -regular graph of order . Since and , it follows that .

#### Acknowledgments

The first author would like to thank University of the Thai Chamber of Commerce for financial support. The second author gratefully acknowledges the financial support provided by the Thailand Research Fund through the Grant number BRG5280015. The authors wish to thank the referee for useful suggestions and comments that led to an improvement of this paper.