An Intermediate Value Theorem for the Arboricities

Abstract

Let 𝐺 be a graph. The vertex (edge) arboricity of 𝐺 denoted by 𝚊(𝐺)(𝚊₁(𝐺)) is the minimum number of subsets into which the vertex (edge) set of 𝐺 can be partitioned so that each subset induces an acyclic subgraph. Let 𝐝 be a graphical sequence and let ℛ(𝐝) be the class of realizations of 𝐝. We prove that if 𝜋∈{𝚊,𝚊₁}, then there exist integers 𝑥(𝜋) and 𝑦(𝜋) such that 𝐝 has a realization 𝐺 with 𝜋(𝐺)=𝑧 if and only if 𝑧 is an integer satisfying 𝑥(𝜋)≤𝑧≤𝑦(𝜋). Thus, for an arbitrary graphical sequence 𝐝 and 𝜋∈{𝚊,𝚊₁}, the two invariants 𝑥(𝜋)=min(𝜋,𝐝)∶=min{𝜋(𝐺)∶𝐺∈ℛ(𝐝)} and  𝑦(𝜋)=max(𝜋,𝐝)∶=max{𝜋(𝐺)∶𝐺∈ℛ(𝐝)} naturally arise and hence 𝜋(𝐝)∶={𝜋(𝐺)∶𝐺∈ℛ(𝐝)}={𝑧∈ℤ∶𝑥(𝜋)≤𝑧≤𝑦(𝜋)}. We write 𝐝=𝑟^𝑛∶=(𝑟,𝑟,…,𝑟) for the degree sequence of an 𝑟-regular graph of order 𝑛. We prove that 𝚊₁(𝑟^𝑛)={⌈(𝑟+1)/2⌉}. We consider the corresponding extremal problem on vertex arboricity and obtain min(𝚊,𝑟^𝑛) in all situations and max(𝚊,𝑟^𝑛) for all 𝑛≥2𝑟+2.

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 𝐺, it is always possible to partition 𝑉(𝐺) into subsets 𝑉_𝑖,1≤𝑖≤𝑘, such that each induced subgraph ⟨𝑉_𝑖⟩ contains no cycle. The vertex arboricity a(𝐺) of a graph 𝐺 is the minimum number of subsets into which 𝑉(𝐺) can be partitioned so that each subset induces an acyclic subgraph. Thus, 𝚊(𝐺)=1 if and only if 𝐺 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, 𝚊(𝐶_𝑛)=2. 𝚊(𝐾_𝑛)=⌈𝑛/2⌉. Also 𝚊(𝐾_𝑟,𝑠)=1 if 𝑟=1 or 𝑠=1 and 𝚊(𝐾_𝑟,𝑠)=2 otherwise. It is clear that, for every graph 𝐺 of order 𝑛, 𝚊(𝐺)≤⌈𝑛/2⌉. We now turn to the second decomposition problem. The edge arboricity or simply the arboricity 𝚊₁(𝐺) of a nonempty graph 𝐺 is the minimum number of subsets into which 𝐸(𝐺) 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 𝚊₁(𝐾_𝑛)=⌈𝑛/2⌉.

A linear forest of a graph 𝐺 is a forest of 𝐺 in which each component is a path. The linear arboricity is the minimum number of subsets into which 𝐸(𝐺) can be partitioned so that each subset induces a linear forest. It was first introduced by Harary in [4] and is denoted by Ξ(𝐺). Note that the Greek letter, capital, 𝑋𝑖, looks like three paths! It was proved in [5] that Ξ(𝐺)=2 for all cubic graphs 𝐺 and Ξ(𝐺)=3 for all 4-regular graphs 𝐺. It was conjectured that Ξ(𝐺)=⌈(𝑟+1)/2⌉ for all 𝑟-regular graphs 𝐺.

A sequence (𝑑₁,𝑑₂,…,𝑑_𝑛) of nonnegative integers is called a degree sequence of a graph 𝐺 if the vertices of 𝐺 can be labeled 𝑣₁,𝑣₂,…,𝑣_𝑛 so that d𝑒𝑔𝑣_𝑖=𝑑_𝑖 for all 𝑖. If a sequence 𝐝=(𝑑₁,𝑑₂,…,𝑑_𝑛) of nonnegative integers is a degree sequence of some graph 𝐺, then 𝐝 is called graphical sequence. In this case, 𝐺 is called a realization of 𝐝. A graph 𝐺 is regular of degree 𝑟 if d𝑒𝑔𝑣=𝑟 for each vertex 𝑣 of 𝐺. Such graphs are called 𝑟-regular. We write 𝐝=𝑟^𝑛 for the sequence (𝑟,𝑟,…,𝑟) of length 𝑛, where 𝑟 is a nonnegative integer and 𝑛 is a positive integer. It is well known that an 𝑟-regular graph of order 𝑛 exists if and only if 𝑛≥𝑟+1 and 𝑛𝑟≡0(mod2). Furthermore, there exists a disconnected 𝑟-regular graph of order 𝑛 if and only if 𝑛≥2𝑟+2.

Let 𝐺 be a graph and 𝑎𝑏,𝑐𝑑∈𝐸(𝐺) be independent where 𝑎𝑐,𝑏𝑑∉𝐸(𝐺). Put𝐺^{𝜎(𝑎,𝑏;𝑐,𝑑)}=(𝐺−{𝑎𝑏,𝑐𝑑})+{𝑎𝑐,𝑏𝑑}. (1.1)

The operation 𝜎(𝑎,𝑏;𝑐,𝑑) is called a switching operation. It is easy to see that the graph obtained from 𝐺 by a switching has the same degree sequence as 𝐺.

Havel [6] and Hakimi [7] independently obtained that if 𝐺₁ and 𝐺₂ are any two realizations of 𝐝, then 𝐺₂ can be obtained from 𝐺₁ by a finite sequence of switchings. Let ℛ(𝐝) denote the set of all realizations of degree sequence 𝐝. As a consequence of this result, Eggleton and Holton [8] defined, in 1978, the graph ℛ(𝐝) of realizations of 𝐝 whose vertex set is the set ℛ(𝐝), two vertices being adjacent in the graph ℛ(𝐝) 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 𝐝 be a graphical sequence. Then, the graph ℛ(𝐝) is connected.

Let 𝜋∈{𝚊,𝚊₁} and 𝐝 be a graphical sequence. Put 𝜋(𝐝)∶={𝜋(𝐺)∶𝐺∈ℛ(𝐝)}. (1.2)

A graph parameter 𝜋 is said to satisfy an intermediate value theorem over a class of graphs 𝒥 if 𝐺,𝐻∈𝒥 with 𝜋(𝐺)<𝜋(𝐻), then, for every integer 𝑘 with 𝜋(𝐺)≤𝑘≤𝜋(𝐻), there is a graph 𝐾∈𝒥 such that 𝜋(𝐾)=𝑘. If a graph parameter 𝜋 satisfies an intermediate value theorem over 𝒥, then we write (𝜋,𝒥)∈IVT. The main purpose of this section is to prove that if 𝜋∈{𝚊,𝚊₁}, then (𝜋,ℛ(𝐝))∈IVT.

Theorem 1.2. Let 𝐝 be a graphical sequence and 𝜋∈{𝚊,𝚊₁}. Then, there exist integers 𝑥(𝜋) and 𝑦(𝜋) such that there exists a graph 𝐺∈ℛ(𝐝) with 𝜋(𝐺)=𝑧 if and only if 𝑧 is an integer satisfying 𝑥(𝜋)≤𝑧≤𝑦(𝜋). That is, 𝜋(𝐝)={𝑧∈ℤ∶𝑥(𝜋)≤𝑧≤𝑦(𝜋)}.

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

Lemma 1.3. Let 𝜋∈{𝚊,𝚊₁} and 𝜎 be a switching on a graph 𝐺. Then, 𝜋(𝐺^𝜎)≤𝜋(𝐺)+1.

Proof. Let 𝐺 be a graph and 𝜎=𝜎(𝑎,𝑏;𝑐,𝑑) be a switching on 𝐺. Then, 𝚊(𝐺^𝜎−{𝑎,𝑏})≤𝚊(𝐺) and hence 𝚊(𝐺^𝜎)≤𝚊(𝐺^𝜎−{𝑎,𝑏})+1≤𝚊(𝐺)+1. Since 𝑎𝑐,𝑏𝑑∈𝐸(𝐺^𝜎), we have that 𝚊₁(𝐺^𝜎−{𝑎𝑐,𝑏𝑑})≤𝚊₁(𝐺) and hence 𝚊₁(𝐺^𝜎)≤𝚊₁(𝐺^𝜎−{𝑎𝑐,𝑏𝑑})+1≤𝚊₁(𝐺)+1.

Let 𝜋∈{𝑎,𝑎₁}. By Lemma 1.3, if 𝜋(𝐺)≤𝜋(𝐺^𝜎), then we have that 𝜋(𝐺)≤𝜋(𝐺^𝜎)≤𝜋(𝐺)+1. If 𝜋(𝐺)≥𝜋(𝐺^𝜎), then, by Lemma 1.3, we see that 𝜋(𝐺)=𝜋((𝐺^𝜎)^𝜎−1)≤𝜋(𝐺^𝜎)+1≤𝜋(𝐺)+1, where 𝜎⁻¹=𝜎(𝑎,𝑐;𝑏,𝑑).

We have the following corollary.

Corollary 1.4. Let 𝐺 be a graph and let 𝜎 be a switching on 𝐺. If 𝜋∈{𝚊,𝚊₁}, then |𝜋(𝐺)−𝜋(𝐺^𝜎)|≤1.

2. Extremal Results

Let 𝜋∈{𝚊,𝚊₁}. By Theorem 1.2, 𝜋(𝐝) is uniquely determined by 𝑥(𝜋) and 𝑦(𝜋), thus it is reasonable to denote 𝑥(𝜋)=min(𝜋,𝐝)∶=min{𝜋(𝐺)∶𝐺∈ℛ(𝐝)} and 𝑦(𝜋)=max(𝜋,𝐝)∶=max{𝜋(𝐺)∶𝐺∈ℛ(𝐝)}. We first state a famous result on edge arboricity of Nash-Williams [3].

Theorem 2.1. For every nonempty graph 𝐺, 𝚊₁(𝐺)=max⌈|𝐸(𝐻)|/(|𝑉(𝐻)|−1)⌉, where the maximum is taken over all nontrivial induced subgraphs 𝐻 of 𝐺.

As a consequence of the theorem, it follows that 𝚊₁(𝐾_𝑛)=⌈𝑛/2⌉ and 𝚊₁(𝐾_𝑟,𝑠)=⌈𝑟𝑠/(𝑟+𝑠−1)⌉.

If 𝜋∈{𝚊,𝚊₁} and 𝐻 is a subgraph of 𝐺, then 𝜋(𝐻)≤𝜋(𝐺). It is well known that if 𝐺 is a graph with a degree sequence 𝐝=(𝑑₁,𝑑₂,…,𝑑_𝑛)(𝑑₁≥𝑑₂≥⋯≥𝑑_𝑛), then 𝐺 can be embedded as an induced subgraph of a 𝑑₁-regular graph 𝐻 and 𝜋(𝐺)≤𝜋(𝐻). Thus, it is reasonable to determine min(𝜋,𝐝) and max(𝜋,𝐝) in the case where 𝐝=𝑟^𝑛. It should be noted that if 𝑟∈{0,1}, then 𝚊₁(𝐺)=1 for all 𝐺∈ℛ(𝑟^𝑛). Therefore, we may assume from now on that 𝑟≥2.

Theorem 2.2. min(𝚊₁,𝑟^𝑛)=max(𝚊₁,𝑟^𝑛)=⌈(𝑟+1)/2⌉.

Proof. Let 𝐺 be an 𝑟-regular graph of order 𝑛. By Theorem 2.1, we see that 𝚊₁(𝐺)≥⌈𝑛𝑟/(2(𝑛−1))⌉ for any 𝑟-regular graph 𝐺 of order 𝑛. Since 𝑛𝑟/(2(𝑛−1))=𝑟/2+𝑟/(2(𝑛−1)), it follows that 𝚊₁(𝐺)≥⌈(𝑟+1)/2⌉. Let 𝐻 be an induced subgraph of 𝐺 of order 𝑚. If 𝑚≥𝑟+1, then |𝐸(𝐻)|/(|𝑉(𝐻)|−1)≤𝑚𝑟/(2(𝑚−1)). It follows that |𝐸(𝐻)|/(|𝑉(𝐻)|−1)≤(𝑟+1)/2. If 𝑚≤𝑟, then |𝐸(𝐻)|≤(^𝑚₂). It is clear that |𝐸(𝐻)|/(|𝑉(𝐻)|−1)≤(𝑟+1)/2. Thus, 𝚊₁(𝐺)≤(𝑟+1)/2. Therefore, min(𝚊₁,𝑟^𝑛)=max(𝚊₁,𝑟^𝑛)=⌈(𝑟+1)/2⌉.

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

Let 𝑋 and 𝑌 be finite nonempty sets. We denote by 𝐾(𝑋,𝑌) the complete bipartite graph with partite sets 𝑋 and 𝑌. Let 𝐺 and 𝐻 be any two graphs. Then, 𝐺∗𝐻 is the graph with 𝑉(𝐺∗𝐻)=𝑉(𝐺)∪𝑉(𝐻) and 𝐸(𝐺∗𝐻)=𝐸(𝐺)∪𝐸(𝐻). It is clear that the binary operation “∗” is associative. If 𝐺₁ and 𝐺₂ are disjoint (i.e., 𝑉(𝐺₁)∩𝑉(𝐺₂)=∅), then 𝐺₁∗𝐺₂=𝐺₁∪𝐺₂. We also use 𝑝𝐺 for the union of 𝑝 disjoint copies of 𝐺.

It should be noted once again that an 𝑟-regular graph 𝐺 of order 𝑛 have 𝚊(𝐺)=1 if and only if 𝑟∈{0,1}. Thus, min(𝚊,𝑟^𝑛)=max(𝚊,𝑟^𝑛)=1 if and only if 𝑟∈{0,1}. Moreover, min(𝚊,2^𝑛)=max(𝚊,2^𝑛)=2.

Let 𝑟 and 𝑛 be integers with 𝑟≥3 and ℛ(𝑟^𝑛)≠∅. Then, (1)if 𝑛 is even and 𝑛≥2𝑟, then there exists an 𝑟-regular bipartite graph 𝐺 of order 𝑛. Therefore, 𝚊(𝐺)=2; (2)if 𝑛 is odd and 𝑛≥2𝑟+1, then 𝑟 is even and there exists an 𝑟-regular bipartite graph 𝐻 of order 𝑛−1. Let 𝐹 be a set of 𝑟/2 independent edges of 𝐻. Then, 𝐻−𝐹 is a bipartite graph of order 𝑛 having 𝑟 vertices of degree 𝑟−1 and 𝑛−1−𝑟 vertices of degree 𝑟. Let 𝐺 be a graph obtained from 𝐻−𝐹 by joining the 𝑟 vertices of 𝐻−𝐹 to a new vertex. Therefore, 𝐺 is an 𝑟-regular graph of order 𝑛 having 𝚊(𝐺)=2; (3)if 𝑛 is even and 𝑛=2(𝑟−1), then 𝐾_{𝑟−1,𝑟−1} is the complete (𝑟−1)-regular bipartite graph of order 𝑛. Let 𝑋={𝑥₁,𝑥₂,…,𝑥_𝑟−1} and 𝑌={𝑦₁,𝑦₂,…,𝑦_𝑟−1} be the corresponding partite sets of 𝐾_{𝑟−1,𝑟−1}. Let 𝐺 be defined by 𝐺=(𝐾_{𝑟−1,𝑟−1}−{𝑥₂𝑦₂,𝑥₃𝑦₃,…,𝑥_𝑟−2𝑦_𝑟−2})+{𝑥₁𝑥₂,𝑥₂𝑥₃,…,𝑥_𝑟−2𝑥_𝑟−1,𝑦₁𝑦₂,𝑦₂𝑦₃,…,𝑦_𝑟−2𝑦_𝑟−1}. Then 𝐺 is an 𝑟-regular graph of order 𝑛 having 𝚊(𝐺)=2; (4)if 𝑛=2𝑟−1, then 𝑟 is even. Let 𝑋 and 𝑌 be as described in the previous case and 𝑢 be a new vertex. Then, 𝐻=𝐾(𝑋,𝑌)+𝐸₁, where 𝐸₁={𝑢𝑥₁,𝑢𝑦₁,𝑢𝑥₂,𝑢𝑦₂,…,𝑢𝑥_𝑟/2,𝑢𝑦_𝑟/2}, is a graph of order 2𝑟−1 having 𝑟+1 vertices of degree 𝑟 and 𝑟−2 vertices of degree 𝑟−1. Put 𝑋₁={𝑥_𝑖∈𝑋∶𝑟/2<𝑖≤𝑟−1} and 𝑌₁={𝑦_𝑖∈𝑌∶𝑟/2<𝑖≤𝑟−1}. Thus, |𝑋₁|=|𝑌₁|=(𝑟−2)/2.

Case 1. If (𝑟−2)/2 is even, then we can construct an 𝑟-regular graph 𝐺 from 𝐻 by adding (𝑟−2)/4 independent edges to each of 𝑋₁ and 𝑌₁. This construction yields 𝚊(𝐺)=2.

Case 2. If (𝑟−2)/2 is odd, then 𝐻−𝑥_𝑟/2𝑦_𝑟/2 is a graph having 𝑟−1 vertices of degree 𝑟 and 𝑟 vertices of degree 𝑟−1. Thus, it has 𝑟/2 vertices of degree 𝑟−1 in 𝑋 and also 𝑟/2 vertices in 𝑌. Since 𝑟/2 is even, we can easily construct an 𝑟-regular graph 𝐺 from 𝐻−𝑥_𝑟/2𝑦_𝑟/2 by adding appropriate independent edges so that 𝚊(𝐺)=2.

With these observations, we easily obtain the following result.

Lemma 2.3. min(𝚊,𝑟^𝑛)=2forall𝑛≥2𝑟−2.

In order to obtain min(𝚊,𝑟^𝑛) in all other cases, we first state some known result concerning the forest number of regular graphs. Let 𝐺 be a graph and 𝐹⊆𝑉(𝐺). 𝐹 is called an induced forest of 𝐺 if the subgraph ⟨𝐹⟩ of 𝐺 contains no cycle. The maximum cardinality of an induced forest of a graph 𝐺 is called the forest number of 𝐺 and is denoted by 𝚏(𝐺). That is,𝚏(𝐺)∶=max{|𝐹|∶𝐹isaninducedforestof𝐺}. (2.1)

The second author proved in [9] the following result: Let 𝑟 and 𝑠 be integers with 1≤𝑠≤𝑟. Then, (1)f(𝐺)≤𝑠+1 for all 𝐺∈ℛ(𝑟^𝑟+𝑠), (2)there exists a graph 𝐻∈ℛ(𝑟^𝑟+𝑠) such that 𝚏(𝐻)=𝑠+1.

With these observations, we have the following result.

Lemma 2.4. If 1≤𝑠≤𝑟−3, then min(𝚊,𝑟^𝑟+𝑠)≥⌈(𝑟+𝑠)/(𝑠+1)⌉.

It is well known that 𝚊(𝐾_𝑟+1)=⌈(𝑟+1)/2⌉. Thus, min(𝚊,𝑟^𝑟+1)=max(𝚊,𝑟^𝑟+1)=⌈(𝑟+1)/2⌉. It is also well known that 𝐺=𝐾_𝑟+2−𝐹, where 𝐹 is a 1-factor of 𝐾_𝑟+2, is the unique 𝑟-regular graph of order 𝑟+2. Since 𝐸(𝐺)=𝐹, it follows that a subgraph of 𝐺 induced by any four vertices of 𝑉(𝐺) contains at most two edges. Equivalently, a subgraph of 𝐺 induced by any four vertices must contain a cycle. Thus, 𝚊(𝐺)≥⌈(𝑟+2)/3⌉. It is easy to show that 𝚊(𝐺)=⌈(𝑟+2)/3⌉. Thus, min(𝚊,𝑟^𝑟+2)=max(𝚊,𝑟^𝑟+2)=⌈(𝑟+2)/3⌉.

Let 𝑇_𝑛,𝑞 be the Turán graph of order 𝑛 containing no clique of order 𝑞+1. Thus, there exists a unique pair of integers 𝑥 and 𝑦 such that 𝑛=𝑞𝑥+𝑦,0≤𝑦<𝑞. Note that 𝑇_𝑛,𝑞 is a complete 𝑞-partite graph of order 𝑛 consisting of 𝑦 partite sets of cardinality 𝑥+1 and 𝑞−𝑦 partite sets of cardinality 𝑥. Therefore, 𝑇_𝑛,𝑞 is an (𝑛−𝑥)-regular graph if and only if 𝑦=0. If 𝑦>0, then 𝑇_𝑛,𝑞 contains 𝑦(𝑥+1) vertices of degree 𝑛−𝑥−1 and (𝑞−𝑦)𝑥 vertices of degree 𝑛−𝑥. A little modification of 𝑇_𝑛,𝑞 yields a regular graph with prescribed arboricity number as in the following theorem.

Theorem 2.5. Let 𝑛,𝑥,𝑞, and 𝑦 be positive integers satisfying 𝑛=𝑞𝑥+𝑦,0<𝑦<𝑞. Then,
(1)there exists an (𝑛−𝑥)-regular graph 𝐺 of order 𝑛 such that 𝚊(𝐺)≤𝑞 if 𝑥 is odd, (2)there exists an (𝑛−𝑥)-regular graph 𝐺 of order 𝑛 such that 𝚊(𝐺)≤𝑞 if 𝑥 and 𝑦 are even, (3)there exists an (𝑛−𝑥−1)-regular graph 𝐻 of order 𝑛 such that 𝚊(𝐺)≤𝑞 if 𝑥 is even and 𝑦 is odd.

Proof. Note that 𝑇_𝑛,𝑞 is a complete 𝑞-partite graph of order 𝑛 consisting of 𝑦 partite sets of cardinality 𝑥+1 and 𝑞−𝑦 partite sets of cardinality 𝑥. Let 𝑉₁,𝑉₂,…,𝑉_𝑦 be partite sets of 𝑇_𝑛,𝑞 of cardinality 𝑥+1 and 𝑈₁,𝑈₂,…,𝑈_𝑞−𝑦 be partite sets of 𝑇_𝑛,𝑞 of cardinality 𝑥. It is clear that deg_{𝑇𝑛,𝑞}𝑣=𝑛−𝑥−1 if 𝑣∈𝑉₁∪𝑉₂∪⋯∪𝑉_𝑦 and deg_{𝑇𝑛,𝑞}𝑢=𝑛−𝑥 if 𝑢∈𝑈₁∪𝑈₂∪⋯∪𝑈_𝑞−𝑦. For each 𝑖=1,2,…𝑦, let 𝑉_𝑖={𝑣_𝑖1,𝑣_𝑖2,…,𝑣_𝑖(𝑥+1)}; similarly, for each 𝑗=1,2,…,𝑞−𝑦, let 𝑈_𝑗={𝑢_𝑗1,𝑢_𝑗2,…,𝑢_𝑗𝑥}.
(1) Suppose that 𝑥 is odd. Since |𝑉_𝑖|=𝑥+1 for each 𝑖=1,2,…,𝑦, an (𝑛−𝑥)-regular graph 𝐺 can be obtained from 𝑇_𝑛,𝑞 by adding (𝑥+1)/2 independent edges to each 𝑉_𝑖,1≤𝑖≤𝑦. It is clear that 𝚊(𝐺)≤𝑞.
(2) Suppose that 𝑥 and 𝑦 are even. Let 𝐻=⋃^𝑦_𝑖=1𝑃(𝑉_𝑖) where 𝑃(𝑉_𝑖) is a path with 𝑉_𝑖 as its vertex set and 𝐸_𝑖={𝑣_𝑖𝑘𝑣_𝑖(𝑘+1)∶1≤𝑘≤𝑥} as its edge set. Let 𝐹₁,𝐹₂,…,𝐹_𝑦/2 be defined by 𝐹_𝑘={𝑣_𝑘𝑡𝑣_{(𝑘+𝑦/2)𝑡}∶2≤𝑡≤𝑥} and put 𝐹=⋃^𝑦/2_𝑘=1𝐹_𝑘. Finally, we obtain a graph 𝐺=(𝑇_𝑛,𝑞−𝐹)∗𝐻 which is an (𝑛−𝑥)-regular graph of order 𝑛 and 𝚊(𝐺)≤𝑞.
(3) Suppose that 𝑥 is even and 𝑦 is odd. Then, 𝑛=𝑞𝑥+𝑦 is odd. If 𝑞−𝑦≥2, then, by Dirac [10], a subgraph of 𝑇_𝑛,𝑞 induced by 𝑈₁∪𝑈₂∪⋯∪𝑈_𝑞−𝑦 contains a Hamiltonian cycle 𝐶. Since 𝑥 is even and 𝐶 is of order 𝑥(𝑞−𝑦), it follows that 𝐶 is of even order. Let 𝐹 be a set of 𝑥(𝑞−𝑦)/2 independent edges of 𝐶. Then, 𝐻=𝑇_𝑛,𝑞−𝐹 is an (𝑛−𝑥−1)-regular graph of order 𝑛 and 𝚊(𝐻)≤𝑞.
Suppose that 𝑞−𝑦=1. Let 𝐹₁={𝑣₁₁𝑢₁₁,𝑣₁₂𝑢₁₂,…,𝑣_1𝑥𝑢_1𝑥} and 𝐹₂={𝑣₁₁𝑣₁₂,𝑣₁₃𝑣₁₄,…,𝑣_1(𝑥−1)𝑣_1𝑥}. Then, 𝐻=(𝑇_𝑛,𝑞−𝐹₁)+𝐹₂ is an (𝑛−𝑥−1)-regular graph of order 𝑛 and 𝚊(𝐻)≤𝑞.

Note that if 𝐺 is an 𝑟-regular graph on 𝑟+𝑠 vertices where 1≤𝑠≤𝑟, then 𝑟≥(𝑟+𝑠)/2. It follows, by Dirac [10], that 𝐺 is Hamiltonian.

Theorem 2.6. Let 𝑟 and 𝑠 be integers with 𝑟≥3 and 1≤𝑠≤𝑟−3. Then, min𝚊,𝑟^𝑟+𝑠=𝑟+𝑠.𝑠+1 (2.2)

Proof. By Lemma 2.4, we have that min(𝚊,𝑟^𝑟+𝑠)≥⌈(𝑟+𝑠)/(𝑠+1)⌉ if 1≤s≤𝑟−3. We have already shown that min(𝚊,𝑟^𝑟+𝑠)=⌈(𝑟+𝑠)/(𝑠+1)⌉ if 𝑠∈{1,2}. Thus, in order to prove this theorem, it suffices to construct an 𝑟-regular graph 𝐺 of order 𝑟+𝑠 such that 𝚊(𝐺)≤⌈(𝑟+𝑠)/(𝑠+1)⌉ for 𝑟≥3 and 3≤𝑠≤𝑟−3. Suppose that 3≤𝑠≤𝑟−3. Let 𝑞=⌈(𝑟+𝑠)/(𝑠+1)⌉ and put 𝑟+𝑠=𝑞𝑥+𝑦,0≤𝑦≤𝑞−1. Then, 𝑥≤𝑠+1 and 𝑥=𝑠+1 if and only if 𝑦=0 and 𝑟+𝑠 is a multiple of 𝑠+1.
Note that 𝑇_𝑛,𝑞 is a complete 𝑞-partite graph of order 𝑛 consisting of 𝑞 partite sets of cardinality 𝑠+1. Suppose that 𝑠 is odd. Let 𝐺 be a graph obtained from 𝑇_𝑛,𝑞 by adding (𝑠+1)/2 independent edges to each partite set. Then, 𝐺 is an 𝑟-regular graph of order 𝑛 having 𝚊(𝐺)=𝑞.
If 𝑠 is even, then 𝑞 is even. By the same argument with 2 in Theorem 2.5, there exists an 𝑟-regular graph 𝐺 obtained from 𝑇_𝑛,𝑞 with 𝚊(𝐺)=𝑞.
Now suppose that 𝑦>0 and therefore 𝑥≤𝑠. Thus, 𝑇_{𝑟+𝑠,𝑞} contains 𝑦 partite sets of cardinality 𝑥+1 and 𝑞−𝑦 partite sets of cardinality 𝑥. By Theorem 2.5, there exists an (𝑟+𝑠−𝑥)-regular graph 𝐺 of order 𝑟+𝑠 with 𝚊(𝐺)≤𝑞 if 𝑥 is odd or both 𝑥 and 𝑦 are even and there exists an (𝑟+𝑠−𝑥−1)-regular graph 𝐻 of order 𝑟+𝑠 with 𝚊(𝐻)≤𝑞 if 𝑥 is even and 𝑦 is odd. Since 𝑥≤𝑠, it follows that 𝑟+𝑠−𝑥≥𝑟. We will consider two cases according to the parity of 𝑥 and 𝑦 as in Theorem 2.5.
Case 1. If 𝑥 is odd or both 𝑥 and 𝑦 are even, then, by Theorem 2.5, there exists an (𝑟+𝑠−𝑥)-regular graph 𝐺 of order 𝑟+𝑠 with 𝚊(𝐺)≤𝑞. Since 𝑥≤𝑠, it follows that 𝑟+𝑠−𝑥≥𝑟. If 𝑠−𝑥=0, then 𝐺∈ℛ(𝑟^𝑟+𝑠) with 𝚊(𝐺)≤𝑞, as required. Suppose that 𝑠−𝑥 is even and 𝑠−𝑥≥2. By Dirac [10], 𝐺 contains enough edge-disjoint Hamiltonian cycles whose removal produces an 𝑟-regular graph 𝐺^′ of order 𝑟+𝑠 and 𝚊(𝐺^′)≤𝚊(𝐺)≤𝑞, as required.
Suppose 𝑠−𝑥 is odd. Then, either 𝑟+𝑠−𝑥 or 𝑟 is odd and therefore 𝑟+𝑠 is even. Also by Dirac [10], 𝐺 contains a 1-factor 𝐹 where 𝐺−𝐹 is an (𝑟+𝑠−𝑥−1)-regular graph and 𝚊(𝐺−𝐹)≤𝚊(𝐺)≤𝑞. Since 𝑠−𝑥−1 is even, the result follows by the same argument as above.
Case 2. If 𝑥 is even and 𝑦 is odd, then, by Theorem 2.5, there exists an (𝑟+𝑠−𝑥−1)-regular graph 𝐻 of order 𝑟+𝑠 and 𝚊(𝐻)≤𝑞. Since 𝑟+𝑠=𝑞𝑥+𝑦 is odd, it follows that both 𝑟+𝑠−𝑥−1 and 𝑟 are even. Since 𝑥≤𝑠, it follows that 𝑟+𝑠−𝑥−1≥𝑟−1. Therefore, it follows by the parity that 𝑟+𝑠−𝑥−1≥𝑟. 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 𝐺, 𝚊(𝐺)≤1+⌊max𝛿(𝐺^′)/2⌋, where the maximum is taking over all induced subgraphs 𝐺^′ of 𝐺. In particular, 𝚊(𝐺)≤1+⌊𝑟/2⌋ if 𝐺 is an 𝑟-regular graph.

In general, the bound given in Theorem 2.7 is not sharp but it is sharp in the class of 𝑟-regular graphs of order 𝑛≥2𝑟+2 as we will prove in the next theorem.

Theorem 2.8. For 𝑛≥2𝑟+2,max(𝚊,𝑟^𝑛)=1+⌊𝑟/2⌋.

Proof. The result follows easily if 𝑟∈{0,1,2}. Suppose that 𝑟≥3. We write 𝑛=(𝑟+1)𝑞+𝑡,0≤𝑡≤𝑟. Thus, 𝑞≥2. Note that an 𝑟-regular graph of order 𝑛 exists implying 𝑟𝑛 must be even. Thus, if 𝑟 is odd, then 𝑛 must be even and hence 𝑡 must be even. Put 𝐺=(𝑞−1)𝐾_𝑟+1∪𝐻, where 𝐻 is an 𝑟-regular graph of order 𝑟+𝑡+1. Since 𝚊(𝐾_𝑟+1)=⌈(𝑟+1)/2⌉=1+⌊𝑟/2⌋ and 𝚊(𝐻)≤1+⌊𝑟/2⌋, it follows that 𝚊(𝐺)=1+⌊𝑟/2⌋.

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.