On the Strong Equitable Vertex 2-Arboricity of Complete Bipartite Graphs

Abstract

An equitable partition of a graph G is a partition of the vertex set of G such that the sizes of any two parts differ by at most one. The strong equitable vertexk-arboricity of G, denoted by $v{a}_k^{\equiv }\left(G\right)$, is the smallest integer t such that G can be equitably partitioned into $t^{\prime }$ induced forests for every $t^{\prime }\ge t$, where the maximum degree of each induced forest is at most k. In this paper, we provide a general upper bound for $v{a}_2^{\equiv }\left(K_{n,n}\right)$. Exact values are obtained in some special cases.

1. Introduction

All graphs considered in this paper are finite, undirected and simple. For a real number x, $\lceil x\rceil$ is the least integer not less than x and $\lfloor x\rfloor$ is the largest integer not larger than x. For a graph G, we use $V\left(G\right)$ and $E\left(G\right)$ to denote the vertex set and edge set, respectively. For a complete bipartite graph $K_{n,n}$, we let $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$ be the partite sets of $K_{n,n}$. For other undefined concepts, we refer the readers to [1].

The vertex arboricity of a graph G, denoted by $va\left(G\right)$, is the minimum number of subsets into which the vertex set $V\left(G\right)$ can be partitioned so that each subset induces a forest. This notion was first introduced by Chartrand and Kronk in 1969 [2].

Wu, Zhang and Li [3] introduced an equitable version of the vertex arboricity in 2013. An equitable partition of a graph G is a partition of the vertex set of G such that the sizes of any two parts differ by at most one. The equitable partition problem was first introduced by Meyer [4], motivated by a practical application to municipal garbage collection [5]. In this context, the vertices of the graph represent garbage collection routes. A pair of vertices share an edge if the corresponding routes should not be run on the same day. It is desirable that the number of routes run on each day be approximately the same. Therefore, the problem of assigning one of the six weekly working days to each route reduces to finding a proper equitable partition which has six parts. The equitable partition problem has many applications such as in scheduling, constructing timetables and load balance in parallel memory systems, please see [6,7,8,9,10,11].

Moreover, Wu, Zhangf and Li [3] consider a restriction on the maximum degree of each induced forest. The equitable vertex k-arboricity of a graph G, denoted by $v{a}_k^{=}\left(G\right)$, is the minimum number of induced forests into which G can be equitably partitioned, where the maximum degree of each induced forest is at most k. The strong equitable vertexk-arboricity of G, denoted by $v{a}_k^{\equiv }\left(G\right)$, is the smallest integer t such that G can be equitably partitioned into $t^{\prime }$ induced forests for every $t^{\prime }\ge t$, where the maximum degree of each induced forest is at most k.

Note that $v{a}_k^{=}\left(G\right)$ and $v{a}_k^{\equiv }\left(G\right)$ may vary greatly. For example, it is easy to see that $v{a}_{\infty }^{=}\left(K_{n,n}\right)=2$, but $v{a}_{\infty }^{\equiv }\left(K_{n,n}\right)=2\lfloor (\sqrt{8n+9}-1)/4\rfloor$ if $2n=t(t+3)$ and t is odd (see [3]).

Wu et al. [3] first investigated the strong equitable vertex 1-arboricity of complete bipartite graphs. They provided a sharp upper bound for $v{a}_1^{\equiv }\left(K_{n,n}\right)$ in general case:

Lemma 1

([3]). $v{a}_1^{\equiv }\left(K_{n,n}\right)\le 2\lfloor \frac{n+1}{3}\rfloor$.

Noting that the upper bound given in Lemma 1 is not very tight for some special graphs, Wu et al. [3] commented that determining the strong equitable 1-arboricity for every $K_{n,n}$ seems not to be an easy task. Furthermore, the exact values of some special cases were studied by Wu et al. [3] and Tao [12]. Concerning $v{a}_{\infty }^{\equiv }\left(G\right)$, Wu et al. [3] posed the following conjectures:

Conjecture 1

([3]). $v{a}_{\infty }^{\equiv }\left(G\right)\le \lceil \frac{\Delta \left(G\right)+1}{2}\rceil$ for every graph.

Conjecture 2

([3]). There is a constant c such that $v{a}_{\infty }^{\equiv }\left(G\right)\le c$ for every planar graph G.

Conjecture 1 was confirmed by Esperet, Lemoine and Maffray [13], who proved that $v{a}_{\infty }^{\equiv }\left(G\right)\le 4$ for every planar graph G. For Conjecture 1, it is still widely open. As far as we know, Conjecture 1 has been verified for graphs with $\Delta \left(G\right)\ge \frac{\left|V\right(G\left)\right|-1}{2}$ [14], subcubic graphs [15], and 5-degenerate graphs [16].

In this paper, we focus on the strong equitable vertex 2-arboricity of $K_{n,n}$. In Section 2, we provide a general upper bound for $v{a}_2^{\equiv }\left(K_{n,n}\right)$. In Section 3, Section 4 and Section 5, we make our efforts to improve this upper bound where $n\equiv 0,1,2\left(\mathrm{mod}3\right)$. In some special cases, the exact values of $v{a}_2^{\equiv }\left(K_{n,n}\right)$ are determined.

2. Upper Bound for ${\mathit{va}}_{\mathbf{2}}^{\equiv }\left({\mathit{K}}_{\mathit{n},\mathit{n}}\right)$

In this section, we provide an upper bound for $v{a}_2^{\equiv }\left(K_{n,n}\right)$.

Lemma 2

([3]). The complete bipartite graph $K_{n,n}$ can be equitably partitioned into t induced forests with the maximum degree at most k for every even integer $t\ge 2$.

Theorem 1.

Let $K_{n,n}$ be a complete bipartite graph. Then,

$$v{a}_2^{\equiv }\left(K_{n,n}\right)\le 2\lfloor \frac{n+1}{3}\rfloor .$$

Proof. 

It suffices to show that $K_{n,n}$ can be equitably partitioned into q induced forests for every $q\ge 2\lfloor \frac{n+1}{3}\rfloor$, where the maximum degree of each forest is at most 2. Let $\lfloor \frac{n+1}{3}\rfloor =m$. We split the proof into three parts:

Case 1.$n+1=3m$.

We partition $V\left(K_{n,n}\right)$ into q subsets equitably, that is, each subset has size $\lfloor 2n/q\rfloor$ or $\lceil 2n/q\rceil$. Since $q\ge 2m$, then $\frac{2n}{q}\le \frac{2n}{2m}=\frac{6m-2}{2m}<3$, hence each subset contains no more than three vertices. It follows that each subset induces a forest with the maximum degree at most 2.

Case 2.$n+1=3m+1$.

We partition $V\left(K_{n,n}\right)$ into q subsets equitably. Since $\frac{2n}{q}\le \frac{2n}{2m}=\frac{6m}{2m}=3$, one can easily check that each subset induces a forest with the maximum degree at most 2.

Case 3.$n+1=3m+2$.

subcase 3.1. If $q\ge 2m+1$, then $\frac{2n}{q}\le \frac{6m+2}{2m+1}<3$. We can partition $V\left(K_{n,n}\right)$ into q subsets equitably. Each subset induces a forest with the maximum degree at most 2.

subcase 3.2. If $q=2m$, By Lemma 2, the conclusion holds. □

It is worth noting that the upper bound for $v{a}_2^{\equiv }\left(K_{n,n}\right)$ is the same as that for $v{a}_1^{\equiv }\left(K_{n,n}\right)$. The upper bound in Theorem 1 is helpful to determine the exact value of $v{a}_2^{\equiv }\left(K_{n,n}\right)$ for small n. For example, we can deduce from Theorem 1 that $v{a}_2^{\equiv }\left(K_{6,6}\right)=4$(It is not difficult to check that $v{a}_2^{\equiv }\left(K_{6,6}\right)>3$). However, for large n, we need to improve this bound.

The following lemma is useful in our proofs.

Lemma 3.

If the cardinality of a subset is at least 4 in any equitable partition of $K_{n,n}$ where each subset induces a forest with the maximum degree at most 2, then the vertices of this subset belong to the same partite set.

Proof. 

Suppose, to the contrary, that all the vertices of this subset belong to different partite sets of $K_{n,n}$. Since it contains at least four vertices, the subgraph induced by all the vertices of this subset either has a cycle or has maximum degree at least 3, a contradiction. □

In Section 3, Section 4 and Section 5, we study the cases where $n\equiv 0,1,2\left(\mathrm{mod}3\right)$ and improve the upper bound.

3. ${\mathit{va}}_{\mathbf{2}}^{\equiv }\left({\mathit{K}}_{\mathbf{3}\mathit{t},\mathbf{3}\mathit{t}}\right)$

Theorem 2.

If $t\ge 4m$ with $m\ge 0$, then $v{a}_2^{\equiv }\left(K_{3t,3t}\right)\le 2t-2m$.

Proof. 

We prove it by induction on m. If $m=0$, the result holds by Theorem 1. Suppose $m\ge 1$. Since $t\ge 4m>4(m-1)$, by Lemma 2 and the induction hypothesis, it suffices to show that $K_{3t,3t}$ can be equitably partitioned into $2t-2m+1$ induced forests with the maximum degree of each forest at most 2. We equitably partition X into $t-m$ subsets and Y into $t-m+1$ subsets, respectively. It is straightforward to check that each subset contains either three or four vertices and each subset forms an independent set. The proof of the theorem is complete. □

Compared with Theorem 1, Theorem 2 improves the upper bound for $v{a}_2^{\equiv }\left(K_{n,n}\right)$ greatly, especially for large n. For example, we can deduce from Theorem 1 that $v{a}_2^{\equiv }\left(K_{48,48}\right)\le 2\times 16=32$. In fact, this upper bound can be improved to 24 using Theorem 2.

In the following, we make our efforts to determine the exact values of $v{a}_2^{\equiv }\left(K_{3t,3t}\right)$ in some special cases.

Theorem 3.

For $t=1\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)=\frac{3t+1}{2}$.

Proof. 

If $t=1$, it is trivial since $v{a}_2^{\equiv }\left(K_{3,3}\right)=2=\frac{3+1}{2}$. Next, suppose $t=4m+1$ with $m\ge 1$. If $K_{3t,3t}$ can be equitably partitioned into $6m+1$ induced forests with maximum degree at most 2, then the size of every subset is at least 4 because $\lfloor \frac{6t}{6m+1}\rfloor =\lfloor \frac{24m+6}{6m+1}\rfloor =4$. By Lemma 3, the vertices of every subset belong to the same partite set. Without loss of generality, suppose that X is partitioned into at least $3m+1$ subsets as required. It follows that $\left|X\right|\ge 4(3m+1)>12m+3=3t=\left|X\right|$, a contradiction. It implies that $v{a}_2^{\equiv }\left(K_{3t,3t}\right)\ge 6m+2=\frac{3t+1}{2}$. By Theorem 2, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)\le 2t-2m=2t-\frac{t-1}{2}=\frac{3t+1}{2}$. This completes the proof of Theorem 3. □

Theorem 4.

For $t=2\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)=\frac{3t+2}{2}$.

Proof. 

Let $t=4m+2$. By Theorem 2, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)\le 2t-2m=2t-\frac{t-2}{2}=\frac{3t+2}{2}$. It suffices to show that $K_{3t,3t}$ cannot be equitably partitioned into $6m+3$ induced forests, where the maximum degree of each forest is at most 2. Thus, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)\ge 6m+4=\frac{3t+2}{2}$. Suppose, to the contrary, $K_{3t,3t}$ can be partitioned into $6m+3$ induced forests as desired. Notice that $\frac{6t}{6m+3}=\frac{24m+12}{6m+3}=4$. Thus, the vertices of every subset belong to the same partite set. On the other hand, $\left|X\right|=\left|Y\right|=3t=12m+6$, which is not a multiple of 4, a contradiction. This completes our proof. □

4. ${\mathit{va}}_{\mathbf{2}}^{\equiv }\left({\mathit{K}}_{\mathbf{3}\mathit{t}+\mathbf{1},\mathbf{3}\mathit{t}+\mathbf{1}}\right)$

Theorem 5.

If $t\ge 4m+1$ with $m\ge 0$, then $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)\le 2t-2m$.

Proof. 

We prove this theorem by induction on m. If $m=0$, the result holds by Theorem 1. Suppose $m\ge 1$. Since $t\ge 4m+1>4(m-1)+1$, by Lemma 2 and the induction hypothesis, it suffices to show that $K_{3t+1,3t+1}$ can be equitably partitioned into $2t-2m+1$ induced forests with the maximum degree of each forest at most 2. We equitably partition X into $t-m$ subsets and Y into $t-m+1$ subsets, respectively. It is straightforward to check that each subset contains either three or four vertices and each subset forms an independent set. The proof of the theorem is complete. □

For some special cases, the exact values of $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)$ can be determined by Theorem 5 and Lemma 3.

Theorem 6.

For $t=2\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)=\frac{3t+2}{2}$.

Proof. 

Let $t=4m+2$. By Theorem 5, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)\le 2t-2m=2t-\frac{t-2}{2}=\frac{3t+2}{2}$. In the following, we show that $K_{3t+1,3t+1}$ cannot be equitably partitioned into $6m+3$ induced forests, where the maximum degree of each forest is at most 2. Thus, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)\ge 6m+4=\frac{3t+2}{2}$, which completes our proof. Suppose, to the contrary, that $K_{3t+1,3t+1}$ can be partitioned into $6m+3$ induced forests as desired. Then, the size of every subset is not less than 4 because $\lfloor \frac{6t+2}{6m+3}\rfloor =\lfloor \frac{24m+14}{6m+3}\rfloor =4$. By Lemma 3, the vertices of every subset belong to the same partite set. Suppose that X is partitioned into at least $3m+2$ subsets as required, without loss of generality. It implies that $\left|X\right|\ge 4(3m+2)>12m+7=3t+1=\left|X\right|$, a contradiction. □

Theorem 7.

For $t=3\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)=\frac{3t+3}{2}$.

Proof. 

Let $t=4m+3$. By Theorem 5, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)\le 2t-2m=2t-\frac{t-3}{2}=\frac{3t+3}{2}$. In the following, we will show that $K_{3t+1,3t+1}$ cannot be equitably partitioned into $6m+5$ induced forests, where the maximum degree of each forest is at most 2. Thus, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)\ge 6m+6=\frac{3t+3}{2}$. Suppose, to the contrary, that $K_{3t+1,3t+1}$ can be partitioned into $6m+5$ induced forests as desired. Note that $\frac{6t+2}{6m+5}=\frac{24m+20}{6m+5}=4$. The vertices of every subset belong to the same partite set. Nevertheless, $\left|X\right|=\left|Y\right|=3t+1=12m+10$, which is not a multiple of 4, a contradiction. □

5. ${\mathit{va}}_{\mathbf{2}}^{\equiv }\left({\mathit{K}}_{\mathbf{3}\mathit{t}+\mathbf{2},\mathbf{3}\mathit{t}+\mathbf{2}}\right)$

We first prove a useful lemma.

Lemma 4.

$v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\le 2t$.

Proof. 

By Theorem 1, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\le 2\lfloor \frac{3t+2+1}{3}\rfloor =2t+2$. By Lemma 2, it suffices to prove that $K_{3t+2,3t+2}$ can be equitably partitioned into $2t+1$ induced forests with the maximum degree of each induced forest at most 2. Let $x\in X$ and $y_1,y_2\in Y$. First, equitably partition each of $X\setminus \left\{x\right\}$ and $Y\setminus \{y_1,y_2\}$ into t subsets respectively. Thus, every subset contains three or four vertices. Next, let $\{x,y_1,y_2\}$ be the $(2t+1)$th subset. We thus obtain an equitable partition of $K_{3t+2,3t+2}$ as required. Hence, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\le 2t$. □

Theorem 8.

If $t\ge 4m+2$ with $m\ge 0$, then $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\le 2t-2m$.

Proof. 

The proof is similar to that of Theorems 2 and 5, which is omitted here. □

Theorem 9.

For $t=3\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)=\frac{3t+3}{2}$.

Proof. 

Let $t=4m+3$. By Theorem 8, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\le \frac{3t+3}{2}$. It suffices to show that $K_{3t+2,3t+2}$ cannot be equitably partitioned into $6m+5$ induced forests, where the maximum degree of each forest is at most 2. Thus $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\ge 6m+6=\frac{3t+3}{2}$. Suppose, to the contrary, that $K_{3t+2,3t+2}$ can be partitioned into $6m+5$ induced forests as desired. Then the size of every subset is at least 4 because $\lfloor \frac{6t+4}{6m+5}\rfloor =\lfloor \frac{24m+22}{6m+5}\rfloor =4$. By Lemma 3, the vertices of every subset belong to the same partite set. We can suppose that X is partitioned into at least $3m+3$ subsets as required. Thus, $\left|X\right|\ge 4(3m+3)>12m+11=3t+2=\left|X\right|$, a contradiction. □

Theorem 10.

For $t=0\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)=\frac{3t+4}{2}$.

Proof. 

If $t=0$, it is easy to see that $v{a}_2^{\equiv }\left(K_{2,2}\right)=2=\frac{0+4}{2}$. Let $t=4m+4$, $m\ge 0$. By Theorem 8, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\le \frac{3t+4}{2}$. In the following, we show that $K_{3t+2,3t+2}$ cannot be equitably partitioned into $6m+7$ induced forests, where the maximum degree of each forest is at most 2. Thus, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)\ge 6m+8=\frac{3t+4}{2}$. Suppose to the contrary that $K_{3k+2,3k+2}$ has a desired partition. Notice that $\frac{6t+4}{6m+7}=\frac{24m+28}{6m+7}=4$. This implies that the vertices of every subset belong to the same partite set. On the other hand, $\left|X\right|=\left|Y\right|=3t+2=12m+14$, which is not a multiple of 4. This contradiction completes our proof. □

6. Conclusions

In this paper, we discuss the strong equitable vertex 2-arboricity of $K_{n,n}$. First, we obtain a general upper bound: $2\lfloor \frac{n+1}{3}\rfloor$. Next, we improve this upper bound by considering three cases.

For the case of $n\equiv 0\left(\mathrm{mod}3\right)$, we obtain the exact value of $v{a}_2^{\equiv }\left(K_{3t,3t}\right)$ for two subcases: (i) For $t=1\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)=\frac{3t+1}{2}$. (ii) For $t=2\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t,3t}\right)=\frac{3t+2}{2}$.

For the case of $n\equiv 1\left(\mathrm{mod}3\right)$, we obtain the exact value of $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)$ for two subcases: (i) For $t=2\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)=\frac{3t+2}{2}$. (ii) For $t=3\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+1,3t+1}\right)=\frac{3t+3}{2}$.

For the case of $n\equiv 2\left(\mathrm{mod}3\right)$, we obtain the exact value of $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)$ for two subcases: (i) For $t=3\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)=\frac{3t+3}{2}$. (ii) For $t=0\left(\mathrm{mod}4\right)$, $v{a}_2^{\equiv }\left(K_{3t+2,3t+2}\right)=\frac{3t+4}{2}$.

It is worth pointing out that the upper bound in Theorem 1 is not very tight for some special graphs. For example, one can deduce from Theorem 3 that: $v{a}_2^{\equiv }\left(K_{63,63}\right)=32<42$. Thus, to determine the exact values of $v{a}_2^{\equiv }\left(K_{n,n}\right)$ for other cases seems not to be an easy task.

Moreover, investigation of $v{a}_2^{\equiv }\left(K_{n,m}\right)$ ($n\ne m$) is a challenging work in the future.