On $(s,t)$-relaxed $L(2,1)$-labeling of graphs

Abstract

We initiate the study of relaxed $L(2,1)$-labelings of graphs. Suppose $G$ is a graph. Let $u$ be a vertex of $G$. A vertex $v$ is called an $i$-neighbor of $u$ if $d_G(u,v)=i$. A $1$-neighbor of $u$ is simply called a neighbor of $u$. Let $s$ and $t$ be two nonnegative integers. Suppose $f$ is an assignment of nonnegative integers to the vertices of $G$. If the following three conditions are satisfied, then $f$ is called an $(s,t)$-relaxed $L(2,1)$-labeling of $G$: (1) for any two adjacent vertices $u$ and $v$ of $G, f(u)\not =f(v)$; (2) for any vertex $u$ of $G$, there are at most $s$ neighbors of $u$ receiving labels from $\{f(u)-1,f(u)+1\}$; (3) for any vertex $u$ of $G$, the number of $2$-neighbors of $u$ assigned the label $f(u)$ is at most $t$. The minimum span of $(s,t)$-relaxed $L(2,1)$-labelings of $G$ is called the $(s,t)$-relaxed $L(2,1)$-labeling number of $G$, denoted by $\lambda ^{s,t}_{2,1}(G)$. It is clear that $\lambda ^{0,0}_{2,1}(G)$ is the so called $L(2,1)$-labeling number of $G$. $\lambda ^{1,0}_{2,1}(G)$ is simply written as $\widetilde{\lambda }(G)$. This paper discusses basic properties of $(s,t)$-relaxed $L(2,1)$-labeling numbers of graphs. For any two nonnegative integers $s$ and $t$, the exact values of $(s,t)$-relaxed $L(2,1)$-labeling numbers of paths, cycles and complete graphs are determined. Tight upper and lower bounds for $(s,t)$-relaxed $L(2,1)$-labeling numbers of complete multipartite graphs and trees are given. The upper bounds for $(s,1)$-relaxed $L(2,1)$-labeling number of general graphs are also investigated. We introduce a new graph parameter called the breaking path covering number of a graph. A breaking path $P$ is a vertex sequence $v_1,v_2,\ldots ,v_k$ in which each $v_i$ is adjacent to at least one vertex of $v_{i-1}$ and $v_{i+1}$ for $i=2,3,\ldots ,k-1$. A breaking path covering of $G$ is a set of disjoint such vertex sequences that cover all vertices of $G$. The breaking path covering number of $G$, denoted by $bpc(G)$, is the minimum number of breaking paths in a breaking path covering of $G$. In this paper, it is proved that $\widetilde{\lambda }(G)= n+bpc(G^{c})-2$ if $bpc(G^{c})\ge 2$ and $\widetilde{\lambda }(G)\le n-1$ if and only if $bpc(G^{c})=1$. The breaking path covering number of a graph is proved to be computable in polynomial time. Thus, if a graph $G$ is of diameter two, then $\widetilde{\lambda }(G)$ can be determined in polynomial time. Several conjectures and problems on relaxed $L(2,1)$-labelings are also proposed.

Appendix: The proof of Theorem 6.2

We use the following four lemmas to prove Theorem 6.2.

Lemma 9.1

Let $G$ be a graph on $n$ vertices. Then $\widetilde{\lambda }(G)\le n+bpc(G)-2$.

Proof

Let $q=bpc(G)$. Suppose $\{P_1,P_2,\ldots ,P_q\}$ is a breaking path covering of $G$ with $P_i=v_{i_1},v_{i_2},\ldots ,v_{i_{p_i}}$ for $i=1,2,\ldots ,q$. For each $i\in \{1,2,\ldots ,q\}$ and $j\in \{1,2,\ldots ,p_i\}$, let

$$\begin{aligned} f(v_{i_j})=i+j-2+\sum \limits _{k=1}^{i-1}p_k \end{aligned}$$

Then it is straightforward to check that $f$ is a relaxed $L(2,1)$-labeling of $G$ with span $n+bpc(G)-2$. Thus the lemma follows. $\square$

Let $f$ be a relaxed $L(2,1)$-labeling of $G$ with span $p$. For $j=0,1,\ldots ,p$, denote by $F_j$ the set of vertices with label $j$ and $f_j$ the order of $F_j$. If $f_j=0$, then $j$ is called a hole of $f$. In a relaxed $L(2,1)$-labeling with minimum span, it is clear that $0$ and $p$ can not be holes. If $f_j\ge 2$, then $j$ is called a multiplicity of $f$. Suppose $g$ is a hole of $f$. If $f_{g-1}=f_{g+1}=1$ and the two vertices with labels $g-1$ and $g+1$ respectively are adjacent, then $g$ is called a gap of $f$. The set of multiplicities of $f$ is denoted by $M(f)$ and the set of gaps of $f$ is denoted by $G(f)$.

A relaxed $L(2,1)$-labeling $f$ of $G$ is called optimal if $span(f)=\widetilde{\lambda }(G)$. We say that an optimal relaxed $L(2,1)$-labeling $f$ of $G$ is minimum if $f$ has the minimum number of holes over all optimal relaxed $L(2,1)$-labelings.

Lemma 9.2

Let $f$ be a minimum relaxed $L(2,1)$-labeling of $G$. If $h$ is a hole of $f$ then $f_{h-1}=f_{h+1}>0$ and $G[F_{h-1}\cup F_{h+1}]$ is $1$-regular. In particular, if $f_{h-1}=f_{h+1}=1$, then $h$ is a gap.

Proof

Since $f$ is an optimal relaxed $L(2,1)$-labeling of $G$, it is clear that $f$ has no consecutive holes. It follows that if $h$ is a hole of $f$, then $f_{h-1}$ and $f_{h+1}$ are positive.

If there is one vertex, say $w$, in $F_{h-1}$ (or $F_{h+1}$) that is not adjacent to any vertex in $F_{h+1}$ (or $F_{h-1}$), then, by relabeling the vertex $w$ with the label $h$, we obtain a new optimal relaxed $L(2,1)$-labeling with fewer holes than $f$, contradicting the choice of $f$. Since vertices with the same label are at distance at least $3$, each vertex in $F_{h-1}$ (respectively, $F_{h+1}$) is adjacent to at most one vertex in $F_{h+1}$ (respectively, $F_{h-1}$). It follows that each vertex in $F_{h-1}$ (respectively, $F_{h+1}$) is adjacent to exact one vertex in $F_{h+1}$ (respectively, $F_{h-1}$). This implies that $f_{h-1}=f_{h+1}$ and $G[F_{h-1}\cup F_{h+1}]$ is $1$-regular. Therefore, if $f_{h-1}=f_{h+1}=1$, then $h$ is a gap. $\square$

Lemma 9.3

At least one of $G(f)$ and $M(f)$ is empty.

Proof

Suppose that $G(f)$ and $M(f)$ are all nonempty. Let $g$ and $m$ be a gap and a multiplicity of $f$ respectively. Suppose $m<g$ and between $m$ and $g$ there is no gaps and multiplicities. Then $m<g-1$ and $f_j=1$ for $m<j<g$. Let $v_{j}$ be the only vertex in $F_j$ for $j=m+1,m+2,\ldots ,g-1,g+1$

We claim that $f_m=2$. Let $w$ be a vertex in $F_m$. If $w$ is not adjacent to both $v_{g-1}$ and $v_{g+1}$, then by relabeling the vertex $w$ with the label $g$ we obtain a new optimal relaxed $L(2,1)$-labeling with fewer number of holes than $f$, contradicting the choice of $f$. Thus each vertex $w$ in $F_m$ is adjacent to at least one vertex of $v_{g-1}$ and $v_{g+1}$, implying $f_m=2$.

Let $F_m=\{w_1,w_2\}$. Then $w_1v_{g-1}\in E(G)$ and $w_2v_{g+1} \in E(G)$, or $w_2v_{g-1}\in E(G)$ and $w_1v_{g+1} \in E(G)$. With no loss of generality, suppose $w_1v_{g-1}\in E(G)$ and $w_2v_{g+1} \in E(G)$. Then $w_1$ is not adjacent to $v_{g+1}$ and $w_2$ is not adjacent to $v_{g-1}$. Define

$$\begin{aligned} f'(v)=\left\{ \begin{array}{ll} f(v), &{}\quad \hbox {if} \,\,\, v\not \in \{w_1,v_{m+1},v_{m+2},\ldots ,v_{g-1}\}\\ g, &{}\quad \hbox {if} \,\,\, v = w_1,\\ g+m-j, &{}\quad \hbox {if}\,\,\, v = v_j \hbox {for} \,\,\, j\in \{m+1,m+2,\ldots ,g-1\}. \end{array} \right. \end{aligned}$$

Since $f$ is a relaxed $L(2,1)$-labeling of $G$, no vertex $v_j\in \{v_{m+2},\ldots ,v_{g-2}\}$ is adjacent to both $v_{j-1}$ and $v_{j+1}$. Note that $w_1$ is not adjacent to $v_{g+1}$ and $w_2$ is not adjacent to $v_{g-1}$, it is obvious that $f'$ is an optimal relaxed $L(2,1)$-labeling with fewer number of holes than $f$. This contradicts the choice of $f$. Thus at least one of $G(f)$ and $M(f)$ is empty. $\square$

Lemma 9.4

Let $G$ be a graph on $n$ vertices. Then $\widetilde{\lambda }(G)\le n-1$ if and only if $bpc(G^c)=1$.

Proof

By Lemma 9.1, if $bpc(G^c)=1$, then $\widetilde{\lambda }(G)\le n-1$.

Now suppose $\widetilde{\lambda }(G)\le n-1$. It suffices to find a Hamiltonian breaking path of $G^c$. Let $f$ be a minimum relaxed $L(2,1)$-labeling of $G$. We claim that $f$ has no gaps. If $f$ has no holes, then of course $f$ has no gaps. If $f$ has at least one hole, then since there are $n$ vertices, by the pigeon-hole principle, $f$ has at least one multiplicity. By Lemma 9.3, $f$ has no gaps.

Let the span of $f$ be $p$ ($\le n-1$). Note that, for $j=0,1,\ldots ,p$, any two vertices in $F_j$ are at distance at least $3$. Thus $F_j$ is an independent set in $G$. Due to the distance two condition, the subgraph induced by $F_j\cup F_{j+1}$ has maximum degree at most $1$ for $j=0,1,\ldots ,p-1$. For $j=0,1,\ldots ,p-2$, if $F_j$ is nonempty but $F_{j+1}$ is empty, then $j+1$ is a hole, by Lemma 9.2, the subgraph induced by $F_j\cup F_{j+2}$ is $1$-regular. Let the vertices in $F_j$ be $v_{j1},v_{j2},\ldots ,v_{j{f_j}}$ for $j=0,1,\ldots ,p$. Then

$$\begin{aligned} v_{01},v_{02},\ldots ,v_{0{f_0}},v_{11},v_{12},\ldots ,v_{1{f_1}},\ldots , v_{p1},v_{p2},\ldots ,v_{p{f_p}} \end{aligned}$$

is a Hamiltonian breaking path of $G^c$ since $f$ is a relaxed $L(2,1)$-labeling of $G$ and so each vertex in $F_j$ is adjacent to at most one vertex in $F_{j-1}\cup F_{j+1}$ for $j=1,2,\ldots ,p-1$. Thus $bpc(G^c)=1$. $\square$

We are now ready to prove Theorem 6.2.

Proof of Theorem 6.2

By Lemma 9.4, to prove Theorem 6.2, it is sufficient to show that $\widetilde{\lambda }(G)=n+k-2$ if and only if $bpc(G^c)=k$ for $k\ge 2$. We prove this by mathematical induction.

Suppose $bpc(G^c)=2$. Then $\widetilde{\lambda }(G)\le n$ by Lemma 9.1, and $\widetilde{\lambda }(G)\ge n$ by Lemma 9.4. Thus $bpc(G^c)=2$ implies $\widetilde{\lambda }(G)=n$. If $\widetilde{\lambda }(G)=n$, then $bpc(G^c)\ge 2$ by Lemma 9.4. Let $G_1$ be the graph obtained from $G$ by adding an isolated vertex $z$. Then $\widetilde{\lambda }(G_1)=\widetilde{\lambda }(G)=n=|V(G_1)|-1$. It follows from Lemma 9.4 that $bpc(G_{1}^{c})=1$. Let $P$ be a Hamiltonian breaking path of $G_{1}^{c}$. By deleting $z$ from $P, P$ is divided into two breaking paths in $G^c$ covering all vertices, implying that $bpc(G^c)\le 2$. Therefore, $\widetilde{\lambda }(G)=n$ implies $bpc(G^{c})=2$.

Suppose $\widetilde{\lambda }(G)=n+i-2$ if and only if $2\le bpc(G^c)=i <k$. If $bpc(G^c)=k$, then, by Lemma 9.1 and the induction hypothesis, $\widetilde{\lambda }(G)= n+k-2$. If $\widetilde{\lambda }(G)= n+k-2$, then by Lemma 9.1, $bpc(G^c)\ge k$. Let $G_1$ be the graph obtained from $G$ by adding an isolated vertex $z$. Then $\widetilde{\lambda }(G_1)=\widetilde{\lambda }(G)=n+k-2=|V(G_1)|+(k-1)-2$. It follows from the induction hypothesis that $bpc(G_1^c)=k-1$. Let $\mathcal {P}$ be a breaking path covering of $G_1^c$ of order $k-1$. By deleting $z$ from the breaking path in $\mathcal {P}$ containing $z$, we obtain a breaking path covering of $G^c$ of order at most $k$, implying $bpc(G^c)\le k$. Thus $bpc(G^c)= k$. The induction step is completed. $\square$