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11institutetext: Department of Mathematical and Natural Sciences, The University of Tokushima, Tokushima 770–8502 Japan. 11email: hasunuma@ias.tokushima-u.ac.jp 22institutetext: Department of Information and Management Science, Otaru University of Commerce, Otaru 047-8501, Japan. 22email: ishii@res.otaru-uc.ac.jp 33institutetext: Department of Computer Science and Communication Engineering, Kyushu University, Fukuoka 812-8581, Japan. 33email: ono@csce.kyushu-u.ac.jp 44institutetext: Department of Mathematics and Information Sciences, Graduate School of Science, Osaka Prefecture University, Sakai 599-8531, Japan. 44email: uno@mi.s.osakafu-u.ac.jp

A tight upper bound on the (2,1)-total labeling number of outerplanar graphs

Toru Hasunuma 11    Toshimasa Ishii 22    Hirotaka Ono and Yushi Uno 3344
Abstract

A (2,1)(2,1)-total labeling of a graph GG is an assignment ff from the vertex set V(G)V(G) and the edge set E(G)E(G) to the set {0,1,,k}\{0,1,\ldots,k\} of nonnegative integers such that |f(x)f(y)|2|f(x)-f(y)|\geq 2 if xx is a vertex and yy is an edge incident to xx, and |f(x)f(y)|1|f(x)-f(y)|\geq 1 if xx and yy are a pair of adjacent vertices or a pair of adjacent edges, for all xx and yy in V(G)E(G)V(G)\cup E(G). The (2,1)(2,1)-total labeling number λ2T(G)\lambda^{T}_{2}(G) of a graph GG is defined as the minimum kk among all possible assignments. In [4], Chen and Wang conjectured that all outerplanar graphs GG satisfy λ2T(G)Δ(G)+2\lambda^{T}_{2}(G)\leq\Delta(G)+2, where Δ(G)\Delta(G) is the maximum degree of GG, while they also showed that it is true for GG with Δ(G)5\Delta(G)\geq 5. In this paper, we solve their conjecture completely, by proving that λ2T(G)Δ(G)+2\lambda^{T}_{2}(G)\leq\Delta(G)+2 even in the case of Δ(G)4\Delta(G)\leq 4.

1 Introduction

In the channel/frequency assignment problems, we need to assign different frequencies to ‘close’ transmitters so that they can avoid interference. The L(p,q)L(p,q)-labelings of a graph have been widely studied as one of important graph theoretical models of this problem. An L(p,q)L(p,q)-labeling of a graph GG is an assignment ff from the vertex set V(G)V(G) to the set {0,1,,k}\{0,1,\ldots,k\} of nonnegative integers such that |f(x)f(y)|p|f(x)-f(y)|\geq p if xx and yy are adjacent and |f(x)f(y)|q|f(x)-f(y)|\geq q if xx and yy are at distance 2, for all xx and yy in V(G)V(G). We call this invariant (i.e., the minimum value kk) the L(p,q)L(p,q)-labeling number and denote it by λp,q(G)\lambda_{p,q}(G). Notice that we can use k+1k+1 different labels when λp,q(G)=k\lambda_{p,q}(G)=k since we can use 0 as a label for conventional reasons. We can find many results on L(p,q)L(p,q)-labelings in comprehensive surveys by Calamoneri [3] and by Yeh [16].

In [15], Whittlesey et al. studied the L(2,1)L(2,1)-labeling number of incidence graphs, where the incidence graph of a graph GG is the graph obtained from GG by replacing each edge (vi,vj)(v_{i},v_{j}) with two edges (vi,vij)(v_{i},v_{ij}) and (vij,vj)(v_{ij},v_{j}) introducing one new vertex vijv_{ij}. Observe that an L(p,1)L(p,1)-labeling of the incidence graph of a given graph GG can be regarded as an assignment ff from V(G)E(G)V(G)\cup E(G) to the set {0,1,,}\{0,1,\ldots,\ell\} of nonnegative integers such that |f(x)f(y)|p|f(x)-f(y)|\geq p if xx is a vertex and yy is an edge incident to xx, and |f(x)f(y)|1|f(x)-f(y)|\geq 1 if xx and yy are a pair of adjacent vertices or a pair of adjacent edges, for all xx and yy in V(G)E(G)V(G)\cup E(G). Such a labeling of GG is called a (p,1)(p,1)-total labeling of GG, introduced by Havet and Yu [7, 8]. The (p,1)(p,1)-total labeling number is defined as the minimum value \ell, and denoted by λpT(G)\lambda_{p}^{T}(G).

Notice that a (1,1)(1,1)-total labeling of GG is equivalent to a total coloring of GG. Generalizing the Total Coloring Conjecture [2, 13], Havet and Yu [7, 8] conjectured that

λpT(G)Δ(G)+2p1\lambda_{p}^{T}(G)\leq\Delta(G)+2p-1 (1)

holds for any graph GG, where Δ(G)\Delta(G) denotes the maximum degree of a vertex in GG. They also showed that (i) λpT(G)min{2Δ(G)+p1,χ(G)+χ(G)+p2}\lambda_{p}^{T}(G)\leq\min\{2\Delta(G)+p-1,\ \chi(G)+\chi^{\prime}(G)+p-2\} for any graph GG where χ(G)\chi(G) and χ(G)\chi^{\prime}(G) denote the chromatic number and the chromatic index of GG, respectively, (ii) λ2T(G)2Δ(G)\lambda_{2}^{T}(G)\leq 2\Delta(G) if p=2p=2 and Δ(G)2\Delta(G)\geq 2, (iii) λ2T(G)2Δ(G)1\lambda_{2}^{T}(G)\leq 2\Delta(G)-1 if p=2p=2 and Δ(G)\Delta(G) is an odd 5\geq 5, and (iv) λpT(G)n+2p2\lambda_{p}^{T}(G)\leq n+2p-2 if GG is the complete graph where n=|V(G)|n=|V(G)|; the conjecture (1) is true if (a) pΔ(G)p\geq\Delta(G), (b) p=2p=2 and Δ(G)3\Delta(G)\leq 3, or (c) GG is the complete graph. By (i), it follows that λpT(G)Δ(G)+p\lambda_{p}^{T}(G)\leq\Delta(G)+p for any bipartite graph [1, 7, 8] (by χ(G)2\chi(G)\leq 2 and χ(G)=Δ(G)\chi^{\prime}(G)=\Delta(G) [9]). Also, Bazzaro et al. [1] investigated that λpT(G)Δ(G)+p+s\lambda_{p}^{T}(G)\leq\Delta(G)+p+s for any ss-degenerated graph (by χ(G)s+1\chi(G)\leq s+1 and χ(G)Δ(G)+1\chi^{\prime}(G)\leq\Delta(G)+1 [12]), where an ss-degenerated graph GG is a graph which can be reduced to a trivial graph by successive removal of vertices with degree at most ss, and that λpT(G)Δ(G)+p+3\lambda_{p}^{T}(G)\leq\Delta(G)+p+3 for any planar graph (by the Four-Color Theorem). They also showed sufficient conditions about Δ(G)\Delta(G) and girth for which the conjecture (1) holds. In [11], Montassier and Raspaud proved that λpT(G)Δ(G)+2p2\lambda_{p}^{T}(G)\leq\Delta(G)+2p-2 when p2p\geq 2 and Δ(G)\Delta(G) and the maximum average of GG degree satisfy some conditions. In [10], Lih et al. showed that λpT(G)3Δ(G)/2+4p3\lambda_{p}^{T}(G)\leq\lfloor 3\Delta(G)/2\rfloor+4p-3 for any graph GG. They also investigated λpT(Km,n)\lambda_{p}^{T}(K_{m,n}) of the complete bipartite graphs Km,nK_{m,n}.

Let GG be an outerplanar graph. In [1], Bazzaro et al. pointed out that λpT(G)Δ(G)+p+1\lambda_{p}^{T}(G)\leq\Delta(G)+p+1 for any outerplanar graph other than an odd cycle, since any outerplaner graph is 2-degenerated, and any outerplaner graph other than an odd cycle satisfies χ(G)=Δ(G)\chi^{\prime}(G)=\Delta(G) [5]. In particular, Chen and Wang [4] conjectured that in the case of p=2p=2, λ2T(G)Δ(G)+2\lambda_{2}^{T}(G)\leq\Delta(G)+2, which is a better bound than the conjecture (1). They also proved that this conjecture is true if (i) Δ(G)5\Delta(G)\geq 5, (ii) Δ(G)=3\Delta(G)=3 and GG is 2-connected, or (iii) Δ(G)=4\Delta(G)=4 and every two faces consisting of three vertices have no vertex in common. In the case of Δ(G)=2\Delta(G)=2 (i.e., GG is a path or a cycle), we can easily see that λ2T(G)4\lambda^{T}_{2}(G)\leq 4, since the incidence graph of a path (resp., a cycle) is also a path (resp., a cycle), and the L(2,1)L(2,1)-labeling number λ2,1(Cn)\lambda_{2,1}(C_{n}) for a cycle CnC_{n} with nn vertices is at most 4 [6]. The cases of Δ(G){0,1}\Delta(G)\in\{0,1\} are clear. However, the general cases of Δ(G){3,4}\Delta(G)\in\{3,4\} were left open. In this paper, we solve Cheng and Wang’s conjecture completely, by showing the remaining cases of Δ(G){3,4}\Delta(G)\in\{3,4\}. On the other hand, the bound Δ(G)+2\Delta(G)+2 is tight, since there exist infinitely many outerplanar graphs GG such that λ2T(G)=Δ(G)+2\lambda_{2}^{T}(G)=\Delta(G)+2 if Δ(G)2\Delta(G)\geq 2, as investigated in [4].

The rest of this paper is organized as follows. Section 2 gives basic definitions. In Sections 3 and 4, we show that λ2T(G)Δ(G)+2\lambda_{2}^{T}(G)\leq\Delta(G)+2 in the cases of Δ(G)=3\Delta(G)=3 and Δ(G)=4\Delta(G)=4, recpectively.

2 Preliminaries

A graph GG is an ordered set of its vertex set V(G)V(G) and edge set E(G)E(G) and is denoted by G=(V(G),E(G))G=(V(G),E(G)). We assume throughout this paper that a graph is undirected, simple and connected, unless otherwise stated. Therefore, an edge eE(G)e\in E(G) is an unordered pair of vertices uu and vv, which are end vertices of ee, and we often denote it by e=(u,v)e=(u,v). For two graphs G1G_{1} and G2G_{2}, we denote (V(G1)V(G2),E(G1)E(G2))(V(G_{1})\cup V(G_{2}),E(G_{1})\cup E(G_{2})) by G1+G2G_{1}+G_{2}. For a vertex set VV(G)V^{\prime}\subseteq V(G), let GVG-V^{\prime} be the subgraph of GG induced by V(G)VV(G)-V^{\prime} . For an edge set EE^{\prime}, let GEG-E^{\prime} (resp., G+EG+E^{\prime}) be the graph (V,EE)(V,E-E^{\prime}) (resp., (VV(E),EE)(V\cup V(E^{\prime}),E\cup E^{\prime}) where V(E)V(E^{\prime}) is the set of end vertices of edges in EE^{\prime}). Two vertices uu and vv are adjacent if (u,v)E(G)(u,v)\in E(G). Let NG(v)N_{G}(v) denote the set of neighbors of a vertex vv in GG; NG(v)={uV(u,v)E(G)}N_{G}(v)=\{u\in V\mid(u,v)\in E(G)\}. The degree of a vertex vv is |NG(v)||N_{G}(v)|, and is denoted by dG(v)d_{G}(v). A vertex vv with dG(v)=kd_{G}(v)=k is called a kk-vertex. We use Δ(G)\Delta(G) (resp., δ(G)\delta(G)) to denote the maximum (resp., minimum) degree of a graph GG. A path PP in GG is a sequence v1,v2,,vv_{1},v_{2},\ldots,v_{\ell} of vertices such that (vi,vi+1)E(G)(v_{i},v_{i+1})\in E(G) for i=1,2,,1i=1,2,\ldots,\ell-1; we may denote PP as (v1,v2,,v)(v_{1},v_{2},\ldots,v_{\ell}) and let V(P)={v1,v2,,v}V(P)=\{v_{1},v_{2},\ldots,v_{\ell}\} and E(P)={(v1,v2),(v2,v3),,(v1,v)}E(P)=\{(v_{1},v_{2}),(v_{2},v_{3}),\ldots,(v_{\ell-1},v_{\ell})\}. We often drop GG in these notations if there are no confusions.

The vertex-connectivity of a graph GG, denoted by κ(G)\kappa(G), is defined as the minimum cardinality of a vertex set VV(G)V^{\prime}\subseteq V(G) such that GVG-V^{\prime} is disconnected or trivial. A vertex vv is called a cut vertex ((of G)G) if GvG-v is disconnected. Notice that a graph GG has no cut vertex if and only if κ(G)2\kappa(G)\geq 2. Two vertices uu and uu^{\prime} are called 2-connected if they belong to the same connected component in GvG-v for any cut vertex vv of GG with v{u,u}v\notin\{u,u^{\prime}\}. A subgraph GG^{\prime} of GG is called a 2-connected component if every two vertices in GG^{\prime} are 2-connected and GG^{\prime} is maximal to this property. It is not difficult to see that if κ(G)=1\kappa(G)=1, then there exists a 2-connected component which contains exactly one cut vertex of GG.

A graph is called planar if it can be drawn in the plane without generating a crossing by two edges. For such a drawing of GG in the plane, after omitting the line segments, the plane is divided into regions; such a region is called a face. A face whose vertex set is {u1,u2,,uk}\{u_{1},u_{2},\ldots,u_{k}\} with (ui,ui+1)E(G)(u_{i},u_{i+1})\in E(G), i=1,2,,ki=1,2,\ldots,k (where uk+1=u1u_{k+1}=u_{1}) is denoted by [u1u2uk][u_{1}u_{2}\cdots u_{k}]. We call a face consisting of kk vertices a kk-face.

A planar graph GG is called outerplanar if it can be drawn in the plane so that all vertices lie on the boundary of some face. Such a drawing is referred to as an outerplane graph. The face whose boundary contains all vertices of GG is called the outer face and all other faces are called inner faces. We call an edge belonging to the boundary of the outer face an outer edge and all other edges inner edges. An inner face is called an endface if its boundary contains exactly one inner edge (note that there exist at least two endfaces if GG contains an inner edge).

Let k={0,1,,k}{\cal L}_{k}=\{0,1,\ldots,k\} and f:V(G)E(G)kf:V(G)\cup E(G)\to{\cal L}_{k} be a (2,1)-total labeling of GG. For an edge e=(u,v)E(G)e=(u,v)\in E(G), we may denote f(e)f(e) by f(uv)f(uv). Let f¯\overline{f} denote the labeling such that f¯(z)=kf(z)\overline{f}(z)=k-f(z) for each zV(G)E(G)z\in V(G)\cup E(G). Notice that f¯\overline{f} is also a (2,1)-total labeling of GG.

3 Case Δ(G)=3\Delta(G)=3

Let G=(V,E)G=(V,E) be an outerplane graph with Δ(G)=3\Delta(G)=3. Chen and Wang [4] showed that if κ(G)2\kappa(G)\geq 2, then λ2T(G)5\lambda_{2}^{T}(G)\leq 5. First review their constructive proof of the case of κ(G)2\kappa(G)\geq 2. Assume that on the boundary of its outerface, each 3-vertex (i.e., each vertex which some inner edge is incident to) is numbered x1,x2,,xpx_{1},x_{2},\ldots,x_{p} in clockwise order. Also assume that for i=1,2,,pi=1,2,\ldots,p, each 2-vertex on the path connecting xix_{i} and xi+1x_{i+1} (on the boundary of its outerface) is numbered y1i,y2i,,yqiiy_{1}^{i},y_{2}^{i},\ldots,y_{q_{i}}^{i} in clockwise order, where xp+1=x1x_{p+1}=x_{1}. A (2,1)-total labeling f:VE5f:V\cup E\to{\cal L}_{5} can be obtained by the following algorithm, which slightly generalizes Chen and Wang’s method.

Algorithm LABEL-K2

Input: An outerplane graph GG with κ(G)2\kappa(G)\geq 2.

Output: A total (2,1)-labeling f:VE5f:V\cup E\to{\cal L}_{5} of GG.

1.  Assign labels 0 and 1 alternately to a sequence x1,x2,,xpx_{1},x_{2},\ldots,x_{p} of 3-vertices.
2.  For each i=1,2,,pi=1,2,\ldots,p, label 2-vertices as follows. Without loss of generality, let f(xi)=0f(x_{i})=0 and f(xi+1)=1f(x_{i+1})=1.

2.1. If qiq_{i} is even, then assign labels 1 and 0 alternately to a sequence y1i,y2i,,yqiiy^{i}_{1},y^{i}_{2},\ldots,y^{i}_{q_{i}} of 2-vertices.

2.2. If qiq_{i} is odd, then assign label 2 to yjiy_{j}^{i} for some jj, and labels 1 and 0 alternately to a sequence y1i,y2i,,yqiiy^{i}_{1},y^{i}_{2},\ldots,y^{i}_{q_{i}} of 2-vertices other than yjiy_{j}^{i}.

3.  Assign label 3 to each inner edge.

4.  Label outer edges as follows.

4.1.  If |V(G)||V(G)| is even, then assign labels 4 and 5 alternately in clockwise order.

4.2.  If |V(G)||V(G)| is odd, then pick up an arbitrary endface F=[xiy1iy2iyqiixi+1]F=[x_{i}y_{1}^{i}y_{2}^{i}\cdots y_{q_{i}}^{i}x_{i+1}]. Without loss of generality, let f(xi)=0f(x_{i})=0 and f(xi+1)=1f(x_{i+1})=1.

4.2.1.  If qiq_{i} is an even or an odd with qi3q_{i}\geq 3, then assign label 3 to (y1i,y2i)(y_{1}^{i},y_{2}^{i}) and labels 4 and 5 alternately to all remaining outer edges. Then if f(y1i)=2f(y_{1}^{i})=2 or f(y2i)=2f(y_{2}^{i})=2, then reassign labels for y1iy_{1}^{i}, y2iy_{2}^{i}, and y3iy_{3}^{i} as f(y1i):=1f(y_{1}^{i}):=1, f(y2i):=0f(y_{2}^{i}):=0, and f(y3i):=2f(y_{3}^{i}):=2.

4.2.2.  If qi=1q_{i}=1, then assign labels 5, 2, and 3 to y1iy_{1}^{i}, (xi,y1i)(x_{i},y_{1}^{i}), and (y1i,xi+1)(y_{1}^{i},x_{i+1}), respectively, and labels 4 and 5 alternately to all remaining outer edges. Then if the label for two outer edges incident to xix_{i} and xi+1x_{i+1} other than (xi,y1i)(x_{i},y_{1}^{i}) and (y1i,xi+1)(y_{1}^{i},x_{i+1}) is 4 (resp., 5), reassign a label for (xi,xi+1)(x_{i},x_{i+1}) as f(xixi+1):=5f(x_{i}x_{i+1}):=5 (resp., 4).

Observe that for each inner edge (xi,xj)(x_{i},x_{j}), we have f(xi)f(xj)f(x_{i})\neq f(x_{j}), since otherwise the subgraph GG^{\prime} of GG induced by V(Pi)V(Pi+1)V(Pj1)V(P_{i})\cup V(P_{i+1})\cup\cdots\cup V(P_{j-1}) would have odd 3-vertices, a contradiction, where PiP_{i} denotes the path (xi,y1i,,yqii,xi+1)(x_{i},y_{1}^{i},\ldots,y_{q_{i}}^{i},x_{i+1}). Notice that the reassignment in Step 4.2.1 is necessary because otherwise |f(y1i)f(y1iy2i))|1|f(y_{1}^{i})-f(y_{1}^{i}y_{2}^{i}))|\leq 1 or |f(y2i)f(y1iy2i))|1|f(y_{2}^{i})-f(y_{1}^{i}y_{2}^{i}))|\leq 1 would hold. Also notice that in Step 4.2.2, f(e)=f(e)f(e)=f(e^{\prime}) holds for two outer edges ee (resp., ee^{\prime}) incident to xix_{i} (resp., xi+1x_{i+1}) other than (xi,y1i)(x_{i},y_{1}^{i}) (resp., (y1i,xi+1)(y_{1}^{i},x_{i+1})), since |V(G)||V(G)| is odd and qi=1q_{i}=1. Thus, a labeling ff obtained from the above procedure is a (2,1)-total labeling of GG. Here, we have the following easy observation:

Observation 1. In Step 4.2, we can start with assigning an arbitrary label in {4,5}\{4,5\} to an arbitrary outer edge for the label assignment of outer edges except (y1i,y2i)(y_{1}^{i},y_{2}^{i}) (resp., (xi,y1i)(x_{i},y_{1}^{i}) and (y1i,xi+1)(y_{1}^{i},x_{i+1})) in Step 4.2.1 (resp., Step 4.2.2).

Moreover, we can see the following property.

Lemma 1

Let G=(V,E)G=(V,E) be an outerplane graph with Δ(G)=3\Delta(G)=3, G1G_{1} be a 2-connected component in GG, and (u,v)(u,v) be an inner edge in G1G_{1}. Assume that G1{u,v}G_{1}-\{u,v\} contains a connected component G2G_{2} with no cut vertex in GG and |V(G2)|2|V(G_{2})|\geq 2. Let uV(G2)u^{\prime}\in V(G_{2}) (resp., vV(G2))v^{\prime}\in V(G_{2})) be the neighbor of uu (resp., vv) in GG (note that by Δ(G)=3\Delta(G)=3, such uu^{\prime} and vv^{\prime} are uniquely determined and that by |V(G2)|2|V(G_{2})|\geq 2, we have uvu^{\prime}\neq v^{\prime}). If G(GV(G2))+{(u,u),(v,v)}G^{\prime}\triangleq(G-V(G_{2}))+\{(u,u^{\prime}),(v,v^{\prime})\} has a ((2,1))-total labeling f:V(G)E(G)5f:V(G^{\prime})\cup E(G^{\prime})\to{\cal L}_{5} such that

f(u)f(v) and (i) {f(u),f(v)}{0,1} and {f(uu),f(vv)}{4,5} or(ii) {f(u),f(v)}{4,5} and {f(uu),f(vv)}{0,1},\begin{array}[]{l}\mbox{$f(u^{\prime})\neq f(v^{\prime})$ and {\rm(i)} $\{f(u^{\prime}),f(v^{\prime})\}\subseteq\{0,1\}$ and $\{f(uu^{\prime}),f(vv^{\prime})\}\subseteq\{4,5\}$ or}\\ \mbox{{\rm(ii)} $\{f(u^{\prime}),f(v^{\prime})\}\subseteq\{4,5\}$ and $\{f(uu^{\prime}),f(vv^{\prime})\}\subseteq\{0,1\}$,}\end{array}

then there exists a ((2,1))-total labeling g:V(G)E(G)5g:V(G)\cup E(G)\to{\cal L}_{5}.

Proof

By symmetry of labelings, we only consider the case where f(u)f(v)f(u^{\prime})\neq f(v^{\prime}), {f(u),f(v)}{0,1}\{f(u^{\prime}),f(v^{\prime})\}\subseteq\{0,1\}, and {f(uu),f(vv)}{4,5}\{f(uu^{\prime}),f(vv^{\prime})\}\subseteq\{4,5\}. Without loss of generality, let f(u)=0f(u^{\prime})=0 and f(v)=1f(v^{\prime})=1. There are the two possible cases (Case-1) f(uu)=f(vv)f(uu^{\prime})=f(vv^{\prime}) and (Case-2) f(uu)f(vv)f(uu^{\prime})\neq f(vv^{\prime}).

(Case-1)  Let G3G_{3} be the graph obtained from G2G_{2} by adding two new vertices xx and yy and three new edges (u,x),(x,y)(u^{\prime},x),(x,y), and (y,v)(y,v^{\prime}). Notice that G3G_{3} is an outerplane graph with κ(G3)2\kappa(G_{3})\geq 2. There are the following two possible cases: (1.1) both of uu^{\prime} and vv^{\prime} are 2-vertices or 3-vertices, (1.2) otherwise.

(1.1)  If uu^{\prime} and vv^{\prime} are both 2-vertices, then add an inner edge (u,v)(u^{\prime},v^{\prime}) to G3G_{3} and redefine G3G_{3} by the resulting graph. By applying Algorithm LABEL-K2 to G3G_{3} as x1:=ux_{1}:=u^{\prime}, we can obtain a ((2,1))-total labeling f1:V(G3)E(G3)5f_{1}:V(G_{3})\cup E(G_{3})\to{\cal L}_{5} such that f1(u)=0f_{1}(u^{\prime})=0 and f1(v)=1f_{1}(v^{\prime})=1. Moreover, by choosing an endface not containing {x,y}\{x,y\} in Step 4.2 of Algorithm LABEL-K2, we can obtain such a labeling f1f_{1} that f1(xu)=f1(yv)=f(uu)(=f(vv))f_{1}(xu^{\prime})=f_{1}(yv^{\prime})=f(uu^{\prime})~(=f(vv^{\prime})) (note that by Observation 1, we can start with assigning label f(uu)f(uu^{\prime}) to (x,u)(x,u^{\prime}) for the label assignment of outer edges). Observe that the labeling gg such that g(z)=f(z)g(z)=f(z) for each zV(G)E(G)z\in V(G^{\prime})\cup E(G^{\prime}) and g(z)=f1(z)g(z)=f_{1}(z) for each zVE(V(G)E(G))z\in V\cup E-(V(G^{\prime})\cup E(G^{\prime})) is a ((2,1))-total labeling of GG.

(1.2)  Assume that uu^{\prime} is a 3-vertex and vv^{\prime} is a 2-vertex (the other case can be treated similarly). Let f1:V(G3)E(G3)5f_{1}:V(G_{3})\cup E(G_{3})\to{\cal L}_{5} be a ((2,1))-total labeling obtained by applying Algorithm LABEL-K2 to G3G_{3} as x1:=ux_{1}:=u^{\prime}; f1(u)=0f_{1}(u^{\prime})=0. Moreover, as observed in (Case-1.1), we can obtain such a f1f_{1} that f1(xu)=f1(yv)=f(uu)f_{1}(xu^{\prime})=f_{1}(yv^{\prime})=f(uu^{\prime}). Also, if vv^{\prime} is not adjacent to a 3-vertex, we can obtain such a f1f_{1} that f1(v)=1f_{1}(v^{\prime})=1 by avoiding assigning label 2 to any vertex in {x,y,v}\{x,y,v^{\prime}\} in Step 2.2. Observe that if f1(v)=1f_{1}(v^{\prime})=1, then a ((2,1))-total labeling gg of GG can be constructed by combining ff and f1f_{1} similarly to (Case-1.1). Hence, we consider the case where the vertex w(y)w~(\neq y) adjacent to vv^{\prime} in G3G_{3} is a 3-vertex; f1(w)=1f_{1}(w)=1. Then we can reassign labels for x,yx,y, and vv^{\prime} as any one of two possible labelings (a) f1(x)=1,f1(y)=0f_{1}(x)=1,f_{1}(y)=0, and f1(v)=2f_{1}(v^{\prime})=2 and (b) f1(x)=1,f1(y)=2f_{1}(x)=1,f_{1}(y)=2, and f1(v)=0f_{1}(v^{\prime})=0 without violating the feasibility. Observe that the labeling gg such that g(z)=f(z)g(z)=f(z) for each zV(G)E(G){v}z\in V(G^{\prime})\cup E(G^{\prime})-\{v^{\prime}\}, g(z)=f1(z)g(z)=f_{1}(z) for each zVE(V(G)E(G))z\in V\cup E-(V(G^{\prime})\cup E(G^{\prime})), and g(v)={0,2}{f(v)}g(v^{\prime})=\{0,2\}-\{f(v)\} is a ((2,1))-total labeling of GG.

(Case-2)  Let G3G_{3} be the graph obtained from G2G_{2} by adding a new vertex xx and two new edges (u,x)(u^{\prime},x) and (x,v)(x,v^{\prime}). Notice that G3G_{3} is an outerplane graph with κ(G3)2\kappa(G_{3})\geq 2. There are the following two possible cases: (2.1) both of uu^{\prime} and vv^{\prime} are 2-vertices or 3-vertices, (2.2) otherwise.

(2.1)  If uu^{\prime} and vv^{\prime} are both 2-vertices, then add an inner edge (u,v)(u^{\prime},v^{\prime}) to G3G_{3} and redefine G3G_{3} by the resulting graph. By applying Algorithm LABEL-K2 to G3G_{3} as x1:=ux_{1}:=u^{\prime}, we can obtain a ((2,1))-total labeling f1:V(G3)E(G3)5f_{1}:V(G_{3})\cup E(G_{3})\to{\cal L}_{5} such that f1(u)=0f_{1}(u^{\prime})=0 and f1(v)=1f_{1}(v^{\prime})=1. Moreover, by choosing an endface not containing xx in Step 4.2 of Algorithm LABEL-K2, we can obtain such a labeling f1f_{1} that f1(xu)=f(uu)f_{1}(xu^{\prime})=f(uu^{\prime}) and f1(xv)=f(vv)f_{1}(xv^{\prime})=f(vv^{\prime}). Observe that the labeling gg such that g(z)=f(z)g(z)=f(z) for each zV(G)E(G)z\in V(G^{\prime})\cup E(G^{\prime}) and g(z)=f1(z)g(z)=f_{1}(z) for each zVE(V(G)E(G))z\in V\cup E-(V(G^{\prime})\cup E(G^{\prime})) is a ((2,1))-total labeling of GG.

(2.2)  Assume that uu^{\prime} is a 3-vertex and vv^{\prime} is a 2-vertex (the other case can be treated similarly). Let f1:V(G3)E(G3)5f_{1}:V(G_{3})\cup E(G_{3})\to{\cal L}_{5} be a ((2,1))-total labeling obtained by applying Algorithm LABEL-K2 to G3G_{3} as x1:=ux_{1}:=u^{\prime}; f1(u)=0f_{1}(u^{\prime})=0. Moreover, as observed in (Case-2.1), we can obtain such a f1f_{1} that f1(xu)=f(uu)f_{1}(xu^{\prime})=f(uu^{\prime}) and f1(xv)=f(vv)f_{1}(xv^{\prime})=f(vv^{\prime}). Also, in the case where Step 2.2 is applied to the sequence of 2-vertices containing xx and vv^{\prime}, we can assign label 1 to vv^{\prime} by assigning label 2 to xx (notice that in this case, vv^{\prime} is not adjacent to any 3-vertex because the sequence of 2-vertices containing vv^{\prime} consists of odd vertices). Observe that if f1(v)=1f_{1}(v^{\prime})=1, then a ((2,1))-total labeling gg of GG can be constructed by combining ff and f1f_{1} similarly to (Case-2.1). Hence, we consider the case where Step 2.1 is applied to the sequence of 2-vertices containing xx and vv^{\prime}; f1(v)=0f_{1}(v^{\prime})=0. Then we can reassign label 2 to vv^{\prime} without violating the feasibility. We can see that the labeling gg such that g(z)=f(z)g(z)=f(z) for each zV(G)E(G){v}z\in V(G^{\prime})\cup E(G^{\prime})-\{v^{\prime}\}, g(z)=f1(z)g(z)=f_{1}(z) for each zVE(V(G)E(G))z\in V\cup E-(V(G^{\prime})\cup E(G^{\prime})), and g(v)={0,2}{f(v)}g(v^{\prime})=\{0,2\}-\{f(v)\} is a ((2,1))-total labeling of GG. ∎

By using this lemma, we can prove that Chen and Wang’s conjecture is true in the case of Δ(G)=3\Delta(G)=3.

Theorem 3.1

If G=(V,E)G=(V,E) is an outerplane graph with Δ(G)=3\Delta(G)=3, then λ2T(G)5\lambda_{2}^{T}(G)\leq 5.

Proof

We prove this by induction on k=|V(G)|+|E(G)|k=|V(G)|+|E(G)|. The theorem clearly holds if k7k\leq 7. Consider the case of k8k\geq 8 and assume that for each k<kk^{\prime}<k, this theorem holds. We also assume that GG is connected, since otherwise we can treat each component separately. Thus, 1δ(G)21\leq\delta(G)\leq 2.

Consider the case where δ(G)=1\delta(G)=1. Let u1u_{1} be a vertex with d(u1)=1d(u_{1})=1. By the induction hypothesis, Gu1G-u_{1} has a (2,1)-total labeling f:V(Gu1)E(Gu1)5f:V(G-u_{1})\cup E(G-u_{1})\to{\cal L}_{5}. Let u2u_{2} be the neighbor of u1u_{1} in GG and u3,u4u_{3},u_{4} be vertices adjacent to u2u_{2} in Gu1G-u_{1} where u3=u4u_{3}=u_{4} may occur (note that Δ(G)=3\Delta(G)=3). Hence we can extend ff to the edge (u1,u2)(u_{1},u_{2}) and the vertex u1u_{1} as follows: assign a label aa\in 5{\cal L}_{5}- {f(u2)1,f(u2),f(u2)+1,f(u2u3),f(u2u4)}\{f(u_{2})-1,f(u_{2}),f(u_{2})+1,f(u_{2}u_{3}),f(u_{2}u_{4})\} to (u1,v1)(u_{1},v_{1}), and then a label in 5{f(u2),a1,a,a+1}{\cal L}_{5}-\{f(u_{2}),a-1,a,a+1\} to u1u_{1}.

Consider the case of δ(G)=2\delta(G)=2. In [4], Chen and Wang showed that if κ(G)2\kappa(G)\geq 2, then λ2T(G)5\lambda_{2}^{T}(G)\leq 5. Hence, we here only consider the case of κ(G)=1\kappa(G)=1. Let G1G_{1} be a 2-connected component in GG which has exactly one cut vertex vcv_{c} of GG. By δ(G)=2\delta(G)=2, we have |V(G1)|3|V(G_{1})|\geq 3 and hence dG1(vc)2d_{G_{1}}(v_{c})\geq 2. By κ(G)=1\kappa(G)=1 and Δ(G)=3\Delta(G)=3, it follows that dG1(vc)=2d_{G_{1}}(v_{c})=2 holds and V(G1)V(G_{1}) and VV(G1)V-V(G_{1}) are connected by a bridge incident to vcv_{c}, where a bridge is an edge whose deletion makes the graph disconnected; denote this bridge by (vc,w)(v_{c},w) where wVV(G1)w\in V-V(G_{1}). By the induction hypothesis, H=(GV(G1))+{(vc,w)}H=(G-V(G_{1}))+\{(v_{c},w)\} has a (2,1)-total labeling f:V(H)E(H)5f:V(H)\cup E(H)\to{\cal L}_{5}. By symmetry, it suffices to consider the following three possible cases: (Case-1) f(vcw)=5f(v_{c}w)=5, (Case-2) f(vcw)=4f(v_{c}w)=4, and (Case-3) f(vcw)=3f(v_{c}w)=3. In each case, we will extend ff to a (2,1)-total labeling f:VE5f^{\prime}:V\cup E\to{\cal L}_{5} of GG. Let f(x):=f(x)f^{\prime}(x):=f(x) for every xV(H)E(H)x\in V(H)\cup E(H).

First consider the case where G1G_{1} has no inner edge. Assume that on the boundary of its outerface in G1G_{1}, each vertex is numbered u1(=vc),u2,,uru_{1}(=v_{c}),u_{2},\ldots,u_{r} in clockwise order. Then, in each case of (Case-1)–(Case-3), we can obtain a (2,1)-total labeling ff^{\prime} of GG in the following manner:

1.  If f(vcw)=3f^{\prime}(v_{c}w)=3, r=3r=3, and f(w)0f^{\prime}(w)\neq 0, then let f(vc)=0f^{\prime}(v_{c})=0, f(u2):=1f^{\prime}(u_{2}):=1, f(u3):=5f^{\prime}(u_{3}):=5, f(vcu2):=5f^{\prime}(v_{c}u_{2}):=5, f(u2u3):=3f^{\prime}(u_{2}u_{3}):=3, and f(u3vc):=2f^{\prime}(u_{3}v_{c}):=2. If f(vcw)=3f^{\prime}(v_{c}w)=3, r=3r=3, and f(w)=0f^{\prime}(w)=0, then let f(vc)=5f^{\prime}(v_{c})=5, f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=0f^{\prime}(u_{3}):=0, f(vcu2):=0f^{\prime}(v_{c}u_{2}):=0, f(u2u3):=5f^{\prime}(u_{2}u_{3}):=5, and f(u3vc):=2f^{\prime}(u_{3}v_{c}):=2.

2. Unless f(vcw)=3f^{\prime}(v_{c}w)=3 and r=3r=3, then assign labels for vertices or edges in G1G_{1} as follows.

2.1.  Assign a label in {0,1}{f(w)}\{0,1\}-\{f^{\prime}(w)\} (say, 0) to vcv_{c}.

2.2.  Assign labels 1 and 0 alternately to u2,u3,,uru_{2},u_{3},\ldots,u_{r}. Then, if rr is odd, then reassign label 2 to uru_{r}.

2.3.  If f(vcw){3,4}f^{\prime}(v_{c}w)\in\{3,4\} (resp., f(vcw)=5f^{\prime}(v_{c}w)=5), then assign label 5 (resp., 4) to (ur,vc)(u_{r},v_{c}), and labels 4 and 5 (resp., 5 and 4) alternately to the sequence (ur,ur1),(ur1,ur2),,(u2,vc)(u_{r},u_{r-1}),(u_{r-1},u_{r-2}),\ldots,(u_{2},v_{c}) of outer edges in counter-clockwise order. Then if f(vcw){4,5}f^{\prime}(v_{c}w)\in\{4,5\}, then reassign label 3 to (u2,vc)(u_{2},v_{c}), and if f(vcw)=3f^{\prime}(v_{c}w)=3 and rr is odd, then reassign label 3 to (u3,u2)(u_{3},u_{2}) and label 4 to (u2,vc)(u_{2},v_{c}).

Next consider the case where G1G_{1} has an inner edge. Assume that on the boundary of its outerface in G1G_{1}, each 3-vertex is numbered x1,x2,,xpx_{1},x_{2},\ldots,x_{p} in clockwise order. Also assume that for i=1,2,,pi=1,2,\ldots,p, each 2-vertex on the path connecting xix_{i} and xi+1x_{i+1} (on the boundary of its outerface) is numbered y1i,y2i,,yqiiy_{1}^{i},y_{2}^{i},\ldots,y_{q_{i}}^{i} in clockwise order, where xp+1=x1x_{p+1}=x_{1}. Also, if qi=0q_{i}=0, then let yqii:=xiy^{i}_{q_{i}}:=x_{i}. Let PiP_{i} denote the path (xi,y1i,,yqii,xi+1)(x_{i},y_{1}^{i},\ldots,y_{q_{i}}^{i},x_{i+1}). Without loss of generality, assume that vcv_{c} is a 2-vertex between x1x_{1} and x2x_{2} in G1G_{1}; vc=y1v_{c}=y_{\ell}^{1} for some \ell. Let f1:V(G1)E(G1)5f_{1}:V(G_{1})\cup E(G_{1})\to{\cal L}_{5} be a (2,1)-total labeling obtained by applying Algorithm LABEL-K2 to G1G_{1}, while (A) choosing yj1y^{1}_{j} as a vertex other than vcv_{c} in Step 2.2 if q13q_{1}\geq 3, and (B) choosing yj1y^{1}_{j} as a vertex not in {vc}NG(vc)\{v_{c}\}\cup N_{G}(v_{c}) in Step 2.2 if (i) q15q_{1}\geq 5 or (ii) q1=3q_{1}=3 and vcy21v_{c}\neq y_{2}^{1}, and (C) choosing an endface not containing vcv_{c} in Step 4.2.

Consider the case where vcv_{c} is adjacent to a 2-vertex in G1G_{1}; q1>1q_{1}>1. Without loss of generality, assume that y11y_{\ell-1}^{1} is a 2-vertex. Then by q1>1q_{1}>1 and the choice of yj1y_{j}^{1} in Step 2.2 of Algorithm LABEL-K2, we can see that f1(vc)2f_{1}(v_{c})\neq 2; f1(vc){0,1}f_{1}(v_{c})\in\{0,1\}. Moreover, we can assume that f1(vc){0,1}{f(w)}f_{1}(v_{c})\in\{0,1\}-\{f^{\prime}(w)\} since otherwise we recompute a (2,1)-total labeling f1f^{\prime}_{1} of G1G_{1} by starting with x2x_{2} instead of x1x_{1} in Step 1 of Algorithm LABEL-K2 and redefine f1f_{1} by f1f_{1}^{\prime}. Also, we can assume that in the case of f(vcw)=4f^{\prime}(v_{c}w)=4 (resp., 5), if q13q_{1}\neq 3 or vcy21v_{c}\neq y_{2}^{1}, then f1(y11vc)=4f_{1}(y_{\ell-1}^{1}v_{c})=4 (resp., 5) and if q1=3q_{1}=3 and vc=y21v_{c}=y_{2}^{1}, then f1(vcy)=4f_{1}(v_{c}y^{\prime})=4 (resp., 5) for y{y11,y31}y^{\prime}\in\{y_{1}^{1},y_{3}^{1}\} with f1(y)2f_{1}(y^{\prime})\neq 2 (note that this is possible by Observation 1 and that in the case of q1=3q_{1}=3 and vc=y21v_{c}=y_{2}^{1}, exactly one of y11y_{1}^{1} and y31y_{3}^{1} has label 2 in f1f_{1}). Let f(z):=f1(z)f^{\prime}(z):=f_{1}(z) for each zV(G1)E(G1)z\in V(G_{1})\cup E(G_{1}). Observe that if f(vcw)=3f^{\prime}(v_{c}w)=3, then ff^{\prime} is a (2,1)-total labeling of GG. In the case of f(vcw){4,5}f^{\prime}(v_{c}w)\in\{4,5\}, if q13q_{1}\neq 3 or vcy21v_{c}\neq y_{2}^{1} (resp., q1=3q_{1}=3 and vc=y21v_{c}=y_{2}^{1}), reassign a label for (y11,vc)(y_{\ell-1}^{1},v_{c}) (resp., (y,vc)(y^{\prime},v_{c})) as f(y11vc):=3f^{\prime}(y_{\ell-1}^{1}v_{c}):=3 (resp., f(yvc):=3f^{\prime}(y^{\prime}v_{c}):=3). Then, we can see that the resulting ff^{\prime} is a (2,1)-total labeling of GG since if q13q_{1}\neq 3 or vcy21v_{c}\neq y_{2}^{1}, then f(y11)2f^{\prime}(y_{\ell-1}^{1})\neq 2 by the choice of yj1y_{j}^{1} in Step 2.2 and otherwise then f(y)2f^{\prime}(y^{\prime})\neq 2.

Finally, consider the case where NG1(vc)={x1,x2}N_{G_{1}}(v_{c})=\{x_{1},x_{2}\}; q1=1q_{1}=1. Let xpx_{p^{\prime}} be a 3-vertex such that (x1,xp)(x_{1},x_{p^{\prime}}) is an inner edge of G1G_{1}. Let G2G_{2} be the subgraph of G1G_{1} induced by V(P1)V(P2)V(Pp1)V(P_{1})\cup V(P_{2})\cup\cdots\cup V(P_{p^{\prime}-1}). Let uV(G1)V(G2)u^{\prime}\in V(G_{1})-V(G_{2}) (resp., vV(G1)V(G2)v^{\prime}\in V(G_{1})-V(G_{2})) be the neighbor of x1x_{1} (resp., xpx_{p^{\prime}}). Notice that the component in G{x1,xp}G-\{x_{1},x_{p^{\prime}}\} containing {u,v}\{u^{\prime},v^{\prime}\} has no cut vertex in GG. Below, we will show that if uvu^{\prime}\neq v^{\prime}, then there exists a (2,1)-total labeling gg^{\prime} of (H+G2)+{(x1,u),(xp,v)}(H+G_{2})+\{(x_{1},u^{\prime}),(x_{p^{\prime}},v^{\prime})\} satisfying the conditions of Lemma 1. Note that in this case, Lemma 1 ensures the existence of a (2,1)-total labeling g:VE5g:V\cup E\to{\cal L}_{5} of GG. Also, in the case of u=vu^{\prime}=v^{\prime}, we will show directly the existence of a (2,1)-total labeling g:VE5g:V\cup E\to{\cal L}_{5} of GG.

(Case-1) f(vcw)=5f(v_{c}w)=5.

(1.1) Assume that f(w)3f(w)\neq 3.

Let f(z):=f1¯(z)f^{\prime}(z):=\overline{f_{1}}(z) for each zV(G1)E(G1)z\in V(G_{1})\cup E(G_{1}). Then, ff^{\prime} is a (2,1)-total labeling of GG (note that f(vc)=3f^{\prime}(v_{c})=3).

(1.2) Assume that f(w)=3f(w)=3.

If p=2p^{\prime}=2 and u=vu^{\prime}=v^{\prime}, then let f(x1):=5f^{\prime}(x_{1}):=5, f(vc):=0f^{\prime}(v_{c}):=0, f(x2):=2f^{\prime}(x_{2}):=2, f(u):=0f^{\prime}(u^{\prime}):=0, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(vcx2):=4f^{\prime}(v_{c}x_{2}):=4, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x2u):=5f^{\prime}(x_{2}u^{\prime}):=5, and f(ux1):=2f^{\prime}(u^{\prime}x_{1}):=2. Observe that the resulting labeling ff^{\prime} is a (2,1)-total labeling of GG.

If p=2p^{\prime}=2 and uvu^{\prime}\neq v^{\prime}, then let f(x1):=5f^{\prime}(x_{1}):=5, f(vc):=0f^{\prime}(v_{c}):=0, f(x2):=4f^{\prime}(x_{2}):=4, f(u):=4f^{\prime}(u^{\prime}):=4, f(v):=5f^{\prime}(v^{\prime}):=5, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(vcx2):=2f^{\prime}(v_{c}x_{2}):=2, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x2v):=1f^{\prime}(x_{2}v^{\prime}):=1, and f(ux1):=1f^{\prime}(u^{\prime}x_{1}):=1. Observe that the resulting labeling ff^{\prime} is feasible for H+G2+{(x1,u),(x2,v)}H+G_{2}+\{(x_{1},u^{\prime}),(x_{2},v^{\prime})\} and satisfies the conditions of Lemma 1.

Assume that p2p^{\prime}\neq 2. Then we can obtain a (2,1)-total labeling of GG in the following manner, where “assigning a label kk to zVEz\in V\cup E” means that f(z):=kf^{\prime}(z):=k.

1. Assign label 3 to each inner edge in G2G_{2}.

2. Let f(vc):=0f^{\prime}(v_{c}):=0. Assign labels 1 and 0 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (note that as a result, f(xp1)=0f^{\prime}(x_{p^{\prime}-1})=0). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of algorithm LABEL-K2.

3. Trace outer edges from (vc,x2)(v_{c},x_{2}) to (yqp2p2,xp1)(y^{p^{\prime}-2}_{q_{p^{\prime}-2}},x_{p^{\prime}-1}) on {(vc,x2)}E(P2)E(Pp2)\{(v_{c},x_{2})\}\cup E(P_{2})\cup\cdots\cup E(P_{p^{\prime}-2}) in clockwise order and assign labels 4 and 5 alternately.

3.1. If qp1>0q_{p^{\prime}-1}>0, then assign labels to the remaining vertices or edges as follows. Let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=4f^{\prime}(x_{p^{\prime}}):=4, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(x1xp):=2f^{\prime}(x_{1}x_{p^{\prime}}):=2, and f(xp1,y1p1):=2f^{\prime}(x_{p^{\prime}-1},y_{1}^{p^{\prime}-1}):=2. Assign labels 0 and 1 alternately to outer edges (y1p1,y2p1),(y2p1,y3p1),,(yqp1p1,xp),(y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1}),(y_{2}^{p^{\prime}-1},y_{3}^{p^{\prime}-1}),\ldots,(y_{q_{p^{\prime}-1}}^{p^{\prime}-1},x_{p^{\prime}}), (xp,v)(x_{p^{\prime}},v^{\prime}). Moreover, if qp1q_{p^{\prime}-1} is even (resp., odd), then assign labels 4 and 5 (resp., 5 and 4) alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}. If u=vu^{\prime}=v^{\prime}, then let f(u):=3f^{\prime}(u^{\prime}):=3 and f(ux1):={0,1}{f(xpu)}f^{\prime}(u^{\prime}x_{1}):=\{0,1\}-\{f^{\prime}(x_{p^{\prime}}u^{\prime})\}. If uvu^{\prime}\neq v^{\prime}, then let f(u):=4f^{\prime}(u^{\prime}):=4, f(v):=5f^{\prime}(v^{\prime}):=5, and f(ux1):=1f^{\prime}(u^{\prime}x_{1}):=1, and extend ff^{\prime} to a (2,1)-total labeling of GG according to Lemma 1.

3.2. If qp1=0q_{p^{\prime}-1}=0 and uvu^{\prime}\neq v^{\prime}, then let f(x1):=f(v):=4f^{\prime}(x_{1}):=f^{\prime}(v^{\prime}):=4, f(xp):=f(u):=5f^{\prime}(x_{p^{\prime}}):=f^{\prime}(u^{\prime}):=5, f(x1xp):=0f^{\prime}(x_{1}x_{p^{\prime}}):=0, f(xp1xp):=f(x1vc):=2f(x_{p^{\prime}-1}x_{p^{\prime}}):=f(x_{1}v_{c}):=2, and f(x1u):=f(xpv):=1f^{\prime}(x_{1}u^{\prime}):=f^{\prime}(x_{p^{\prime}}v^{\prime}):=1, and extend ff^{\prime} to a (2,1)-total labeling of GG according to Lemma 1.

3.3. If qp1=0q_{p^{\prime}-1}=0, u=vu^{\prime}=v^{\prime} and f(yqp2p2xp1)=4f^{\prime}(y_{q_{p^{\prime}-2}}^{p^{\prime}-2}x_{p^{\prime}-1})=4, then let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=3f^{\prime}(x_{p^{\prime}}):=3, f(u):=4f^{\prime}(u^{\prime}):=4, f(x1xp):=1f^{\prime}(x_{1}x_{p^{\prime}}):=1, f(xp1xp):=5f(x_{p^{\prime}-1}x_{p^{\prime}}):=5, f(x1vc):=3f(x_{1}v_{c}):=3, and f(x1u):=2f^{\prime}(x_{1}u^{\prime}):=2, and f(xpu):=0f^{\prime}(x_{p^{\prime}}u^{\prime}):=0.

3.4. If qp1=0q_{p^{\prime}-1}=0, u=vu^{\prime}=v^{\prime} and f(yqp2p2xp1)=5f^{\prime}(y_{q_{p^{\prime}-2}}^{p^{\prime}-2}x_{p^{\prime}-1})=5, then let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=2f^{\prime}(x_{p^{\prime}}):=2, f(u):=0f^{\prime}(u^{\prime}):=0, f(x1xp):=0f^{\prime}(x_{1}x_{p^{\prime}}):=0, f(xp1xp):=4f(x_{p^{\prime}-1}x_{p^{\prime}}):=4, f(x1vc):=3f(x_{1}v_{c}):=3, and f(x1u):=2f^{\prime}(x_{1}u^{\prime}):=2, and f(xpu):=5f^{\prime}(x_{p^{\prime}}u^{\prime}):=5.

(Case-2) f(vcw)=4f(v_{c}w)=4.

(2.1) Assume that f(w)0f(w)\neq 0.

If p=2p^{\prime}=2 and u=vu^{\prime}=v^{\prime}, then let f(x1):=5f^{\prime}(x_{1}):=5, f(vc):=0f^{\prime}(v_{c}):=0, f(x2):=2f^{\prime}(x_{2}):=2, f(u)=0f^{\prime}(u^{\prime})=0, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(vcx2):=5f^{\prime}(v_{c}x_{2}):=5, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x2u):=4f^{\prime}(x_{2}u^{\prime}):=4, and f(ux1):=2f^{\prime}(u^{\prime}x_{1}):=2.

If p=2p^{\prime}=2 and uvu^{\prime}\neq v^{\prime}, then let f(x1):=f(v):=5f^{\prime}(x_{1}):=f^{\prime}(v^{\prime}):=5, f(vc):=0f^{\prime}(v_{c}):=0, f(x2):=f(u)=:4f^{\prime}(x_{2}):=f^{\prime}(u^{\prime})=:4, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(vcx2):=2f^{\prime}(v_{c}x_{2}):=2, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x2v):=1f^{\prime}(x_{2}v^{\prime}):=1, f(ux1):=1f^{\prime}(u^{\prime}x_{1}):=1. Observe that the resulting labeling ff^{\prime} is feasible for H+G2+{(x1,u),(x2,v)}H+G_{2}+\{(x_{1},u^{\prime}),(x_{2},v^{\prime})\} and satisfies the conditions of Lemma 1.

The case of p2p^{\prime}\neq 2 can be treated similarly to (Case-1.2), by starting with assigning to (vc,x2)(v_{c},x_{2}) label 5 instead of label 4.

(2.2) Assume that f(w)=0f(w)=0.

If p=2p^{\prime}=2 and u=vu^{\prime}=v^{\prime}, then let f(vc):=1f^{\prime}(v_{c}):=1, f(x1):=5f^{\prime}(x_{1}):=5, f(x2):=3f^{\prime}(x_{2}):=3, f(u):=4f^{\prime}(u^{\prime}):=4, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(vcx2):=5f^{\prime}(v_{c}x_{2}):=5, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x1u):=2f^{\prime}(x_{1}u^{\prime}):=2, and f(x2u):=1f^{\prime}(x_{2}u^{\prime}):=1.

If p=2p^{\prime}=2 and uvu^{\prime}\neq v^{\prime}, then f(vc):=1f^{\prime}(v_{c}):=1, f(x1):=5f^{\prime}(x_{1}):=5, f(x2):=3f^{\prime}(x_{2}):=3, f(u):=4f^{\prime}(u^{\prime}):=4, f(v):=5f^{\prime}(v^{\prime}):=5, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(vcx2):=5f^{\prime}(v_{c}x_{2}):=5, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x1u):=1f^{\prime}(x_{1}u^{\prime}):=1, and f(x2v):=1f^{\prime}(x_{2}v^{\prime}):=1. Observe that the resulting labeling is feasible for H+G2+{(x1,u),(x2,v)}H+G_{2}+\{(x_{1},u^{\prime}),(x_{2},v^{\prime})\} and satisfies the conditions of Lemma 1.

Assume that p2p^{\prime}\neq 2. Then we can obtain a (2,1)-total labeling of GG in the following manner.

1. Assign label 3 to each inner edge in G2G_{2}.

2. Trace outer edges from (vc,x2)(v_{c},x_{2}) to (yqp1p1,xp)(y^{p^{\prime}-1}_{q_{p^{\prime}-1}},x_{p^{\prime}}) on {(vc,x2)}E(P2)E(Pp1)\{(v_{c},x_{2})\}\cup E(P_{2})\cup\cdots\cup E(P_{p^{\prime}-1}) in clockwise order and assign labels 5 and 4 alternately.

2.1. If f(yqp1p1xp)=4f^{\prime}(y^{p^{\prime}-1}_{q_{p^{\prime}-1}}x_{p^{\prime}})=4 and uvu^{\prime}\neq v^{\prime}, then assign labels for the remaining vertices and edges as follows. Let f(vc):=2f^{\prime}(v_{c}):=2. Assign labels 1 and 0 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (notice that f(xp1)=0f^{\prime}(x_{p^{\prime}-1})=0). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of algorithm LABEL-K2. Let f(x1):=3f^{\prime}(x_{1}):=3, f(xp):=2f^{\prime}(x_{p^{\prime}}):=2, f(u):=4f^{\prime}(u^{\prime}):=4, f(v):=5f^{\prime}(v^{\prime}):=5, f(x1vc):=0f^{\prime}(x_{1}v_{c}):=0, f(x1xp):=5f^{\prime}(x_{1}x_{p^{\prime}}):=5, f(x1u):=1f^{\prime}(x_{1}u^{\prime}):=1, and f(xpv):=0f^{\prime}(x_{p^{\prime}}v^{\prime}):=0. Assign labels 1 and 0 alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}. Then, extend ff^{\prime} to a (2,1)-total labeling of GG according to Lemma 1.

2.2. If f(yqp1p1xp)=5f^{\prime}(y^{p^{\prime}-1}_{q_{p^{\prime}-1}}x_{p^{\prime}})=5 and uvu^{\prime}\neq v^{\prime}, then assign labels for the remaining vertices and edges as follows. Let f(vc):=1f^{\prime}(v_{c}):=1. Assign labels 0 and 1 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (notice that f(xp1)=1f^{\prime}(x_{p^{\prime}-1})=1). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of algorithm LABEL-K2. Let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=3f^{\prime}(x_{p^{\prime}}):=3, f(u):=4f^{\prime}(u^{\prime}):=4, f(v):=5f^{\prime}(v^{\prime}):=5, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(x1xp):=0f^{\prime}(x_{1}x_{p^{\prime}}):=0, f(x1u):=1f^{\prime}(x_{1}u^{\prime}):=1, and f(xpv):=1f^{\prime}(x_{p^{\prime}}v^{\prime}):=1. Assign labels 0 and 1 alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}. Then, extend ff^{\prime} to a (2,1)-total labeling of GG according to Lemma 1.

2.3. If f(yqp1p1xp)=4f^{\prime}(y^{p^{\prime}-1}_{q_{p^{\prime}-1}}x_{p^{\prime}})=4 and u=vu^{\prime}=v^{\prime}, then assign labels for the remaining vertices and edges as follows. Let f(vc):=2f^{\prime}(v_{c}):=2. Assign labels 1 and 0 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (notice that f(xp1)=0f^{\prime}(x_{p^{\prime}-1})=0). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of algorithm LABEL-K2. Let f(x1):=3f^{\prime}(x_{1}):=3, f(xp):=2f^{\prime}(x_{p^{\prime}}):=2, f(u):=5f^{\prime}(u^{\prime}):=5, f(x1vc):=0f^{\prime}(x_{1}v_{c}):=0, f(x1xp):=5f^{\prime}(x_{1}x_{p^{\prime}}):=5, f(x1u):=1f^{\prime}(x_{1}u^{\prime}):=1, and f(xpu):=0f^{\prime}(x_{p^{\prime}}u^{\prime}):=0. Assign labels 1 and 0 alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}.

2.4. If f(yqp1p1xp)=5f^{\prime}(y^{p^{\prime}-1}_{q_{p^{\prime}-1}}x_{p^{\prime}})=5 and u=vu^{\prime}=v^{\prime}, then assign labels for the remaining vertices and edges as follows. Let f(vc):=1f^{\prime}(v_{c}):=1. Assign labels 0 and 1 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (notice that f(xp1)=1f^{\prime}(x_{p^{\prime}-1})=1). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of algorithm LABEL-K2. Let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=3f^{\prime}(x_{p^{\prime}}):=3, f(u):=4f^{\prime}(u^{\prime}):=4, f(x1vc):=3f^{\prime}(x_{1}v_{c}):=3, f(x1xp):=0f^{\prime}(x_{1}x_{p^{\prime}}):=0, f(x1u):=2f^{\prime}(x_{1}u^{\prime}):=2, and f(xpu):=1f^{\prime}(x_{p^{\prime}}u^{\prime}):=1. Assign labels 0 and 1 alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}.

(Case-3) f(vcw)=3f(v_{c}w)=3.

First consider the case where p=2p^{\prime}=2 and u=vu^{\prime}=v^{\prime}. If f(w)5f(w)\neq 5, then let f(x1):=0f^{\prime}(x_{1}):=0, f(vc):=5f^{\prime}(v_{c}):=5, f(x2):=2f^{\prime}(x_{2}):=2, f(u):=1f^{\prime}(u^{\prime}):=1, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(vcx2):=0f^{\prime}(v_{c}x_{2}):=0, f(x1x2):=5f^{\prime}(x_{1}x_{2}):=5, f(x2u):=4f^{\prime}(x_{2}u^{\prime}):=4, and f(ux1):=3f^{\prime}(u^{\prime}x_{1}):=3. If f(w)=5f(w)=5, then let f(x1):=4f^{\prime}(x_{1}):=4, f(vc):=0f^{\prime}(v_{c}):=0, f(x2):=2f^{\prime}(x_{2}):=2, f(u):=3f^{\prime}(u^{\prime}):=3, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(vcx2):=4f^{\prime}(v_{c}x_{2}):=4, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x2u):=5f^{\prime}(x_{2}u^{\prime}):=5, and f(ux1):=1f^{\prime}(u^{\prime}x_{1}):=1.

Next consider the case where p=2p^{\prime}=2 and uvu^{\prime}\neq v^{\prime}. If f(w)5f(w)\neq 5, then let f(x1):=0f^{\prime}(x_{1}):=0, f(vc):=5f^{\prime}(v_{c}):=5, f(x2):=2f^{\prime}(x_{2}):=2, f(u):=1f^{\prime}(u^{\prime}):=1, f(v):=0f^{\prime}(v^{\prime}):=0, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(vcx2):=0f^{\prime}(v_{c}x_{2}):=0, f(x1x2):=5f^{\prime}(x_{1}x_{2}):=5, f(x2v):=4f^{\prime}(x_{2}v^{\prime}):=4, and f(ux1):=4f^{\prime}(u^{\prime}x_{1}):=4. If f(w)=5f(w)=5, then let f(x1):=3f^{\prime}(x_{1}):=3, f(vc):=0f^{\prime}(v_{c}):=0, f(x2):=5f^{\prime}(x_{2}):=5, f(u):=5f^{\prime}(u^{\prime}):=5, f(v):=4f^{\prime}(v^{\prime}):=4, f(x1vc):=5f^{\prime}(x_{1}v_{c}):=5, f(vcx2):=2f^{\prime}(v_{c}x_{2}):=2, f(x1x2):=0f^{\prime}(x_{1}x_{2}):=0, f(x2v):=1f^{\prime}(x_{2}v^{\prime}):=1, and f(ux1):=1f^{\prime}(u^{\prime}x_{1}):=1. Observe that in both cases, the resulting labeling is feasible for H+G2+{(x1,u),(xp,v)}H+G_{2}+\{(x_{1},u^{\prime}),(x_{p^{\prime}},v^{\prime})\} and satisfies the conditions of Lemma 1.

Finally consider the case where p2p^{\prime}\neq 2. We divide this case into the following two subcases: (3.1) f(w)0f^{\prime}(w)\neq 0 and (3.2) f(w)=0f^{\prime}(w)=0.

(3.1) Assume that f(w)0f^{\prime}(w)\neq 0.

We can obtain a (2,1)-total labeling of GG in the following manner.

1. Assign label 3 to each inner edge in G2G_{2}.

2. Let f(vc):=0f^{\prime}(v_{c}):=0. Assign labels 1 and 0 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (note that as a result, f(xp1)=0f^{\prime}(x_{p^{\prime}-1})=0). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of algorithm LABEL-K2. Assign labels 1 and 0 alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}.

3. Trace outer edges from (vc,x2)(v_{c},x_{2}) to (yqp1p1,xp)(y^{p^{\prime}-1}_{q_{p^{\prime}-1}},x_{p^{\prime}}) on {(vc,x2)}E(P2)E(Pp1)\{(v_{c},x_{2})\}\cup E(P_{2})\cup\cdots\cup E(P_{p^{\prime}-1}) in clockwise order and assign labels 4 and 5 (resp., 5 and 4) alternately if the number of outer edges in G2G_{2} is even (resp., odd). Notice that as a result, the edge (yqp1p1,xp)(y^{p^{\prime}-1}_{q_{p^{\prime}-1}},x_{p^{\prime}}) has label 5.

3.1. If uvu^{\prime}\neq v^{\prime}, then let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=3f^{\prime}(x_{p^{\prime}}):=3, f(u):=4f^{\prime}(u^{\prime}):=4, f(v):=5f^{\prime}(v^{\prime}):=5, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(x1xp):=0f^{\prime}(x_{1}x_{p^{\prime}}):=0, f(x1u):=1f^{\prime}(x_{1}u^{\prime}):=1, and f(xpv):=1f^{\prime}(x_{p^{\prime}}v^{\prime}):=1. Then, extend ff^{\prime} to a (2,1)-total labeling of GG according to Lemma 1.

3.2. If u=vu^{\prime}=v^{\prime}, then let f(x1):=5f^{\prime}(x_{1}):=5, f(xp):=2f^{\prime}(x_{p^{\prime}}):=2, f(u):=0f^{\prime}(u^{\prime}):=0, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(x1xp):=0f^{\prime}(x_{1}x_{p^{\prime}}):=0, f(x1u):=3f^{\prime}(x_{1}u^{\prime}):=3, and f(xpu):=4f^{\prime}(x_{p^{\prime}}u^{\prime}):=4.

(3.2) Assume that f(w)=0f^{\prime}(w)=0.

We can obtain a (2,1)-total labeling of GG in the following manner.

1. Assign label 2 to each inner edge in G2G_{2}.

2. Let f(vc):=5f^{\prime}(v_{c}):=5. Assign labels 4 and 5 alternately to 3-vertices x2,x3,,xp1x_{2},x_{3},\ldots,x_{p^{\prime}-1} (note that as a result, f(xp1)=5f^{\prime}(x_{p^{\prime}-1})=5). Assign labels to 2-vertices yjiy_{j}^{i} (2ip22\leq i\leq p^{\prime}-2) according to Step 2 of Algorithm LABEL-K2’, where Algorithm LABEL-K2’ denotes the algorithm obtained from Algorithm LABEL-K2 by replacing each label ii to be assigned with 5i5-i. Assign labels 4 and 5 alternately to 2-vertices y1p1,y2p1,,yqp1p1y_{1}^{p^{\prime}-1},y_{2}^{p^{\prime}-1},\ldots,y_{q_{p^{\prime}-1}}^{p^{\prime}-1}.

3. Trace outer edges from (vc,x2)(v_{c},x_{2}) to (yqp1p1,xp)(y^{p^{\prime}-1}_{q_{p^{\prime}-1}},x_{p^{\prime}}) on {(vc,x2)}E(P2)E(Pp1)\{(v_{c},x_{2})\}\cup E(P_{2})\cup\cdots\cup E(P_{p^{\prime}-1}) in clockwise order and assign labels 1 and 0 (resp., 0 and 1) alternately if the number of outer edges in G2G_{2} is even (resp., odd). Notice that as a result, the edge (yqp1p1,xp)(y^{p^{\prime}-1}_{q_{p^{\prime}-1}},x_{p^{\prime}}) has label 0.

3.1. If uvu^{\prime}\neq v^{\prime}, then let f(x1):=0f^{\prime}(x_{1}):=0, f(xp):=2f^{\prime}(x_{p^{\prime}}):=2, f(u):=1f^{\prime}(u^{\prime}):=1, f(v):=0f^{\prime}(v^{\prime}):=0, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(x1xp):=4f^{\prime}(x_{1}x_{p^{\prime}}):=4, f(x1u):=5f^{\prime}(x_{1}u^{\prime}):=5, and f(xpv):=5f^{\prime}(x_{p^{\prime}}v^{\prime}):=5. Then, extend ff^{\prime} to a (2,1)-total labeling of GG according to Lemma 1.

3.2. If u=vu^{\prime}=v^{\prime}, then let f(x1):=0f^{\prime}(x_{1}):=0, f(xp):=2f^{\prime}(x_{p^{\prime}}):=2, f(u):=1f^{\prime}(u^{\prime}):=1, f(x1vc):=2f^{\prime}(x_{1}v_{c}):=2, f(x1xp):=4f^{\prime}(x_{1}x_{p^{\prime}}):=4, f(x1u):=3f^{\prime}(x_{1}u^{\prime}):=3, and f(xpu):=5f^{\prime}(x_{p^{\prime}}u^{\prime}):=5.

4 Case Δ(G)=4\Delta(G)=4

Let G=(V,E)G=(V,E) be an outerplane graph. The following structural properties of outerplane graphs are known.

Theorem 4.1

[14] Every outerplane graph GG with δ(G)=2\delta(G)=2 contains one of the following configurations:
(C1) two adjacent 2-vertices u1u_{1} and u2u_{2}.
(C2) a 3-face [u1u2u3][u_{1}u_{2}u_{3}] with d(u1)=2d(u_{1})=2 and d(u2)=3d(u_{2})=3.
(C3) two 3-faces [u1u2u3][u_{1}u_{2}u_{3}] and [u3u4u5][u_{3}u_{4}u_{5}] such that d(u3)=4d(u_{3})=4 and d(u2)=d(u4)=2d(u_{2})=d(u_{4})=2.

In [4], Chen and Wang proved that if GG does not contain (C3) and Δ(G)=4\Delta(G)=4 holds, then λ2T(G)6\lambda_{2}^{T}(G)\leq 6. Their proof utilizes the property that if GG contains (C1) or (C2), then a (2,1)-total labeling of GG can be extended from a (2,1)-total labeling of some proper subgraph of GG. Here, we investigate the case where GG contains neither (C1) nor (C2), and derive a new structural property, i.e, if GG contains neither (C1) nor (C2) and Δ(G)=4\Delta(G)=4 holds, then GG contains a new configuration as shown in Lemma 2 below, which includes (C3) as a subconfiguration.

Assume that Δ(G)=4\Delta(G)=4. Define a chain of 3-faces as a sequence

𝒞={[u1u2u3],[u3u4u5],,[u2t1u2tu2t+1]}(t2){\cal C}=\{[u_{1}u_{2}u_{3}],[u_{3}u_{4}u_{5}],\ldots,[u_{2t-1}u_{2t}u_{2t+1}]\}~(t\geq 2)

of 3-faces such that (u2i1,u2i)(u_{2i-1},u_{2i}) and (u2i,u2i+1)(u_{2i},u_{2i+1}) are on the outer face of GG and (u2i1,u2i+1)(u_{2i-1},u_{2i+1}) is an inner edge for each i=1,2,,ti=1,2,\ldots,t (see Fig. 1).

Refer to caption
Figure 1: A chain of 3-faces.

Notice that d(u2i)=2d(u_{2i})=2 holds for each i=1,2,,ti=1,2,\ldots,t and that d(u2i+1)=4d(u_{2i+1})=4 holds for each i=1,2,,t1i=1,2,\ldots,t-1 by Δ(G)=4\Delta(G)=4; no vertex other than u2i1,u2i,u2i+2u_{2i-1},u_{2i},u_{2i+2}, and u2i+3u_{2i+3} is adjacent to u2i+1u_{2i+1}.

Lemma 2

If GG is an outerphane graph with Δ(G)=4\Delta(G)=4 and δ(G)=2\delta(G)=2 which contains neither (C1) nor (C2), then GG has a chain {[u1u2u3],[u3u4u5],,[u2t1u2tu2t+1]}\{[u_{1}u_{2}u_{3}],[u_{3}u_{4}u_{5}],\ldots,[u_{2t-1}u_{2t}u_{2t+1}]\} of 3-faces such that (u1,u2t+1)(u_{1},u_{2t+1}) is an inner edge.

Proof

Let GG be an outerphane graph with δ(G)=2\delta(G)=2 which contains neither (C1) nor (C2). Let GG^{\prime} be a 2-connected component of GG which contains exactly one cut vertex vcv_{c} of GG if GG is not 2-connected, G=GG^{\prime}=G otherwise. Notice that |V(G)|3|V(G^{\prime})|\geq 3 by δ(G)=2\delta(G)=2. Then GG^{\prime} has an inner edge since otherwise GG contains (C1). There are the following two possible cases: (Case-I) every inner edge in GG^{\prime} belongs to some endface, (Case-II) otherwise. Assume that each vertex on the boundary of the outerface of GG^{\prime} is numbered v1,v2,,vtv^{\prime}_{1},v^{\prime}_{2},\ldots,v^{\prime}_{t} in clockwise order. Notice that each endface not containing vcv_{c} as a 2-vertex in GG^{\prime} is a 3-face since otherwise it would contain (C1).

(Case-I) Notice that there are at least two end faces in GG^{\prime}. Without loss of generality, let [v1v2v3][v_{1}^{\prime}v_{2}^{\prime}v_{3}^{\prime}] be an endface of GG^{\prime} with vc{v2,v3}v_{c}\notin\{v_{2}^{\prime},v_{3}^{\prime}\}; dG(v2)=dG(v2)=2d_{G}(v_{2}^{\prime})=d_{G^{\prime}}(v_{2}^{\prime})=2. Then an inner edge other than (v1,v3)(v_{1}^{\prime},v_{3}^{\prime}) is incident to v3v_{3}^{\prime}, since otherwise dG(v3)=3d_{G}(v_{3}^{\prime})=3 holds and the 3-face [v1v2v3][v_{1}^{\prime}v_{2}^{\prime}v_{3}^{\prime}] would satisfy the conditions of (C2). By the assumption of Case-I, the inner edge belongs to some endface, and hence we can see that another 3-face [v3v4v5][v_{3}^{\prime}v_{4}^{\prime}v_{5}^{\prime}] exists; {[v1v2v3],[v3v4v5]}\{[v_{1}^{\prime}v_{2}^{\prime}v_{3}^{\prime}],[v_{3}^{\prime}v_{4}^{\prime}v_{5}^{\prime}]\} is a chain. Moreover, observe that if v1vcv_{1}^{\prime}\neq v_{c} also holds, then [vt1vtv1][v_{t-1}^{\prime}v_{t}^{\prime}v_{1}^{\prime}] is a 3-face and {[vt1vtv1],[v1v2v3],[v3v4v5]}\{[v_{t-1}^{\prime}v_{t}^{\prime}v_{1}^{\prime}],[v_{1}^{\prime}v_{2}^{\prime}v_{3}^{\prime}],[v_{3}^{\prime}v_{4}^{\prime}v_{5}^{\prime}]\} is a chain. By repeating similar observations, it follows that there exists a chain 𝒞={[v1v2v3],[v3v4v5],,[vt1vtv1]}{\cal C}=\{[v_{1}^{\prime}v_{2}^{\prime}v_{3}^{\prime}],[v_{3}^{\prime}v_{4}^{\prime}v_{5}^{\prime}],\ldots,[v_{t-1}^{\prime}v_{t}^{\prime}v_{1}^{\prime}]\}. Then notice that dG(vi)=4d_{G^{\prime}}(v_{i}^{\prime})=4 (resp., 2) if ii is odd (resp., even); by Δ(G)=4\Delta(G)=4, if vcv_{c} exists, then vcv_{c} is some vertex vjv_{j}^{\prime} with dG(vj)=2d_{G^{\prime}}(v_{j}^{\prime})=2. Hence, we can see that if vcv_{c} exists, then GvcG^{\prime}-v_{c} contains a chain with an inner edge (vj1,vj+1)(v^{\prime}_{j-1},v_{j+1}^{\prime}), and that otherwise 𝒞{[vt1vtv1]}{\cal C}-\{[v_{t-1}^{\prime}v_{t}^{\prime}v_{1}^{\prime}]\} is a chain with an inner edge (v1,vt1)(v_{1}^{\prime},v_{t-1}^{\prime}) as required.

(Case-II) For an inner edge (vi,vj)E(v_{i}^{\prime},v_{j}^{\prime})\in E, let G[vi,vj]G[v_{i}^{\prime},v_{j}^{\prime}] denote the subgraph of GG induced by {vi,vi+1,,vj1,\{v_{i}^{\prime},v_{i+1}^{\prime},\ldots,v_{j-1}^{\prime}, vj}v_{j}^{\prime}\}. Without loss of generality, let (v1,vi)(v^{\prime}_{1},v^{\prime}_{i}) be an inner edge in GG^{\prime} which does not belong to any endface such that G[v2,vi1]G[v^{\prime}_{2},v^{\prime}_{i-1}] does not contain vcv_{c}. Moreover, we can choose such an edge (v1,vi)(v^{\prime}_{1},v^{\prime}_{i}) that any inner edge in G[v1,vi]G[v^{\prime}_{1},v^{\prime}_{i}] belongs to some endface of GG. Note that if G[v1,vi]G[v_{1}^{\prime},v_{i}^{\prime}] contains an inner edge e=(vj,vk)e^{\prime}=(v_{j}^{\prime},v_{k}^{\prime}) (1jki)(1\leq j\leq k\leq i) not in any endface, then the number of inner edges not in any endface in G[vj,vk]G[v_{j}^{\prime},v_{k}^{\prime}] is less than that in G[v1,vi]G[v_{1}^{\prime},v_{i}^{\prime}] and this observation indicates that this choice of (v1,vi)(v_{1}^{\prime},v_{i}^{\prime}) is possible. Then by applying the similar arguments in Case-I to endfaces in G[v1,vi]G[v^{\prime}_{1},v^{\prime}_{i}] (also in GG), we can observe that there exists a chain {[v1v2v3],[v3v4v5],,[vi2vi1vi]}\{[v_{1}^{\prime}v_{2}^{\prime}v_{3}^{\prime}],[v_{3}^{\prime}v_{4}^{\prime}v_{5}^{\prime}],\ldots,[v_{i-2}^{\prime}v_{i-1}^{\prime}v_{i}^{\prime}]\}, which is a required chain. ∎

By utilizing this lemma, we show that λ2T(G)Δ(G)+2\lambda_{2}^{T}(G)\leq\Delta(G)+2 holds even if Δ(G)=4\Delta(G)=4.

Theorem 4.2

If G=(V,E)G=(V,E) is an outerplane graph with Δ(G)=4\Delta(G)=4, then λ2T(G)6\lambda^{T}_{2}(G)\leq 6.

Proof

We prove this by induction on k=|V(G)|+|E(G)|k=|V(G)|+|E(G)|. The theorem clearly holds if k9k\leq 9. Consider the case of k10k\geq 10 and assume that for each k<kk^{\prime}<k, this theorem holds. We also assume that GG is connected, since otherwise we can treat each component separately. Thus, 1δ(G)21\leq\delta(G)\leq 2.

In the case of δ(G)=1\delta(G)=1, similarly to the proof of Theorem 3.1, we can observe that GG has a (2,1)-total labeling f:V(G)E(G)6f:V(G)\cup E(G)\to{\cal L}_{6}.

Consider the case where δ(G)=2\delta(G)=2. By Theorem 4.1, GG contains at least one of the configurations (C1)–(C3). In [4, Theorem 13], Chen and Wang showed that if GG contains (C1) or (C2), then a (2,1)-total labeling of GG can be obtained from a feaible labeling of some subgraph HH of GG. Here we only consider the case where GG contains neither (C1) nor (C2). By Lemma 2, GG contains a chain 𝒞={[u1u2u3],[u3u4u5],,[u2t1u2tu2t+1]}{\cal C}=\{[u_{1}u_{2}u_{3}],[u_{3}u_{4}u_{5}],\ldots,[u_{2t-1}u_{2t}u_{2t+1}]\} of 3-faces such that (u1,u2t+1)(u_{1},u_{2t+1}) is an inner edge. Let w1w_{1} (resp., w2t+1w_{2t+1}) denote the vertex not in 𝒞{\cal C} which is adjacent to u1u_{1} (resp., u2t+1u_{2t+1}) (note that by Δ(G)=4\Delta(G)=4, no edge other than (u1,w1)(u_{1},w_{1}) and (u2t+1,w2t+1)(u_{2t+1},w_{2t+1}) connects 𝒞{\cal C} and the remaining part of GG). By the induction hypothesis, H:=(G{u2,u3,,u2t}){(u1,u2t+1)}H:=(G-\{u_{2},u_{3},\ldots,u_{2t}\})-\{(u_{1},u_{2t+1})\} has a (2,1)-total labeling f:V(H)E(H)6f:V(H)\cup E(H)\to{\cal L}_{6}. Then by dH(u1)=dH(u2t+1)=1d_{H}(u_{1})=d_{H}(u_{2t+1})=1, we can assign any label in 6{f(w1),f(u1w1)1,f(u1w1),f(u1w1)+1}{\cal L}_{6}-\{f(w_{1}),f(u_{1}w_{1})-1,f(u_{1}w_{1}),f(u_{1}w_{1})+1\} (resp., 6{f(w2t+1),f(u2t+1w2t+1)1,f(u2t+1w2t+1),f(u2t+1w2t+1)+1}{\cal L}_{6}-\{f(w_{2t+1}),f(u_{2t+1}w_{2t+1})-1,f(u_{2t+1}w_{2t+1}),f(u_{2t+1}w_{2t+1})+1\}) to u1u_{1} (resp., u2t+1u_{2t+1}) without violating the feasibility. Let L(u1;f)L(u_{1};f) and L(u2t+1;f)L(u_{2t+1};f) denote the sets of such possible labels with respect to ff for u1u_{1} and u2t+1u_{2t+1}, respectively; L(u1;f)=6{f(w1),L(u_{1};f)={\cal L}_{6}-\{f(w_{1}), f(u1w1)f(u_{1}w_{1}) 1,f(u1w1),f(u1w1)+1}-1,f(u_{1}w_{1}),f(u_{1}w_{1})+1\} and L(u2t+1;f)=6{f(w2t+1),f(u2t+1w2t+1)1,L(u_{2t+1};f)={\cal L}_{6}-\{f(w_{2t+1}),f(u_{2t+1}w_{2t+1})-1, f(u2t+1w2t+1),f(u_{2t+1}w_{2t+1}), f(u2t+1w2t+1)+1}f(u_{2t+1}w_{2t+1})+1\}. Notice that |L(u1;f)|3|L(u_{1};f)|\geq 3 and |L(u2t+1;f))|3|L(u_{2t+1};f))|\geq 3.

Claim

L(u1;f)×L(u2t+1;f)L(u_{1};f)\times L(u_{2t+1};f) contains one of (0,6),(0,1),(0,6),(0,1), (1,2),(2,3),(3,(1,2),(2,3),(3, 4),(4,5),(5,4),(4,5),(5, 6),6), (6,0),(6,5),(5,4),(4,(6,0),(6,5),(5,4),(4, 3),3), (3,2),(2,1)(3,2),(2,1), and (1,0)(1,0).

Proof

Assume for contradiction that this claim does not hold. First consider the case of {0,6}L(u1;f)\{0,6\}\subseteq L(u_{1};f). By assumption, {0,6}L(u2t+1;f)=\{0,6\}\cap L(u_{2t+1};f)=\emptyset. This inidcates that (a1) f(w2t+1)=0f(w_{2t+1})=0 and f(u2t+1w2t+1){5,6}f(u_{2t+1}w_{2t+1})\in\{5,6\} or (a2) f(w2t+1)=6f(w_{2t+1})=6 and f(u2t+1w2t+1){0,1}f(u_{2t+1}w_{2t+1})\in\{0,1\}. In the case of (a1), we have 1L(u2t+1;f)1\in L(u_{2t+1};f) (i.e., (0,1)L(u1;f)×L(u2t+1;f))(0,1)\in L(u_{1};f)\times L(u_{2t+1};f)) and in the case of (a2), we have 5L(u2t+1;f)5\in L(u_{2t+1};f) (i.e., (6,5)L(u1;f)×L(u2t+1;f))(6,5)\in L(u_{1};f)\times L(u_{2t+1};f))), a contradiction.

Next consider the case of 0L(u1;f)0\in L(u_{1};f) and 6L(u1;f)6\notin L(u_{1};f). By assumption, {1,6}L(u2t+1;f)=\{1,6\}\cap L(u_{2t+1};f)=\emptyset. This indicates that (b1) f(w2t+1)=1f(w_{2t+1})=1 and f(u2t+1w2t+1)=5f(u_{2t+1}w_{2t+1})=5, (b2) f(w2t+1)=1f(w_{2t+1})=1 and f(u2t+1w2t+1)=6f(u_{2t+1}w_{2t+1})=6, (b3) f(w2t+1)=6f(w_{2t+1})=6 and f(u2t+1w2t+1)=0f(u_{2t+1}w_{2t+1})=0, (b4) f(w2t+1)=6f(w_{2t+1})=6 and f(u2t+1w2t+1)=1f(u_{2t+1}w_{2t+1})=1, or (b5) f(w2t+1)=6f(w_{2t+1})=6 and f(u2t+1w2t+1)=2f(u_{2t+1}w_{2t+1})=2. In the case of (b1) (resp., (b2), (b3), (b4), (b5)), we have L(u2t+1;f)={0,2,3}L(u_{2t+1};f)=\{0,2,3\} (resp., {0,2,3,4}\{0,2,3,4\}, {2,3,4,5}\{2,3,4,5\}, {3,4,5}\{3,4,5\}, {0,4,5}\{0,4,5\}). By assumption that ff does not satisfy this claim, it follows that in the case of (b1) (resp., (b2), (b3), (b4), (b5)), L(u1;f){0,5}L(u_{1};f)\subseteq\{0,5\} (resp., {0}\{0\}, {0}\{0\} {0,1}\{0,1\}, {0,2}\{0,2\}). All of these cases contraict |L(u1;f)|3|L(u_{1};f)|\geq 3.

By symmetry of labelings, the case of 6L(u1;f)6\in L(u_{1};f) and 0L(u1;f)0\notin L(u_{1};f) can be treated similarly to the previous case. Finally, consider the case of {0,6}L(u1;f)=\{0,6\}\cap L(u_{1};f)=\emptyset. Assume that {0,6}L(u2t+1;f)=\{0,6\}\cap L(u_{2t+1};f)=\emptyset, since any other case can be treated similarly by exchanging parts of u1u_{1} and u2t+1u_{2t+1}. Then we can observe that 3L(u1;f)3\in L(u_{1};f), {2,4}L(u1;f)\{2,4\}\cap L(u_{1};f)\neq\emptyset, 3L(u2t+1;f)3\in L(u_{2t+1};f), and {2,4}L(u2t+1;f)\{2,4\}\cap L(u_{2t+1};f)\neq\emptyset. It follows that (2,3)(2,3) or (4,3)(4,3) are contained in L(u1;f)×L(u2t+1;f)L(u_{1};f)\times L(u_{2t+1};f), a contradiction. ∎

By symmetry, it suffices to consider the following four cases: (Case-1) 0L(u1;f)0\in L(u_{1};f) and 6L(u2t+1;f)6\in L(u_{2t+1};f), (Case-2) 0L(u1;f)0\in L(u_{1};f) and 1L(u2t+1;f)1\in L(u_{2t+1};f), (Case-3) 1L(u1;f)1\in L(u_{1};f) and 2L(u2t+1;f)2\in L(u_{2t+1};f), and (Case-4) 2L(u1;f)2\in L(u_{1};f) and 3L(u2t+1;f)3\in L(u_{2t+1};f). In each case, we will extend ff to a (2,1)-total labeling f:VE6f^{\prime}:V\cup E\to{\cal L}_{6} of GG. Let f(x):=f(x)f^{\prime}(x):=f(x) for every xV(H)E(H)x\in V(H)\cup E(H).

(Case-1) Let f(u1):=0f^{\prime}(u_{1}):=0 and f(u2t+1):=6f^{\prime}(u_{2t+1}):=6.

(1.1) Assume that tt is even; t=2kt=2k.

(1.1.1) Assume that f(u1w1)6f^{\prime}(u_{1}w_{1})\neq 6 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f(u2):=6f^{\prime}(u_{2}):=6, f(u3):=3f^{\prime}(u_{3}):=3, f(u4):=0f^{\prime}(u_{4}):=0, f(u5):=6f^{\prime}(u_{5}):=6, f(u2u3):=1f^{\prime}(u_{2}u_{3}):=1, f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6, f(u3u4):=5f^{\prime}(u_{3}u_{4}):=5, f(u4u5):=4f^{\prime}(u_{4}u_{5}):=4, and f(u3u5):=0f^{\prime}(u_{3}u_{5}):=0. For i=2,3,,ki=2,3,\ldots,k, let f(u4i2):=1f^{\prime}(u_{4i-2}):=1, f(u4i1):=3f^{\prime}(u_{4i-1}):=3, f(u4i):=0f^{\prime}(u_{4i}):=0, f(u4i+1):=6f^{\prime}(u_{4i+1}):=6, f(u4i3u4i2):=3f^{\prime}(u_{4i-3}u_{4i-2}):=3, f(u4i2u4i1):=6f^{\prime}(u_{4i-2}u_{4i-1}):=6, f(u4i3u4i1):=1f^{\prime}(u_{4i-3}u_{4i-1}):=1, f(u4i1u4i):=5f^{\prime}(u_{4i-1}u_{4i}):=5, f(u4iu4i+1):=4f^{\prime}(u_{4i}u_{4i+1}):=4, and f(u4i1u4i+1)f^{\prime}(u_{4i-1}u_{4i+1}) :=0:=0. Denote this labeling ff^{\prime} by f1f_{1} (see Fig. 2).

Refer to caption
Figure 2: A labeling f1f_{1} on 𝒞{\cal C}.

We next assign a label {2,3,4}{f(u1w1),f(u2t+1w2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}), and a label in {2,3,4}{f(u1w1),f(u1u2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {2,3,4}{f(u2t+1w2t+1),\{2,3,4\}-\{f^{\prime}(u_{2t+1}w_{2t+1}), f(u1u2t+1)}f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then the resulting labeling ff^{\prime} is a a (2,1)-total labeling of GG.

(1.1.2) Assume that f(u1w1)=6f^{\prime}(u_{1}w_{1})=6 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f:=f1f^{\prime}:=f_{1}. We reassign labels for some vertices and edges as follows. Let f(u2):=1f^{\prime}(u_{2}):=1, f(u3):=2f^{\prime}(u_{3}):=2, f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u2u3):=6f^{\prime}(u_{2}u_{3}):=6 and f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4. Next we assign a label in {2,3}{f(u2t+1w2t+1)}\{2,3\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and reassign a label in {2,3,4}{f(u1u2t+1),f(u2t+1w2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then the resulting labeling ff^{\prime} is a a (2,1)-total labeling of GG.

(1.1.3) Assume that f(u1w1)6f^{\prime}(u_{1}w_{1})\neq 6 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

Let f:=f1f^{\prime}:=f_{1}. We reassign labels for some vertices and edges as follows. Let f(u2t1):=4f^{\prime}(u_{2t-1}):=4, f(u2t):=5f^{\prime}(u_{2t}):=5, f(u2t1u2t):=0f^{\prime}(u_{2t-1}u_{2t}):=0, f(u2tu2t+1):=1f^{\prime}(u_{2t}u_{2t+1}):=1, and f(u2t1u2t+1):=2f^{\prime}(u_{2t-1}u_{2t+1}):=2. Next we assign a label in {3,4}{f(u1w1)}\{3,4\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {2,3,4}{f(u1u2t+1),f(u1w1)}\{2,3,4\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Then the resulting labeling ff^{\prime} is a (2,1)-total labeling of GG.

(1.1.4) Assume that f(u1w1)=6f^{\prime}(u_{1}w_{1})=6 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

We can obtain a (2,1)-total labeling ff^{\prime} of GG as follows. If t=2t=2, then let f(u2):=3f^{\prime}(u_{2}):=3, f(u3):=1f^{\prime}(u_{3}):=1, f(u4):=3f^{\prime}(u_{4}):=3, f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u2u3):=6f^{\prime}(u_{2}u_{3}):=6, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u3u4):=5f^{\prime}(u_{3}u_{4}):=5, f(u4u5):=1f^{\prime}(u_{4}u_{5}):=1, f(u3u5)=3f^{\prime}(u_{3}u_{5})=3, and f(u1u5):=2f^{\prime}(u_{1}u_{5}):=2.

Assume that t4t\geq 4. Let f:=f1f^{\prime}:=f_{1}. We reassign labels for some vertices and edges similarly to (1.1.2) and (1.1.3). Namely, let f(u2):=1f^{\prime}(u_{2}):=1, f(u3):=2f^{\prime}(u_{3}):=2, f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u2u3):=6f^{\prime}(u_{2}u_{3}):=6, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u2t1):=4f^{\prime}(u_{2t-1}):=4, f(u2t):=5f^{\prime}(u_{2t}):=5, f(u2t1u2t):=0f^{\prime}(u_{2t-1}u_{2t}):=0, f(u2tu2t+1):=1f^{\prime}(u_{2t}u_{2t+1}):=1, f(u2t1u2t+1):=2f^{\prime}(u_{2t-1}u_{2t+1}):=2, and f(u1u2t+1):=3f^{\prime}(u_{1}u_{2t+1}):=3.

(1.2) Assume that tt is odd; t=2k+1(k1)t=2k+1~(k\geq 1) .

(1.2.1) Assume that f(u1w1)6f^{\prime}(u_{1}w_{1})\neq 6 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f(u2):=6f^{\prime}(u_{2}):=6, f(u3):=3f^{\prime}(u_{3}):=3, f(u4):=0f^{\prime}(u_{4}):=0, f(u5):=4f^{\prime}(u_{5}):=4, f(u6):=0f^{\prime}(u_{6}):=0, f(u7):=6f^{\prime}(u_{7}):=6, f(u2u3):=0f^{\prime}(u_{2}u_{3}):=0, f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6, f(u3u4):=5f^{\prime}(u_{3}u_{4}):=5, f(u4u5):=2f^{\prime}(u_{4}u_{5}):=2, f(u3u5):=1f^{\prime}(u_{3}u_{5}):=1, f(u5u6):=6f^{\prime}(u_{5}u_{6}):=6, f(u6u7):=4f^{\prime}(u_{6}u_{7}):=4, and f(u5u7):=0f^{\prime}(u_{5}u_{7}):=0. For i=2,3,,ki=2,3,\ldots,k, let f(u4i):=0f^{\prime}(u_{4i}):=0, f(u4i+1):=4f^{\prime}(u_{4i+1}):=4, f(u4i+2):=0f^{\prime}(u_{4i+2}):=0, f(u4i+3):=6f^{\prime}(u_{4i+3}):=6, f(u4i1u4i):=3f^{\prime}(u_{4i-1}u_{4i}):=3, f(u4iu4i+1):=2f^{\prime}(u_{4i}u_{4i+1}):=2, f(u4i1u4i+1):=1f^{\prime}(u_{4i-1}u_{4i+1}):=1, f(u4i+1u4i+2):=6f^{\prime}(u_{4i+1}u_{4i+2}):=6, f(u4i+2u4i+3):=4f^{\prime}(u_{4i+2}u_{4i+3}):=4, and f(u4i+1u4i+3):=0f^{\prime}(u_{4i+1}u_{4i+3}):=0. Denote this labeling ff^{\prime} by f2f_{2} (see Fig. 3).

Refer to caption
Figure 3: A labeling f2f_{2} on 𝒞{\cal C}.

We next assign a label {2,3,4}{f(u1w1),f(u2t+1w2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and assign a label in {2,3,4}{f(u1w1),f(u1u2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {2,3,4}{f(u2t+1w2t+1),f(u1u2t+1)}\{2,3,4\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}).

(1.2.2) Assume that f(u1w1)=6f^{\prime}(u_{1}w_{1})=6 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f:=f2f^{\prime}:=f_{2}. We reassign labels for some vertices and edges as follows. Let f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=6f^{\prime}(u_{3}):=6, f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, and f(u3u4):=3f^{\prime}(u_{3}u_{4}):=3. Next we assign a label in {2,3}{f(u2t+1w2t+1)}\{2,3\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}). Reassign a label in {2,3,4}{f(u1u2t+1),f(u2t+1w2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}).

(1.2.3) Assume that f(u1w1)6f^{\prime}(u_{1}w_{1})\neq 6 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

Let f:=f2f^{\prime}:=f_{2}. We reassign labels for some vertices and edges as follows. Let f(u2t1):=5f^{\prime}(u_{2t-1}):=5, f(u2t):=3f^{\prime}(u_{2t}):=3, f(u2t1u2t):=0f^{\prime}(u_{2t-1}u_{2t}):=0, f(u2tu2t+1):=1f^{\prime}(u_{2t}u_{2t+1}):=1, and f(u2t1u2t+1):=3f^{\prime}(u_{2t-1}u_{2t+1}):=3. Next we assign a label in {2,4}{f(u1w1)}\{2,4\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {2,3,4}{f(u1w1),f(u1u2t+1)}\{2,3,4\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}).

(1.2.4) Assume that f(u1w1)=6f^{\prime}(u_{1}w_{1})=6 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

Let f:=f2f^{\prime}:=f_{2}. We reassign labels for some vertices and edges similarly to (1.2.2) and (1.2.3). Namely, let f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=6f^{\prime}(u_{3}):=6, f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u3u4):=3f^{\prime}(u_{3}u_{4}):=3, f(u2t1):=5f^{\prime}(u_{2t-1}):=5, f(u2t):=3f^{\prime}(u_{2t}):=3, f(u2t1u2t):=0f^{\prime}(u_{2t-1}u_{2t}):=0, f(u2tu2t+1):=1f^{\prime}(u_{2t}u_{2t+1}):=1, f(u2t1u2t+1):=3f^{\prime}(u_{2t-1}u_{2t+1}):=3, and f(u1u2t+1):=2f^{\prime}(u_{1}u_{2t+1}):=2.

(Case-2) Let f(u1):=0f^{\prime}(u_{1}):=0 and f(u2t+1):=1f^{\prime}(u_{2t+1}):=1.

(2.1) Assume that tt is even; t=2kt=2k.

(2.1.1) Assume that f(u1w1)2f^{\prime}(u_{1}w_{1})\neq 2 and f(u2t+1w2t+1){3,4}f^{\prime}(u_{2t+1}w_{2t+1})\notin\{3,4\}.

Let f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=6f^{\prime}(u_{3}):=6, f(u4):=5f^{\prime}(u_{4}):=5, f(u5):=1f^{\prime}(u_{5}):=1, f(u2u3):=0f^{\prime}(u_{2}u_{3}):=0, f(u1u3):=2f^{\prime}(u_{1}u_{3}):=2, f(u3u4):=1f^{\prime}(u_{3}u_{4}):=1, f(u4u5):=3f^{\prime}(u_{4}u_{5}):=3, and f(u3u5):=4f^{\prime}(u_{3}u_{5}):=4. For i=2,3,,ki=2,3,\ldots,k, let f(u4i2):=4f^{\prime}(u_{4i-2}):=4, f(u4i1):=0f^{\prime}(u_{4i-1}):=0, f(u4i):=2f^{\prime}(u_{4i}):=2, f(u4i+1):=1f^{\prime}(u_{4i+1}):=1, f(u4i3u4i2):=6f^{\prime}(u_{4i-3}u_{4i-2}):=6, f(u4i2u4i1):=2f^{\prime}(u_{4i-2}u_{4i-1}):=2, f(u4i3u4i1):=5f^{\prime}(u_{4i-3}u_{4i-1}):=5, f(u4i1u4i):=6f^{\prime}(u_{4i-1}u_{4i}):=6, f(u4iu4i+1):=4f^{\prime}(u_{4i}u_{4i+1}):=4, and f(u4i1u4i+1)f^{\prime}(u_{4i-1}u_{4i+1}) :=3:=3. Denote this labeling ff^{\prime} by f3f_{3} (see Fig. 4).

Refer to caption
Figure 4: A labeling f3f_{3} on 𝒞{\cal C}.

Consider the case of t=2t=2. We assign a label {3,5,6}{f(u1w1),f(u5w5)}\{3,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{5}w_{5})\} to (u1,u5)(u_{1},u_{5}) and a label in {4,5,6}{f(u1w1),f(u1u5)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{5})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {3,5,6}{f(u5w5),f(u1u5)}\{3,5,6\}-\{f^{\prime}(u_{5}w_{5}),f^{\prime}(u_{1}u_{5})\} to (u4,u5)(u_{4},u_{5}). Then if f(u4u5){5,6}f^{\prime}(u_{4}u_{5})\in\{5,6\}, then reassign a label for u4u_{4} as f(u4):=3f^{\prime}(u_{4}):=3.

Consider the case of t4t\geq 4. Assign a label {4,5,6}{f(u1w1),f(u2t+1w2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1w1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5,6}{f(u2t+1w2t+1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=6f^{\prime}(u_{2t}u_{2t+1})=6, then reassign a label for (u2t1,u2t)(u_{2t-1},u_{2t}) as f(u2t1u2t):=4f^{\prime}(u_{2t-1}u_{2t}):=4.

(2.1.2) Assume that f(u1w1)=2f^{\prime}(u_{1}w_{1})=2 and f(u2t+1w2t+1){3,4}f^{\prime}(u_{2t+1}w_{2t+1})\notin\{3,4\}.

Let f:=f3f^{\prime}:=f_{3}. We reassign a label for (u1,u3)(u_{1},u_{3}) to 3.

Consider the case of t=2t=2. Assign a label in {5,6}{f(u5w5)}\{5,6\}-\{f^{\prime}(u_{5}w_{5})\} to (u1,u5)(u_{1},u_{5}) and a label in {4,5,6}{f(u1u5)}\{4,5,6\}-\{f^{\prime}(u_{1}u_{5})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {3,5,6}{f(u5w5),f(u1u5)}\{3,5,6\}-\{f^{\prime}(u_{5}w_{5}),f^{\prime}(u_{1}u_{5})\} to (u4,u5)(u_{4},u_{5}). Then if f(u4u5){5,6}f^{\prime}(u_{4}u_{5})\in\{5,6\}, then let f(u4):=3f^{\prime}(u_{4}):=3.

Consider the case of t4t\geq 4. Assign a label in {4,5,6}{f(u2t+1w2t+1)}\{4,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}\{4,5,6\} {f(u1u2t+1)}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5,6}{f(u2t+1w2t+1),\{4,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}), f(u1u2t+1)}f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then, if f(u2tu2t+1)f^{\prime}(u_{2t}u_{2t+1}) =6=6, then let f(u2t1u2t):=4f^{\prime}(u_{2t-1}u_{2t}):=4.

(2.1.3) Assume that f(u1w1)2f^{\prime}(u_{1}w_{1})\neq 2 and f(u2t+1w2t+1)=3f^{\prime}(u_{2t+1}w_{2t+1})=3.

If t=2t=2, then we can obtain a feasible labeling similarly to (2.1.1). Consider the case of t4t\geq 4. Let f:=f3f^{\prime}:=f_{3}. Reassign a label for (u2t1,u2t+1)(u_{2t-1},u_{2t+1}) as f(u2t1u2t+1):=6f^{\prime}(u_{2t-1}u_{2t+1}):=6. Next we assign a label in {4,5}{f(u1w1)}\{4,5\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1u2t+1),f(u1w1)}\{4,5,6\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5}{f(u1u2t+1)}\{4,5\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=4f^{\prime}(u_{2t}u_{2t+1})=4, then let f(u2t):=6f^{\prime}(u_{2t}):=6 and f(u2t1u2t):=3f^{\prime}(u_{2t-1}u_{2t}):=3, and if f(u2tu2t+1)=5f^{\prime}(u_{2t}u_{2t+1})=5, then let f(u2t1u2t):=4f^{\prime}(u_{2t-1}u_{2t}):=4.

(2.1.4) Assume that f(u1w1)2f^{\prime}(u_{1}w_{1})\neq 2 and f(u2t+1w2t+1)=4f^{\prime}(u_{2t+1}w_{2t+1})=4.

If t4t\geq 4, then we can obtain a feasible labeling similarly to (2.1.1). Consider the case of t=2t=2. Let f:=f3f^{\prime}:=f_{3}. We reassign labels for (u3,u5)(u_{3},u_{5}) and u4u_{4} as f(u3u5):=3f^{\prime}(u_{3}u_{5}):=3 and f(u4):=3f^{\prime}(u_{4}):=3. Next we assign a label in {5,6}{f(u1w1)}\{5,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u5)(u_{1},u_{5}) and a label in {4,5,6}{f(u1u5),f(u1w1)}\{4,5,6\}-\{f^{\prime}(u_{1}u_{5}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {5,6}{f(u1u5)}\{5,6\}-\{f^{\prime}(u_{1}u_{5})\} to (u4,u5)(u_{4},u_{5}).

(2.1.5) Assume that f(u1w1)=2f^{\prime}(u_{1}w_{1})=2 and f(u2t+1w2t+1)=3f^{\prime}(u_{2t+1}w_{2t+1})=3.

If t=2t=2, then we can obtain a feasible labeling similarly to (2.1.2). Consider the case of t4t\geq 4. Let f:=f3f^{\prime}:=f_{3}. We reassign labels for some vertices and edges similarly to (2.1.2) and (2.1.3). Namely, let f(u1u2):=4f^{\prime}(u_{1}u_{2}):=4, f(u1u3):=3f^{\prime}(u_{1}u_{3}):=3, f(u2t):=6f^{\prime}(u_{2t}):=6, f(u2t1u2t):=3f^{\prime}(u_{2t-1}u_{2t}):=3, f(u2tu2t+1):=4f^{\prime}(u_{2t}u_{2t+1}):=4, f(u2t1u2t+1):=6f^{\prime}(u_{2t-1}u_{2t+1}):=6, and f(u1u2t+1):=5f^{\prime}(u_{1}u_{2t+1}):=5.

(2.1.6) Assume that f(u1w1)=2f^{\prime}(u_{1}w_{1})=2 and f(u2t+1w2t+1)=4f^{\prime}(u_{2t+1}w_{2t+1})=4.

If t4t\geq 4, then we can obtain a feasible labeling similarly to (2.1.2). If t=2t=2, let f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=6f^{\prime}(u_{3}):=6, f(u4):=3f^{\prime}(u_{4}):=3, f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u2u3):=0f^{\prime}(u_{2}u_{3}):=0, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u3u4):=1f^{\prime}(u_{3}u_{4}):=1, f(u4u5):=5f^{\prime}(u_{4}u_{5}):=5, f(u3u5)=3f^{\prime}(u_{3}u_{5})=3, and f(u1u5):=6f^{\prime}(u_{1}u_{5}):=6.

(2.2) Assume that tt is odd; t=2k+1(k1)t=2k+1~(k\geq 1) .

(2.2.1) Assume that f(u1w1)2f^{\prime}(u_{1}w_{1})\neq 2 and f(u2t+1w2t+1)5f^{\prime}(u_{2t+1}w_{2t+1})\neq 5.

Let f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=6f^{\prime}(u_{3}):=6, f(u4):=4f^{\prime}(u_{4}):=4, f(u5):=0f^{\prime}(u_{5}):=0, f(u6):=6f^{\prime}(u_{6}):=6, f(u7):=1f^{\prime}(u_{7}):=1, f(u2u3):=0f^{\prime}(u_{2}u_{3}):=0, f(u1u3):=2f^{\prime}(u_{1}u_{3}):=2, f(u3u4):=1f^{\prime}(u_{3}u_{4}):=1, f(u4u5):=6f^{\prime}(u_{4}u_{5}):=6, f(u3u5):=4f^{\prime}(u_{3}u_{5}):=4, f(u5u6):=2f^{\prime}(u_{5}u_{6}):=2, f(u6u7):=3f^{\prime}(u_{6}u_{7}):=3, and f(u5u7):=5f^{\prime}(u_{5}u_{7}):=5. For i=2,3,,ki=2,3,\ldots,k, let f(u4i):=6f^{\prime}(u_{4i}):=6, f(u4i+1):=0f^{\prime}(u_{4i+1}):=0, f(u4i+2):=6f^{\prime}(u_{4i+2}):=6, f(u4i+3):=1f^{\prime}(u_{4i+3}):=1, f(u4i1u4i):=4f^{\prime}(u_{4i-1}u_{4i}):=4, f(u4iu4i+1):=3f^{\prime}(u_{4i}u_{4i+1}):=3, f(u4i1u4i+1):=6f^{\prime}(u_{4i-1}u_{4i+1}):=6, f(u4i+1u4i+2):=2f^{\prime}(u_{4i+1}u_{4i+2}):=2, f(u4i+2u4i+3):=3f^{\prime}(u_{4i+2}u_{4i+3}):=3, and f(u4i+1u4i+3):=5f^{\prime}(u_{4i+1}u_{4i+3}):=5. Denote this labeling ff^{\prime} by f4f_{4} (see Fig. 5).

Refer to caption
Figure 5: A labeling f4f_{4} on 𝒞{\cal C}.

We next assign a label {3,4,6}{f(u1w1),f(u2t+1w2t+1)}\{3,4,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}). and a label in {4,5,6}{f(u1w1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {3,4,6}{f(u2t+1w2t+1),f(u1u2t+1)}\{3,4,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=6f^{\prime}(u_{2t}u_{2t+1})=6, then let f(u2t):=4f^{\prime}(u_{2t}):=4.

(2.2.2) Assume that f(u1w1)=2f^{\prime}(u_{1}w_{1})=2 and f(u2t+1w2t+1)5f^{\prime}(u_{2t+1}w_{2t+1})\neq 5.

Let f:=f4f^{\prime}:=f_{4}. We reassign labels for some vertices and edges as follows. Let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5 and f(u1u3):=3f^{\prime}(u_{1}u_{3}):=3. Next we assign a label in {4,6}{f(u2t+1w2t+1)}\{4,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}). Reassign a label in {3,4,6}{f(u1u2t+1),\{3,4,6\}-\{f^{\prime}(u_{1}u_{2t+1}), f(u2t+1w2t+1)}f^{\prime}(u_{2t+1}w_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=6f^{\prime}(u_{2t}u_{2t+1})=6, then let f(u2t):=4f^{\prime}(u_{2t}):=4.

(2.2.3) Assume that f(u1w1)2f^{\prime}(u_{1}w_{1})\neq 2 and f(u2t+1w2t+1)=5f^{\prime}(u_{2t+1}w_{2t+1})=5.

Let f:=f4f^{\prime}:=f_{4}.

First consider the case of t=3t=3. Let f(u6):=2f^{\prime}(u_{6}):=2, f(u5u6):=5f^{\prime}(u_{5}u_{6}):=5, and f(u5u7):=3f^{\prime}(u_{5}u_{7}):=3. Next we assign a label in {4,6}{f(u1w1)}\{4,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u7)(u_{1},u_{7}) and a label in {4,5,6}{f(u1w1),f(u1u7)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{7})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,6}{f(u1u7)}\{4,6\}-\{f^{\prime}(u_{1}u_{7})\} to (u6,u7)(u_{6},u_{7}).

Next consider the case of t5t\geq 5. Let f(u2t1u2t+1):=4f^{\prime}(u_{2t-1}u_{2t+1}):=4. Next we assign a label in {3,6}{f(u1w1)}\{3,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1w1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {3,6}{f(u1u2t+1)}\{3,6\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=6f^{\prime}(u_{2t}u_{2t+1})=6, then f(u2t):=4f^{\prime}(u_{2t}):=4.

(2.2.4) Assume that f(u1w1)=2f^{\prime}(u_{1}w_{1})=2 and f(u2t+1w2t+1)=5f^{\prime}(u_{2t+1}w_{2t+1})=5.

Let f:=f4f^{\prime}:=f_{4}. We reassign labels for some vertices and edges similarly to (2.2.2) and (2.2.3). Namely, if t=3t=3, let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=3f^{\prime}(u_{1}u_{3}):=3, f(u6):=2f^{\prime}(u_{6}):=2, f(u5u6):=5f^{\prime}(u_{5}u_{6}):=5, f(u6u7):=6f^{\prime}(u_{6}u_{7}):=6, f(u5u7):=3f^{\prime}(u_{5}u_{7}):=3, and f(u1u7):=4f^{\prime}(u_{1}u_{7}):=4. If t5t\geq 5, then let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=3f^{\prime}(u_{1}u_{3}):=3, f(u2t1u2t+1):=4f^{\prime}(u_{2t-1}u_{2t+1}):=4, and f(u1u2t+1):=6f^{\prime}(u_{1}u_{2t+1}):=6.

(Case-3) Let f(u1):=1f^{\prime}(u_{1}):=1 and f(u2t+1):=2f^{\prime}(u_{2t+1}):=2.

(3.1) Assume that tt is even; t=2kt=2k.

(3.1.1) Assume that f(u1w1)3f^{\prime}(u_{1}w_{1})\neq 3 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f(u2):=0f^{\prime}(u_{2}):=0, f(u3):=5f^{\prime}(u_{3}):=5, f(u4):=3f^{\prime}(u_{4}):=3, f(u5):=2f^{\prime}(u_{5}):=2, f(u2u3):=2f^{\prime}(u_{2}u_{3}):=2, f(u1u3):=3f^{\prime}(u_{1}u_{3}):=3, f(u3u4):=1f^{\prime}(u_{3}u_{4}):=1, f(u4u5):=6f^{\prime}(u_{4}u_{5}):=6, and f(u3u5):=0f^{\prime}(u_{3}u_{5}):=0. For i=2,3,,ki=2,3,\ldots,k, let f(u4i2):=0f^{\prime}(u_{4i-2}):=0, f(u4i1):=6f^{\prime}(u_{4i-1}):=6, f(u4i):=0f^{\prime}(u_{4i}):=0, f(u4i+1):=2f^{\prime}(u_{4i+1}):=2, f(u4i3u4i2):=5f^{\prime}(u_{4i-3}u_{4i-2}):=5, f(u4i2u4i1):=3f^{\prime}(u_{4i-2}u_{4i-1}):=3, f(u4i3u4i1):=4f^{\prime}(u_{4i-3}u_{4i-1}):=4, f(u4i1u4i):=2f^{\prime}(u_{4i-1}u_{4i}):=2, f(u4iu4i+1):=6f^{\prime}(u_{4i}u_{4i+1}):=6, and f(u4i1u4i+1)f^{\prime}(u_{4i-1}u_{4i+1}) :=0:=0. Denote this labeling ff^{\prime} by f5f_{5} (see Fig. 6).

Refer to caption
Figure 6: A labeling f5f_{5} on 𝒞{\cal C}.

Assign a label {4,5,6}{f(u1w1),f(u2t+1w2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1w1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5,6}{f(u2t+1w2t+1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u4u5)=4f^{\prime}(u_{4}u_{5})=4, then let f(u4):=6f^{\prime}(u_{4}):=6.

(3.1.2) Assume that f(u1w1)=3f^{\prime}(u_{1}w_{1})=3 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f:=f5f^{\prime}:=f_{5}. We reassign labels for some vertices and edges as follows. Let f(u3):=6f^{\prime}(u_{3}):=6, f(u2u3):=3f^{\prime}(u_{2}u_{3}):=3, and f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4. Assign a label in {5,6}{f(u2t+1w2t+1)}\{5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {5,6}{f(u1u2t+1)}\{5,6\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5,6}{f(u2t+1w2t+1),\{4,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}), f(u1u2t+1)}f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u4u5)=4f^{\prime}(u_{4}u_{5})=4, then let f(u4):=0f^{\prime}(u_{4}):=0 and f(u3u4):=2f^{\prime}(u_{3}u_{4}):=2.

(3.1.3) Assume that f(u1w1)3f^{\prime}(u_{1}w_{1})\neq 3 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

Let f:=f5f^{\prime}:=f_{5}.

Consider the case of t=2t=2. Let f(u3u5):=4f^{\prime}(u_{3}u_{5}):=4 and f(u3):=6f^{\prime}(u_{3}):=6. Next we assign a label in {5,6}{f(u1w1)}\{5,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u5)(u_{1},u_{5}) and a label in {4,5,6}{f(u1u5),f(u1w1)}\{4,5,6\}-\{f^{\prime}(u_{1}u_{5}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {5,6}{f(u1u5)}\{5,6\}-\{f^{\prime}(u_{1}u_{5})\} to (u4,u5)(u_{4},u_{5}).

Consider the case of t4t\geq 4. Let f(u2t1u2t+1):=5f^{\prime}(u_{2t-1}u_{2t+1}):=5, f(u2t2):=1f^{\prime}(u_{2t-2}):=1, and f(u2t1):=0f^{\prime}(u_{2t-1}):=0. Next we assign a label in {4,6}{f(u1w1)}\{4,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1u2t+1),f(u1w1)}\{4,5,6\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,6}{f(u1u2t+1)}\{4,6\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=4f^{\prime}(u_{2t}u_{2t+1})=4, then let f(u2t):=6f^{\prime}(u_{2t}):=6, and if f(u2tu2t+1)=6f^{\prime}(u_{2t}u_{2t+1})=6, then let f(u2t):=4f^{\prime}(u_{2t}):=4.

(3.1.4) Assume that f(u1w1)=3f^{\prime}(u_{1}w_{1})=3 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

If t=2t=2, let f(u2):=2f^{\prime}(u_{2}):=2, f(u3):=0f^{\prime}(u_{3}):=0, f(u4):=6f^{\prime}(u_{4}):=6, f(u1u2):=6f^{\prime}(u_{1}u_{2}):=6, f(u2u3):=5f^{\prime}(u_{2}u_{3}):=5, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u3u4):=2f^{\prime}(u_{3}u_{4}):=2, f(u4u5):=4f^{\prime}(u_{4}u_{5}):=4, f(u3u5)=6f^{\prime}(u_{3}u_{5})=6, and f(u1u5):=5f^{\prime}(u_{1}u_{5}):=5.

Consider the case of t4t\geq 4. Let f:=f5f^{\prime}:=f_{5}. We reassign labels for some vertices and edges similarly to (3.1.2) and (3.1.3). Namely, let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u3):=6f^{\prime}(u_{3}):=6, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u2t2):=1f^{\prime}(u_{2t-2}):=1, f(u2t1):=0f^{\prime}(u_{2t-1}):=0, f(u2t):=6f^{\prime}(u_{2t}):=6, f(u2tu2t+1):=4f^{\prime}(u_{2t}u_{2t+1}):=4, f(u2t1u2t+1):=5f^{\prime}(u_{2t-1}u_{2t+1}):=5, and f(u1u2t+1):=6f^{\prime}(u_{1}u_{2t+1}):=6.

(3.2) Assume that tt is odd; t=2k+1(k1)t=2k+1~(k\geq 1) .

(3.2.1) Assume that f(u1w1)3f^{\prime}(u_{1}w_{1})\neq 3 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f(u2):=0f^{\prime}(u_{2}):=0, f(u3):=5f^{\prime}(u_{3}):=5, f(u4):=2f^{\prime}(u_{4}):=2, f(u5):=6f^{\prime}(u_{5}):=6, f(u6):=0f^{\prime}(u_{6}):=0, f(u7):=2f^{\prime}(u_{7}):=2, f(u2u3):=2f^{\prime}(u_{2}u_{3}):=2, f(u1u3):=3f^{\prime}(u_{1}u_{3}):=3, f(u3u4):=0f^{\prime}(u_{3}u_{4}):=0, f(u4u5):=4f^{\prime}(u_{4}u_{5}):=4, f(u3u5):=1f^{\prime}(u_{3}u_{5}):=1, f(u5u6):=2f^{\prime}(u_{5}u_{6}):=2, f(u6u7):=6f^{\prime}(u_{6}u_{7}):=6, and f(u5u7):=0f^{\prime}(u_{5}u_{7}):=0. For i=2,3,,ki=2,3,\ldots,k, let f(u4i):=0f^{\prime}(u_{4i}):=0, f(u4i+1):=6f^{\prime}(u_{4i+1}):=6, f(u4i+2):=0f^{\prime}(u_{4i+2}):=0, f(u4i+3):=2f^{\prime}(u_{4i+3}):=2, f(u4i1u4i):=5f^{\prime}(u_{4i-1}u_{4i}):=5, f(u4iu4i+1):=3f^{\prime}(u_{4i}u_{4i+1}):=3, f(u4i1u4i+1):=4f^{\prime}(u_{4i-1}u_{4i+1}):=4, f(u4i+1u4i+2):=2f^{\prime}(u_{4i+1}u_{4i+2}):=2, f(u4i+2u4i+3):=6f^{\prime}(u_{4i+2}u_{4i+3}):=6, and f(u4i+1u4i+3):=0f^{\prime}(u_{4i+1}u_{4i+3}):=0. Denote this labeling ff^{\prime} by f6f_{6} (see Fig. 7).

Refer to caption
Figure 7: A labeling f6f_{6} on 𝒞{\cal C}.

We next assign a label {4,5,6}{f(u1w1),f(u2t+1w2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1w1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5,6}{f(u2t+1w2t+1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}).

(3.2.2) Assume that f(u1w1)=3f^{\prime}(u_{1}w_{1})=3 and f(u2t+1w2t+1)0f^{\prime}(u_{2t+1}w_{2t+1})\neq 0.

Let f:=f6f^{\prime}:=f_{6}. We reassign labels for some vertices and edges as follows. Let f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6 and f(u3):=4f^{\prime}(u_{3}):=4. Next we assign a label in {4,5}{f(u2t+1w2t+1)}\{4,5\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5}{f(u1u2t+1)}\{4,5\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5,6}{f(u1u2t+1),f(u2t+1w2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}).

(3.2.3) Assume that f(u1w1)3f^{\prime}(u_{1}w_{1})\neq 3 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

Let f:=f6f^{\prime}:=f_{6}.

First consider the case of t=3t=3. Let f(u4):=4f^{\prime}(u_{4}):=4, f(u4u5):=2f^{\prime}(u_{4}u_{5}):=2, f(u5u6):=3f^{\prime}(u_{5}u_{6}):=3, and f(u5u7):=4f^{\prime}(u_{5}u_{7}):=4. Next we assign a label in {5,6}{f(u1w1)}\{5,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u7)(u_{1},u_{7}) and a label in {4,5,6}{f(u1w1),f(u1u7)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{7})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {5,6}{f(u1u7)}\{5,6\}-\{f^{\prime}(u_{1}u_{7})\} to (u6,u7)(u_{6},u_{7}).

Next consider the case of t5t\geq 5. Let f(u2t2):=6f^{\prime}(u_{2t-2}):=6, f(u2t1):=0f^{\prime}(u_{2t-1}):=0, f(u2t3u2t2):=4f^{\prime}(u_{2t-3}u_{2t-2}):=4, f(u2t3u2t1):=5f^{\prime}(u_{2t-3}u_{2t-1}):=5, and f(u2t1u2t+1):=6f^{\prime}(u_{2t-1}u_{2t+1}):=6. Next we assign a label in {4,5}{f(u1w1)}\{4,5\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {4,5,6}{f(u1w1),f(u1u2t+1)}\{4,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {4,5}{f(u1u2t+1)}\{4,5\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1)=4f^{\prime}(u_{2t}u_{2t+1})=4 (resp., 5), then let f(u2t):=6f^{\prime}(u_{2t}):=6 (resp., 1) and f(u2t1u2t):=2f^{\prime}(u_{2t-1}u_{2t}):=2 (resp., 4).

(3.2.4) Assume that f(u1w1)=3f^{\prime}(u_{1}w_{1})=3 and f(u2t+1w2t+1)=0f^{\prime}(u_{2t+1}w_{2t+1})=0.

Let f:=f6f^{\prime}:=f_{6}. We reassign labels for some vertices and edges similarly to (3.2.2) and (3.2.3). Namely, if t=3t=3, let f(u1u2):=4f^{\prime}(u_{1}u_{2}):=4, f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6, f(u3):=4f^{\prime}(u_{3}):=4, f(u4):=5f^{\prime}(u_{4}):=5, f(u3u4):=1f^{\prime}(u_{3}u_{4}):=1, f(u4u5):=2f^{\prime}(u_{4}u_{5}):=2, f(u3u5):=0f^{\prime}(u_{3}u_{5}):=0, f(u5u6):=3f^{\prime}(u_{5}u_{6}):=3, f(u6u7):=6f^{\prime}(u_{6}u_{7}):=6, f(u5u7):=4f^{\prime}(u_{5}u_{7}):=4, and f(u1u7):=5f^{\prime}(u_{1}u_{7}):=5. If t5t\geq 5, then let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6, f(u3):=4f^{\prime}(u_{3}):=4, f(u2t2):=6f^{\prime}(u_{2t-2}):=6, f(u2t1):=0f^{\prime}(u_{2t-1}):=0, f(u2t):=1f^{\prime}(u_{2t}):=1, f(u2t3u2t2):=4f^{\prime}(u_{2t-3}u_{2t-2}):=4, f(u2t3u2t1):=5f^{\prime}(u_{2t-3}u_{2t-1}):=5, f(u2t1u2t):=4f^{\prime}(u_{2t-1}u_{2t}):=4, f(u2tu2t+1):=5f^{\prime}(u_{2t}u_{2t+1}):=5, f(u2t1u2t+1):=6f^{\prime}(u_{2t-1}u_{2t+1}):=6, and f(u1u2t+1):=4f^{\prime}(u_{1}u_{2t+1}):=4.

(Case-4) Let f(u1):=2f^{\prime}(u_{1}):=2 and f(u2t+1):=3f^{\prime}(u_{2t+1}):=3.

(4.1) Assume that tt is even; t=2kt=2k.

(4.1.1) Assume that f(u1w1)4f^{\prime}(u_{1}w_{1})\neq 4 and f(u2t+1w2t+1)1f^{\prime}(u_{2t+1}w_{2t+1})\neq 1.

Let f(u2):=1f^{\prime}(u_{2}):=1, f(u3):=6f^{\prime}(u_{3}):=6, f(u4):=4f^{\prime}(u_{4}):=4, f(u5):=3f^{\prime}(u_{5}):=3, f(u2u3):=3f^{\prime}(u_{2}u_{3}):=3, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u3u4):=2f^{\prime}(u_{3}u_{4}):=2, f(u4u5):=0f^{\prime}(u_{4}u_{5}):=0, and f(u3u5):=1f^{\prime}(u_{3}u_{5}):=1. For i=2,3,,ki=2,3,\ldots,k, let f(u4i2):=2f^{\prime}(u_{4i-2}):=2, f(u4i1):=4f^{\prime}(u_{4i-1}):=4, f(u4i):=5f^{\prime}(u_{4i}):=5, f(u4i+1):=3f^{\prime}(u_{4i+1}):=3, f(u4i3u4i2):=5f^{\prime}(u_{4i-3}u_{4i-2}):=5, f(u4i2u4i1):=0f^{\prime}(u_{4i-2}u_{4i-1}):=0, f(u4i3u4i1):=6f^{\prime}(u_{4i-3}u_{4i-1}):=6, f(u4i1u4i):=2f^{\prime}(u_{4i-1}u_{4i}):=2, f(u4iu4i+1):=0f^{\prime}(u_{4i}u_{4i+1}):=0, and f(u4i1u4i+1)f^{\prime}(u_{4i-1}u_{4i+1}) :=1:=1. Denote this labeling ff^{\prime} by f7f_{7} (see Fig. 8).

Refer to caption
Figure 8: A labeling f7f_{7} on 𝒞{\cal C}.

Assign a label {0,5,6}{f(u1w1),f(u2t+1w2t+1)}\{0,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {0,5,6}{f(u1w1),f(u1u2t+1)}\{0,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {0,5,6}{f(u2t+1w2t+1),f(u1u2t+1)}\{0,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5, and if f(u2tu2t+1){5,6}f^{\prime}(u_{2t}u_{2t+1})\in\{5,6\}, then let f(u2t):=0f^{\prime}(u_{2t}):=0.

(4.1.2) Assume that f(u1w1)=4f^{\prime}(u_{1}w_{1})=4 and f(u2t+1w2t+1)1f^{\prime}(u_{2t+1}w_{2t+1})\neq 1.

Let f:=f7f^{\prime}:=f_{7}. We reassign a label for (u1,u3)(u_{1},u_{3}) as f(u1u3):=0f^{\prime}(u_{1}u_{3}):=0. Assign a label in {5,6}{f(u2t+1w2t+1)}\{5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {5,6}{f(u1u2t+1)}\{5,6\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {0,5,6}{f(u2t+1w2t+1),\{0,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}), f(u1u2t+1)}f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u2tu2t+1){5,6}f^{\prime}(u_{2t}u_{2t+1})\in\{5,6\}, then let f(u2t):=0f^{\prime}(u_{2t}):=0.

(4.1.3) Assume that f(u1w1)4f^{\prime}(u_{1}w_{1})\neq 4 and f(u2t+1w2t+1)=1f^{\prime}(u_{2t+1}w_{2t+1})=1.

Let f:=f7f^{\prime}:=f_{7}.

Consider the case of t=2t=2. Let f(u3u5):=0f^{\prime}(u_{3}u_{5}):=0 and f(u4):=0f^{\prime}(u_{4}):=0. Next we assign a label in {5,6}{f(u1w1)}\{5,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u5)(u_{1},u_{5}) and a label in {0,5,6}{f(u1u5),f(u1w1)}\{0,5,6\}-\{f^{\prime}(u_{1}u_{5}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {5,6}{f(u1u5)}\{5,6\}-\{f^{\prime}(u_{1}u_{5})\} to (u4,u5)(u_{4},u_{5}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5.

Consider the case of t4t\geq 4. Let f(u2t1u2t+1):=5f^{\prime}(u_{2t-1}u_{2t+1}):=5, f(u2t2):=1f^{\prime}(u_{2t-2}):=1, f(u2t1):=0f^{\prime}(u_{2t-1}):=0, f(u2t2u2t1):=3f^{\prime}(u_{2t-2}u_{2t-1}):=3, and f(u2t):=4f^{\prime}(u_{2t}):=4. Next we assign a label in {0,6}{f(u1w1)}\{0,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {0,5,6}{f(u1u2t+1),f(u1w1)}\{0,5,6\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{1}w_{1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {0,6}{f(u1u2t+1)}\{0,6\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5.

(4.1.4) Assume that f(u1w1)=4f^{\prime}(u_{1}w_{1})=4 and f(u2t+1w2t+1)=1f^{\prime}(u_{2t+1}w_{2t+1})=1.

If t=2t=2, let f(u2):=4f^{\prime}(u_{2}):=4, f(u3):=0f^{\prime}(u_{3}):=0, f(u4):=1f^{\prime}(u_{4}):=1, f(u1u2):=6f^{\prime}(u_{1}u_{2}):=6, f(u2u3):=2f^{\prime}(u_{2}u_{3}):=2, f(u1u3):=5f^{\prime}(u_{1}u_{3}):=5, f(u3u4):=3f^{\prime}(u_{3}u_{4}):=3, f(u4u5):=5f^{\prime}(u_{4}u_{5}):=5, f(u3u5)=6f^{\prime}(u_{3}u_{5})=6, and f(u1u5):=0f^{\prime}(u_{1}u_{5}):=0.

Consider the case of t4t\geq 4. Let f:=f7f^{\prime}:=f_{7}. We reassign labels for some vertices and edges similarly to (4.1.2) and (4.1.3). Namely, let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=0f^{\prime}(u_{1}u_{3}):=0, f(u2t2):=1f^{\prime}(u_{2t-2}):=1, f(u2t1):=0f^{\prime}(u_{2t-1}):=0, f(u2t):=4f^{\prime}(u_{2t}):=4, f(u2t2u2t1):=3f^{\prime}(u_{2t-2}u_{2t-1}):=3, f(u2t1u2t+1):=5f^{\prime}(u_{2t-1}u_{2t+1}):=5, and f(u1u2t+1):=6f^{\prime}(u_{1}u_{2t+1}):=6.

Refer to caption
Figure 9: A labeling f8f_{8} on 𝒞{\cal C}.

(4.2) Assume that tt is odd; t=2k+1(k1)t=2k+1~(k\geq 1) .

(4.2.1) Assume that f(u1w1)4f^{\prime}(u_{1}w_{1})\neq 4 and f(u2t+1w2t+1)1f^{\prime}(u_{2t+1}w_{2t+1})\neq 1.

Let f(u2):=1f^{\prime}(u_{2}):=1, f(u3):=0f^{\prime}(u_{3}):=0, f(u4):=1f^{\prime}(u_{4}):=1, f(u5):=6f^{\prime}(u_{5}):=6, f(u6):=2f^{\prime}(u_{6}):=2, f(u7):=3f^{\prime}(u_{7}):=3, f(u2u3):=3f^{\prime}(u_{2}u_{3}):=3, f(u1u3):=4f^{\prime}(u_{1}u_{3}):=4, f(u3u4):=5f^{\prime}(u_{3}u_{4}):=5, f(u4u5):=3f^{\prime}(u_{4}u_{5}):=3, f(u3u5):=2f^{\prime}(u_{3}u_{5}):=2, f(u5u6):=4f^{\prime}(u_{5}u_{6}):=4, f(u6u7):=0f^{\prime}(u_{6}u_{7}):=0, and f(u5u7):=1f^{\prime}(u_{5}u_{7}):=1. For i=2,3,,ki=2,3,\ldots,k, let f(u4i):=2f^{\prime}(u_{4i}):=2, f(u4i+1):=4f^{\prime}(u_{4i+1}):=4, f(u4i+2):=5f^{\prime}(u_{4i+2}):=5, f(u4i+3):=3f^{\prime}(u_{4i+3}):=3, f(u4i1u4i):=5f^{\prime}(u_{4i-1}u_{4i}):=5, f(u4iu4i+1):=0f^{\prime}(u_{4i}u_{4i+1}):=0, f(u4i1u4i+1):=6f^{\prime}(u_{4i-1}u_{4i+1}):=6, f(u4i+1u4i+2):=2f^{\prime}(u_{4i+1}u_{4i+2}):=2, f(u4i+2u4i+3):=0f^{\prime}(u_{4i+2}u_{4i+3}):=0, and f(u4i+1u4i+3):=1f^{\prime}(u_{4i+1}u_{4i+3}):=1. Denote this labeling ff^{\prime} by f8f_{8} (see Fig. 9). We next assign a label {0,5,6}{f(u1w1),f(u2t+1w2t+1)}\{0,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {0,5,6}{f(u1w1),f(u1u2t+1)}\{0,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {0,5,6}{f(u2t+1w2t+1),f(u1u2t+1)}\{0,5,6\}-\{f^{\prime}(u_{2t+1}w_{2t+1}),f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5, and if f(u2tu2t+1){5,6}f^{\prime}(u_{2t}u_{2t+1})\in\{5,6\}, then let f(u2t):=0f^{\prime}(u_{2t}):=0.

(4.2.2) Assume that f(u1w1)=4f^{\prime}(u_{1}w_{1})=4 and f(u2t+1w2t+1)1f^{\prime}(u_{2t+1}w_{2t+1})\neq 1.

Let f:=f8f^{\prime}:=f_{8}. We reassign a label for (u1,u3)(u_{1},u_{3}) as f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6. Next we assign a label in {0,5}{f(u2t+1w2t+1)}\{0,5\}-\{f^{\prime}(u_{2t+1}w_{2t+1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {0,5}{f(u1u2t+1)}\{0,5\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {0,5,6}{f(u1u2t+1),f(u2t+1w2t+1)}\{0,5,6\}-\{f^{\prime}(u_{1}u_{2t+1}),f^{\prime}(u_{2t+1}w_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5, and if f(u2tu2t+1){5,6}f^{\prime}(u_{2t}u_{2t+1})\in\{5,6\}, then let f(u2t):=0f^{\prime}(u_{2t}):=0.

(4.2.3) Assume that f(u1w1)4f^{\prime}(u_{1}w_{1})\neq 4 and f(u2t+1w2t+1)=1f^{\prime}(u_{2t+1}w_{2t+1})=1.

Let f:=f8f^{\prime}:=f_{8}.

First consider the case of t=3t=3. Let f(u5u7):=0f^{\prime}(u_{5}u_{7}):=0. Next we assign a label in {5,6}{f(u1w1)}\{5,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u7)(u_{1},u_{7}) and a label in {0,5,6}{f(u1w1),f(u1u7)}\{0,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{7})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {5,6}{f(u1u7)}\{5,6\}-\{f^{\prime}(u_{1}u_{7})\} to (u6,u7)(u_{6},u_{7}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5.

Next consider the case of t5t\geq 5. Let f(u2t2):=0f^{\prime}(u_{2t-2}):=0, f(u2t1):=1f^{\prime}(u_{2t-1}):=1, f(u2t2u2t1):=4f^{\prime}(u_{2t-2}u_{2t-1}):=4, f(u2t1u2t):=3f^{\prime}(u_{2t-1}u_{2t}):=3, and f(u2t1u2t+1):=5f^{\prime}(u_{2t-1}u_{2t+1}):=5. Next we assign a label in {0,6}{f(u1w1)}\{0,6\}-\{f^{\prime}(u_{1}w_{1})\} to (u1,u2t+1)(u_{1},u_{2t+1}) and a label in {0,5,6}{f(u1w1),f(u1u2t+1)}\{0,5,6\}-\{f^{\prime}(u_{1}w_{1}),f^{\prime}(u_{1}u_{2t+1})\} to (u1,u2)(u_{1},u_{2}). Reassign a label in {0,6}{f(u1u2t+1)}\{0,6\}-\{f^{\prime}(u_{1}u_{2t+1})\} to (u2t,u2t+1)(u_{2t},u_{2t+1}). Then if f(u1u2)=0f^{\prime}(u_{1}u_{2})=0, then let f(u2):=5f^{\prime}(u_{2}):=5, and if f(u2tu2t+1)=6f^{\prime}(u_{2t}u_{2t+1})=6, then let f(u2t):=0f^{\prime}(u_{2t}):=0.

(4.2.4) Assume that f(u1w1)=4f^{\prime}(u_{1}w_{1})=4 and f(u2t+1w2t+1)=1f^{\prime}(u_{2t+1}w_{2t+1})=1.

Let f:=f8f^{\prime}:=f_{8}. We reassign labels for some vertices and edges similarly to (4.2.2) and (4.2.3). Namely, if t=3t=3, let f(u1u2):=0f^{\prime}(u_{1}u_{2}):=0, f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6, f(u2):=5f^{\prime}(u_{2}):=5, f(u6u7):=6f^{\prime}(u_{6}u_{7}):=6, f(u5u7):=0f^{\prime}(u_{5}u_{7}):=0, and f(u1u7):=5f^{\prime}(u_{1}u_{7}):=5. If t5t\geq 5, then let f(u1u2):=5f^{\prime}(u_{1}u_{2}):=5, f(u1u3):=6f^{\prime}(u_{1}u_{3}):=6, f(u2t2):=0f^{\prime}(u_{2t-2}):=0, f(u2t1):=1f^{\prime}(u_{2t-1}):=1, f(u2t):=0f^{\prime}(u_{2t}):=0, f(u2t2u2t1):=4f^{\prime}(u_{2t-2}u_{2t-1}):=4, f(u2t1u2t):=3f^{\prime}(u_{2t-1}u_{2t}):=3, f(u2tu2t+1):=6f^{\prime}(u_{2t}u_{2t+1}):=6, f(u2t1u2t+1):=5f^{\prime}(u_{2t-1}u_{2t+1}):=5, and f(u1u2t+1):=0f^{\prime}(u_{1}u_{2t+1}):=0. ∎

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