Let G be a simple graph with diameter four, if G does not contain complete
subgraph K3 of order three

Lin, Zhao

WeiLi,  He

Xiang,  Ren

English

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International J.Math. Combin. Vol.1 (2007), No.1, 87-93A Note onthe Maximum Genus of Graphs with Diameter 4Xiang Ren, WeiLi He and Lin Zhao(Department of Mathematics of Beijing Jiaotong University, Beijing 100044, P.R.China)Email: 05121759@bjtu.edu.cnAbstract: Let G be a simple graph with diameter four,if G does not contain completesubgraph K3 of order three. We prove that the Betti deficient number of G, ξ(G) ≤ 2.i.e. the maximum genus of G, γM (G) ≥ 12β(G) − 1 in this paper, which is related withSmarandache 2-manifolds with minimum faces.Key words: Diameter, Betti deficiency number, maximum genus.AMS (2000): 05C10.§1. Preliminaries and known resultsIn this paper, G is a finite undirected simple connected graph. The maximum genus γM (G) ofG is the largest genus of an orientable surface on which G has a 2-cell embedding, and ξ(G) isthe Betti deficiency of G. To determine the maximum genus γM (G) of a graph G on orientablesurfaces is related with map geometries, i.e., Smarandache 2-manifolds (see [1] for details) withminimum faces.By Xuong’s theory on the maximum genus of a connected graph, ξ(G) equal to β(G) −2γM (G), where β(G) = |E(G)| − |V (G)|+1 is the Betti number of G. For convenience, we usedeficiency to replace the words Betti deficiency in this paper. Nebesky´[2] showed that if G is aconnected graph and A ⊆ E(G), let υ(G,A) = c(G−A) + b(G−A)− |A| − 1, where c(G−A)denotes the number of components in G−A and b(G−A) denotes the number of componentsin G−A with an odd Betti number, then we have ξ(G) = max{υ(G,A)|A ⊆ E(G)}.Clearly, the maximum genus of a graph can be determined by its deficiency. In case of thatξ(G) ≤ 1, the graph G is said to be upper embeddable. As we known, following theorems arethe main results on relations of the maximum genus with diameter of a graph.Theorem 1.1 Let G be a multigraph of diameter 2. Then ξ(G) ≤ 1.Skoviera proved Theorem 1.1 by a different method in [3]− [4].Hunglin Fu and Minchu Tsai considered multigraphs of diameter 3 and proved the followingtheorem in [5].1Received July 24, 2007. Accepted August 25, 20072Supported by the NNSF of China under the granted No.1057101388 Xiang Ren, WeiLi He and Lin ZhaooTheorem 1.2 Let G be a multigraph of diameter 3. Then ξ(G) ≤ 2.When the diameter of graphs is larger than 3, the Betti deficiency of G is unbounded. Thefollowing investigations have focused on graphs with a given diameter and some characters.Some results in this direction are presented in the following.Theorem 1.3([16]) Let G be a 3-connected multigraph of diameter 4, then ξ(G) ≤ 4.Theorem 1.4([16]) Let G be a 3-connected simple graph of diameter 5. Then ξ(G) ≤ 18.Yuanqiu Huang and Yanpei Liu proved the following result in [6].Theorem 1.5 Let G be a simple, K3-free graph of diameter 4, then ξ(G) ≤ 4, where K3-freegraph means that there are no spanning subgraphs K3 in G.The main purpose of this paper is to improve this result.§2. Main result and its proofNebesky´’s method is useful and the minimality property of the edge subset A in this methodplays an important role. For convenience, we call a graph with ξ(G) ≥ 2 a deficient graph.Any set A ⊆ E(G) such that υ(G,A) = ξ(G) will be called a Nebesky´ set. Furthermore, if A isminimal, then it will be called a minimal Nebesky´ set.Lemma 2.1([5]) Let G be a deficient graph and A a minimal Nebesky´ set of G. Then(a) b(G−A) = c(G−A) ≥ 2. More, if G is a simple graph then every component of G−Acontains at least three vertices;(b) the end vertices of every edge in A belong to distinct components of G−A;(c) any two components of G−A are joined by at most one edge of A;(d) ξ(G) = 2c(G−A)− |A| − 1.With the support of Lemma 2.1, we are able to construct a new graph based on the choiceof A. Let G be a deficient graph and A a minimal Nebesky´ set of G. GA is called a testablegraph of G if V (GA) is the set of components of G−A and two vertices in GA are adjacent ifand only if they are joined in G by an edge of A. We shall refer the vertices of GA to as thenodes of GA, and uA, υA,...are typical notation for the nodes.Lemma 2.2 Let G be a deficient graph and A a minimal Nebesky´ set of G. Thenξ(G) = 2p(GA)− q(GA)− 1,where p(GA) and q(GA) are the numbers of nodes and edges of GA, respectively.Proof By the definition of GA, we know that p(GA) = c(G−A) and q(GA) = |A|. ApplyingLemma 2.1, we find thatξ(G) = 2c(G−A)− |A| − 1 = 2p(GA)− q(GA)− 1.A Note on the Maximum Genus of Graphs with Diameter 4 89Lemma 2.3 If G is triangle-free, there exist a ωA ∈ V (GA) such that 2 ≤ |E(ωA, A)| ≤ 3,where E(ωA, A) denotes the set of edges of GA incident with ωA.Proof Let TωA denote the component of G−A which corresponds to ωA in GA. By Lemma2.1 |V (GA)| ≥ 2. If for all ωA ∈ V (GA), there is |E(ωA, A)| ≥ 4, then|A| = 12∑ωA∈V (GA)|E(ωA, A)| ≥ 2|V (GA)|.Applying Lemma 2.1 and the definition of GA, ξ(G) = 2V (GA)−|A|−1 ≤ −1, a contradiction.For G is connected, |E(ωA, A)| ≥ 1. If |E(ωA, A)| = 1, let E(ωA, A) = {e}, e = fh, f ∈V (TωA), h ∈ V (TσA), σA ∈ V (GA). By Lemma 2.1, β(TωA) is odd and TωA is simple andtriangle-free, there exists f′ ∈ V (TωA) such that f′ 6= f, ff ′ 6∈ E(G). Similarly, there existsh′ ∈ V (TσA) such that h′ 6= h, hh′ 6∈ E(G). Since e is a bridge, dG(f ′ , h′) ≥ 5, a contradiction.So we get that 2 ≤ |E(ωA, A)| ≤ 3.Theorem Let G be a simple, triangle-free graph of diameter 4, then ξ(G) ≤ 2, i.e., themaximum genus of G, γM (G) ≥ 12β(G) − 1.Proof Let Π={H |H is a simple graph of diameter 4 and does not contain a spanningsubgraph K3 with ξ(G) > 2 }. We claim that Π is an empty set. Suppose it is not true, letG ∈ Π be with minimum order. Clearly, G is a deficient graph. Now let A be a minimal Nebesky´set. Applying Lemma 2.1(a), each component of G − A has odd Betti number. Thus, eachcomponent of G−A must be a quadrangle. Otherwise, there exists a graph |V (G′)| < |V (G)|.Now let TxA denote the component of G − A which corresponds to xA in GA for each nodexA ∈ V (GA).By Lemma 2.3, choose zA ∈ V (GA) with 2 ≤ |E(zA, A)| ≤ 3, and define D0 = {zA}, D1 =N(zA) and D2 = V (GA) − N(zA). We call x ∈ V (G) a distance k vertex, if min {d(x, z)|z ∈V (TzA)} = k and denote E(Di, Dj) = {xAyA ∈ E(GA)|xA ∈ Di and yA ∈ Dj}, where 0 ≤i, j ≤ 2 (Note that the order of xA and yA is important throughout of the proof). We also needthe following definitions.A1={xAyA ∈ E(D2, D1)| there exists a distance 1 vertex of TyA adjacent to a distance 2vertex of TxA , or a distance 2 vertex of TyA adjacent to a distance 3 vertex of TxA and a distance1 vertex of TωA for some ωA ∈ D1 − {yA}}.A2={xAyA ∈ E(D2, D2)| xA is not incident with any edge of A1 and yA is incident withone edge of A1 and TyA contains a vertex both adjacent to a vertex of TxA and a vertex of TuAfor some uA ∈ D1} ∪ {xAyA ∈ E(D2, D2)| xA is not incident with any edge of A1 and yA isincident with at least two edges of A1}.A3={xAyA ∈ E(D1, D1)| there exists a distance 2 vertex of TxA adjacent to a distance 1vertex of TyA}.Now, according to these edge subsets A1 − A3 of E(GA), we define a directed graph −→GAbased on GA:(i) V (−→GA) = V (GA);(ii) if xAyA ∈ E′ = (⋃3i=1 Ai)⋃(D1, D0), then join two arcs from yA to xA;90 Xiang Ren, WeiLi He and Lin Zhaoo(iii) if xAyA ∈ E(GA)− E′ , then let (xA, yA) and (yA, xA) be arcs of −→GA.By this definition, it is easy to see that∑xA∈V (GA)deg(xA) =∑xA∈V (−→GA)deg−(xA),where deg−(xA) denotes the in-degree of xA in−→GA. Therefore, the in-degree sum of−→GA gives2q(GA).Now, we count the in-degree sum of−→GA. Let xA be an arbitrary node in V (−→GA).(1) xA ∈ D0. Then deg−(xA) = 0 clearly.(2) xA ∈ D2. The situation is divided into the discussions (i)-(iv) following.(i) xA is not incident with edges of A1, but incident with edges of A2.Case 1 xA is incident with at least two edges of A2, then deg−(xA) ≥ 4.Case 2 xA is incident with one edge e of A2. Let x1y1 be an edge of E(G) which correspondsto the edge e. Accordingly, TzA is a quadrangle and 2 ≤ |E(zA, A)| ≤ 3. Then there existz1 ∈ V (TzA) and deg(z1) = 2. We know that d(x1, z1) = 4 in G. Let V (TxA) = {x1, x2, x3, x4}.In TxA , x2 must be incident with an edge of E(GA) − E′such that d(x2, z1) ≤ 4(in factd(x2, z1) = 4). Similar discussion can be done done for vertices x3 and x4. So deg−(xA) ≥ 4.(ii) xA is not incident with edges of A1⋃A2.Let V (TxA) = {x1, x2, x3, x4}. In TxA , x1 must be incident with an edge of E(GA) − E′such that d(x1, z1) ≤ 4(in fact d(x2, z1) = 4). Similar discussion can be done done for verticesx2,x3 and x4. So deg−(xA) ≥ 4.(iii) xA is incident with edges of A1, but not incident with edges of A2.Case 1 xA is incident with at least two edges of A1, then deg−(xA) ≥ 4.Case 2 xA is incident with one edge e of A1. Let x1y1 be an edge of E(G) which correspondsto the edge e. Let V (TxA) = {x1, x2, x3, x4} and d(x1, z1) ≥ 3. In TxA , it supposes that x3 isnot incident with x1, then x3 must be incident an edge of E(GA). Let this edge be e′. Thene′ ∈ E(GA)−E′ , and e′ contributes one de-agree. So deg−(xA) ≥ 3(in fact, when deg−(xA) = 3,e′ ∈ E(D2, D1)).(iv) xA is incident with edges of A1 and A2.Case 1 xA is incident with at least two edges of A1, then deg−(xA) ≥ 4.Case 2 xA is incident with one edge e of A1. Let x1y1 be an edge of E(G) which correspondsto the edge e. Let V (TxA) = {x1, x2, x3, x4} and d(x1, z1) ≥ 3. In TxA , it supposes that x3is not incident with x1, then x3 must be incident with an edge of E(GA). Let this edge bee′. Then e′ ∈ A2 or e′ ∈ E(GA) − E′ . In the former, e′ must contributes two de-agree forxA. In the latter, e′contributes one de-agree. So deg−(xA) ≥ 3(in fact,when deg−(xA) = 3,e′ ∈ E(D2, D1)).Hence, for xA ∈ D2, deg−(xA) ≥ 3.A Note on the Maximum Genus of Graphs with Diameter 4 91Let M={xA ∈ D2|deg−(xA) = 3}. We get that∑xA∈D2deg−(xA) ≥ 4|D2| − |M |.(3) xA ∈ D1.By the definition of−→GA, the edge connects D0 and D1 contributes two de-agree for xA.Let x1y1 be an edge of E(G) corresponds to this edge(y1 ∈ E(TzA)).Let V (TxA) = {x1, x2, x3, x4}. In TxA , it supposes that x3 is not incident with x1. In TzA ,there exists z2 ∈ V (TzA) so that if d(x3, z2) ≤ 4. If x3 does not connect z2 though x2 or x4,x3 must be incident with one edge of E(GA). Let that edge be e. Then e ∈ E(GA) − E′ anddeg(xA) ≥ 3. If x3 connects z2 though x2 or x4, x2 or x4 is incident with one edge of E(GA).Let that edge be e. Then e ∈ E(GA) − E′ or e ∈ A3, and e contributes at least one de-agree.So deg(xA) ≥ 3.Hence, for all xA ∈ D1,∑xA∈D1deg−(xA) ≥ 3|D1|+ |M |.Now by discussions (1) and (2), we get that2q(GA) =∑xA∈V (−→GA)deg−(xA)≥ 4|D2| − |M |+ 3|D1|+ |M |= 4p(GA)− |D1| − 4≥ 4P (GA)− 7.Applying Lemma 2.2 again, we get that ξ(G) = 2p(GA)− 1− q(GA) ≤ 2, also a contradic-tion. This completes the proof.To see that the upper bound presented in our theorem is best possible, let us consider thefollowing family of infinite graphs, as depicts in Fig. 1. There are even paths with length 2from m to n. Thus, this graph is triangle-free with diameter 4. 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A Note on the Maximum Genus of Graphs with Diameter 4

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A Note on the Maximum Genus of Graphs with Diameter 4

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