Discrete Mathematics 274 (2004) 299–302www.elsevier.com/locate/disc

Note

On maximum number of edges in a spanningeulerian subgraph�

Dengxin Lia ;∗ , Deying Lib , Jingzhong MaobaDepartment of Mathematics, Chongqing Technology & Business University, Chongqing 400020,

People’s Republic of ChinabDepartment of Mathematics, Central China Normal University, Wuhan, Hubei 430079,

People’s Republic of China

Received 15 January 2002; received in revised form 26 March 2003; accepted 2 April 2003

Abstract

A graph is supereulerian if it has a spanning eulerian subgraph. It has been an open problemto determine the in1mum L of the ratio of the maximum size of a spanning eulerian subgraphH of a graph G to the size of G, among all nontrivial supereulerian graphs. Catlin once thoughtthat L could be 2

3 . In this note, we present in1nite families of graphs to show that L should beless than 2

3 . Moreover, we show that when restricted to r-regular supereulerian graphs, if r �= 5,then L¿ 2

3 and if r = 5, then L¿ 35 .

c© 2003 Elsevier B.V. All rights reserved.

Keywords: Number of edges; Spanning euleriah subgraph; Supereulerian

We follow the notation of Bondy and Murty [1]. Graphs in this note are 1nite andloopless. For a graph G;O(G) denotes the set of all vertices of odd degree in G. Aconnected graph G with O(G)=� is called an eulerian graph. A graph is supereulerianif it has a spanning eulerian subgraph. The collection of all supereulerian graphs willbe denote by SL [2].In 1990, Catlin noted that if a cubic graph G admits a spanning eulerian subgraph

H (which is also a Hamilton cycle as G is cubic), then |E(H)| = 23 |E(G)|. Thus, he

proposed a conjecture [4].

� Partially supported by National Natural Science Foundation of China (No. 10171074), and by theChongqing Education Committee.

∗ Corresponding author. Department of Mathematics, Yuzhou University, Chongqing 400033, China.E-mail address: denxinli@sina.com (D. Li).

0012-365X/03/$ - see front matter c© 2003 Elsevier B.V. All rights reserved.doi:10.1016/S0012-365X(03)00202-4

300 D. Li et al. / Discrete Mathematics 274 (2004) 299–302

Fig. 1.

…

G2

…

G1

V1

V2

Vk

...

Fig. 2.

Catlin’s 23 -Conjecture. If G ∈ SL and G �= K1, then G has a spanning eulerian subgraphH with |E(H)|¿ 2

3 |E(G)|.

In 1995, Zhi-Hong Chen and Hong-Jian Lai presented the following open problem[3, Problem 8.8]:

Problem. Determine

L= minmaxG∈SL−{K1}

{ |E(H)||E(G)| : H is a spanning eulerian subgraph of G

}:

The main objective of this note is to show that 23 is not the right bound for L. We

will present two in1nite families of graphs for this purpose.

First, we suppose that G is a simple graph.Let G be the graph depicted in Fig. 1.The graph G has a maximum spanning eulerian subgraph H with 11 edges, but

|E(G)|= 17. Thus, we have that |E(H)|=|E(G)|= 1117 ¡

23 .

Let G1 and G2 be the two graphs depicted in Fig. 2.Suppose that G1 contains x subgraphs isomorphic to G2 and a path of length k + 1

(see the graph G1 of Fig. 2), then G1 has a maximum spanning eulerian subgraph Hwith 3x+k+1 edges. If k+1¡x and 0¡k, then |E(H)|=|E(G)|=(3x+k+1)=(5x+k + 1)¡ 2

3 .

D. Li et al. / Discrete Mathematics 274 (2004) 299–302 301

As the inequality holds for every positive integer k, we obtain

limx→∞

3x + k + 15x + k + 1

=35:

The limit above indicates that L6 35 ¡

23 .

Another in1nite family of graphs allows multiple edges, as constructed below.Let Cn = v1v2v3 · · · vnv1 be an n-cycle for some n¿ 4. Let Gn be obtained from

Cn by adding a parallel edge to each of vivi+1; i = 2; 3; : : : ; n − 1. Cn is a maximumspanning eulerian subgraph of Gn.

|E(Cn)||E(Gn)| =

n2n− 2

¡23

(n¿ 4):

The two families of examples show that in general, L6 12 and that when restricted

to simple graphs, L6 35 .

Next, we present a partial solution to the problem. De1ne

L′ = minmaxG∈SL−{K1}

{|E(H)|=|E(G)| : |E(G)|¿ 3|V (G)|

and H is a spanning eulerian subgraph of G}.

Theorem. L′¿ 23 .

Proof. Let G ∈ SL and H be a maximum spanning eulerian subgraph of G. If |E(H)|¡23 |E(G)|, then |E(G)−E(H)|¿ 1

3 |E(G)|¿ |V (G)|. Thus G−E(H) contains a cycle C,so H ∪C is also a spanning eulerian subgraph of G, which contradicts the maximalityof |E(H)|. Therefore, we have |E(H)|¿ 2

3 |E(G)|, i.e. L′¿ 23 .

Corollary. Let G be an r-regular graph, if r �= 5, then L¿ 23 ; if r = 5, then L¿ 3

5 .

Proof. It is trivial for r=2; 3; 4. If r¿ 6, then |E(G)|6 3|V (G)|. If r=5, we assumethat H is a maximum spanning eulerian subgraph of G. If |E(H)|6 3

5 |E(G)|, then|E(G) − E(H)|¿ 2

5 |E(G)| = |V (G)|. Thus G − E(H) contains a cycle, a contradic-tion.

Conjecture. Let G be a simple graph. If G ∈ SL, and G �= K1, then G has a spanningeulerian subgraph H with |E(H)|¿ 3

5 |E(G)|.

The authors would like to thank the referees for their helpful suggestions whichhelped to improve the presentation of this paper.

302 D. Li et al. / Discrete Mathematics 274 (2004) 299–302

References

[1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Elsevier, New York, 1976.[2] P.A. Catlin, Supereulerian graphs: a survey, J. Graph Theory 16 (1992) 177–196.[3] Z.-H. Chen, H.-J. Lai, Reduction techniques for supereulerian graphs and related topics—an update, in:

Ku Tung-Hsin (Ed.), Combinatorics and Graph Theory ’95, World Scienti1c, Singapore, London, 1995,pp. 53–69.

[4] H.-J. Lai, Lecture Notes on Supereulerian Graphs and Related Topics, 1996, unpublished notes.