JOURNAL OF GRAPH THEORY, VOL. I, 79-84 (1977)

The Spanning Subgraphs of Eulerian Graphs

F. T. BOESCH Bell Telephone Laboratories

C . SUFFEL AND R. TINDELL Stevens Institute of Technology

ABSTRACT It is shown that a connected graph G spans an eulerian graph if and only if G is not

spanned by an odd complete bigraph K ( 2 m + 1,2n + 1). A disconnected graph spans an eulerian graph if and only if it is not the union of the trivial graph with a complete graph of odd order. Exact formulas are obtained for the number of lines which must be added to such graphs in order to get eulerian graphs.

1. Introduction

Although the study of eulerian graphs* gave birth to the discipline of graph theory, there remain interesting open questions concerning related concepts. For example, recent publications by Goodman and Hedetniemi [2] and by Edmonds and Johnson [l] consider the question of the minimum length of a closed covering walk in a noneulerian graph. In this paper we consider the problem of when a noneulerian graph can be made eulerian by the addition of lines. Equivalently, we characterize sub- eulerian graphs which are spanning subgraphs of eulerian graphs. Moreover, by using a generalized form of parameters introduced by Goodman and Hedetniemi, we specify the minimum number eu'(G) of lines which must be added to a subeulerian graph G to obtain an eulerian spanning supergraph.

The reader will notice that the question of when a graph G is subeulerian is trivial if G has odd order p , since G spans the eulerian graph Kp. However, it is easily verified that the complete bipartite graph K(1,3) spans no eulerian graph. The structure of graphs which are not subeulerian is captured in this example. Call a complete bigraph K(r , s) odd if both r and s are odd; in this case all points have odd degree. A graph is euen if all points have even degree. Our main theorem is: A connecied graph G is subeulerian i f and only i f G is not spanned by an odd

* Our notation and terminology follows that used in the book by Harary [3]. @ 1977 by John Wiley & Sons, Inc. 79

80 BOESCH, SUFFEL, AND TINDELL

complete bipartite graph. For disconnected graphs, we prove that all graphs but those of the form K1 U KZlcl are subeulerian.

2. Subeulerian multigraphs

Before considering the problem of subeulerian graphs, we digress slightly to study the problem for multigraphs. In this setting the question, although still interesting, is considerably easier to answer. First, observe that any multigraph on p points is subeulerian, since it spans the multig- raph Kp(2) obtained by doubling each line of Kp.

We now determine the eu'(M) for noneulerian multigraph M. If M is connected, eu'(M) is clearly one-half the number of points of odd degree. In the disconnected case, decompose the multigraph M into its k ( M ) components and add k ( M ) lines according to the following scheme: The components are to be connected by adding lines so that the new lines form a cycle when the original components are collapsed to points. The two new lines incident at an even component are to have a common end-point in that component; the two new lines incident at 'a component containing at least two odd points are attached at distinct odd points of that component. This creates a connected multigraph, and the situation reduces to the previous case. By considering the increase in the degrees of the points, it is easy to verify that this construction utilizes a minimum of new lines. A simple explicit formula for eu'(M) can be given as follows: Let k,(M) denote the number of even components of M and let p o be the number of odd points. Then for any noneulerian multigraph M,

eu+ ( M ) = k, ( M ) + po/2.

3. Preeulerian graphs

Let 8 =(PI, Pz, * - - , pk} be a set of paths in G, and let dPi denote the set of the two end-points of Pi. If S is a set of points of a graph G, then a

pairing of S on G is a set 8 of paths in G with S = U dPi and

aPi n6'4 = 0 when i # j . For a pairing of S on G, we define 181 = 1 lPil,

where \Pi[ denotes path length. If there is at least one pairing of S on G, we define m ( S ; G) to be the minimum value of 181 for all pairings 9; a pairing realizing this value will be called a minimum pairing. Note that if 8 ={PI, Pz, * * . , pk} is a pairing of S on G, then each u in S determines a unique i for which u is an end-point of Pi.

k

i = l k

i=1

THE SPANNING SUBGRAPHS OF EULERIAN GRAPHS 81

Remark 1. There is a pairing of S on G if and only if S intersects each component of G in evenly many points.

Three lemmas and a corollary are now derived to facilitate the proof of the first theorem. The next statement is essentially given as Corollary 5a in [2].

LEMMA 1. The paths constituting a minimum pairing are pairwise edge - disjoint.

Proof. Suppose distinct paths Pi and Pi of a pairing 9 of S on G have a line in common. We may order the points of paths Pi as uo, ul, . * - , u, and of Pi as uo, u1; * . , un so that for some k and 1, u k = U I and U k + l = U l c l . We can then define a new pairing 9’ by setting P:= P, if r# i and r f j , Pi as given by uo, u l , - * , u k J u l - l , U I - ~ , * , uo, and Pj as given by Urn, U,,,-l, - * , u k + l , u l + 2 , * * - , u,. Clearly 19’1 = 191 - 2, so that 9 is not a minimum pairing. I

LEMMA 2. I f {P1, Pz, * - * , Pk} is a minimum pairing of S on G, then the k

set of odd points of the graph U Pi is exactly the set S. i = l

k

Proof. Let U Pi =Lo. Since & is the union of pairwise line-disjoint

paths, the degree in LO of a point u is the sum of the degrees of u in the

paths Pi so that deg(u, L9)= C deg (u, Pi). By convention, we set

deg (u, Pi) = 0 if u is not in Pi. Now if u is in S, then it is an end-point of exactly one path in 9; hence, precisely one term in the above degree sum for u has value 1 and the remaining terms are 0 or 2, giving deg (u, LO) as odd. If u is not in S , then it is not an end-point of path P i ; hence, it has even degree in LO. I

i = l

k

i -1

LEMMA 3. If G has euen order and some component of has odd order, then G is spanned by an odd complete bigraph.

Proof. Let V={ul , uz, - * a , u ~ ~ + ~ } be the set of points of some odd order component of G, and let U={uI, , u ~ , , + ~ } be the remaining points. By hypothesis,G, and hence G, has even order. No point of V is adjacent in to any point of U, so all the lines between the two point sets must lie in G and thus define a K ( 2 m + 1 , 2 n + 1 ) spanning subgraph of G. I

COROLLARY 1. Let 8(G) denote the set of odd points of G. If G is not spanned by any K ( 2 m + 1 , 2 n + l ) , then there is a pairing of O(G) on c.

82 BOESCH, SUFFEL, AND TINDELL

Proof. By Remark 1, we need only show that each component of G intersects O(G) in evenly many points. When G has odd order, 8(G)= O(G) and the desired result is immediate. When G has even order, the desired result follows from the contrapositive of Lemma 3. I

THEOREM 1. A connected graph G is subeulerian if and only i f G is not spanned by an odd complete bigraph.

Proof. To prove the necessity, we need only show that K ( 2 m + 1,2n + 1 ) does not span an eulerian graph. Suppose H is a spanning supergraph of Ki., with parts Vl and V,, each having an odd number of points. Letting H1 be the induced subgraph (V,) of H, we note that deg (u, H ) = deg(u, H1)+deg (u, Ki,i), for u E V1. Since H , has an odd number of points, it contains an even degree vertex u. Let u be in V1 so that deg (u, = I V21 which is odd. From this we see that u has odd degree in H ; hence, H is not eulerian.

To complete the proof of Theorem 1, assume G is a connected graph which is not spanned by any K(2m+1,2n+1) . By Corollary 1 there exists a minimum pairing 9 of O(G) on G, and by Lemma 2, O(&-.)= B(G). Defining H = G U L9, we see that H is a spanning supergraph of G having only even vertices, and H is connected since G is connected. Thus H is an eulerian graph. 1

By the proof of Theorem 1, we know that eu'(G)Sm(O(G), c) when G is a connected subeulerian graph. We now show that equality holds.

THEOREM 2. If G is a connected subeulerian graph, then eu'(G)=

Suppose H is a spanning eulerian supergraph of G and let L be the intersection of H with G, that is, the lines of L are the lines added to G to obtain H. Clearly O(L) = O(G) and each component of L contains evenly many points of O(L). Hence, there exists a minimum pairing 9 of 8(L) on L, which is also a pairing of O(L)=O(G) on G. Thus, the number of lines of L is at least 191, and 191 L m(B(G), G), i.e., one must add at least m(O(G), G) lines to G to produce a spanning eulerian supergraph of G. Hence m(O(G), G) d euf( G). I

Remark 2. When G has odd order, O(G)= 8(G) and rn(O(G), G) is just the parameter m(G) defined by Goodman and Hedetniemi [2]. More precisely, m(O(G), e) is sum of the values of m on the components of G.

m(O(G), GI. Proof.

THEOREM 3. not K1 U K2m+l.

A disconnected graph G is subeulerian if and only if G is

THE SPANNING SUBGRAPHS OF EULERIAN GRAPHS 83

Proof. First notice that any connected spanning supergraph of KIU K2,,r is spanned by a K ( l , 2 m + 1). Since an eulerian graph is necessarily connected, this establishes the necessity.

We now prove the sufficiency. By adding new lines as needed, we may clearly reduce the situation to the case where G has exactly two compo- nents G, and G2. Form a spanning supergraph G by connecting GI and G2 with a line x. We now apply Theorem 1 to G. Clearly, x must lie in any connected spanning subgraph of G. Thus, x will be a bridge for any such subgraph. However, the only complete bipartite graph with line connec- tivity 1 is K1,l. A necessary condition for 6 to be spanned by a KIJ is that one of the components of G, say G,, contains only one point u. But our hypotheses guarantee that in this situation there is a point u in G2 which does not have full degree in G2. If we choose x to connect u to such a point u, the resulting graph is cleariy not spanned by K1,l. Thus, we can always construct a connected spanning supergraph which is itself not spanned by any K(2m+1,2n+1) . I

LEMMA 4. If a minimum pairing 9 of 8(G) on (? contains a path with more than one line, then some component of G U b contains 8(G).

Proof. This lemma follows easily by a “substitution argument.” If Pi and Pi are paths in a pairing 9 of 8(G) on e, with }Pi1Z2, and with Pi and Pi in different components of G U b , replace Pi and Pi by two disjoint lines connecting the end-points of Pi to end-points of Pi. This will reduce the total number of lines in the pairing, so 9 could not have been minimal. I

Observing that this substitution process also reduces the number of components of G U L8, even when lPil = 1, we are led to the following consequence of the proof of Lemma 4.

COROLLARY 2. If G is a subeulerian graph, then there is a pairing 9 of 8(G) on (? with 8(G) contained in some component of G U Ls.

We are now able to formulate an expression for eu+(G) which also covers the disconnected case. Again, let po be the number of odd points and k, be the number of even components of G.

THEOREM 4. If G is a subeulerian graph, which is not eulerian, then eu’(G)=max{m(O(G), G), (k,+po/2)}.

Since the formula for eu’(G) reduces to m(e(G), (?) when G is connected, we may appeal to Theorem 2 in this instance, and proceed

Proof.

84 BOESCH, SUFFEL, AND TINDELL

under the assumption that G is disconnected. Let 9 be a minimum pairing of B(G) on G with B(G) contained in a component of GU L. If k, = 0, we are done, so assume k, Z 1. List the even order components G1, - * , Gk. of G, and choose points ul, * * , Uk. with u, in Gi, 1 S i S k,. Because G is disconnected, G has diameter 2 and hence p 0 / 2 5 rn(B(G>, d>Sp,. A minimum pairing 9 of B(G) on G would contain only paths of length 1 or 2 .

If m(O(G), c) =po/2, we merely choose a path P in the pairing 9, with dP={u, u'}, and replace P by the path u, ul, u2,. . a , uk,, u' . This results in a pairing 9' with 19'1 = p0 /2+: k,, and G U L , connected, hence eulerian. In this instance it is clear that e u + ( G ) = ( 9 ' ( = max{m(B(G), G), ( k , + p 0 / 2 ) } , as claimed. If rn(e(G), @ > p 0 / 2 , then list the paths in the pairing 9 so that P1, Pz, * * - , P, are those of length 2 ; clearly r = m(B(G), d ) - p o / 2 . If r L k, and aP, ={u , , d}, replace P, by the path u,, u,, u:, 15 i 5 k,. The result will be a minimal pairing 9' of O(G) on G with G U & connected, verifying the theorem in this case (note that B(G) is contained in a component of G U Lp, by Lemma 4). If r < k, and 1 S i S r - 1, replace PI as above, and replace P, by u,, u,, u,+~, . - * , uk,, u:; the result will be pairing 9' of B(G) on with G U LB connected, hence eulerian. The minimality is apparent, concluding the proof of Theorem 4. I

We note that the calculation of m(B(G), G) in the disconnected case is equivalent to finding the size of a maximum matching. In the connected case, it has been shown by Edmonds and Johnson [l] that the determi- nation of a minimum pairing is equivalent to finding a minimum weight- maximum matching in an even complete graph with line weights.

4. Unresolved problem

The related question of characterizing those graphs which are spanned by eulerian subgraphs appears very difficult. Likewise, the determination of the related minimum number of lines to remove, eu-(G), appears to be nontrivial.

REFERENCES

1. J. Edmonds and E. L. Johnson, Matching, euler tours and the Chinese postman. Math.

2. S . Goodman and S. Hedetniemi, Eulerian walks in graphs. S I A M J. Compuf. 2 (1973)

3. F. Harary, Graph Theory. Addison-Wesley, Reading, Mass. (1969).

Prog. 5 (1973) 88-128.

16-27.