What is the smallest number of dicycles in a dicycle decomposition of an eulerian digraph?

What is the Smallest Number of Dicycles in a Dicycle Decomposition of an Eulerian Digraph?

Nathaniel Dean DEPARTMENT OF MATHEMATKS

VANDERBlL T U N l V f R S I N NASHVILLE, TENNESSEE 37240

ABSTRACT

A digraph D with n vertices is said to be decomposable into a set S of dicycles if every arc of D is contained in exactly one member of S. Counterexamples are given to the following conjectures which are generalizations of three well-known conjectures by G. Hajos, P. Erdos, and P. J. Kelly: (1) [B. Jackson] Every eulerian-oriented graph is decomposable into at most $ dicycles. (2) [W. Bienia & H. Meyniel] Every eulerian digraph is decomposable into at most n dicycles.

Certain observations lead us to make three other conjectures: (a) Every eulerian-oriented graph is decomposable into a t most ? dicycles. (b) Every symmetric digraph with n > 1 is decomposable into at most 2n - 3 dicycles. (c) Every eulerian digraph with n > 1 is decomposable into at most $ - 3 dicycles.

1. INTRODUCTION

Graphs in which every vertex is even are called even graphs. An eulerian graph then is a connected even graph. Let m ( G ) and n(G) denote the number of edges and vertices in a graph G . The function LxJ is used to denote the greatest integer S x . For terms not defined here, see Bondy and Murty [4].

A cover of a graph G is a set of subgraphs of G whose union is G. Viewing the elements of the cover as subsets of E ( G ) , we call the cover a decornposi-

Supported by a fellowship from the Dorothy Danforth Compton Foundation.

Journal of Graph Theory, Vol. 10. No. 3, 299-308 (1986) 0 1986 by John Wiley & Sons, Inc. CCC 0364-9024/86/030299-10$04.00

300 JOURNAL OF GRAPH THEORY

tion of G if its members partition E(G). The cycle number cy(G) and cycle- edge number c e ( G ) of G denote the cardinality of a minimum cycle and cycle-edge decomposition of G. If G cannot be decomposed into cycles, define cy(G) = s. Let G(G) and Z(G) denote the cardinality of the corresponding minimum covers of G (that is, some elements of the cover may share edges). Trivally, if G is 2-edge-connected, G(G) = G(G).

A digraph D is a directed graph in which there arc no loops, and no two arcs with the same ends have the same orientation. The underlying graph U(D) of D is the graph on the same vertex set as D where there is exactly one edge of U(D) corresponding to the arcs of D with the same ends as the edge. We say D is k-connected if U(D) is k-connected. If D is connected 2nd d ' ( u ) = d - ( u ) for each u E V ( D ) , then D is an eulerian digraph. An oriented graph is a digraph obtained from a graph by orienting each edge. We write D, for an eulerian orientation of K, , provided such an orientation exists, and K,* for the complete symmetric digraph on n vertices.

In general, we use the same notation for covers of digraphs as for graphs. Of course, when referring to a cycle cover of a digraph, we mean dicycles. With this in mind, the terms already defined for graphs (for example, cycle number) have interesting applications to digraphs as well.

It is well-known (for example, see exercise 4.1.4 of [4]) that a connected graph (digraph) is eulerian if and only if it can be decomposed into cycles (dicycles). There has been much interest in just how few cycles (dicycles) are needed in such a decomposition. This paper presents counterexamples to certain generalizations of the following three conjectures.

HC: EGPC:

n(G) - 1. KC: (Kelly [ l l ] ) cy(Dzk, = k.

(Hajos [lo]) If G is eulerian, then cy(G) 5 Ln(G)/2J. (Erdos, Goodman, and P6sa [7]) For every graph G , Z ( G ) 5

EGPC has been solved (see [12j), but HC and KC are still unsettled. The following two generalizations are discussed in Sections 2 and 3 of this paper, respectively.

JC: (Jackson [9]) For any eulerian oriented graph D , cy(D) 5 Ln(D)/2J. BMC: (Bienia and Meyniel [3]) If D is an eulerian digraph, then

cy(D) I n(D).

2. THE CYCLE NUMBER OF EULERIAN-ORIENTED GRAPHS

Clearly JC generalizes HC and KC. The variety of counterexamples presented here lead us to believe that there is no simple characterization of the counterexamples to JC.

DICYCLES IN AN EULERIAN GRAPH 301

We begin our investigation by first constructing counterexamples to JC with a cut vertex. Let D be an eulerian oriented graph, and let be the class of graphs obtained by identifying vertices from k distinct copies of D so that each block of the resulting digraph is isomorphic to D (see Figure 1). Let G E %@). Then cy(G) = cy(D)k. If JC holds for G and cy(D) = n(D)/2, then

n(G) - (n(D) - 1)k + 1 = cy(D)k = cy(G) I - - kn(D)

2 2 2

Thus, k = 1. Therefore, any eulerian oriented graph D with cy(D) = n(D)/2 yields a finite class %@) of counterexamples (k 2 2). Hence U km,2 glk (0) is an infinite class of counterexamples. Certainly any block of G could be replaced by another eulerian-oriented graph D ‘ satisfying cy(D ’) = n(D ’)/2 to yield even more counterexamples. The underlying graph of the digraph D drawn in Figure 2 is the smallest graph which can be oriented to have this property.

The digraph D is also the basis for a 2-connected counterexample to the conjecture. Consider the digraph D ’ of Figure 3 which is constructed from three copies D, , D2, D3 of D. Since cy(D) = 3, cy(D’) I 9. It must be shown that cy(D’) 2 8.


D 2 D,

4 7

FIGURE 3. 2-connected counterexample D' to JC.

Let '4' be a minimum dicycle decomposition of D '. If each element of V containing y (see Figure 3) is contained in D , U D 2 , then 191 = 9 since cy(D, U D2) = 6 and cy(D3) = 3. Otherwise, there exists C, E q such that y E V ( C , ) and A ( C l ) r l A(D3) # 8. Let PI be the ub- (or ba-) path of C, lying in D3. Now D3 - A ( P l ) is not eulerian, so there must be a ba- (or ab-) path Pz of some C2 E q such that C2 contains y. From the diagram, it is clear that PI U P2 is a dicycle of D 3 , and ( C , U C,) - A(P, U P2) consists of two dicycles joined at y. Since dd,(a) = d ; , ( ~ ) = dd,(b) = d&(b) = 1 , these are the only elements of q containing a (or 6 ) and y . Hence, I"( can be no smaller than one less than the size of the previous decomposition; that is,


2 8. Since

D’ = 1231 + 4564 + 7897 + uybwa + u2b3a + abzyxu + x5y6x + y8z9y

cy(D’) = 8, contradicting JC. An infinite class of 2-connected counterexamples can be obtained by replac-

ing D3 with digraphs of the form shown in Figure 4. B. Jackson (personal communication) has constructed the digraph Jk of

Figure 5 which yields a much simpler inifinte class of 2-connected counterexamples than D ‘ . Observing that cy(Jk) = $&Ik) - 1 (proof given by Dean [ 5 ] ) , he suggested the following

Con:ecture 2.1. If D is an eulerian oriented graph, then cy(D) 5 L;n(D)J.

Jackson also posed another conjecture related to HC. Define a digraph to be 2-cyclic if any two vertices lie on a common dicycle. He conjectured that cy(D) I Ln(D)/2J for every 2-cyclic eulerian-oriented graph D. This approach to investigating HC rules out the previous counterexamples and the author’s idea of using a repeating dicycle of length 4 to construct counterexamples. However, the author has found a counterexample to this conjecture to be presented elsewhere.

3. THE CYCLE NUMBER OF EULERIAN DIGRAPHS

Now we consider digraphs which are not necessarily orientations of simple graphs; that is, dicycles of length 2 are allowed. However, in keeping with our

FIGURE 4

1 2 3 4

FIGURE 5. 2-connected counterexample Jk to JC


definition of digraphs, loops are not allowed and no two arcs with the same ends have the same orientation.

According to Bermond and Thomassen (198 1 , see [ 2 ) ) , Meyniel stated

BMCA: If D is an eulerian digraph, then cy(D) 5 n(D) - 1.

More recently, Bienia and Meyniel 131 weakened this conjecture without explanation to the conjecture we have called BMC. Since BMCA implies BMC, the following theorem shows that they are equivalent and that they generalize EGPC.

Theorem 3.1. BMC 3 BMCA 3 EGPC.

Proof. For the first implication, assume BMC holds and BMCA does not. Then there exists a digraph D such that cy(D) = n ( D ) . Let H be the digraph obtained by joining two copies of D at a single vertex. By BMC

a contradiction. Hence, BMC implies BMCA. For the second implication, let G be a graph and let H be the digraph

obtained from G by replacing each edge with a dicycle of length two. Let q be a minimum dicycle decomposition of H. For each C E 'if, let C ' be the cycle or edge (if m ( C ) = 2) obtained from C by replacing each arc of C by the corresponding edge in G . By BMCA, this modification of 9 yields a cycle-edge cover qr of G of size = [*I I n ( H ) - 1 = n(G) - 1. I

We will present a variety of counterexamples to BMC and BMCA. Indeed, two counterexamples to BMCA were published before the conjecture was. By considering all possible dicycle decompositions, Bermond and Faber (1976, see [ll) verified that K ? and KZ cannot be decomposed into Hamilton cycles. It is not difficult to construct cycle decompositions of K S and KZ of size 4 and 6 , respectively. Hence, c y ( K ? ) = 4 and c y ( K Z ) = 6 , contradicting BMCA. Tillson has shown that c y ( K 2 ) = 2m - 1 for 2m 2 8. Hence, there are no more counterexamples of the form K ; . Since cy(K2,, I ) = m , cy(K; ,+ , ) = 2m. So K? and KZ are the only complete symmetric counterexamples to BMCA.

No 2-connected counterexamples to BMC have been found, but counterexamples to BMC (hence, BMCA) with a cut vertex may be constructed in a manner analogous to those constructed for JC (see Figure I ) . In this case, we let D be K? or K Z . For G E Yie(K?), n(G) = k(n(K:) - 1) + 1. So c y ( G ) = cy(K,*)k = n(K,*)k = n(G) + k - 1 where k can be made arbitrarily large. In other words, if the cycle number is a linear function of the number of vertices, then the slope of this function must be greater than one. Since n(G) = 3k + 1


4 n(G) 3k + I 3

< -. 4k - - - - CY(G I

It is tempting to conjecture that c y ( D ) 5 Gn(D )J for all eulerian digraphs D . The assertion would be consistent with Conjecture 2. I , and (if true) the above inequality would show that the bound is sharp.

Another 2-connected counterexample to BMCA is shown in Figure 6 where

be a minimum cycle decomposition of H . If uuwxu E q , = 1 + cy(H,) + cy(H,) = 9 and we're done. So suppose uuwxu e q. We may assume that some element C of 'JJ has an arc in HI and contains u . Since x is a cut vertex of H - A(C), every dicycle in H - A(C) containing x has arcs in HI and none in H 2 . Since cy(H - A(H2)) h A - ( H - A(H2)) = 4, c y ( H ) 2 cy(H - A(H2)) + c y ( H z ) 1 8. In fact, we can construct a decomposition of size 8, so c y ( H ) = 8. It should be clear that using a sequence of dicycles of length 4 would yield an infinite class of 2-connected counterexamples to BMCA (proof given by Dean IS]). This idea is illustrated in Figure 7. Inciden- tally, these digraphs are not 2-cyclic. Consequently, K? and KZ are the only known 2-cyclic counterexamples to BMCA.

We prefer to replace BMC with a conjecture derived from HC (see Theo- rem 3.5).

H I = H - = - K ? . Now n ( H ) = 8, so it must be shown that c y ( H ) 2 8. Let V

Conjecture 3.1. n(D) > 1 .

c y ( D ) 5 2n(D) - 3 for all symmetric digraphs D with

U V

FIGURE 7. 2-connected counterexamples to BMCA.


Lemma 3.2. any 2-connected symmetric digraph G, cy(G) cc 2n(G) - 3.

Conjecture 3. I is equivalent to the following statement: For

Proof. Obviously, Conjecture 3.1 implies the statement. The converse is proven by induction on n(G). Since it is true for the graph on 2 vertices, let H be a graph with n ( H ) > 2 , and suppose the implication holds for all graphs on fewer than n ( H ) vertices. If H is 2-connected, there is nothing to prove. If H has a cut vertex x, decompose H into two subgraphs G, and G? which intersect only at x . By induction, c y ( H ) = cy(G,) + cy(Gz) 5 2n(GI) - 3 + 2n(G2) - 3 = 2(n(GI) + n(G2) - I ) - 4 = 2 n ( H ) - 4. I

Lemma 3.3. cy(G) 5 Ln(G) - 1)/2J.

HC is equivalent to the following statement: If C is even, then

Proof. Obviously, the statement implies HC. For the converse, notice first that the inequality in HC is satisfied for eulerian graphs if and only if i t is satisfied for even graphs. Let G be an even graph. If n(G) is odd, it is obvious that HC implies cy(G) 5 ( n ( G ) - 1) /2 . If n ( G ) is even, let H be the graph obtained by intersecting two copies of G at one vertex. Since n ( H ) is odd, cy(G) = fcy(H) 9 ( n ( H ) - 1)/4 = (2n(G) - 2)/4 = ( n ( G ) - 1)/2. I

The first sentence in the statement of the following lemma is proven in IS]. A rewording of its proof is given here only because it is essential to the proof of the second sentence (not found in [8]).

Lemma 3.4. Every connected graph G contains a spanning forest F such that G - E ( F ) is an even graph. Moreover, if G is 2-connected and contains an even vertex, then F may be chosen so that m(F) 5 n ( G ) - 2 .

Proof. Let T be a spanning tree of G. Initially, let H = G and F = T - E ( T ) . F will be constructed from the edges of T successively deleted from H . Perform a depth-first search (DFS) on T. Each edge uu of T is traversed twice, once in the forward direction and once backwards. When uu is traversed backwards from u to u, if the degree of u is odd in the current graph H , then delete uu from H and add it to F. On termination, the only vertex which could con- ceivably be odd in H is the final vertex (same as initial vertex) of the DFS. Since H must have an even number of odd vertices, H is even.

Suppose G is 2-connected and contains an even vertex x . Let T ' he a spanning tree of G - x and xw € E(G) . Then T = T' + xw is a spanning tree of G where x has degree one in T. Perform a DFS of T beginning at x . Since x is even, xw is not added to F, so m(F) 5 m(T) - 1 5 n ( G ) - 2 . I

Theorem 3.5. HC Conjecture 3.1


Proof. Let D be a symmetric digraph with n(D) > I . By Lemma 3.2 we may assume D is 2-connected. By Lemma 3.4, we see that D contains a digraph F* such that each arc of F* lies in a dicycle of length 2, the underlying graph F of F* is a forest, and each vertex u of F - A ( F * ) has d ' ( u ) = d - ( u ) even in D - A ( F * ) . Moreover, F * may be chosen so that m ( F ) 5 n(D) - 2, if n(D) is odd because then U ( D ) contains an even vertex. Orient the cycles in a minimum cycle decomposition of the underlying graph H of D - A ( F * ) . Using these dicycles, their converse, Lemma 3.3, and HC, one obtains

+ n(D) - 2 = 2n(D) - 3, if n(D) is odd;

+ n(D) - 1 = 2n(D) - 3, otherwise. I

Realizing that every eulerian digraph can be decomposed into a symmetric digraph and an oriented even graph, it is easy to see that Conjecture 2.1 com- bined with Conjecture 3 .1 implies that c y ( D ) 5 $ n ( D ) + 2n(D) - 3 = sn(D) - 3 for every eulerian digraph D with n(D) > 1. 8

Conjecture 3.2. n(D) > 1.

c y ( D ) 5 t n ( D ) - 3 for every eulerian digraph D with

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What is the smallest number of dicycles in a dicycle decomposition of an eulerian digraph?

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