SHORT CYCLE COVERS OF GRAPHS WITH MINIMUM DEGREE THREE


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1 SHOT YLE OVES OF PHS WITH MINIMUM DEEE THEE TOMÁŠ KISE, DNIEL KÁL, END LIDIKÝ, PVEL NEJEDLÝ OET ŠÁML, ND bstract. The Shortest ycle over onjecture of lon and Tarsi asserts that the edges of every bridgeless graph with m edges can be covered by cycles of total length at most 7m/5 = 1.400m. We show that every cubic bridgeless graph has a cycle cover of total length at most 34m/ m and every bridgeless graph with minimum degree three has a cycle cover of total length at most 44m/ m. Key words. ycle cover, cycle double cover, shortest cycle cover MS subject classifications. 0521, Introduction. ycle covers of graphs form a prominent topic in graph theory which is closely related to several deep and open problems. cycle in a graph is a subgraph with all degrees even. cycle cover is a collection of cycles such that each edge is contained in at least one of the cycles; we say that each edge is covered. The ycle Double over onjecture of Seymour [24] and Szekeres [25] asserts that every bridgeless graph has a collection of cycles containing each edge of exactly twice which is called a cycle double cover. In fact, it was conjectured by elmins [3] and Preissmann [22] that every graph has a cycle cover comprised of five cycles. The ycle Double over onjecture is related to other deep conjectures in graph theory. In particular, it is known to be implied by the Shortest ycle over onjecture of lon and Tarsi [1] which asserts that every bridgeless graph with m edges has a cycle cover of total length at most 7m/5. ecall that the length of a cycle is the number of edges contained in it and the length of a cycle cover is the sum of the lengths of its cycles. The reduction of the ycle Double over onjecture to the Shortest ycle over onjecture can be found in the paper of Jamshy and Tarsi [15]. The best known general result on short cycle covers is due to lon and Tarsi [1] and ermond, Jackson and Jaeger [2]: every bridgeless graph with m edges has a cycle cover of total length at most 5m/ m. There are numerous results on short cycle covers for special classes of graphs, e.g., graphs with no short cycles, highly connected graphs or graphs admitting a nowherezero 4/5flow, see e.g. [6, 7, 12, 16, 23]. The reader is referred to the monograph of Zhang [27] for further exposition of such results where an entire chapter is devoted to results on the Shortest ycle over onjecture. The least restrictive of refinements of the general bound of lon and Tarsi [1] and ermond, Jackson and Jaeger [2] are the bounds for cubic graphs. Jackson [13] The work of the first three authors was partially supported by the grant 201/09/0197. Institute for Theoretical omputer Science (ITI) and Department of Mathematics, University of West ohemia, Univerzitní 8, Plzeň, zech epublic. Supported by esearch Plan MSM of the zech Ministry of Education. Institute for Theoretical omputer Science, Faculty of Mathematics and Physics, harles University, Malostranské náměstí 25, Prague, zech epublic. The Institute for Theoretical omputer Science (ITI) is supported by Ministry of Education of the zech epublic as project 1M0545. Department of pplied Mathematics, Faculty of Mathematics and Physics, harles University, Malostranské náměstí 25, Prague, zech epublic. 1
2 2 T. KISE ET L. showed that every cubic bridgeless graph with m edges has a cycle cover of total length at most 64m/ m and Fan [7] later showed that every such graph has a cycle cover of total length at most 44m/ m. In this paper, we strengthen this bound in two different ways. First, we improve it to 34m/ m for cubic bridgeless graphs. nd second, we show that the bound 44m/27 also holds for medge bridgeless graphs with minimum degree three, i.e., we extend the result from [7] on cubic graphs to all graphs with minimum degree three. s in [7], the cycle covers that we construct consist of at most three cycles. Though the improvements of the original bound of 5m/3 = 1.667m on the length of a shortest cycle cover of an medge bridgeless graph can seem to be rather minor, obtaining a bound below 8m/5 = 1.600m for a significant class of graphs might be quite challenging since the bound of 8m/5 is implied by Tutte s 5Flow onjecture [16]. The paper is organized as follows. In Section 2, we introduce auxiliary notions and results needed for our main results. We demonstrate the introduced notation in Section 3 where we reprove the bound 5m/3 of lon and Tarsi [1] and ermond, Jackson and Jaeger [2] using our terminology. In Sections 4 and 5, we prove our bounds for cubic graphs and graphs with minimum degree three. 2. Preliminaries. In this section, we introduce notation and auxiliary concepts used throughout the paper. We focus on those terms where confusion could arise and refer the reader to standard graph theory textbooks, e.g. [4], for exposition of other notions. raphs considered in this paper can have loops and multiple (parallel) edges. If E is a set of edges of a graph, \ E denotes the graph with the same vertex set with the edges of E removed. If E = {e}, we simply write \ e instead of \ {e}. For an edge e of, /e is the graph obtained by contracting the edge e, i.e., /e is the graph with the endvertices of e identified, the edge e removed and all the other edges, including new loops and parallel edges, preserved. Note that if e is a loop, then /e = \ e. Finally, for a set E of edges of a graph, /E denotes the graph obtained by contracting all edges contained in E. If is a graph and v is a vertex of of degree two, then the graph obtained from by suppressing the vertex v is the graph obtained from by contracting one of the edges incident with v, i.e., the graph obtained by replacing the twoedge path with the inner vertex v by a single edge. n edgecut in a graph is a set E of edges such that the vertices of can be partitioned into two sets and such that E contains exactly the edges with one endvertex in and the other in. Such an edgecut is also denoted by E(, ). Note that edgecuts need not be minimal sets of edges whose removal increases the number of components of. n edge forming an edgecut of size one is called a bridge and graphs with no edgecuts of size one are said to be bridgeless. Note that we do not require bridgeless graphs to be connected. The quantity e(x, Y ) denotes the number of edges with one endvertex in X and the other vertex in Y ; in particular, if E(, ) is an edgecut of, then e(, ) is its size. graph with no edgecuts of odd size less than k is said to be koddconnected. For every set F of edges of, edgecuts in /F correspond to edgecuts (of the same size) in. Therefore, if has no edgecuts of size k, then /F also has no edgecuts of size k. s said before, a cycle of a graph is a subgraph of with all vertices of even degree. circuit is a connected subgraph with all vertices of degree two and a 2factor is a spanning subgraph with all vertices of degree two Splitting and expanding vertices. In the proof of our main result, we will need to construct a nowherezero Z 2 2flow of a special type. In order to exclude
3 SHOT YLE OVES 3 v 1 v 2 v 1 v 2 v v Fig Splitting the vertices v 1 and v 2 from the vertex v. some bad nowherezero flows, we will first modify the graph under consideration in such a way that some of its edges must get the same flow value. This goal will be achieved by splitting some of the vertices of the graph. Let be a graph, v a vertex of and v 1 and v 2 some of the neighbors of v in. The graph.v 1 vv 2 that is obtained by removing the edges vv 1 and vv 2 from and adding a twoedge path between v 1 and v 2 (see Figure 2.1) is said to be obtained by splitting the vertices v 1 and v 2 from the vertex v. Note that if v 1 = v v 2, i.e., the edge vv 1 is a loop, the graph.v 1 vv 2 is the graph obtained from by removing the loop vv 1 and subdividing the edge vv 2. Similarly, if v 1 v = v 2,.v 1 vv 2 is obtained by removing the loop vv 2 and subdividing the edge vv 1. Finally, if v 1 = v = v 2, then the graph.v 1 vv 2 is obtained from by removing the loops vv 1 and vv 2 and introducing a new vertex joined by two parallel edges to v. lassical (and deep) results of Fleischner [9], Mader [21] and Lovász [19] assert that it is possible to split vertices without creating new small edgecuts. Let us now formulate one of the corollaries of their results. Lemma 2.1. Let be a 5oddconnected graph. For every vertex v of of degree four, six or more, there exist two neighbors v 1 and v 2 of the vertex v such that the graph.v 1 vv 2 is also 5oddconnected. Zhang [28] proved a version of Lemma 2.1 where only some pairs of vertices are allowed to be split off (also see [26] for related results). Lemma 2.2. Let be an loddconnected graph for an odd integer l. For every vertex v of with neighbors v 1,..., v k, k 2, l, there exist two neighbors v i and v i+1 such that the graph.v i vv i+1 is also loddconnected (indices are modulo k). However, none of these results is sufficient for our purposes since we need to specify more precisely which pair of the neighbors of v should be split from v. This is guaranteed by the lemmas we establish in the rest of this section. Let us remark that Lemma 2.2 can be obtained as a consequence of our results. We start with a modification of the wellknown fact that minimal odd cuts do not cross for the situation where small even cuts can exist. Lemma 2.3. Let be an loddconnected graph (for some odd integer l) and let E( 1, 2 ) and E( 1, 2 ) be two cuts of of size l. Further, let W ij = i j for i, j {1, 2}. If the sets W ij are nonempty for all i, j {1, 2}, then there exist integers a, b and c such that a + b + c = l and one of the following holds: e(w 11, W 12 ) = e(w 12, W 22 ) = a, e(w 11, W 21 ) = e(w 21, W 22 ) = b, e(w 11, W 22 ) = c and e(w 12, W 21 ) = 0, or e(w 12, W 11 ) = e(w 11, W 21 ) = a, e(w 12, W 22 ) = e(w 22, W 21 ) = b, e(w 12, W 21 ) = c and e(w 11, W 22 ) = 0. See Figure 2.2 for an illustration of the two possibilities. Proof. Let w ij be the number of edges with exactly one endvertex in W ij. Observe that w 11 + w 12 = e( 1, 2 ) + 2e(W 11, W 12 ) = l + 2e(W 11, W 12 ) and (2.1)
4 4 T. KISE ET L W 22 W 22 a W 12 c b W 21 b W 12 c b W 21 a W 11 b a W 11 a Fig The two configurations described in the statement of Lemma 2.3. w 12 + w 22 = e( 1, 2 ) + 2e(W 12, W 22 ) = l + 2e(W 12, W 22 ). In particular, one of the numbers w 11 and w 12 is even and the other is odd. ssume that w 11 is odd. Hence, w 22 is also odd. Since is loddconnected, both w 11 and w 22 are at least l. If e(w 11, W 12 ) e(w 12, W 22 ), then w 12 = e(w 11, W 12 ) + e(w 12, W 21 ) + e(w 12, W 22 ) e(w 11, W 12 ) + e(w 12, W 22 ) 2e(W 11, W 12 ). (2.2) The equation (2.1), the inequality (2.2) and the inequality w 11 l imply that w 11 = l. Hence, the inequality (2.2) is an equality; in particular, e(w 11, W 12 ) = e(w 12, W 22 ) and e(w 12, W 21 ) = 0. If e(w 11, W 12 ) e(w 12, W 22 ), we obtain the same conclusion. Since the sizes of the cuts E( 1, 2 ) and E( 1, 2 ) are the same, it follows that e(w 11, W 21 ) = e(w 21, W 22 ). We conclude that the graph and the cuts have the structure as described in the first part of the lemma. The case that w 11 is even (and thus w 12 is odd) leads to the other configuration described in the statement of the lemma. Next, we use Lemma 2.3 to characterize graphs where some splittings of neighbors of a given vertex decrease the oddconnectivity. Lemma 2.4. Let be an loddconnected graph for an odd integer l 3, v a vertex of and v 1,..., v k some neighbors of v. If every graph.v i vv i+1, i = 1,...,k 1, contains an edgecut of odd size smaller than l, the vertex set V () can be partitioned into two sets V 1 and V 2 such that v V 1, v i V 2 for i = 1,...,k and the size of the edgecut E(V 1, V 2 ) is l. Proof. First observe that it is enough to prove the statement of the lemma for simple graphs; if is not simple, then subdivide each edge of and apply the lemma to the resulting graph in the natural way. The assumption that is simple allows us to avoid unpleasant technical complications in the proof. The proof proceeds by induction on k. The base case of the induction is that k = 2. Let E(V 1, V 2 ) be an edgecut of.v 1 vv 2 of odd size less than l and let v be the new degreetwo vertex of.v 1 vv 2. y symmetry, we can assume that v V 1. We can also assume that v 1 and v are in the same set V i ; otherwise, moving v to the set V i containing v 1 either preserves the size of the cut or decreases the size by two. If both v 1 V 1 and v 2 V 1, then E(V 1 \ {v }, V 2 ) as an edgecut of has the
5 SHOT YLE OVES 5 V 22 V 2 v 2 v 3 V 1 v 1 v k V 21 V 12 v V 11 V 1 V 2 Fig Notation used in the proof of Lemma 2.4. same size as in.v 1 vv 2 which contradicts the assumption that is loddconnected. If v 1 V 1 and v 2 V 2, then E(V 1 \ {v }, V 2 ) is also an edgecut of of the same size as in.v 1 vv 2 which is again impossible. Hence, both v 1 and v 2 must be contained in V 2, and the size of the edgecut E(V 1, V 2 \ {v }) in is larger by two compared to the size of E(V 1, V 2 ) in.v 1 vv 2. Since has no edgecuts of size l 2, the size of the edgecut E(V 1, V 2 ) in.v 1 vv 2 is l 2 and the size of E(V 1, V 2 \ {v }) in is l. Hence, V 1 and V 2 \ {v } form the partition of the vertices as in the statement of the lemma. We now consider the case that k > 2. y the induction assumption, contains a cut E(V 1, V 2 ) of size l such that v V 1 and v 1,..., v k 1 V 2. Similarly, there is a cut E(V 1, V 2 ) of size l such that v V 1 and v 2,...,v k V 2. Let V ij = V i V j for i, j {1, 2} (see Figure 2.3). pply Lemma 2.3 for the graph with i = V i and i = V i. Since e(v 11, V 22 ) = e(v 1 V 1, V 2 V 2 ) k 2 > 0, the first case described in Lemma 2.3 applies and the size of the cut E(V 1 V 1, V 2 V 2 ) is l. Hence, the cut E(V 1 V 1, V 2 V 2 ) is a cut of size l in. We now present a series of lemmas that we need later in the paper. ll these lemmas are simple corollaries of Lemma 2.4. Lemma 2.5. Let be a 5oddconnected graph, and let v be a vertex of degree four and v 1, v 2, v 3 and v 4 its four neighbors. Then, the graph.v 1 vv 2 or the graph.v 2 vv 3 is also 5oddconnected graph. Proof. Observe that the graph.v 1 vv 2 is 5oddconnected if and only if the graph.v 3 vv 4 is 5oddconnected. Lemma 2.4 applied for l = 5, the vertex v, k = 4 and the vertices v 1, v 2, v 3 and v 4 yields that the vertices of can be partitioned into two sets V 1 and V 2 such that v V 1, {v 1, v 2, v 3, v 4 } V 2 and e(v 1, V 2 ) = 5. Hence, e(v 1 \ {v}, V 2 {v}) = e(v 1, V 2 ) 4 = 5 4 = 1. This contradicts our assumption that has no edgecuts of size one. Lemma 2.6. Let be a 5oddconnected graph, and let v be a vertex of degree six and v 1,...,v 6 its neighbors. t least one of the graphs.v 1 vv 2,.v 2 vv 3 and.v 3 vv 4 is also 5oddconnected. Proof. Lemma 2.4 applied for l = 5, the vertex v, k = 4 and v 1, v 2, v 3 and v 4 yields that the vertices of can be partitioned into two sets V 1 and V 2 such that v V 1, {v 1, v 2, v 3, v 4 } V 2 and e(v 1, V 2 ) = 5. If V 2 contains σ neighbors of v (note
6 6 T. KISE ET L. V 1 v V 1 V 1 v 1 v 2 V 1 : v : V 1 Fig n example of the expansion a vertex v with respect to set V 1. that σ 4), then e(v 1 \ {v}, V 2 {v}) = e(v 1, V 2 ) σ + (6 σ) = 11 2σ which is equal to 1 or 3 contradicting the fact that is 5oddconnected. Lemma 2.7. Let be a 5oddconnected graph, and let v be a vertex of degree d 6 and v 1,..., v d its neighbors. t least one of the graphs.v i vv i+1, i = 1,...,5, is also 5oddconnected. Proof. Since there is no partition of the vertices of into two parts V 1 and V 2 such that v V 1, v i V 2 for i = 1,...,6, and e(v 1, V 2 ) = 5, Lemma 2.4 applied for l = 5, the vertex v, k = 6 and the vertices v i, i = 1,..., 6 yields the statement of the lemma. We will also need the following corollary. Lemma 2.8. Let be a 5oddconnected graph, and let v be a vertex of degree eight and v 1,...,v 8 its neighbors. Suppose that.v 1 vv 2 is 5oddconnected. t least one of the following graphs is also 5oddconnected:.v 1 vv 2.v 3 vv 4,.v 1 vv 2.v 7 vv 8,.v 1 vv 2.v 3 vv 8.v 4 vv 5 and.v 1 vv 2.v 3 vv 8.v 4 vv 6. Proof. The degree of the vertex v in.v 1 vv 2 is six. y Lemma 2.6, at least one of the graphs.v 1 vv 2.v 3 vv 4,.v 1 vv 2.v 3 vv 8 and.v 1 vv 2.v 7 vv 8 is 5oddconnected. If.v 1 vv 2.v 3 vv 8 is 5oddconnected, we apply Lemma 2.5 for the vertex v and its neighbors v 5, v 4, v 6 and v 7 (in this order). Hence, the graph.v 1 vv 2.v 3 vv 8.v 4 vv 5 (which is isomorphic to.v 1 vv 2.v 3 vv 8.v 6 vv 7 ) or the graph.v 1 vv 2.v 3 vv 8.v 4 vv 6 is 5 oddconnected. We need one more vertex operation in our arguments vertex expansion. If is a graph, v a vertex of and V 1 a subset of its neighbors, then the graph : v : V 1 is the graph obtained from by removing the vertex v and introducing two new vertices v 1 and v 2, joining v 1 to the vertices of V 1, v 2 to the neighbors of v not contained in V 1, and adding an edge v 1 v 2. We say that : v : V 1 is obtained by expanding the vertex v with respect to the set V 1. See Figure 2.4 for an example. Let us remark that this operation will be applied only to vertices v incident with no parallel edges. The following auxiliary lemma directly follows from results of Fleischner [9]: Lemma 2.9. Let be a bridgeless graph and v a vertex of degree four in incident with no parallel edges. Further, let v 1, v 2, v 3 and v 4 be the four neighbors of v. The graph : v : {v 1, v 2 } or the graph : v : {v 2, v 3 } is also bridgeless Special types of Z 2 2flows. We start with recalling a classical result of Jaeger on the existence of nowherezero Z 2 2flows. Theorem 2.10 (Jaeger [14]). If is a graph that contains no edgecuts of size one or three, then has a nowherezero Z 2 2flow. In our arguments, we will need to construct Z 2 2flows that have some additional properties. Some of these properties will be guaranteed by vertex splittings introduced
7 SHOT YLE OVES 7 Fig ad vertices of degree five and six (symmetric cases are omitted). The letters indicate edges contained in the sets, and. Different flow values (colors) are denoted by different types of edges. earlier. In this section, we establish an auxiliary lemma that guarantees the existence of a nowherezero Z 2 2flow which we were not able to prove with vertex splittings only. We use the following notation in Lemma 2.11: if ϕ is a flow on a graph and E is a subset of the edges of, then ϕ(e) is the set of the values of ϕ on E. Lemma Let be a bridgeless graph admitting a nowherezero Z 2 2 flow. ssume that for every vertex v of degree five, there are given two multisets v and v of three edges incident with v such that v v = 2 (a loop incident with v can appear twice in the same set), and for every vertex v of degree six, the incident edges are partitioned into three multisets v, v and v of size two each (every loop incident with v appears twice in the sets v, v and v, possibly in the same set). The graph has a nowherezero Z 2 2flow ϕ such that for every vertex v of degree five, ϕ( v ) 2 and ϕ( v ) 2, and for every vertex v of degree six, ϕ( v v v ) = 3, i.e., the edges incident with v have all the three possible flow values, or ϕ( v ) = 1, or ϕ( v ) = 1, or ϕ( v ) = 2. Proof. For simplicity, we refer to edges with the flow value 01 red, 10 green and 11 blue. Note that each vertex of odd degree is incident with odd numbers of red, green and blue edges and each vertex of even degree is incident with even numbers of red, green and blue edges (counting loops twice). We say that a vertex v of degree five is bad if ϕ( v ) = 1 or ϕ( v ) = 1, and it is good, otherwise. Similarly, a vertex v of degree six is bad if ϕ has only two possible flow values at v, ϕ( v ) = 2, ϕ( v ) = 2 and ϕ( v ) = 1; otherwise, v is good. hoose a Z 2 2flow ϕ of with the least number of bad vertices. If there are no bad vertices, then there is nothing to prove. ssume that there is a bad vertex v. Let us first analyze the case that the degree of v is five. Let e 1,..., e 5 be the edges incident with v. y symmetry, we can assume that v = {e 1, e 2, e 3 }, v = {e 2, e 3, e 4 }, the edges e 1, e 2 and e 3 are red, the edge e 4 is green and the edge e 5 is blue (see Figure 2.5). We now define a closed trail W in formed by red and blue edges. The first edge of W is e 1. Let f = ww be the last edge of W defined so far. If w = v, then f is one of the edges e 2, e 3 and e 5 and the definition of W is finished. ssume that w v. If w is not a vertex of degree five or six or w is a bad vertex, add to the trail W any red or blue edge incident with w that is not already contained in W. If w is a good vertex of degree five, let f 1,..., f 5 be the edges incident with w, w = {f 1, f 2, f 3 } and w = {f 2, f 3, f 4 }. If w is incident with a single red and a
8 8 T. KISE ET L. Fig outing the trail W (indicated by dashed edges) through a good vertex of degree five with three red edges. The letters indicate edges contained in the sets and and different types of edges represent colors (thick edges are red, thin ones are green and dotted ones are blue). Symmetric cases are omitted. single blue edge, leave w through the other edge that is red or blue. Otherwise, there are three red edges and one blue edge or vice versa. The next edge f of the trail W is determined as follows (note that the role of red and blue can be swapped): ed edges lue edge f = f 1 f = f 2 f = f 3 f = f 4 f = f 5 f 1, f 2, f 4 f 3 f = f 4 f = f 3 f = f 2 f = f 1 N/ f 1, f 2, f 4 f 5 f = f 2 f = f 1 N/ f = f 5 f = f 4 f 1, f 2, f 5 f 3 f = f 5 f = f 3 f = f 2 N/ f = f 1 f 1, f 2, f 5 f 4 f = f 2 f = f 1 N/ f = f 5 f = f 4 f 1, f 4, f 5 f 2 f = f 2 f = f 1 N/ f = f 5 f = f 4 f 2, f 3, f 5 f 1 f = f 2 f = f 1 f = f 5 N/ f = f 3 See Figure 2.6 for an illustration of these rules. If w is a good vertex of degree six, proceed as follows. If ϕ( w ) = 1 and f w, let the next edge f of W be the other edge contained in w ; if ϕ( w ) = 1 and f w, let f be any red or blue edge not contained in w or in W. symmetric rule applies if ϕ( w ) = 1, i.e., f is the other edge of w if f w and f is a red or blue edge not contained in w or W, otherwise. If ϕ( w ) = 2 and f w and the other edge of w is red or blue, set f to be the other edge of w ; if f w and the other edge of w is green, choose f to be any red or blue edge incident with w that is not contained in W. If f w (and ϕ( w ) = 2), choose f to be a red or blue edge incident with w not contained in W that is also not contained in w. If such an edge does not exist, choose f to be the red or blue edge contained in w (note that the other edge of w is green since w is incident with an even number of red, green and blue edges). It remains to consider the case that w is incident with two edges of each color and ϕ( w ) = ϕ( w ) = 2 and ϕ( w ) = 1. If f is blue, set f to be any red edge incident with w not contained in W and if f is red, set f to be any such blue edge. See Figure 2.7 for an illustration of these rules. The definition of the trail W is now finished. Swap the red and blue colors on W. It is straightforward to verify that all good vertices remain good and the vertex v becomes good (see Figures ). In particular, the number of bad vertices is decreased which contradicts the choice of ϕ. ssume that there is a bad vertex v of degree six, i.e., the colors of the edges of v are distinct, the colors of the edges of v are distinct, the colors of the edges of v are
9 SHOT YLE OVES 9 Fig outing the trail W (indicated by dashed edges) through a good vertex of degree six. The letters indicate edges contained in the sets, and, and different types of edges represent colors (thick edges are red, thin ones are green and dotted ones are blue). Symmetric cases are omitted. the same and not all the flow values are present at the vertex v (see Figure 2.5). y symmetry, we can assume that the two edges of v are red and green, the two edges of v are also red and green, and the two edges of v are both red (recall that the vertex v is incident with even numbers of red, green and blue edges). s in the case of vertices of degree five, we find a trail formed by red and blue edges and swap the colors of the edges on the trail. The first edge of the trail is any red edge incident with v and the trail W is finished when it reaches again the vertex v. fter swapping red and blue colors on the trail W, the vertex v is incident with two edges of each of the three colors. gain, the number of bad vertices has been decreased which contradicts our choice of the flow ϕ ainbow Lemma. In this section, we present a generalization of an auxiliary lemma referred to as the ainbow Lemma. Lemma 2.12 (ainbow Lemma). Every cubic bridgeless graph contains a 2 factor F such that the edges of not contained in F can be colored with three colors,
10 10 T. KISE ET L. red, green and blue in the following way: every even circuit of F contains an even number of vertices incident with red edges, an even number of vertices incident with green edges and an even number number of vertices incident with blue edges, and every odd circuit of F contains an odd number of vertices incident with red edges, an odd number of vertices incident with green edges and an odd number number of vertices incident with blue edges. In the rest, a 2factor F with an edgecoloring satisfying the constraints given in Lemma 2.12 will be called a rainbow 2factor. ainbow 2factors implicitly appear in, e.g., [7, 18, 20], and are related to the notion of parity 3edgecolorings from the Ph.D. thesis of oddyn [11]. The lemma follows from Theorem 2.10 and the folklore statement that every cubic bridgeless graph has a 2factor F such that /F is 5oddconnected (see [17, 27]). The following strengthening of this statement appears in [5] which we use in the proof of Lemma Lemma Let be a bridgeless cubic graph with edges assigned weights and let w 0 be the total weight of all the edges of. The graph contains a 2factor F such that /F is 5oddconnected and the total weight of the edges of F is at most 2w 0 /3. For the proofs of our bounds, we need two strengthenings of the ainbow Lemma. Some additional notation must be presented. The pattern of a circuit = v 1... v k of a rainbow 2factor F is X 1... X k where X i is the color of the edge incident with the vertex v i ; we use to represent the red color, the green color and the blue color. Two patterns are said to be symmetric if one of them can be obtained from the other by a rotation, a reflection and/or a permutation of the red, green and blue colors. For example, the patterns and are symmetric but the patterns and are not. pattern P is compatible with a pattern P if P is obtained from P by replacement of some of the colors with the letter x (which represents a wildcard); note that rotating and reflecting P are not allowed. For example, the pattern is compatible with xxxx. The first modification of the ainbow Lemma is the following. Lemma Every cubic bridgeless graph contains a rainbow 2factor F such that no circuit of the 2factor F has length three, every circuit of length four has a pattern symmetric to or, and every circuit of length eight has a pattern symmetric to one of the following 16 patterns:,,,,,,,,,,,,,, and. Proof. Let F be a 2factor of such that the graph /F is 5oddconnected. Since is cubic and /F 5oddconnected, the 2factor F contains no circuits of length three. Let H 0 be the graph obtained from the graph /F by subdividing each edge. learly, H 0 is also 5oddconnected. This modification of /F to H 0 is needed only to simplify our arguments later in the proof since it guarantees that the graph is simple and makes it more convenient to apply Lemmas In a series of steps, we iteratively modify the graph H 0 to graphs H 1, H 2, etc. ll the graphs H 1, H 2,... will be simple and 5oddconnected. If the graph H i contains a vertex v of degree four, then the vertex v corresponds
11 SHOT YLE OVES 11 v 3 v 2 v 2 v 3 v 3 v 2 v or v 1 v 4 v 1 v 4 v 1 v 4 Fig eduction of a vertex of degree four in a graph H i in the proof of Lemma The newly created vertices of degree two are not drawn in the figure. v 3 v 2 v 4 v 5 v 3 v 2 v 4 v 5 v 3 v 2 v 4 v 5 v 3 v 2 v 4 v 5 v 3 v 2 v 4 v 5 v or or or v 1 v 8 v 7 v 6 v 1 v 8 v 7 v 6 v 1 v 8 v 7 v 6 v 1 v 8 v 7 v 6 v 1 v 8 Fig eduction of a vertex of degree eight in a graph H i in the proof of Lemma The newly created vertices of degree two are not drawn in the figure. v 7 v 6 to a circuit of length four in F. Let v 1, v 2, v 3 and v 4 be the neighbors of v in the order in which the edges vv 1, vv 2, vv 3 and vv 4 correspond to edges incident with the circuit. The graph H i+1 is a 5oddconnected graph among H i.v 1 vv 2 and H i.v 2 vv 3 (by Lemma 2.5 at least one of them is 5oddconnected); see Figure 2.8. No new vertices of degree four or eight were introduced at this stage. If the graph H i contains a vertex v of degree eight, we proceed as follows. The vertex v corresponds to a circuit of F of length eight; let v 1,..., v 8 be the neighbors of v in H i in the order in that they correspond to the edges incident with the circuit. y Lemma 2.7, we can assume that the graph H i.v 1 vv 2 is 5oddconnected (for a suitable choice of the cyclic rotation of the neighbors of v). y Lemma 2.8, at least one of the following graphs is 5oddconnected: H i.v 1 vv 2.v 3 vv 4, H i.v 1 vv 2.v 7 vv 8, H i.v 1 vv 2.v 3 vv 8.v 4 vv 5 and H i.v 1 vv 2.v 3 vv 8.v 4 vv 6. If the graph H i.v 1 vv 2.v 3 vv 4 is 5oddconnected, we then apply Lemma 2.5 to the graph H i.v 1 vv 2.v 3 vv 4 and conclude that the graph H i.v 1 vv 2.v 3 vv 4.v 5 vv 6 or the graph H i.v 1 vv 2.v 3 vv 4.v 6 vv 7 is 5 oddconnected. Since the case that the graph H i.v 1 vv 2.v 7 vv 8 is 5oddconnected is symmetric to the case of the graph H i.v 1 vv 2.v 3 vv 4, it can be assumed that (at least) one of the following four graphs is 5oddconnected: H i.v 1 vv 2.v 3 vv 4.v 5 vv 6, H i.v 1 vv 2.v 3 vv 4.v 6 vv 7, H i.v 1 vv 2.v 3 vv 8.v 4 vv 5 and H i.v 1 vv 2.v 3 vv 8.v 4 vv 6. Let H i+1 be a graph among these graphs that is 5oddconnected (see Figure 2.9). We eventually reach a 5oddconnected graph H k with no vertices of degree four or eight. y Theorem 2.10, the graph H k has a nowherezero Z 2 2flow. This flow yields a nowherezero Z 2 2 flow in H k 1,..., H 0 and eventually in /F. Note that the pairs of edges split away from a vertex are assigned the same flow value. The nowherezero Z 2 2flow of /F gives the coloring of the edges with red, green and blue. Since the colors of the edges of /F correspond to a nowherezero 4flow of /F, F is a rainbow 2factor with respect to this edgecoloring. We now verify that the edgecoloring of /F also satisfies the additional constraints given in the statement. Let us start with the first constraint, and let be a circuit of F of length four, v the vertex of /F corresponding to and e 1, e 2, e 3 and e 4 the four (not necessarily distinct) edges leaving in. In H 0, the edge e i corresponds to an edge vv i for a neighbor v i of v. During the construction of H k, either the vertices v 1 and v 2 or the vertices v 2 and v 3 are split away from v. In the
12 12 T. KISE ET L. former case, the colors of the edges e 1 and e 2 are the same and the colors of the edges e 3 and e 4 are the same; in the latter case, the colors of the edges e 1 and e 4 and the colors of the edges e 2 and e 3 are the same. In both cases, the pattern of is symmetric to or. Let be a circuit of F of length eight, v the vertex of /F corresponding to and e 1,..., e 8 the eight edges leaving in (note that some of the edges e 1,..., e 8 can be the same). Let c i be the color of the edge e i. ased on the splitting, one of the following four cases (up to symmetry) applies: 1. c 1 = c 2, c 3 = c 4, c 5 = c 6 and c 7 = c 8, 2. c 1 = c 2, c 3 = c 4, c 5 = c 8 and c 6 = c 7, 3. c 1 = c 2, c 3 = c 8, c 4 = c 5 and c 6 = c 7, and 4. c 1 = c 2, c 3 = c 8, c 4 = c 6 and c 5 = c 7. In the first case, the pattern of the circuit is symmetric to,,,, or. In the second case, the pattern of is symmetric to,,,,,,, or. The third case is symmetric to the second one (see Figure 2.9). In the last case, the pattern of is symmetric to,,,,,,,, or . In all the four cases, the pattern of is one of the patterns listed in statement of the lemma. We now present the second modification of the rainbow lemma. Lemma Let be a bridgeless cubic graph with edges assigned nonnegative integer weights and w 0 be the total weight of the edges. In addition, suppose that no two edges with weight zero have a vertex in common. The graph contains a rainbow 2factor F such that the total weight of the edges of F is at most 2w 0 /3. Moreover, the patterns of circuits with four edges of weight one are restricted as follows. Every circuit = v 1... v k of F that consists of four edges of weight one and at most four edges of weight zero (and no other edges) has a pattern: compatible with xx or xx if has no edges of weight zero (and thus k = 4), compatible with xxx or if the only edge of of weight zero is v 4 v 5 (and thus k = 5), compatible with xxxx, xxxx, xx or xx if the only edges of of weight zero are v 3 v 4 and v 5 v 6 (and thus k = 6), not compatible with,, or if the only edges of of weight zero are v 2 v 3 and v 5 v 6 (and thus k = 6), compatible with xxxxx, xxxxx, xxxxx, xxx, xxx, xxx, xxx, xxx or xxx if the only edges of of weight zero are v 2 v 3, v 4 v 5 and v 6 v 7 (and thus k = 7), and compatible with xxxxxx, xxxxxx, xxxxxx, xxxxxx, xxxx, xxxx, xxxx or xxxx if the edges v 1 v 2, v 3 v 4, v 5 v 6 and v 7 v 8 of have weight zero (and thus k = 8). Proof. Let F be the 2factor as in Lemma 2.13 and M the complementary perfect matching. We modify the graph H = /F in such a way that an application of Lemma 2.11 will yield a Z 2 2flow that yields an edgecoloring satisfying the conditions from the statement of the lemma. Let w be a vertex of H corresponding to a circuit v 1... v k of F consisting of four edges with weight one and some edges with weight zero, and let e i be the edge of M incident with v i. Finally, let w i be the neighbor of
13 SHOT YLE OVES 13 or or or or or or or or or or or or Fig Modifications of the graph H performed in the proof of Lemma The edges of weight one are solid and the edges of weight zero are dashed. The sets w, w and w are indicated by letters near the edges. Vertices of degree two obtained through splittings are not depicted and some symmetric cases are omitted in the case of a circuit of length eight. w in H that corresponds to the circuit containing the other endvertex of the edge e i. The graph H is modified as follows (see Figure 2.10): if k = 4, split the pair w 1 and w 2 or the pair w 2 and w 3 from w in such a way that the resulting graph is 5oddconnected (at least one of the two splittings works by Lemma 2.5).
14 14 T. KISE ET L. if k = 5 and the weight of the edge v 4 v 5 is zero, set w = {e 1, e 3, e 5 } and w = {e 1, e 3, e 4 }. if k = 6 and the weights of the edges v 3 v 4 and v 5 v 6 are zero, split the pair w 3 and w 4, w 4 and w 5, or w 5 and w 6 from w without creating edgecuts of size one or three (one of the splitting works by Lemma 2.6). If the pair w 4 and w 5 is split off, split further the pair w 2 and w 6, or the pair w 3 and w 6 from w again without creating edgecuts of size one or three (one of the splitting works by Lemma 2.5). if k = 6 and the weights of the edges v 2 v 3 and v 5 v 6 are zero, set w = {e 2, e 3 }, w = {e 5, e 6 } and w = {e 1, e 4 }. if k = 7 and the weights of the edges v 2 v 3, v 4 v 5 and v 6 v 7 are zero, split one of the pairs w i and w i+1 from w for i {2, 3, 4, 5, 6} (the existence of such a splitting is guaranteed by Lemma 2.7). If w 3 and w 4 is split off, set w = {e 1, e 2, e 6 } and w = {e 1, e 2, e 7 }. If w 5 and w 6 is split off, set w = {e 1, e 2, e 7 } and w = {e 1, e 3, e 7 }. if k = 8 and the weights of the edges v 1 v 2, v 3 v 4, v 5 v 6 and v 7 v 8 are equal to zero, split one of the pairs w i and w i+1 from w for some i {1,..., 8} (indices taken modulo eight) without creating edgecuts of size one or three. This is possible by Lemma 2.7. If i is odd, then there are no further modifications to be performed. If i is even, one of the pairs w i+3 and w i+4, w i+4 and w i+5, and w i+5 and w i+6 is further split off from the vertex w in such a way that the graph stays 5oddconnected (one of the splittings has this property by Lemma 2.6). In case that the vertices w i+4 and w i+5 are split off, split further the pair of vertices w i+2 and w i+3 or the pair of vertices w i+3 and w i+6, again, keeping the graph 5oddconnected (and do not split off other pairs of vertices in the other cases). Lemma 2.5 guarantees that one of the two splittings works. Fix a nowherezero Z 2 2flow ϕ with the properties described in Lemma 2.11 with respect to the sets w, w and w as defined before (and where the sets w, w and w are undefined, choose them arbitrarily). The edges of ϕ 1 (01) are colored with red, the edges of ϕ 1 (10) with green and the edges of ϕ 1 (11) with blue. This defines the coloring of the edges of not contained in F. learly, F is a rainbow 2factor. It remains to verify that the patterns of circuits with four edges of weight one are as described in the statement of the lemma. Let = v 1... v k be a circuit of F consisting of four edges with weight one and some edges with weight zero, and let c i be the color of the edge of M incident with v i. We distinguish six cases based on the value of k and the position of zeroweight edges (symmetric cases are omitted): if k = 4, then all the edges of have weight one. y the modification of H, it holds that c 1 = c 2 or c 2 = c 3. Hence, the pattern of is compatible with xx or xx. if k = 5 and the weight of v 4 v 5 is zero, then either c 1 c 3, or c 1 = c 3 {c 4, c 5 }. Since is incident with an odd number of edges of each color, its pattern is compatible with xxx or. if k = 6 and the weights of v 3 v 4 and v 5 v 6 are zero, then c 3 = c 4, or c 4 = c 5 and c 2 = c 6, or c 4 = c 5 and c 3 = c 6, or c 5 = c 6. Hence, the pattern of is compatible with xxxx, xx or xx, xx or xx, or xxxx. if k = 6 and the weights of v 2 v 3 and v 5 v 6 are zero, then the pattern of
15 SHOT YLE OVES 15 contains all three possible colors or it is compatible with xxxx, xxxx or xxxx. In particular, it is not compatible with any of the patterns listed in the statement of the lemma. if k = 7 and the weights of v 2 v 3, v 4 v 5 and v 6 v 7 are zero, then c i = c i+1 for some i {2, 3, 4, 5, 6} by the modification of H. If i is even, then the pattern of is compatible with xxxxx, xxxxx or xxxxx. If i = 3, then c 1 c 2 or c 1 = c 2 {c 6, c 7 }. Hence, the pattern of is compatible with xxx, xxx, x or x (unless c 2 = c 3 ). Since is incident with an odd number of edges of each colors, its pattern is compatible with one of the patterns listed in the statement of the lemma. symmetric argument applies if i = 5 and either c 1 c 7 or c 1 = c 7 {c 2, c 3 }. if k = 8 and the weights v i v i+1, i = 1, 3, 5, 7, then c i = c i+1 for i {1,...,8} by the modification of H. If there is such odd i, the pattern of is compatible with xxxxxx, xxxxxx, xxxxxx or xxxxxx. Otherwise, at least one of the following holds for some even i: c i+3 = c i+4, c i+4 = c i+5 or c i+5 = c i+6. In the first and the last case, the pattern is again compatible with xxxxxx, xxxxxx, xxxxxx or xxxxxx. If c i+4 = c i+5, then c i+2 = c i+3 or c i+3 = c i+6. Hence, the pattern of is compatible with xx, xx, xx, xx, xx or one of the patterns rotated by two, four or six positions. ll these patterns are listed in the statement of the lemma. 3. Intermezzo. In order to help the reader to follow our arguments, we present a proof that every bridgeless graph with m edges has a cycle cover of length at most 5m/3 based on the ainbow Lemma. The proof differs both from the proof of lon and Tarsi [1] which is based on 6flows and the proof of ermond, Jackson and Jaeger [2] based on 8flows; on the other hand, its main idea resembles the proof of Fan [7] for cubic graphs. Let F be a set of disjoint circuits of a graph, e.g., F can be a 2factor of a cubic graph. For a circuit of F and for a set of edges of E such that E =, we define (E) to be the set of vertices of incident with an odd number of edges of E. If (E) has even cardinality, it is possible to partition the edges of into two sets (E) and (E) such that each vertex of (E) is incident with one edge of (E) and one edge of (E), and each vertex of not contained in (E) is incident with either two edges of (E) or two edges of (E). We will always assume that the number of edges of (E) does not exceed the number of edges of (E), i.e., (E) (E). Note that if (E) =, then (E) contains no edges of and (E) contains all the edges of. Theorem 3.1. Let be a bridgeless graph with m edges. has a cycle cover of length at most 5m/3. Proof. If has a vertex v of degree four or more, then, by Lemma 2.2, v has two neighbors v 1 and v 2 such that the graph.v 1 vv 2 is also bridgeless. Let be the graph.v 1 vv 2. The number of edges of is the same as the number of edges of and every cycle of corresponds to a cycle of of the same length. Hence, a cycle cover of corresponds to a cycle cover of of the same length. Through this process we can reduce any bridgeless graph to a bridgeless graph with maximum degree three. In particular, we can assume without loss of generality that the graph
16 16 T. KISE ET L. has maximum degree three and is connected (if not, consider each component separately). If is a circuit, the statement is trivial. Otherwise, we proceed as we now describe. First, we suppress all vertices of degree two in. Let 0 be the resulting cubic (bridgeless) graph. We next assign each edge e of 0 the weight equal to the number of edges in the path corresponding to e in. In particular, the total weight of the edges of 0 is equal to m. Let F 0 be a rainbow 2factor with weight at least 2m/3; the existence of F 0 is guaranteed by Lemma 2.13 applied with the weight function w. The 2factor F 0 corresponds to a set F of disjoint circuits of the graph which do not necessarily cover all the vertices of. Let w F be the weight of the edges contained in the 2factor F 0, and r, g and b the weight of red, green and blue edges, respectively. y symmetry, we can assume that r g b. Since the weight w F of the edges contained in the 2factor F 0 is at least 2m/3, the sum r + g + b is at most m/3. Finally, let be the set of edges of corresponding to red edges of 0, the set of edges corresponding to green edges, and the set of edges corresponding to blue edges. y the choice of edgeweights, the cardinality of is r, the cardinality of is g and the cardinality of is b. The desired cycle cover of which is comprised of three cycles can now be defined. The first cycle 1 consists of all the red and green edges and the edges of ( ) for all circuits of the 2factor F. The second cycle 2 consists of all the red and green edges and the edges of ( ) for all circuits of F. Finally, the third cycle 3 consists of all the red and blue edges and the edges of ( ) for all circuits of F. Let us first verify that the cycles 1, 2 and 3 cover the edges of. learly, every edge not contained in F, i.e., a red, green or blue edge, is covered by at least one of the cycles. On the other hand, every edge of F is contained either in the cycle 1 or the cycle 2. Hence, the cycles 1, 2 and 3 form a cycle cover of. It remains to estimate the lengths of the cycles 1, 2 and 3. Each edge of F is covered once by the cycles 1 and 2 ; since (E) (E) for every circuit of F, at most half of the edges of F is also covered by the cycle 3. We conclude that the total length of the constructed cycle cover is at most: 3r + 2g + b + F + F /2 2(r + g + b) + 3w F /2 = 3(r + g + b + w F )/2 + (r + g + b)/2 3m/2 + m/6 = 5m/3. This finishes the proof of the theorem. 4. ubic graphs. We are now ready to improve the bound for cubic bridgeless graphs. Theorem 4.1. Every cubic bridgeless graph with m edges has a cycle cover comprised of three cycles of total length at most 34m/21. Proof. We present two bounds on the length of a cycle cover of and the bound claimed in the statement of the theorem is eventually obtained as their combination. In both bounds, the constructed cycle cover will consist of three cycles. Fix a rainbow 2factor F and an edgecoloring of the edges not contained in F with the red, green and blue colors as described in Lemma Let, and be the sets of the red, green and blue edges, respectively, and let r, g and b be their numbers. Finally, let d l be the number of circuits of lengths l contained in F. y Lemma 2.14, d 3 = 0.
17 SHOT YLE OVES 17 The first cycle cover. efore we proceed with constructing the first cycle cover, recall the notation of (E) and (E) introduced before Theorem 3.1. The cycle cover is comprised of three cycles 1, 2 and 3 which we next define. Let be a circuit of the 2factor F. Lemma 2.14 allows us to assume that if the length of is four, then the pattern of is either or (otherwise, we can for the purpose of extending the cycles 1, 2 and 3 to switch the roles of the red, green and blue colors and the roles of the cycles 1, 2 and 3 in the remaining analysis; note that we are not recoloring the edges, just apply the arguments presented in the next paragraphs with respect to a different permutation of colors). Similarly, we can assume that the pattern of the circuit of length eight is one of the following 16 patterns:,,,,,,,,,,,,,, and. For every circuit of F, we define subsets 1, 2 and 3 of its edges. The subset 1 is just ( ). The subset 2 is either ( ) or ( ) we choose the set with smaller intersection with ( ). Finally, 3 is chosen so that every edge of is contained in an odd number of the sets i ; explicitly, 3 = 1 2. Note that 3 is either ( ) or ( ). The cycle 1 consists of all red and green edges and the edges of 1 for every circuit of F. The cycle 2 consists of all red and blue edges and the edges of 2 for every circuit of F. Finally, the cycle 3 consists of all green and blue edges and the edges of 3 for every circuit of F. learly, the sets 1, 2 and 3 form cycles. We now estimate the number of the edges of contained in 1, 2 and 3. Let l be the length of. The sum of the numbers edges contained in each of the cycles is: = \ ( 1 2 ) = = l Since 1 = ( ) ( ), the number of edges of 1 is at most l/2. y the choice of 2, /2 l/4. Hence, the sets 1, 2 and 3 contain at most l + 2 l/4 edges of the circuit. The estimate on the number of edges of contained in the cycles 1, 2 and 3 can further be improved if the length of the circuit is four: if the pattern of is, then 1 = ( ) = and thus 1 2 =. If the pattern is, then 1 2 = ( ) ( ) =. In both the cases, it holds that 1 2 = and thus the cycles 1, 2 and 3 contain (at most) = = l = 4 edges of. Similarly, the estimate on the number of edges of contained in the cycles can be improved if the length of is eight. s indicated in Figure 4.1, it holds that 1 = ( ) 3. Hence, /2 3/2. onsequently, and the number of edges of included in the cycles 1, 2 and 3 is at most = 10. ased on the analysis above, we can conclude that the cycles 1, 2 and 3 contain
18 18 T. KISE ET L. Fig The sets 1 = ( ) for circuits with length eight; the edges contained in the set are drawn bold. at most the following number of edges of the 2factor F in total: = 3 2 2d 2 + 4d 4 + 7d 5 + 8d 6 + 9d d d d d 11 + l=12 3l 2 d l ld l d 2 2d d 5 d d 7 2d d 9 d d 11. (4.1) Since the 2factor F contains 2m/3 edges, the estimate (4.1) translates to: m d 2 2d d 5 d d 7 2d d 9 d d 11. (4.2) Since each red, green or blue edge is contained in exactly two of the cycles 1, 2 and 3 and there are m/3 such edges, the total length of the cycle cover of formed by 1, 2 and 3 does not exceed: 5m 3 d 2 2d d 5 d d 7 2d d 9 d d 11. (4.3) This finishes the construction and the analysis of the first cycle cover of.
19 SHOT YLE OVES 19 The second cycle cover. We reuse the 2factor F and the coloring of the edges of by red, green and blue colors from the construction of the first cycle cover. s long as the graph H = /F contains a red circuit, choose a red circuit of H = /F and recolor its edges with blue. Similarly, recolor edges of green circuits with blue. The modified edgecoloring still gives a rainbow 2factor. Let, and be the sets of red, green and blue in the modified edgecoloring and r, g and b their cardinalities. The construction of the cycle cover now follows the lines of the proof of Theorem 3.1. The first cycle 1 is formed by the red and green edges and the edges of ( ) for every circuit of the 2factor F. The cycle 2 is also formed by the red and green edges and, in addition, the edges of ( ) for every circuit of F. Finally, the cycle 3 is formed by the red and blue edges and the edges of ( ) for every circuit of F. learly, the sets 1, 2 and 3 are cycles of and they cover all the edges of. Let us now estimate the lengths of the cycles 1, 2 and 3. Each red edge is contained in all the three cycles, each green edge in two cycles and each blue edge in one cycle. Each edge of a circuit of length l of the 2factor F is contained either in 1 or in 2 and at most half of the edges of is also contained in the cycle 3. Hence, the total length of the cycles 1, 2 and 3 is at most: 3r + 2g + b + 3l d l. (4.4) 2 Since the red edges form an acyclic subgraph of /F, the number of red edges is at most the number of the cycles of F, i.e., d 2 + d 3 + d 4 + d Similarly, the number of green edges does not exceed the number of the cycles of F. Since r +g +b = m/3, the expression (4.4) can be estimated from above by m 3 +2r +g + 3l d l m d l + 3l d l = m l d l. (4.5) Since the 2factor F contains 2m/3 = 2d 2 + 3d 3 + 4d edges, the bound (4.5) on the number of edges contained in the constructed cycle cover can be rewritten to m m ( 3l l )d l 3m ( 3l l ) d l. (4.6) 4 Note that the last inequality follows from the fact that 3l l 4 3l 2 7l = 3 l 4 0 for l 12 and the expression 3l l 4 = 1/4 is also nonpositive for l = 11. The estimate (4.6) can be expanded to the following form (recall that d 3 = 0): 3m d 2 + 2d d d d 7 + d d d 10. (4.7) We remark that the bound (4.7) could also be obtained by a suitable substitution to the bounds presented in [7]. However, we decided to present the construction to make the paper selfcontained. The length of the shortest cycle cover of with three cycles exceeds neither the bound given in (4.3) nor the bound given in (4.7). Hence, the length of such cycle
20 20 T. KISE ET L. cover of is bounded by any convex combination of the two bounds, in particular, by the following: ( 5 5m 7 3 d 2 2d d 5 d d 7 2d d 9 d 10 3 ) 2 d 11 + ( 2 3m d 2 + 2d d d d 7 + d d ) 2 d 10 = 34m d d d d d d d 11 34m 21. The proof of Theorem 4.1 is now completed. 5. raphs with minimum degree three. We first show that it is enough to prove the main theorem of this section (Theorem 5.5) for graphs that do not contain parallel edges of certain type. We state four auxiliary lemmas. The first two lemmas deal with the cases when there is a vertex incident only with parallel edges leading to the same vertex, and the next two with the case that each of the two vertices joined by parallel edges is adjacent to another vertex. Since their proofs are similar, we prove only one of them (the one we consider most difficult) as an example and leave the proofs of the remaining proofs to the reader. Lemma 5.1. Let be an medge bridgeless graph with vertices v 1 and v 2 joined by k 3 parallel edges. If the degree of v 1 is k, the degree of v 2 is at least k+3 and the graph = \ v 1 has a cycle cover with three cycles of length at most 44(m k)/27, then has a cycle cover with three cycles of length at most 44m/27. Lemma 5.2. Let be an medge bridgeless graph with vertices v 1 and v 2 joined by k 4 parallel edges. If the degree of v 1 is k, the degree of v 2 is k +2 and the graph obtained from by removing all the edges between v 1 and v 2 and suppressing the vertex v 2 has a cycle cover with three cycles of length at most 44(m k 1)/27, then has a cycle cover with three cycles of length at most 44m/27. Lemma 5.3. Let be an medge bridgeless graph with vertices v 1 and v 2 joined by k 2 parallel edges. If the degree of v 1 is at least k + 1, the degree of v 2 is at least k + 2 and the graph obtained by contracting all the edges between v 1 and v 2 has a cycle cover with three cycles of length at most 44(m k)/27, then has a cycle cover with three cycles of length at most 44m/27. Proof. Let 1, 2 and 3 be the cycles of total length at most 44(m k)/27 covering the edges of and e 1,...,e k the k parallel edges between the vertices v 1 and v 2. y symmetry, we can assume that the cycles 1,..., i0 contain an odd number of edges incident with v 1 and the cycles i0+1,..., 3 contain an even number of such edges for some i 0 {0, 1, 2, 3}. Since 1,..., 3 form a cycle cover of, if v 1 is incident with an odd number of edges of i, i = 1, 2, 3, then v 2 is incident with an odd number of edges of i and vice versa. The edges are added to the cycles 1, 2 and 3 as follows based on the value of i 0 and the parity of k:
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