Small Cycle Cover of 2-Connected Cubic Graphs

. Small Cycle Cover of 2-Connected Cubic Graphs Hong-Jian Lai and Xiangwen Li 1 Department of Mathematics West Virginia University, Morgantown WV 26505 Abstract Every 2-connected simple cubic graph of order n 6 has a cycle cover with at most n cycles. This bound is best possible. 1. Introduction Graphs considered in this note are finite and undirected graphs. Undefined terms and notation can be found in [3]. The ceiling of a number x, denoted by x, is the smallest integer that is not less than x. A cycle cover of a graph G is a collection C of cycles of G such that every edge of G lies in at least one member of C. A cycle double cover of G is a cycle cover C of G such that each edge of G lies in exactly two members of C. The cycle double conjecture ([7], [6], and [8]) states that every bridgeless graph has a cycle double cover. It is known that if the cycle double cover conjecture is false, then a minimal counterexample would be a 2-edge-connected simple cubic graph. Bondy [2] conjectured that every 2-connected simple cubic graph on n vertices admits a double cycle cover C with C n 2. Bondy [2] also conjectured that if G is a 2-connected simple graph with n 3 vertices, then the edges of G can be covered by at most 2n 3 3 cycles. Earlier, Y. X. Luo and R. S. Chen [5] proved that this conjecture holds for 2-connected simple cubic graphs. Recently Fan [] proved this conjecture by showing that it holds for all simple 2-connected graphs. Barnette [1] proved that if G is a 3-connected simple planar graph of order n, then the edges of G can be covered by at most n+1 2 cycles. In this note, the following is proved. 1 Present address is Department of Mathematics and Statistics, University of Regina, Canada, SS 0A2 1

Theorem 1.1 Every 2-connected simple cubic graph of order n 6 admits a cycle cover C with C n. The result of Theorem 1.1 is sharp in the sense that there exists an infinite family of 2-connected cubic graphs such that each graph G in the family requires at least cycles in any cycle cover of G (see Figure 1). V (G) Let L 1, L 2 be two graphs isomorphic to K e, i.e. K minus an edge. Let G denote the graph in Figure 1. Note that n = V (G) = 2µ + 8. Since a cycle of G contains at most 2 edges in {e 1, e 2,..., e µ }, and since it takes at least 2 cycles to cover E(L 1 ) E(L 2 ), it follows that any cycle cover of G must have at least µ 2 + 2 = n cycles.......... L 1 e 1 e 2 e µ 1 e µ L 2......... Figure 1 2. Lemmas Let G be a 2-connected simple cubic graph and let T = {u 1, u 2, u 3 } be a 3-cut of G. If there is no vertex v V (G) such that N(v) = T, T is called a nontrivial 3-cut. Similarly we can define nontrivial 3-edge cut of G. A set of subgraphs of G is said to be independent if no two of them have a common vertex. The removal of an edge e = uv in a cubic graph G with N(u) = {x 1, y 1, v} and N(v) = {x 2, y 2, u} is to remove e and to replace the paths x 1 uy 1 and x 2 vy 2 by the edges x 1 y 1 and x 2 y 2 respectively. Denote by (G e) the resulting graph. Note that when G is a simple graph, (G e) may be not a simple graph. The following lemma is straightforward. Lemma 2.1 Let G be a simple cubic graph. Then (i) G is k-edge-connected if and only if G is k-connected, where 1 k 3. (ii) Suppose that G is a 3-connected graph of order n 6. Then G does not contain two distinct triangles T and T such that E(T ) E(T ). 2

Lemma 2.2 Let G be a 3-connected simple cubic graph of order n > and let T = x 1 x 2 x 3 x 1 be a triangle of G and let y i be the neighbor of x i for 1 i 3 such that {y 1, y 2, y 3 } V (T ) =. Suppose that T is contracted into one vertex x and that G is the resulting graph. If G has a cycle cover C, then G has a cycle cover C such that C = C. Proof. By Lemma 2.1 (ii), all triangles of G must be independent. Since G is 3-connected and cubic, it follows that G is a 3-connected cubic graph with V (G ) = V (G) 2. Let C 1, C 2 and C 3 denote the sets of cycles in C containing the paths y 1 xy 2, y 1 xy 3 and y 2 xy 3, respectively. Now replace the paths y 1 xy 2, y 1 xy 3 and y 2 xy 3 on all cycles containing them by the paths y 1 x 1 x 3 x 2 y 2, y 1 x 1 x 2 x 3 y 3 and y 2 x 2 x 1 x 3 y 3, respectively. Since C is a cycle cover of G, at least two of C 1, C 2, C 3 are nonempty. It follows that all edges of T can be covered by the modified cycles. These modified cycles, together with the unmodified cycles, form a cycle cover C of G with C = C. A graph G is essentially -edge-connected if it is 3-edge-connected and, if G S is disconnected for some set S of three edges of G, then G S has exactly two components, one of which is a single vertex. It follows that if G is a 3-connected cubic graph and G does not contain any nontrivial 3-edge cuts, then G is essentially -edge-connected. Lemma 2.3 Let G be a triangle free simple cubic graph. Then (i) Suppose that G is a 3-edge-connected graph and e E(G). 2-connected simple cubic graph. Then (G e) is a (ii) Suppose that G is an essentially -edge-connected graph and e E(G). (G e) is a 3-connected simple cubic graph. Proof. Then (i) By the assumption that G is a 3-edge-connected triangle free simple cubic graph, V (G) 8. Let e = uv E(G) and N(u) = {z 1, z 2, v} and N(v) = {w 1, w 2, u}. Since G is triangle free, z 1 z 2 / E(G), w 1 w 2 / E(G) and {z 1, z 2 } {w 1, w 2 } =. Since G is 3-edge-connected, (G e) is a connected simple cubic graph. Assume that e is an edge cut of (G e). Then {e, e} is an edge cut of G, contrary to the assumption that κ (G) 3. (ii) Let T = {e 1, e 2,..., e k } be a minimum nontrivial edge cut of G containing e = e 1, k. By (i), (G e 1 ) is a 2-connected simple cubic graph. We claim that (G e 1 ) is a 3-connected simple cubic graph. By contradiction, assume that (G e 1 ) has a 2-edge cut {e 1, e 2 }. Then G has an edge cut T {e 1, e 1, e 2 }. Since G is essentially -edge-connected, T = {e 1, e 1, e 2 } and there is a vertex w such that w is 3

an end of all edges of e 1, e 1 and e 2. On the other hand, by the definition of (G e 1), e 1, e 2 / E(G e 1), a contradiction. 3. Proof of Theorem 1.1 We argue by induction on n = V (G). As G is cubic, V (G) is even. When n = 6, G is one of the two graphs in Figure 2. It is easy to verify that each of G 1 and G 2 has a cycle cover C with C = 2. G 1 G 2 Figure 2: G 1 and G 2 Now we assume that n 8. Consider the following three cases. Case 1 G has a 2-cut {u, v} such that uv E(G). Then G has two vertex disjoint subgraphs G 1 and G 2 with nonadjacent vertices x i, y i V (G i ), (1 i 2), and a ladder G 3 whose ends are x 1, y 1, x 2 and y 2, such that E(G) = E(G 1 ) E(G 2 ) E(G 3 ), as shown in Figure 3. x 1 u 1 u 2 u µ 2 u µ 1 u µ......... x 2 G 1 G 2 y 1......... v 1 v 2 v µ 2 v µ 1 v µ y 2 Figure 3: The Graph G in the Proof for Case 1

Let u 1, v 1, u µ, v µ V (G 3 ) with x 1 u 1, y 1 v 1, x 2 u µ, y 2 v µ E(G). We define the three graphs G 1, G 2 and G 3 from G as follows: G i = G i +x i y i, i = 1, 2 and G 3 = G V (G 1 G 2 ). Let V (G i ) = n i, i = 1, 2, 3. It follows that n 1 + n 2 + n 3 = n, that both G 1 and G 2 are two 2-connected simple cubic graphs and that both n 1 and n 2 are even. By induction, for each i = 1, 2, if G = i K, then G i has a cycle cover C i with C i n i. We will distinguish the following subcases. Subcase 1.1 G i = K, i = 1, 2. Suppose first that µ = 1. On those cycles of C 1 containing x 1 y 1, replace x 1 y 1 by the path x 1 u 1 v 1 y 1 and on those cycles of C 2 containing x 2 y 2, replace by the path x 2 u 1 v 1 y 2. The modified and unmodified cycles then form a cycle cover C of G with C = C 1 + C 2 n 1 + n 2 n. In the case µ 2, let C be one of the cycles of C 1 containing x 1 y 1 and let D be one of the cycles of C 2 containing x 2 y 2. On those cycles of C 1 containing x 1 y 1, replace x 1 y 1 by the paths x 1 u 1 u 2... u µ x 2, D x 2 y 2 and y 2 v µ v µ 1... v 1 y 1, juxtaposed in the obvious way. On the cycles of C 2 containing x 2 y 2 replace x 2 y 2 by the paths x 2 u µ u µ 1... u 1 x 1, C x 1 y 1 and y 1 v 1 v 2... v µ y 2. We now have C 1 + C 2 1 cycles (two of the modified cycles are identical) which together cover all edges of G except the rungs of the ladder, which can be covered by µ 2 cycles of length. Thus we have a cycle cover C of G with C = C 1 + C 2 1+ µ 2 n. Subcase 1.2 Exactly one of G 1 and G 2 is isomorphic to K. We may assume that G 1 = K, G 2 = K. Since G 1 = K, E(G 1 ) has a cycle cover C 1 = {C 1, C 2 } such that x 1 y 1 E(C 1 ) E(C 2 ). By induction, G 2 has a cycle cover C 2 with at most n 2 cycles. Let D be a cycle of C 2 containing x 2 y 2. In case µ = 1, replace x 1 y 1 on C 1 by the path x 1 u 1 v 1 y 1 and replace x 1 y 1 on C 2 by the paths x 1 u 1 x 2, D x 2 y 2 and y 2 v 1 y 1, juxtaposed in the obvious way. On those cycles of C 2 containing x 2 y 2 ( other than D) replace x 2 y 2 by the path y 2 v 1 u 1 x 2. The modified and unmodified cycles together form a cycle cover C of G satisfying C = C 2 + 1 n 2 + 1 n İn case µ 2, replace x 1y 1 on C 1 by the path x 1 u 1 v 1 y 1 and replace x 1 y 1 on C 2 by the paths x 1 u 1 u 2... u µ x 2, D x 2 y 2, y 2 v µ v µ 1... v 1 y 1. On those cycles of C 2 (other than D) that contain x 2 y 2, replace x 2 y 2 by the path x 2 u µ v µ y 2. The modified and unmodified cycles cover all edges of G except µ 1 rungs of the ladder, which can be covered by µ 1 2 cycles of length 5

. These results a cycle cover C of G such that C = C 2 +1+ µ 1 2 n 2 +1+ µ 1 2 n. Subcase 1.3 G i = K, i = 1, 2. Then G is the graph in Figure 1. It is easy to check that G has a cycle cover C with C n if n = 8 or n = 10. So we assume that n 12. It follows that µ 2. All edges of G except the rungs of the ladder can be covered by two cycles and the rungs of the ladder can be covered by µ 2 cycles of length. Thus G has a cycle cover C such that C = 2 + µ 2 = n. Case 2. G has a 2-cut {u, v} but Case 1 does not occur. Then G has an edge cut X with X = 2 such that G X is the disjoint union of two subgraphs G 1 and G 2, (see Figure ). Since Case 1 does not occur, we must have both x 1 y 1 / E(G) and x 2 y 2 / E(G). G 1 x 1 x 2 G 2 y 1 y 2 Figure : The structure of G in Case 2 For i = 1, 2, define G i = G i + x i y i, and let n i = V (G i ). Note that n = n 1 + n 2. By induction, if G i = K, then G i has a cycle covers C i with C i n i for i = 1, 2. Since n 8, we assume first that G i = K, i = 1, 2. Then G is the graph in Figure 5. It is easy to verify that G has a cycle cover C with C = 2. 6

Figure 5 We assume then that G 1 = K and G 2 = K. G 1 can be covered C = {C 1, C 2 } such that x 1 y 1 E(C 1 ) E(C 2 ). Let D 1 be a cycle of C 2 containing x 2 y 2. On those cycles of C 2 containing x 2 y 2, replace x 2 y 2 by the paths x 1 x 2, C 1 x 1 y 1 and y 1 y 2. On two cycles of C 1 containing x 1 y 1, replace x 1 y 1 by the paths x 1 x 2, D 1 x 2 y 2 and y 1 y 2. The modified and unmodified cycles together form a cycle cover C of G. Since two of the modified cycles are identical, C = C 1 + C 2 1 2 + n 2 1 = n. Finally we assume that G i = K, i = 1, 2. Let C be a cycle of C 1 containing x 1 y 1 and D a cycle of C 2 containing x 2 y 2. On those cycles of C 1 containing x 1 y 1, replace x 1 y 1 by the paths x 1 x 2, D x 2 y 2 and y 1 y 2. On those cycles of C 2 containing x 2 y 2, replace x 2 y 2 by the paths x 1 x 2, C x 1 y 1 and y 1 y 2. Thus the modified cycles and unmodified cycles form a cycle cover C of G. Since two of the modified cycles are identical, C = C 1 + C 2 1 n 1 + n 2 1 n. Case 3. κ(g) 3. By Lemmas 2.1 and 2.2, we may assume that G is triangle free. If n = 8, then G must be one of the two graphs in Figure 6. It is easy to verify that each of which has a cycle cover C with C = 2. Figure 6 We now assume that n 10. We will distinguish the following two subcases. Subcase 3.1 G has a nontrivial 3-cut. 7

Let T = {e 1, e 2, e 3 } be a nontrivial 3 edge cut of G. Since G is a triangle free simple cubic graph, G T has only two components and each component of G T has at least 5 vertices. Let H 1, H 2 be the components of G T and assume that for i = 1, 2, 3, e i = x i y i with x 1, x 2, x 3 V (H 1 ) and y 1, y 2, y 3 V (H 2 ). Let u and v be two new vertices not in V (G), let H1 = H 1 {x 1 u, x 2 u, x 3 u} and H2 = H 2 {y 1 v, y 2 v, y 3 v}, and let n i = V (Hi ), i = 1, 2. It follows that for i = 1, 2, H i is a 2-connected (in fact, 3-connected) simple cubic graph with n i 6 and Hi = K. Thus, by induction, Hi has a cycle cover C i with C i n i. Let C(1) 1, C(2) 1 and C (3) 1 denote the sets of cycles of C 1 containing the paths x 1 ux 2, x 1 ux 3 and x 2 ux 3, respectively. Let C (1) 2, C(2) 2 and C (3) 2 denote the sets of cycles in C 2 containing the paths y 1 vy 2, y 1 vy 3 and y 2 vy 3, respectively. At least two of C (1) i, C (2) i and C (3) i are nonempty for each i = 1, 2. We may assume, without loss of generality, that C (3) 1 and C (3) 2 are nonempty. Let C (3) 1 C (3) 1 and C (3) 2 C (3) 2. Let P 1 be a path in H 1 joining x 1 and x 2 and Q 1 a path in H 1 joining x 1 and x 3. Let P 2 be a path in H 2 joining y 1 and y 2 and Q 2 a path in H 2 joining y 1 and y 3. Step 1. Modify the cycles in C 1 containing u as follows. From each cycle of C (1) 1 form a new cycle by deleting the path x 1 ux 2 and inserting the edges e 1 and e 2 and the path P 2. From each cycle of C (2) 1 form a new cycle by deleting the path x 1 ux 3 and inserting the edges e 1 and e 3 and the path Q 2. From each cycle of C (3) 1 form a new cycle by the deleting the path x 2 ux 3 and inserting the edges e 2 and e 3 and the path obtained from C (3) 2 by deleting y 2 vy 3. These modified cycles, together with the unmodified cycles of C 1 cover all edges of H 1, the edges e 1 e 2 and e 3 and some edges of H 2. Step 2. In exactly the same manner (with P 1, Q 1 and C (3) 1 playing the roles of P 2, Q 2 and C (3) 2 ) we modify all of the cycles of C 2 containing v. The resulting modified cycles, together with the unmodified cycles cover all of the edges of H 2 and some other edges. One of the modified cycles here is the same as one of those obtained in Step 1. These results a cycle cover C of G such that C = C 1 + C 2 1 n 1 + n 2 1 n. Subcase 3.2 G has no nontrivial 3-cut. graph. By Lemma 2.3, there is an edge e 1 such that (G e 1 ) is a 3-connected simple cubic Suppose first that (G e 1 ) contains a triangle T. Let G 1 be the graph obtained from (G e 1 ) by contracting T. It follows that G 1 is a 2-connected simple cubic graph. By induction, G 1 has a cycle cover C with C n. By Lemma 2.2 (G e 1) has a cycle cover C with C = C n. Since G is 3-connected, there is a cycle containing the 8

edge e 1. Therefore G has a cycle cover C with C + 1 n + 1 n. We then assume that (G e 1 ) is triangle free. Let e 1 = u 1 v 1, N(u 1 ) = {x 1, y 1, v 1 } and N(v 1 ) = {x 2, y 2, u 1 }. Then x 1 y 1, x 2 y 2 E((G e 1 ) ). Claim. (G e 1 ) contains an edge e 2 = u 2 v 2 such that {u 2, v 2 } {x 1, y 1, x 2, y 2 } =. Proof. Since G is triangle free, x 1, x 2, y 1, y 2 are four distinct vertices. Thus x 1 y 1 is independent of x 2 y 2. Since (G e 1 ) is 3-edge-connected, there is a cycle containing both x 1 y 1 and x 2 y 2. Let T be the minimal edge cut of (G e 1 ) which contains both x 1 y 1 and x 2 y 2. It follows that (G e 1 ) T contains only two connected components H 1 and H 2. Since (G e 1 ) is 3-edge-connected, there is an edge e = x 3 y 3 T {x 1 y 1, x 2 y 2 }. Without loss of generality, we assume that x 1, x 2, x 3 V (H 1 ) and y 1, y 2, y 3 V (H 2 ). If x 3 / {x 1, x 2 } and y 3 / {y 1, y 2 }, then e 2 = e is a desired edge. We now assume, without loss of generality, that y 1 = y 3. Then x 1 x 3. Suppose that x 3 = x 2. Since (G e 1 ) is a 3-edge-connected simple graph, let z 3 be a vertex such that z 3 y 3 E(G) and z 3 / {x 1, x 2 }. Since (G e 1 ) is cubic and simple, z 3 has two neighbors z 1, z 2 such that y 3 / {z 1, z 2 }. Since (G e 1 ) is triangle free, {z 1, z 2 } {x 1, x 3 } =. We assume, without loss of generality, that z 1 y 2. Then e 2 = z 1 z 3 is a desired edge. Now we assume that x 3 x 2. Observe the vertex x 3. Let a 1 and a 2 be two neighbors of x 3 such that y 3 / {a 1, a 2 }. Since (G e 1 ) is triangle free, x 1 / {a 1, a 2 } and {a 1, a 2 } {x 2, y 2 } 1. Without loss of generality, we assume that a 1 / {x 2, y 2 }. Then the edge a 1 x 3 is a desired edge. By Claim and by Lemma 2.3, ((G e 1 ) e 2 ) is a 2-connected simple cubic graph. Let N(u 2 ) = {s 1, t 1, v 2 }, N(v 2 ) = {s 2, t 2, u 2 } ( Note that {s 1, t 1, s 2, t 2 } {x 1, y 1, x 2, y 2 } can be not empty). Let G = ((G e 1 ) e 2 ). Since V (G ) = n 6, by induction, G has a cycle cover C 1 such that C 1 n. Replace the edges s 1 t 1 and s 2 t 2 on any cycles of C 1 containing them by the paths s 1 u 2 t 1 and s 2 v 2 t 2 respectively. Let C be the set of all modified and unmodified cycles of C 1. Then all edges of (G e 1 ) except e 1 lie in at least one member of C. Replace the edges x 1 y 1 and x 2 y 2 on any cycles of C containing them by the paths x 1 u 1 y 1 and x 2 v 1 y 2 respectively. Then the modified and unmodified cycles of C cover all edges of G except e 1 and e 2. Since G is 3-connected, there is a cycle in G containing e 1 and e 2, and so G has a cycle cover C of G such that C = C 1 + 1 n + 1 n. This completes the proof of Theorem 1.1. 9

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