The Chvátal-Erdős condition for supereulerian graphs and the hamiltonian index

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1 The Chvátal-Erdős condition for supereulerian graphs and the hamiltonian index Hong-Jian Lai Department of Mathematics West Virginia University Morgantown, WV 6506, U.S.A. Huiya Yan Department of Mathematics West Virginia University Morgantown, WV 6506, U.S.A. November, 005 Liming Xiong Department of Mathematics, Beijing Institute of Technology Beijing , P.R. of China Abstract A classical result of Chvátal and Erdős says that the graph G with connectivity κ(g) not less than its independent number α(g) (i.e. κ(g) α(g)) is hamiltonian. In this paper, we show that the graph G with κ(g) α(g) 1 is either supereulerian, or the Petersen graph, or the graphs obtained from K, by adding at most one vertex in one edge of K, and by replacing exactly one vertex whose neighbors have degree three in the resulting graph with a complete subgraph. We also show that the hamiltonian index of the graph G with κ(g) α(g) t is at most for any nonnegative integer t. t+ 1 Introduction We denote by κ(g) and α(g) the connectivity and the independent number of G respectively, considering only the simple graph with κ(g) in the following. Chvátal and Erdős gave the following well-known sufficient condition for a graph to be hamiltonian. Theorem 1. (Chvátal-Erdős, [8]) If κ(g) α(g), then G is hamiltonian. Corresponding author, This research has been supported by the Fund of Basic Research of Beijing Institute of Technology and The Project Sponsored by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry 1

2 There have existed many extensions of Theorem 1, one direction of the extensions is to study what better property a graph has when it satisfies a stronger Chvátal-Erdős condition. Theorem. (Häggkvist and Thomassen, [9]) Let t be an nonnegative integer. If κ(g) α(g) + t, then any system of disjoint paths of total length at most t is contained in a hamiltonian cycle of G. An immediately consequence of Theorem is that G is hamiltonian connected if κ(g) α(g) + 1. A graph is called supereulerian if it contains a spanning eulerian subgraph. Obviously a hamiltonian graph is supereulerian, but the reverse is not true. Our first aim of this paper is to consider whether the graph with a slight weaker Chvátal-Erdős condition is supereulerian. Theorem. If κ(g) α(g) 1, then G is either supereulerian, or the Petersen graph, or the graphs obtained from K, by adding at most one vertex in one edge of K, and by replacing exactly one vertex whose neighbors have degree three in the resulting graph with a complete subgraph. The line graph of G = (V (G), E(G)) has E(G) as its vertex set, and two vertices are adjacent in L(G) if and only if the corresponding edges share an common end vertex in G. The m-iterated line graph L m (G) is defined recursively by L 0 (G) = G, L 1 (G) = L(G) and L m (G) = L(L m 1 (G)). The hamiltonian index of a graph G, denoted by h(g), is the smallest integer m such that L m (G) is hamiltonian. Since every supereulerian graph has a hamiltonian line graph, the following known result is a consequence of Theorem. Theorem 4. (Benhocine and Fouquet, [4]) If κ(g) α(g) 1, then L(G) is hamiltonian, i.e., h(g) 1. The following conjecture extents Theorem 1 which indicates a weaker property of a graph with a weaker Chvátal-Erdős condition. Conjecture 5. (Fournier, see [1]) Let t be an nonnegative integer. If κ(g) = α(g) t, then G has a cycle C such that α(g C) t. Conjecture 5 has been verified for t = 1 [] and for t = []. There are many other directions of extension of Theorem 1, see the survey paper [1] and update papers [1], [11], [14] and [15]. Our second aim of this paper is to extend Theorems 1 and 4 by presenting an upper bound for h(g). Theorem 6. Let t be an nonnegative integer. If κ(g) α(g) t, then h(g) t+. In Section, we will give some auxiliary results which will be applied in Sections and 4 to prove our main results. The sharpness of Theorem 6 is presented in the last section.

3 Preliminaries.1 Reduced methods for supereulerian graphs For a graph G, let O(G) denote the set of odd degree vertices in G. In [5] Catlin defined collapsible graphs. Given a subset R V (G) with R is even, a subgraph Γ of G is an R-subgraph if both O(Γ) = R and G E(Γ) is connected. A graph G is collapsible if for any even subset R of V (G), G has an R-subgraph. Catlin showed in [5] that every vertex of G lies in a unique maximal collapsible subgraph of G. The reduction of G, denoted by G, is obtained from G by contracting all maximal collapsible subgraphs of G. Theorem 7. (Catlin, [5]) Let G be a connected graph and G the reduction of G. Then each of the following holds. (a) G is supereulerian if and only if G is supereulerian; (b) G is triangle-free with δ(g). Theorem 8. (Jaeger, [1]) If G is a 4-edge-connected graph, then G is supereulerian. The following theorem is obvious, where κ (G) denotes the edge-connectivity of G. Theorem 9. Let G be the reduction of G. Then each of the following holds. (a) α(g ) α(g); (b) If G is not K 1, then κ (G ) κ (G). Theorem 10. (Chen, [6]) If G is a reduced graph and α(g) 4, then we have (δ(g)α(g) + 4)/ V (G) 4α(G) 5. Theorem 11. (Chen, [7]) Let G be the graph of order n 11. If κ (G), then G is either collapsible or the Petersen graph.. Hamiltonian iterated line graphs Let G be a graph. A connected subgraph of G is called eulerian if it has only even degree. For any two subgraphs H 1 and H of G, the distance d G (H 1, H ) between H 1 and H is defined to be the minimum of the distances d G (v 1, v ) over all pairs with v 1 V (H 1 ) and v V (H ). If d G (e, H) = 0 for en edge e of G then we say that H dominates e. A subgraph H of G is called dominating if it dominates all edges of G. There is a characterization of a graph G with h(g) 1 which involves the existence of a dominating eulerian subgraph in G. Theorem 1. (Harary and Nash-Williams, [10]) Let G be a graph with at least three edges. Then h(g) 1 if and only if G has a dominating eulerian subgraph.

4 For each integer i 0, define V i (G) = {v V (G) : d G (v) = i}. A branch in G is a nontrivial path whose end vertices are not in V (G) and whose internal vertices, if any, are in V (G). We denote by B(G) the set of branches of G and by B 1 (G) the subset of B(G) in which every branch has an end vertex in V 1 (G). The following theorem can be considered as an analogue of Theorem 1 for L m (G). Theorem 1. (Xiong and Liu, [16]) Let G be a connected graph that is not a - cycle and let m be an integer. Then h(g) m if and only if EU m (G) where EU m (G) denotes the set of these subgraphs H of G which satisfy the following five conditions: (I) any vertex of H has even degree in H; (II) V 0 (H) (G) i= V i (G) V (H); (III) d G (H 1, H H 1 ) m 1 for any subgraph H 1 of H; (IV) E(b) m + 1 for any branch b with E(b) E(H) = ; (V) E(b) m for any branch in B 1 (G). Note that if we only consider -connected graphs then the condition (V) in the definition of EU m (G) is superfluous. Proof of Theorem Before presenting our proof, we start an useful lemma. Lemma 14. Let G be the reduced graph such that κ(g ) = and α(g ) =. Then G is isomorphic to these graphs obtained from K, by replacing at most two branches of K, with a branch of length three respectively. Proof of Lemma 14. Let k = κ(g ), and C = u 1 u u 1 be a closed trail with a maximal number of vertices in it (i.e., V (G ) \ V (C)) is minimal). We denote by u + (u respectively) the successor (the predecessor respectively) of u on C and let u ++ = (u + ) +, u = (u ) and so on. By Theorem 7, we obtain that u u + E(G) for any vertex u in C. Since δ(g ) κ(g ), and every graph with δ(g ) has a cycle of length at least δ(g ) + 1, C has at least k + 1 vertices. If C is not a spanning eulerian subgraph of G, then there is a component H, of G V (C). We will prove that G is isomorphic to these graphs obtained from K, by replacing at most two branches of K, with a branch of length three respectively. 4

5 Since κ(g ) =, C has at least k vertices with edges to H. Let x, y be the vertices with edges to H. We first claim that x + y + and x y since otherwise (say x + = y + ) the closed trail obtained by adding a x, y-path through H and by deleting the two edges in the path xx + y contains more vertices than C. If one of {x, x +, y, y + }, say, x +, has a neighbor in H, then we can get an eulerian subgraph from C by adding a x, x + -path through H and by deleting the edge xx +, which contains more vertices than C, a contradiction. Claim 1. x + y + E(G ) and x y E(G ). Proof of Claim 1. By the symmetry, we only need to prove that x + y + E(G ). If x + y + E(G ) \ E(C), then we can get a closed trail which contains more vertices than C, by using x + y +, the portions of C from x + to y and y + to x, and a x, y-path through H, a contradiction. If x + y + E(C), then there is at least one element of {x +, y + }, say, y +, has degree two in C, and S 0 = {x ++, y +, y +++ } is -vertices independent set since G is triangle-free. Noting that y ++ = x + and neither x ++ nor y ++ is adjacent to H, we get a four-vertices independent set from S 0 by adding a vertex of H to S 0, a contradiction. Claim 1 and the fact that α(g ) imply that either x + = y or x + y E(C) and either x = y + or x y + E(C). If there is an another vertex z V (C) different from x and y such that z has a neighbor in H, then one can easily find a four-vertices independent set, a contradiction. This implies that {x, y} is a -vertices cut set of G. We now claim that G is isomorphic to these graphs obtained from K, by replacing at most two branches of K, with a branch of length three respectively. This completes the proof of Lemma 14. Now we finish the proof of Theorem. Proof of Theorem. Let G be the reduced graph of G. If G = K 1, then obviously G is supereulerian by Theorem 7. So we only need to consider the case that G is not K 1. If κ(g) 4, then obviously G is supereulerian by Theorem 8. If κ(g) =, then by the hypothesis we have that α(g) 4. We let α(g) = 4 since otherwise G is hamiltonian by Theorem 1. Theorem 9 gives that α(g ) 4 and κ (G ) κ (G) κ(g). If α(g ) = 4, then from Theorem 10 we deduce that V (G ) 11. Hence G is either collapsible or the Petersen graph by Theorem 11. Now Theorem 7 and the fact that α(g) = 4 implies that G must be either collapsible (hence supereulerian) or the Petersen graph. Suppose α(g ). We claim that κ(g ) by Theorem 7. In the case that κ(g ), we have that G is hamiltonian (hence supereulerian) by Theo- 5

6 rems 1. So G is supereulerian by Theorem 7. It remains the case that κ(g ) = and α(g ) =. By Lemma 14, G is isomorphic to these graphs obtained from K, by replacing at most two branches of K, with a branch of length three respectively. Now the assumption that α(g) = 4 forces G must be the graphs obtained from K, by adding at most one vertex in one edge of K, and by replacing exactly one vertex whose neighbors have degree three in the resulting graph with a complete subgraph. It remains to consider the case that κ(g) = and α(g) =. Theorem 9 gives that α(g ). Hence by Theorem 7 we deduce that κ(g ) since otherwise α(g ) 4. We only need to consider the case that κ(g ) = and α(g ) = since in all other cases G is hamiltonian. By Lemma 14, G is isomorphic to these graphs obtained from K, by replacing at most two branches of K, with a branch of length three respectively. Now the assumption that α(g) = makes G must be isomorphic to G. This completes the proof of Theorem. 4 Proof of Theorem 6 Before presenting our proof of Theorem 6, we need some additional terminologies and notations. A subset S of B(G) is called a branch cut if the deletion of all internal vertices (of degree two) in any branch of S will produce more components than G. A minimal branch cut is called a branch-bond. It is easily shown that, for a connected graph G, a subset S of B(G) is a branch-bond if and only if the deletion of all internal vertices (of degree two) in any branch of S will produce exactly two components. We denote by BB(G) the set of branch-bonds of G. A branch-bond is called odd if it consists of an odd number of branches. The length of a branch-bond S BB(G), denoted by l(s), is the length of a shortest branch in it. Define BB (G) = {S BB(G) : S and S is odd}. Define h (G) = max{l(s) : S BB (G)} if BB (G) is not empty; 0, otherwise. From a result of [17] one can easily obtain the following result. Theorem 15. Let G be a -connected graph that is not a path. Then h(g) h (G) + 1. Now we present the proof of our second main result. Proof of Theorem 6. By Theorem 1 and Corollary 4, we only need to consider the case that t. Let m = t+. Then m. Since obviously h(g) for the graph G with κ(g) by Theorem 1, we can assume that κ(g) =. If h (G) m 1 then h(g) m by Theorem 15. So it remains to consider the case that h (G) m. Let B 0 be the branch-bond in BB (G) such that h (G) = min{ E(b) : b B 0 }. Noting that there are at least h (G) vertices in G[ b B0 V (b)] consisting of an independent set of G, which contains at most one end vertex of B 0, and hence α(g) h (G). We claim that h (G) m+1 6

7 since otherwise α(g) h (G) h (G) 1 m+5 which contradicts the fact that α(g) κ(g) + t = + t < + m + = m + 5. (4.1) We will prove that there is a subgraph H EU m (G) which implies that h(g) t+ by Theorem 1. Then there is an eulerian subgraph which contains exactly B 0 1 branches of B 0. Let H be the subgraph containing exactly B 0 1 branches of B 0 such that (1) H satisfies (I) and (II); () subject to (1), max H1 H max{d G (H 1, H H 1 ), } is minimized; () subject to (1) and (), H contains as few subgraph F with d G (F, H F ) = max H1 H max{d G (H 1, H H 1 ), } as possible; (4) subject to (1), () and (), H contains as many branches of length at least m + 1 as possible. We claim that H EU m (G). It only need to check H satisfies (III) and (IV). This follows from the following two claims. Claim 1. d G (H 1, H H 1 ) m 1 for any subgraph H 1 of H. Proof of Claim 1. Suppose otherwise there is a subgraph H 1 of H with d G (H 1, H H 1 ) m, then obviously any shortest path between H 1 and H H 1 is a branch of G. We claim that there are at least two branches of length at least m that are not in B 0 by the assumptions () and (). If any path between H 1 and H H 1 is not in B 0 then there are at least two such branches that are not in B 0 since k(g) = ; otherwise, say, b 0 is a branch of B 0 that is a path between H and H H 1, then there is another branch b 1 B 0, which is not a path between H 1 and H H 1, such that it has length at least m since otherwise we can obtain a new subgraph H by replacing the branch of B 0 H with a path containing b 0, contradicting the assumptions () and (). Hence there are at least ( m + 1) vertices consisting of an independent set, any element of which is not adjacent to any inner vertex of any branch of B 0, and so α(g) h (G) + ( m m 1 + 1) + ( m 1 + 1) m+5, which contradicts (4.1). This proves Claim 1. Claim. E(b) m + 1 for any branch b of G with E(b) E(H) =. Proof of Claim. Suppose otherwise there is a branch b 0 of G with E(b 0 ) E(E) = and E(b 0 ) m +. If h (G) = m, then we claim that there are at least h (G) + vertices consisting of an independent set, which implies α(g) m+5, a contradiction. If b 0 B 0 then one can easily find three vertices consisting of an independent set, such that any element of which is not an inner vertex of any branch of B 0 since b 0 has length at least four; if b 0 B 0 then either there is a branch 7

8 b 1 B 0 \{b 0 } of length at least m+ or there are at least two vertices consisting of an independent set, any element of which is not adjacent to any inner vertex of any branch of B 0, in either case we can find at least h (G) + vertices consisting of an independent set. It remains to consider the case that h (G) = m + 1. In this case, we claim that there are at least h (G) + vertices consisting of an independent set, which also implies α(g) m+5, a contradiction. Let b be a branch of B 0 with the length h (G). It is easy to see that the claim is true for the case that b 0 B 0 ; assuming next that b 0 B 0, we obtain that there is a branch-bond B 1 such that b B 1 and b 0 B 1 by the choice of (4) and let F be the component that any vertex of F is not the ends of these branches of B 0 \ {b 1 }. By the choice of H, G(F, H F ) and hence there are at least two nonadjacent vertices that are all not adjacent to any inner vertex of any branch of B 0. So α(g) h (G) +. This proves Claim. Thus we, in fact, have proved that G has a subgraph in EU m (G). Hence h(g) m = t+ by Theorem 1. This also completes the proof of Theorem 6. 5 Sharpness In this section, we discuss the sharpness. We will show that Theorem 6 is sharp by giving a graph satisfying the conditions of Theorem 6 with hamiltonian index exactly t+. By Theorems 1, and 1, we only need to show the sharpness of the case that t. Let H be a complete graph of order at least four and P 1, P, P be three edgedisjoint pathes with length m + 1. Now obtain the graph G by identifying one end of P i and the other end of P i with three vertices of H respectively. It is easy to see that if m is even then t = α(g) = m. It is easy to check that P 1 P H EU m (G). So h(g) m by Theorem 1, and obviously h(g) m, hence h(g) = m = t+. Moreover, the reduced graph G, of G, which is obtained from G by contracting H, has the property that t = α(g ) and h(g ) = t+ = m for every integer m 1. References [1] A. Ainouche, A common generalization of Chvátal-Erdős and Fraisse s sufficient conditions for hamiltonian graphs, Discrete Math. 14 (1995) [] D. Amar, I. Fournier and A. Germa, Structure des graphs de connexité k et de stabilité α = κ +, Ann. Discrete Math. 17 (198) [] D. Amar, I. Fournier, A. Germa and R. Häggkvist, Covering of vertices of a simple graph with given connectivity and independent number, Ann. Discrete Math. 0 (1984)

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