Chords in Graphs. Department of Mathematics Texas State University-San Marcos San Marcos, TX Haidong Wu

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1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volme 32 (2005), Pages Chords in Graphs Weizhen G Xingde Jia Department of Mathematics Texas State Uniersity-San Marcos San Marcos, TX Haidong W Department of Mathematics Uniersity of Mississippi Uniersity, MS Abstract We characterize all simple graphs sch that each edge is a chord of some cycle. As a conseqence, we characterize all simple 2-connected graphs sch that, for any two adjacent ertices x and y, the local connectiity k(x, y) 3. We also make a conjectre abot chords for 3-connected graphs. 1 Introdction All graphs considered in the paper are ndirected and simple. Sppose that e is an edge of a graph G. For simplicity, we will se E(G)\e to denote E(G)\{e}, for example. An edge e is a chord of a cycle C if E(C) can be partitioned into two sets C 1 and C 2 sch that both C 1 e and C 2 e are cycles. We say that an edge is a chord of the graph if it is a chord of some cycle in the graph. A graph is chordal if each cycle of size at least for has a chord. The following is a characterization for chordal graphs de to Hajnal and Srányi, and Dirac, respectiely (see [3, Theorem 8.11]). Theorem 1.1 A graph G is chordal if and only if either G is complete or G can be obtained from two chordal graphs G 1 and G 2 (haing fewer ertices than that of G) by identifying two complete sbgraphs of the same order in G 1 and G 2. This athor s research was partially spported by the Smmer Research Grant of the College of Liberal Arts at the Uniersity of Mississippi in 2003.

2 118 WEIZHEN GU, XINGDE JIA, AND HAIDONG WU A graph G is minimally k-connected if G is k-connected, bt for each edge e of G, the deletion G\e is not k-connected. Dirac [4] and Plmmer [6] independently proed the following reslt. Theorem 1.2 Let G be a 2-connected simple graph. Then no cycle has a chord if and only if G is minimally 2-connected. If a graph G has minimm degree at least three, then it is shown that it has a cycle with a chord. In [1], Ali and Staton inestigated minimm degree conditions to force the existence of a cycle with k chords. Theorem 1.1 determines all graphs sch that each cycle of size at least for has a chord. Theorem 1.2 determines all 2-connected graphs sch that no cycle has a chord. In this paper, we answer the following natral qestion. Problem 1: Determine all simple graphs sch that each edge is a chord of some cycle. We will se D to denote the class of graphs sch that each edge is a chord. By Whitney s theorem [8], a simple graph with at least for ertices is 3-connected if and only if eery two ertices are connected by at least three internally disjoint paths. Ths each edge in a 3-connected graph is a chord of some cycle. Hence the class of graphs D contains all 3-connected graphs. Or main reslt shows that we can generate all sch graphs based on 3-connected graphs. First we describe two well-known sefl graph operations (see, for example, [5]). Let G 1 and G 2 be two graphs with no common ertices or edges. Let e 1 = 1 1 be an edge of G 1 and e 2 = 2 2 be an edge of G 2. Identify 1 and 2 and relabel as. Identify 1 and 2 and relabel as (the edges e 1 and e 2 are identified too and relabelled as e). We obtain a graph P (G 1, G 2 ), called the parallel connection of G 1 and G 2 with respect to the edges e 1 and e 2. The graph P (G 1, G 2 )\e is called the 2-sm of G 1 and G 2, denoted by G 1 2 G 2. Let S be a sbset of V (G). We se G[S] to denote the sbgraph indced by S. Before stating or main reslt, we define a class of graphs G and a graph operation first. T 0 w w T 1 w T 2 w T H H Figre 1:

3 CHORDS IN GRAPHS 119 Or class of graphs G is constrcted from K 2, K 3, or any 3-connected graph G. First sppose that we start with a 3-connected graph G. Let S be a non-empty set of edges of G. Take a graph N = K 4. For each edge e in S, label an edge of N as e first, then perform either the parallel connection or 2-sm operation between G and N along the edge e. We obtain a graph in G by repeating the operation for each element of S, while replacing G with the newly-obtained graph. There are six more graphs in G which we will describe next. Start with a triangle. Let S be the set of all edges in the triangle in the aboe constrction. One of the graphs constrcted is T 0, shown in Figre 1. The other graphs T 1, T 2, and T 3 can be obtained from T 0 by deleting one, two, and three edges in the set {, w, w}. Finally, sticking three copies of K 4 along a common edge (K 2 ), we get the the graph H 0. Remoe from H 0, we get the graph H 1 (see Figre 1.) Next we define a graph operation. We se K4 to denote the graph obtained by deleting an edge from K 4. Definition 1.3 Let G 1 and G 2 be two simple graphs. Sppose that for i = 1, 2, the graph G i has a 2-element ertex-ct { i, i }, sch that G i { i, i } has a component with ertex set S i where the sbgraph G[S i { i, i }] i i is isomorphic to K4. We define a new graph G 1 G 2 as follows (see Figre 2). From G 1 S 1 and G 2 S 2, identify 1 and 2 and relabel as, then identify 1 and 2 and relabel as, and remoe any mltiple edges if there are any (when i i E(G i ) for i = 1, 2). 1 w 1 t 1 t 2 w G 1 G 2 G 1 G 2 Figre 2: The next reslt is the main reslt of this paper. It has a similar flaor to that of Theorem 1. A proof of the theorem will be gien in the next section. Theorem 1.4 Let G be a simple graph. Then each edge of G is a chord of some cycle if and only if each block of any component of G is either 3-connected, or is a graph in G, or can be constrcted by a seqence of operations starting from graphs in G. For any two non-adjacent ertices x, y V (G), the local connectiity k(x, y) is defined to be the minimm nmber of ertices that separates x and y. For xy E(G), k(x, y) = k G xy (x, y) + 1. A graph is 3-connected if and only if the local connectiity

4 120 WEIZHEN GU, XINGDE JIA, AND HAIDONG WU is at least three for any two ertices of G. Or main reslt characterizes all simple 2-connected graphs sch that the local connectiity of any two adjacent ertices is at least three. Corollary 1.5 Let G be a simple 2-connected graph. Then for each edge xy E(G), the local connectiity k(x, y) 3 if and only if G is either 3-connected, or is a graph in G, or can be constrcted by a seqence of operations starting from graphs in G. Next we introdce some notation which will be needed in the proof. Let S be a ertex ct of G. We se G S to denote the sbgraph obtained by deleting S from G. A minimm component of a graph G is a component of G with minimm size. 2 Proof of the main reslt In this section, we will proe or main reslt. First we proe seeral lemmas. Lemma 2.1 Let G be a graph sch that each edge is a chord. Then each block of any component of G is both 2-connected and 3-edge-connected. Proof. Let H be a block of a component of G. Then each edge of H is a chord of some cycle in H. Hence H is not a single edge and therefore is 2-connected. For each edge e = in H, there are at least two internally disjoint paths connecting and in the graph H. Sppose that H is not 3-edge-connected and T = {, 1 1 } is a 2-element edge-ct. Then H has a ct edge. Ths H cannot hae two internally disjoint paths connecting and, a contradiction. Lemma 2.2 Let G be a 2-connected graph sch that each edge is a chord. Then each edge of H = G 2 K 4 (or P (G, K 4 )) is a chord of some cycle of H. Proof. We will gie a proof for H = G 2 K 4 only. The case for P (G, K 4 ) is obios. Sppose that H is obtained by first identifying a common edge of G and K = K 4, then deleting the edge. Now let e be any edge of H. If e E(G), then e is a chord of some cycle C of G. If C is also a cycle of H, then e is a chord of a cycle of H. If C is not a cycle of H, then C. Let w V (K) {, }, then C\ {w, w} is a cycle of H. Clearly, e is still a chord of this cycle. Now sppose that f / E(G). Then f is a chord of cycle C 1 in K. If / C 1, then C 1 is still a cycle of H. If C 1, then take a path P in G. This path exists as G is 2-connected. Now C 2 = C 1 P is also a cycle haing f as a chord. This completes the proof of the lemma. Lemma 2.3 Let G 1 and G 2 be two 2-connected graphs and G = G 1 G 2. Sppose that each edge of G i is a chord of G i for i = 1, 2. Then each edge of G is a chord of some cycle.

5 CHORDS IN GRAPHS 121 Proof. We se the same notation as in Definition 1.3. Let e be any edge of H. Withot loss of generality, assme that e E(G 1 ). Ths e is a chord of a cycle C of G 1. First sppose that e. If neither w 1 nor t 1 is in C (see Figre 2), then C is also a cycle of G haing e as a chord.. If either w 1 or t 1 is in C, then as { 1, 1 } is a 2-element ertex-ct, we dedce that both 1 and 1 are in C. The ertices 1 and 1 diides the cycle C into two 1 1 paths. As e is a chord of C, we conclde that one path, denoted by P 1 [, ], mst be in G. As G 2 is 2-connected, there is a path P 2 [ 2, 2 ] in the graph G 2 {w 2, t 2 } (otherwise, both 2 and 2 are ct-ertices of G 2, a contradiction). Now in G, P 1 [ 1, 1 ] P 2 [ 2, 2 ] is a cycle. Clearly this cycle has e as a chord. Now sppose that e =. Pick a i i -path P [ i, i ] in G i {w i, t i } for i = 1, 2. Then P 1 [ 1, 1 ] P 2 [ 2, 2 ] is a cycle haing e as a chord. This completes the proof of the lemma. Next we proe or main theorem. Recall that we se D to denote the class of graphs sch that each edge is a chord of some cycle. Proof of Theorem 1.4. Clearly, each edge of G is a chord if and only if each edge of all blocks of any component of G is a chord. Ths we may assme that G is a block. If G is 3-connected, then from Whitney s theorem, each edge is a chord. It is easy to check that each graph T 0, T 1, T 2, T 3, H 0, and H 1 is in D. By Lemma 2.2, each graph of G is in D. By Lemma 2.3, if a graph G is constrcted by a seqence of operations starting from graphs in G, then each edge of G is a chord. Now we proe the conerse. Sppose that eery edge of G is a chord and that G is connected. Clearly G has at least for ertices. Ths G is 2-connected as it is a block. We se indction on n = V (G). If n 4, it is straightforward to see that G mst be K 4 and the theorem holds. Sppose that G is not 3-connected, and G is not a member of G. By Lemma 2.1, G is 3-edge-connected. As G is not 3-connected, there is a 2-element ertex-ct S = {, } of G. Then G S has at least two components. Claim 1. Let T be the ertex set of a component of G S. If T = 2, then G[(S T )] = K 4. Proof. The claim follows from the fact that d G (x) 3 for each x T since, by Lemma 2.1, G is 3-edge-connected. Claim 2. Let S = {, } be a 2-element ertex-ct of G. Sppose that there are two sbgraphs H 1 and H 2 of G sch that V (H 1 ) V (H 2 ) = V (G) and V ((H 1 ) V (H 2 ) = S. If min{ V (H 1 ), V (H 2 ) } 5, then there are two graphs G 1 D, G 2 D sch that G = G 1 G 2. Moreoer, both G 1 and G 2 hae fewer nmber of ertices than G.

6 122 WEIZHEN GU, XINGDE JIA, AND HAIDONG WU 1 w 1 t 1 t 2 w G G 1 G 2 Figre 3: Proof. If E(G), let G 1 = P (H 1, K 4 ) and G 2 = P (H 2, K 4 ), where is the commonly identified edge. If / E(G), add to G first then define G 1 and G 2 similarly with the only difference being deleting the edge in the constrction of G 1 and G 2 (see Figre 2). Then as V (H 1 ), V (H 2 ) 5, both G 1 and G 2 hae fewer nmber of ertices than G has. Moreoer, it is straightforward to check that G 1 D, G 2 D by the constrction of G 1 and G 2. Now we contine the proof of the theorem. If G has no 2-element ertex-ct, then G is 3-connected and the theorem holds. Sppose that G has a 2-element ertex-ct S. Then there are two sbgraphs H 1 and H 2 of G sch that V ((H 1 ) V (H 2 ) = V (G) and V (H 1 ) V (H 2 ) = S. Sppose that there exist sch sbgraphs H 1 and H 2 sch that min{ V (H 1 ), V (H 2 ) } 5. By Claim 2, there are two graphs G 1 D, G 2 D sch that G = G 1 G 2. Moreoer, both G 1 and G 2 hae fewer nmber of ertices than G has. By indction, the theorem holds. Otherwise, for each 2-element ertex-ct S and each sbgraphs H 1 and H 2 of G sch that V (H 1 ) V (H 2 ) = V (G) and V ((H 1 ) V (H 2 ) = S, we hae min{ V (H 1 ), V (H 2 ) } = 4. Next we complete the proof of the theorem by showing that G G. By Lemma 2.1, G is 3-edge connected. Ths δ(g) 3. Hence each component of G S has at least two ertices. By or assmption, it is straightforward to see that G S has at most three components. Moreoer, if G S has exactly three components, then each sch component has exactly two ertices. Using Claim 1, it is easy to see that G is either H 0 or H 1. Therefore, we may assme that G W has exactly two components for each 2- element ertex-ct W of G. For each minimm component T of G W, we dedce that V (T ) = 2 by or assmption. Sppose that W = {p, q}. By Claim 1, G[(W T )] = K4. Remoe V (T ) from G, then add an edge pq. Repeat this process for each 2-ertex ct of G. Finally remoe any mltiple edges to get a graph H. Dring this process, all sbgraphs of the form K4 has been redced. If H has exactly one edge, then G = P (K 4, K 4 ), or K 4 2 K 4 and the theorem holds. Now sppose that H is a triangle with edges labelled as a, b, and c. Then G can be obtained from H and copies of K 4 by performing the operation of parallel connection or 2-sm along the edges a, b, and c seqentially. As each edge is a chord of some cycle, it is straightforward to check now that G is one of T 0, T 1, T 2 or T 3. Sppose that H has at least for ertices.

7 CHORDS IN GRAPHS 123 Clearly H is 2-connected. Now we show that H mst be 3-connected. Sppose that H has 2-element ertex-ct W. Then W is clearly also a 2-element ertex-ct of G. Ths G W has exactly two components and has one 2-element component T 1. By Claim 1, we conclde that G[(S T 1 ] = K4 and therefore T 1 is remoed in H and W cannot be a ertex-ct of H, a contradiction. This completes the proof of the theorem. 3 A conjectre A well-known conjectre of Thomassen (see, for example, [2]) is the following: Conjectre 1 Each longest cycle in a 3-connected graph has a chord. We make the following conjectre which is stronger than the aboe conjectre. Conjectre 2 Let G be a 3-connected graph and e be an edge of G. Then among all the cycles containing e, each longest sch cycle has a chord. While the aboe two conjectres seem to be ery hard, the following weaker reslt can be proed. Theorem 3.1 Let G be a 2-connected simple graph with minimm degree at least three. Then each edge lies in a cycle which contains a chord. This reslt follows from a stronger reslt of Voss [7]. Bt the proof of that reslt is long. Here we gie a simple proof of this weaker reslt. Proof. Sppose that the theorem fails. As δ(g) 3, G has at least 4 ertices. Let e be any edge of G. Take a longest cycle C containing e. Then clearly C 3. Sppose that C = 3. Let f E(G)\C. As G is 2-connected, there is a cycle C 1 containing both e and f. As G is simple and C = 3, we dedce that C 1 = 3 and C C 1 = {e}. Ths C C 1 \e is a 4-element cycle, a contradiction. Hence C 4. Let C = a 1 a 2... a n a 1, where n 4 and e = a n a 1. As G is 2-connected with minimm degree at least three, for each ertex a i, there is an a ki for some 1 k i n, k i i and a path P [a i, a ki ] sch that V (P [a i, a ki ]) V (C) = {a i, a ki }. Althogh there might be more than one path, we assme that path from a ki to the cycle is also P [a i, a ki ]. In other words, we assme that a kki = a i. As C is a longest cycle of G containing e, we conclde that (a) k i i 1, i + 1 for all 2 i n 1. (b) For 1 i n 1, P [a i, a ki ] and P [a i+1, a ki+1 ] do not meet except possibly at the end ertex (i.e., the only possible ertex they cold meet is when a ki = a ki+1.) (c) Sppose that k i > i for some i where 2 i n. Then k i+1 k i. Proof. by (b), V (P [a i, a ki ]) V (P [a i+1, a ki+1 ]) =. Sppose that k i+1 > k i. We dedce that a n a 1 a 2... a i P [a i, a ki ]a ki 1... a i+1 P [a i+1, a ki+1 ]a ki a n is a cycle

8 124 WEIZHEN GU, XINGDE JIA, AND HAIDONG WU containing e. Clearly this cycle has a chord a i a i+1, a contradiction. Now we complete the proof of the theorem. Clearly, k 2 > 2 by (a). Hence k 3 k 2. If k 3 > 3, then k 4 k 3. Contine the process and sppose that t (where k 2 t 2) is the largest nmber sch that k i > i for all 2 i t. Using (c) and the fact that P [a 2, a k2 ] is a path, it is straightforward to see that t exists and that 2 t k 2 1. By (c), we dedce that k 2 k 3... k t. By (a), t k t 2. Hence t + 1 < k t. By the choice of t and (a), we hae 1 k t+1 t 1. Denote k t+1 by j. As 2 j + 1 t, k j+1 k t t + 2. By (b) again, V (P [a j, a kj ]) V (P [a j+1, a kj+1 ]) =, noting that k j = t + 1. Ths a n a 1... a j P [a j, a kj ]a t... a j+1 P [a j+1, a kj+1 ]a kj a n is a cycle containing e. Clearly this cycle has a chord a j a j+1, a contradiction. This completes the proof of the theorem. References [1] Ali, A. and Staton, W., The extremal qestion for cycles with chords, Ars Combinatoria 51 (1999), [2] Bondy, J. A. Basic graph theory: paths and circits, in Handbook of combinatorics, Vol. 1, 3 110, Elseier, Amsterdam, [3] Chartrand, G. and Lesniak, L., Graphs & Digraphs, Chapman & Hall/CRC, 3rd edition, [4] Dirac, G. A., Minimally 2-connected graphs, J. Reine Angew. Math. 228 (1967), [5] Oxley, J. G., Matroid Theory, Oxford Uniersity Press, New York, [6] Plmmer, M. D., On minimal blocks, Trans. Amer. Math. Soc. 134 (1968), [7] Voss, H., Graphs haing circits with at least two chords, J. Combin. Theory, Ser. B 32 (1982), [8] Whitney, H., Congrent graphs and the connectiity of graphs, Amer. J. Math. 54 (1932), (Receied 15 Ag 2003)