Czech Republic. Ingo Schiermeyer. Germany. June 28, A generalized (i; j)-bull B i;j is a graph obtained by identifying each of some two

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1 Claw-free and generalized bull-free graphs of large diameer are hamilonian RJ Faudree Deparmen of Mahemaical Sciences The Universiy of Memphis Memphis, TN USA Zdenek Ryjacek Kaedra maemaiky Zapadoceska univerzia Univerzin Plzen Czech Republic Ingo Schiermeyer Lehrsuhl fur Diskree Mahemaik und Grundlagen der Informaik TU Cobus D Cobus Germany June 28, 1998 Absrac A generalized (i; j)-bull B i;j is a graph obained by idenifying each of some wo disinc verices of a riangle wih an endverex of one of wo verex-disjoin pahs of lenghs i; j We prove ha every 2-conneced claw-free B 2;j -free graph of diameer a leas maxf7; 2jg (j 2) is hamilonian Keywords: hamilonian graphs, forbidden subgraphs, claw-free graphs 1991 Mahemaics Subjec Classicaion: 05C45 Research suppored by gran GA CR No 201/97/0407 1

2 1 Inroducion In his paper we consider nie simple undireced graphs G = (V (G); E(G)) and for deniions no dened here we refer o [2] For a se S V (G) we denoe by N(S) he neighborhood of S, ie he se of all verices of G which have a neighbor in S If S = fxg, we simply wrie N(x) for N(fxg) For any subse M V (G), we denoe N M (S) = N(S) \ M If H is a subgraph of G, we wrie N H (S) for N V (H) (S) The induced subgraph on a se M V (G) will be denoed by hm i By diam(g) we denoe he diameer of G, ie he larges disance of a pair of verices x; y 2 V (G) A pah wih endverices x; y will be referred someimes o as an xy-pah If x; z are verices a disance diam(g), hen any shores xz-pah will be called a diameer pah of G If H 1 ; : : : ; H k (k 1) are graphs, hen a graph G is said o be H 1 ; : : : ; H k -free if G conains no copy of any of he graphs H 1 ; : : : ; H k as an induced subgraph; he graphs H 1 ; : : : ; H k will be also referred o in his conex as forbidden subgraphs Specically, he four-verex sar K 1;3 will be also denoed by C and called he claw and in his case we say ha G is claw-free Whenever verices of an induced claw are lised, is cener, (ie is only verex of degree 3) is always he rs verex of he lis Furher graphs ha will be considered as forbidden subgraphs are shown in Fig 1 The claw C Z 1 Z 2 The bull B The wounded W The ne N Figure 1 There are many resuls dealing wih hamilonian properies in classes of graphs dened in erms of forbidden induced subgraphs (see eg [9], [7], [10], [3], [4]) Bedrossian [1] (see also [8]) characerized all pairs X; Y of conneced forbidden subgraphs implying hamiloniciy Theorem A [1] Le X and Y be conneced graphs wih X; Y 6= P 3, and le G be a 2-conneced graph ha is no a cycle Then, G being XY -free implies G is hamilonian if and only if (up o symmery) X = C and Y = P 4 ; P 5 ; P 6 ; C 3 ; Z 1 ; Z 2 ; B; N or W The resuls on hamiloniciy in CP i -free, CZ i -free and CN-free graphs were exended o larger classes (by characerizing he classes of nonhamilonian excepions) in [5] and 2

3 [6] by using he closure concep inroduced in [11] A similar exension is possible in he class of CB-free graphs by inroducing he class of CB i;j -free graphs, where by B i;j (i; j 1) we denoe he generalized bull, ie he graph obained by idenifying each of some wo disinc verices of a riangle wih an endverex of one of wo verex-disjoin pahs of lenghs i; j (see Fig 2) a i 1 a 2 1 a 1 1 b 1 b 2 a 1 2 a 2 2 a j 2 u : : : u u u u u u : : : u JJ J J J u b 3 B i;j Figure 2 However, as shown in [5], he closure mehod is no applicable in his class since here are CB i;j -free graphs such ha heir closure [11] is no CB i;j -free I is easy o see ha here are CB i;j -free graphs of arbirarily large diameer (a simple example can be obained by aking d + 1 verex-disjoin cliques K 0 ; K 1 ; : : : ; K d and by adding all of he edges beween consecuive cliques, namely fxyj x 2 K i ; y 2 K i+1 ; i = 0; 1; : : : ; d? 1g) In he main resul of his paper we show ha, for any j 2, all 2-conneced nonhamilonian CB 2;j -free graphs have small diameer 2 Main resul Before we prove he main resul of he paper, Theorem 2, we rs make some preliminary observaions on shores pahs and heir neighborhoods Le G be a claw-free graph, le x; y 2 V (G) and le P : x = v 0 v 1 v 2 : : : v k = y (k 3) be a shores xy-pah in G Le z 2 V (G) n V (P ) 1 If jn P (z)j = 1, hen, since G is claw-free, z is adjacen o x or o y 2 If jn P (z)j 2 and fv i ; v j g N P (z), hen, since P is a shores pah, ji? jj 2 3 By (1) and (2), jn P (z)j 3 for every verex z 2 V (G) n V (P ) Moreover, if 2 jn P (z)j 3, hen he verices of N P (z) are consecuive on P This moivaes he following noaion: N i := fz 2 V (G) n V (P )j N P (z) = fv i?1 ; v i ; v i+1 gg for 1 i k? 1, 3

4 M i := fz 2 V (G) n V (P )j N P (z) = fv i?1 ; v i gg for 1 i k, M 0 := fz 2 V (G) n V (P )j N P (z) = fv 0 gg, M k+1 := fz 2 V (G) n V (P )j N P (z) = fv k gg Thus, by (1), (2) and (3), N(P ) [ V (P ) = ( S k i=1 N i) [ ( S k+1 i=0 M i) [ V (P ) We furher denoe S = V (P ) [ N(P ) and R = V (G) n S Lemma 1 Le j 2, le G be a CB 2;j -free graph of diameer a leas maxf7; 2jg and le P : v 0 v 1 v 2 : : : v d be a diameer pah in G Then (i) hn i i is complee for 1 i d? 1 and hm j i is complee for 0 j d + 1, (ii) M i = ; for 3 i d? 2, (iii) hn i [ N i+1 i is complee for 1 i d? 2, (iv) for every verex z 2 R we have N P (z) = ; and N S (z) M 0 [ M 1 [ M 2 [ M d?1 [ M d [ M d+1 Proof (i) If some N i or M i is no complee, hen some v j, j 2 fi? 1; i; i + 1g, is a cener of an induced claw, a conradicion (ii) Suppose M i 6= ; for some i, 3 i d? 2 Then, since d 2j, for any verex x 2 M i we have hfv i?3 ; v i?2 ; v i?1 ; x; v i ; v i+1 ; : : : ; v i+j gi ' B 2;j or hfv i?j?1 ; v i?j ; : : : ; v i?1 ; x; v i ; v i+1 ; v i+2 gi ' B 2;j, a conradicion (iii) Suppose xy =2 E(G) for some i wih 1 i d? 2 and wo verices x 2 N i, y 2 N i+1 Then hfv i?1 ; x; v i+1 ; y; v i+2 ; v i+3 ; : : : ; v i+2+j gi ' B 2;j or hfv i+2 ; y; v i ; x; v i?1 ; v i?2 ; : : : ; v i?1?j gi ' B 2;j, a conradicion (iv) By he deniion of P and R, we have N P (z) = ; for every verex z 2 R Since G is claw-free, we have also N Ni (z) = ; for 1 i d? 1 We can now sae he main resul of he paper Theorem 2 Le j 2 be an ineger and le G be a 2-conneced CB 2;j -free graph of diameer d maxf7; 2jg Then G is hamilonian Remark From [1] we know ha every 2-conneced CB 1;1 -free or CB 2;1 -free graph is hamilonian The graph in Fig 3 indicaes ha here are 2-conneced nonhamilonian graphs of diameer d = 6 ha are CB 2;j -free for any j 2 The example in Fig 4 shows ha here are 2-conneced nonhamilonian graphs which are CB 2;j -free and have diameer d = 2j? 1 for any j 3 Hence he requiremen d maxf7; 2jg in Theorem 2 is sharp Moreover, he example in Figure 5 indicaes ha here are 2-conneced nonhamilonian graphs of arbirary diameer d 3 which are CB i;j -free for any pair i; j such ha i 3, j i Hence he requiremen i = 2 in Theorem 2 is also bes possible 4

5 I is easy o see ha, in fac, each of he examples in Figures 3 { 5 yields an innie family, since each of he verical edges (marked in he gure by K i ) can be blown up o a clique of arbirary order K K K K 1 2 K 3 K Figure 3 K 0 K j?2 K j?1 K j K j+1 K 2j?1 Figure 4 K 0 K 1 K 2 K 3 K 4 K d Figure 5 Proof of Theorem 2 Le G be a 2-conneced CB 2;j -free graph of diameer d maxf7; 2jg, j 2, and le P : v 0 v 1 v 2 : : : v d be a diameer pah in G Le M i ; N i ; S; R be as in Lemma 1 For i 2 f0; 1; 2; d? 1; d; d + 1g furher denoe M i = N R (M i ) We rs make he following observaion concerning he srucure of G "close" o he ends of P Denoe S 0 = [ 2 i=0 (N i [ M i [ M i [ fv ig); R 0 = V (G) n (S [ S 0 ) 5

6 and S d = [ d i=d?2 (N i [ M i+1 [ M i+1 [ fv ig); R d = V (G) n (S [ S d ) (where we se N 0 = N d = ;) Then we have he following claims Claim 1a The subgraph hs 0 i saises one of he following: (i) N R0 (S 0 ) = ;, N 2 6= ;, and for any x 2 2 N 2 here is a hamilonian x 2 v 2 -pah P 0 in hs 0 i, (ii) N R0 (S 0 ) 6= ;, M 1 = M 2 = M 2 = ;, and N S0 (R 0 ) M 0 Claim 1b The subgraph hs d i saises one of he following: (i) N Rd (S d ) = ;, N d?2 6= ;, and for any x d?2 2 N d?2 here is a hamilonian x d?2 v d?2 - pah P d in hs d i, (ii) N Rd (S d ) 6= ;, M d = M d?1 = M d?1 = ;, and N S d (R d ) M d+1 By symmery, i is sucien o prove Claim 1a We disinguish wo cases Case 1: M 2 6= ; We rs show ha M 0 M 2 If M 0 = ;, hen obviously M 0 = ; M 2 Hence we may assume ha M 0 6= ; Then hm 0 [ M 2 i is complee, since oherwise, for some wo verices x 2 M 0, y 2 M 2 such ha xy =2 E(G) we have hfx; v 0 ; v 1 ; y; v 2 ; v 3 ; : : : ; v 2+j gi ' B 2;j, a conradicion (noe ha boh hm 0 i and hm 2 i are complee by Lemma 1 (i)) Suppose now ha yz =2 E(G) for wo verices y 2 M 2, z 2 M Then hfx; z; y; v 0 0gi ' C for a verex x 2 M 0, a conradicion This implies ha yz 2 E(G) for every y 2 M 2, z 2 M 0 Bu hen every verex in M 0 has a neighbor in M 2, ie M 0 M 2, as required Nex we show ha M 1 M 2 We may assume ha M 6= ; Le z 1 2 M 1, ie xz 2 E(G) for some x 2 M 1, and suppose ha zy =2 E(G) for some y 2 M 2 If xy =2 E(G), hen hfz; x; v 1 ; y; v 2 ; v 3 ; : : : ; v 2+j gi ' B 2;j, and if xy 2 E(G), hen hfx; z; y; v 0 gi ' C Hence zy 2 E(G), implying M 1 M 2 Thus, we conclude ha (M 0 [ M 1 ) M 2 Now, if N R0 (S 0 ) 6= ;, hen yz 2 E(G) for some wo verices y 2 M 2 and z 2 R 0, bu hen, for a verex x 2 M 2, hfz; y; x; v 1 ; v 2 ; v 3 ; : : : ; v 2+j gi ' B 2;j, a conradicion Hence N R0 (S 0 ) = ; There is also no edge from M o any of M 2 i, i 3, since M i = ; for 3 i d? 2 by Lemma 1(ii), and an edge from M 2 o any of M d?1, M d, M d+1 yields a v 0 v d -pah of lengh a mos 6, conradicing he fac ha P is a diameer pah and d 7 Consequenly, N 2 [ fv 2 g is a cuse of G Since G is 2-conneced, N 2 6= ; Summarizing, we already know ha hm 0 [ M 2 i is complee, (M 0 [ M ) 1 M and, 2 by Lemma 1(iii), hn 1 [ N 2 i is complee Moreover, i is easy o see ha N(x) \ M 2 is complee or empy for all x 2 M 2, and if M 0 6= ;, hen M 0 = M 2 (oherwise we have a claw wih cener in M 2 ) Bu hen i is sraighforward o check ha in each of he 6

7 possible cases (according o wheher M 0, M 1, N 1 and M 2 are empy or nonempy) here is a hamilonian x 2 v 2 -pah in hs 0 i for any x 2 2 N 2 Thus, we are in siuaion (i) of Claim 1a Case 2: M 2 = ; We rs consider he subcase when M 0 = M 1 = ; This immediaely implies ha N R0 (S 0 ) = ; and, since G is 2-conneced, N 2 6= ; Moreover, if M 0 6= ;, hen, since v 0 canno be a cuverex, here is an edge xy wih x 2 M 0 and y 2 M 1 [ N 1 In all hese cases, i is easy o nd a hamilonian x 2 v 2 -pah in hs 0 i for any x 2 2 N 2, ie we are again in siuaion (i) of Claim 1a Hence we suppose ha M [M 6= ; Now, if M 6= ;, hen for any verex v M 1, any shores vv d -pah hrough M 1 has lengh d + 1 (noe ha here is no pah vu 1 u 2 v 3 : : : v d of lengh d wih u 1 2 M 1 and u 2 2 N 2, since oherwise we have hfu 1 ; v; v 0 ; u 2 gi ' C, a conradicion) Since G has diameer d, here mus be anoher vv d -pah P 0 of lengh a mos d Le w be he successor of v on P 0 If w 2 M d?1 [ M d [ M d+1, hen we ge a v 0 v d -pah of lengh a mos 5; hence w 2 R 0 Bu hen, for a verex x 2 M 1, hfw; v; x; v 0 ; v 1 ; : : : ; v 1+j gi ' B 2;j, a conradicion Hence M = ;, implying M 1 0 6= ; Summarizing, we now have M 2 = ;, M 1 = ; and M 6= ; By he deniion of R 0 0 and M i (i = 0; 1; 2), here is no edge beween R 0 and M 0 [ M 1 [ M 2, which implies ha N S0 (R 0 ) M 0 (if nonempy) Thus, if N R0 (S 0 ) 6= ;, we are in siuaion (ii) of Claim 1a Hence nally suppose ha N R0 (S 0 ) = ; Then N(M ) 0 M 0 [ M 0 [ M d?1 [ M d [ M d+1 If N Sd (M ) 6= ;, we obain a v 0 0v d -pah of lengh ` < 7, a conradicion Hence N(M ) 0 M 0 [ M 0, bu hen any verex x 2 M 0 is a disance a leas d + 2 from v d, conradicing he fac ha P is a diameer pah Hence he claim follows 2 Suppose now ha S 0 saises (i) of Claim 1a and S d saises (i) of Claim 1b Then every fv i g[n i is a cuse of G, and since G is 2-conneced, N i 6= ; for 3 i d?3 Le P i be a hamilonian pah in hn i i, 3 i d?3 Then x 2 P 0 v 2 P 3 P 4 : : : P d?3 x d?2 P d v d?2 v d?3 : : : v 3 x 2 is a hamilonian cycle in G By symmery, i remains o consider he case when hs 0 i saises (ii) of Claim 1a Le x 2 M 0 If xy 2 E(G) for some y 2 N 1, hen hfz; x; y; v 1 ; v 2 ; v 3 ; : : : ; v 2+j gi ' B 2;j for a verex z 2 M 0, a conradicion Hence N N1 (M 0 ) = ; If N(x) \ (M d?1 [ M d [ M d+1 ) 6= ;, we ge a v 0 v d -pah of lengh a mos d? 1 Since M i = ; for 2 i d? 2, here is no xv d -pah in hsi of lengh a mos d Hence here is a shores v d x-pah P 0 of lengh ` such ha d? 1 ` d and V (P 0 ) \ R 6= ; This immediaely implies ha N Rd (S d ) 6= ;, ie hs d i saises (ii) of Claim 1b (specically, he successor of v d on P 0 is in M d+1 ) Le v d ; v d+1 ; : : : ; v d+` = x be he verices of P 0 If ` = d, denoe by P 00 he pah v d+1 v d+2 : : : v 2d v 0 ; oherwise se P 00 = P 0 I is apparen ha P 00 is also a diameer pah 7

8 Denoe by! C he cycle (wih an orienaion) v 0 v 1 : : : v d v d+1 : : : v 2d (v 0 ) (of lengh 2d + 1 or 2d, respecively) We show ha! C has no chord Since P, P 0 and P 00 are shores pahs, here is no chord xy wih x; y 2 V (P ) or x; y 2 V (P 00 ) By Lemma 1(ii) (applied on P, P 0 and P 00 ), and since all he pahs P, P 0, P 00 saisfy (ii) of Claim 1a, 1b, he only possible chords in C! are he edges v d?1 v d+1 and xv 1 Bu, if eg v d?1 v d+1 2 E(G), hen v d+1 2 M d, implying v d+2 2 M d, which conradics Claim 1b (ii) Hence v d?1 v d+1 =2 E(G) and, by symmery, xv 1 =2 E(G) Now we observe ha, for any x 2 V (G)nV ( C),! jn! (x)j 3 Immediaely jn! (x)j 4, C C since G is claw-free and C! is chordless If jn! (x)j = 4, hen x has neighbors on boh P C and P 00 Since M i = ; for 2 i d? 1 and no verex in any N i can have a neighbor ouside S, he only possibiliy (up o a symmery) is N! (x) = fv d?2; v d?1 ; v d ; v d+1 g, C bu hen hfv d+2 ; v d+1 ; x; v d?1 ; v d?2 ; : : : ; v d?2?j gi ' B 2;j, a conradicion Hence for every x 2 V (G) n V ( C)! wih N! (x) 6= ; we have jn! (x)j 3 and, since C! is chordless and G C C is claw-free, jn! (x)j 2 We can hus dene analogously as before: C N C i M C i := fz 2 V (G) n V (! C)j zv i?1 ; zv i ; zv i+1 2 E(G)g, := fz 2 V (G) n V (! C)j zv i?1 ; zv i 2 E(G)g for 1 i jv ( C)j! (indices are considered modulo jv ( C)j)! By Lemma 1(ii) (applied on P, P 0 and P 00 ), and by Claim 1a, 1b(ii), we have M i = ; for 2 i d? 1 and d + 2 i 2d? 1 (and also i = 2d if x = v 2d ) Bu his and he fac ha C! has no chords implies, ogeher wih Lemma 1(iv), ha, for = d d e, 2 he pah P 000 : v v +1 : : : v +d is a shores v v +d -pah in G Since P 000 has lengh d, P 000 is a diameer pah, implying ha, by Lemma 1 (iv) (applied o P 000 ), M d = M d+1 = ; By symmery, we also have M 2d?1 = M 2d = M 1 = ; Bu hen, by Lemma 1 (iv),! V (G) = [ jv ( C )j i=0 (fv i g [ N i ) Le P i be a hamilonian pah in hn i i, i = 0; : : : ; jv ( C)j! Then v 0 P 0 v 1 P 1 : : : v 2d?1 P 2d?1 v 2d (P 2d v 0 ) is a hamilonian cycle in G References [1] Bedrossian, P: Forbidden subgraph and minimum degree condiions for hamiloniciy Thesis, Memphis Sae Universiy, USA, 1991 [2] Bondy, JA; Mury, USR: Graph heory wih applicaions Macmillan, London and Elsevier, New York, 1976 [3] Broersma, HJ; Veldman, HJ: Resricions on induced subgraphs ensuring hamiloniciy or pancycliciy of K 1;3 -free graphs Conemporary mehods in Graph Theory (R Bodendiek), BI-Wiss-Verl, Mannheim-Wien-Zurich, 1990,

9 [4] Brousek, J: Minimal 2-conneced non-hamilonian claw-free graphs Discree Mah (o appear) [5] Brousek, J, Favaron, O, Ryjacek, Z: Forbidden subgraphs, hamiloniciy and closure in claw-free graphs Preprin No 97, Univ of Wes Bohemia, Pilsen, 1997 (submied) [6] Brousek, J, Ryjacek, Z, Schiermeyer, I: Forbidden subgraphs, sabiliy and hamiloniciy Preprin No 98, Univ of Wes Bohemia, Pilsen, 1997 (submied) [7] Duus, D; Gould, RJ; Jacobson, MS: Forbidden subgraphs and he hamilonian heme The heory and applicaions of graphs (Kalamazoo, Mich 1980), Wiley, New York, [8] Faudree, RJ; Gould, RJ: Characerizing forbidden pairs for hamilonian properies Discree Mah 173(1997), [9] Goodman, S; Hedeniemi, S: Sucien condiions for a graph o be hamilonian J Combin Theory Ser B 16(1974) [10] Gould, RJ; Jacobson, MS: Forbidden subgraphs and hamilonian properies of graphs Discree Mah 42(1982), [11] Ryjacek, Z: On a closure concep in claw-free graphs J Combin Theory Ser B 70(1997),