WHEN DO THREE LONGEST PATHS HAVE A COMMON VERTEX?

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1 WHEN DO THREE LONGEST PATHS HAVE A COMMON VERTEX? MARIA AXENOVICH Abstract. It is well known that any two longest paths in a connected graph share a vertex. It is also known that there are connected graphs where 7 longest paths do not share a common vertex. It was conjectured that any three longest paths in a connected graph have a vertex in common. In this note we prove the conjecture or outerplanar graphs and give suicient conditions or the conjecture to hold in general.. Introduction Gallai asked in 966 whether every connected graph has a vertex that appears in all its longest paths. Zamirescu [6] ound a graph with vertices in which there is no common vertex to all longest paths. See Voss, [5] or related problems. It is well-known that every two longest paths in a connected graph have a common vertex. Skupien [4] obtained, or k 7, a connected graph in which some k longest paths have no common vertex, but every k longest paths have a common vertex. Klavzar et al. [3] showed that in a connected graph G, all longest paths have a vertex in common i and only i or every block B o G all longest paths in G which use at least one edge o B have a vertex in common. Thus, i every block o a graph G is Hamilton-connected, almost Hamilton-connected, or a cycle, then all longest paths in G have a vertex in common. It was also proved in [3] that in a split graph all longest paths intersect. Balister et al. [] showed that all longest path o a circular arc graphs share a common vertex. Still, the ollowing conjecture remains open in general: Conjecture For any three longest paths in a connected graph, there is a vertex which belongs to all three o them. In this note we prove the conjecture or special classes o graphs, more speciicaly or triples o longest paths whose union belongs to special classes o graphs. One o our main results is the ollowing. Theorem. Let G be a connected graph and P 0, P, P be its longest paths. I P 0 P P orms an outerplanar graph then there is a vertex v V (P 0 ) V (P ) V (P ). Note that using Kuratowski theorem and the above result, we see that i the union o three longest paths in a connected graph does not contain K 4 or K,3 as a topological minor, or as a minor then there is a vertex common to all three o these paths.. Deinitions and Results We say that a amily F, o graphs is monotone i or any G F, and any e E(G), G e F. For a path P = x 0, x,..., x k, we say that path P = x i, x i+,..., x j, j > i, is a segment o P; we say that P is an end-segment o P i i = 0 or j = k. The addition o indeces will be taken modulo 3. For a path P, P denotes its lengths. The ollowing deinition describes two conigurations orbidden in the union o three longest path with no common vertex, see Figure. Keywords: longest paths, intersection.

2 MARIA AXENOVICH S 0 x S S Q Q Deinition. Figure. Conigurations impossible in the union o three longest paths o a connected graph. Coniguration Q is a cycle which is the union o internally disjoint segments S 0, S, S o P 0, P, P, such that a) interior o S 0 and interior o S do not contain any vertices o P and b) interior o S and interior o S do not contain any vertices o P 0. Coniguration Q is deined by a segment P o P 0 with end-points x, y, such that a) x V (P ), y V (P ), b) internal vertices o P belong to P 0 only, and c) P {x} is the union o two paths P, P ; P {y} is the union o two paths P, P ; such that V (P P ) V (P P ) = or V (P P ) V (P P ) =. Lemma. Let G be the union o its longest paths P 0, P, P. Then G does not contain either coniguration Q or Q. Proo. Assume that G contains coniguration Q. Let the length o S i be l i, i = 0,,. Replacing S 0 with S S in path P 0 requires that l 0 l + l. Similarly, replacing S with S 0 S gives l l 0 + l. Adding these inequalitied gives l 0, a contradition to the act that S is a segment o positive length. Assume that G has a coniguration Q, let c = P. Let y split P into segments o lengths l and l. Let x split P into segments o lengths l and l, respectively, so l = l +l = l +l = P i, i = 0,,. Then, without loss o generality, we have paths o length l +c+l and l +c+l. So, c+l +l l +l, c+l +l l +l. By adding these last two inequalities, we get that c = 0, a contradiction. Lemma. For a monotone amily F, let G F be a connected graph with smallest V (G) + E(G) having three longest paths with no common vertex. Then G has exactly one nontrivial block. Proo. Assume that G has at least two nontrivial blocks and there are three longest paths P 0, P, P which do not have a vertex in common. By minimality we have then that G = P 0 P P since otherwise we could delete edges not in P 0 P P. I there are two nontrivial blocks each containing the vertices o all three paths, then clearly there is a cutvertex contained in all these paths. Thus there is a nontrivial block, say B, containing vertices o only two paths, say o P 0 and P. Thus P is in a component, X, o G B. Since P

3 WHEN DO THREE LONGEST PATHS HAVE A COMMON VERTEX? 3 intersects P 0 and P, X must contain vertices o P 0 and P, in particular, X contains one endpoints o P 0 and one endpoint o P. Thus G has a cut-vertex, u V (B), such that G {u} has disjoint graphs with vertex sets X, X, where X = X {u} has vertices o P 0 and P only, and X = X {u} has vertices rom all three paths, satisying an additional property that one endpoint o P j is in X and another endpoint o P j in in X, j = 0,. Let l = P 0 [X ] and l = P [X ]. Then l = l, otherwise, i say l > l, then P [X ] P 0 [V \X ] is a path o length greater than the maximum length, a contradiction. Thus one can replace G[X ] with a single path P 0 [X ]. The resulting graph has less edges and it has three longest paths P 0, P and P = P [V X ] P 0 [X ]. Moreover, P 0, P, P do not share a common vertex. In addition, i G is rom a monotone amily F then the resulting graph is rom F too. A contradiction to minimality o G. I G has no nontrivial blocks, i.e., G is a tree, then the result ollows instantly since each longest path must pass through the center. Lemma 3. Let G be a connected graph which is the union o its longest paths, P 0, P, P. I P P has at most one cycle then there is a vertex common to all three paths P 0, P, P. Proo. The path P 0 intersect both P and P. Thus, there is a segment o P 0 with one endpoint, x, in P and another endpoint, y, in P such that all other vertices on this segment belong only to P 0. I P P is acyclic then we have orbidden coniguration Q. So we assume that P P has a unique cycle C. I x, y V (C), or x, y V (C) then we have a orbidden coniguration Q. I x V (C), y V (C), we have a orbidden coniguration Q. Now, we are ready to prove the main theorem. Proo o Theorem. Throughout the proo we shall say that an edge e is in the ace F, or that F has an edge e, i e is an edge on the boundary o F. Assume that there is a graph with three longest paths orming an outerplanar graph and such that these paths do not have a vertex in common. Let G be the union o these paths, P 0, P, P and assume urther that G is minimum such graph. By Lemma, G has exactly one nontrivial block, B. Consider the outer-planar embedding o G. Let C be the cycle o the unbounded ace. We reer to the edges o G which are not in C but have endpoints in C as chords. Clearly, the vertices o each chord orm a cutset o G, and any two bounded aces are separated by some chord. Each chord e splits G into two new outerplanar graphs, G(e), G(e), such that G(e) G(e) = G and G(e) G(e) = e. Observe also that or each i {0,, } there is a chord which belongs to P i, otherwise, i, or example, there is no chord rom P 0, we have that P P has at most one cycle, a contradiction to Lemma 3. We call a ace using edges o all three paths P 0, P, P three-colored. Claim There is a bounded three-colored ace. Assume that there is no bounded three-colored ace. We have rom above observation that or each i {0,, } there is a bounded ace containing edges rom P i. Then, there are two bounded aces which share a chord e such that, without loss o generality, one o these aces has edges rom P 0 P and another has edges rom P P. Then e = {x, y} E(P ). Moreover, P G(e) and P 0 G(e). Since P and P 0 intersect, we have that V (P 0 ) V (P ) {x, y}, but x, y V (P ), so there is a vertex which belongs to all three paths, a contradiction. Claim For every chord e, either G(e) or G(e) has edges rom at most two paths, P i, P j, i {0,, }. Assume that e = {x, y} E(P 0 ), a chord such that G(e) and G(e) both have edges o all three paths P 0, P, P. We have then that P and P must pass through {x, y}. Because no vertex belongs to all three paths, we have that x V (P ) and y V (P ), which gives a orbidden coniguration Q.

4 4 MARIA AXENOVICH F Figure. Three-colored ace F with chords o types {0, }, {, }, {, 0} on its boundary. This implies that there is exactly one bounded three-colored ace, call it F. Let F have chords e 0, e,...,e k, k 3 on its boundary, in order. Then let G(e i ), i = 0,...,k not contain F. Then, G(e i ) uses vertices o only two paths, P i, P i, i, i {0,, }, by Claim. We say that e i has type {i, i }, see Figure. Claim 3 For each i, i {0,, } there is an e i o type {i, i } or soem i {0,,..., k }. Assume that there is no e i o type {0, }. Since there are chords rom all three paths, we have that there are e j, e l, j, l {0,,..., k } such that e j is o type {, } and e l is o type {0, }. Let, without loss o generality l =, j =. Thus G (e ) P 0 P and G (e ) P P. Then there is an edge ẽ on G(e ) G(e ) path such that ẽ E(P ) \ E(P 0 P ). Thus C P 0 P. Thereore P 0 P is acyclic, a contradiction to Lemma 3. Claim 4 The chords e i s o the same type appear consecutively along C. Assume that e i, e i3 are o type {0, }, e i is or type {0, }, and e i4 is o type {, }, or some i < i < i 3 < i 4, in cyclic order on C. Let vertex x be incident to e i and vertex y be incident to e i3. We have that x, y V (P 0 ) V (P ), moreover, {x, y} is a cutset o G, and G (e i ) and G (e i4 ) are in the dierent components o G {x, y}. Since G (e i ) and G (e i4 ) contain edges o P, it implies that P is disconnected, a contradiction. To conclude the proo, let e 0, e,...,e S be o type {0, }, e s+,...,e t be o type {, } and e t+,..., e k be o type {0, }. There is an edge E(C) between G(e s ) and G(e s+ ), so that E(P )\E(P 0 P ). Similarly, there is E(C) E(P ) \ E(P 0 P ), 0 E(C) E(P 0 ) \ E(P P ), see Figure. Consider the longest segment o P o P 0 containing 0 such that all its vertices except or the endvertices are not in V (P ) V (P ), one endvertex is in V (P ) and another endvertex is in V (P ). Then this segment deines coniguration Q, a contradiction. Acknowledgments The author would like to thank Douglas West or a nice open problem web-page stating the problem considered in this note, Richard McBride and Xiaoyang Gu or useul discussions. Reerences [] Balister, P., Györi, E., Lehel, J., Schelp, R., Longest paths in circular arc graphs. Combin. Probab. Comput. 3 (004), no. 3, [] Havet, F. Stable set meeting every longest path. Discrete Math. 89 (004), no. -3, [3] Klavžar, S., Petkovšek, M., Graphs with nonempty intersection o longest paths. Ars Combin. 9 (990), 43 5.

5 WHEN DO THREE LONGEST PATHS HAVE A COMMON VERTEX? 5 [4] Skupień, Z., Smallest sets o longest paths with empty intersection. Combin. Probab. Comput. 5 (996), no. 4, [5] Voss, H.-J. Cycles and bridges in graphs, Mathematics and its Applications (East European Series), 49. Kluwer Academic Publishers Group, Dordrecht; VEB Deutscher Verlag der Wissenschaten, Berlin, 99. [6] Zamirescu, T., Intersecting longest paths or cycles: a short survey. An. Univ. Craiova Ser. Mat. Inorm. 8 (00), 9. 4 Carver Hall, Iowa State University, Ames, IA 500, axenovic@iastate.edu