Solution of a conjecture of Volkmann on the number of vertices in longest paths and cycles of strong semicomplete multipartite digraphs

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1 Solution of a conjecture of Volkmann on the number of vertices in longest paths and cycles of strong semicomplete multipartite digraphs Gregory Gutin Department of Computer Science Royal Holloway University of London Egham, Surrey, TW0 0EX, UK G.Gutin@rhul.ac.uk Anders Yeo Department of Computer Science University of Aarhus, Ny Munkegade, Building Aarhus C, Denmark yeo@daimi.au.dk Abstract A digraph obtained by replacing each edge of a complete multipartite graph by an arc or a pair of mutually opposite arcs with the same end vertices is called a semicomplete multipartite digraph. L. Volkmann conjectured that l c 1, where l (c, respectively) is the number of vertices in a longest path (longest cycle) of a strong semicomplete multipartite digraph. The bound on l is sharp. We settle this conjecture in affirmative. Running title: Solution of a conjecture of Volkmann 1 Introduction, terminology and notation L. Volkmann [6] conjectured that l c 1, where l (c, respectively) is the number of vertices in a longest path (longest cycle) of a strong semicomplete multipartite digraph. An example taken from [] shows that the bound on l is sharp. (We describe this example in the end of this paper.) We settle Volkmann s conjecture in affirmative. We will assume that the reader is familiar with the standard terminology on graphs and digraphs and refer the reader to [3]. By a cycle and a path in a directed graph we mean a directed simple cycle and path, respectively. Let D be a digraph. V (D) (A(D)) denotes 1

2 the vertex (arc) set of D. A digraph D is strong if there exists a path from x to y for every choice of distinct vertices x, y of D. A path from x to y is an (x, y)-path. The subpath of a cycle C from a vertex v to a vertex w will be denoted by C[v, w]. Analogously, C[v, w[ stands for the subpath of C from v to the predecessor of w (in the path C[v, w]). If D has an arc xy A(D), then we often use the notation x y and say that x dominates y and y is dominated by x. For disjoint sets X and Y of vertices in D, we say that X strongly dominates Y, and use the notation X Y, if there is no arc from Y to X. This means that for every pair x X, y Y of adjacent vertices x dominates y, but y does not dominate x. For a subset X of V (D), D X is the subdigraph of D induced by X. A collection B 1, B,..., B m of sets is a partition of a set B, if and only if, m i=1 B i = B and B i B j = for all 1 i < j m. Note that we allow some B i s to be empty sets. A digraph D is called semicomplete p-partite (or, just, semicomplete multipartite) if there is a partition V 1,..., V p of V (D) such that no set of the partition is empty, no arc has both end vertices in the same V i, and, for every pair x i V i, x j V j (i j) at least one of the arcs x i x j and x j x i is in D. The sets V i are the partite sets of D. Semicomplete multipartite digraphs are well studied, see, e.g., survey papers [1, 4, 5, 7]. Bound Lemma.1 Let D be a semicomplete multipartite digraph. Let non-empty sets Q 1, Q,..., Q l form a partition of V (D) such that Q i Q j for every 1 i < j l. Assume that V (D) > l and D Q i has a Hamilton path q i 1 qi...qi Q i for every i = 1,,..., l. Then, D has a (q1 1, ql Q l )-path with at least V (D) l + 1 vertices. Proof: We use the induction on l. Clearly the theorem holds when l = 1, so assume that l > 1. If V (D) Q l > (l 1) then, by the induction hypothesis, there is a (q1 1, ql 1 Q l 1 )- path, p 1 p..p k, in D Q l which contains k V (D) Q l (l 1) + 1 vertices. As {p k 1, p k } q1 l and p k 1 and p k belong to different partite sets, p k 1 q1 l or p k q1 l. Therefore, the path p 1 p...p s q1 l ql...ql Q l, where s = k 1 or k, is of the desired type. If V (D) Q l (l 1), then clearly Q l > 1. As q1 1 {ql 1, ql } and ql 1 and ql belong to different partite sets, q1 1 ql 1 or q1 1 ql. Therefore, the path q1 1 ql sqs+1 l...ql Q l, where s {1, }, is of the desired type. Theorem. Let D be a strong semicomplete multipartite digraph and let l be the number of vertices in a longest path in D and let c be the number of vertices in a longest cycle in D. Then l c 1.

3 Proof: Let P = p 1 p..p l be a path in D of maximum length and let R = V (D) V (P ). Let x 0 = p l and define S i, x i and y i recursively as follows (i = 1,,...). First let S 1 be a (p l, p k )-path in D, such that k is chosen as small as possible. Let x 1 = p k, let y 1 = p l and let S 1 = S 1 {x 1, y 1 } (note that S 1 =, by the maximality of l). Now for i =, 3, 4,... let S i be a (p t, p k )-path in D {p t, p k } R (V (S 1 ) V (S )... V (S i 1 ), such that p t V (P [x i 1, p l ]) and p k V (P [p 1, x i 1 [), and firstly k is chosen as small as possible, thereafter t is chosen as large as possible. Let also x i = p k, y i = p t and S i = S i {x i, y i }. (Some paths S i can be empty) We continue the above process until x i = p 1. Let the last value of i found above be denoted by m (i.e. x m = p 1 ). Observe that the paths S i always exist as D is strong. Observe also that y 1 = p l and that T below is a path in D: T = y 1 S 1 P [x 1, y ]S P [x, y 3 ]S 3...P [x m 1, y m ]S m x m Let U 0 = P [x m, x m 1 ] {x m, x m 1 } and let U i = P [y m i+1, x m i 1 ] {y m i+1, x m i 1 } for i = 1,,.., m 1. Note that some of the U i s (i = 0, 1,,.., m 1) can be empty. Observe that V (T ), U 0, U 1,..., U m 1 partitions the set V (P ) V (S 1 )... V (S m ). Let Z 0, Z 1,..., Z m 1 be the non-empty sets in U 0, U 1,..., U m 1, where the relative ordering has been kept (i.e. if Z i = U i, Z j = U j and i < j then i < j). Let B 0 = Z 0 Z... Z f and B 1 = Z 1 Z 3... Z g, where f (g, respectively) is the maximum even integer (odd integer, respectively) not exceeding m 1. If p l p 1, then we are done (the cycle P p 1 is of length l). Thus, we may assume that p 1 is not dominated by p l. As x m = p 1, we obtain the following: By the definitions of x i, y i and S i, p 1 Z 1 Z... Z m 1 {p l }. (1) Z i Z i+ Z i+3... Z m 1 {p l }, i = 0, 1,.., m. () As {x 0, x 1,..., x m } V (T ) and m m, we have V (T ) m + 1. (3) As V (T ), B 0, B 1 partitions the set V (P ) V (S 1 )... V (S m ), V (T ) + B 0 + B 1 l. (4) All Z i {x i } are disjoint sets containing at least two vertices. Thus, there are at most l/ such sets. Hence, we obtain 3

4 m l. (5) We only consider the case when m is odd since the case of even m can be treated similarly. If B 0 + > m +1 then, by (1), () and Lemma.1, there is a path W 0 from p 1 to p l in D {p 1, p l } B 0 containing at least B 0 + m vertices (let V (Q 1 ) = {p 1 } Z 0, V (Q ) = Z,..., V (Q (m 1)/) = Z m 3, V (Q (m +1)/) = Z m 1 {p l }). Analogously if B 1 > m 1, then there is a path W 1 from p 1 to p l in D {p 1, p l } B 1 containing at least B 1 m vertices (let V (Q 1 ) = {p 1 }, V (Q ) = Z 1,..., V (Q (m 1)/) = Z m, V (Q (m +1)/) = {p l }). We now consider the cases where none, one or two of the paths W 0 and W 1 exist. Case 1: Both W 0 and W 1 exist. The cycle C 0 = W 0 T contains V (T ) + V (W 0 ) vertices (as p 1 and p l are counted twice). The cycle C 1 = W 1 T contains V (T ) + V (W 1 ) vertices. By (3) and (4), this implies the following: V (C 0 ) + V (C 1 ) = V (T ) + V (W 0 ) + V (W 1 ) 4 V (T ) + ( B 0 + m ) + ( B 1 m 1 + 1) 4 = V (T ) + ( V (T ) + B 0 + B 1 ) m V (T ) + l m l + 1. This implies that the largest cycle of C 0 and C 1 contains at least (l + 1)/ vertices. Thus, we are done. Case : Exactly one of W 0 or W 1 exists. Let j {0, 1} be defined such that W j exists, but W 1 j does not exist. Using (4) and (5), and observing that either B 0 + m or B 1 m , we obtain the following (C j = W j T, as above): V (C j ) V (T ) + ( B 0 + m ) + ( B 1 m 1 + 1) 1 V (T ) + B 0 + B 1 m + 1 l m + 1 l l + 1 l+. This is the desired result. Case 3: Neither W 0 nor W 1 exists. This means that B 0 + m +1 0 and B 1 m 1 0. Thus, 4

5 V (T ) V (T ) + ( B 0 + m +1 ) + ( B 1 m 1 = V (T ) + B 0 + B 1 + m l m + l If p 1 p l then there is a cycle of length at least l+ + 1 (using all vertices in V (T )). By (1), p l does not dominate p 1, so we may assume that p 1 and p l are in the same partite set. We have S m = as otherwise S m P is longer than P which is impossible, hence y m and p l are in different partite sets. If p l y m then either P [y m, p l ]y m or P [p 1, y m ]p 1 is a cycle with at least l+1 vertices. Therefore, we may assume that y m p l. Then the cycle T [p l, y m ]p l contains at least l+ vertices. We are done. ) 3 Bondy s example This example originates from []. Let H be a semicomplete m-partite digraph with partite sets V 1, V,..., V m (m 3) such that V 1 = {v} and V i for every i n. Let also V v, v V j for 3 j m, and V k V i for i < k m. It is easy to verify that a longest path (cycle, respectively) in H contains m 1 (m, respectively) vertices. References [1] Bang-Jensen, J., Gutin, G.: Generalizations of tournaments: a survey. J. Graph Theory 8, (1998). [] Bondy, J.A.: Diconnected orientation and a conjecture of Las Vergnas. J. London Math. Soc. 14, 77-8 (1976). [3] Bondy, J.A., Murty, U.R.S.: Graph Theory with Applications, MacMillan Press (1976). [4] Gutin, G.: Cycles and paths in semicomplete multipartite digraphs, theorems and algorithms: a survey. J. Graph Theory 19, (1995). [5] Reid, K.B.: Tournaments: scores, kings, generalizations and special topics. Congressus Nummerantium 115, (1996). [6] Volkmann, L.: Spanning multipartite tournaments of semicomplete multipartite digraphs. Manuscript, June, [7] Volkmann, L.: Cycles in multipartite tournaments: results and problems. Submitted. 5