Diregularc-partite tournaments are vertex-pancyclic whenc ? 5

Diregular c-PartiteTournaments areVertex-PancyclicWhen c ≥ 5

Anders YeoDEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE

ODENSE UNIVERSITY, DK-5230ODENSE, DENMARK

E-mail: [email protected]

Received June 20, 1998; revised January 17, 1999

Abstract: In [Volkmann, to appear] it is conjectured that all diregular c-partitetournaments, with c ≥ 4, are pancyclic. In this article, we show that all diregularc-partite tournaments, with c ≥ 5, are in fact vertex-pancyclic. c© 1999 John Wiley &

Sons, Inc. J Graph Theory 32: 137–152, 1999

Keywords: multipartite tournaments, regular, vertex-pancyclic, pancyclic

1. INTRODUCTION

A semicomplete c-partite digraph is a digraph obtained from a complete c-partitegraph by substituting each edge with an arc, or a pair of mutually opposite arcs withthe same end vertices. A semicomplete multipartite digraph (SMD) is a semicom-plete c-partite digraph with c ≥ 2. Special cases of SMDs are semicomplete bipartitedigraphs (c = 2) and semicomplete digraphs (c = n, the number of vertices). Ac-partite tournament is a semicomplete c-partite digraph with no cycles of length2, and, analogously, a multipartite tournament (MT) is an SMD with no cycles oflength 2. A cycle subgraph in a digraph is a collection of vertex disjoint cycles. Afactor in a digraph is a spanning cycle subgraph, and a minimal factor is a factor con-sisting of the minimum number of cycles (e.g., a Hamilton cycle is a minimal factor).

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/020137-16

138 JOURNAL OF GRAPH THEORY

The local irregularity of a digraph D is defined as il(D) = max |d+(x)− d−(x)|over all vertices x ∈ V (D). A digraph, D, is pancyclic, if it contains cycles oflengths 3, 4, . . . , |V (D)|. A digraph,D, is vertex-pancyclic, if for everyw ∈ V (D)it contains cycles of lengths 3, 4, . . . , |V (D)|, which all include the vertex w. Adigraph D is diregular, if there is a constant k such that d+(v) = d−(v) = k forevery v ∈ V (D). Note that, if D is a diregular MT, then all partition classes (alsocalled the color classes) must have the same size, which we denote by v∗D. IfX andY are subsets of vertices in a digraph, D, then X⇒Y denotes the fact that thereare no arcs from Y to X . An (X,Y )-path is a directed path from a vertex inX to avertex in Y , such that the first and last vertex on the path are the only vertices fromX ∪ Y . We write (x, y)-path instead of ({x}, {y})-path. A digraph, D, is strongif, for each pair of vertices x and y, there is a (x, y)-path (and a (y, x)-path) inD. A digraph, D, is k-strong, if D −X is strong for every set of vertices X , with|X| < k.

In [4], it is conjectured that all diregular c-partite tournaments, with c ≥ 4, arepancyclic. In this article, we prove that all diregular c-partite tournaments, withc ≥ 5, are in fact vertex-pancyclic. In order to prove this, we first show a fewgeneral results for MTs, which are of interest in themselves. We prove that any

diregular MT with v∗D vertices in each color class is d |V (D)|−v∗D3 e-strong. We show

that, if there is a cycle subgraph, F , in a diregular MT, D, with w ∈ V (F ), thenthere is a cycle inDwith the same number of vertices asF , and which also includesthe vertex w. Bondy (see [1]) showed that there is a cycle of length 3, 4, . . . , c inevery strong c-partite tournament. We extend this result by showing that there isa pancyclic subgraph of order 3, 4, . . . , c in every strong c-partite tournament. Inorder to prove our main result, we also use a result on minimal factors (see [5]) anda result on local irregularity (see [6]).

We can also show that all diregular 4-partite tournaments with more than N0vertices (for some constant N0) are vertex-pancyclic. As the proof of this result isquite long and uses a different, probabilistic approach, we have decided to have aseparate paper [7] devoted to the 4-partite case.

Clearly our main result, together with the result from [7], implies that the con-jecture from [4] is true, except possibly for a finite number of exceptions. Fur-thermore the conjecture from [4] cannot be extended to 3-partite tournaments, asthere are diregular 3-partite tournaments, which are not pancyclic (e.g., if V (D) ={a1, a2, b1, b2, c1, c2} and A(D) = {aibj , bicj , ciaj |i = 1, 2 j = 1, 2}).

The pancyclicity and vertex-pancyclicity problems for SMDs seem to be diffi-cult problems in general. Therefore, there are relatively few articles on this topic.However, for additional results on SMDs, see [3].

2. TERMINOLOGY AND NOTATION

We assume that the reader is familiar with the standard terminology on graphs anddigraphs, and refer the reader to [2].

DIREGULAR c-PARTITE TOURNAMENTS 139

A digraph D = (V (D), A(D)) is determined by its set of vertices, V (D), andits set of arcs, A(D). An arc in D is an ordered pair of vertices. If xy ∈ A(D)(x, y ∈ V (D)), then we say that there is an arc fromx to y inD, and thatx dominatesy and y is dominated by x. We also denote this by x→ y. If X,Y ⊆ V (D) andthere is no arc from Y to X , then we say that X⇒Y (called X strongly dominatesY ). For a vertex x ∈ V (D) the in-degree, d−D(x) (out-degree, d+

D(x)) of x is thenumber of vertices dominating x (dominated by x) in D. We sometimes omit thesubscript, if it is clear from the context which digraph we are considering. By acycle (path) we mean a simple directed cycle (path, respectively). A Hamilton cycleof D is a cycle that contains all vertices of D. A k-cycle is a cycle of length k (i.e.,it include k arcs). IfC is a cycle including the vertices u and v, thenC[u, v] denotesthe path starting at u and ending at v and using the relevant arcs from C. If Q isa subgraph in a digraph D, then V (Q) denotes the vertices in Q and D 〈Q〉 is thesubgraph induced by V (Q).

Multipartite tournaments (MT), semicomplete multipartite digraphs (SMD), anddiregular digraphs are all defined in the Introduction. Whenever we consider anSMD (or MT),D, we use the term ‘color classes’ to denote the uniquely determinedpartition classes of V (D). For each x ∈ V (D), V c(x) denotes the color class thatx belongs to. Factors, minimal factors, pancyclicity, vertex-pancyclicity, and localirregularity (il(D)) are also defined in the Introduction.

Let D be a digraph and let X,Y ⊆ V (D). We use the following notationE(X,Y ) = |{(x, y) ∈ X × Y : xy ∈ A(D) or yx ∈ A(D)}| and d(X,Y ) =|{(x, y) ∈ X × Y : xy ∈ A(D)}|. Clearly, d(X,Y ) + d(Y,X) ≥ E(X,Y ).A collection of subsets, X1, X2, . . . , Xl, is a partition of a set X if and only if∪li=1Xi = X and Xi ∩Xj = ∅ for all i 6= j.

Let D be any digraph and let {x, y} ⊆ V (D) be arbitrary. We now define thefunctions κv and κ in the following way. Let κv(x,D) = d+

D(x) − d−D(x) andlet κ(x, y,D) = κv(x,D) − κv(y,D). Note that il(D) = max{|κv(x,D)| : x∈ V (D)}.

Let D be a digraph and let C be a cycle in D. If x ∈ V (C), then x+ denotes thesuccessor of x on the cycle C. Analogously, x− denotes the predecessor of x onC.We define x+ = x(+1) and x(+δ) = (x+(δ−1))+ for δ ≥ 2 (e.g., x+++ = x(+3)).

Let δ be an integer in {1, 2, . . . , |V (C)|}, and let {x, y} ⊆ V (D) − V (C) bearbitrary. A δ-partner of (x, y) on C is a vertex z ∈ V (C) such that z→x andy→z(+δ).

Let D be a digraph, and let k be some integer. A cycle C0 is k-reducible if thereare cycles C1, C2, . . . , Ck such that for every i = 0, 1, . . . , k − 1 there is a vertexwi in Ci, such that Ci+1 = Ci[w+

i , w−i ]w+

i (in each step w−i →w+i ). If w ∈ V (D),

then a cycle C0 is (w, k)-reducible if it is k-reducible, and w belongs to all thecycles C0, C1, . . . , Ck (i.e., wi 6= w for all i = 0, 1, . . . , k − 1).


3. GENERAL RESULTS ON SMDS AND MTS

We recall the following theorem from [6].

Theorem 3.1. [6] Let D be a SMD with color classes V1, V2, . . . , Vc. For all∅ ⊂ X ⊂ V (D) we have the following.

E(X,V (D)−X)|X| +

E(X,V (D)−X)|V (D)−X| ≥ |V (D)| −max{|Vi| : i = 1, 2, . . . , c}

(1)

Furthermore, if equality holds above, then |V1| = |V2| = . . . = |Vc|.The above theorem is formulated in [6] without the statement that if equality

holds, then |V1| = |V2| = . . . = |Vc|. However, this is easily seen from the proofof the theorem.

We can use Theorem 3.1 to show the following lemma.

Lemma 3.1. Let D be a c-partite tournament with color classes V1, V2, . . . , Vcand letS ⊆ V (D) such thatD′ = D−S is not strong. IfQ1 andQ2 are a partition ofV (D′) such thatQ1⇒Q2, then the following holds, where v′ = max{|Vi∩V (D′)| :i = 1, 2, . . . , c}.

il(D) ≥ (|V (D′)| − v′)d|V (D′)|/2e|V (D′)| − |S| ≥ |V (D′)| − v′

2− |S| (2)

Furthermore, if min{|Vi ∩ V (D′)| : i = 1, 2, . . . , c} < v′, then il(D) is strictlygreater than the values above.

Proof. By Theorem 3.1, we obtain the following equation:

E(Q1, Q2)|Q1| +

E(Q1, Q2)|Q2| ≥ |V (D′)| − v′ (3)

This implies that |V (D′)|E(Q1, Q2)/(|Q1||Q2|) ≥ (|V (D′)| − v′). Assumewithout loss of generality that |Q1| ≤ |Q2|. This implies that |Q2| ≥ d|V (D′)|/2e.Let Z = (|V (D′)| − v′)d|V (D′)|/2e/|V (D)′| and observe that E(Q1, Q2)/|Q1| ≥ Z. We observe that

∑x∈Q1

κv(x,D′) =∑x∈Q1

(d(x,Q1) − d(Q1, x))+∑x∈Q1

d(x,Q2) = 0 + E(Q1, Q2), which implies that there must be somex ∈ Q1 with κv(x,D′) ≥ E(Q1, Q2)/|Q1|. Therefore, we observe that il(D) ≥κv(x,D) ≥ κv(x,D′) − |S| ≥ Z − |S| ≥ |V (D′)|−v′

2 − |S|, which proves thefirst part.

Part two is proved analogously by observing that by Theorem 3.1 we haveE(Q1, Q2)/|Q1|+ E(Q1, Q2)/|Q2| > |V (D′)| − v′ in this case.

Corollary 3.1. IfD is a diregular MT with color classes V1, V2, . . . , Vc, thenD

is d |V (D)|−v∗D3 e-strong.


Proof. Let S be a minimum separating set inD. By Lemma 3.1, we obtain that0 = il(D) ≥ (|V (D)|−|S|v∗D)/2−|S|, which implies that 3|S| ≥ (|V (D)|−v∗D).

In [1], Bondy proved the following theorem.

Theorem 3.2. [1] IfD is a strong c-partite tournament, then there is a p-cycle inD for all p = 3, 4, . . . , c.

We generalize this in the following lemma and corollary.

Lemma 3.2. If D is a strong c-partite tournament (c ≥ 3), then there is a(k − 3)-reducible k-cycle in D, for every k = 3, 4, . . . , c.

Proof. We prove this by induction over k. When k = 3, there is a 3-cycle inD by Theorem 3.2. Assume that 4 ≤ k ≤ c and that there is (k − 4)-reducible(k−1)-cycle,Ck−1, inD. We will now construct a (k−3)-reducible k-cycle inD.

Let A = {x ∈ V (D) : V c(x) ∩ V (Ck−1) = ∅} and note that A 6= ∅, sincek − 1 < c. If there is a vertex a ∈ A with a6⇒V (Ck−1) and V (Ck−1)6⇒a, then,since a is adjacent with every vertex in V (C), there exists a vertex z ∈ V (C) suchthatC[z, z−]az is a k-cycle, which is (k−3)-reducible. Therefore, we assume thatfor every a ∈ A either a⇒V (Ck−1) or V (Ck−1)⇒a.

Let A1 = {a ∈ A : a⇒V (Ck−1)}, and let A2 = {a ∈ A : V (Ck−1)⇒a}.Let l be the length of a shortest path of all (Ck−1, A1)-paths and (A2, Ck−1)-paths. Without loss of generality, assume that P = p0p1 . . . pl is a (Ck−1, A1)-pathof length l (as (A2, Ck−1)-paths can be handled analogously). Assume that pl ∈V c(p1). Then l ≥ 3, as p1 ∈ A2. Let x ∈ V (Ck−1)−V c(p2) be arbitrary. If x→p2,then the path xP [p2, pl] is a contradiction against the minimality of l. If p2→x, thenthe pathp1p2x is a contradiction against the minimality of l. Therefore,pl 6∈ V c(p1).Furthermore, from the minimality of l, we obtain thatV c(pl)∩{p1, p2, . . . , pl−1} =∅ andpl→{p0, p1, . . . , pl−2}. LetCk = PCk−1[p(+l)

0 , p−0 ]. Aspl→V (Ck)−{pl−1},we conclude that Ck is a (k − 3)-reducible k-cycle.

The following corollary now follows immediately.

Corollary 3.2. If D is a strong c-partite tournament (c ≥ 3), then there is apancyclic subgraph of order k, for all k = 3, 4, . . . , c.

4. PRELIMINARY RESULTS

We recall the following two lemmas from [6].

Lemma 4.1. [6] Let D be a digraph with no factor. Then we can partition V (D)into subsets Y , Z, R1, R2 such that R1⇒Y , (R1 ∪ Y )⇒R2, |Y | > |Z| and Y is aset of independent vertices.


Lemma 4.2. [6] Let D be a SMD with color classes V1, V2, . . . , Vc, such that|V1| ≤ |V2| ≤ . . . ≤ |Vc| ≤ |V1|+ 1. If il(D) ≤ (|V (D)| − |Vc1| − 2|Vc|+ 2)/2,then D is Hamiltonian.

We now improve Theorem 5.4 from [6], by showing the following lemma.

Lemma 4.3. Let D be a diregular c-partite tournament and let w ∈ V (D) bearbitrary. There exists a cycle C of length p in D with w ∈ V (C), for all integersp such that |V (D)|2c−2

3c−5 + 2c3c−5 ≤ p ≤ |V (D)|.

Proof. Let n = |V (D)| and let D have color classes V1, V2, . . . , Vc. Assumewithout loss of generality thatw ∈ V1. Let p be an integer such that n(2c−2)/(3c−5) + 2c/(3c − 5) ≤ p ≤ n. Let k = bp/cc and r = p mod c, which implies thatp = kc+ r and 0 ≤ r < c. Let V ′i ⊆ Vi, such that |V ′i | = k + 1 for i = 1, 2, . . . , rand |V ′i | = k for i = r+ 1, r+ 2, . . . , c, and such that w ∈ V ′1 . The following nowholds:

n2c−23c−5 + 2c

3c−5 ≤ p

⇓ Subtract p2c−23c−5 from both sides

(n− p)2c−23c−5 + 2c

3c−5 ≤ p(1− 2c−2

3c−5

)⇓ Multiply both sides with 3c−5

2c(n− p) c−1

c + 1 ≤ p c−32c⇓ As p = kc+ r

n− p− n−pc + 1 ≤ p−3k

2 − 3r2c

⇓ As n−pc − 1 < bn−pc c and 3r

2c ≥ 0n− p− bn−pc c < p−3k

2⇓ As n, p, k and bn−pc c are integers

n− p− bn−pc c ≤ p−3k−12

Let D′ = D 〈∪ci=1V′i 〉 and note that |V (D′)| = p and that we have deleted

at least b(n − p)/cc vertices from each color class. This implies that il(D′) ≤n − p − b(n − p)/cc ≤ (p − 3(k + 1) + 2)/2, since D is diregular and we havedeleted at most n− p− b(n− p)/cc neighbors of any vertex. By Lemma 4.2, D′has a Hamilton cycle, which corresponds to a p-cycle C in D with w ∈ V (C).

Lemma 4.4. LetD be a diregular c-partite tournament, with c ≥ 4. For all vertexsets X ⊆ V (D) and Y ⊆ V (D), there is a path of length at most 3 from X to Yin D.

Proof. Let x ∈ X and y ∈ Y be arbitrary. Clearly, we are done, if we showthat there is an (x, y)-path of length at most 3 in D. Assume that this is not thecase, for the sake of contradiction. Note that N−(y) ∪ {y}⇒N+(x) ∪ {x} and(N−(y) ∪ {y}) ∩ (N+(x) ∪ {x}) = ∅, since there is no (x, y)-path of length atmost 3 inD. This implies thatS = V (D)−N+(x)−N−(y)−{x, y} is a separatingset in D. Since D is diregular, we have |N+(x)| = |N−(y)| = (c− 1)v∗D/2, and,thus, |S| = v∗D − 2 < (cv∗D − v∗D)/3 when c ≥ 4. This contradicts Corollary 3.1.


In [5], we prove the following theorem.

Theorem 4.1. [5] Let D be a MT and let F = C1 ∪ C2 ∪ . . . ∪ Cl be a factorin D with the minimum number of cycles. Then there exists a color class χ and anordering of the cycles C1, C2, . . . , Cl in F , such that the following holds. For allarcs xy with x ∈ V (Cj), y ∈ V (C1) and j > 1, we have {x+, y−} ⊆ χ.

We can now prove the following lemma.

Lemma 4.5. Let D be a diregular c-partite tournament, with c ≥ 4, and letw ∈ V (D) be arbitrary. If F = C1 ∪ C2 ∪ . . . ∪ Cl is a cycle subgraph in D withw ∈ V (F ), then there is a cycle C in D with |V (C)| = |V (F )| and w ∈ V (C).

Proof. Let F ′ = C ′1 ∪ C ′2 ∪ . . . ∪ C ′m be a cycle subgraph, with w ∈ V (F ′)and |V (F ′)| = V (F )|, and assume that m is as small as possible. If m = 1, weare done, so assume that m ≥ 2. By looking at D′ = D 〈V (F ′)〉, there exists acolor class χ such that the conditions of Theorem 4.1 hold. This implies that allarcs xy with x ∈ V (C ′j), y ∈ V (C ′1) and j > 1 have {x+, y−} ⊆ χ. Assumewithout loss of generality that w ∈ V (F ′) − V (C ′1), as we otherwise reverse allarcs (this gives us a different cycle called C ′1). LetR = C ′2 ∪C ′3 ∪ . . .∪C ′m and letP = p1p2 . . . pk be a shortest possible path fromR toC ′1. Assume that p1 ∈ V (C ′j),where j ∈ {2, 3, . . . ,m}. By Lemma 4.4, we have that 2 ≤ k ≤ 4. We now showclaim (i) and cases 1–3 below.

Claim (i). If z ∈ V (C ′1), v ∈ V (R), z⇒V (R), V (C ′1)⇒v and V c(z) = V c(v),then there exist two distinct vertices u1 and u2 such that v→ui→z and ui ∈ V (D−F ) for i = 1, 2.

Let χ1 = V c(z) = V c(v) and note that d+D′(z)− d−D′(z) + d−D′(v)− d+

D′(v) ≥(|V (R)−χ1|+1)−(|V (C ′1)−χ1|−1)+(|V (C ′1)−χ1|+1)−(|V (R)−χ1|−1) = 4.Now define A2 = {a ∈ V (D) − V (F ′) : v→a→z}, A−2 = {a ∈ V (D) −V (F ′) : z→a→v} and A0 = V (D) − V (F ′) − A2 − A−2. This implies that0 = κ(z, v,D) = κ(z, v,D′) + 2|A−2|+ 0|A0| − 2|A2| ≥ 4− 2|A2|. Therefore,|A2| ≥ 2, which implies the desired result.

Case 1. k = 2. By Theorem 4.1, we have {p+1 , p

−2 } ⊆ χ and p−2⇒V (R) and

V (C ′1)⇒p+1 . By Claim (i) there is a vertex z ∈ V (D)−V (F ′) such thatp+

1→z→p−2 .If p−−−2 ∈ χ, then the cycle subgraph F ′′ = C ′1[p−2 , p

−−−2 ]C ′j [p

++1 , p+

1 ]zp−2 ∪(F ′ − C ′1 − C ′j) has |V (F ′′)| = |V (F ′)| = |V (F )|, and as p−−2 6= w, we havew ∈ V (F ′′). Therefore, F ′′ is a contradiction against the minimality of m.

Ifp−−−2 6∈ χ, then the cycle subgraphF ′′ = C ′1[p2, p−−−2 ]p+

1 zp−2 C′j [p

++1 , p1]p2∪

(F ′ − C ′1 − C ′j) has |V (F ′′) = |V (F ′)| = |V (F )|, and as p−−2 6= w, we havew ∈ V (F ′′). Therefore, F ′′ is a contradiction against the minimality of m.

Case 2. k = 3. We note that by the minimality of kwe must haveV (C ′1)⇒V (R).We look at the following subcases, which each construct a cycle subgraph F ′′ with|V (F ′′)| = |V (F ′)| = |V (F )| and w ∈ V (F ′′), which contradicts the minimalityof m.


Subcase 2.a. p++1 = w and V c(p++

1 ) 6= V c(p−3 ). The cycle subgraph F ′′ =C ′1[p3, p

−3 ]C ′j [p

++1 , p1]p2p3∪(F ′−C ′1−C ′j) hasw ∈ V (F ′′), since p+

1 6= w.

Subcase 2.b. p++1 = w, V c(p++

1 ) = V c(p−3 ) and V c(p+++1 ) 6= V c(p−−−3 ). By

Claim (i) there is a vertex z ∈ V (D) − V (F ′) such that p++1 →z→p−3 . The

cycle subgraph F ′′ = C ′1[p−3 , p−−−3 ]C ′j [p

+++1 , p++

1 ]zp−3 ∪ (F ′ − C ′1 − C ′j)has w ∈ V (F ′′), since p−−3 6= w (w 6∈ V (C ′1)).

Subcase 2.c. p++1 = w, V c(p++

1 ) = V c(p−3 ) and V c(p+++1 ) = V c(p−−−3 ). By

Claim (i) there is a vertex z ∈ V (D)−V (F ′)−{p2} such that p++1 →z→p−3 .

The cycle subgraph F ′′ = C ′1[p3, p−−−3 ]p++

1 zp−3 C′j [p

+++1 , p1]p2p3 ∪ (F ′ −

C ′1 − C ′j) has w ∈ V (F ′′), since w 6∈ {p−−3 , p+1 }.

Subcase 2.d. p++1 6= w and V c(p+

1 ) 6= V c(p−−3 ). The cycle subgraphF ′′ = C ′1[p3, p

−−3 ]C ′j [p

+1 , p1]p2p3 ∪ (F ′ − C ′1 − C ′j) has w ∈ V (F ′′), since

p−3 6= w.Subcase 2.e. p++

1 6= w, V c(p+1 ) = V c(p−−3 ) and V c(p+++

1 ) 6= V c(p−−−3 ). ByClaim (i) there is a vertex z ∈ V (D) − V (F ′) such that p+

1→z→p−−3 . Thecycle subgraph F ′′ = C ′1[p−−3 , p−−−3 ]C ′j [p

+++1 , p+

1 ]zp−−3 ∪ (F ′ −C ′1 −C ′j)has w ∈ V (F ′′), since p++

1 6= w.Subcase 2.f. p++

1 6= w, V c(p+1 ) = V c(p−−3 ) and V c(p+++

1 ) = V c(p−−−3 ). ByClaim (i) there is a vertex z ∈ V (D)−V (F ′)−{p2} such that p+

1→z→p−−3 .The cycle subgraph F ′′ = C ′1[p3, p

−−−3 ]p+

1 zp−−3 C ′j [p

+++1 , p1]p2p3 ∪ (F ′ −

C ′1 − C ′j) has w ∈ V (F ′′), since w 6∈ {p−3 , p++1 }.

Case 3.k = 4. We note that, by the minimality ofk, we must haveV (C ′1)⇒V (R).Furthermore, if V c(p−−−4 ) = V c(p+

1 ), then Claim (i) gives us a contradictionagainst the minimality of k, so we must have V c(p−−−4 ) 6= V c(p+

1 ). Now the cyclesubgraph F ′′ = C ′1[p4, p

−−−4 ]C ′j [p

+1 , p1]p2p3p4 ∪ (F ′ − C ′1 − C ′j) has |V (F ′′)| =

|V (F )| and w ∈ V (F ′′) as w 6∈ V (C ′1). This is a contradiction against the mini-mality of m.

Lemma 4.6. Let D be a diregular c-partite tournament and let δ ∈ {1, 2, 3} bearbitrary. Let F be a cycle subgraph in D, let D′ = D − V (F ), and let {x, y} ⊆V (D′) be arbitrary. Then there are at least (κ(x, y,D′)−|V c(x)∩V (F )|−|V c(y)∩V (F )|)/2 distinct δ-partners of (x, y) in F .

Proof. We define the following nine sets, which are a partition of V (F ):

A1 = {z ∈ V (F ) : z→x & y→z(+δ)} A5 = {z ∈ V (F ) : z ∈ V c(x) & y→z(+δ)}A2 = {z ∈ V (F ) : z→x & z(+δ)→y} A6 = {z ∈ V (F ) : z ∈ V c(x) & z(+δ)→y}A3 = {z ∈ V (F ) : x→z & y→z(+δ)} A7 = {z ∈ V (F ) : z→x & z(+δ) ∈ V c(y)}A4 = {z ∈ V (F ) : x→z & z(+δ)→y} A8 = {z ∈ V (F ) : x→z & z(+δ) ∈ V c(y)}

A9 = {z ∈ V (F ) : z ∈ V c(x) & y(+δ) ∈ V c(y)}.

Note that κv(x,D) = κv(x,D′) + |A3| + |A4| + |A8| − |A1| − |A2| − |A7|and κv(y,D) = κv(y,D′) + |A1| + |A3| + |A5| − |A2| − |A4| − |A6|. Thus,0 = κ(x, y,D) = κ(x, y,D′) + 2|A4|+ |A6|+ |A8| − 2|A1| − |A5| − |A7|, which


implies that 2|A1| ≥ κ(x, y,D′)− |A5| − |A7| ≥ κ(x, y,D′)− |V c(x)∩V (F )| −|V c(y) ∩ V (F )|). By dividing by two, we get the desired property.

Lemma 4.7. Let D be a diregular c-partite tournament with c ≥ 4. For allw ∈ V (D) there exists a (w, 1)-reducible 4-cycle in D and a (w, 2)-reduciblek-cycle in D for some k ∈ {5, 6}.

Proof. Let A = N+(w) and let B = N−(w). Let a ∈ A belong to the set Alif and only if the longest path in D 〈A〉 that ends in a has length l. Analogously,define the sets Bl such that b ∈ Bl if and only if the longest path in D 〈B〉 that

begins in b has length l. Let A∗ = ∪|A|l=2Al, and let B∗ = ∪|B|l=2Bl.If A1 is not independent, then let {x, y} ⊆ A1 be chosen such that x→y. Since

x ∈ A1, there is a vertex z ∈ A such that z→x. Since D is a MT, we have z 6= y,so z→x→y is a simple path in D 〈A〉. However, this is a contradiction againsty ∈ A1, which implies that A1 is independent. Analogously, A0, B1, and B0 areindependent, which implies that |A0|, |A1|, |B0|, |B1| ≤ v∗D.

Clearly, A0⇒A1 ∪ A∗, since d−D〈A〉(a) = 0 for all a ∈ A0. If there is a vertexa2 ∈ A∗ and a1 ∈ A1 such that a2→a1, then we can find a path of length 2 endingin a1, a contradiction against a1 ∈ A1. Therefore,A1⇒A∗. Analogously, we obtainthat B∗ ∪B1⇒B0 and B∗⇒B1. We now consider the following three cases.

Case 1. A∗ 6= ∅. If B 6⇒A∗, then let a3 ∈ A∗ and b ∈ B be chosen suchthat a3→b. Let a1a2a3 be a path of length 2 in D 〈A〉 ending in a3, and observethat C = wa1a2a3bw is a cycle of length 5. Since wa2a3bw and wa3bw also arecycles, we have our desired cycles in this case. Therefore, assume that B⇒A∗.However, this implies that S = V c(w) − {w} is a separating set in D, sinceA∪B∪{w}−A∗⇒A∗. However, |S| = v∗D−1 < (|V (D)|−v∗D)/3 when c ≥ 4,which is a contradiction against Corollary 3.1.

Case 2. B∗ 6= ∅. This is proved analogously to case 1.Case 3. A∗ = ∅ and B∗ = ∅. If B1 6⇒A1, then let b1 ∈ B1 and a1 ∈ A1 be

chosen such that a1→b1. Let a0a1 be a path in A and let b1b0 be a path in B. Nowwa0a1b1b0w, wa1b1b0w, and wa1b1w are all cycles in D, which implies that thelemma is true in this case. Therefore, we assume that B1⇒A1.

If A1 = ∅, then d+(w) ≤ v∗D, which is a contradiction when c ≥ 4. Therefore,A1 6= ∅ and, analogously, A0 6= ∅, B1 6= ∅ and B0 6= ∅. Let V c(A0) be thecolor class that includes all the vertices fromA0, and, analogously, define V c(A1),V c(B0), V c(B1). Clearly, V c(A1) 6= V c(A0), since otherwise A1 ∪ A0 would beindependent and A1 = ∅. Analogously, V c(B1) 6= V c(B0).

If B⇒A1, then V c(w)− {w} is a separating set in D, which is a contradictionagainst Corollary 3.1. Therefore, assume thatB 6⇒A1, and let a1 ∈ A1 and b0 ∈ B0be chosen such that a1→b0. Let a0 ∈ A0 be arbitrary and observe that C =wa0a1b0w is a (w, 1)-reducible 4-cycle, since wa1b0w is a cycle. We now provethe existence of a (w, 2)-reducible 6-cycle in D.

Let a1 ∈ A1, a0 ∈ A0, b1 ∈ B1 and b0 ∈ B0 be chosen arbitrarily. We observethat d(a1, A ∪ B ∪ {w}) ≤ |B0| ≤ v∗D and d(A ∪ B ∪ {w}, b1) ≤ |A0| ≤ v∗D.Since d+(a1) = d−(b1) = (c − 1)v∗D/2, we have |N+(a1) ∩ (V c(w) − {w})| ≥


v∗D/2 and |N−(b1) ∩ (V c(w)− {w})| ≥ v∗D/2. Therefore, there must be a vertexw∗ ∈ V c(w)− {w} with a1→w∗→b1. Now the cycle wa0a1w

∗b1b0w is a (w, 2)-reducible 6-cycle in D, since wa1w

∗b1b0w and wa1w∗b1w also are cycles.

Lemma 4.8. Let D be a diregular c-partite tournament with c ≥ 4, and letw ∈ V (D) be arbitrary. For all integers p with 3 ≤ p ≤ (c−2)v∗D + 2, there existsa cycle C in D with w ∈ V (C) and |V (C)| = p.

Proof. LetD be a diregular c-partite tournament with c ≥ 4 and let w ∈ V (D)be arbitrary. Let F be a cycle subgraph inD, such that |V (F )| ≤ p,w ∈ V (F ) and|V (F )| mod 3 = p mod 3. Such a cycle subgraph exists, since by Lemma 4.7 thereexists a 3-cycle, a 4-cycle, and a 5-cycle all including the vertex w. Furthermore,let F have the maximum number of vertices of all cycle subgraphs with the desiredproperties. If |V (F )| = p, we are done by Lemma 4.5, so assume for the sake ofcontradiction that |V (F )| < p, which implies that |V (F )| ≤ p−3 ≤ (c−2)v∗D−1.

Let D′ = D − V (F ), and note that, if D′ has a strong component with verticesfrom three or more color classes, then by Theorem 3.2 there exists a 3-cycle C inD. However, F ∪C would be a contradiction against the maximality of |V (F )|, soclearly there are vertices from at most two color classes in any strong componentin D′. As |V (F )| ≤ (c − 2)v∗D − 1, there are vertices from at least three colorclasses in D′, which implies that D′ is not strong. Let Q1, Q2, . . . , Qm be thestrong components of D′, such that Qi⇒Qj for all 1 ≤ i < j ≤ m.

Let x ∈ Q1 be chosen such that κv(x,D 〈Q1〉) ≥ 0, and let y ∈ Qm bechosen such that κv(y,D 〈Qm〉) ≤ 0. As |V (D′)| ≥ 2v∗D + 1, there is a vertexz ∈ V (D′)−V c(x)−V c(y). If z ∈ Q1, then asQ1 is strong and contains verticesfrom only two color classes, there must be a vertex z′ ∈ V c(z) such that x→z′. Thisimplies that x→z′→y is a (x, y)-path of length 2 inD′. Analogously, if z ∈ Qm wecan obtain a (x, y)-path of length 2 in D′. Finally, if z 6∈ Q1 ∪Qm, then x→z→yis a (x, y)-path of length 2 in D′. Therefore, there exists a (x, y)-path of length 2in D′, which we shall denote by R.

Clearly, κv(x,D′) = κv(x,D 〈Q1〉) + |V (D′) − V c(x) − Q1| and, analo-gously, κv(y,D′) = κv(y,D 〈Qm〉) − |V (D′) − V c(y) − Qm|. This implies thefollowing:

κ(x, y,D′) ≥ |V (D′)− V c(x)−Q1|+ |V (D′)− V c(y)−Qm|≥ |V (D′)| − |Q1| − |V c(x) ∩ V (D′)|+ |V (D′)|−|Qm| − |V c(y) ∩ V (D′)|≥ |V (D′)| − |V c(x) ∩ V (D′)| − |V c(y) ∩ V (D′)|.

By Lemma 4.6, (x, y) has at least (|V (D′)| − 2v∗D)/2 > 0 1-partners in F . Letz be any 1-partner of (x, y) in F , and let C ∈ F be the cycle with z ∈ V (C). Nowthe cycle subgraph (F − C) ∪ C[z+, z]R is the desired contradiction against themaximality of |V (F )|.

We can in most cases improve the bound in Lemma 4.8. However we first needthe following lemma on acyclic MTs.


Lemma 4.9. Let D be an acyclic c-partite tournament (c ≥ 2), and let P =p0p1 . . . pl be a longest path in D with l ≥ 1. For all integers k with 1 ≤ k ≤ l,there exists a (p0, pl)-path of length k or k + 1 in D 〈V (P )〉.

Furthermore, if x ∈ V (D) has d−(x) = 0 and y ∈ V (D) has d+(y) = 0, thenthere exists an (x, y)-path in D of length l.

Proof. LetD be an acyclic c-partite tournament (c ≥ 2), and letP = p0p1 . . . plbe a longest path in D. If l ≤ 2, then the first part of the lemma is clearly true, soassume that l ≥ 3. Let k be an integer with 1 ≤ k ≤ l. As V c(pk−1) 6= V c(pk),we must have either pl 6∈ V c(pk−1) or pl 6∈ V c(pk). If pl 6∈ V c(pk−1), then thepath p0p1 . . . pk−1pl is a (p0, pl)-path of length k; and if pl 6∈ V c(pk), then the pathp0p1 . . . pkpl is a (p0, pl)-path of length k + 1.

Let x ∈ V (D) have d−(x) = 0 and let y ∈ V (D) have d+(y) = 0. By themaximality of l, we must have d−(p0) = 0 and d+(pl) = 0; therefore, V c(x) =V c(p0) and V c(y) = V c(pl). The path xp1p2 . . . pl−1y is now an (x, y)-path oflength l.

Lemma 4.10. Let D be a diregular c-partite tournament with c ≥ 4. Let w ∈V (D) be arbitrary, and let an integer p with 3 ≤ p ≤ (c − 1)v∗D be arbitrary.Define the following properties of a cycle subgraph F :

(α) There exists two cycles in F , say C1 and C2, such that either C1 is (w, 1)-reducible and C2 is 2-reducible, or C1 is 1-reducible and C2 is (w, 2)-reducible.

(β) There exist two cycles in F , say C1 and C2, such that C1 is (w, 1)-reducibleand C2 is 1-reducible.

If there is a cycle subgraph F , |V (F )| ≤ p, inD such that (α) is true, then thereis a cycle C in D with w ∈ V (C) and |V (C)| = p.

If there is a cycle subgraph F , |V (F )| ≤ p, in D such that (β) is true and|V (F )| = p− 3, then there is a cycle C in D with w ∈ V (C) and |V (C)| = p.

Proof. Let F be a cycle subgraph in D with the maximum number of verticessuch that |V (F )| ≤ p and one of the statements (1)–(10) below holds, whereD′ = D − V (F ) and l is the length of a longest path in D′:

(1) p = |V (F )| and w ∈ V (F ).(2) p = |V (F )|+ 1, D′ is acyclic, l ≤ 1, and (β) holds.(3) p = |V (F )|+ 1, D′ is acyclic, l ≥ 2, and (α) holds.(4) p = |V (F )|+ 2, D′ is acyclic, and (β) holds.(5) p = |V (F )|+ 3, D′ is acyclic, and (β) holds.(6) p = |V (F )|+ 4, D′ is acyclic, and (α) holds.(7) p = |V (F )|+ 5, D′ is acyclic, and (α) holds.(8) p ≥ |V (F )|+ 6, D′ is acyclic, and (α) holds.(9) p > |V (F )|, p 6= |V (F )|+ 3, D′ is not acyclic, and (α) holds.(10) p = |V (F )|+ 3, D′ is not acyclic, and (β) holds.

Such an F exists, since (α) implies (β). If (1) holds, we are done, so assumethat (1) does not hold, for the sake of contradiction.


If (9) or (10) holds, then let C be a smallest possible cycle in D′. Clearly,3 ≤ |V (C)| ≤ 4, since any cycle of length 5 or more in a MT has a chord that impliesthe existence of a smaller cycle. If |V (F )|+ |V (C)| ≤ p, then the cycle subgraphF ∪C is a contradiction against the maximality of |V (F )|. If |V (F )|+ |V (C)| > p,then by (α) (or (β) if (10) holds), we can delete |V (F )|+ |V (C)|−p vertices fromF to obtain a new cycle subgraph F ′ with w ∈ V (F ′) and |V (F ′)| = p− |V (C)|.The cycle subgraph F ′ ∪ C contradicts the maximality of |V (F )|.

If one of (2)–(8) holds, then let P = p0p1p2 . . . pl be a longest path in D′ andlet R = r0r1 . . . rl′ be a longest path in D′ − V (P ). Observe that d−D′(p0) = 0and d+

D′(pl) = 0 by the maximality of l. By Lemma 4.6, (p0, pl) has (|V (D′)| −|V c(p0)∩ V (D′)|+ |V (D′)| − |V c(pl)∩ V (D′)| − |V c(p0)∩ V (F )| − |V c(pl)∩V (F )|)/2 = |V (D′)|−v∗D δ-partners, for all δ = 1, 2, 3. LetC1 ∈ F andC2 ∈ F bedefined as in (α) or (β). Let u1 6= w be the vertex onC1 such thatD 〈V (C1)− u1〉is Hamiltonian, with Hamilton cycle C ′1. Let u2 6= w be the vertex on C2 suchthat D 〈V (C2)− u2〉 is Hamiltonian, with Hamilton cycle C ′2. If (α) holds, thenlet u3 6= w be the vertex on C2 such that D 〈V (C2)− {u2, u3}〉 is Hamiltonian,with Hamilton cycle C ′′2 . Note that, if |V (F )| < p, then |V (D′)| > v∗D so l ≥ 1.

We need the following two claims before we look at each of the cases (2)–(8)separately.

Claim (i). If one of (2)–(8) holds, then l ≥ 2.Assume that one of (2)–(8) holds and l = 1 (l = 0 is impossible as |V (D′)| >

v∗D). By the definition of l, we can partition D′ into the independent vertex setsQ1 and Q2 such that Q1⇒Q2. Let q = p − |V (F )| and note that |Q1| ≥ q and|Q2| ≥ q. Let Q1 = {q′1, q′2, . . . , q′|Q1|} and let Q2 = {q′′1 , q′′2 , . . . , q′′|Q1|}. ByLemma 4.6, every pair (q′i, q′′j ) has q 1-partners in F . If q is even, then we insertthe paths q′1q′′1 , q′2q′′2 , . . . , q′q/2q

′′q/2 into F using different 1-partners (i.e., if z is a

1-partner for q′1q′′1 and z ∈ C ∈ F , then we insert q′1q′′1 into C by creating the newcycle C[z+, z]q′1q′′1z+). The resulting cycle subgraph has size p, a contradictionagainst the maximality of |V (F )|. If q = 1, we insert q′1q′′1 in F and afterwardseither delete u1 or u2 (depending on which cycle q′1q′′1 was inserted into).

If q = 3, then without loss of generality assume that (q′1, q′′1) has two 1-partnersthat are different from u1 and u−1 (otherwise it has two 1-partners that are differentfrom u2 and u−2 ). As (q′2, q′′2) has a 1-partner that is different from both u1 and u−1 ,we can insert q′2q′′2 using such a 1-partner, then we can insert q′1q′′1 using a 1-partnerthat is different from (q′2, q′′2)’s 1-partner and different from u1 and u−1 . Then wedelete u1 to obtain a cycle subgraph of size p, contradicting the maximality of|V (F )|.

If q ≥ 5 and q is odd then we can insert the paths q′1q′′1 , q′2q′′2 , . . . , q′(q+1)/2q′′(q+1)/2

into F using different 1-partners which all are different from u1 and u−1 . Then wedelete u1 to obtain a cycle subgraph of size p, contradicting the maximality of|V (F )|.

This completes the proof of the claim (i).

Claim (ii). If one of (2)–(8) holds, then l ≤ p− |V (F )| − 2.


Since one of (2)–(8) holds, clearly (β) holds. Assume that l ≥ p− |V (F )| − 1and, by claim (i), that l ≥ 2. Now look at the following two cases separately.

Case 1: p − |V (F )| ≥ 2. By Lemma 4.9, there is a (p0, pl)-path, P ′, of lengthp− |V (F )| − 1 or of length p− |V (F )|. As (p0, pl) has a 1-partner in F , we insertP ′ in F to obtain a new cycle subgraph F ′ of size p or p+1. If |V (F ′)| = p, we geta contradiction against the maximality of |V (F )|, so assume that |V (F ′)| = p+ 1.IfP ′ is inserted intoC1, then we can delete u2 fromF ′, and ifP ′ is not inserted intoC1, then we can delete u1 from F ′ instead. In both cases, we get a cycle subgraphof size p, which contradicts the maximality of |V (F )|.

Case 2: p − |V (F )| = 1. Clearly (3) is true, which implies that (α) holds. LetF ′ = C ′2 ∪ (F − C2) and let D′′ = D 〈V (D)− V (F ′)〉. If D′′ is acyclic, then,since a longest path in D′′ is at least as long as in D′, we are done analogouslyto case 1 above. Therefore, assume that D′′ is not acyclic. Let C be a cycle inD′′ with |V (C)| ∈ {3, 4}. If |V (C)| = 3, then delete the vertex u1 from C ∪ F ′to obtain a cycle subgraph of size p, contradicting the maximality of |V (F )|. If|V (C)| = 4, then delete both u1 and u3 from C ∪F ′ to obtain a cycle subgraph ofsize p, contradicting the maximality of |V (F )|.

This completes the proof of the claim (ii).

If (2), (3), (4), or (5) holds, we are done by claim (i) or claim (ii), since l ≥ 2and l ≤ p− |V (F )| − 2 is impossible when (2), (3), (4), or (5) hold.

If (6) holds, then by claim (i) and claim (ii) we must have l = 2. We now lookat the following three cases.

Case 1: l′ = 0. This is impossible, as |V (F ) ∪ V (P )| < p, so |V (D′) −V (P )| > v∗D.

Case 2: l′ = 1. P has four 1-partners in F , so let z1 be one of them. Insert Pin F using the 1-partner z1 and denote the new cycle subgraph by F ′. By Lemma4.6, (r0, r1) has a 1-partner in F ′, say z. By the maximality of l, we have z 6∈{z1, p0, p1, p2}. Therefore, we could insertR into F using the 1-partner z to obtaina cycle subgraph F ′′. If z ∈ V (C1), then let Z = {u2, u

−2 }, and if z 6∈ V (C1), then

letZ = {u1, u−1 }. Now let z2 be a 1-partner of (p0, p2) inF such that z2 6∈ {z}∪Z.

We can now insert the path P into F ′′ and delete either u1 or u2 to obtain a cyclesubgraph with size p, a contradiction.

Case 3: l′ = 2. It is easy to see that d−D′(p0) = d−D′(r0) = d+D′(p2) = d+

D′(r2) =0, which implies that both (p0, p2) and (r0, r2) have four 1-partners in F . We nowlook at the cases when u2 and u3 are neighbors on C2, and when they are not.

Ifu2 andu3 are neighbors on the cycleC2, then assume without loss of generalitythat u2 = u−3 and do the following. If either (p0, p2) or (r0, r2) have a 1-partnerthat is different from u1, u−1 , u−2 , u3 and u2 (= u−3 ), then let z be such a 1-partner.Without loss of generality assume that z is a 1-partner for (p0, p2). Let z1 be a1-partner for (r0, r2) such that z1 6∈ {u2, u

−2 , z}. We can now insert P using the

1-partner z, and R using the 1-partner z1, to obtain a new cycle subgraph F ′. Wecan delete u2 from F ′, and one of the vertices in {u1, u3} (depending on z1), toobtain a cycle subgraph of size p, a contradiction.


If both (p0, p2) and (r0, r2) only have 1-partners in {u1, u−1 , u

−2 , u2, u3}, then

do the following. Clearly, (p0, p2) must have a 1-partner in {u1, u−1 }, and assume

without loss of generality that u1 is a 1-partner for (p0, p2). If u−1 is a 1-partner for(r0, r2), then we can insert P and R into F , and then delete u2 and u3 to obtaina cycle subgraph of size p, a contradiction. If u−1 is not a partner for (r0, r1), thenu−2 , u2, u3 and u1 are all partners of (r0, r2). This implies that u−2 is a 3-partnerof (r0, r2), because u−2→r0 and r2→u++

2 = u+3 . We can now insert P using the

1-partner u1, and R using the 3-partner u−2 resulting in a cycle subgraph of sizep including the vertex w as we have only removed u2 and u3 from F . This is acontradiction against the maximality of |V (F )|.

If u2 and u3 are not neighbors on C2, then note that D 〈V (C2)− u3〉 is Hamil-tonian and do the following. If either (p0, p2) or (r0, r2) have a 1-partner that isdifferent from u1, u−1 , u2, u−2 , u3, and u−3 , then use that 1-partner to insert one ofthe paths in F and use any other 1-partner to insert the other path. We can thendelete two of the vertices {u1, u2, u3} from the resulting cycle subgraph to obtaina cycle subgraph with p vertices, a contradiction.

If both (p0, p2) and (r0, r2) only have 1-partners among{u1, u−1 , u2, u

−2 , u3, u

−3 },

then there must be an i ∈ {1, 2, 3} such that ui is a partner for either P or R, andu−i is a partner for the other path. Insert P andR into F using the 1-partners ui andu−i , and then delete the vertices {u1, u2, u3} − {ui}, to obtain a cycle subgraph ofsize p, a contradiction.

If (7) holds, then by claim (i) l ≥ 2. Lemma 4.9, therefore, implies that thereis a (p0, pl)-path P ′ in D′ of length 1 or of length 2. Since (p0, pl) has at least 51-partners on F , it has a 1-partner, v, which is different from u1, u−1 , u2, and u−2 .We insert P ′ into F using the 1-partner v to obtain the new cycle subgraph F ′′.Observe that p−3 ≤ |V (F ′′)| ≤ p−2, and that (β) holds. Therefore, the existenceof F ′′ is a contradiction against the maximality of |V (F )|.

If (8) holds, then by claim (i) l ≥ 2. If l ≥ 3, then we proceed analogously asin the proof when (7) held (where P ′ has length 2 or 3). If l = 2, then note that(p0, pl) has 6 1-partners on F , by Lemma 4.6. If there is a 1-partner different fromu1, u−1 , u2, u−2 , u3, and u−3 , then we can use such a 1-partner to insert P , and get acontradiction against the maximality of |V (F )|. Otherwise, u1, u−1 , u2, u−2 , u3, andu−3 are all 1-partners of (p0, pl) and all six vertices are distinct. Therefore, we mayuse u3 as a 1-partner and insert P in F , which, since u−3→p0 implies that (α) stillholds, and we have obtained a contradiction against the maximality of |V (F )|.Corollary 4.1. Let D be a diregular c-partite tournament with c ≥ 5, and letw ∈ V (D). For all integers p with 3 ≤ p ≤ (c− 1)v∗D, there exists a cycle C in Dwith w ∈ V (C) and |V (C)| = p.

Proof. If v∗D ≤ 2, we are done by Lemma 4.8, so assume that v∗D ≥ 3. ByLemma 4.7, there exists a (w, 1)-reducible 4-cycle, C1, in D.

We will now show that D is 5-strong. Clearly, this is true when v∗D ≥ 4 orc ≥ 6, by Corollary 3.1. Therefore, assume that v∗D = 3 and c = 5. If D is not5-strong, then there is a separating set S in D with |S| ≤ 4. By Corollary 3.1, we


must have |S| = 4. Let X = V (D)− S and observe that Lemma 3.1 implies that0 = il(D) ≥ (|X| − v∗D) d|X|/2e|X| − |S| = (11− 3) d11/2e

11 − 4 = 8 ∗ 6/11− 4 > 0,a contradiction.

Therefore, D− V (C1) is strong and, and by Lemma 3.2 and the fact that c ≥ 5,it contains a 2-reducible 5-cycle. Lemma 4.10 now implies this corollary.

5. MAIN RESULT

Theorem 5.1. All diregular c-partite tournaments with c ≥ 5 are vertex pancyclic.Proof. Let D be a diregular c-partite tournament with c ≥ 5. By Lemma 4.3

and Corollary 4.1, D is vertex pancyclic if v∗Dc(2c− 2)/(3c− 5) + 2c/(3c− 5) ≤v∗D(c− 1) + 1, since v∗D(c− 1) + 1 is the smallest integer that is strictly larger thanv∗D(c − 1). However, this implies the theorem, because the following holds whenc ≥ 5:

v∗Dc2c−23c−5 + 2c

3c−5 ≤ v∗D(c− 1) + 1m

2c− (3c− 5) ≤ v∗D((c− 1)(3c− 5)− c(2c− 2))m

5− c ≤ v∗D(c2 − 6c+ 5)

6. OPEN PROBLEMS

We would like to state the following related conjectures.

Conjecture 6.1. All diregular semicomplete c-partite digraphs with c ≥ 4 arepancyclic.

Conjecture 6.2. All diregular c-partite tournaments with c ≥ 4 have a pair ofvertex disjoint cycles of length p and |V (D)| − p for every p ∈ {3, 4, . . . , |V (D)|− 3}.

Clearly the results in this article, and in [7], give support for Conjecture 6.1. In[8], the following related conjecture is given: Any diregular bipartite tournament,D, not belonging to a given class of bipartite tournaments, has a pair of vertexdisjoint cycles of length p and |V (D)|− p for every p ∈ {4, 6, 8, . . . , |V (D)|− 4}.

References

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[2] J. A. Bondy and U. S. R. Murty, Graph theory with applications, MacMillan,New York, 1976.

[3] G. Gutin, Cycles and paths in semicomplete multipartite digraphs, theoremsand algorithms: a survey, J Graph Theory 19 (1995), 481–505.

[4] L. Volkmann, Cycles in multipartite tournaments: results and problems, toappear.

[5] A. Yeo, One-diregular subgraphs in semicomplete multipartite digraphs, JGraph Theory 24 (1997), 175–185.

[6] A. Yeo, How close to regular must a semicomplete multipartite digraph be tosecure Hamiltonicity? to appear.

[7] A. Yeo, Large diregular 4-partite tournaments are vertex-pancyclic, to appear.[8] K. M. Zhang, Y. Manoussakis, and Z. M. Song, Complementary cycles con-

taining a fixed arc in diregular bipartite tournaments, Discrete Math 133(1994), 325–328.

Diregularc-partite tournaments are vertex-pancyclic whenc ? 5

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