The triangle-free graphs which are competition graphs of multipartite tournaments
aa r X i v : . [ m a t h . C O ] S e p The triangle-free graphs which are competition graphsof multipartite tournaments
Myungho Choi , Minki Kwak , and Suh-Ryung Kim Department of Mathematics Education, Seoul National University, Seoul 08826, Republic of Korea
September 22, 2020
Abstract
In this paper, we show that a connected triangle-free graph is the competitiongraph of a k -partite tournament if and only if k ∈ { , , } , and a disconnectedtriangle-free graph is the competition graph of a k -partite tournament if and onlyif k ∈ { , , } . Then we list all the triangle-free graphs in each case. Given a digraph D , N + D ( x ) and N − D ( x ) denote the sets of out-neighbors and in-neighbors,respectively, of a vertex x in D . The nonnegative integers | N + D ( x ) | and | N − D ( x ) | are calledthe outdegree and the indegree , respectively, of x and denoted by d + D ( x ) and d − D ( x ), re-spectively. When no confusion is likely, we omit D in N + D ( x ), N − D ( x ), d + D ( x ), and d − D ( x ),to just write N + ( x ), N − ( x ), d + ( x ), and d − ( x ), respectively.The competition graph C ( D ) of a digraph D is the (simple undirected) graph G definedby V ( G ) = V ( D ) and E ( G ) = { uv | u, v ∈ V ( D ) , u = v, N + D ( u ) ∩ N + D ( v ) = ∅} . Compe-tition graphs arose in connection with an application in ecology (see [4]) and also haveapplications in coding, radio transmission, and modeling of complex economic systems.Early literature of the study on competition graphs is summarized in the survey papersby Kim [11] and Lundgren [13].For a digraph D , the underlying graph of D is the graph G such that V ( G ) = V ( D ) and E ( G ) = { uv | ( u, v ) ∈ A ( D ) } . An orientation of a graph G is a digraph having no directed2-cycles, no loops, and no multiple arcs whose underlying graph is G . A tournament is anorientation of a complete graph. A k -partite tournament is an orientation of a complete k -partite graph for some positive integer k ≥ k -partitetournaments for k ≥ k -partite tournament if and only if k ∈ { , , } , andlist all the connected triangle-free graphs which are competition graphs of multipartitetournaments (Theorem 2.19).We also show that a disconnected triangle-free graph is the competition graph of a k -partite tournament if and only if k ∈ { , , } , and list all the disconnected triangle-freegraphs which are competition graphs of multipartite tournaments (Theorems 3.4, 3.9,and 3.10). Lemma 2.1.
Let D be an orientation of K n ,n ,n whose competition graph has no isolatedvertex for some positive integers n , n , and n . Then at least two of n , n , and n aregreater than .Proof. Suppose, to the contrary, that at most one of n , n , and n is greater than 1,that is, at least two of n , n , and n equal 1. Without loss of generality, we may assumethat n = n = 1. Let { u } , { v } , and V be the partite sets of D with | V | = n . Since D is an orientation of K n ,n ,n , either ( u, v ) ∈ A ( D ) or ( v, u ) ∈ A ( D ). By symmetry,we may assume that ( u, v ) ∈ A ( D ). Since C ( D ) has no isolated vertex, v is adjacent tosome vertex. Since ( u, v ) ∈ A ( D ), v is not adjacent to any vertex in V . Thus u and v areadjacent in C ( D ) and so u and v have a common out-neighbor w in V . Then neither u nor v is an out-neighbor of w and so N + ( w ) = ∅ . Therefore w is isolated in C ( D ), whichis a contradiction. Lemma 2.2.
Let D be a digraph with n vertices for a positive integer n . If the competitiongraph C ( D ) of D is triangle-free, then | E ( C ( D )) | ≤ | A ( D ) | / ≤ | V ( D ) | .Proof. Suppose that the competition graph C ( D ) of D is triangle-free. Then d − ( v ) ≤ v in D . Therefore | E ( C ( D )) | ≤ |{ v ∈ V ( D ) | d − ( v ) = 2 }| ≤ | A ( D ) | ≤ | V ( D ) | . Lemma 2.3.
There is no orientation of K , , whose competition graph is triangle-free. roof. Suppose, to the contrary, that there exists an orientation D of K , , whose com-petition graph is triangle-free. Then | A ( D ) | = 14. Since C ( D ) is triangle-free, each vertexhas indegree at most 2. Then, since | V ( D ) | = 7 and | A ( D ) | = 14, d − ( v ) = 2 (1)for each vertex v in D . Let V = { x , x , x , x } , V = { y , y } , and V = { z } be thepartite sets of D . By (1), each vertex in V is a common out-neighbor of two vertices in V ∪ V and so has outdegree 1. We note that if a vertex a in V ∪ V is an out-neighbor ofa vertex b in V , then b is a common out-neighbor of the two vertices in V ∪ V \ { a } andso they are adjacent in C ( D ). Therefore there must be a vertex in V ∪ V which is notan out-neighbor of any vertex in V to prevent from creating a triangle y y z in C ( D ).By (1), such a vertex in V ∪ V must be z and N − ( z ) = { y , y } . Then N + ( z ) = V . By (1) again, each of y and y is a common out-neighbor of two verticesin V . Since each vertex in V has outdegree 1, N − ( y ) ∩ N − ( y ) = ∅ . Without loss ofgenerality, we may assume N − ( y ) = { x , x } and N − ( y ) = { x , x } . Then N + ( x ) = N + ( x ) = { y } and N + ( x ) = N + ( x ) = { y } , so N − ( x ) = { y , z } and N − ( x ) = { y , z } . Hence { y , y , z } forms a triangle in C ( D ), which is a contradiction. Lemma 2.4.
Let n , n , and n be positive integers such that n ≥ n ≥ n . Suppose thatthere exists an orientation D of K n ,n ,n whose competition graph C ( D ) is triangle-free.Then one of the following holds: (a) n = n = n = 2 ; (b) n ≤ , n = 2 , and n = 1 ;(c) n = n = 1 . In particular, if C ( D ) is connected, then the case (c) does not occur.Proof. It is easy to check that | A ( D ) | = n n + n n + n n . Then, by Lemma 2.2, n n + n n + n n ≤ n + n + n ) . Thus n ( n −
2) + n ( n −
2) + n ( n − ≤ n − n −
2, and n − n ≥ n ≥ n , n − ≤
0. Suppose n = 2. Then, by (2), n n ≤ n , n , n ) = (2 , , n = 1. Then, by (2), ( n − n − ≤
3. Since n ≥ n , n ≤
2. Suppose n = 2. Then n ≤
4. If n = 4, then ( n , n , n ) = (4 , , n ≤ n = 1, then n = 1 and so (c) holds. If C ( D )is connected, then C ( D ) has no isolated vertices and so none of n and n equals 1 byLemma 2.1 and so the “in particular” part is true.3 C ( D )Figure 1: A digraph D which is an orientation of K , , and whose competition graph isisomorphic to C The following lemma is an immediate consequence of Lemma 2.4.
Lemma 2.5.
For a connected triangle-free graph G of order n , if G is the competitiongraph of a tripartite tournament, then n ∈ { , } . Lemma 2.6.
For a positive integer n ≥ , a cycle C n of length n is the competition graphof a tripartite tournament if and only if n = 6 .Proof. Let D be the digraph in Figure 1 which is an orientation of K , , . It is easy tocheck that C ( D ) ∼ = C . Therefore the “if” part is true.Now suppose that a cycle C n is the competition graph of a tripartite tournament T for a positive integer n ≥ n = 3, then the only possible size of partite sets is (1 , , n ≥
4. Thus n = 5 or n = 6 by Lemma 2.5.Suppose, to the contrary, that n = 5. Then T is an orientation of K , , by Lemma 2.4.Since C ( T ) is triangle-free, d − ( v ) ≤ v ∈ V ( T ). Moreover, since each edgeis a maximal clique and there are five edges in C ( T ), each vertex has indegree 2 in T .Therefore 8 = | A ( T ) | = X v ∈ V ( T ) d − ( v ) = 10and we reach a contradiction. Thus n = 6. Lemma 2.7.
For a positive integer n ≥ , a path P n of length n − is the competitiongraph of a tripartite tournament if and only if n = 6 .Proof. Let D be the digraph in Figure 2 which is an orientation of K , , . It is easy tocheck that C ( D ) ∼ = P . Therefore the “if” part is true.Now suppose that a path P n is the competition graph of a tripartite tournament T for a positive integer n ≥
3. Since P n is connected and triangle-free, by Lemma 2.5, n ∈ { , } . Suppose, to the contrary, that n = 5. Then T is an orientation of K , , byLemma 2.4. Let V , V , and V be the partite sets with | V | = | V | = 2 and | V | = 1. Since4 C ( D )Figure 2: A digraph D which is an orientation of K , , and whose competition graph isisomorphic to P C ( T ) is triangle-free, d − ( v ) ≤ v ∈ V ( T ). Since there are four edges in C ( T ),there are four vertices of indegree 2 in T . Since | A ( T ) | = 8 and n = 5, there exists exactlyone vertex of indegree 0 in T . Let u be the vertex of indegree 0 in T . If V = { u } , then N + ( u ) = V ∪ V . Otherwise, either N + ( u ) = V ∪ V or N + ( u ) = V ∪ V . Therefore d + ( u ) = 3 or 4. Thus u is incident to at least three edges in C ( T ), which is impossible ona path. Hence n = 6. Lemma 2.8.
Let D be an orientation of K , , whose competition graph is connected andtriangle-free. Then the following are true:(1) There is no vertex of indegree in D ;(2) There are exactly five vertices with indegree in D no two of which have the samein-neighborhood.Proof. Since C ( D ) is triangle-free, d − ( v ) ≤ v ∈ V ( D ). If there is a vertex ofindegree 0 in D , then 11 = P v ∈ V ( D ) d − ( v ) ≤
10 and we reach a contradiction. Thereforethe statement (1) is true and so the indegree sequence of D is (2 , , , , , C ( D ) is connected, the number of edges of C ( D ) is at least 5. Thus there are exactlyfive vertices with indegree 2 in D no two of which have the same in-neighborhood. Lemma 2.9.
Let G be a connected and triangle-free graph with n vertices. Then G is thecompetition graph of a tripartite tournament if and only if G is isomorphic to a graphbelonging to the following set: ( { G , G } if n = 5 ; { G , G , P , C } if n = 6 where G i is the graph given in Figure 3 for each ≤ i ≤ . G G G Figure 3: Connected triangle-free graphs mentioned in Lemma 2.9 D D Figure 4: Two digraphs D and D which are orientations of K , , and whose competitiongraphs are isomorphic to G and G , respectively D D Figure 5: Two digraphs D and D which are orientations of K , , and whose competitiongraphs are isomorphic to G and G , respectively6igure 6: A graph considered in the proof of Lemma 2.9 Proof.
Let D be a tripartite tournament whose competition graph is G . Since G is triangle-free, d − ( v ) ≤ v ∈ V ( D ) and n ∈ { , } by Lemma 2.5. If G is a path or a cycle, then, byLemmas 2.6 and 2.7, G is isomorphic to P or C . Now we suppose that G is neither apath nor a cycle. Then, there exists a vertex of degree at least three in C ( D ). Case 1. n = 5. Then, by Lemma 2.4, D is an orientation of K , , . Since | A ( D ) | = 8and C ( D ) is connected, by (3), there are exactly four edges in C ( D ). Therefore C ( D ) isisomorphic to G or G in Figure 3. Thus the “only if” part is true in this case. To showthe “if” part, let D and D be the digraphs in Figure 4 which are some orientations of K , , . It is easy to check that C ( D ) ∼ = G and C ( D ) ∼ = G . Hence the “if” part is true. Case 2. n = 6. Then, by Lemma 2.4, D is an orientation of K , , or K , , . Supposethat D is an orientation of K , , . Since P v ∈ V ( D ) d − ( v ) = 12, by (3), d − ( v ) = 2 for each v ∈ V ( D ) and so d + ( v ) = 2 for each v ∈ V ( D ). Therefore every vertex has degree at most2 in C ( D ), which is a contradiction to the assumption that G is neither a path nor cycle.Thus D is an orientation of K , , . By Lemma 2.8, there are exactly five edges in C ( D ).Let V , V , and V be the partite sets of D with | V i | = i for each i = 1 , , and 3.Suppose that there is a vertex w of degree at least 4 in C ( D ). Then, by (3), w hasoutdegree at least 4 in D . Thus w belongs to V or V . If w belongs to V , then theindegree of w is 0, which contradicts Lemma 2.8(1). Therefore w ∈ V and so V = { w } .Moreover, the outdegree of w in D is 4 by Lemma 2.8(1). Then the indegree of each vertexin D except w is exactly 2 by Lemma 2.8(2). If three out-neighbors of w belong to thesame partite set, then two of them share the same in-neighborhood, which contradictsLemma 2.8(2). Therefore two of the out-neighbors of w belong to V and the remainingout-neighbors belong to V . Since the indegree of each vertex in D except w is exactly2 by Lemma 2.8(2), each vertex in V has exactly one in-neighbor in V . Thus there isone vertex in V which is not an in-neighbor of any vertex in V . Then w is the only itsout-neighbor and so it is isolated in C ( D ). Hence we have reached a contradiction and sothe degree of each vertex of C ( D ) is at most 3.Now suppose that there are at least two vertices x and y of degree 3 in C ( D ). Then,since the number of edges in C ( D ) is exactly 5, C ( D ) is isomorphic to the tree given inFigure 6. By (3), d + ( x ) ≥ d + ( y ) ≥
3. If x or y belongs to V , then d − ( x ) = 0 or d − ( y ) = 0, which contradicts Lemma 2.8(1). Therefore x and y belong to V or V . Suppose7hat V = { x, y } . Then, since | V ∪ V | = 4, there are at least two vertices of indegree 2in V ∪ V which have the same in-neighborhood { x, y } , which contradicts Lemma 2.8(2).Thus one of x and y belongs to V and the other belongs to V . Without loss of generality,we may assume that x ∈ V and y ∈ V . Then V = { x } and, by Lemma 2.8(1), d − ( y ) = 0,so d + ( y ) = 3. If N + ( y ) = V , then y is adjacent to at most two vertices in C ( D ), whichis a contradiction. Therefore N + ( y ) ∩ V = { z , z } for some vertices z and z in V and( y, x ) ∈ A ( D ). Since C ( D ) is isomorphic to the tree given in Figure 6, x and y have acommon out-neighbor in V . By Lemma 2.8(2), exactly one of z and z can be a commonout-neighbor of x and y in D . By symmetry, we may assume that z is a common out-neighbor of x and y and ( z , x ) ∈ A ( D ). Then N + ( x ) = { y ′ , z , z } for the vertices y ′ other than y in V and z other than z and z in V . Therefore, by (3), ( z , y ′ ) ∈ A ( D )and so, by (3) again, N + ( y ′ ) = { z , z } . Then y ′ is adjacent to y and x in C ( D ) and so { x, y, y ′ } forms a triangle in C ( D ), which is a contradiction. Thus we have shown thatthere is the only one vertex of degree 3 in C ( D ) and so C ( D ) is isomorphic to G or G in Figure 3. Hence the “only if” part is true. To show the “if” part is true, let D and D be two digraphs given in Figure 5 which are isomorphic to some orientations of K , , . Itis easy to check that C ( D ) ∼ = G and C ( D ) ∼ = G . Hence the “if” part is true.The following lemma is immediately true by the definition of the competition graph. Lemma 2.10.
Let D be a digraph and D ′ be a subdigraph of D . Then the competitiongraph of D ′ is a subgraph of the competition graph of D . Lemma 2.11.
For a positive integer k ≥ , each competition graph of a k -partite tour-nament contains a triangle.Proof. Suppose that D is a k -partite tournament for a positive integer k ≥
6. Let V , V , . . . , V k be the partite sets of D . Then we take a vertex v i in V i for each 1 ≤ i ≤ { v , . . . , v } forms a 6-tournament T . Since T has 15 arcs, there exists a vertex in T whose indegree is at least 3. Therefore C ( T ) has a triangle and so, by Lemma 2.10, C ( D )contains a triangle. Proposition 2.12. (Fisher [9]) For n ≥ , the minimum possible number of edges in thecompetition graph of an n -tournament is (cid:0) n (cid:1) − n . An n -tournament is regular if n is odd and every vertex has outdegree ( n − / n -tournament ( n ≥ dominationgraph of a tournament T is the complement of the competition graph of the tournamentformed by reversing the arcs of T . Accordingly, their results can be restated as follows. Proposition 2.13. (Fisher [10]) A path on four or more vertices is not the complementof the competition graph of a tournament.
Proposition 2.14. (Cho [1]) If T is a regular n -tournament ( n ≥ , then the complementof the competition graph of T is either an odd cycle or a forest of two or more paths. Lemma 2.15.
If the competition graph C ( D ) of a -partite tournament D is triangle-free,then D is a regualr -tournament and C ( D ) is isomorphic to a cycle of length .Proof. Suppose that D is a 5-partite tournament whose competition graph is triangle-free. Let V , . . . , V be the partite sets of D . To show | V ( D ) | = 5 by contrary, suppose | V ( D ) | ≥
6. Then there exists a partite set whose size is at least 2. Without loss ofgenerality, we may assume | V | ≥
2. We take v i in V i for each 1 ≤ i ≤
5. Then we may takea vertex v ′ distinct from v in V so that the subdigraph T induced by { v , v ′ , v , . . . , v } is a 5-partite tournament. Since T has 14 arcs and | V ( T ) | = 6, there exists a vertex ofindegree at least 3 in T , which is a contradiction. Thus V i = { v i } for each 1 ≤ i ≤ D is a tournament.Since | V ( D ) | = 5, | E ( C ( D )) | ≥ | E ( C ( D )) | = 5. Then, since | V ( D ) | = 5, each vertex has indegree exactly 2 and soeach vertex has outdegree 2. Thus D is a regular 5-tournament. Since it is easy to checkthat a regular 5-tournament is unique up to isomorphism as shown in Figure 7, C ( D ) isisomorphic to a cycle of length 5. Lemma 2.16.
Let D be a multipartite tournament whose competition graph is triangle-free. If two vertices u and v with outdegree at least one have the same out-neighborhood orin-neighborhood, then u and v belong to the same partite set of D and form a componentin C ( D ) .Proof. Suppose, to the contrary, that there are two vertices u and v with outdegree atleast one such that N + ( u ) = N + ( v ) or N − ( u ) = N − ( v ) but u and v belong to the distinctpartite sets. Then ( u, v ) ∈ A ( D ) or ( v, u ) ∈ A ( D ). Without loss of generality, we mayassume ( u, v ) ∈ A ( D ). Then u ∈ N − ( v ) but u / ∈ N − ( u ). Therefore N − ( u ) = N − ( v ) and N + ( u ) = N + ( v ), which is a contradiction. Thus u and v belong to the same partite set.Then, since D is a multipartite tournament, N + ( u ) = N + ( v ) if and only if N − ( u ) = N − ( v ). Therefore N + ( u ) = N + ( v ) = ∅ by the hypothesis. Since C ( D ) is triangle-free, u and v are the only in-neighbors of each vertex in N + ( u ) and so they form a componentin C ( D ). 9 emma 2.17. Let n , n , n , n be positive integers such that n ≥ · · · ≥ n . If D isan orientation of K n ,n ,n ,n whose competition graph is triangle-free, then n ≤ and n = n = n = 1 .Proof. Suppose that there exists an orientation D of K n ,n ,n ,n whose competition graphis triangle-free. Let V , . . . , V be the partite sets of D with | V i | = n i for each 1 ≤ i ≤ v i in V i for each 1 ≤ i ≤
4. Suppose, to the contrary, that n ≥
2. Then n ≥
2. We may take a vertex v ′ (resp. v ′ ) distinct from v (resp. v ) in V (resp. V ) sothat the subdigraph induced by { v , v ′ , v , v ′ , v , v } is a 4-partite tournament T . Then T has 13 arcs. Since | V ( T ) | = 6, at least one vertex of T has indegree at least 3, which isa contradiction. Thus at most one partite set of D has size at least 2. Hence n = n = n = 1. Therefore | A ( D ) | = 3 n + 3. By Lemma 2.2, | A ( D ) | ≤ | V ( D ) | ) = 2( n + 3).Thus 3 n + 3 ≤ n + 3), so n ≤ n = 3. Then D is an orientation of K , , , . Since | V ( D ) | = 6 and | A ( D ) | = 12, d − ( v ) = 2 (4)for each vertex v in D . We note that | E ( C ( D )) | ≤ | E ( C ( D )) | =6. Then, since | V ( D ) | = 6, each pair of vertices shares at most one common out-neighborin D . Since n = 3 and each vertex in V has indegree 2 by (4), each vertex in V is acommon out-neighbor of v i and v j for some i, j ∈ { , , } . Therefore v i and v j have acommon out-neighbor in V for each 2 ≤ i = j ≤
4. Thus { v , v , v } forms a trianglein C ( D ), which is a contradiction. Hence | E ( C ( D )) | 6 = 6 and so | E ( C ( D )) | ≤
5. Then,there exists at least one pair of vertices which has two distinct common out-neighborsby (4). Since each vertex in V has outdegree 1 by (4), such a pair of vertices belongs to { v , v , v } . Without loss of generality, we may assume { v , v } is such a pair. Let x and y be their distinct common out-neighbors of v and v . Then N − ( x ) = N − ( y ) = { v , v } by (4). Since each vertex in D has outdegree at least 1 by (4), x and y belong to thesame partite set by Lemma 2.16 and so { x, y } ⊂ V . Thus N + ( x ) = N + ( y ) = { v } . Hence { x, y } ⊆ N − ( v ) and so, by (4), N − ( v ) = { x, y } . Therefore N + ( v ) = { v , v , z } where z is a vertex in D distinct from x and y in V . Without loss of generality, we may assume( v , v ) ∈ A ( D ) . Then v is a common out-neighbor of v and v . Therefore by (4), ( v , z ) ∈ A ( D ). Hence z is a common out-neighbor of v and v . Thus { v , v , v } forms a triangle in C ( D ), whichis a contradiction. Therefore | V | 6 = 3 and so | V | ≤ Proposition 2.18. (Kim [12]). Let D be an orientation of a bipartite graph with bipar-tition ( V , V ) . Then the competition graph of D has no edges between the vertices in V and the vertices in V . heorem 2.19. Let G be a connected and triangle-free graph. Then G is the competitiongraph of a k -partite tournament for some k ≥ if and only if k ∈ { , , } and G isisomorphic to a graph belonging to the following set: { G , G , G , G , P , C } if k = 3 ; { P , K , , G } if k = 4 ; { C } if k = 5 ,where K , is a star graph with four vertices and G , G , G , and G are the graphs givenin Figure 3.Proof. Let D be a k -partite tournament whose competition graph is connected andtriangle-free for some k ≥
2. Then, k ∈ { , , } by Proposition 2.18 and Lemma 2.11. If k = 3, then C ( D ) is isomorphic to a graph in { G , G , G , G , P , C } by Lemma 2.9. If k = 5, then C ( D ) is isomorphic to C by Lemma 2.15.Suppose k = 4. Let V , V , V , and V be the partite sets of D . Without loss of gener-ality, we may assume n ≥ n ≥ n ≥ n where | V i | = n i for each 1 ≤ i ≤
4. Then n ≤ n = n = n = 1 by Lemma 2.17. Case 1. n = 2. Then | V ( D ) | = 5 and | A ( D ) | = 9. Therefore | E ( C ( D )) | ≤ C ( D ) is connected, | E ( C ( D )) | ≥ | E ( C ( D )) | = 4. Therefore C ( D ) is a tree. Thus C ( D ) is isomorphic to a path graph, G , or a star graph. Suppose, tothe contrary, that C ( D ) is a star graph. Then there exists a center v in C ( D ). Since v hasdegree 4 in C ( D ), d + ( v ) ≥
4. Then v ∈ V ∪ V ∪ V and so d + ( v ) = 4 and d − ( v ) = 0. Since C ( D ) is triangle-free, each vertex in D has indegree at most 2. Therefore | A ( D ) | ≤ C ( D ) is isomorphic to P or G . Case 2. n = 1. Then | A ( D ) | = 6 and so, by Lemma 2.2, | E ( C ( D )) | ≤
3. Since C ( D )is connected, | E ( C ( D )) | ≥ | E ( C ( D )) | = 3. Therefore C ( D ) is a path graph or astar graph. If C ( D ) is a path graph, then the complement of C ( D ) is a path graph, whichcontradicts Proposition 2.13. Therefore C ( D ) is a star graph K , .Now we show the “if” part. The competition graph of the 5-tournament given inFigure 7 is C . For the 4-partite tournaments D , D , and D given in Figure 8, itis easy to check that C ( D ) ∼ = K , , C ( D ) ∼ = P , and C ( D ) ∼ = G . Each graph in { G , G , G , G , P , C } is the competition graph of a tripartite tournament by Lemma 2.9.Hence we have shown that the “if” part is true. Let n and n be positive integers such that n ≥ n . Suppose that thereexists an orientation D of K n ,n whose competition graph is triangle-free. Then one of D D Figure 8: Three digraphs D , D , and D which are orientations of K , , , , K , , , , and K , , , , respectively, and whose competition graphs are isomorphic to K , , P , and G ,respectively the following holds: (a) n = 1 ; (b) n = 2 ; (c) n ≤ and n = 3 ; (d) n = 4 and n = 4 .Proof. It is easy to check that | A ( D ) | = n n . Then, by Lemma 2.2, n n ≤ n + n ).Thus ( n − n − ≤ . (5)Then it is easy to check that n ≤
4. If n = 1 or n = 2, then n can be any positivenumber satisfying the inequality n ≥ n . If n = 3, then n ≤
6. If n = 4, then n = 4. Proposition 3.2. (Kim [12]). Let m and n be positive integers such that m ≥ n . Then P m ∪ P n is the competition graph of a bipartite tournament if and only if ( m, n ) is one of (1 , , (2 , , (3 , , and (4 , . Proposition 3.3. (Kim [12]). Let m and n be positive integers greater than or equalto . Then C m ∪ C n is the competition graph of a bipartite tournament if and only if ( m, n ) = (4 , . We give a complete characterization for a triangle-free which is a competition graphof a bipartite tournament. We denote the set of k isolated vertices in a graph by I k . Theorem 3.4.
Let G be a triangle-free graph. Then G is the competition graph of abipartite tournament if and only if G is isomorphic to one of the followings:(a) an empty graph of order at least (b) P with at least one isolated vertex(c) P ∪ P with at least one isolated vertex d) P ∪ P with at least one isolated vertex(e) P ∪ P ∪ P with at least one isolated vertex(f ) P ∪ I (g) P ∪ P (h) P ∪ P (i) P ∪ P ∪ P (j) C ∪ C (k) P ∪ P ∪ P ∪ P .Proof. We first show the “only if” part. Let D be an orientation of K n ,n with n ≥ n whose competition graph is G . Let V = { u , . . . , u n } and V = { v , . . . , v n } be thepartite sets of D . By Proposition 2.18, G is disconnected and there is no edge betweenthe vertices in V and the vertices in V . Since G is triangle-free, by Lemma 3.1, there arefour cases to consider; n = 1; n = 2; n ≤ n = 3; n = 4 and n = 4. Case 1 . n = 1. Then, since each vertex in D has indegree at most 2, G is an emptygraph of order at least 2 or P with at least one isolated vertex. Case 2 . n = 2. Then G has at most two edges between the vertices in V . We denoteby H the subgraph obtained from the subgraph G [ V ] induced by V by removing isolatedvertices in it, if any. Then H is isomorphic to P or P or P ∪ P . Suppose n ≥
5. Then,since each vertex in V has indegree at most 2, each vertex in V has outdegree at least n −
2. Therefore v and v have a common out-neighbor u i in D for some i ∈ { , . . . , n } .Thus v and v are adjacent and u i is isolated in G . Hence G is isomorphic to P ∪ I n − or P ∪ P ∪ I n − or P ∪ P ∪ I n − or P ∪ P ∪ P ∪ I n − .Now we suppose n ≤
4. If H ∼ = P ∪ P , then G [ V ] ∼ = P ∪ P and so G ∼ = P ∪ P ∪ I since the two vertices in V has no common out-neighbor. Suppose H ∼ = P . If G [ V ] hastwo isolated vertices, then at least one of them is a common out-neighbor of v and v and so G ∼ = P ∪ P ∪ I . If G [ V ] has exactly one isolated vertex, then G ∼ = P ∪ P ∪ I or G ∼ = P ∪ I . If H ∼ = G [ V ], then G ∼ = P ∪ I . Suppose H ∼ = P . If G [ V ] has an isolatedvertex, then it must be a common out-neighbor of v and v and so G ∼ = P ∪ P ∪ I . If H ∼ = G [ V ], then G ∼ = P ∪ I . Case 3 . n ≤ n = 3. Suppose, to the contrary, that n ≥
5. Since each vertexin D has indegree at most 2, each vertex in V has outdegree at least n −
2. Since n − > n /
2, any pair of vertices in V has a common out-neighbor in V . Therefore thevertices in V form a triangle, which is a contradiction. Thus n = 3 or n = 4. Subcase 3-1. n = 3. Then, since each vertex in D has indegree at most 2, d + ( v ) ≥ ′ v v v u u u u D ′′ v v v u u u u Figure 9: Digraphs D ′ and D ′′ in the proof of Theorem 3.4.for each vertex v in D . To reach a contradiction, we suppose that G has at least threeisolated vertices. Then at least two isolated vertices belong to the same partite set in D .Without loss of generality, we may assume that V has two isolated vertices u and u .Since | V | = 3, u is also isolated in G . Then, since | V | = 3, each vertex in V has exactlyone out-neighbor by (6) and the out-neighbors of the vertices in V are distinct. Thereforeany pair of the vertices in V has a common out-neighbor in V , which implies that thevertices in V form a triangle in G . Thus G has at most two isolated vertices. Hence G isisomorphic to P ∪ P or P ∪ P ∪ I or P ∪ P ∪ I . Subcase 3-2. n = 4. Then, since each vertex in D has indegree at most 2, d + ( v ) ≥ v in V . We first suppose that there exists a vertex in V which is isolatedin G . Without loss of generality, we may assume v is an isolated vertex in G . Then, since n = 4, d + ( v ) = 2 by (7). Without loss of generality, we may assume N + ( v ) = { u , u } .Then, since v is isolated in G , N + ( v ) = N + ( v ) = { u , u } by (7). Therefore G [ V ]is isomorphic to I ∪ P and G [ V ] is isomorphic to P ∪ P . Thus G is isomorphic to I ∪ P ∪ P ∪ P .Now we suppose that each vertex in V is not isolated in G . Then G [ V ] is isomorphicto P . Without loss of generality, we may assume that G [ V ] is the path v v v . Then D contains a subdigraph isomorphic to D ′ given in Figure 9. We may assume that D ′ itself isa subdigraph of D . Then, by (7), N + ( v ) ∩ { u , u } 6 = ∅ and N + ( v ) ∩ { u , u } 6 = ∅ . Since v and v are not adjacent in G , those intersections are disjoint. We may assume that N + ( v ) ∩ { u , u } = { u } and N + ( v ) ∩ { u , u } = { u } . Then D contains the subdigraph D ′′ given in Figure 9. Then v (resp. v ) is a common out-neighbor of u and u (resp. u and u ). If v is a common out-neighbor of u and u , then G [ V ] is the path u u u u and so G is isomorphic to P ∪ P . If v is not a common out-neighbor of u and u , then G [ V ] is the union of two paths u u and u u , and so G is isomorphic to P ∪ P ∪ P . Case 4 . n = 4 and n = 4. Then | A ( D ) | = 16. Noting that | V ( D ) | = 8 and eachvertex has indegree at most 2, we have d − ( v ) = 2 (8)14or each vertex v in D . Then, for each vertex v in D , d + ( v ) = 2 (9)since v is adjacent to four vertices in D . Subcase 4-1. | E ( G [ V ]) | ≥
4. Then | E ( G [ V ]) | = 4 by (8) and G [ V ] is isomorphic to C since G has no triangle. Without loss of generality, we may assume G [ V ] = u u u u u .Without loss of generality, we may assume that N − ( v ) = { u , u } , N − ( v ) = { u , u } , N − ( v ) = { u , u } , and N − ( v ) = { u , u } by (8). Therefore all arcs in D are determinedand so G [ V ] is a 4-cycle v v v v v . Thus G is isomorphic to C ∪ C . Subcase 4-2. | E ( G [ V ]) | ≤
3. Since | V | = 4, there exists a pair of vertices in V which shares the same in-neighborhood by (8). Without loss of generality, we may assume N − ( v ) = N − ( v ) = { u , u } . Then N + ( u ) = N + ( u ) = { v , v } by (9). Therefore N + ( u ) = N + ( u ) = { v , v } by (8) and (9). Then N − ( u ) = N − ( u ) = { v , v } . Thus itis easy to check that G is isomorphic to P ∪ P ∪ P ∪ P . Hence we have shown that the“only if” part is true.To show the “if” part, we fix a positive integer k . Let D be a bipartite tournamentwith the partite sets { u , . . . , u k } and { v } , and the arc set A ( D ) = { ( v, u i ) | ≤ i ≤ k } (see the digraph D given in Figure 10 for an illustration). Then C ( D ) is an empty graphof order k + 1.Let D be a bipartite tournament with the partite sets { u , . . . , u k +1 } and { v } , andthe arc set A ( D ) = { ( u , v ) , ( u , v ) } ∪ { ( v, u i ) | < i ≤ k + 1 } (see the digraph D given in Figure 10 for an illustration). Then C ( D ) is the path u u with k isolated vertices.Let D be a bipartite tournament with the partite sets { u , . . . , u k +2 } and { v , v } ,and the arc set A ( D ) = { ( u i , v j ) | ≤ i, j ≤ } ∪ { ( v i , u j ) | ≤ i ≤ , ≤ j ≤ k + 2 } (see the digraph D given in Figure 10 for an illustration). Then C ( D ) is the paths u u and v v with k isolated vertices.Let D be a bipartite tournament with the partite sets { u , . . . , u k +3 } and { v , v } ,and the arc set A ( D ) = { ( u , v ) , ( u , v ) , ( u , v ) , ( u , v ) , ( v , u ) , ( v , u ) }∪ { ( v i , u j ) | ≤ i ≤ , ≤ j ≤ k + 3 } (see the digraph D given in Figure 10 for an illustration). Then C ( D ) is the paths u u u and v v with k isolated vertices. 15et D be a bipartite tournament with the partite sets { u , . . . , u k +4 } and { v , v } ,and the arc set A ( D ) = { ( u i , v ) , ( v , u i ) | i = 1 , } ∪ { ( u i , v ) , ( v , u i ) | i = 3 , }∪ { ( v i , u j ) | ≤ i ≤ , ≤ j ≤ k + 4 } (see the digraph D given in Figure 10 for an illustration). Then C ( D ) is the paths u u , u u , and v v with k isolated vertices.The competition graph of the digraph D given in Figure 10 is isomorphic to P ∪ I . By Proposition 3.2, there exists a bipartite tournament whose competition graph isisomorphic to P ∪ P . By the way, bipartite tournaments whose competition graphs areisomorphic to P ∪ P and P ∪ P ∪ P , respectively, were constructed in the subcase3-2. By Proposition 3.3, there exists a bipartite tournament whose competition graph isisomorphic to C ∪ C . It is easy to check that the competition graph of the bipartitetournament D given in Figure 10 is the disjoint union of the paths u u , u u , v v , and v v . Hence we have shown that the “if” part is true. k -partite tournaments for k ≥ Lemma 3.5.
If the competition graph of a k -partite tournament is triangle-free and dis-connected for some positive integer k ≥ , then k = 3 or k = 4 . By Lemma 3.5, it is sufficient to consider tripartite tournaments and 4-partite tour-naments for studying disconnected triangle-free competition graphs of multipartite tour-naments.
Lemma 3.6.
Let D be a multipartite tournament whose competition graph is triangle-free.Suppose that a vertex v is contained in a partite set X of D . Then | V ( D ) |−| X |− ≤ d + ( v ) .Proof. Since C ( D ) is triangle-free, d − ( v ) ≤
2. Then, since D is a multipartite tournament, d − ( v ) = | V ( D ) | − | X | − d + ( v ) and so | V ( D ) | − | X | − ≤ d + ( v ).The following is immediately true by Lemma 3.6. Corollary 3.7.
If the competition graph of a -partite tournament D is triangle-free, theneach vertex has outdegree at least in D . Lemma 3.8.
Let D be a multipartite tournament whose competition graph is triangle-free.If m is the number of vertices of indegree in D , then | V ( D ) | − | A ( D ) | ≥ m . vu u k D vu u u k +1 D v v u u u u k +2 D v v u u u u u k +3 D v v u u u u u u k +4 D v u u u v D v v v v u u u u Figure 10: Bipartite tournaments in the proof of Theorem 3.417 roof.
Let m be the number of vertices of indegree 1 in D . Since C ( D ) is triangle-free,each vertex has indegree at most 2. Therefore | A ( D ) | = X v ∈ V ( D ) d − ( v ) ≤ | V ( D ) | − m ) + m = 2 | V ( D ) | − m. Now we are ready to introduce one of our main theorems.
Theorem 3.9.
Let G be a disconnected and triangle-free graph. Then G is the competitiongraph of a -partite tournament if and only if G is isomorphic to P ∪ P or P ∪ I .Proof. We first show the “only if” part. Let D be an orientation of K n ,n ,n ,n with n ≥ · · · ≥ n whose competition graph C ( D ) is disconnected and triangle-free. Let V , . . . , V be the partite sets of D with | V i | = n i for each 1 ≤ i ≤
4. Then n ≤ n = n = n = 1 by Lemma 2.17. Case 1 . n = 2. Then | A ( D ) | = 9. Therefore | E ( C ( D )) | ≤ l and m be the number of isolated vertices in C ( D ) and the number of vertices of indegree 1 in D , respectively. By Corollary 3.7, each vertex has outdegree at least 1, so each isolatedvertex in C ( D ) has an out-neighbor in D . Yet, since each out-neighbor of an isolatedvertex has indegree 1, l ≤ m . By Lemma 3.8, m ≤
1. Therefore l ≤ . Suppose, to the contrary, that l = 1. Then m = 1. Let w be the isolated vertex in C ( D ). Since each vertex in N + ( w ) has indegree 1, d + ( w ) ≤
1. Since each vertex hasoutdegree at least 1, d + ( w ) = 1. Since C ( D ) is triangle-free, d − ( w ) ≤ w ∈ V .Let V = { v , w } , V = { v } , V = { v } and V = { v } . Without loss of generality, we mayassume N + ( w ) = { v } . Then N − ( w ) = { v , v } . Since w is an isolated vertex in C ( D ), N − ( v ) = { w } and so N + ( v ) = { v , v , v } .Without loss of generality, we may assume ( v , v ) ∈ A ( D ). Then, since d − ( v ) ≤ N − ( v ) = { v , v } , and so ( v , v ) ∈ A ( D ). Therefore N − ( v ) = { v , v } . Thus { v , v , v } forms a triangle in C ( D ), which is a contradiction. Hence l = 0. Since C ( D ) is disconnected and | V ( D ) | = 5, C ( D ) has two components each of which has 2and 3 vertices, respectively. Then, one of the components must be P . On the other hand,since C ( D ) is triangle-free, the other component is isomorphic to P . Therefore C ( D ) isisomorphic to P ∪ P . Case 2 . n = 1. Then D is an orientation of K , , , , which is a tournament. ByProposition 2.12, | E ( C ( D )) | ≥
2. By the way, since | A ( D ) | = 6, | E ( C ( D )) | ≤ D Figure 11: The digraphs D and D in the proof of Theorem 3.9Lemma 2.2. Therefore | E ( C ( D )) | = 2 or 3. Thus C ( D ) has exactly two components andso is isomorphic to I ∪ P or P ∪ P . If C ( D ) is isomorphic to P ∪ P , then D has twovertices a and b such that d − ( a ) = d − ( b ) = 2 and N − ( a ) ∩ N − ( b ) = ∅ , which is impossiblefor a digraph of order four. Therefore C ( D ) is isomorphic to I ∪ P .To show the “if” part, we consider the 4-partite tournaments D and D given inFigure 11. It is easy to check that C ( D ) ∼ = P ∪ P , and C ( D ) ∼ = I ∪ P . Therefore wehave shown that the “if” part is true.By Lemma 3.5, it only remains to characterize disconnected and triangle-free compe-tition graphs of tripartite tournaments. The following theorem lists all the disconnectedand triangle-free competition graphs of tripartite tournaments Theorem 3.10.
Let G be a disconnected and triangle-free graph. Then G is the com-petition graph of a tripartite tournament if and only if G is isomorphic to one of thefollowings:(a) an empty graph of order (b) P with at least one isolated vertex(c) P with at least one isolated vertex(d) P with at least one isolated vertex(e) K , ∪ I (f ) K , ∪ P (g) P ∪ P (h) P ∪ P with at least one isolated vertex(i) P ∪ P with or without isolated vertices j) P ∪ P ∪ P .Proof. To show the “only if” part, suppose that D is an orientation of K n ,n ,n whose com-petition graph is disconnected and triangle-free where n , n , and n are positive integerssuch that n ≥ n ≥ n . Then, by Lemma 2.4, ( n , n , n ) ∈ A ∪{ (2 , , , (2 , , , (3 , , } where A = { ( m, , | m is a positive integer } . Case 1 . ( n , n , n ) ∈ A . Let V := { x , . . . , x n } , V := { y } , and V := { z } be thepartite sets of D . Without loss of generality, we may assume( y, z ) ∈ A ( D ) . Suppose that C ( D ) is an empty graph. Then y is an isolated vertex in C ( D ). If d + ( y ) ≥ V should be an out-neighbor of y and so y is adjacent to one of the vertexand z in C ( D ), which is a contradiction. Therefore d + ( y ) = 1 and so N + ( y ) = { z } . If n ≥
2, then the in-neighbors of y are adjacent in C ( D ), which is a contradiction. Therefore n = 1. Thus C ( D ) is isomorphic to three isolated vertices.Now we suppose that C ( D ) is not an empty graph. Then y is the only possible neighborof z in C ( D ). Moreover, z and at most one vertex in V are the only possible neighbors of y in C ( D ). Since z has indegree at most 2 in D , y is the only possible common out-neighborof two vertices in V . Therefore only one pair of vertices in V is possibly adjacent in C ( D ).Hence the following are the only possible graphs isomorphic to C ( D ): • an empty graph of order 3 • P with at least one isolated vertex • P ∪ P with at least one isolated vertex • P with at least one isolated vertex • P ∪ P with at least one isolated vertex • P with at least one isolated vertex. Case 2 . ( n , n , n ) = (2 , , C ( D ) has l isolatedvertices for some positive integer l ≥
2. By Lemma 3.6, each vertex in D has outdegreeat least 1. By the way, since 2 | V ( D ) | − | A ( D ) | = 2, D has at most two vertices ofindegree 1 by Lemma 3.8. Then, since each out-neighbor of isolated vertices has indegree1, l ≤ l = 2. Let u and u be the isolated vertices in C ( D ). Then, d + ( u ) = d + ( u ) = 1. Suppose that u and u are contained in distinct partite sets in D . Withoutloss of generality, we may assume ( u , u ) ∈ A ( D ). Then d − ( u ) = 1. However, since d + ( u ) = 1, d + ( u ) + d − ( u ) = 2 = | V ( D ) \ X | where X is a partite set containing u , which is impossible. Therefore u and u belong to the same partite set of D . Then { u , u } ⊆ V or { u , u } ⊆ V . Without loss of generality, we may assume { u , u } ⊆ V .20hen V = { u , u } . Let v and v be the out-neighbors of u and u , respectively. Then N − ( v ) = { u } and N − ( v ) = { u } . Therefore there is no arc between v and v , and so v and v belong to the same partite set V . Then V = { v , v } . Since d − ( v ) = d − ( v ) = 1,the vertex z , in the remaining partite set of D , is a common out-neighbor of v and v .Then N − ( z ) = { v , v } and so N + ( z ) = { u , u } . Therefore u (resp. u ) is a commonout-neighbor of v (resp. v ) and z . Thus { v , v , z } forms a triangle in C ( D ), which is acontradiction. Hence C ( D ) has at most one isolated vertex. Therefore C ( D ) has at mostthree components.Since | A ( D ) | = 8, | E ( C ( D )) | ≤ | E ( C ( D )) | ≤
1, then C ( D ) hasat least 2 isolated vertices, which is a contradiction. Therefore 2 ≤ | E ( C ( D )) | ≤
4. If | E ( C ( D )) | = 2, then, C ( D ) is isomorphic to P ∪ P ∪ I . If | E ( C ( D )) | = 3, then C ( D ) isisomorphic to P ∪ I or K , ∪ I or P ∪ P . Suppose | E ( C ( D )) | = 4. Since | A ( D ) | = 8 and | E ( C ( D )) | = 4, there exists a vertex w of indegree 0 in D and each vertex in V ( D ) \ { w } has indegree 2. Moreover, for distinct vertices a and b of indegree 2, N − ( a ) = N − ( b ).Then, since w has outdegree at least 3, w has degree at least 3 in C ( D ). Since C ( D ) isdisconnected and triangle-free, K , ∪ I is the only possible graph isomorphic to C ( D ),which contradicts the assumption that | E ( C ( D )) | = 4. Therefore C ( D ) is isomorphic toone of P ∪ P ∪ I or P ∪ I or K , ∪ I or P ∪ P . Case 3 . ( n , n , n ) = (2 , , | A ( D ) | = 12. Since | V ( D ) | = 6 and each vertexof D has indegree at most 2, d − ( v ) = 2 (10)and so d + ( v ) = 2 (11)for each vertex v in D . Therefore C ( D ) has no isolated vertex. Thus each component of C ( D ) contains at least two vertices. Let t be the number of the components of C ( D ).Then t ≤
3. Since C ( D ) is disconnected, t = 2 or t = 3. Suppose, to the contrary, that t = 2. Then, since C ( D ) is triangle-free, it is easy to check that | E ( C ( D )) | ≤
5. Since | V ( D ) | = 6, there exist at least two vertices a and a sharing the same in-neighborhoodby (10). Then a and a are contained in the same partite set and form a component in C ( D ) by Lemma 2.16. Then the other component must contain four vertices. Withoutloss of generality, we may assume V := { a , a } is a partite set of D . Let { b , b } = N − ( a ) = N − ( a ) for some vertices b and b in D . Then N + ( b ) = N + ( b ) = { a , a } by (11). Therefore b and b are contained in the same partite sets and { b , b } forms acomponent in C ( D ) by Lemma 2.16, which is a contradiction. Therefore t = 2 and so t = 3. Then, since each component of C ( D ) contains at least two vertices, C ( D ) must beisomorphic to P ∪ P ∪ P . Case 4 . ( n , n , n ) = (3 , , | A ( D ) | = 11. Since each vertex has indegree atmost 2 in D , one vertex has indegree 1 and the other vertices have indegree 2. Let V , V , and V be the partite sets of D satisfying | V | = 3, | V | = 2, and | V | = 1 and v ∗ bethe vertex of indegree 1 in D . Then, since ( n , n , n ) = (3 , , V has21utdegree at least 1 and each vertex in V ∪ V has outdegree at least 2.Suppose, to the contrary, that C ( D ) has an isolated vertex u . Since v ∗ is the onlyvertex of indegree 1, N + ( u ) = { v ∗ } . Therefore u ∈ V . Then v ∗ ∈ V ∪ V . Suppose v ∗ ∈ V . Since d − ( v ∗ ) = 1, N + ( v ∗ ) = ( V ∪ V ) \ { u } . Moreover, since V ⊂ N + ( v ∗ ) andeach vertex in D other than v ∗ has indegree 2, each vertex in V has an out-neighborin V \ { u } . Therefore each vertex in V is adjacent to v ∗ in C ( D ). By the way, since N + ( u ) = { v ∗ } , u is a common out-neighbor of the two vertices in V . Therefore V ∪ { v ∗ } forms a triangle in C ( D ), which is a contradiction. Thus v ∗ ∈ V . Let V = { u, x , x } , V = { v ∗ , y } , and V = { z } . Then N + ( v ∗ ) = { x , x , z } and N − ( u ) = { y, z } , so y and z are adjacent in C ( D ). Since d − ( y ) = 2, d + ( y ) = 2. Thus N + ( v ∗ ) ∩ N + ( y ) = ∅ and so v ∗ and y are adjacent in C ( D ). Since d − ( z ) = 2 and v ∗ ∈ N − ( z ), { x , x } 6⊂ N − ( z ).Therefore N + ( z ) ∩ { x , x } 6 = ∅ and so N + ( v ∗ ) ∩ N + ( z ) = ∅ . Thus v ∗ and z are adjacent in C ( D ). Hence { v ∗ , y, z } forms a triangle in C ( D ), whichis a contradiction. Consequently, we have shown that C ( D ) has no isolated vertex, soeach component in C ( D ) has size at least two. Then, since | V ( D ) | = 6, C ( D ) has twoor three components. If C ( D ) has three components, then C ( D ) must be isomorphic to P ∪ P ∪ P .Now we suppose that C ( D ) has two components. Then it is easy to check that 4 ≤| E ( C ( D )) | ≤ C ( D ) is triangle-free and has no isolated vertices. Suppose, to thecontrary, that | E ( C ( D )) | = 5. Then, since the vertices in C ( D ) except the five verticesof indegree 2 have indegree less than 2, no pair of adjacent vertices in C ( D ) have twodistinct common out-neighbors in D . Moreover, C ( D ) must be isomorphic to P ∪ C where C is a cycle of length 4. Since only one vertex has indegree 1 and the othervertices have indegree 2 in D , there exist two vertices a and b in V which have outdegree1 in D . Then a and b have at most degree 1 in C ( D ), so { a, b } is a path componentin C ( D ). Therefore a and b have a common out-neighbor c , in D . Then a and b arecommon out-neighbors of the two vertices in V ∪ V \ { c } , which is a contradiction. Hence | E ( C ( D )) | 6 = 5 and so | E ( C ( D )) | = 4. Since five vertices have indegree 2 in D , thereexists a pair of adjacent vertices which have two common out-neighbors. Therefore thereexist two vertices whose in-neighbors are the same. Then, since each vertex has outdegreeat least 1 by Lemma 3.6, the two vertices form a component in C ( D ) by Lemma 2.16.Thus C ( D ) must be isomorphic to P ∪ K , or P ∪ P . Hence we have shown that the“only if” part is true. 22ow we show the “if” part. The competition graph of the digraph D given in Fig-ure 12 is an empty graph of order 3.Now we fix a positive integer k . Let D be a tripartite tournament with the partitesets { w , . . . , w k } , { x } , { y } and the arc set A ( D ) = { ( x, y ) } ∪ { ( x, w i ) , ( y, w i ) | ≤ i ≤ k } (see the digraph D given in Figure 12 for an illustration). Then C ( D ) is isomorphicto P with k isolated vertices.Let D be a tripartite tournament with the partite sets { v, w , . . . , w k } , { x } , { y } andthe arc set A ( D ) = { ( v, x ) , ( v, y ) , ( x, y ) } ∪ { ( x, w i ) , ( y, w i ) | ≤ i ≤ k } (see the digraph D given in Figure 12 for an illustration). Then C ( D ) is the path vxy together with k isolated vertices.Let D be a tripartite tournament with the partite sets { v , v , w , . . . , w k } , { x } , { y } and the arc set A ( D ) = { ( v , x ) , ( v , x ) , ( v , y ) , ( x, y ) , ( y, v ) } ∪ { ( x, w i ) , ( y, w i ) | ≤ i ≤ k } (see the digraph D given in Figure 12 for an illustration). Then C ( D ) is the path v v xy with k isolated vertices.The competition graphs of the digraphs D , D and D given in Figure 12 areisomorphic to K , ∪ I , K , ∪ P , and P ∪ P , respectively.Let D be a tripartite tournament with the partite sets { v , v , w , . . . , w k } , { x } , { y } and the arc set A ( D ) = { ( v , x ) , ( v , x ) , ( x, y ) , ( y, v ) , ( y, v ) } ∪ { ( x, w i ) , ( y, w i ) | ≤ i ≤ k } (see the digraph D given in Figure 12 for an illustration). Then C ( D ) is isomorphicto P ∪ P with k isolated vertices.Let D be a tripartite tournament with the partite sets { v , v , v , w , . . . , w k } , { x } , { y } and the arc set A ( D ) = { ( v , x ) , ( v , y ) , ( v , x ) , ( x, v ) , ( x, y ) , ( y, v ) , ( y, v ) }∪{ ( x, w i ) , ( y, w i ) | ≤ i ≤ k } (see the digraph D given in Figure 12 for an illustration). Then C ( D ) is isomorphicto P ∪ P with k isolated vertices.The competition graphs of the digraphs D and D given in Figure 12 are isomorphicto P ∪ P and P ∪ P ∪ P , respectively. Hence we have shown that the “if” part is true.23 D yxw w k D yxw v w k D yxw v v w k D D D D yxv v w w k D yxw v v v w k D D Figure 12: The digraphs in the proof of Theorem 3.1024
Closing remarks
In this paper, we completely identified the triangle-free graphs which are the competitiongraphs of k -partite tournaments for k ≥
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