The (1, 2)-step competition graph of a hypertournament

Ruijuan Li; Xiaoting An; Xinhong Zhang

doi:10.1515/math-2021-0047

Open Access Published by De Gruyter Open Access June 10, 2021

The (1, 2)-step competition graph of a hypertournament

Ruijuan Li , Xiaoting An and Xinhong Zhang

From the journal Open Mathematics

https://doi.org/10.1515/math-2021-0047

Abstract

In 2011, Factor and Merz [Discrete Appl. Math. 159 (2011), 100–103] defined the ( 1 , 2 ) -step competition graph of a digraph. Given a digraph D = ( V , A ) , the ( 1 , 2 ) -step competition graph of D, denoted C 1 , 2 ( D ) , is a graph on V ( D ) , where x y ∈ E ( C 1 , 2 ( D ) ) if and only if there exists a vertex z ≠ x , y such that either d D − y ( x , z ) = 1 and d D − x ( y , z ) ≤ 2 or d D − x ( y , z ) = 1 and d D − y ( x , z ) ≤ 2 . They also characterized the (1, 2)-step competition graphs of tournaments and extended some results to the ( i , j ) -step competition graphs of tournaments. In this paper, the definition of the (1, 2)-step competition graph of a digraph is generalized to a hypertournament and the (1, 2)-step competition graph of a k-hypertournament is characterized. Also, the results are extended to ( i , j ) -step competition graphs of k-hypertournaments.

Keywords: k-hypertournament; (1, 2)-step competition graph; (i, j)-step competition graph

MSC 2010: 05C65; 05C12; 05C20

1 Terminology and introduction

Let G = ( V , E ) be an undirected graph, or a graph for short. V ( G ) and E ( G ) are the vertex set and the edge set of G, respectively. The complement G c of a graph G is the graph with vertex set V ( G ) in which two vertices are adjacent if and only if they are not adjacent in G. Let G 1 and G 2 be two graphs. The union of G 1 and G 2 , denoted by G 1 ∪ G 2 , is the graph with vertex set V ( G 1 ) ∪ V ( G 2 ) and edge set E ( G 1 ) ∪ E ( G 2 ) .

Let D = ( V , A ) be a directed graph, or a digraph for short. V ( D ) and A ( D ) are the vertex set and the arc set of D, respectively. Let i ≥ 1 , j ≥ 1 . The ( i , j ) -step competition graph of D, denoted C i , j ( D ) , is a graph on V ( D ) , where x y ∈ E ( C i , j ( D ) ) if and only if there exists a vertex z ≠ x , y such that either d D − y ( x , z ) ≤ i and d D − x ( y , z ) ≤ j or d D − x ( y , z ) ≤ i and d D − y ( x , z ) ≤ j . When ( i , j ) = ( 1 , 1 ) , it is also called the competition graph of D.

The notion of competition graph was introduced by Cohen [1] as a means of determining the smallest dimension of ecological phase space. In recent years, many researchers investigated m-step competition graphs of some special digraphs and the competition numbers of some graphs etc. (see [2–4]). Particularly, in 1998, Fisher, Lundgren, Merz and Reid [5] studied the domination graphs and competition graphs of a tournament. Recall that a tournament is an orientation of a complete graph. In 2011, Factor and Merz [6] gave the definition of the ( i , j ) -step competition graph of a digraph. They also characterized the ( 1 , 2 ) -step competition graph of a tournament and extended some results to the ( i , j ) -step competition graph of a tournament. They proved the following theorems related to this paper.

Theorem 1.1

[6] A graph G on n ≥ 5 vertices is the ( 1 , 2 ) -step competition graph of some strong tournament if and only if G is K n , K n − E ( P 2 ) , or K n − E ( P 3 ) .

Theorem 1.2

[6] G, a graph on n vertices, is the ( 1 , 2 ) -step competition graph of some tournament if and only if G is one of the following graphs:

K n , where n ≠ 2 , 3 , 4 ;
K n − 1 ∪ K 1 , where n > 1 ;
K n − E ( P 3 ) , where n > 2 ;
K n − E ( P 2 ) , where n ≠ 1 , 4 , or
K n − E ( K 3 ) , where n ≥ 3 .

Theorem 1.3

[6] If T is a tournament with n vertices, i ≥ 1 and j ≥ 2 , then C i , j ( T ) = C 1 , 2 ( T ) .

Given two integers n and k , n ≥ k > 1 , a k-hypertournament T on n vertices is a pair ( V , A ) , where V is a set of vertices, ∣ V ∣ = n and A is a set of k-tuples of vertices, called arcs, so that for any k-subset S of V , A contains exactly one of the k ! k-tuples whose entries belong to S . As usual, we use V ( T ) and A ( T ) to denote the vertex set and the arc set of T, respectively. Clearly, a 2-hypertournament is merely a tournament. When k = n , the hypertournament has only one arc and it does not have much significance to study. Thus, in what follows, we consider 3 ≤ k ≤ n − 1 .

Let T = ( V , A ) be a k-hypertournament on n vertices. For an arc a of T, T − a denotes a hyperdigraph obtained from T by removing the arc a and a ¯ denotes the set of vertices contained in a . If v i , v j ∈ a ¯ and v i precedes v j in a , we say that v i dominates v j in a . We also say the vertex v j is an out-neighbour of v i and use the following notation:

N T + ( v i ) = { v j ∈ V ⧹ { v i } : v i precedes v j in some arc a ∈ A ( T ) } .

We will omit the subscript T if the k-hypertournament T is known from the context.

A path P in a k-hypertournament T is a sequence v 1 a 1 v 2 a 2 v 3 ⋯ v t − 1 a t − 1 v t of distinct vertices v 1 , v 2 , … , v t , t ≥ 1 and distinct arcs a 1 , a 2 , … , a t − 1 such that v i precedes v i + 1 in a i , 1 ≤ i ≤ t − 1 . Meanwhile, let the vertex set V ( P ) = { v 1 , v 2 , … , v t } and the arc set A ( P ) = { a 1 , a 2 , … , a t − 1 } . The length of a path P is the number of its arcs, denoted ℓ ( P ) . A path from x to y is an ( x , y ) -path. The k-hypertournament T is called strong if T has an ( x , y ) -path for every pair x , y of distinct vertices in T.

A k-hypertournament T is said to be transitive if its vertices are labeled v 1 , v 2 , … , v n in such an order so that i < j if and only if v i precedes v j in each arc containing v i and v j .

Now we generalize the ( 1 , 2 ) -step competition graph of a digraph to a k-hypertournament.

Definition 1.4

The ( i , j ) -step competition graph of a k-hypertournament T with i ≥ 1 and j ≥ 1 , denoted C i , j ( T ) , is a graph on V ( T ) , where x y ∈ E ( C i , j ( T ) ) if and only if there exist a vertex z ≠ x , y and an ( x , z ) -path P and a ( y , z ) -path Q satisfying the following:

y ∉ V ( P ) , x ∉ V ( Q ) ;
ℓ ( P ) ≤ i and ℓ ( Q ) ≤ j , or ℓ ( Q ) ≤ i and ℓ ( P ) ≤ j ;
P and Q are arc-disjoint.

If x y ∈ E ( C i , j ( T ) ) , we say that x and y ( i , j ) -step compete. In particular, we say that x and y compete if ℓ ( P ) = 1 and ℓ ( Q ) = 1 . C 1 , 1 ( T ) is also called the competition graph of the k-hypertournament T. Clearly, when k = 2 , T is a tournament and C i , j ( T ) is the ( i , j ) -step competition graph of T.

The k-hypertournaments form one of the most interesting class of digraphs. For the class of k-hypertournaments, the popular topics are the Hamiltonicity and vertex-pancyclicity (see [7–10]). Besides, some researchers investigated the degree sequences and score sequences of k-hypertournaments (see [11,12]). Recently, the H -force set of a hypertournament was also studied (see [13]). In this paper, we study the ( 1 , 2 ) -step competition graph of a k-hypertournament and extend Theorems 1.1–1.3 to k-hypertournaments.

In Sections 2 and 3, useful lemmas are provided to make the proof of the main results easier. In Sections 4 and 5, the ( 1 , 2 ) -step competition graph of a (strong) k-hypertournament is characterized. In Section 6, the main results are extended to the ( i , j ) -step competition graph of a k-hypertournament.

2 The missing edges of C 1 , 2 ( T )

Let T = ( V , A ) be a k-hypertournament. For a pair of distinct vertices x and y in T, A T ( x , y ) denotes the set of all arcs of T in which x precedes y , A T { x , y } denotes the set of all arcs containing x , y in T and A T ∗ { x , y } denotes the set of all arcs containing x , y in T and in which neither x nor y is the last entry.

Lemma 2.1

Let T be a k-hypertournament with n vertices, where 3 ≤ k ≤ n − 1 . Then x y ∉ E ( C 1 , 2 ( T ) ) if and only if one of the following holds:

N + ( x ) = ∅ ;
N + ( y ) = ∅ ;
N + ( x ) = { y } ;
N + ( y ) = { x } ;
A T ∗ { x , y } contains exactly an arc a , and N T − a + ( x ) ⊆ { y } , N T − a + ( y ) ⊆ { x } .

Proof

First, we show the “if” part. Clearly, if one of (a)–(d) holds, we have x y ∉ E ( C 1 , 2 ( T ) ) . Now we assume that the argument (e) holds. Since A T ∗ { x , y } contains exactly an arc a , and N T − a + ( x ) ⊆ { y } , N T − a + ( y ) ⊆ { x } , we have to use the unique arc a to obtain the out-neighbour except y of x and the out-neighbour except x of y . So x and y are impossible to ( 1 , 2 ) -step compete and hence x y ∉ E ( C 1 , 2 ( T ) ) .

Now we show the “only if” part. Assume that x y ∉ E ( C 1 , 2 ( T ) ) . Also, assume that x and y do not satisfy (a)–(d). That means N + ( x ) ⧹ { y } ≠ ∅ , N + ( y ) ⧹ { x } ≠ ∅ . Now we show that x and y satisfy ( e ) . Suppose A T ∗ { x , y } consists of at least two arcs, say a 1 , a 2 ∈ A T ∗ { x , y } . Let w i be the last entry of a i for i = 1 , 2 . If w 1 = w 2 , then x and y compete, a contradiction. So assume w 1 ≠ w 2 . Note that n − 2 k − 2 − 2 ≤ ∣ A T { w 1 , w 2 } ⧹ { a 1 , a 2 } ∣ ≤ n − 2 k − 2 . For ( n , k ) ≠ ( 4 , 3 ) , we have ∣ A T { w 1 , w 2 } ⧹ { a 1 , a 2 } ∣ ≥ 1 . For ( n , k ) = ( 4 , 3 ) , since both a 1 and a 2 contain x , y and a 1 ≠ a 2 , we have a 1 , a 2 ∉ A T { w 1 , w 2 } and hence ∣ A T { w 1 , w 2 } ⧹ { a 1 , a 2 } ∣ = ∣ A T { w 1 , w 2 } ∣ = 2 . Let a 3 ∈ A T { w 1 , w 2 } ⧹ { a 1 , a 2 } . W.l.o.g., a 3 ∈ A T ( w 1 , w 2 ) . Then P = x a 1 w 1 a 3 w 2 and Q = y a 2 w 2 are the paths such that x and y ( 1 , 2 )-step compete, a contradiction. So A T ∗ { x , y } consists of exactly an arc a . Suppose N T − a + ( x ) ⧹ { y } ≠ ∅ . Then there exists an arc b distinct from a such that x has an out-neighbour distinct from y in b . Similar to the aforementioned proof, whether or not the last entries of a and b are same, we always have x y ∈ E ( C 1 , 2 ( T ) ) , a contradiction. So N T − a + ( x ) ⊆ { y } . Similarly, N T − a + ( y ) ⊆ { x } . Thus, A T ∗ { x , y } contains exactly an arc a , and N T − a + ( x ) ⊆ { y } , N T − a + ( y ) ⊆ { x } .

The lemma holds.□

By the proof of Lemma 2.1, we obtain the following result.

Corollary 2.2

Let T be a strong k-hypertournament with n vertices, where 3 ≤ k ≤ n − 1 . Then x y ∉ E ( C 1 , 2 ( T ) ) if and only if one of the following holds:

N + ( x ) = { y } ;
N + ( y ) = { x } ;
A T ∗ { x , y } contains exactly an arc a , and N T − a + ( x ) ⊆ { y } , N T − a + ( y ) ⊆ { x } .

3 The forbidden subgraphs of ( C 1 , 2 ( T ) ) c

Lemma 3.1

Let G on n vertices be the ( 1 , 2 ) -step competition graph of some k-hypertournament T, where 3 ≤ k ≤ n − 1 . Then the complement G c of G does not contain a pair of disjoint edges.

Proof

Suppose the complement G c of G contains a pair of disjoint edges, say x y and z w . So x y , z w ∉ E ( G ) and x , y , z , w are distinct. By Lemma 2.1, we have x y satisfies one of the cases (a)–(e) and z w satisfies one of the cases (a)–(e).

Suppose that at least one of x y and z w satisfies one of the cases (a)–(d). W.l.o.g., we assume that N + ( x ) ⊆ { y } . Then it must be true that the vertex z dominates x in each arc containing x , z but not containing w . Meanwhile, it must be true that the vertex w dominates x in each arc containing x , w but not containing z . So z and w compete and hence z w ∈ E ( C 1 , 2 ( T ) ) = E ( G ) , a contradiction. Thus, both x y and z w satisfy ( e ) .

However, since A T ∗ { x , y } contains exactly an arc a , and N T − a + ( x ) ⊆ { y } , N T − a + ( y ) ⊆ { x } , we have the vertex x must be the last entry in each arc containing x , z , w but not containing y and the vertex y must be the last entry in each arc containing y , z , w but not containing x . Thus, A T ∗ { z , w } contains at least two arcs, which contradicts the fact that z w satisfies ( e ) .

The lemma holds.□

Lemma 3.2

Let G on n vertices be the ( 1 , 2 ) -step competition graph of some k-hypertournament T, where 3 ≤ k ≤ n − 1 . Then the complement G c of G does not contain 3-cycle.

Proof

Suppose to the contrary that the complement G c of G contains 3-cycle, say x y z x . So x y , x z , y z ∉ E ( G ) and x , y , z are distinct. By Lemma 2.1, we have x y , x z and y z satisfy one of the cases (a)–(e), respectively.

Claim 1

None of x y , x z and y z satisfies the case ( a ) or ( b ) .

Proof

Suppose at least one of x y , x z and y z satisfies the case ( a ) or ( b ) . W.l.o.g., we assume that x y satisfies ( a ) , i.e. N + ( x ) = ∅ . Then it must be true that the vertex y dominates x in each arc containing x , y but not containing z . Also, the vertex z dominates x in each arc containing x , z but not containing y . So y and z compete and y z ∈ E ( C 1 , 2 ( T ) ) = E ( G ) , a contradiction. Thus, none of x y , x z and y z satisfies the case ( a ) or ( b ) .□

Claim 2

At most one of x y , x z and y z satisfies the case ( c ) or ( d ) .

Proof

Suppose at least two edges among x y , x z and y z satisfy the case ( c ) or ( d ) . W.l.o.g., we assume that both x y and y z satisfy ( c ) or ( d ) . We consider the following four cases.

Case 1: Both x y and y z satisfy ( c ) . It means that N + ( x ) = { y } , N + ( y ) = { z } . If x z satisfies ( c ) , i.e. N + ( x ) = { z } , it contradicts N + ( x ) = { y } . If x z satisfies ( d ) , i.e. N + ( z ) = { x } , the arcs containing simultaneously x , y , z do not satisfy N + ( x ) = { y } , N + ( y ) = { z } and N + ( z ) = { x } , a contradiction. If x z satisfies ( e ) , i.e. A T ∗ { x , z } contains exactly an arc a , then there exists a vertex w such that w ∈ N + ( x ) . Since N + ( x ) = { y } , we have w = y . So x is the second last entry, y is the last entry and z is any other entry in a . Then the vertex z dominates y in a . Also, the vertex x dominates y in each arc containing x , y but not containing z . So x z ∈ E ( C 1 , 2 ( T ) ) = E ( G ) , a contradiction.

Case 2: Both x y and y z satisfy ( d ) . It means that N + ( y ) = { x } , N + ( z ) = { y } . Similarly to Case 1, we can also get a contradiction.

Case 3: x y satisfies ( c ) and y z satisfies ( d ) . It means that N + ( x ) = { y } , N + ( z ) = { y } . Now the arcs containing simultaneously x , y , z do not satisfy N + ( x ) = { y } , N + ( z ) = { y } , a contradiction.

Case 4: x y satisfies ( d ) and y z satisfies ( c ) . It means that N + ( y ) = { x } and N + ( y ) = { z } . Then x = z , a contradiction.

Thus, at most one of x y , x z and y z satisfies the case ( c ) or ( d ) .□

Claim 3

At most one of x y , x z and y z satisfies the case ( e ) .

Proof

Suppose at least two edges among x y , x z and y z satisfy the case ( e ) . W.l.o.g., we assume that both x z and y z satisfy ( e ) . From the assumption that x z satisfies ( e ) , we get the vertex y dominates x in each arc containing x , y but not containing z . From the assumption that y z satisfies ( e ) , we get the vertex x dominates y in each arc containing x , y but not containing z . This is a contradiction. Thus, at most one of x y , x z and y z satisfies the case ( e ) .□

By Claims 1–3, it is impossible that x y , x z , y z ∉ E ( G ) hold simultaneously. Thus, the complement G c of G does not contain 3-cycle. The lemma holds.□

Lemma 3.3

Let G on n vertices be the ( 1 , 2 ) -step competition graph of some k-hypertournament T, where 3 ≤ k ≤ n − 1 . Then the complement G c of G does not contain K 1 , 3 , unless G = K n − 1 ∪ K 1 .

Proof

Let T be a k-hypertournament on n vertices, where 3 ≤ k ≤ n − 1 , and G the ( 1 , 2 ) -step competition graph of T. Assume G ≠ K n − 1 ∪ K 1 . Now we show that the complement G c of G does not contain K 1 , 3 . Suppose not. Let { x , y , z , w } and { x y , x z , x w } be the vertex set and edge set of the subgraph K 1 , 3 , respectively. So x y , x z , x w ∉ E ( G ) . By Lemma 2.1, we have x y , x z and x w satisfy one of the cases (a)–(e).

Claim 1

None of x y , x z and x w satisfies the case ( a ) or ( b ) .

Proof

Suppose at least one of x y , x z and x w satisfies the case ( a ) or ( b ) . W.l.o.g., we assume that x y satisfies ( a ) , i.e. N + ( x ) = ∅ . Let V ( T ) = { v 1 , v 2 , … , v n } and x = v n . By Lemma 2.1, for all 1 ≤ i ≤ n − 1 , we have v i v n ∉ E ( C 1 , 2 ( T ) ) . We claim that for all 1 ≤ i < j ≤ n − 1 , v i v j ∈ E ( C 1 , 2 ( T ) ) . Indeed,

For 1 ≤ i < j ≤ n − ( k − 1 ) , the vertex v i dominates v n by the arc consisting of v i , … , v i + ( k − 3 ) , v n − 1 , v n and the vertex v j dominates v n by the arc consisting of v j , … , v j + ( k − 3 ) , v n − 1 , v n . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T ) ) .
For n − ( k − 2 ) ≤ i < j ≤ n − 1 , the vertex v i dominates v n by the arc consisting of v n − k , … , v n − 3 , v n − 1 , v n for n − ( k − 2 ) ≤ i ≤ n − 3 and by the arc consisting of v 1 , … , v 1 + ( k − 3 ) , v n − 2 , v n for i = n − 2 and the vertex v j dominates v n by the arc consisting of v n − ( k − 1 ) , … , v n − 2 , v n − 1 , v n . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T ) ) .
For 1 ≤ i ≤ n − k and n − ( k − 2 ) ≤ j ≤ n − 1 , the vertex v i dominates v n by the arc consisting of v i , … , v i + ( k − 3 ) , v n − 1 , v n and the vertex v j dominates v n by the arc consisting of v n − ( k − 1 ) , … , v n − 2 , v n − 1 , v n . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T ) ) .
For i = n − ( k − 1 ) and n − ( k − 2 ) ≤ j ≤ n − 1 , the vertex v n − ( k − 1 ) dominates v n by the arc consisting of v 1 , … , v 1 + ( k − 3 ) , v n − 2 , v n for k = 3 and by the arc consisting of v n − k , v n − ( k − 1 ) , … , v n − 3 , v n − 1 , v n for 4 ≤ k ≤ n − 1 and the vertex v j dominates v n by the arc consisting of v n − ( k − 1 ) , … , v n − 2 , v n − 1 , v n . Then v j and v n − ( k − 1 ) compete and v j v n − ( k − 1 ) ∈ E ( C 1 , 2 ( T ) ) .

Then C 1 , 2 ( T ) = K n − 1 ∪ K 1 , a contradiction. Thus, none of x y , x z and x w satisfies the case ( a ) or ( b ) .□

Claim 2

At most one of x y , x z and x w satisfies the case ( c ) or ( d ) .

Proof

Suppose at least two edges among x y , x z and x w satisfy the case ( c ) or ( d ) . W.l.o.g., we assume that both x y and x z satisfy the case ( c ) or ( d ) . We consider the following four cases.

Case 1: Both x y and x z satisfy ( c ) . It means that N + ( x ) = { y } and N + ( x ) = { z } . Then y = z , a contradiction.

Case 2: Both x y and x z satisfy ( d ) . It means that N + ( y ) = { x } and N + ( z ) = { x } . Now the arcs containing simultaneously x , y , z do not satisfy N + ( y ) = { x } , N + ( z ) = { x } , a contradiction.

Case 3: x y satisfies ( c ) and x z satisfies ( d ) . It means that N + ( x ) = { y } and N + ( z ) = { x } . If x w satisfies ( c ) , then N + ( x ) = { w } , contradicting N + ( x ) = { y } . If x w satisfies ( d ) , then N + ( w ) = { x } . Now the arcs containing simultaneously x , z , w do not satisfy N + ( z ) = { x } , N + ( w ) = { x } , a contradiction. If x w satisfies ( e ) . Let b be an arc containing x , z , w . N + ( x ) = { y } yields z dominates x in b . N + ( z ) = { x } yields z is the second last entry, x is the last entry and w is any other entry of b . Clearly, b ∉ A T ∗ { x , w } , i.e. z ∈ N T − a + ( w ) , which contradicts the fact that N T − a + ( w ) ⊆ { x } .

Case 4: x y satisfies ( d ) and x z satisfies ( c ) . Similarly to Case 3, we can also get a contradiction.

Thus, at most one of x y , x z and x w satisfies the case ( c ) or ( d ) .□

Claim 3

At most one of x y , x z and x w satisfies the case ( e ) .

Proof

Suppose at least two edges among x y , x z and x w satisfy the case ( e ) . Assume that x y and x z satisfy ( e ) . From the assumption that x y satisfies ( e ) , we get the vertex z dominates y in each arc containing y , z but not containing x . From the assumption that x z satisfies ( e ) , we get the vertex y dominates z in each arc containing y , z but not containing x . This is a contradiction. Thus, at most one of x y , x z and x w satisfies the case ( e ) .□

By Claims 1–3, it is impossible that x y , x z , x w ∉ E ( G ) hold simultaneously. Thus, the complement G c of G does not contain K 1 , 3 unless G = K n − 1 ∪ K 1 . The lemma holds.□

By Corollary 2.2 and the proof of Lemma 3.3, we obtain the following result.

Corollary 3.4

Let G on n vertices be the ( 1 , 2 ) -step competition graph of some strong k-hypertournament T, where 3 ≤ k ≤ n − 1 . Then the complement G c of G does not contain K 1 , 3 .

4 Strong k-hypertournaments

Theorem 4.1

A graph G on n vertices is the ( 1 , 2 ) -step competition graph of some strong k-hypertournament T with 3 ≤ k ≤ n − 1 if and only if G is K n , K n − E ( P 2 ) , or K n − E ( P 3 ) .

Proof

We first show the “if” part. Let T be a transitive k-hypertournament with the vertices v 1 , v 2 , … , v n . Let T 1 be a k-hypertournament obtained from T by replacing the arc ( v 1 , v 2 , v n − ( k − 3 ) , … , v n ) with ( v n , … , v n − ( k − 3 ) , v 2 , v 1 ) . It is easy to check that T 1 is strong. Now we show that C 1 , 2 ( T 1 ) = K n − E ( P 2 ) . For convenience, let a = ( v n , … , v n − ( k − 3 ) , v 2 , v 1 ) . We claim that v n − 1 v n ∉ E ( C 1 , 2 ( T 1 ) ) . Indeed, for k = 3 , the vertex v n − 1 has a unique out-neighbour v n and Corollary 2.2 ( a ) implies v n − 1 v n ∉ E ( C 1 , 2 ( T 1 ) ) . For 4 ≤ k ≤ n − 1 , A T 1 ∗ { v n − 1 , v n } contains exactly an arc a , and N T 1 − a + ( v n − 1 ) = { v n } , N T 1 − a + ( v n ) = ∅ . Corollary 2.2 ( c ) implies v n − 1 v n ∉ E ( C 1 , 2 ( T 1 ) ) . We also claim v i v j ∈ E ( C 1 , 2 ( T 1 ) ) for all { i , j } ≠ { n − 1 , n } . W.l.o.g., we assume i < j .

For 1 ≤ i < j ≤ n − ( k − 1 ) , v i dominates v n by the arc ( v i , v n − ( k − 2 ) , … , v n − 1 , v n ) and v j dominates v n by the arc ( v j , v n − ( k − 2 ) , … , v n − 1 , v n ) . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T 1 ) ) .
For n − ( k − 2 ) ≤ i < j ≤ n − 1 , v i dominates v n by the arc ( v 1 , v n − ( k − 2 ) , … , v i , … , v j , … , v n − 1 , v n ) and v j dominates v n by the arc ( v 2 , v n − ( k − 2 ) , … , v i , … , v j , … , v n − 1 , v n ) . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T 1 ) ) .
For i = 1 and n − ( k − 2 ) ≤ j ≤ n − 1 , v 1 dominates v n by the arc ( v 1 , v n − ( k − 2 ) , … , v n − 1 , v n ) and v j dominates v n by the arc ( v 2 , v n − ( k − 2 ) , … , v j , … , v n − 1 , v n ) . Then v 1 and v j compete and v 1 v j ∈ E ( C 1 , 2 ( T 1 ) ) .
For 2 ≤ i ≤ n − ( k − 1 ) and n − ( k − 2 ) ≤ j ≤ n − 1 , v i dominates v n by the arc ( v i , v n − ( k − 2 ) , … , v n − 1 , v n ) and v j dominates v n by the arc ( v 1 , v n − ( k − 2 ) , … , v j , … , v n − 1 , v n ) . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T 1 ) ) .
For i = 1 and j = n , v n dominates v 2 by the arc a , v 2 dominates v n − 1 by the arc ( v 2 , v n − ( k − 2 ) , … , v n − 1 , v n ) and v 1 dominates v n − 1 by the arc ( v 1 , v n − ( k − 2 ) , … , v n − 1 , v n ) . Then v 1 and v n ( 1 , 2 )-step compete and v 1 v n ∈ E ( C 1 , 2 ( T 1 ) ) .
For 2 ≤ i ≤ n − 2 and j = n , v n dominates v 1 by the arc a , v 1 dominates v n − 1 by the arc ( v 1 , v n − ( k − 2 ) , … , v n − 1 , v n ) and v i dominates v n − 1 by ( v i , v n − ( k − 2 ) , … , v n − 1 , v n ) for 2 ≤ i ≤ n − ( k − 1 ) and by ( v 2 , v n − ( k − 2 ) , … , v i , … , v n − 1 , v n ) for n − ( k − 2 ) ≤ i ≤ n − 2 . Then v i and v n ( 1 , 2 )-step compete and v i v n ∈ E ( C 1 , 2 ( T 1 ) ) .

Thus, C 1 , 2 ( T 1 ) = K n − E ( P 2 ) .

Let T 2 be a k-hypertournament obtained from T 1 above by replacing the arc ( v 1 , v 2 , v n − ( k − 2 ) , … , v n − 1 ) with ( v n − 1 , … , v n − ( k − 2 ) , v 2 , v 1 ) . It is easy to check that T 2 is strong. Now we show that C 1 , 2 ( T 2 ) = K n .

For { i , j } ≠ { n − 1 , n } , similarly to the proof of T 1 , we have v i v j ∈ E ( C 1 , 2 ( T 2 ) ) .
For i = n − 1 and j = n , v n − 1 dominates v 1 by the arc ( v n − 1 , … , v n − ( k − 2 ) , v 2 , v 1 ) and v n dominates v 1 by the arc ( v n , … , v n − ( k − 3 ) , v 2 , v 1 ) . Then v n − 1 and v n compete and v n − 1 v n ∈ E ( C 1 , 2 ( T 2 ) ) .

Thus, C 1 , 2 ( T 2 ) = K n .

Let T 3 be a k-hypertournament with the vertices v 1 , v 2 , … , v n satisfying the following:

Each arc excluding v 1 , v 2 satisfies i < j if and only if v i precedes v j ;
Each arc including v 1 , v 2 satisfies that v 1 is the second last entry, v 2 is the last entry and the remaining k − 2 entries satisfy i < j if and only if v i precedes v j ;
Each arc including v 1 but excluding v 2 satisfies that v 1 is the last entry and the remaining k − 1 entries satisfy i < j if and only if v i precedes v j ;
Each arc including v 2 but excluding v 1 , v 3 satisfies that v 2 is the last entry and the remaining k − 1 entries satisfy i < j if and only if v i precedes v j ;
Each arc including v 2 , v 3 but excluding v 1 satisfies that v 2 is the second last entry, v 3 is the last entry and the remaining k − 2 entries satisfy i < j if and only if v i precedes v j .

It is easy to check that T 3 is strong. Now we show that C 1 , 2 ( T 3 ) = K n − E ( P 3 ) . Note that N + ( v 1 ) = { v 2 } and N + ( v 2 ) = { v 3 } . By Corollary 2.2 ( a ) , we have v 1 v 2 , v 2 v 3 ∉ E ( C 1 , 2 ( T 3 ) ) . Now we consider the arc v i v j for { i , j } ≠ { 1 , 2 } and { i , j } ≠ { 2 , 3 } . W.l.o.g., we assume i < j .

For 3 ≤ i < j ≤ n − ( k − 3 ) , v i dominates v 2 by the arc ( v i , … , v i + ( k − 3 ) , v 1 , v 2 ) and v j dominates v 2 by the arc ( v j , … , v j + ( k − 3 ) , v 1 , v 2 ) . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T 3 ) ) .
For n − ( k − 4 ) ≤ i < j ≤ n , v i dominates v 2 by the arc ( v n − ( k − 2 ) , … , v i , … , v j , … , v n − 1 , v 1 , v 2 ) and v j dominates v 2 by the arc ( v n − ( k − 3 ) , … , v i , … , v j , … , v n , v 1 , v 2 ) . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T 3 ) ) .
For 3 ≤ i ≤ n − ( k − 3 ) and n − ( k − 4 ) ≤ j ≤ n , v i dominates v 2 by the arc ( v i , … , v i + ( k − 3 ) , v 1 , v 2 ) for 3 ≤ i ≤ n − ( k − 2 ) and by the arc ( v n − ( k − 2 ) , v n − ( k − 3 ) , … , v n − 1 , v 1 , v 2 ) for i = n − ( k − 3 ) and v j dominates v 2 by the arc ( v n − ( k − 3 ) , … , v j , … , v n , v 1 , v 2 ) . Then v i and v j compete and v i v j ∈ E ( C 1 , 2 ( T 3 ) ) .
For i = 1 and 3 ≤ j ≤ n − ( k − 2 ) , v 1 dominates v 2 by the arc ( v n − ( k − 3 ) , … , v n , v 1 , v 2 ) and v j dominates v 2 by the arc ( v j , … , v j + ( k − 3 ) , v 1 , v 2 ) . Then v 1 and v j compete and v 1 v j ∈ E ( C 1 , 2 ( T 3 ) ) .
For i = 1 and n − ( k − 3 ) ≤ j ≤ n , v 1 dominates v 2 by the arc ( v 3 , … , v k , v 1 , v 2 ) and v j dominates v 2 by the arc ( v n − ( k − 3 ) , … , v j , … , v n , v 1 , v 2 ) . Then v 1 and v j compete and v 1 v j ∈ E ( C 1 , 2 ( T 3 ) ) .
For i = 2 and 4 ≤ j ≤ n − ( k − 3 ) , v 2 dominates v 3 by the arc ( v n − ( k − 3 ) , … , v n , v 2 , v 3 ) , v 3 dominates v 1 by the arc ( v 3 , … , v k , v 1 , v 2 ) and v j dominates v 1 by the arc ( v j , … , v j + ( k − 3 ) , v 1 , v 2 ) . Then v 2 and v j ( 1 , 2 )-step compete and v 2 v j ∈ E ( C 1 , 2 ( T 3 ) ) .
For i = 2 and n − ( k − 4 ) ≤ j ≤ n , v 2 dominates v 3 by the arc ( v n − ( k − 3 ) , … , v n , v 2 , v 3 ) , v 3 dominates v 1 by the arc ( v 3 , … , v k , v 1 , v 2 ) and v j dominates v 1 by the arc ( v n − ( k − 3 ) , … , v j , … , v n , v 1 , v 2 ) . Then v 2 and v j ( 1 , 2 )-step compete and v 2 v j ∈ E ( C 1 , 2 ( T 3 ) ) .

Thus, C 1 , 2 ( T 3 ) = K n − E ( P 3 ) .

Now we show the “only if” part. Let T be a strong k-hypertournament and G the ( 1 , 2 ) -step competition graph of T. We show that G is K n , K n − E ( P 2 ) , or K n − E ( P 3 ) . We claim that G c contains at most two edges. Suppose to the contrary that G c contains at least three edges, say e 1 , e 2 , e 3 ∈ E ( G c ) . Let e i = x i y i for i = 1 , 2 , 3 . By Lemma 3.1, e 1 and e 2 have a common end-point. W.l.o.g., we assume that y 1 = x 2 . By Lemma 3.1, e 3 and e 1 have a common end-point, and e 3 and e 2 have also a common end-point. So either e 3 = x 1 y 2 or x 2 is an end-point of e 3 . However, this implies G c contains 3-cycle or K 1 , 3 , which contradicts Lemma 3.2 and Corollary 3.4. So G c contains at most two edges. Thus, if G c contains two edges, Lemma 3.1 implies G = K n − E ( P 3 ) ; if G c contains one edge, then G = K n − E ( P 2 ) ; if G c contains no edge, then G = K n .

Therefore, the theorem holds.□

5 Remaining k-hypertournaments

Theorem 5.1

A graph G on n vertices is the ( 1 , 2 ) -step competition graph of some k-hypertournament T with 3 ≤ k ≤ n − 1 if and only if G is K n , K n − E ( P 2 ) , K n − E ( P 3 ) , or K n − 1 ∪ K 1 .

Proof

The “if” part follows from the proof of Lemma 3.3 and Theorem 4.1. Now we show the “only if” part. Let T be a k-hypertournament and G the ( 1 , 2 ) -step competition graph of T. We show that G is K n , K n − E ( P 2 ) , K n − E ( P 3 ) , or K n − 1 ∪ K 1 . Similarly to the proof of “only if” of Theorem 4.1, we get G c contains at most two edges unless G = K n − 1 ∪ K 1 . Thus, if G c contains two edges, Lemma 3.1 implies G = K n − E ( P 3 ) ; if G c contains one edge, then G = K n − E ( P 2 ) ; if G c contains no edge, then G = K n .

Therefore, the theorem holds.□

6 The ( i , j ) -step competition graph of a k-hypertournament

We generalize the ( 1 , 2 ) -step competition graph to the ( i , j ) -step competition graph as follows. By the definition of the ( i , j ) -step competition graph for a k-hypertournament T, we obtain that if i ≥ 1 , j ≥ 2 , then E ( C 1 , 2 ( T ) ) ⊆ E ( C i , j ( T ) ) . It is easy to see that the proof of Lemma 2.1 implies the following corollary.

Corollary 6.1

Let T be a k-hypertournament with n vertices satisfying 3 ≤ k ≤ n − 1 and i ≥ 1 , j ≥ 2 integers. Then x y ∉ E ( C i , j ( T ) ) if and only if one of the following holds:

N + ( x ) = ∅ ;
N + ( y ) = ∅ ;
N + ( x ) = { y } ;
N + ( y ) = { x } ;
A T ∗ { x , y } contains exactly an arc a , and N T − a + ( x ) ⊆ { y } , N T − a + ( y ) ⊆ { x } .

Theorem 6.2

Let T be a k-hypertournament with n vertices satisfying 3 ≤ k ≤ n − 1 and i ≥ 1 , j ≥ 2 integers. Then C i , j ( T ) = C 1 , 2 ( T ) .

Proof

Clearly, V ( C i , j ( T ) ) = V ( C 1 , 2 ( T ) ) = V ( T ) . Since E ( C 1 , 2 ( T ) ) ⊆ E ( C i , j ( T ) ) , it suffices to show that E ( C i , j ( T ) ) ⊆ E ( C 1 , 2 ( T ) ) . Let x y ∈ E ( C i , j ( T ) ) . Suppose x y ∉ E ( C 1 , 2 ( T ) ) . By Lemma 2.1, x and y must satisfy one of the cases (a)–(e). This contradicts Corollary 6.1. Thus, x y ∈ E ( C 1 , 2 ( T ) ) and E ( C i , j ( T ) ) ⊆ E ( C 1 , 2 ( T ) ) .□

Funding information: This research was partially supported by the Youth Foundation of Shanxi Province (201901D211197).
Conflict of interest: Authors state no conflict of interest.

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Received: 2020-10-12

Accepted: 2021-05-07

Published Online: 2021-06-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

The (1, 2)-step competition graph of a hypertournament

Abstract

1 Terminology and introduction

Theorem 1.1

Theorem 1.2

Theorem 1.3

Definition 1.4

2 The missing edges of C 1 , 2 ( T )

Lemma 2.1

Proof

Corollary 2.2

3 The forbidden subgraphs of ( C 1 , 2 ( T ) ) c

Lemma 3.1

Proof

Lemma 3.2

Proof

Claim 1

Proof

Claim 2

Proof

Claim 3

Proof

Lemma 3.3

Proof

Claim 1

Proof

Claim 2

Proof

Claim 3

Proof

Corollary 3.4

4 Strong k-hypertournaments

Theorem 4.1

Proof

5 Remaining k-hypertournaments

Theorem 5.1

Proof

6 The ( i , j ) -step competition graph of a k-hypertournament

Corollary 6.1

Theorem 6.2

Proof

References

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