The (1,2) -step competition graph of a pure local tournament that is not round decomposable

Discrete Applied Mathematics 205 (2016) 180–190

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The (1, 2)-step competition graph of a pure local tournament that is not round decomposable Xinhong Zhang a , Ruijuan Li b,∗ a

Department of Applied Mathematics, Taiyuan University of Science and Technology, 030024 Taiyuan, PR China

b

School of Mathematical Sciences, Shanxi University, 030006 Taiyuan, PR China

article

info

Article history: Received 19 January 2015 Received in revised form 3 September 2015 Accepted 16 December 2015 Available online 3 February 2016 Keywords: Competition graph (1, 2)-step competition graph Locally semicomplete digraph Local tournament Round-decomposable

abstract The competition graph of a digraph was introduced by Cohen in 1968 associated with the study of ecosystems. In 2011, Factor et al. (2011), defined the (1, 2)-step competition graph of a digraph which is a generalization of the competition graph and gave a characterization of the (1, 2)-step competition graph of a tournament. In this paper, we characterize the (1, 2)-step competition graph of a pure local tournament that is not round decomposable. © 2016 Elsevier B.V. All rights reserved.

1. Terminology and introduction Competition graphs were first introduced by Joel Cohen [4] in connection to a problem in ecology in 1968 and have since been extensively studied. An ecological food web can be modeled by a digraph D, where the vertices of D denote the species of the ecosystem, and there is an arc from vertex x to y if x preys on y. The competition graph C (D), of this ecological food web has the same vertex set as D, and xy is an edge in C (D) if there is a vertex z such that (x, z ) and (y, z ) are arcs in D. That means z is the common prey of x and y. The competition graph can be used to reflect the competition relations among the predators in the food web. Furthermore, the study of competition graphs has been applied widely to the coding, channel assignment in communications, modeling of complex systems arising from study of energy and economic systems, etc. For a comprehensive introduction to competition graphs, see [5,7,12,13]. Recent works in competition graph theory include [9,11]. In 2011, the (1, 2)-step competition graph was defined by Factor and Merz in [6]. In this paper, we characterize the (1, 2)-step competition graphs of the pure local tournaments that are not round decomposable. It will be assumed that the reader is familiar with the concepts of graphs and digraphs. Other terminology can be found in [3]. In this paper, all digraphs are finite and have no parallel arcs and loops. Let D be a digraph on n vertices. We denote by V (D) and A(D) the vertex set and the arc set respectively. If (x, y) is an arc of D, then we write x → y and say x dominates y. More generally, if A and B are two disjoint subdigraphs of D such that every vertex of A dominates every vertex of B, then we say that A dominates B, denoted by A → B. If A → B, but there is no arc from B to A, then we say that A strictly dominates B, denoted by A → B. Furthermore, A B denotes the property that there is no arc from B to A. An arc (x, y) of a digraph D is ordinary if (y, x) is not in D. A cycle Q of a digraph D is ordinary if all arcs of Q are ordinary. If x and y are vertices of D, then the distance from x to y in D, denoted dD (x, y) = d(x, y), is the minimum length of an (x, y)-path, if y is reachable from x, and

∗

Corresponding author. E-mail address: [email protected] (R. Li).

http://dx.doi.org/10.1016/j.dam.2015.12.013 0166-218X/© 2016 Elsevier B.V. All rights reserved.

X. Zhang, R. Li / Discrete Applied Mathematics 205 (2016) 180–190

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Fig. 1. (a) and (a1 ): A strong digraph and its (1, 2)-step competition graph, where Si is a complete subdigraph (subgraph, respectively) with |V (Si )| ≥ 3 for i ∈ {1, 2, . . . , 5}; (b) and (b1 ): A connected digraph which is not strong and its (1, 2)-step competition graph, where Di is a complete subdigraph (subgraph, respectively) with |V (Dj )| ≥ 3 for j ∈ {1, 2, 3, 4} and |V (D5 )| = 1.

otherwise d(x, y) = ∞. Furthermore, let H be a subdigraph of D. Then the distance from x to H, denoted dD (x, H ) = d(x, H ), is the minimum length of an (x, z )-path for all z ∈ V (H ). The outset of a vertex x ∈ V (D) is the set ND+ (x) = {y|(x, y) ∈ A(D)}. Similarly, ND− (x) = {y|(y, x) ∈ A(D)} is the inset of x. More generally, for subdigraphs A and B of D, we define the outset(inset) of B in A by NA+ (B) = {z |(x, z ) ∈ A(D), x ∈ V (B), z ∈ V (A)}(NA− (B) = {z |(z , x) ∈ A(D), x ∈ V (B), z ∈ V (A)}). A subdigraph induced by a subset U ⊆ V (D) is denoted by D[U ]. In addition, D − U = D[V (D) \ U ] for any U ⊆ V (D). A digraph D is strong if every vertex of D is reachable by a path from every other vertex of D. For a strong digraph D, a set S ⊆ V is a separating set if D − S is not strong. A strong component of a digraph D is a maximal induced subdigraph of D which is strong. If D1 , D2 , . . . , Dt are the strong components of D, then clearly V (D1 ) ∪ V (D2 ) ∪ · · · ∪ V (Dt ) = V (D). We call V (D1 ) ∪ V (D2 ) ∪ · · · ∪ V (Dt ) the strong decomposition of D. It is known that the strong components of D can be labeled D1 , D2 , . . . , Dt such that there is no arc from Dj to Di unless j < i. We call such an ordering an acyclic ordering of the strong components of D. The underlying graph of D, denoted by UG(D), is the graph obtained by ignoring the orientations of arcs in D and deleting parallel edges. We say that D is connected if its underlying graph is connected. In this paper, we only consider connected digraphs. A digraph D is semicomplete, if for any pair of vertices x, y ∈ V (D), either (x, y) ∈ A(D), or (y, x) ∈ A(D), or both. A tournament is a semicomplete digraph without a cycle of length 2. A digraph D is locally semicomplete (or a locally semicomplete digraph), if D[ND+ (x)] and D[ND− (x)] are both semicomplete for every vertex x of D. A locally semicomplete digraph containing no cycle of length 2 is called a local tournament. In others words, a local tournament is a locally semicomplete digraph without cycle of length 2. A pure local tournament is a local tournament which is not a tournament. Local semicomplete digraphs were introduced in 1990 by Bang-Jensen (see [1]). Local tournaments are the most important subclass of the local semicomplete digraphs. This class of digraphs has many nice properties in common with its subclass, tournaments. The (1, 2)-step competition graph of a digraph D, denoted by C1,2 (D), is a graph on V (D) where xy ∈ E (C1,2 (D)) if and only if there exists a vertex z ̸= x, y, such that either dD−y (x, z ) ≤ 1 and dD−x (y, z ) ≤ 2 or dD−x (y, z ) ≤ 1 and dD−y (x, z ) ≤ 2. For example, two digraphs and their respective (1, 2)-step competition graphs are shown in Fig. 1. Let G be a graph. Pi is a path on i vertices in this paper. As for digraphs, a subgraph induced by a subset X ⊆ V (G) is denoted by G[X ] and G − X = G[V (D) \ X ] for X ⊆ V (D). The graph G − E (H ) is obtained from G by removing the edges from a subgraph of G that is isomorphic to H. In [6], Factor et al. completely characterized the (1, 2)-step competition graphs of tournaments. Theorem 1.1 (Factor & Merz [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: 1. 2. 3. 4. 5.

Kn where n ̸= 2, 3, 4, Kn−1 ∪ K1 where n > 1, Kn − E (P3 ) where n > 2, Kn − E (P2 ) where n ̸= 1, 4, or Kn − E (K3 ) where n ≥ 3.

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In this paper, we characterize the (1, 2)-step competition graphs of pure local tournaments that are not round decomposable. Moreover, the characterization of the (1, 2)-step competition graphs of pure local tournaments that are round decomposable can be found in [14]. 2. The structure of locally semicomplete digraphs In this section, we recall a useful structural classification of locally semicomplete digraphs (see Theorem 2.8) which will play an essential role in our proofs. First, we give the definition of a round digraph. A digraph on n vertices is called a round digraph if we can label its vertices v0 , v1 , . . . , vn such that for each i, ND+ (vi ) = {vi+1 , . . . , vi+d+ (vi ) } and ND− (vi ) = {vi−d− (vi ) , . . . , vi−1 }, where the subscripts are taken modulo n. Theorem 2.1 (Huang [10]). A connected locally semicomplete digraph D is round if and only if the following holds for each vertex x of D: (a) N + (x) \ N − (x) and N − (x) \ N + (x) induce transitive tournaments and (b) N + (x) ∩ N − (x) induces a (semicomplete) subdigraph containing no ordinary cycle. It follows from Theorem 2.1 that if a local tournament D is round then there exists a unique (up to cyclic permutations) labeling of vertices of D which satisfies the properties in the definition. We refer to this as the round labeling of D. Let D be a digraph with the vertex set {v1 , v2 , . . . , vr }, and G1 , G2 , . . . , Gr be digraphs which are pairwise vertexdisjoint. The composition D[G1 , G2 , . . . , Gr ] is the digraph with vertex set V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gr ) and arc set r ( i=1 A(Gi )) ∪ {(gi , gj )|gi ∈ V (Gi ), gj ∈ V (Gj ), (vi , vj ) ∈ A(D)}. A locally semicomplete digraph D is round decomposable if there exists a round local tournament R on p ≥ 2 vertices such that D = R[D1 , D2 , . . . , Dp ], where each Di is a strong semicomplete subdigraph of D for i = 1, 2, . . . , p. We call R[D1 , D2 , . . . , Dp ] a round decomposition of D. Lemma 2.2 (Bang-Jensen et al. [2]). If a locally semicomplete digraph D is round decomposable, then it has a unique round decomposition D = R[D1 , D2 , . . . , Dα ]. Lemma 2.3. Let D be a round decomposable digraph with round decomposition R[D0 , D1 , . . . , Dp−1 ]. Let vi ∈ V (Di ) for i = 0, 1, . . . , p − 1. If vi → vj , then Dα → Dβ holds for α, β ∈ {1, 2, . . . , p} with the property that Di , Dα , Dβ , Dj (possibly Dα = Di , Dβ = Dj ) are in the order of the round labeling of D, where the subscripts of all of Di are taken modulo p. Proof. Since vi → vj , Di → Dj according to the definition of a round decomposable digraph. We assume w.l.o.g. that 0 ≤ i < j ≤ p. For any i < α < β < j, V (Dβ ) ⊆ N + (Di ) since V (Dj ) ⊆ N + (Di ) and hence V (Dα ) ⊆ N − (Dβ ) since V (Di ) ⊆ N − (Dβ ) according to the definition of a round decomposable digraph. Thus Dα → Dβ .  Subsequently, we begin with the structure of non-strong locally semicomplete digraphs. Lemma 2.4 (Bang-Jensen et al. [2]). Every non-strong locally semicomplete digraph D has a unique round decomposition R[D1 , D2 , . . . , Dp ], where D1 , D2 , . . . , Dp is the unique ordering of the strong components of D and R is a round local tournament containing no cycle. Theorem 2.5 (Guo & Volkmann [8]). If D is a connected locally semicomplete digraph that is not strong where D1 , D2 , . . . , Dp is the unique ordering of the strong components of D, then D can be decomposed into r ≥ 2 semicomplete subdigraphs D′1 , D′2 , . . . , D′r as follows: D′1 = Dp , λ1 = p

λi+1 = min{j|ND+ (Dj ) ∩ V (D′i ) ̸= ∅}, and D′i+1 = D[V (Dλi+1 ) ∪ V (Dλi+1 +1 ) ∪ · · · ∪ V (Dλi −1 )]. The subdigraphs D′1 , D′2 , . . . , D′r satisfy the properties below: (a) D′1 is the terminal component of D and D′i consists of some strong components of D for i ≥ 2; (b) D′i+1 dominates the initial component of D′i and there exists no arc from D′i to D′i+1 for i = 1, 2, . . . , r − 1; (c) if r ≥ 3, then there is no arc between D′i and D′j for i, j satisfying |j − i| ≥ 2. The unique sequence D′1 , D′2 , . . . , D′r defined in Theorem 2.5 will be referred to as the semicomplete decomposition of D. We now turn to the structure of strong locally semicomplete digraphs. Lemma 2.6 (Bang-Jensen et al. [2]). If a strong locally semicomplete digraph D is not semicomplete, then there exists a minimal separating set S ⊂ V (D) such that D − S is not semicomplete. Furthermore, if D1 , D2 , . . . , Dp is the acyclic ordering of the strong components of D − S and D′1 , D′2 , . . . , D′r is the semicomplete decomposition of D − S, then r ≥ 3, D[S ] is semicomplete and we have Dp → S → D1 .

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Proposition 2.7 (Bang-Jensen et al. [2]). Let D be a strong locally semicomplete digraph that is not round decomposable and let S be a minimal separating set of D such that D − S is not semicomplete. Let D1 , D2 , . . . , Dp be the acyclic ordering of the strong components of D − S and Dp+1 , . . . , Dp+q be the acyclic ordering of the strong components of D[S ]. Suppose that there is an arc sv from S to D′2 with s ∈ V (Di ) and v ∈ V (Dj ), then Di ∪ Di+1 ∪ · · · ∪ Dp+q → D′3 → Dλ2 ∪ · · · ∪ Dj . Theorem 2.8 (Bang-Jensen et al. [2]). Let D be a strong locally semicomplete digraph which is not semicomplete. Then D is not round decomposable if and only if the following conditions are satisfied: (a) There is a minimal separating set S such that D − S is not semicomplete and for each such S, D[S ] is semicomplete and the semicomplete decomposition of D − S has exactly three semicomplete subdigraphs D′1 , D′2 , D′3 ; (b) There are integers α, β, µ, ν with λ ≤ α ≤ β ≤ p − 1 and p + 1 ≤ µ ≤ ν ≤ p + q such that N − (Dα ) ∩ V (Dµ ) ̸= ∅ and or

N + (Dα ) ∩ V (Dν ) ̸= ∅,

N − (Dµ ) ∩ V (Dα ) ̸= ∅ and

N + (Dµ ) ∩ V (Dβ ) ̸= ∅,

where D1 , D2 , . . . , Dp and Dp+1 , . . . , Dp+q are the strong decompositions of D − S and D[S ], respectively, and Dλ is the initial component of D′2 . Since D is a local tournament and there is no arc between D′1 and D′3 , we can obtain the following result immediately. Proposition 2.9. Let D be a local tournament and S be a minimal separating set satisfying Theorem 2.8. Then there is no arc from D′3 to S. Now, we can state a full classification of locally semicomplete digraphs. Theorem 2.10 (Bang-Jensen et al. [2]). Let D be a connected locally semicomplete digraph. Then exactly one of the following possibilities holds. (a) D is round decomposable with a unique round decomposition given by D = R[D1 , D2 , . . . , Dα ], where R is a round local tournament on α ≥ 2 vertices and Di is a strong semicomplete digraph for i = 1, 2, . . . , α ; (b) D is not round decomposable and not semicomplete and it has the structure as described in Theorem 2.8; (c) D is a semicomplete digraph which is not round decomposable. It is clear that all of theorems in this section are correct if we replace ‘‘local semicomplete digraph’’ with ‘‘local tournament’’ and ‘‘semicomplete digraph’’ with ‘‘tournament’’. 3. Main results According to the definition of the (1, 2)-step competition graph, the following result is clear. Lemma 3.1. Let D be a connected local tournament and S be a minimal separating set of D when D is strong or an empty set when D is not strong. Let D1 , D2 , . . . , Dr be the acyclic ordering of the strong components of D − S. Then (a) every vertex of Di is adjacent to every vertex Dj in C1,2 (D) if Di → Dj and |V (Dj )| ≥ 3 for i, j ∈ {1, 2, . . . , t }; (b) every vertex of Di is adjacent to every vertex Dk in C1,2 (D) if Di → Dj → Dk and |V (Dk )| ≥ 3 for i, j, k ∈ {1, 2, . . . , t }. Theorem 3.2. A graph G on n vertices is the (1, 2)-step competition graph of some pure local tournament D that is not round decomposable if and only if G is one of the following graphs: 1. 2. 3. 4.

Kn , when n ≥ 8, Kn − E (P2 ), when n ≥ 7, Kn − E (P3 ), when n ≥ 6, Kn − E (K3 ), when n ≥ 5.

Proof. (⇒) Let D be a pure local tournament that is not round decomposable and G be the (1, 2)-step competition graph of D, i.e. G = C1,2 (D). By Theorem 2.8, we consider D with a minimal separating set S such that the semicomplete decomposition of D − S has exactly three components D′1 , D′2 , D′3 . Let S be the separating set of minimum cardinality among all the separating sets satisfying Theorem 2.8. Let D1 , D2 , . . . , Dλ−1 denote the strong components of D′3 , Dλ , . . . , Dp−1 denote the strong components of D′2 and Dp denote the strong component of D′1 . Let Dp+1 , . . . , Dp+q denote the strong components of D[S ]. By Lemma 2.6 and Theorem 2.8(a), there is no arc between D′1 and D′3 , D′3 → Dλ and D′2 → D′1 → S → D1 (see Fig. 3(a)). Since D′3 → Dλ , D′2 → D′1 , D′1 → S , S → D1 , each of the sets V (D′i )(i = 1, 2, 3) and S induce a complete subgraph in G. Subsequently, we will complete the proof of the necessity by several claims.

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Claim 1. Every vertex of D′3 is adjacent to every vertex of D′2 in G. Proof. Since D′3 → Dλ → D′1 and D′2 → D′1 , x is adjacent to y for any x ∈ V (D′3 ) and y ∈ V (D′2 ) \ {Dλ } in G. If |V (Dλ )| ≥ 3, then x is adjacent to y for any x ∈ V (D′3 ) and y ∈ V (Dλ ) in G by Lemma 3.1. We consider the case when |V (Dλ )| = 1. Now we claim that ND+′ (D′3 ) \ V (Dλ ) ̸= ∅. Suppose not. Then we have 2

|ND+′ (D′3 )| = |V (Dλ )| = 1. Thus V (Dλ ) is a separating set of D such that D − Dλ is not a tournament according to the 2 structure of D. Since S is a minimum separating set which satisfying Theorem 2.8, we have |S | = 1. By Theorem 2.8(b), ND+′ (S ) ̸= ∅. Furthermore, we have that ND+′ (S ) = V (Dλ ). Instead, suppose not. There exists one vertex v ∈ V (Dj0 ) for some 2

2

j0 ∈ {λ + 1, . . . , p − 1} such that S → v . Then Proposition 2.7 yields that S → D′3 → v , which contradicts our assumption that ND+′ (D′3 ) = V (Dλ ). So ND+′ (S ) = V (Dλ ), and hence there is no structure of Theorem 2.8(b), which contradicts that D 2

2

is not round decomposable. Therefore, ND+′ (D′3 ) \ V (Dλ ) ̸= ∅, i.e. there is an arc (u, v) from D′3 to D′2 with u ∈ V (Di0 ) and 2

v ∈ V (D′2 ) \ V (Dλ ) for 1 ≤ i0 ≤ λ − 1. It is obvious Di0 → v since D1 , D2 , . . . , Dp are the acyclic ordering of D − S. Because Di0 → Dk for i0 + 1 ≤ k ≤ λ − 1 and Di0 → v , we have Dk → v . Then, for a strong component Dl of D′3 ,  Dl → v, when i0 ≤ l ≤ λ − 1, Dl → Di0 → v, when 1 ≤ l ≤ i0 − 1. Note that Dλ → v . We have for any vertex x ∈ V (D′3 ), y ∈ V (Dλ ), xy is an edge of G. Thus, we have that every vertex of D3 is adjacent to every vertex of D′2 in G.  ′

Claim 2. Every vertex of D′3 is adjacent to every vertex of D′1 in G. Proof. There is an arc (w, v) from S to D′2 with w ∈ S and v ∈ V (D′2 ) by Theorem 2.8(b). So Proposition 2.7 yields that D′3 → v . Note that D′1 → w → v . Hence, every vertex of D′3 is adjacent to every vertex of D′1 in G.  Claim 3. Every vertex of D′2 is adjacent to every vertex of D′1 in G or there is at most one edge xy ̸∈ E (G) such that x ∈ V (D′2 ) and y ∈ V (D′1 ). Proof. In this claim, we distinguish two cases. Case 1: |V (D′1 )| ≥ 3. Since D′2 → D′1 and |V (D′1 )| ≥ 3, by Lemma 3.1(a), x is adjacent to y for any x ∈ V (D′2 ) and y ∈ V (D′1 ) in G. Case 2: |V (D′1 )| = 1. Let V (D′1 ) = {y0 }. There exists an arc from D′2 to S in D by Theorem 2.8(b). Subcase 2.1: There exists an arc (v, w) satisfying v ∈ V (Dp−1 ) and w ∈ S, i.e. ND+[S ] (Dp−1 ) ̸= ∅. Since Dj → v → w for λ ≤ j ≤ p − 2 and y0 → w , we have y0 is adjacent to every vertex of V (D′2 ) \ V (Dp−1 ) in G. If + ND[S ] (u) ̸= ∅ for every vertex u ∈ V (Dp−1 ), then it is obvious y0 is adjacent to every vertex of Dp−1 in G. If not, then there exists at least one vertex in Dp−1 such that it does not dominate any vertex of S. Since v → w and y0 → w , we see that y0 is adjacent to every vertex of ND−p−1 (v) in G. Since D is a local tournament, w is adjacent to every vertex of ND+p−1 (v) in D. If x → w for x ∈ ND+p−1 (v), then y0 x ∈ E (G) because y0 → w . Otherwise, we have w → x. If ND+p−1 (x) ∩ ND+p−1 (v) ̸= ∅, then we

consider any vertex x0 ∈ ND+p−1 (x) ∩ ND+p−1 (v). If x0 → w , then we have y0 x ∈ E (G) because x → x0 and y0 → w . If w → x0 , then y0 x ∈ E (G) because y0 → w and x → x0 . Otherwise, ND+p−1 (x) ∩ ND+p−1 (v) = ∅, i.e. ND+p−1 (x) ⊆ ND−p−1 (v). Notice that

there is at most one vertex x ∈ ND+p−1 (v) satisfying ND+p−1 (x) ⊆ ND−p−1 (v). We claim that either y0 x ∈ E (G) or there is no arc

between ND+p−1 (x) and S. Indeed, if there exist two vertices x1 ∈ ND+p−1 (x) and w1 ∈ S such that x1 → w1 or w1 → x, then

y0 x ∈ E (G) because x → x1 → w1 and y0 → w1 , or x → x1 and y0 → w1 → x1 . So there is no arc between ND+p−1 (x) and

S, i.e. ND+p−1 (x) ∩ S = ∅. It follows that there is no vertex z such that dD−y0 (x, z ) = 1 and dD−x (y0 , z ) ≤ 2, or dD−y0 (x, z ) ≤ 2

and dD−x (y0 , z ) = 1, i.e. y0 x ̸∈ E (G).

Subcase 2.2: There exists no arc from Dp−1 to S, i.e. ND+[S ] (Dp−1 ) = ∅. Thus, there exists an arc (v, w) from D′2 to S satisfying v ∈ V (Dj ) for some λ ≤ j ≤ p − 2 and w ∈ S by Theorem 2.8(b). Since v → w and v → Dp−1 , we have w is adjacent to every vertex of Dp−1 in D. So w → Dp−1 according to ND+[S ] (Dp−1 ) = ∅. Then y0 is adjacent to x in G for any vertex x ∈ V (D′2 ) \ V (Dp−1 ) because x → Dp−1 and y0 → w → Dp−1 . If |V (Dp−1 )| > 1, then y0 is adjacent to every vertex of Dp−1 in G. If |V (Dp−1 )| = 1, then y0 is not adjacent to Dp−1 in G because ND+ (Dp−1 ) = {y0 }. In any case, there exist at most two vertices x ∈ V (D′2 ) and y ∈ V (D′1 ) such that xy ̸∈ E (G).  Claim 4. Every vertex of S is adjacent to every vertex of D′1 in G or there is at most one edge xy ̸∈ E (G) such that x ∈ S and y ∈ V (D′1 ).

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Proof. If |S | ̸= 1, then every vertex of S is adjacent to every vertex of D′1 in G because D′1 → S → D1 . Otherwise, |S | = 1. Let S = {w0 }. If |V (D′1 )| ̸= 1, then every vertex of S is adjacent to every vertex of D′1 in G because there exists one vertex v ∈ V (D′2 ) satisfying w0 → v by Theorem 2.8(b) such that w0 → v → D′1 . Otherwise, |V (D′1 )| = 1 and let V (D′1 ) = {y0 }. Since ND+ (y0 ) = {w0 }, we see that y0 w0 ̸∈ E (G) according to the definition of C1,2 (D). So there exist at most two vertices x ∈ S and y ∈ V (D′1 ) such that xy ̸∈ E (G).  Claim 5. Every vertex of S is adjacent to every vertex of D′2 in G or there is at most one edge xy ̸∈ E (G) such that x ∈ S and y ∈ V (D′2 ). Proof. If |V (Dp+q )| ̸= 1, then every vertex of S is adjacent to every vertex of D′2 in G by Lemma 3.1 because D′2 → D′1 → S and Di → Dj for p + 1 ≤ i < j ≤ p + q. So assume |V (Dp+q )| = 1 and let V (Dp+q ) = {w0 }. Since D′2 → D′1 → w0 and Di → w0 for p + 1 ≤ i ≤ p + q − 1, every vertex of S \ {w0 } is adjacent to every vertex of D′2 in G. Subsequently, the relation between {w0 } and D′2 in G will be discussed by three cases. Case 1: |ND+′ (w0 )| ≥ 2. 2

For any vertices x ∈ ND+′ (w0 ), we have that w0 is adjacent to every vertex of D′2 − x in G because w0 → x → D′1 and 2

D′2 → D′1 . Since there exists the vertex y ∈ ND+′ (w0 ) \ {x}, it follows that w0 x ∈ E (G). 2

Case 2: |ND′ (w0 )| = 1. +

2

Let ND+′ (w0 ) = {v1 } with v1 ∈ V (Dj1 ) for λ ≤ j1 ≤ p − 1. Since D′2 → D′1 and w0 → v1 → D′1 , we know w0 is adjacent to 2

every vertex of D′2 − v1 in G. Since w0 → v1 , where w0 → D′3 → Dj1 by Proposition 2.7. If |V (Dj1 )| ̸= 1, then w0 v1 ∈ E (G) by Lemma 3.1. Otherwise, |V (Dj1 )| = 1 and V (Dj1 ) = {v1 }. If there exists an arc (u, v2 ) from D′3 to Dj2 with u ∈ V (D′3 ) and v2 ∈ V (Dj2 ) for λ ≤ j1 < j2 ≤ p − 1, then w0 v1 ∈ E (G) because w0 → u → v2 and v1 → v2 . Otherwise, there is no arc from D′3 to Dj2 for λ ≤ j1 < j2 ≤ p − 1. If ND+[S \{w0 }] (v1 ) ̸= ∅, then w0 v1 ∈ E (G) because S → D1 . Otherwise, w0 v1 ̸∈ E (G). Thus either v ′ is adjacent to v ′′ for any v ′ ∈ S and v ′′ ∈ V (D′2 ) in G or there exist at most two vertices v1 ∈ S and v2 ∈ V (D′2 ) such that v1 v2 ̸∈ E (G). Case 3: |ND+′ (w0 )| = 0. 2

Let β = max{j|w0 → u → v, u ∈ V (D′3 ), v ∈ V (Dj ), λ ≤ j ≤ p − 1}. Then there exists an arc from S to D′2 by Theorem 2.8(b). So there exist at least two strong components in D[S ] because |ND+′ (w0 )| = 0, i.e. q > 1. Since w0 → u → v 2

with v ∈ V (Dβ ) and Dk → Dβ for k ∈ {λ, . . . , β − 1}, w0 is adjacent to every vertex of will discuss the possible edges between w0 and every vertex of

p−1

j=β V (Dj ) in G.

β−1 i=λ

V (Di ) in G. Subsequently, we

Subcase 3.1: β = p − 1. If |V (Dβ )| ̸= 1, then u → Dβ . We see that w0 v ∈ E (G) for any v ∈ V (Dβ ) by Lemma 3.1 because w0 → u →  Dβ . If not, then we have |V (Dβ )| = 1 and let V (Dβ ) = {v0 }. If ND+[S \{w0 }] (v0 ) ̸= ∅, then w0 v0 ∈ E (G) because w0 → D1 and

v0 → w → D1 for w ∈ ND+[S \{w0 }] (v0 ). Otherwise, ND+[S \{w0 }] (v0 ) = ∅. Then ND+ (v0 ) = V (D′1 ) or ND+ (v0 ) = V (D′1 ) ∪ {w0 }. So ND+ (D′1 ) = S, ND+ (w0 ) ⊆ V (D′3 ) and ND+ (D′3 ) ⊆ V (D′2 ). This means w0 v0 ̸∈ E (G), since the length of a path from w0 to any vertex in ND+ (v0 ) is at least 3 and the length of a path from w0 to any vertex in the outsets of ND+ (v0 ) is at least 2. Thus either w0 v ∈ E (G) for any vertex v ∈ V (Dβ ) or there exists at most one vertex v ′ ∈ V (Dβ ) such that w0 v ̸∈ E (G). Subcase 3.2: λ ≤ β < p − 1. By Theorem 2.8(b), there exists an arc (w1 , v1 ) from S to D′2 with v1 ∈ V (Di ) and w1 ∈ V (Dj ) by Theorem 2.8(b). Thus, we see that w1 → D′3 → Di by Proposition 2.7. Hence, λ ≤ i ≤ β since β is maximal. Since |ND+′ (w0 )| = 0, w1 ̸= w0 . Then 2  p−1

we have v1 → w0 because w1 → w0 and w1 → v1 . Hence, w0 is adjacent to every vertex of k=β+1 V (Dk ) in D because v1 → Dk for k ∈ {β + 1, . . . , p − 1} and v1 → w0 . Furthermore, we see that Dk → w0 for k ∈ {β + 1, . . . , p − 1} because  −1 |ND+′ (w0 )| = 0. Similarly, we have that w1 is adjacent to every vertex of pk=β+ 1 V (Dk ) in D. Thus we get that Dk → w1 2

p−1

since β is maximal. So w0 is adjacent to every vertex of k=β+1 V (Dk ) in G because w0 → D1 and Dk → w1 → D1 for k ∈ {β + 1, . . . , p − 1}. Subsequently, the possible edges between w0 and V (Dβ ) in G will be discussed. By Proposition 2.7, w0 → D1 → Dβ . If |V (Dβ )| > 1, then w0 is adjacent to every vertex of V (Dβ ) in G. If not, then we have |V (Dβ )| = 1. If there exists one vertex w ∈ S \ {w0 } such that Dβ → w, then w0 is adjacent to Dβ in G because w0 → D1 and Dβ → w → D1 . Otherwise, ND+−w0 (Dβ ) =

p−1

k=β+1

V (Dk ) ∪ V (D′1 ). According to the structure of D in this case (see Fig. 2),

w0 is not adjacent to Dβ in G because ND (w0 ) ⊆ V (D′3 ). +

Hence, every vertex of S is adjacent to every vertex of D′2 in G or there exist at most two vertices x ∈ S and y ∈ D′2 such that xy ̸∈ E (G). 

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Fig. 2. The structure of D in Subcase 3.2 of Claim 5.

Claim 6. Every vertex of S is adjacent to every vertex of D′3 in G or there is at most one edge xy ̸∈ E (G) such that x ∈ S and y ∈ V (D′3 ). Proof. Since S → D1 → Dλ and Di → Dλ for 2 ≤ i ≤ λ − 1, v ′ is adjacent to v ′′ for any v ′ ∈ S and v ′′ ∈ Subsequently, the possible edges between S and V (D1 ) in G will be discussed.

λ−1 i =2

V (Di ) in G.

Case 1: |V (D1 )| ̸= 1. Since S → D1 , we have v ′ is adjacent to v ′′ for any v ′ ∈ S and v ′′ ∈ V (D1 ) in G by Lemma 3.1. Case 2: |V (D1 )| = 1. Let V (D1 ) = {u0 }. Subcase 2.1: |ND+′ (S )| > 1. 3

Note that there exists an arc (w, u) from S to D′3 with u ∈ V (Di ) and w ∈ V (Dj ) for 2 ≤ i ≤ λ − 1 and p + 1 ≤ j ≤ p + q. If |V (Dj )| = 1, then it is obvious that Dj → u. If |V (Dj )| > 1, then there exists a hamiltonian cycle C in Dj because Dj is a strong tournament. For the vertex v ∈ ND+j (w) ∩ V (C ), it is easy to see v → u because there is no arc from D′3 to S in D. Continuing along the hamiltonian cycle C , we see that Dj → u. In any case, we have that Dj → u. Since Dj → Dk , we get that Dk → u for j + 1 ≤ k ≤ p + q. Since D1 (=u0 ) → u and Dk → Dj → u, Dk → u,



for p + 1 ≤ k ≤ j − 1, for j ≤ k ≤ p + q,

we see that v ′ is adjacent to v ′′ for any v ′ ∈ S and v ′′ ∈ V (D1 ) in G. Subcase 2.2: |ND+′ (S )| = 1. 3

We claim that λ = 2, i.e. there is only one strong component D1 in D′3 . Suppose not. Since there exists an arc from S to D′2 by Theorem 2.8(b), there is at least one vertex w ∈ S such that w → D′3 by Proposition 2.7, which contradicts |ND+′ (S )| = 1. 3

So there is only one vertex in V (D′3 ). By Theorem 2.8(b), there exists an arc (w, v) from S to D′2 satisfying w ∈ V (Dk ) and v ∈ V (Dl ) for p + 1 ≤ k ≤ p + q and λ ≤ l ≤ p − 1. Let M = max{k|w → v, v ∈ V (D′2 ) and w ∈ V (Dk ) for p + 1 ≤ k ≤ p + q}. If M = p + q, then dD−u0 (x, v) ≤ 2 for any vertex x ∈ S \ V (Dp+q ) because Dk → Dp+q for p + 1 ≤ k ≤ p + q − 1 and w → v . Thus u0 is adjacent to every vertex of Dk for p + 1 ≤ k < p + q in G because u0 → v by Proposition 2.7. If p + 1 ≤ M ≤ p + q − 1, i.e. there is an arc (w, v) for w ∈ V (DM ) and v ∈ V (D′2 ), then u0 → v by Proposition 2.7 and v → Dp+q because w → Dp+q , w → v and M is maximal. Thus u0 is adjacent to every vertex of Dk for p + 1 ≤ k ≤ p + q − 1 in G because Dk → Dp+q . Subsequently, the possible edges between {u0 } and V (Dp+q ) in G will be discussed. If ND+′ (Dp+q ) ̸= ∅, i.e. there is an arc (w, v) from Dp+q to D′2 , then we have u0 → v by Proposition 2.7. Then u0 is 2

adjacent to Dp+q in G if |V (Dp+q )| = 1. Otherwise, |V (Dp+q )| ≥ 3. It is easy to know that u0 is adjacent to every vertex of ND−p+q (w) ∪ {w} in G because ND−p+q (w) → w → v and u0 → v . According to the definition of a local tournament, v is adjacent to every vertex of ND+p+q (w) in D. If for x ∈ ND+p+q (w), x → v , then u0 x ∈ E (G). If not, then we have v → x. Since Dp+q

is strong, it is obvious ND+p+q (x) ̸= ∅. If ND+p+q (x) ∩ ND+p+q (w) ̸= ∅, i.e. there is a vertex x1 ∈ ND+p+q (w) such that x → x1 , then

u0 x ∈ E (G) because either u0 → v → x1 and x → x1 , or u0 → v and x → x1 → v . Otherwise, ND+p+q (x) ∩ ND+p+q (w) = ∅,

i.e. ND+p+q (x) ⊆ ND−p+q (w). Clearly, there is at most one such vertex x ∈ ND+p+q (w) with ND+p+q (x) ∩ ND+p+q (w) = ∅. For any

vertex z ∈ ND+p+q (x), if ND+′ (z ) ̸= ∅ then it is easy to see u0 x ∈ E (G) because u0 → v and x → z → v for v ∈ V (D′2 ). If not, 2

we have ND+′ (z ) = ∅ for any vertex z ∈ ND+p+q (x). Thus ND+ (x) = ND+p+q (x) ∪ {u0 } and the outset of ND+ (x) is u0 . In this case, 2

u0 x ̸∈ E (G). So either u0 is adjacent to every vertex of D′3 or there exists at most one vertex w ∈ S such that u0 w ̸∈ E (G).

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187

If ND+′ (Dp+q ) = ∅, then there exist two vertices v ∈ V (D′2 ) and w ∈ V (Dk ) for p + 1 ≤ k ≤ p + q − 1 such that w → v 2

by Theorem 2.8(b). Furthermore, we know that u0 → v by Proposition 2.7. Since w → Dp+q , we have v → Dp+q . Thus we see d(u0 , x) = 2 for any vertex x ∈ V (Dp+q ). If |V (Dp+q )| ≥ 3, then u0 is adjacent to every vertex of Dp+q in G by Lemma 3.1 because u0 → v → Dp+q . If |V (Dp+q )| = 1, then it is easy to know u0 is not adjacent to Dp+q in G because ND+ (Dp+q ) = {u0 }. Thus either every vertex of D′3 is adjacent to every vertex of S or there are at most two vertices u ∈ V (D′3 ) and w ∈ S such that uw ̸∈ E (G).  According to the discussion above, we know that either every vertex of S is adjacent to every vertex of V (D′i ) for i = 1, 2, 3 or there exist at most two vertices x ∈ S and y ∈ V (D′i ) for i = 1, 2, 3 such that xy ̸∈ E (G) and there is at most one edge missing between V (D′1 ) and V (D′2 ) in G. Furthermore, every vertex of V (D′j ) for j = 1, 2 is adjacent to every vertex of V (D′3 ) in G. Let ei (i = 1, 2, 3, 4) be the edge missing possibly in G between S and V (D′3 ), S and V (D′2 ), S and V (D′1 ), V (D′1 ) and V (D′2 ), respectively. Next, we consider the edges missing from G. Claim 7. e1 and e3 cannot be missing simultaneously in G. Proof. We assume there exist two vertices w ∈ S and y ∈ V (D′1 ) such that w y ̸∈ E (G). Then we have |V (D′1 )| = |S | = 1 by Claim 4, i.e. V (D′1 ) = {y} and S = {w}. By Theorem 2.8(b), we see there exists one vertex z ∈ V (D′2 ) such that w → z. By Proposition 2.7, D′3 → z. Thus w is adjacent to every vertex of D′3 in G. So there is no arc missing between S and V (D′3 ) in G.  Claim 8. e1 and e4 cannot be missing simultaneously in G. Proof. Suppose e1 and e4 are missing simultaneously in G. By Claim 3, we have |V (D′1 )| = 1. According to Case 1 of Claim 6, we see |V (D′3 )| = 1 because e1 is missing in G. Let V (D′3 ) = {u0 } and V (D′1 ) = {y0 }. Next, we distinguish two cases. Case 1: There is an arc (v, w) satisfying v ∈ V (Dp−1 ) and w ∈ S, i.e. ND+[S ] (Dp−1 ) ̸= ∅.

By Claim 3, the end-vertices of the missing arc e4 are x and y0 for some x ∈ Dp−1 and the vertex x ∈ ND+p−1 (v),

NDp−1 (x) ⊆ ND−p−1 (v), ND+ (x) ∩ S = ∅ and there is no arc between ND+p−1 (x) and S in D. Then w → x because v → w, v → x +

and ND+ (x) ∩ S = ∅. By Proposition 2.7, u0 → D′2 , and in particular, u0 → x. Furthermore, if w ∈ V (Dp+q ), then Dp+q → x because w → x and ND+ (x) ∩ S = ∅. If w ∈ V (Dk ) for p + 1 ≤ k ≤ p + q − 1, then we have that Dp+q → x because w → Dp+q , w → x and ND+ (x) ∩ S = ∅. Thus we see that u0 is adjacent to every vertex of Dp+q in G because u0 → x and Dp+q → x. Since the missing arc between S and D′3 is only possibly between Dp+q and u0 , every vertex of D′3 is adjacent to every vertex of S in G, a contradiction.

Case 2: There is no arc from Dp−1 to S, i.e. ND+[S ] (Dp−1 ) = ∅. By Theorem 2.8(b), there exists an arc (v, w) from D′2 to S satisfying v ∈ V (Dj ) for λ ≤ j ≤ p − 2 and w ∈ V (Dk ) for p + 1 ≤ k ≤ p + q. Then Dk → Dp−1 because v → w, v → Dp−1 and ND+[S ] (Dp−1 ) = ∅. Furthermore, we have that

Dα → Dp−1 for k + 1 ≤ α ≤ p + q because Dk → Dα , Dk → Dp−1 and ND+[S ] (Dp−1 ) = ∅. Thus we see that u0 is adjacent to every vertex of S in G because u0 → Dp−1 , Di → Dk → Dp−1 for p + 1 ≤ i ≤ k − 1 and Dj → Dp−1 for k ≤ j ≤ p + q, i.e. every vertex of D′3 is adjacent to every vertex of S, a contradiction. 

Claim 9. The edges ei and ej for i, j ∈ {2, 3, 4} and i ̸= j can be missing simultaneously in G and the two missing edges can exactly form one path of length 2. Proof. According to Fig. 3(b), we can see ei and ej for i, j ∈ {2, 3, 4} and i ̸= j can be missing simultaneously in G. By Claim 4, if e3 is missing in G then |V (D′1 )| = |S | = 1. Then it is easy to know that they will form a path of length 2 if e2 and e3 (or e3 and e4 ) are absent simultaneously in G. Next, we will discuss the case when e2 and e4 are missing simultaneously in G. Since there exist two vertices w ∈ S and z ∈ V (D′2 ) such that w z ̸∈ E (G), there must be |V (Dp+q )| = 1 by Claim 5. Let V (Dp+q ) = {w0 }. Then we only consider the cases |ND+′ (w0 )| = 1 and |ND+′ (w0 )| = 0 by Claim 5. We claim that |ND+′ (w0 )| = 1. 2

2

2

Suppose |ND+′ (w0 )| = 0. Then there exists one arc (x, y) from S to D′2 satisfying x ∈ V (Dj ) for some j with p + 1 ≤ j ≤ p + q − 1 2

and y ∈ V (Dk ) for some k with λ ≤ k ≤ p − 1 by Theorem 2.8(b). Since x → y, x → w0 and |ND+′ (w0 )| = 0, we have y → w0 . 2

If |V (Dk )| > 1, then ND+k (y) ̸= ∅ because Dk is a strong component. Thus z → w0 for any vertex z ∈ ND+k (y). So Dk → w0 in D. Since Dk → Dβ for k ≤ β ≤ p − 1 and |ND+′ (w0 )| = 0, we have Dβ → w0 for k + 1 ≤ β ≤ p − 1. Then every vertex of 2

V (D′2 ) is adjacent to every vertex of V (D′1 ) in G because Dα → Dk → w0 for λ ≤ α ≤ k − 1, Dβ → w0 for k + 1 ≤ β ≤ p − 1 and D′1 → w0 , which contradicts w z ̸∈ E (G). So |ND+′ (w0 )| = 1 when e2 and e4 are simultaneous in G. Subsequently, we will 2

prove that ND+′ (w0 ) = V (Dp−1 ) when e2 and e4 are missing simultaneously in G. 2

Since |ND+′ (w0 )| = 1, there exists one vertex v0 ∈ V (Dj ) for some j ∈ {λ, λ + 1, . . . , p − 1} such that w0 → v0 . By 2

Claim 5, V (Dj ) = {v0 }. If λ + 1 ≤ j ≤ p − 2, then Dλ → w0 because Dλ → v0 , w0 → v0 and |ND+′ (w0 )| = 1. Similarly, we 2

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X. Zhang, R. Li / Discrete Applied Mathematics 205 (2016) 180–190

Fig. 3. (a) The structure of a pure local tournament that is not round decomposable. (b) One example of Claim 10 in the proof of the necessity in Theorem 3.2.

have Dp−1 → w0 because Dλ → Dp−1 and Dλ → w0 . Thus every vertex of V (D′2 ) is adjacent to every vertex of V (D′1 ) in G because Di → Dp−1 → w0 for λ ≤ i ≤ p − 2 and D′1 → w0 , which contradicts that there exist two vertices u ∈ V (D′1 ) and v ∈ V (D′2 ) such that uv ̸∈ E (G). Next, we consider j = λ ≤ p − 2. By Theorem 2.8(b), there exists one arc from D′2 to S. If there is one vertex w ∈ S \ {w0 } such that v0 → w , then w0 v0 ∈ E (G) because w0 → D1 and v0 → w → D1 . In addition, we have w0 is adjacent to every vertex of D′2 − v0 in G because w0 → v0 → D′1 and v → D′1 for any vertex v ∈ V (D′2 ) \ {v0 }. Thus every vertex of S is adjacent to every vertex of V (D′2 ) in G, a contradiction. So ND+[S ] (v0 ) = ∅. If ND+′ −v (S ) ̸= ∅, i.e. w → v for some w ∈ S and v ∈ V (D′2 ) \ {v0 }, then w0 → D′3 → v by Proposition 2.7. Since v0 = Dλ , 2

0

v0 → Di for λ+ 1 ≤ i ≤ p − 1. Thus w0 v0 ∈ E (G) because v0 → v . So ND+′ −v (S ) = ∅. Since there exists one arc (x, y) ∈ A(D) 2

0

from D′2 to S satisfying x ∈ V (Di ) for some i ∈ {λ, λ + 1, . . . , p − 1} and y ∈ V (Dj ) for some j ∈ {p + 1, . . . , p + q}, it is easy to know that Dα → Dβ for any α ∈ {i, i + 1, . . . , p − 1} and β ∈ {p + 1, . . . , j}. Using Theorem 2.8, it is not difficult to verify D is a round digraph now, a contradiction. At last, we consider j = p − 1, i.e. ND+′ (w0 ) = V (Dp−1 ) = {v0 }. If ND+[S ] (Dp−1 ) ̸= ∅, then every vertex of D′2 is adjacent 2

to every vertex of D′1 in G because Di → Dp−1 → w for λ ≤ i ≤ p − 2 with w ∈ S and D′1 → S, a contradiction. So ND+ (v0 ) = {y0 } = V (D′1 ). Then v0 y0 ̸∈ E (G). Thus e2 (=w0 v0 )e4 (=v0 y0 ) is exactly a path of length 2 when they are simultaneous in G.  Claim 10. If e2 , e3 and e4 are simultaneously missing in G, then the three edges form a cycle of length 3. Proof. Assume e2 , e3 and e4 are missing. Since e2 and e3 are missing simultaneously in G, we see that |V (D′1 )| = |S | = 1 by Claim 4. Since e2 and e4 are missing simultaneously in G, it follows that |V (Dp−1 )| = 1 by Claim 9. Let V (D′1 ) = {y0 }, S = {w0 } and V (Dp−1 ) = {z0 }. Note that ND+′ (w0 ) = {z0 } and w0 is not adjacent to z0 in G by Claim 9. Also ND+ (z0 ) = {y0 } and hence z0 is not adjacent to y0 in G. Thus 2

the missing edges w0 z0 , z0 y0 , y0 w0 in G form exactly a cycle of length 3.



Claim 11. If the edges e1 and e2 can be simultaneously missing in G, then the two edges can exactly form one path of length 2. Proof. Firstly, it is easy to know e1 and e2 can be missing simultaneously in G according to the proofs of Claim 5 and Claim 6. Furthermore, we see that |V (Dp+q )| = 1 when e2 is missing in G by Claim 5 and the unique vertex of Dp+q must be one of the end-vertices of the missing edges when e1 and e2 are missing simultaneously in G. Thus the missing edges can exactly form one path of length 2.  Thus we complete the proof of necessity. (⇐) To complete the proof and make it more explicit we construct four kinds of corresponding pure local tournaments that are not round decomposable such that their (1, 2)-step competition graphs are Kn − E (K3 ), Kn − E (P3 ), Kn − E (P2 ) and Kn respectively, in this order. If D is a pure local tournament that is not round decomposable, then D is satisfied with Theorem 2.8. Thus we see that |S | ≥ 1, |V (D′3 )| ≥ 1 and V (D′1 ) ≥ 1. However, we note |V (D′2 )| ≥ 2. If not, then D is round decomposable, a contradiction. So |V (D)| ≥ 5. Case 1: Consider a pure local tournament D that is not round decomposable with |S | = 1, |V (D′1 )| = 1, |V (D′3 )| = 1, |V (Dp−1 )| = 1 and |V (Dλ )| = 1. Let p = λ + 2, which implies there exist exactly two strong components in D′2 and |V (D′2 )| = 2. So |V (D)| = 5. Let S = {w0 }, V (D′1 ) = {z0 }, V (D′3 ) = {x0 }, V (Dλ ) = {y1 } and V (Dp−1 ) = {y2 }. Let y1 → w0 and w0 → y2 (see Fig. 4(a)). Since x0 → D′2 and w0 → y2 , we have S ∪V (D′3 ) induces a complete subgraph in G. Since x0 → y2 and z0 → w0 → y2 , we see that V (D′3 ) ∪ V (D′1 ) induces a complete subgraph in G. Similarly, V (Dλ ) ∪ S , V (D′2 ) ∪ V (D′3 ) and V (Dλ ) ∪ V (D′1 ) induce complete subgraphs in G respectively. Note ND+ (w0 ) = {y2 } ∪ V (D′3 ), ND+ (y2 ) = {z0 } and ND+ (z0 ) = {w0 }, so we see w0 y2 ̸∈ E (G), y2 z0 ̸∈ E (G) and z0 w0 ̸∈ E (G) according to the definition of the (1, 2)-step competition graph. Thus we get C1,2 (D) = Kn − E (K3 ) for n = 5.

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Fig. 4. The four cases of the sufficiency for Theorem 3.2.

Without loss of generality, we add one strong component Dλ1 between Dλ and Dp−1 in D′2 above. Let |V (Dλ1 )| = 1 and Dλ1 → S. Thus we get a new pure local tournament D′ that is not round decomposable. Let V (Dλ1 ) = {y3 }. Since Dλ1 → Dp−1 and D′3 → Dp−1 , we have V (Dλ1 ) ∪ V (D′3 ) induces a complete subgraph in G. Since D′2 → D′1 , we have V (Dλ1 ) ∪ V (Dλ ) ∪ V (Dp−1 ) induces a complete subgraph in G. Similarly, we see V (Dλ1 ) ∪ S and V (Dλ1 ) ∪ V (D′1 ) induce complete subgraphs in G respectively because Dλ1 → Dp−1 and D′1 → S → Dp−1 in D′ . Thus we see C1,2 (D′ ) = Kn − E (K3 ) for n = 6. Continuing to add the strong components Dj in D′2 as above such that |V (Dj )| = 1 and Dj → S, we can get a pure local tournament D′′ that is not round decomposable such that C1,2 (D′′ ) = Kn − E (K3 ) for n ≥ 7. In conclusion, we can construct a pure local tournament D that is not round decomposable such that C1,2 (D) = Kn − E (K3 ) for n ≥ 5. Case 2: Consider D with |S | = 1, |V (D′1 )| = 1, |V (D′3 )| = 1, |V (Dλ )| = 1, |V (Dλ+1 )| = 1, |V (Dp−1 )| = 1 and p = λ + 3 which implies there exist exactly three strong components in D′2 . Then |V (D)| = 6. Let S = {w0 }, V (D′1 ) = {z0 }, V (D′3 ) = {x0 }, V (Dλ ) = {y0 }, V (Dλ+1 ) = {y1 } and V (Dp−1 ) = {y2 }. Let y0 → S , S → y1 and S → y2 (see Fig. 4(b)). According to the proof of Case 1, we have S ∪ V (D′3 ), V (D′3 ) ∪ V (D′2 ) and V (D′3 ) ∪ V (D′1 ) induce complete subgraphs in G respectively. Since w0 → y1 → z0 and w0 → y2 → z0 , we have S ∪ V (D′2 ) induces a complete subgraph in G. Since ND+ (z0 ) = {w0 } and ND+ (y2 ) = {z0 }, we see y2 z0 ̸∈ E (G) and z0 w0 ̸∈ E (G). So C1,2 (D) = Kn − E (P3 ) for n = 6. Note that the unique pure local tournament with exactly 5 vertices that is not round decomposable is the digraph described in Case 1 and its (1, 2)-step competition graph is Kn − E (K3 ). So we must have n ≥ 6 if Kn − E (P3 ) is the C1,2 (D) of some pure local tournament D that is not round decomposable.

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Construct a pure local tournament that is not round decomposition by adding one strong component Dλ1 between Dλ+1 and Dp−1 in D′2 from D such that |V (Dλ1 )| = 1 and S → Dλ1 . Thus we get a new pure local tournament D′ that is not round decomposable. Let V (Dλ1 ) = {y3 }. Since D′2 → z0 , we have yi yj ∈ E (G) for i, j ∈ {0, 1, 2, 3}. Since y3 → y2 , z0 → w0 → y2 and x0 → y2 , we have y3 z0 , y3 w0 and y3 x0 are the edges of G. So we see y3 v ∈ E (G) for any vertex v ∈ V (D′ ) \ {y3 }. According to the proof above, we have C1,2 (D′ ) = Kn − E (P3 ) for n = 7. Continuing to add the strong components Dj in D′2 as above such that |V (Dj )| = 1 and S → Dj , we can get a pure local tournament D′′ that is not round decomposable such that C1,2 (D′′ ) = Kn − E (P3 ) for n ≥ 8. In conclusion, we can construct a pure local tournament D that is not round decomposable such that C1,2 (D) = Kn − E (P3 ) for n ≥ 6. Case 3: Consider D with |S | = 1, |V (D′3 )| = 1, |V (Dλ )| = 1, |V (Dp−1 )| = 1, p = λ + 2 and |V (D′1 )| = 3 which is a 3-cycle. So |V (D)| = 7. Let S = {w0 }, V (D′3 ) = {x0 }, V (Dλ ) = {y1 }, V (Dp−1 ) = {y2 } and V (D′1 ) = {z0 , z1 , z2 }. Let y1 → S and S → y2 (see Fig. 4(c)). Since x0 → D′2 and D′1 → S → y2 → D′1 , we see S ∪ V (D′3 ), S ∪ V (D′1 ) and V (D′1 ) ∪ V (D′3 ) induce complete subgraphs in G respectively. Since x0 → D′2 → D′1 , we have V (D′1 ) ∪ V (D′2 ) and V (D′2 ) ∪ V (D′3 ) induce complete subgraphs in G respectively. Note w0 → y2 and y1 → y2 , we have w0 y1 ∈ E (G). Since ND+ (y2 ) = V (D′1 ), ND+ (D′1 ) = {w0 } and ND+ (w0 ) = {x0 , y2 }, we see w0 y2 ̸∈ E (G). Hence we see C1,2 (D) = Kn − E (P2 ) for n = 7. Furthermore, it is similar to Case 1 and Case 2 that we can add the strong components Dλj between Dλ and Dp−1 in D′2 such that |V (Dj )| = 1 and Dλj → S for j = 1, 2, . . . , and then get the pure local tournament D′′ that is not round decomposable such that C1,2 (D′′ ) = Kn − E (P2 ) for n ≥ 8. From the proof of necessity, we must have n ≥ 7 if Kn − E (P2 ) is the C1,2 (D) of some pure local tournament D that is not round decomposable. In conclusion, we see Kn − E (P2 ) is the C1,2 (D) of some pure local tournament D that is not decomposable for |V (D)| ≥ 7. Case 4: Consider D with |S | = 1, |V (D′1 )| = 3, |V (D′3 )| = 1, |V (Dλ )| = 1, |V (Dλ+1 )| = 1, |V (Dp−1 )| = 1 and p = λ + 2 which means there exist exactly three strong components in D′2 . At this time, we have |V (D)| = 8. Let S = {w0 }, V (D′3 ) = {x0 }, V (Dλ ) = {y0 }, V (Dλ+1 ) = {y1 }, V (Dp−1 ) = {y2 }. Let y0 → w0 , w0 → y1 and w0 → y2 (see Fig. 4(d)). According to the proof of Cases 1, 2, and 3, we have S ∪ V (D′3 ), S ∪ V (D′1 ) and V (D′1 ) ∪ V (D′2 ) ∪ V (D′3 ) induce complete subgraphs in G respectively. Since w0 → y1 → D′1 and w0 → y2 → D′1 , we have S ∪ V (D′2 ) induces a complete subgraph in G. Thus we see C1,2 (D) = Kn for n = 8. Continuing to add the new vertices in D′1 , we can get a pure local tournament D′ that is not round decomposable such that C1,2 (D′ ) = Kn for n ≥ 9. From the proof of necessity, we must have n ≥ 8 if Kn is the C1,2 (D) of some pure local tournament D that is not round decomposable. In conclusion, we can construct a pure local tournament D that is not decomposable satisfying C1,2 (D) = Kn for |V (D)| ≥ 8. Thus we complete the proof of sufficiency.  We complete the characterization of the (1, 2)-step competition graph of a pure local tournament that is not round decomposable by Theorem 3.2. Acknowledgments We express our sincere thanks to the referees for their valuable suggestions and detailed comments. Research is partially supported by the National Natural Science Foundation of China (11401353), Program for the Top Young Academic Leaders of Higher Learning Institutions of Shanxi, Youth Foundation of Shanxi Province (2013021001-5) and Shanxi Scholarship Council of China (2013-017). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

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