The domination number of round digraphs

Lemma 2.1

Let D be a round purely local tournament which is non-strong on n vertices. Let P = v 0 v 1 … v n − 1 be a Hamilton path in D. If P m = v i 0 v i 0 + 1 … v i 0 + ( m − 1 ) is a maximal pure subpath on P, then γ ( P m ) = m 2 .

Proof

Since P m is a directed path, v j → v j + 1 for j ∈ { i 0 , i 0 + 1 , … , i 0 + ( m − 2 ) } .

When m is even, let T = { v i 0 , v i 0 + 2 , v i 0 + 4 , … , v i 0 + ( m − 2 ) } . Since v j → v j + 1 for j ∈ { i 0 , i 0 + 1 , … , i 0 + ( m − 2 ) } , we have T → P m . Thus, γ ( P m ) ≤ | T | = m 2 . Choosing any vertex set M = { v k 1 , v k 2 , … , v k m 2 − 1 } ⊆ V ( D ) , there must exist a vertex v τ ∈ V ( P m ) such that { v τ , v τ + 1 } ⊈ M . By maximal pure subpath P m , we see that v ↛ v τ + 1 for any vertex v ∈ M since N D − ( v τ + 1 ) = { v τ } . Therefore, M is not a dominating set of P m . For the arbitrariness of M, we have γ ( P m ) ≥ m 2 . Thus, γ ( P m ) = m 2 = m 2 .

When m is odd, let T = { v i 0 , v i 0 + 1 , v i 0 + 3 … , v i 0 + ( m − 2 ) } . It is easy to see that T → P m and | T | = m 2 . So γ ( P m ) ≤ m 2 . Choosing any vertex set M ′ = { v j 1 , v j 2 , … , v j m 2 − 1 } ⊆ V ( D ) , there must exist a vertex v τ ∈ V ( P m ) such that { v τ , v τ + 1 } ⊈ M ′ . By maximal pure subpath P m , we have v ↛ v τ + 1 for any vertex v ∈ M ′ . Then M ′ is not a dominating set of P m . It implies γ ( P m ) ≥ m 2 . Thus, γ ( P m ) = m 2 .□

Lemma 2.2

Let D be a round purely local tournament which is non-strong on n vertices. Let v 0 , v 1 , … , v n − 1 be a round labelling of D. P = v 0 v 1 … v n − 1 is the Hamiltonian path in D. If B = ⋃ t = 0 k − 1 { v i t , v i t + 1 , … , v j t } ⊆ V ( D ) is covered by the maximal cross-arc chain G = { v i t v j t | t = 0 , 1 , 2 , … , k − 1 } , then γ ( D 〈 B 〉 ) = τ , where τ is the number of all valid arcs in G, and all subscripts are ordered in the round ordering.

Proof

Now we distinguish two cases to prove this lemma.

Case 1. There is no invalid cross arc in G.

According to the definitions of the maximal cross arc and the round digraph, there must be v i t → v l where l ∈ { i t + 1 , i t + 2 , … , j t } for t ∈ { 0 , 1 , 2 , … , k − 1 } . Since i t < i t + 1 ≤ j t < j t + 1 for t ∈ { 0 , 1 , … , k − 2 } , we have { v i t | t = 0 , 1 , … , k − 1 } → D 〈 B 〉 . Thus, γ ( D 〈 B 〉 ) ≤ k = τ .

For arbitrary t ∈ { 1 , 2 , … , k − 2 } , v i t v j t is a valid cross arc of G. By the definition of a valid cross arc, | i t + 1 − j t − 1 | ≥ 2 . Since G is a maximal cross-arc chain, | i 1 − i 0 | ≥ 2 and | j k − 1 − j k − 2 | ≥ 2 . Thus, there is at least a vertex v i t 0 in the vertex set { v i t , v i t + 1 , … , v j t } such that v i t 0 is only covered by the maximal cross arc v i t v j t in G, where t ∈ { 0 , 1 , 2 , … , k − 1 } . Let A t = { v i t , v i t + 1 , … , v i t 0 } , t ∈ { 0 , 1 , … , k − 1 } . It is easy to see A i ∩ A j = ∅ , i , j ∈ { 0 , 1 , … , k − 1 } . For any vertex set M ⊆ B for | M | = k − 1 , there exists α ∈ { 0 , 1 , … , k − 1 } such that A α ∩ M = ∅ . By the round purely local tournament D which is non-strong, we have v ↛ v α ∈ A α for any v ∈ A β with β ∈ { α + 1 , α + 2 , … , k − 1 } . Since v α 0 ∈ A α is only covered by the maximal cross arc v i α v j α , we have v γ ↛ v α 0 for γ ∈ { i 0 , i 0 + 1 , … , i α − 1 } . Thus, M ↛ v α 0 . According to the arbitrariness of M, γ ( D 〈 B 〉 ) ≥ k = τ . So γ ( D 〈 B 〉 ) = k = τ (Figure 3(a)).

Case 2. There is at least one invalid cross arc in G.

Let v i α v j α be any invalid cross arc of G. According to the definition of an invalid cross arc, | i α + 1 − j α − 1 | = 1 . It implies that { v i α , v i α + 1 , … , v j α − 1 } is covered by the maximal cross arc v i α − 1 v j α − 1 and { v j α − 1 + 1 , v j α − 1 + 2 , … , v j α } is covered by the maximal cross arc v i α + 1 v j α + 1 . By the definitions of the cover and the round digraph, we obtain { v i α − 1 , v i α + 1 } → { v i α , v i α + 1 , … , v j α } .

For the arbitrariness of v i α v j α , all vertices covered by the invalid cross arcs in G can be covered by the valid cross arcs in G. This means γ ( D 〈 B 〉 ) ≤ τ , where τ is the number of valid arcs in G. According to Case 1, γ ( D 〈 B 〉 ) ≥ τ can be proved similarly. Then, γ ( D 〈 B 〉 ) = τ (Figure 3(b)).□

Lemma 2.3

Let D be a round purely local tournament which is non-strong on n vertices. Let v 0 , v 1 , … , v n − 1 be a round labelling of D. Let P = v 0 v 1 … v n − 1 be the Hamilton path in D, P i = v i 0 v i 0 + 1 … v i 1 − 1 ( P j = v j t − 1 + 1 v j t − 1 + 2 … v j t ) be a maximal subpath on P, and G i = { v i k v j k | k = 1 , 2 , … , t } ( G j = { v i k v j k | k = 0 , 1 , … , t − 1 } ) be a maximal cross-arc chain on P adjacent to P i ( P j ) . If B and C represent the set of vertices covered by P i ( P j ) and G i ( G j ) , respectively, then γ ( D 〈 B ∪ C 〉 ) = γ ( D 〈 B 〉 ) + γ ( D 〈 C 〉 ) .

Proof

Let T B and T C be the minimum dominating set of D 〈 B 〉 and D 〈 C 〉 , respectively. It implies γ ( D 〈 B 〉 ) = | T B | , γ ( D 〈 C 〉 ) = | T C | . It is easy to see T B ∪ T C → D ( 〈 B ∪ C 〉 ) . Thus, we have γ ( D 〈 B ∪ C 〉 ) ≤ | T B | + | T C | . Without loss of generality, suppose that all the cross arcs in G i ( G j ) are valid. Choose any vertex set M ⊆ V ( D ) satisfying | M | = | T B | + | T C | − 1 . According to the proof of Lemma 2.2 and the structure of the round purely local tournament which is non-strong, the minimum dominating set of D 〈 C 〉 must be E = { v τ k | k = 1 , 2 , … , t } ( E = { v τ k | k = 0 , 1 , … , t − 1 } ) , where v τ k ∈ { v i k , v i k + 1 , … , v i k + m } , v i k + m is a vertex in C with the smallest subscript which is only covered by the valid cross arc v i k v j k . It is not difficult to see that | E | = | T C | . Thus, if | M ∩ E | ≤ | T C | − 1 , then M cannot dominate D 〈 C 〉 , which means M is not a dominating set of D. Otherwise, M contains | T B | − 1 vertices in B at most. As we know from Lemma 2.1, there exists v α ∈ { v i 0 , v i 0 + 1 , … , v i 1 − 2 } ( { v j t − 1 + 1 , v j t − 1 + 2 , … , v j t − 1 } ) such that { v α , v α + 1 } ∩ M = ∅ . By the structure of D and the pure subpath D 〈 B 〉 , we have M ↛ v α + 1 , i.e. M is not a dominating set of D 〈 B ∪ C 〉 . Due to the arbitrariness of M, γ ( D 〈 B ∪ C 〉 ) ≥ | T B | + | T C | . Thus, γ ( D 〈 B ∪ C 〉 ) = | T B | + | T C | = γ ( D 〈 B 〉 ) + γ ( D 〈 C 〉 ) (Figure 4).□

Theorem 2.4

Let D be a round purely local tournament which is non-strong on n vertices. Let v 0 , v 1 , … , v n − 1 be a round labelling of D. P = v 0 v 1 … v n − 1 is the Hamiltonian path of D. If there exist k maximal cross-arc chains G i ( i = 1 , 2 , … , k ) and l maximal pure subpaths P j ( j = 1 , 2 , … , l ) on P, then γ ( D ) = ∑ i = 1 k τ i + ∑ j = 1 l n j 2 , where l ∈ { k − 1 , k , k + 1 } , τ i is the number of valid cross arcs in G i , and n j is the number of vertices contained in P j .

Proof

According to the structure of D, we obtain the vertices of D are either covered by a maximal cross-arc chain or covered by a maximal pure subpath. According to the definition of the maximal cross-arc chain and the maximal pure subpath, one can get γ ( D ) ≤ ∑ i = 1 k τ i + ∑ j = 1 l ⌈ n j 2 ⌉ by Lemmas 2.1 and 2.2.

We give a proof for γ ( D ) ≥ ∑ i = 1 k τ i + ∑ j = 1 l ⌈ n j 2 ⌉ as follows. Choose any vertex set M ⊆ V ( D ) for | M | ∑ i = 1 k τ i + ∑ j = 1 l ⌈ n j 2 ⌉ , then one of the following two cases holds at least:

Case 1. There is a maximal cross-arc chain G α = { v i t v j t | t = 1 , 2 , … , τ α } such that | M ∩ { v β | β = i 1 , i 1 + 1 , … , j τ α } | ≤ τ α − 1 .

By the proof of Case 1 in Lemma 2.2, there must be v α 0 ∈ { v β | β = i 1 , i 1 + 1 , … , j τ α } such that M ↛ v α 0 . Then M cannot dominate D.

Case 2. There is a maximal pure subpath P β such that | M ∩ V ( P β ) | ≤ n β 2 − 1 , where n β = | V ( P β ) | .

According to Lemma 2.1, there must be a vertex set { v i β , v i β + 1 } ⊆ V ( P β ) such that M ∩ { v i β , v i β + 1 } = ∅ . Due to the structure of D, it is obvious N D − ( v i β + 1 ) = { v i β } , which means M ↛ v i β + 1 . Then M cannot dominate D.

Therefore, M is not a dominating set of D anyway. By the arbitrariness of M, γ ( D ) ≥ ∑ i = 1 k τ i + ∑ j = 1 l n j 2 . Thus, γ ( D ) = ∑ i = 1 k τ i + ∑ j = 1 l n j 2 .□

Theorem 2.5

Let D be a directed cycle v 0 v 1 … v n − 1 v 0 , then γ ( D ) = ⌈ n 2 ⌉ , where all subscripts are taken modulo n.

Proof

Since D is a directed cycle, we have v i → v i + 1 for 0 ≤ i ≤ n − 1 .

When n is even, let T = { v 0 , v 2 , v 4 , … , v n − 2 } . By the condition v i → v i + 1 for 0 ≤ i ≤ n − 1 , we obtain T → D , and then γ ( D ) ≤ n 2 . Choose arbitrary vertex set M = { v i 1 , v i 2 , … , v i n 2 − 1 } , there must exist α ∈ { 0 , 1 , 2 , … , n − 1 } such that { v α , v α + 1 } ⊈ M . Since D is a cycle, we have M ↛ v α + 1 . So M is not a dominating set of D. By the arbitrariness of M, we have γ ( D ) ≥ n 2 . Therefore, γ ( D ) = n 2 .

When n is odd, it is easy to see the vertex set T = { v 0 , v 1 , v 3 , … , v n − 2 } is a dominating set of D because D is a directed cycle. Thus, γ ( D ) ≤ ⌈ n 2 ⌉ . Choose any vertex set M = { v j 1 , v j 2 , … , v j ⌈ n 2 ⌉ − 1 } , there must exist β ∈ { 0 , 1 , 2 , … , n − 1 } such that { v β , v β + 1 } ⊈ M . Since D is a cycle, M ↛ v β + 1 . Therefore, M is not a dominating set of D. For the arbitrariness of M, we get γ ( D ) ≥ ⌈ n 2 ⌉ . Thus, γ ( D ) = ⌈ n 2 ⌉ .□