The construction of graph G k in the proof of Proposition 5, with an optimal LTD-set (black vertices). Let C be an LTD-set of G k . Consider the k − 1 closed neighborhoods N [s i ] for 1 ≤ i < k. If we have |N [s i ] ∩ C| ≥ 2 for all i with 1 ≤ i < k, then 1≤i<k (N [s i ] ∩ C) ≥ 2k − 2, and at least one of the remaining vertices s k , q k , q ′ k must belong to C, as otherwise N (q k ) ∩ C = N (q ′ k ) ∩ C would follow, a contradiction. This implies |C| ≥ 2k − 1. If, however, for some i with 1 ≤ i < k, we have |N [s i ] ∩ C| = 1, then s i / ∈ C since otherwise s i is not totally dominated by the set C. If N [s i ] ∩ C = {q i }, then N (q ′ i ) ∩ C = Q ∩ C. If N [s i ] ∩ C = {q ′ i }, then N (q i ) ∩ C = (Q ∪ {s k }) ∩ C. The two possibilities can occur at most once each. Assume that they both occur once each, with N [s a ] ∩ C = {q ′ a } and N [s b ] ∩ C = {q b } (with 1 ≤ a < b < k). Note that s k ∈ C, otherwise q a and q ′ b are not located. Moreover, C must contain q ′ k (otherwise q ′ b and q ′ k are not located) and q k (otherwise q a and q k are not located), and so |C| ≥ 2k − 1, as claimed. Similarly, if we have |N [s i ] ∩ C| ≥ 2 for all i with 1 ≤ i < k except that N [s a ] ∩ C = {q ′ a }, if s k ∈ C, then q k ∈ C, otherwise q a and q k are not located. If s k / ∈ C, then both q k , q ′ k are in C to locate the vertices q a , q k , q ′ k . Thus, again |C| ≥ 2k − 1. Finally, if we have |N [s i ] ∩ C| ≥ 2 for all i with 1 ≤ i < k except that N [s b ] ∩ C = {q b }, if s k ∈ C, then q ′ k ∈ C, otherwise q ′ b and q ′ k are not located. If s k / ∈ C, then both q k , q ′ k are in C to locate the vertices q ′ b , q k , q ′ k , and again |C| ≥ 2k − 1. Thus, in all the above cases, we have |C| ≥ 2k − 1 and, together with the upper bound γ L t (G k ) <

Contexts in source publication

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... that q ′ k has no neighbor in S and that the sets N (s i ) are disjoint for 1 ≤ i < k. See Figure 1 for an illustration. ...

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... n ≥ 7. In what follows, we adopt the notation that if there is a (d 1 , d 2 , . . . , d k )-sequence in G, then P : v 1 v 2 . . . ...

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... of subclaim. (a) Suppose that there is a (1, 3, 1)-sequence in G. In this case, n ′ = n − 3 ≥ 4. ...

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... that there is a (1, 2, 2)-sequence in G. Let v ′ be the second neighbor of v 3 . As in the previous case, n ′ = n − 3 ≥ 4 and G ′ is connected. ...

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... in the previous case, n ′ = n − 3 ≥ 4 and G ′ is connected. By part (a), there is no (1, 3, 1)-sequence, implying that G ′ / ∈ F tdom and γ L t (G ′ ) ≤ 2 3 n ′ = 2 3 n − 2. As before every γ L t -set of G ′ can be extended to a LTD-set of G by adding to it the vertices v 2 and v 3 , implying that γ L t (G) ≤ 2 3 n, a contradiction. (c) Suppose that there is a (1, 2, 3, 1)-sequence in G. ...

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... part (a), there is no (1, 3, 1)-sequence, implying that G ′ / ∈ F tdom and γ L t (G ′ ) ≤ 2 3 n ′ = 2 3 n − 2. As before every γ L t -set of G ′ can be extended to a LTD-set of G by adding to it the vertices v 2 and v 3 , implying that γ L t (G) ≤ 2 3 n, a contradiction. (c) Suppose that there is a (1, 2, 3, 1)-sequence in G. In this case, n ′ = n − 4 ≥ 3 and G ′ is connected. ...

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... this case, n ′ = n − 4 ≥ 3 and G ′ is connected. By part (a), there is no (1, 3, 1)-sequence, implying that G ′ / ∈ F tdom and γ L t (G ′ ) ≤ 2 3 n ′ = 2 3 (n−4) < 2 3 n−2. Every γ L t -set of G ′ can be extended to a LTD-set of G by adding to it the vertices v 2 and v 3 , implying that γ L t (G) ≤ 2 3 n, a contradiction. ...

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... γ L t -set of G ′ can be extended to a LTD-set of G by adding to it the vertices v 2 and v 3 , implying that γ L t (G) ≤ 2 3 n, a contradiction. (d) Suppose that there is a (1, 2, 3, 2, 1)-sequence in G. In this case, G ′ is connected and n ′ = n−5 ≥ 2. If G ′ ∈ F tdom , then G ′ ∼ = K 2 by the fact that there is no (1, 3, 1)-sequence in G by part (a). ...

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... Suppose that there is a (1, 2, 3, 2, 1)-sequence in G. In this case, G ′ is connected and n ′ = n−5 ≥ 2. If G ′ ∈ F tdom , then G ′ ∼ = K 2 by the fact that there is no (1, 3, 1)-sequence in G by part (a). The graph G is therefore determined, and is obtained from a star K 1,3 by subdividing every edge once. ...

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... Suppose that there is a (1, 2, 3)-sequence in G. In this case, n ′ = n− 3 ≥ 4 and G ′ contains at most two components. ...

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... γ L t -set of G ′ can be extended to a LTD-set of G by adding to it the vertices v 2 and v 3 , implying that γ L t (G) ≤ 2 3 n, a contradiction. (f) Since there is no (1, 2, 1)-sequence (since n ≥ 7), no (1, 2, 2)-sequence by (b) and no (1, 2, 3)-sequence by (e), there can be no (1, 2)-sequence in G. Hence, part (f) follows immediately from parts (b) and (e). ...

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... Suppose that there is a (1, 3, 2)-sequence in G. In this case, n ′ = n − 3 ≥ 4. If G ′ is disconnected, then by parts (a)-(f), neither component of G ′ belongs to F tdom . ...

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... Suppose that there is a (1, 3, 3, 1)-sequence in G. In this case, n ′ = n − 4 ≥ 3 and G ′ contains at most two components. ...

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... Subclaim H.1(f), the neighbor of every vertex of degree 1 has degree 3 in G. Further by Subclaim H.1(g), such a vertex of degree 3 has both its other two neighbors of degree 3. Therefore the existence of a vertex of degree 1 implies that there is a (1, 3, 3)-sequence in G. In this case, n ′ = n − 3 ≥ 4. Let u 2 be the neighbor of v 2 not on P , and let u 3 and w 3 be the two neighbors of v 3 not on P . ...