On the Hull Number of Triangle-Free Graphs

On the Hull Number of Triangle-free Graphs Mitre C. Dourado1 , Fábio Protti2 , Dieter Rautenbach3 , and Jayme L. Szwarcfiter4 1 2 3 4 ICE, Universidade Federal Rural do Rio de Janeiro and NCE - UFRJ, Brazil, email: [email protected] Universidade Federal do Rio de Janeiro, Instituto de Matemática, NCE, Caixa Postal 2324, 20001-970 Rio de Janeiro, RJ, Brasil, email: [email protected] Institut für Mathematik, Technische Universität Ilmenau, Postfach 100565, D-98684 Ilmenau, Germany, email: [email protected] Universidade Federal do Rio de Janeiro, Instituto de Matemática, NCE and COPPE, Caixa Postal 2324, 20001-970 Rio de Janeiro, RJ, Brasil, email: [email protected] Abstract. A set of vertices C in a graph is convex if it contains all vertices which lie on shortest paths between vertices in C. The convex hull of a set of vertices S is the smallest convex set containing S. The hull number h(G) of a graph G is the smallest cardinality of a set of vertices whose convex hull is the vertex set of G. For a connected triangle-free graph G of order n and diameter d ≥ 3, we prove that h(G) ≤ (n − d + 3)/3, if G has minimum degree at least 3 and that h(G) ≤ 2(n − d + 5)/7, if G is cubic. Furthermore, for a connected graph G of order n, girth  g ≥ 4, minimum degree at least 2, and diameter d, we prove h(G) ≤ 2 + (n − d − 1)/ g−1 . All bounds are 2 best possible. Keywords. Convex hull; convex set; geodetic number; graph; hull number 1 Introduction We consider finite, simple, and undirected graphs G with vertex set V and edge set E. The degree of a vertex u in G is the number of neighbours of u in G. The graph G is cubic if every vertex has degree 3. The distance dist(u, v) between two vertices u and v in G is the minimum number of edges of a path in G between u and v or ∞, if no such path exists. The diameter of G is the largest distance between two vertices in G. The graph G is triangle-free if it does not contain three pairwise adjacent vertices. The girth of G is the minimum length of a cycle in G or ∞, if G has no cycle. For two sets of vertices C and D let I[C, D] denote the set of all vertices which lie on a path of length dist(C, D) = min{dist(u, v) | u ∈ C and v ∈ D} between a vertex in 1 C and a vertex in D. Furthermore, let I(C, D) = I[C, D] \ (C ∪ D). We write I[u, v], I(u, v), I[u, D], and I(u, D) instead of I[{u}, {v}], I({u}, {v}), I[{u}, D], and I({u}, D), respectively. S For a set of vertices S let I[S] = I[u, v]. The set S is convex if I[S] = S. The u,v∈S convex hull of S is the smallest convex set C with S ⊆ C and is denoted by H[S]. Since the intersection of two convex sets is convex, the convex hull is well defined. In [9] Everett and Seidman introduce the hull number h(G) of a graph G as the minimum cardinality of a set of vertices whose convex hull contains all vertices of G. Similarly, Harary et al. [11] define the geodetic number g(G) of a graph G as the minimum cardinality of a set of vertices S for which I[S] contains all vertices of G. It was observed in [5, 9] that, if u and v are two vertices at maximum distance in a connected graph G of order n and diameter d, then I(u, v) contains at least d − 1 elements and I[V \ I(u, v)] contains all vertices of G which immediately implies h(G) ≤ g(G) ≤ n − d + 1. Everett and Seidman [9] strengthened this simple bound as follows. Theorem 1 (Everett and Seidman [9]) Let G be a connected graph of order n and diameter d. If there is no clique K in G such that every vertex of G is at distance at most d − 1 from K, then h(G) ≤ n−d+3 . 2 (1) The hypothesis of Theorem 1 is obviously rather restrictive. If a graph G of diameter d satisfies the hypothesis of Theorem 1, then its radius equals its diameter. Even more strongly, for every breadth first search tree T of G rooted at a vertex u, there are two non-adjacent neighbours of u such that both have descendants in T at depth d. Our first contribution in Section 2 is to show that (1) still holds under a less restrictive hypothesis. In Section 3, we prove best possible upper bounds on the hull number of triangle-free graphs which are either of minimum degree at least two or three or cubic. Finally, in Section 4, we prove a best possible upper bound on the hull number of graphs of large girth and minimum degree at least two. For further results concerning the hull number and convexity in graphs, we refer the reader to [1, 3, 4, 6–8, 10, 12, 13]. 2 A weaker hypothesis for Theorem 1 A natural strategy already adopted by Everett and Seidman [9], to construct a small set of vertices in a graph whose convex hull equals the entire vertex set, is to start with a small initial set having a large convex hull and to extend this set iteratively while its convex 2 hull still does not contain all vertices. A reasonable choice for the elements of the initial set are two vertices at maximum distance. To extend the set, one can iteratively add vertices greedily trying to maximize the cardinality increase of the convex hull divided by the number of added vertices. As a first and simple illustration of this strategy, we prove that the bound in Theorem 1 still holds under a less restrictive hypothesis. We say that a graph G has property P if there are no three sets of vertices A, B and C such that • A is the vertex set of a component of G − C, • B is a clique and, for every vertex v in A, the set of neighbours of v in C is B, and • C is convex. To check that property P is more general than the hypothesis of Theorem 1, simply note that the existence of A, B and C as above for some graph G of diameter d implies that every vertex of G is at distance at most d − 1 from B. The following lemma is somewhat implicit in the proof of Theorem 9 in [9]. Lemma 2 Let G be a connected graph and let S be a non-empty set of vertices whose convex hull does not contain all vertices of G. If G has property P, then there is a vertex u ∈ V \S such that |H[S ∪{u}]| ≥ |H[S]|+2. Proof: Let G and S be as in the statement of the lemma. For contradiction, we assume that a vertex with the desired property does not exist. Let C = H[S]. Let A be the vertex set of a component of G − C. If there is a vertex u ∈ A with dist(u, C) ≥ 2, then u has the desired property. Hence all vertices in A have a neighbour in C. If a vertex u ∈ A has two neighbours v, w ∈ C which are not adjacent, then u ∈ I[v, w] ⊆ H[S] which is a contradiction. Hence the set of neighbours in C of every vertex in A is a clique. If there are two adjacent vertices u, v ∈ A such that v has a neighbour v ′ ∈ C which is not adjacent to u, then v ∈ I[u, v ′ ] and u has the desired property. Hence the sets of neighbours in C of the vertices in A are all equal to some set B. In view of the sets A, B and C, the graph G does not satisfy property P. This final contradiction completes the proof. ✷ With Lemma 2 at hand, the proof of the following result is straightforward. Theorem 3 Let G be a connected graph of order n and diameter d. If G has property P, then n−d+3 . h(G) ≤ 2 Proof: Since n − d ≥ 1, we may assume that h(G) ≥ 3. Let S0 = {u, v} for two vertices u and v with dist(u, v) = d. H[S0 ] contains at least d + 1 vertices. By Lemma 2, there is a sequence S0 ⊆ S1 ⊆ S2 ⊆ . . . of sets of vertices such that for i ≥ 1, |Si | = 2 + i and either H[Si−1 ] = V or |H[Si ]| ≥ (d + 1) + 2i. If i∗ ≥ 1 is maximum such that H[Si∗ −1 ] 6= V , then (d + 1) + 2(i∗ − 1) ≤ n − 2. Hence ∗ i ≤ n−d−1 and h(G) ≤ |Si∗ | = 2 + i∗ ≤ n−d+3 which completes the proof. ✷ 2 2 3 3 Triangle-free graphs Our first result in this section describes a natural and less abstract hypothesis which implies that a graph has property P. Corollary 4 If G is a connected triangle-free graph of order n, minimum degree at least 2 and diameter d, then G has property P and — hence — h(G) ≤ n−d+3 . 2 Proof: In view of Theorem 3, we may assume that G does not have property P. Let the three sets of vertices A, B and C certify this. Since A is connected, B is a clique and G is triangle-free, we obtain that A and B both contain exactly one vertex. This clearly implies the contradiction that the vertex in A has degree 1 which completes the proof. ✷ The graphs whose structure is illustrated in Figure 1 are connected, triangle-free, and of minimum degree at least 2. Their order n, diameter d, and hull number h satisfy n = 5d−7 and Theorem 3 and Corollary 4 are best possible. and h = 2d − 2, i.e. h = n−d+3 2 s s ✁ ❆ ❆s s✁ ❅ ❅s ❅ ❅s s ❆ ✁ ❆s s✁ s s ✁ ❆ ❆s s✁ ❅ ❅s s s ✁ ❆ ✁s ❆s ❅ ❅s ... s s ✁ ❆ ❆s s✁ ❅ ❅s ❅ ❅s s ❆ ✁ ❆s s✁ Figure 1 Extremal graphs for Corollary 4. We proceed to graphs of minimum degree at least 3 and cubic graphs. For these graphs, the simple choice for the initial set described at the beginning of Section 2 would not yield best possible results. Therefore, we first prove the existence of a better initial set. Lemma 5 If G is a connected triangle-free graph of minimum degree at least 3 and diameter d ≥ 3, then there is a set of vertices S0 such that 2 ≤ |S0 | ≤ 4 and |H[S0 ]| ≥ d+3|S0 |−3. Proof: Let x and y be two vertices such that dist(x, y) = d. Clearly, |H[{x, y}]| ≥ d + 1. Depending on properties of x and y, we define two disjoint sets X and Y such that S0 = X ∪ Y has the desired properties. Since X and Y are defined analogously, we will only describe the definition of X in detail. If x has at least two neighbours at distance (d − 1) from y, then let X = {x}. Note that |H[X ∪ {y}]| ≥ (d + 1) + 1. In what follows, we assume that x has exactly one neighbour x′ at distance (d − 1) from y. 4 Let u and w be two neighbours of x different from x′ . Since dist(u, y) = dist(w, y) = d, we may assume, by symmetry with x, that each of u and w has exactly one neighbour u′ and w′ at distance (d − 1) from y. If u′ and w′ are distinct, then let X = {u, w}. Note that u′ , w′ ∈ I[y, X], x ∈ I[u, w], and x′ ∈ I[x, y] which implies that |H[X ∪ {y}]| ≥ (d + 1) + 4. In what follows, we assume that u′ = w′ . Let v be a neighbour of u different from x and u′ . We may assume, by symmetry, that x′ is the unique neighbour of v at distance (d − 1) from y. Let X = {u, w}. Note that u′ ∈ I[y, X], x ∈ I[u, w], x′ ∈ I[x, y], and v ∈ I[u, x′ ] which implies that |H[X ∪ {y}]| ≥ (d + 1) + 4. By defining Y analogously, we obtain a set S0 = X ∪ Y such that either |S0 | = 2 and |H[S0 ]| ≥ (d + 1) + 1 + 1, or |S0 | = 3 and |H[S0 ]| ≥ (d + 1) + 4 + 1, or |S0 | = 4 and |H[S0 ]| ≥ (d + 1) + 4 + 4. This completes the proof. ✷ As a further ingredient, we need a strengthened extension lemma. Lemma 6 If G is a connected triangle-free graph of minimum degree at least 3 and S is a non-empty set of vertices whose convex hull does not contain all vertices of G, then there is a set T ⊆ V \ S such that 1 ≤ |T | ≤ 2 and |H[S ∪ T ]| ≥ |H[S]| + 3|T |. Proof: Let G and S be as in the statement of the lemma. For contradiction, we assume that a set T with the desired properties does not exist. Let C = H[S]. If there is a vertex u with dist(u, C) ≥ 3, then T = {u} has the desired property. Hence V \ C = D1 ∪ D2 for Di = {u ∈ V | dist(u, C) = i} for 1 ≤ i ≤ 2. Since C is convex and G is triangle-free, every vertex in D1 has exactly one neighbour in C. If D2 is empty, then there are three vertices u, v, w ∈ D1 with uv, vw ∈ E. If u′ , w′ ∈ C are such that uu′ , ww′ ∈ E, then u ∈ I[v, u′ ] and w ∈ I[v, w′ ] which implies that T = {v} has the desired property. Hence D2 is not empty. If a vertex u ∈ D2 has two neighbours in D1 , then T = {u} has the desired property. Hence every vertex in D2 has exactly one neighbour in D1 . Let u, v, w ∈ D2 and u′ , v ′ , w′ ∈ D1 be such that uv, vw, uu′ , vv ′ , ww′ ∈ E. If u′ 6= w′ , then v ∈ I[u, w], u′ ∈ I[u, C], v ′ ∈ I[v, C], and w′ ∈ I[w, C] which implies that T = {u, w} has the desired property. Hence u′ = w′ . Since G has minimum degree at least 3 and every vertex in D2 has exactly one neighbour in D1 , there is a vertex x ∈ D2 different from v with wx ∈ E. Since G is triangle-free, vx 6∈ E. In view of the previous observation, we obtain, by symmetry, that v ′ is the unique neighbour of x in D1 . Now v ∈ I[u, w], u′ ∈ I[u, C], v ′ ∈ I[v, C], and x ∈ I[w, v ′ ] which implies that T = {u, w} has the desired property. This final contradiction completes the proof. ✷ As before, with Lemma 5 and Lemma 6 at hand, the proof of the following bound for graphs of minimum degree at least 3 is straightforward. 5 Theorem 7 If G is a connected triangle-free graph of order n, minimum degree at least 3, and diameter d ≥ 3, then n−d+3 h(G) ≤ . 3 Proof: Since n − d ≥ 5, we may assume that h(G) ≥ 3. By Lemma 5, there is a set S0 ⊆ V such that 2 ≤ |S0 | ≤ 4 and |H[S0 ]| ≥ d + 3|S0 | − 3. If H[S0 ] = V , then n−d+3 |H[S0 ]| − d + 3 d + 3|S0 | − 3 − d + 3 = ≥ = |S0 | 3 3 3 and the desired result follows. Hence we may assume that H[S0 ] 6= V . By Lemma 6, there is a sequence S0 ⊆ S1 ⊆ S2 ⊆ . . . of sets of vertices such that for i ≥ 1, either H[Si−1 ] = V or 1 ≤ |Si | − |Si−1 | ≤ 2 and |H[Si ]| ≥ |H[S0 ]| + 3(|Si | − |S0 |) ≥ (d + 3|S0 | − 3) + 3(|Si | − |S0 |) = d − 3 + 3|Si |. Let i∗ ≥ 1 be maximum such that H[Si∗ −1 ] 6= V . If |Si∗ | − |Si∗ −1 | = 1, then d − 3 + 3|Si∗ −1 | ≤ n − 3 which implies h(G) ≤ |Si∗ | = |Si∗ −1 | + 1 ≤ n−d+3 . 3 If |Si∗ | − |Si∗ −1 | = 2, then d − 3 + 3|Si∗ −1 | ≤ n − 6 which implies h(G) ≤ |Si∗ | = |Si∗ −1 | + 2 ≤ n−d+3 . 3 This completes the proof. ✷ The graphs whose structure is illustrated in Figure 2 are connected, triangle-free, and of minimum degree at least 3. Their order n, diameter d, and hull number h satisfy n = 7d−9 and Theorem 7 is best possible. and h = 2d − 2, i.e. h = n−d+3 3 s s s s ❆ ✁ ❆ ✁ ❆s✁ ❆s✁ ❅ ❅s ❅ ❅s s ✁❆ ✁❆ s✁ ❆s s✁ ❆s s s s s ❆ ✁ ❆ ✁ ❆✁s ❆✁s ❅ ❅s s s s s ❆ ✁ ❆ ✁ ❆s✁ ❆s✁ ❅ ❅s ... s s s s ❆ ✁ ❆ ✁ ❆s✁ ❆s✁ ❅ ❅s ❅ ❅s s ✁❆ ✁❆ s✁ ❆s s✁ ❆s Figure 2 Extremal graphs for Theorem 7. 6 For triangle-free graphs G of order n, minimum degree at least 3, and diameter 2, applying the reasoning from the proof of Theorem 7 to a set S0 containing two vertices at distance . 2 easily implies h(G) ≤ n+3 3 For our next result concerning cubic graphs, it suffices to strengthen the extension lemma. Lemma 8 If G is a connected cubic triangle-free graph and S is a set of vertices of order at least 2 whose convex hull does not contain all vertices of G, then there is a set S ′ ⊆ V such that (i) either |S ′ | = |S| + 1 and |H[S ′ ]| ≥ |H[S]| + 4, (ii) or |S ′ | = |S| + 2 and |H[S ′ ]| ≥ |H[S]| + 7, (iii) or |S ′ | = |S| and |H[S ′ ]| > |H[S]|. Proof: Let G and S be as in the statement of the lemma. For contradiction, we assume that a set S ′ with one of the desired properties does not exist. Let C = H[S]. If there is a vertex u with dist(u, C) ≥ 4, then S ′ = S ∪ {u} satisfies (i). Hence V \ C = D1 ∪ D2 ∪ D3 for Di = {u ∈ V | dist(u, C) = i} for 1 ≤ i ≤ 3. Since G is connected and triangle-free, every vertex in D1 has exactly one neighbour in C. Furthermore, since G is cubic and C is a convex set of order at least 2, no vertex in C has three neighbours in D1 . First, we assume that D3 is not empty. Let v ∈ D3 . If |I(v, C)| ≥ 3, then S ′ = S ∪ {v} satisfies (i). Hence I(v, C) contains exactly two vertices v ′ ∈ D2 and v ′′ ∈ D1 such that vv ′ , v ′ v ′′ ∈ E. Furthermore, there are vertices u, w ∈ D3 such that uv, vw ∈ E. If |I({u, v, w}, C)| ≥ 4, then v ∈ I[u, w] and S ′ = S ∪ {u, w} satisfies (ii). Hence I({u, v, w}, C)| ≤ 3. Let u′ ∈ D2 be a neighbour of u. Since u′ 6= v ′ , we obtain I({u, v, w}, C) = {u′ , v ′ , v ′′ } and wu′ , u′ v ′′ ∈ E. Since, by symmetry, |I(w, C)| ≤ 2, there is a vertex x ∈ D3 different from v with wx ∈ E. Since, by symmetry, I|({v, w, x}, C)| ≤ 3, we obtain xv ′ ∈ E, w ∈ I[v, x], u′ , v ′ , v ′′ ∈ I[{v, w}, C], u ∈ I[v, u′ ], and S ′ = S ∪ {v, x} satisfies (ii). This contradiction implies that we may assume that D3 is empty. Next, we assume that D2 is not empty. Let v ∈ D2 . If |I(v, C)| ≥ 3, then S ′ = S ∪ {v} satisfies (i). Hence |I(v, C)| ≤ 2. Let u ∈ D2 be such that uv ∈ E. Let u′ , v ′ ∈ D1 and u′′ , v ′′ ∈ C be such that uu′ , u′ u′′ , vv ′ , v ′ v ′′ ∈ E. If u′′ 6= v ′′ , then u ∈ I[v, u′′ ] and S ′ = S ∪ {v} satisfies (i). Hence u′′ = v ′′ . If there is a vertex y ′ ∈ D1 different from v ′ such that vy ′ ∈ E and y ′′ ∈ C is such that ′ ′′ y y ∈ E, then, because no vertex in C has three neighbours in D1 , we obtain y ′′ 6= v ′′ and S ′ = S ∪ {u} satisfies (i). Hence I(v, C) = {v ′ } and there is a vertex w ∈ D2 different from u such that vw ∈ E. 7 By symmetry, we obtain wu′ ∈ E and there is a vertex x ∈ D2 different from v such that ux ∈ E, xv ′ ∈ E, and I({u, v, w, x}, C) = {u′ , v ′ }. If there is a vertex y ∈ D2 different from u and w such that xy ∈ E, then there are vertices y ′ ∈ D1 and y ′′ ∈ C such that yy ′ , y ′ y ′′ ∈ E, y ′′ 6= v ′′ and S ′ = S ∪ {x} satisfies (i). Hence wx ∈ E. Furthermore, v ′′ is a cutvertex of G and G − {v ′′ } has a component with vertex set {u, v, w, x, u′ , v ′ }. Since v ′′ has two neighbours in D1 and C is convex, we obtain that v ′′ ∈ S. Since S contains at least two elements, this implies that S ′ = (S \ {v ′′ }) ∪ {v ′ } satisfies (iii). This contradiction implies that we may assume that D2 is empty. Since every vertex in D1 has exactly one neighbour in C, there are vertices u, v, w, x ∈ D1 such that uv, vw, wx ∈ E. If v ′ , w′ , x′ ∈ C are such that vv ′ , ww′ , xx′ ∈ E, then v ∈ I[u, v ′ ], w ∈ I[v, w′ ], x ∈ I[w, x′ ], and S ′ = S ∪ {u} satisfies (i). This final contradiction completes the proof. ✷ We proceed to our next result. Theorem 9 If G is a connected cubic triangle-free graph of order n and diameter d ≥ 3, then 2(n − d + 5) h(G) ≤ . 7 Proof: Since n − d ≥ 5, we may assume that h(G) ≥ 3. By Lemma 5, there is a set S0 ⊆ V such that 2 ≤ |S0 | ≤ 4 and |H[S0 ]| ≥ d + 3|S0 | − 3. If H[S0 ] = V , then 2(|H[S0 ]| − d + 5) 2(d + 3|S0 | − 3 − d + 5) 6|S0 | + 4 2(n − d + 5) = ≥ = ≥ |S0 | 7 7 7 7 and the desired result follows. Hence we may assume that H[S0 ] 6= V . By Lemma 8, there is a sequence S0 , S1 , S2 , . . . of sets of vertices such that for i ≥ 1, (i) either H[Si−1 ] = V , (ii) or |Si | = |Si−1 | + 1 and |H[Si ]| ≥ |H[Si−1 ]| + 4, (iii) or |Si | = |Si−1 | + 2 and |H[Si ]| ≥ |H[Si−1 ]| + 7. Note that this implies 7 |H[Si ]| ≥ |H[S0 ]| + (|Si | − |S0 |) 2 7 ≥ (d + 3|S0 | − 3) + (|Si | − |S0 |) 2 1 7 = d − 3 + |Si | − |S0 | 2 2 7 ≥ d − 5 + |Si | 2 8 for i ≥ 1. Let i∗ ≥ 1 be maximum such that H[Si∗ −1 ] 6= V . If |Si | = |Si−1 | + 1, then d − 5 + 72 |Si∗ −1 | ≤ n − 4 which implies h(G) ≤ |Si∗ | = |Si∗ −1 | + 1 ≤ 2(n − d) + 9 . 7 If |Si | = |Si−1 | + 2, then d − 5 + 72 |Si∗ −1 | ≤ n − 7 which implies h(G) ≤ |Si∗ | = |Si∗ −1 | + 2 ≤ 2(n − d) + 10 . 7 This completes the proof. ✷ The graphs whose structure is illustrated in Figure 3 are connected, cubic, and trianglefree. Their order n, diameter d, and hull number h satisfy n = 8d − 26 and h = 2d − 6, i.e. h = 2(n−d+5) and Theorem 9 is best possible. 7 s s s s ❆ ✁ ❆ ✁ ❆s✁ ❆s✁ ❅ ❅s s s s s ❆ ✁ ❆ ✁ ❆✁s ❆✁s ❅ ❅s s s s s ❆ ✁ ❆ ✁ ❆s✁ ❆s✁ ❅ ❅s s s s s ❅ s ❅s ✁❆ ✁❆ ✁s ❆s s✁ ❆s s s s s ❆ ✁ ❆ ✁ ❆✁s ❆✁s ❅ ❅s ... s s ❅ s ❅s ✁❆ ✁❆ ✁s ❆s s✁ ❆s Figure 3 Extremal graphs for Theorem 9. For the finitely many cubic triangle-free graphs G of order n and diameter 2, applying the reasoning from the proof of Theorem 9 to a set S0 containing two vertices at distance 2 . easily implies h(G) ≤ 2(n+4) 7 ∗ For r ≥ 4, let the graph Kr,r arise from the complete bipartite graph Kr,r by subdividing one edge. If a connected triangle-free graph G of order n, minimum degree at least r, and diameter d arises by suitably identifying the vertices of a sufficiently long path with the ∗ vertices of degree 2 in disjoint copies of Kr,r , then h(G) = n−d +O(1). Note that the graphs r in Figures 1 and 2 are obtained in this way and that the graphs in Figure 3 are obtained by a variation of this construction. We believe that this construction is essentially best possible and pose the following conjecture. 9 Conjecture 10 If G is a connected triangle-free graph of order n, minimum degree δ, and diameter d, then n−d h(G) ≤ + O(1). δ Note that for triangle-free graphs, the 2-domination number is an upper bound on the geodetic number and hence also on the hull number. Therefore, Caro and Yuster’s [2] results imply that the hull number of a connected triangle-free graph of order n and minimum degree δ is at most (1 + oδ (1)) n ln(δ) . δ 4 Graphs of large girth All presented examples of graphs showing that our bounds are best possible contained short cycles. In this section, we show that a lower bound on the girth allows to improve the bounds on the hull number. Once again the key ingredient is an extension lemma. Lemma 11 If G is a connected graph of girth at least g ≥ 5 and minimum degree at least 2 and S is a non-empty set of vertices whose convex hull does not contain  g−1  all vertices of G, then there is a vertex z ∈ V \ S such that |H[S ∪ {z}]| ≥ |H[S]| + 2 . Proof: Let G and S be as in the statement of the lemma. For contradiction, we assume that a vertex with the desired properties does not exist. Let C = H[S]. C.  g  Let R = V\g−1 Let ge = 2 2 and go = 2 2 + 1. Note that ge and go are the smallest even and odd integers which are at least g, respectively. Since G has minimum degree at least 2, there is a cycle which intersects R. We assume that K is a cycle in G which intersects R such that, if lK denotes the length of K and dK denotes dist(V (K), C), then lK + 2dK is smallest possible. First, we assume that K contains at least two vertices of C. By the choice of K and the convexity of C, we obtain that there are two vertices x and y in C such that K consists of two paths PC and PR between x and y, all internal vertices of PC belong to C, and all internal vertices of PR belong to R. Furthermore, since C is convex, the length lC of PC is less than the length lR of PR . Letz be  the internal vertex of PR such that the distance with   lR respect to PR between x and z is 2 . By the choice of K, we obtain that dist(x, z) = l2R   and dist(y, z) = l2R . If lK is even, then |H[S ∪ {z}]| − |H[S]| ≥ lR − 1 ≥ lK ≥ g2e . If lK 2  go −1 ge go −1 lK −1 = go2−1 = odd,  then |H[S ∪ {z}]| − |H[S]| ≥ lR − 1 ≥ 2 ≥ 2 . Since min 2 , 2 isg−1 , we obtain a contradiction. Hence K contains at most one vertex of C. 2 Let PK be a path of length dK between a vertex x in V (K) and a vertex y in C. (Note that x = y is possible for dK = 0.) Let z be a vertex of K such that the distance with . (Note that z is unique if lK is even.) By the choice respect to K between x and z is lK 2 l  K of K, we obtain that dist(z, C) = 2 + dK . If lK is even, then |H[S ∪ {z}]| − |H[S]| ≥ (lK − ge − 1. If lK is odd, then |H[S ∪ {z}]| − |H[S]| ≥ lK2−1 + dK ≥ go2−1 . Since  1) + dKgo≥  min ge − 1, 2−1 = go2−1 = g−1 , we obtain a contradiction. 2 10 This final contradiction completes the proof. ✷ We proceed to our final result. Theorem 12 If G is a connected graph of order n, girth g ≥ 5, minimum degree at least 2, and diameter d, then n−d−1 h(G) ≤ 2 +  g−1  . 2 Proof: Since n − d ≥ 1, we may assume that h(G) ≥ 3. Let S0 = {u, v} for two vertices u and v with dist(u, v) = d. H[S0 ] contains at least d + 1 vertices. By Lemma 11, there is a sequence S0 ⊆ S1 ⊆ S2 ⊆ . . . of setsof vertices such that for  g−1 i ≥ 1, |Si | = 2 + i and either H[Si−1 ] = V or |H[Si ]| ≥ (d + 1) + 2 i.   If i∗ ≥ 1 is maximum such that H[Si∗ −1 ] 6= V , then (d + 1) + g−1 (i∗ − 1) ≤ n − g−1 . 2 2 n−d−1 n−d−1 ∗ ∗ Hence i ≤ g−1 and h(G) ≤ |Si∗ | = 2 + i ≤ 2 + g−1 which completes the proof. ✷ ⌈ 2 ⌉ ⌈ 2 ⌉ As before it is easy to construct examples of graphs — similar to the graphs in Figure 1 — which show that Theorem 12 is best possible. For instance, the graphs in Figure 1 are connected, of girth 5, and minimum degree at least 2. Recall that their order n, diameter   d, and hull number h satisfy n = 5d − 7 and h = 2d − 2, i.e. h = 2 + (n − d − 1)/ g−1 . 2 References [1] S.R. Canoy Jr. and G.B. Cagaanan, On the hull number of the composition of graphs, Ars Combinatoria 75 (2005), 113-119. [2] Y. Caro and R. Yuster, Dominating a family of graphs with small connected subgraphs, Comb. Probab. Comput. 9 (2000), 309-313. [3] G. Chartrand, J.F. Fink, and P. Zhang, The hull number of an oriented graph, International Journal of Mathematics and Mathematical Sciences 2003 (2003), 2265-2275. [4] G. 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