Geodomination in graphs

International Mathematical Forum, 2, 2007, no. 35, 1729 - 1736 Geodomination in Graphs Adel P. Kazemi and Doost Ali Mojdeh Department of Mathematics University of Mazandaran P. O. Box 47416-1467, Babolsar, Iran a.kazemi@umz.ac.ir, dmojdeh@umz.ac.ir Abstract In this paper we will define the new concepts independent geodomination, independent k-geodomination numbers and then state some results for graphs and Cartesian product of them. Finally we verify how independent geodomination numbers are affected by adding a vertex. Mathematics Subject Classification: 05C Keywords: Independent (k-)geodomination number; subdivision 1 Introduction and Preliminaries For vertices x and y in a connected graph G = (V, E), the distance d(x, y) is the length of a shortest x-y path in G. The diameter of G is the maximum distance between any two vertices of G and is denoted by diam G. An x-y path of length d(x, y) is called an x-y geodesic. A vertex v is said to lie in an x-y geodesic P if v is an internal vertex of P , that is, v is a vertex of P distinct from x and y. The closed interval I[x, y] consists of x, y and all vertices lying in some x-y geodesic of G, while for S ⊆ V , I[S] = ∪x,y∈S I[x, y]. A set S of vertices is a geodominating (or geodesic) set if I[S] = V , and the minimum cardinality of a geodominating set is the geodomination (or geodetic) number g(G). A geodominating set of cardinality g(G) is called a g-set. For a graph G and integer k ≥ 1, a vertex v of G is k-geodominated by a pair x, y of distinct vertices in G if v is geodominated by x, y and d(x, y) = k. A set S of distinct vertices in G is a k-geodominating set of G if each vertex v ∈ G − S is k-geodominated by some pair of distinct vertices of S. The 1730 Adel P. Kazemi and Doost Ali Mojdeh minimum cardinality of a k-geodominating set of G is its k-geodomination number gk (G). A k-geodominating set of cardinality gk (G) is called a gk -set [1, 2, 3, 4, 5]. We now define other concepts. Let G is a non-complete connected graph and k ≥ 2 is an integer. A set S of vertices of G is an independent geodominating (or independent geodesic) set if S is an independent set and I[S] = V. The minimum cardinality of an independent geodominating set is the independent geodomination (or independent geodetic) number, ig(G). An independent geod-ominating set of cardinality ig(G) is called an ig-set. An independent set S of vertices in G that is an k-geodominating set of G is called an independent k-geodominating set of G. The minimum cardinality of an independent k-geodominating set of G is its independent k-geodomination number igk (G). Also an independent k-geodominating set of cardinality igk (G) is called an igk -set. It is clear that if k ≥ 2 is an integer and G is a graph that has g, ig, gk , igk -sets, then g(G) ≤ ig(G), gk (G) ≤ igk (G). For two graphs G and H of orders m and n, respectively, the cartesian product graph G × H is a graph with V (G × H) = {(g, h) | g ∈ V (G), h ∈ V (H)} and for each two vertices (g, h) and (g ′ , h′ ) of G × H, (g, h) ↔ (g ′, h′ ), that is, this two vertices are adjacent, if and only if ”g = g ′ and h ←→ h′ ” or ”h = h′ and g ←→ g ′ ”. Also a subdivision graph H of a graph G is a graph that is obtained by the operation of replacing at least an edge uv of G with a path u, w1, ..., wn , v, for n ≥ 1, through new vertex or vertices w1 , ..., wn . 2 Main Results In the following theorem we give an independent geodomination number for cycles. Observation 1 Let Cn be a cycle on n vertices. Then  2 n ≥ 4, n is even ig(Cn ) = . 3 n ≥ 7, n is odd Proof. Let V (Cn ) = {1, 2, .., n}. If n = 2k, k ≥ 2, the set {1, k} is an ig-set. For n = 2k + 1, k ≥ 3, trivially ig(Cn ) ≥ 3 and the set {1, k, k + 2} is an ig-set. Now we give a lower bound for independent geodomination number for cartesian product G × H. 1731 Geodomination in graphs Theorem 2 Let G, H and G × H be connected graphs such that they have ig-sets. Then max{ig(G), ig(H)} ≤ ig(G × H). Proof. Let ig(G × H) = t and ig(H) ≤ ig(G) = m. We choose subset {a1 , a2 , ..., at } of V (G) and define a function θ on {1, 2, ..., t} such that S = {(ai , bθ(i) ) | ai ∈ V (G), bθ(i) ∈ V (H), 1 ≤ i ≤ t} be an ig-set of G × H. From the vertices a1 , a2 , ..., at of V (G), we choose the vertices that are not equal. ′ ′ Let S = {a1 , a2 , ..., al } is the same set. Suppose x ∈ V (G) − S . Let l = t. For any vertex y ∈ V (H), since the vertex (x, y) lies on an (ai , bθ(i) )-(aj , bθ(j) ) ′ geodesic, hence x lies on an ai -aj geodesic. This means S is an ig-set for G and so m ≤ t. Finally, let l < t. This means that for each l + 1 ≤ j ≤ t there ′ exists i, 1 ≤ i ≤ l, such that aj = ai . Since x ∈ / S then x = aj , for each l + 1 ≤ j ≤ t. For any vertex y ∈ V (H), if (x, y) lies on an (ai , bθ(i) )-(aj , bθ(j) ) ′ geodesic P in G × H and P is the imaging of P in G, then x lies on ai -aj ′ geodesic for some ai , aj ∈ S . Therefore m ≤ l < t and our proof is completed. In the two following we give equality in the above theorem. Theorem 3 Let G be a connected graph such that has ig-set. If Pm is a path with m ≥ 2 vertices, then ig(Pm × G) = ig(G). Proof. Let V (G) = {1, 2, ..., n}, V (Pm ) = {1, 2, ..., m} and V (Pm × G) = {(i, j) | 1 ≤ i ≤ m, 1 ≤ j ≤ n}. For m = 2 obviously ig(P2 × G) ≥ ig(G). If S0 = {x1 , x2 , ..., xt } be an ig-set of G, obviously {(1, x1 ), (1, x2 ), ..., (1, xt−1 ), (2, xt )} is an ig-set of P2 × G and this proves that ig(P2 × = ig(G). Finally let m ≥ 3. We know ig(Pm × G) ≥ max{ig(Pm ), ig(G)} = ig(G), by Theorem 8. It suffices to show ig(Pm × G) ≤ ig(G). Let ig(G) = t and S = {x1 , x2 , ..., xt } be an ′ ig-set of G. The set S = {(1, x1 ), (1, x2 ), ..., (1, xt−1 ),(m, xt )} is an indepen′ dent set because S is independent. Choose (l, k) ∈ V (Pm × G) − S . (l, k) lies in the (1, xi )-(j, xk )-geodesic: P : (1, Ω1 )(1, k)...(l, k)...(m, k)(m, Ω2 ) in Pm × G such hat Ω2 is a k-xt geodesic in G and Ω1 is a member of {kxi -geodesic| 1 ≤ i ≤ k − 1} with minimum length.We note that if Ωi : r1 r2 ...rq , then (j, Ωi ) : (j, r1 )(j, r2 )...(j, rq ), for (i, j) ∈ {(1, 1), (m, 2)}. Therefore ig(Pm × G) ≤ ig(G). Now the Theorem 2 completes the proof. 1732 Adel P. Kazemi and Doost Ali Mojdeh Theorem 4 Let G be a connected graph such that has ig-set. If C2m is an even cycle with 2m vertices, m ≥ 2, then ig(C2m × G) = ig(G). Proof. Let V (G) = {1, 2, ..., n}, V (C2m ) = {1, 2, ..., 2m} and V (C2m × G) = {(i, j) | 1 ≤ i ≤ 2m, 1 ≤ j ≤ n}. We choose the set S = {x1 , x2 , ..., xt } as an ig-set of G. The set S ′ = {(1, x1 ), (1, x2 ), ..., (1, xt−1 ), (m, xt )} is independent, because S is independent. Similar to the proof of Theorem 3, we can prove that each vertex of V (C2m × G) − S ′ lies on an (a, xi )-(b, xj ) geodesic, for some (a, xi ), (b, xj ) ∈ S ′ and so ig(C2m × G) ≤ ig(G). Hence ig(C2m × G) = ig(G), by Theorem 3. Obviously the following Observation holds. Observation 5 Let k ≥ 2 be an integer. Let G be a graph of order n ≥ 3 that has ig, igk -sets. Then 2 ≤ ig(G), igk (G) < n. Theorem 6 If G is a connected graph of diameter 2 that has independent geodominating and independent 2-geodominating sets, then ig2 (G) = ig(G). Proof. Since ig(G) ≤ ig2 (G), by Observation 5, we need only show that ig2 (G) ≤ ig(G), that is, that every ig-set in G is an ig2 -set. Let S be an ig-set of G. We may assume that S = V . Let v ∈ V − S. Since S is an ig-set, it follows that v is geodominated by some x, y ∈ S. So v lies in an x-y geodesic in G. Since diam G = 2, it follows that d(x, y) = 2. Thus S is an independent 2-geodominating set of G and so S is an ig2 -set, by Observation 5. We note that the graph in Theorem 6 has diameter 2 is necessary. See the following graph. 1 2 3 1733 Geodomination in graphs 1 2 figure 1 The converse of Theorem 6 is failed. For example the following graph G has ig(G) = ig2 (G) = 4 with vertices 1, 2, 3, 4, but diam G = 3. 2 3 1 4 figure 2 By Theorem 6, if G is a connected graph with diam G = 2, then ig(G) = ig2 (G). However, in general , ig(G) = igk (G) for an integer k with 2 ≤ k ≤ diam G. Theorem 7 Let G be a connected graph of order n ≥ 3, with diam G ≥ 3 and ig(G) = 2, and let k ≥ 2 be an integer. Then igk (G) = ig(G) ⇔ k = diam(G). 1734 Adel P. Kazemi and Doost Ali Mojdeh Proof. Let diam G = d and S = {x, y} be an ig-set of G . Then x and y are antipodal , that is , d(x, y) = d. Thus S is an independent d-geodominating set and so igd (G) = 2(= ig(G)). For converse, let igk (G) = ig(G) = 2. For k > d, independent k-geodomination number is not defined. Now, let 2 ≤ k ≤ d − 1. We note that every ig-set S of size 2 of G has two antipodal vertices and so S is an independent d-geodominating set. Since k < d, it follows that S is not an independent k-geodominating set. Thus ig(G) = 2, which is a contradiction. therefore k = d. We know that if independent geodominating and independent k-geodominating sets exist for G, then 2 ≤ ig(G), igk (G), by Observation 5 We show that for each integers a, b with 2 ≤ a ≤ b, and k ≥ 2 there exists a tree T such that ig(T ) = a and igk (T ) = b. First a lemma. Lemma 8 If k ≥ 2, j ≥ 1 are integers and Pjk+1 be a path with jk+1 vertices, then ig(Pjk+1) = j + 1. Proof. If V (Pjk+1) = {1, 2, ..., jk +1}, then the set {1, k +1, 2k +1, ..., jk + 1} is unique ig-set of Pjk+1. Theorem 9 Let a, b, k are integers such that 2 ≤ a ≤ b, k ≥ 2. Then there exists a tree T such that ig(T ) = a and igk (T ) = b. Proof. If (a, k) = (b, 2), then we choose T = K1,a . Since diam T = 2, it follows from Theorem 6 that ig(T ) = igk (T ) = a. So we assume that (a, k) = (b, 2). Let T is a path P : v1 , v2 , ..., vp , by joining the a − 1 vertices u1 , u2 , ..., ua−1 to v1 . Let first a = b and k ≥ 3. Choose p = k. The set S = {u1 , u2, ..., ua−1 , vk } of end-vertices is a g-set of T and we obtain g(T ) = gk (T ) = a by Observation 1.1 [4] and since igk (T ) = gk (T ) and ig(T ) = g(T ), so igk (T ) = ig(T ) = a. We second assume that b = a+j, j ≥ 1. Then we choose p = jk + 1. We know igk (Pjk+1) = j + 1. The tree T has unique ig-set from end-vertices and so ig(T ) = a. Moreover, the set S = {u1 , u2, ..., ua−1 }∪{vji+1 | 0 ≤ i ≤ j} is the unique igk -set and igk (T ) = b. Since for each pair a, n of integers with 2 ≤ a ≤ n, there is not a connected graph G of order n and gk (G) = a, by [4], therefore there exists any connected graph G of order n with igk (G) = a. Theorem 10 Let k ≥ 2 be an integer. A pair (a, n) of integers is realizable as the independent k-geodomination number and order n of some nontrivial connected graph if 2 ≤ a ≤ n − k + 1. Geodomination in graphs 1735 Proof. Let 2 ≤ a ≤ n − k + 1 and let G be the graph obtained from path Pk+1 : v1 , v2, ..., vk+1 of order k + 1 by (1) adding the a − 2 new vertices u1 , u2, ..., ua−2 and joining each of ui (1  i  a − 2) to vk and (2) adding each of wj (1  j  n − k − a + 1) to both v2 and v4 . Since the set {u1 , u2, ..., ua−2 , v1 , vk+1 } of k-extreme vertices of G is an igk -set, it follows igk (G) = a. We know that for every graph G of order n if G has ig-number, then 2 ≤ ig(G) ≤ n − 1. This upper bond is sharp. Because for the star graph G = K1,n−1 this bound holds. For converse we have the following theorem. Theorem 11 For integers k, n if 2 ≤ k ≤ n − 1, then there exists a connected graph G such that ig(G) = k. Proof. If k = n − 1, then say G = K1,n−1 . For k < n − 1 suppose that G is a subdivision of star K1,n−1 of order n. Then ig(G) = k. Finally we show that how independent geodomination numbers are affected by adding a vertex. Theorem 12 Let G be a connected graph such that it has an ig-set. If G′ is a graph obtained by adding a pendant edge to G, then G′ also has an ig-set and ig(G)  ig(G′)  ig(G) + 1. Proof. Let G′ = G + uv such that v ∈ V (G) and u ∈ / V (G). We first prove ′ ′ that ig(G)  ig(G ). Let ig(G) = n and ig(G )  n − 1. Let S ′ = {u} ∪ S be an ig-set of G′ . So S ⊆ V (G), | S | n − 2, v ∈ / S. If NG (v) ∩ S = ∅, then S ∪ {v} is an ig-set of G and so n = ig(G) | S ∪ {v} | n − 1, that is a contradiction. Let now NG (v) ∩ S = {v1 , ..., vm }, (m  1). (We note that vi and vj are not adjacent for i = j if m  2.) If m  2, then S is an ig-set of G and if m = 1, then (S − {v1 }) ∪ {v}is an ig-set of G. In this case we note that v1 is not an end-vertex. So ig(G)  n − 2, that is a contradiction. For converse we consider two cases. If v belongs to each ig-set of G and S be an ig-set of G, then S ′ = (S − {v}) ∪ {u} is an independent geodominating set of G and so ig(G′ )  ig(G) ≤ ig(G) + 1. If there exists an ig-set S0 of G such that v ∈ / S0 , then S0 ∪ {u} is an ig-set of G′ and so ig(G′ )  ig(G) + 1. References [1] G. Chartrand, F. Harary and P. Zhang, Geodetic Sets in Graphs, Discussiones Mathematicae Graph Theory, 20 (2000), 129-138. [2] G. Chartrand, F. Harary, H. C. Swart and P. Zhang, Geodomination in Graphs, Bulletin of the ICA, 31 (2001), 51-59. 1736 Adel P. Kazemi and Doost Ali Mojdeh [3] D. Donovan, E. S. Mahmoodian, R. Colin and P. Street, Defining sets in combinatorices: a survay, London Mathematical Society Lecture Note Series 307 (2003). [4] R. Muntean and P. Zhang, k-Geodomination in Graphs, Ars Combinatoria 63 (2002), 33-47. [5] D. B. West, Introduction to gragh theory, (2nd edition), Prentice Hall USA (2001). Received: November 30, 2006