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The general position problem and strong resolving graphs

  • Sandi Klavžar and Ismael G. Yero EMAIL logo
Published/Copyright: October 13, 2019

Abstract

The general position number gp(G) of a connected graph G is the cardinality of a largest set S of vertices such that no three pairwise distinct vertices from S lie on a common geodesic. It is proved that gp(G) ≥ ω(GSR), where GSR is the strong resolving graph of G, and ω(GSR) is its clique number. That the bound is sharp is demonstrated with numerous constructions including for instance direct products of complete graphs and different families of strong products, of generalized lexicographic products, and of rooted product graphs. For the strong product it is proved that gp(GH) ≥ gp(G)gp(H), and asked whether the equality holds for arbitrary connected graphs G and H. It is proved that the answer is in particular positive for strong products with a complete factor, for strong products of complete bipartite graphs, and for certain strong cylinders.

MSC 2010: 05C12; 05C76

1 Introduction

The general position problem was recently and independently introduced in [1, 2]. If G = (V(G), E(G)) is a graph, then SV(G) is a general position set if no triple of vertices from S lie on a common geodesic in G. The general position problem is to find a largest general position set of G, the order of such a set is the general position number gp(G) of G. A general position set of G of order gp(G) is shortly called gp-set. The general position problem has been further studied in a sequence of very recent papers [3, 4, 5, 6, 7].

A vertex u of a connected graph G is maximally distant from a vertex v if every wN(u) satisfies dG(v, w) ≤ dG(u, v), where N(u) is the open neighborhood of u. If u is maximally distant from v, and v is maximally distant from u, then u and v are mutually maximally distant (MMD for short). The strong resolving graph GSR of G has V(G) as the vertex set, two vertices being adjacent in GSR if they are MMD in G. The notion of the strong resolving graph was introduced in [8] as a tool to study the strong metric dimension. There it was proved that the problem of finding the strong metric dimension of a graph G can be transformed to the problem of finding the vertex cover number of GSR. Further on, the strong resolving graph itself was remarked as a kind of graph transformation in [9], and several characterizations and realizations of it were described.

Now, one of the open problems presented in [9] concerns finding applications for the strong resolving graph construction, other than that of computing the strong metric dimension of graphs. In this paper we give a partial answer to this problem by establishing a connection between the general position number of a graph G and the clique number of the graph GSR. More precisely, in Theorem 3.1 we prove that gp(G) ≥ ω(GSR) holds for any connected graph G. Then we demonstrate with different infinite families of graphs, including direct products of complete graphs, that the bound is sharp. We also show that for any integers rt ≥ 2, there exists a graph G such that gp(G) = r and ω(GSR) = t. In Section 4 we focus on strong products of graphs. We prove that gp(GH) ≥ gp(G)gp(H) holds for connected graphs G and H and that the bound is again sharp. In particular, if gp(G) = ω(GSR), then gp(GKn) = n ⋅ gp(G) = ω((GKn)SR). We close the section with a question on whether actually the equality gp(GH) = gp(G)gp(H) holds for arbitrary connected graphs G and H. In Section 5 we give additional large families of graphs, based on the generalized lexicographic product, for which the equality in Theorem 3.1 holds. In the final section we determine the general position number for different rooted product graphs and relate the values with the corresponding clique numbers of strong resolving graphs.

Before giving our results, we list in the next section definitions and concepts not yet given, as well as some results needed later.

2 Preliminaries

For a positive integer k we will use the notation [k] = {1, …, k}. If G = (V(G), E(G)) is a graph, then n(G) = |V(G)| and m(G) = |E(G)|. If XV(G), then the subgraph of G induced by X is denoted 〈X〉.

The distance dG(u, v) between vertices u and v of a graph G is the number of edges on a shortest u, v-path. A subgraph H of a graph G is isometric if dH(u, v) = dG(u, v) holds for all u, vV(H). A set of subgraphs {H1, …, Hk} of a graph G is an isometric cover of G if each Hi, i ∈ [k], is isometric in G and i=1k V(Hi) = V(G). With this concept in hand we can recall the following.

Theorem 2.1

[1, Theorem 3.1] If {H1, …, Hk} is an isometric cover of G, then

gp(G)i=1kgp(Hi).

If G is a connected graph, SV(G), and 𝓟 = {S1, …, Sp} a partition of S, then 𝓟 is distance-constant (alias “distance-regular” [10, p. 331]) if for any i, j ∈ [p], ij, the distance dG(u, v), where uSi and vSj, is independent of the selection of u and v. This distance is then the distance dG(Si, Sj) between the parts Si and Sj. A distance-constant partition 𝓟 is in-transitive if dG(Si, Sk) ≠ dG(Si, Sj) + dG(Sj, Sk) holds for pairwise different indices i, j, k ∈ [p]. With these concepts, general position sets can be characterized as follows.

Theorem 2.2

[3, Theorem 3.1] Let G be a connected graph. Then SV(G) is a general position set if and only if the components ofSare complete subgraphs, the vertices of which form an in-transitive, distance-constant partition of S.

Let η(G) denote the maximum order of an induced complete multipartite subgraph of the complement G of a graph G. Then we have:

Theorem 2.3

[3, Theorem 4.1] If diam(G) = 2, then gp(G) = max{ω(G), η(G)}.

Vertices u and v of a graph G are true twins if N[u] = N[v], where N[u] is the closed neighborhood of u. Note that only adjacent vertices can be true twins.

Proposition 2.4

If G has no true twins and diam(G) = 2, then gp(G) = ω(GSR) if and only if gp(G) = α(G).

Proof

Since G is true-twin free, GSR is isomorphic to the complement G of G. Hence ω(GSR) = α(G) and thus the conclusion.□

The Petersen graph P is a sporadic example of a graph without true twins and of diameter 2 for which gp(P) ≠ ω(PSR). Indeed, gp(P) = 6 and ω(PSR) = α(P) = 4.

Let G and H be graphs. Among the four standard graph products, we will consider the direct product G × H, the strong product GH, and the lexicographic product G[H]. The vertex set of all these products is V(G) × V(H). Let (g, h), (g′, h′) ∈ V(G) × V(H). In G × H, the vertices (g, h) and (g′, h′) are adjacent if gg′ ∈ E(G) and hh′ ∈ E(H). In GH, the vertices (g, h) and (g′, h′) are adjacent if one of the following three conditions hold: (i) gg′ ∈ E(G) and h = h′, (ii) g = g′ and hh′ ∈ E(H), (iii) gg′ ∈ E(G) and hh′ ∈ E(H). Finally, in G[H] the vertices (g, h) and (g′, h′) are adjacent if either gg′ ∈ E(G), or g = g′ and hh′ ∈ E(H). We note that the lexicographic product is also denoted with GH to emphasize the associativity of the operation, but here we use G[H] to be consistent with the generalized lexicographic product (to be defined later). If GH is one of the above products and hV(H), then the subgraph of GH induced by {(g, h) : gV(G)} is called a G-layer. Analogously H-layers are defined. In G × H, each G-layer is an edgeless graph of order n(G). In all other above products, each G-layer is isomorphic to G. If X is a set of vertices of GH, then the projection of X to G is the set {gV(G) : (g, h) ∈ X for some hV(H)}. Analogously the projection of X to H is defined. For more information on the standard graph products see the book [11], here we just recall the following well-known result (cf. [11, Proposition 5.4]) needed later.

Proposition 2.5

If (g, h) and (g′, h′) are vertices of a strong product GH, then

dGH(g,h),(g,h)=max{dG(g,g),dH(h,h)}.

3 The lower bound and equality cases

In this section we first prove the key result that connects the general position problem with the strong resolving graphs.

Theorem 3.1

If G is a connected graph, then gp(G) ≥ ω(GSR). Moreover, equality holds if and only if G contains a gp-set that induces a complete subgraph of GSR.

Proof

Let SV(GSR) induce a complete subgraph of GSR. This means that any two vertices x, yS are MMD in G. We now consider the vertices of S in the graph G. If there are three distinct vertices x, y, zS lying on a common geodesic, say y lies in an x, z-geodesic, then neither x, y nor y, z are MMD in G, which is a contradiction. Thus, any three vertices of S do not lie in a common geodesic of G, and therefore, S is a general position set in G. Selecting S to be a complete subgraph GSR of order ω(GSR) leads to the desired bound.

Suppose now that gp(G) = ω(GSR). By the above, any complete subgraph of GSR of order ω(GSR) yields a gp-set. Conversely, let S be a gp-set of G that forms a complete subgraph of GSR. Then, using the already proven inequality gp(G) ≥ ω(GSR), we have

|S|ω(GSR)gp(G)=|S|,

from which we conclude that ω(GSR) = gp(G).□

One would immediately think of characterizing the class of graphs achieving the equality in Theorem 3.1. However, such a characterization seems to be elusive because of the great variety of different structures that can appear. In the following we justify this variety and begin a couple of simple examples that were implicitly known previously.

  1. Block graphs, in particular complete graphs and trees.

    Indeed, in [1] it was observed that in block graphs the set of simplicial vertices forms a gp-set. Since simplicial vertices of a graph G also form a set of MMD vertices of a graph (equivalently, they form a complete subgraph of GSR), Theorem 3.1 implies that gp(G) = ω(GSR) if G is a block graph.

  2. Complete multipartite graphs.

    Let G = Kn1,…,nk, where k ≥ 2, n1n2 ≥ ⋯ ≥ nk ≥ 2, and n1k. Then it is easy to see that the vertices of the n1-partite set form a maximum general position set. Moreover, the vertices of this set also form a set of mutually maximally distant vertices of G. Hence, gp(G) = ω(GSR) by Theorem 3.1.

Let G and H be graphs where V(G) = {v1, … , vn}. The corona GH of graphs G and H is obtained from the disjoint union of G and n disjoint copies of H, say H1, …, Hn, where for all i ∈ [n], the vertex viV(G) is adjacent to each vertex of Hi. Then we have another equality case:

Proposition 3.2

If H = ⋃i Kni, ni ≥ 1, then for every graph G, gp(GH) = ω((GH)SR).

Proof

From [4, Theorem 4.3], it can be noticed that gp(GH) = n(G)∑i ni, and also that the union of the sets of vertices of all copies of H in GH form a gp-set S of GH. Every two vertices belonging to one copy of H are MMD, as well as are MMD every two vertices belonging to two different copies of H. Hence S forms a complete subgraph of (GH)SR. Thus we deduce the equality by Theorem 3.1.□

We note in passing that Proposition 3.2 remains valid in a more general setting when different disjoint unions of complete graphs are attached to the vertices of G.

In the next result we provide a family of direct product graphs for which the equality holds in Theorem 3.1. To this end, we locally use the Cartesian product of graphs. As the other standard graph products, the Cartesian product GH of graphs G and H has vertex set V(G) × V(H), and two vertices (g, h) and (g′, h′) are adjacent in GH if one of the following two conditions hold: (i) gg′ ∈ E(G) and h = h′, (ii) g = g′ and hh′ ∈ E(H).

Proposition 3.3

If n1n2 ≥ 3, then

gp(Kn1×Kn2)=ω((Kn1×Kn2)SR)=n1=α((Kn1×Kn2)SR).

Proof

We first note that ω(Kn1 × Kn2) = min{n1, n2} = n2. On the other hand, since every two vertices of Kn1 × Kn2 belonging to two different copies of Kn1 and of Kn2 are adjacent, every maximal induced complete multipartite subgraph of Kn1 × Kn2 is formed by the set of vertices of one copy of Kn1 or of Kn2. Thus, η(Kn1 × Kn2) = max{n1, n2} = n1. Now, since n1n2 ≥ 3, it follows that diam(Kn1 × Kn2) = 2 and hence Theorem 2.3 yields gp(Kn1 × Kn2) = max{η(Kn1 × Kn2), ω(Kn1 × Kn2)} = n1. From [9, 12] it is known that (Kn1 × Kn2)SRKn1Kn2 and since ω(Kn1Kn2) = max{n1, n2} = n1, the first two equalities follows. The last equality then follows by Proposition 2.4.□

Note that if we consider n1 > n2 = 2 in the result above, then Kn1 × K2 is of diameter 3, and its strong resolving graph is Kn1K2. Thus, ω((Kn1 × K2)SR) = 2. Since gp(Kn1 × K2) = n1 > 2, there is no equality as in the proposition.

Another example of direct products for which the equality in Theorem 3.1 does not hold is Kr,t × Kn, where rt ≥ 2 and n ≥ 3. Since diam(Kr,t × Kn) = 3, [3, Theorem 5.1] implies that gp(Kr,t × Kn) = α(Kr,t × Kn). Since it is not difficult to verify that α(Kr,t × Kn) = rn, we get gp(Kr,t × Kn) = rn. On the other hand, from [9, Theorem 35] we know that (Kr,t × Kn)SR i=1n Kr+t, and so ω((Kr,t × Kn)SR) = r + t. As rt ≥ 2 and n ≥ 3 we have rn > r + t.

Based on the above special cases we pose the following question about a possible dichotomy in direct product.

Problem 3.4

Is it true that gp(G × H) = ω((G × H)SR) can only hold in the case when diam(G × H) = 2?

To conclude the section we give the following realization result which intuitively indicates that one cannot expect some natural upper bound on gp(G) in terms of ω(GSR).

Proposition 3.5

For any integers rt ≥ 2, there exists a graph G such that gp(G) = r and ω(GSR) = t.

Proof

Since rt, there exists a non-negative integer q such that r = t + q. We now consider a graph Gq defined as follows. We begin with q copies of the cycle graph C4 and tq copies of the graph P2. Then we add an extra vertex z and one edge between z and exactly one vertex of each copy of C4 and of P2. We observe that the components of the strong resolving graph (Gq)SR are: one complete graph of order t, q complete graphs K2, and t + 1 isolated vertices. Thus ω((Gq)SR) = t. On the other hand, a set formed by two non-adjacent vertices of each copy of the cycle C4 (those ones not adjacent to z), and one vertex of each copy of the path P2, used to construct Gq, is a general position set of Gq, and so, gp(Gq) ≥ 2q + tq = t + q = r. We can readily observe that such set is indeed a gp-set of Gq, and therefore gp(G) = r, which completes the proof.□

4 Strong products

If G and H are connected graphs, then each G-layer and each H-layer of GH is an isometric subgraph of GH. Hence Theorem 2.1 gives the following upper bound.

Corollary 4.1

If G and H are connected graphs, then

gp(GH)min{n(G)gp(H),n(H)gp(G)}.

We will later see that the bound of Corollary 4.1 is tight. On the other hand we have the following lower bound.

Theorem 4.2

If G and H are connected graphs, then gp(GH) ≥ gp(G)gp(H).

Proof

Let SG and SH be gp-sets of G and H, respectively. We claim that SG × SH is a general position set of GH. To prove it, consider arbitrary pairwise different vertices of SG × SH, say (g, h), (g′, h′), (g″, h″), and assume on the contrary that in GH there exists a shortest (g, h), (g″, h″)-path P that passes through (g′, h′). We now distinguish several cases.

Suppose first that g = g′ = g″. Since (g, h), (g′, h′), (g″, h″) are pairwise different vertices of GH, the vertices h, h′, h″ are then pairwise different. But then the projection of P to H is a shortest h, h″-path that contains h′, a contradiction. Similarly, if g, g′, g″ are pairwise different, then the projection of P to G is a shortest g, g″-path that contains g′.

Suppose next that g = g′ and g″ ≠ g. Then clearly hh′. If h″ is different from both h and h′, then, as above, consider the projection of P to H to get a contradiction. The other subcase is that h″ = h′ (the subcase h″ = h is treated analogously). Let dG(g, g″) = k and dH(h, h″) = . By Proposition 2.5, dGH((g, h), (g″, h″)) = max{k, }. Denoting by P′ the (g, h), (g, h′)-subpath of P and by P″ the (g, h′), (g″, h″)-subpath of P, we get that max{k, } = |P| = |P′| + |P″| ≥ + k, a contradiction since k ≥ 1 and ≥ 1. The case g = g″, g′ ≠ g, and the case g′ = g″, gg′, are treated analogously.□

In [6, Theorem 3.3] it was proved that gp(PP) = 4. Since the strong grid PnPm is an isometric subgraph of PP for each n, m ≥ 2, it follows that gp(PnPm) ≤ 4. On the other hand, as PnPm contains K4 we also have gp(PnPm) ≥ 4. We conclude that

gp(PnPm)=4,n,m2. (1)

This result shows that the bound in Theorem 3.1 is sharp. More sharpness examples are provided with the next result which also shows the tightness of Corollary 4.1.

Proposition 4.3

If G is a connected graph and n ≥ 1, then gp(GKn) = n ⋅ gp(G). Moreover, if gp(G) = ω(GSR), then gp(GKn) = ω((GKn)SR).

Proof

The first assertion follows by combining Theorem 4.2 with Corollary 4.1.

Suppose now that in addition gp(G) = ω(GSR) holds. In view of Theorem 3.1, there exists a gp-set SG of G that induces a complete subgraph of GSR. By the proof of Theorem 4.2, SG × V(Kn) is a gp-set of GKn. The components of the subgraph of GKn induced by SG × V(Kn) are of the form QKn, where Q is a complete component induced by SG.

If (g, x) and (g′, x′) belong to different components QKn and Q′ ⊠ Kn induced by SG × V(Kn), then with Proposition 2.5 in mind we have dGKn((g, x), (g′, x′)) = dG(g, g). Since g and g′ are MMD, this implies that also (g, x) and (g′, x′) are MMD.

Suppose next that (g, x) and (g′, x′) belong to the same component QKn. If g = g′, then (g, x) and (g′, x′) are clearly MMD. Suppose now that gg′. Then, since g and g′ are adjacent and MMD in G, the vertices g and g′ are true twins. But then it follows that (g, x) and (g′, x′) are MMD in GKn. The second assertion now follows from Theorem 2.2.□

Since gp(T) = t for every tree T with t leaves, we have gp(TPn) ≥ 2t by Theorem 4.2. We next show that this becomes an equality for an infinite number of trees. To this end, we say that a tree T belongs to a family 𝓣 if there exits a finite sequence T1, …, Tr, r ≥ 1, of trees such that,

  1. T1 is a path on at least three vertices;

  2. T2 is obtained from T1 by adding a path P of order at least 3 and joining by an edge one not leaf vertex of P with one not leaf vertex of T1;

  3. for every i ∈ {3, …, r}, Ti is obtained from Ti–1 by adding a path P of order at least 3 and joining by an edge one not leaf vertex of P with one vertex of degree larger than two of Ti–1; and

  4. T = Tr.

Note that if T ∈ 𝓣 is obtained by the above construction in r steps, then T has exactly 2r leaves.

Proposition 4.4

If T ∈ 𝓣 and has 2r leaves, then gp(TPn) = 4r = ω((TPn)SR).

Proof

From Theorem 4.2 we get gp(TPn) ≥ 4r. We next show that this is also the exact value.

Let Pni, i ∈ [r], be the path used to generate T in the ith step of the construction of T. Let Si = V(Pni) × V(Pn), i ∈ [r]. Then note that S1, …, Sr form a partition of V(TPn), where each Si induces a graph isomorphic to the strong grid graph PniPn.

Since {S1, …, Sr} form an isometric cover of TPn, Theorem 2.1 and (1) imply that

gp(TPn)i=1rgp(Si)=4r,

hence the first equality follows.

From [9, Theorem 40] we know that TSR ⊠ (Pn)SR is a subgraph of (TPn)SR. Since TSR ⊠ (Pn)SR contains a clique of size 4r, we then also have such a clique in (TPn)SR and so ω((TPn)SR) ≥ 4r. Theorem 3.1 completes the argument.□

Proposition 4.5

If r1t1 ≥ 1 and r2t2 ≥ 1, then

gp(Kr1,t1Kr2,t2)=r1r2=ω((Kr1,t1Kr2,t2)SR)=α(Kr1,t1Kr2,t2).

Proof

We first observe that diam(Kr1,t1Kr2,t2) = 2 and thus Theorem 2.3 applies.

The set obtained from the Cartesian product of the partite sets of cardinality r1 and r2 of Kr1,t1 and Kr2,t2, respectively, forms a maximal induced complete multipartite subgraph of the complement of Kr1,t1Kr2,t2 of cardinality r1r2. Since ω(Kr1,t1Kr2,t2) = 4, we deduce that gp(Kr1,t1Kr2,t2) = r1r2.

On the other hand, since Kr1,t1Kr2,t2 has diameter two and has not true twin vertices, the strong resolving graph (Kr1,t1Kr2,t2)SR is just the complement of Kr1,t1Kr2,t2. Thus, we obtain that ω((Kr1,t1Kr2,t2)SR) = α(Kr1,t1Kr2,t2) = r1r2, hence the last two equalities.□

Theorem 4.6

If r ≥ 2 and t ≥ 1, then 6 ≤ gp(PrC2t+1) ≤ 7. Moreover, if t ∈ [2] or r = 2, then gp(PrC2t+1) = 6.

Proof

If t = 1, then by Proposition 4.3, gp(PrC3) = gp(PrK3) = 3gp(Pr) = 6. Hence, from now on we may assume t ≥ 2.

Let U = {u1, …, ur} and V = {v1, …, v2t+1} be the vertex sets of Pr and C2t+1, respectively, with natural adjacencies. From Theorem 4.2, we know that gp(PrC2t+1) ≥ 6. A subpath P of C2t+1 which is of length at most t is an isometric subgraph of C2t+1, hence U × P induces an isometric subgraph of PrC2t+1. In particular this implies that the set {U × {v1, …, vt+1}, U × {vt+2, …, v2t+1}} forms an isometric cover of PrC2t+1 consisting of two strong grids. Hence, again using Theorem 2.1 together with (1) we infer that gp(PrC2t+1) ≤ 8.

We now suppose that gp(PrC2t+1) = 8 and let S be a gp-set of PrC2t+1. Let S′ be the projection of S onto C2t+1 and consider the following situations.

  1. |S′| = 8.

    This means that for each viS′ we have |(U × {vi}) ∩ S| = 1. Without loss of generality we can assume that v1S′. Consider now a partition of V given by the sets V1 = {v1, …, vt+1} and V2 = {vt+2, …, v2t+1}. As noted above, U × V1 and U × V2 induce strong grids that are isometric subgraphs of PrC2t+1. Thus, by (1) and since we have assumed gp(PrC2t+1) = 8, we deduce |S ∩ (U × V1)| = 4 and |S ∩ (U × V2)| = 4. Analogously, if V1 = {v2, …, vt+1} and V2 = {vt+2, …, v2t+1, v1}, then also U × V1 and U × V2 induce two strong grids that are isometric subgraphs of PrC2t+1. Since |S ∩ (U × V2 )| = 5, we get a contradiction.

  2. 4 ≤ |S′| ≤ 7.

    This means that there exists at least one vertex viS′ such that |(U × {vi}) ∩ S| = 2. Note that for every viS′, it must happen |(U × {vi}) ∩ S| ≤ 2, otherwise we find a geodesic containing three vertices of S. Without loss of generality we can assume that v1S′ satisfies that |(U × {v1}) ∩ S| = 2. A similar argument as in Case 1 leads to the partition of V given by V1 and V2 (as in Case 1), and such that U × V1 and U × V2 induce strong grids that are isometric subgraphs of PrC2t+1 for which |S ∩ (U × V2 )| = 6, which is again not possible.

  3. |S′| ≤ 3.

    Since for every viS′, it must happen |(U × {vi}) ∩ S| ≤ 2, we deduce that |S| = ∑viS|(U × {vi}) ∩ S| ≤ 2|S′| ≤ 6. This is a final contradiction proving that |S| = 8 is not possible.

    We have thus proved that gp(PrC2t+1) ≤ 7. Let next t = 2. Then we consider again the projection S′ as defined above, but in this case we clearly have |S′| ≤ 5. Now, if |S| = 7, then we get a contradiction along the same lines as above. Hence gp(PrC5) = 6. Finally, if r = 2, then the situation in which |S′| = 7 leads to the existence of seven vertices lying in different layers of the factor graph P2. But then there are three of such vertices lying on the same geodesic, which is not possible and so gp(P2C2t+1) = 6.□

Upper bounds on the general position number of the cylinder PrC2t and of the torus CrCt, can be deduced by using similar techniques as in the proof above, except that in the last two cases we split the torus into two cylinders. On the other hand, lower bounds can be obtained from Theorem 4.2. That is next stated.

Remark 4.7

Let r, t be two integers.

  1. If r ≥ 2 and t ≥ 3, then 6 ≤ gp(PrC2t) ≤ 8.

  2. If r ≥ 5 and t ≥ 3, then 9 ≤ gp(CrC2t) ≤ 16.

  3. If r ≥ 4 and t ≥ 2, then 9 ≤ gp(CrC2t+1) ≤ 14.

Using similar approach as in the proof of Theorem 4.6, the last upper bound 14 from Remark 4.7 can be lowered to 13.

Since C4 is a complete bipartite graph and satisfies gp(C4) = 2, from Proposition 4.5, we obtain that gp(C4C4) = 4. Note that it also occurs the equality gp(C4C4) = 4 = ω((C4C4)SR) (for information on the structure of (C4C4)SR see [9]).

Based on the results of this section we pose the following:

Problem 4.8

Is it true that if G and H are arbitrary connected graphs, then

gp(GH)=gp(G)gp(H)?

Assuming that the answer to the problem is positive, if gp(G) = ω(GSR) and gp(H) = ω(HSR), then gp(GH) = ω((GH)SR).

5 Generalized lexicographic products

Let G be a graph with V(G) = {g1, …, gn} and let Hi, i ∈ [n], be pairwise disjoint graphs. Then the generalized lexicographic product G[H1, …, Hn] has the vertex set

i[n]{(gi,h):hV(Hi)},

and the edge set

{(gi,h)(gj,h):gigjE(G),hE(Hi),hE(Hj)}i[n]{(gi,h)(gi,h):hhE(Hi)}.

In words, G[H1, …, Hn] is obtained from G by replacing each vertex viV(G) with the graph Hi, and each edge gigjE(G) with all possible edges between Hi and Hj. From this reason we will say that viV(G) expands to Hi in G[H1, …, Hn].

The generalized lexicographic product was introduced by Sabidussi back in [13]. If all the graphs Hi, i ∈ [n], are isomorphic to a graph H, then the generalized lexicographic product G[H1, …, Hn] = G[H, …, H] becomes the standard lexicographic product G[H].

Theorem 5.1

Let G be a graph with V(G) = {v1, …, vn} and let ki, i ∈ [n], be positive integers. If S is a gp-set of G that induces a complete subgraph of GSR, and min{ki : viS} ≥ max{ki : viS}, then

gp(G[Kk1,,Kkn])=i:viSki=ω((G[Kk1,,Kkn])SR).

Proof

Let G and its gp-set S be as stated in the theorem. Then gp(G) = ω(GSR) by Theorem 3.1. To simplify the notation, let Ĝ = G[Kk1, …, Kkn] in the rest of the proof. Moreover, if a vertex viV(G) expands to R = Kki in Ĝ, and V(R), then we will write vi = g(). That is, if Ĝ, then g() is the vertex of G that expands to the complete subgraph of Ĝ to which belongs.

If , ŷV(Ĝ), ŷ, then by the construction of Ĝ we infer that

dG^(x^,y^)=1;g(x^)=g(y^),dG(g(x^),g(y^));g(x^)g(y^). (2)

By Theorem 2.2, the components of 〈S〉 are complete subgraphs of G, denote them with Q1, …, Qr. Then gp(G) = i=1r |V(Qi)|. Since each vertex of Qi expands to a complete subgraph of Ĝ, the complete subgraph Qi expands to a complete subgraph of Ĝ, we will denote it with i.

We first claim that Ŝ = i=1r V(i) is a general position set of Ĝ. If V(i) and ′ ∈ V(i), where i, i′ ∈ [r], ii′, then dĜ(, ′) = dG(g(), g(′)) holds by (2). Therefore, since {Q1, …, Qr} form an in-transitive, distance-constant partition of S, the complete subgraphs {1, …, r} form an in-transitive, distance-constant partition of Ŝ. Hence, in view of Theorem 2.2, Ŝ is a general position set of Ĝ.

We next claim that Ŝ is a gp-set of Ĝ. Assume on the contrary that there exists a general position set of Ĝ such that || > |Ŝ|. Applying Theorem 2.2 again we know that the components of 〈〉 are complete graphs. Let T = {g() : }. Since is a general position set and because of (2) we infer that T is a general position set of G. But since min{ki : viS} ≥ max{ki : viS} and || > |Ŝ| it follows that |T| > |S| = gp(G), a contradiction.

We have thus proved that gp(Ĝ) = ∑i:viS ki. To complete the proof we need to show that also ω(ĜSR) = ∑i:viS ki. Since S is a complete subgraph of GSR and because of (2) we get that Ŝ is a set of MMD vertices of Ĝ. By the equality part of Theorem 3.1 we thus have ω(ĜSR) = gp(Ĝ) = ∑i:viS ki.□

6 Rooted product graphs

By a rooted graph we mean a connected graph having one fixed vertex called the root of the graph. Consider now a connected graph G of order n, and let H be a rooted graph with root v. The rooted product graph Gv H is the graph obtained from G and n copies of H, say H1, …, Hn, by identifying the root of Hi with the ith vertex of G, see [14, 15]. To formulate the following result, the notion of an interval between vertices u and v of a graph G, defined as IG(u, v) = {w : dG(u, v) = dG(u, w) + dG(w, v)}, will be useful.

Theorem 6.1

Let G be any connected graph of order n ≥ 2, and let H be a rooted graph with root v.

  1. gp(Gv H) = n = ω((Gv H)SR) if and only if H is a path and v is a leaf of H.

  2. If H contains a gp-set S not containing v and such that for each pair of vertices u, wS neither uIH(v, w) nor wIH(v, u), then gp(Gv H) = n ⋅ gp(H). Moreover, if in addition S is a maximum clique in HSR, then gp(Gv H) = ω((Gv H)SR).

  3. Suppose H is not a path rooted in one of its leaves. If every gp-set S of H either contains the root v, or contains two vertices x, y such that (xIH(v, y) or yIH(v, x)), then 2n ≤ gp(Gv H) ≤ n(gp(H) – 1). Particularly, if every gp-set of H contains the root v, then gp(Gv H) = n(gp(H) – 1).

Proof

  1. If G is P2 and H is a path rooted in a leaf v, then Gv H is also a path, and so gp(Gv H) = 2 = ω((Gv H)SR). In this sense, from now we may assume G is different from P2.

    If H is a path and v is a leaf, then clearly the set formed by the remaining leaves of all copies of H forms a general position set of Gv H, and so gp(Gv H) ≥ n. Now, suppose gp(Gv H) > n and let S be gp-set of Gv H. In consequence, by the pigeon hole principle there exists a copy, say Hi, of H such that |SV(Hi)| ≥ 2, and indeed, it must happen |SV(Hi)| = 2. But then, the two vertices of SV(Hi) and any other distinct vertex of S lie on a common geodesic, which is not possible. Therefore, gp(Gv H) ≤ n and the first equality follows. On the other hand, it can be easily observed that the strong resolving graph of Gv H is formed by a component isomorphic to a complete graph Kn, and the remaining vertices of it are isolated ones. Thus, ω((Gv H)SR) = n, which gives the second equality.

    On the other hand, assume gp(Gv H) = n = ω((Gv H)SR). If H is not a path rooted in one of its leaves, then there are at least two vertices of H, say a, b, such that dH(a, v) = dH(b, v). In consequence, the set formed by the union of the copies of a and b in each copy of H is a general position set of Gv H of cardinality 2n, which is not possible. Thus, H must be a path rooted in one of its leaves.

  2. Let Ai, i ∈ [n], be a gp-set of Hi satisfying the statement of the item, and let A = i=1n Ai. Then A is a general position set of Gv H, and so gp(Gv H) ≥ n ⋅ gp(H). Hence, suppose gp(Gv H) > n ⋅ gp(H) and let B be a gp-set of Gv H. Thus, again by the pigeon hole principle, there must be a copy Hj of H such that |BV(Hj)| > gp(H), but this is impossible since each copy of H is an isometric subgraph of Gv H and BV(Hj) is a general position set of the graph induced by Hj. Consequently, gp(Gv H) ≤ n ⋅ gp(H) and the equality follows.

    On the other hand, assume that Ai is a maximum clique in HSR. Hence gp(H) = ω(HSR). Thus, from the above we get that gp(Gv H) = nω(HSR). It remains only to prove that ω((Gv H)SR) = nω(HSR). Since any two vertices u, wAi satisfy that neither uIH(w, v) nor wIH(u, v), we see that A = i=1n Ai (defined as above) is also a clique in (Gv H)SR, and so ω((Gv H)SR) ≥ nω(HSR). Clearly, if we suppose that ω((Gv H)SR) > nω(HSR), then we obtain that some (Hj)SR contains a clique of cardinality larger than ω(HSR), which is not possible. Therefore, the required equality follows.

  3. If every gp-set S of H either contains v or contains two vertices x, y such that without loss of generality v belongs to an x, y-geodesic, then in order to construct a general position set of Gv H from the union of the gp-sets S in each copy of H, we need to remove some vertices from each copy of S including v if it is the case. Clearly, the maximum number of vertices we may remove from S is |S| – 1, since a set formed by one vertex from each copy of H is a general position set of Gv H. However, as we next show by removing from S at most |S| – 2 vertices or removing |S| – 1 and adding one other vertex not from S, we also obtain a general position set. Since H is not a path rooted in one of it leaves, there are two vertices xi, yiV(Hi) such that dHi(xi, v) = dHi(yi, v), i ∈ [n]. Thus, the set Q = i=1n {xi, yi} is a general position set of cardinality 2n in Gv H, and the lower bound follows. On the other hand, let D be a gp-set of Gv H and for every i ∈ [n], let Di = DV(Hi). If there is a set Dj such that |Dj| = gp(H) (note that |Dj| ≤ gp(H) since V(Hj) induces an isometric subgraph of Gv H), then either vDj or for any two vertices x, y of Dj it must happen that v does not belong to an x, y-geodesic nor to a y, x-geodesic, but this is a contradiction with our assumption. Therefore, for every i ∈ [n], |Di| ≤ gp(H) – 1, which implies the upper bound.

    We now consider the particular case in which every gp-set of H contains the root v. Let Si be a gp-set of the copy Hi of H and let S = i=1n (Si ∖ {v}). Since v belongs to Si, it happens that v does not belong to any x, y-geodesic for every x, yS ∖ {v}. Thus, no three vertices of S lie on the same geodesic of Gv H, and so, S is a general position set of Gv H. Therefore, gp(Gv H) = n(gp(H) – 1).□

Equality in the lower bound given in item (iii) of Theorem 6.1 above can be noticed if H is a path rooted in a vertex of degree two, where gp(Gv H) = 2n. On the other hand, as we next observe the value of gp(Gv H) can be very far from both bounds given above.

Proposition 6.2

There is a graph G of order n and a graph H rooted in a vertex v such that n ≪ gp(Gv H) ≪ n(gp(H) – 1).

Proof

We consider a graph H obtained as follows. We begin with a complete graph Kr. Next we add a vertex v and join it by an edge to exactly t vertices of Kr, where 2 ≤ tr – 2, and choose v as the root of this graph. Note that H has only one gp-set S formed by the set of vertices of the complete graph Kr. Also, note that if x is adjacent to v, then xIH(y, v) for any y not adjacent to v.

Now, let G be a connected graph and let A be a gp-set of Gv H. Thus, if Ai = AV(Hi), then either every vertex of Ai is adjacent to v or no vertex of Ai is adjacent to v, and so |Ai| ≤ max{t, rt}. As a consequence, gp(Gv H) = |A| = i=1n |Ai| ≤ n ⋅ max{t, rt}. On the other hand, the union of all neighbors of v in each copy of H in Gv H, or the union of all not neighbors of v in each copy of H in Gv H is clearly a general position set of Gv H, and so gp(Gv H) ≥ n ⋅ max{t, rt}, which implies the equality gp(Gv H) = n ⋅ max{t, rt}. Since gp(H) = r, the difference n(gp(H) – 1) – gp(Gv H) = n(r – 1 – max{t, rt}) can be arbitrarily large, as well as the difference gp(Gv H) – n = n(max{t, rt} – 1).□

Acknowledgements

We acknowledge the financial support from the Slovenian Research Agency (research core funding No. P1-0297 and projects J1-1693, J1-9109, N1-0095, N1-0108). This research was done while the second author was visiting the University of Ljubljana, Slovenia, supported by "Ministerio de Educación, Cultura y Deporte", Spain, under the "José Castillejo" program for young researchers (reference number: CAS18/00030).

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Received: 2019-04-24
Accepted: 2019-08-23
Published Online: 2019-10-13

© 2019 Klavžar and Yero, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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