Characterizing extremal graphs for open neighbourhood location-domination
Florent Foucaud, Narges Ghareghani, Aida Roshany-Tabrizi, Pouyeh Sharifani
aa r X i v : . [ m a t h . C O ] J a n Characterizing extremal graphs for open neighbourhoodlocation-domination
Florent Foucaud ∗† Narges Ghareghani ‡ Aida Roshany Tabrizi § Pouyeh Sharifani ¶ January 15, 2021
Abstract
An open neighbourhood locating-dominating set is a set S of vertices of a graph G such that eachvertex of G has a neighbour in S , and for any two vertices u, v of G , there is at least one vertex in S thatis a neighbour of exactly one of u and v . We characterize those graphs whose only open neighbourhoodlocating-dominating set is the whole set of vertices. More precisely, we prove that these graphs areexactly the graphs all whose connected components are half-graphs (a half-graph is a special bipartitegraph with both parts of the same size, where each part can be ordered so that the open neighbourhoodsof consecutive vertices differ by exactly one vertex). This corrects a wrong characterization from theliterature. We consider only simple, finite and undirected graphs. Many graph problems are motivated by the task ofuniquely identifying the vertices using a small set of vertices. In one type of such problems, the identificationis done by the neighbourhood within the solution set. More precisely, denoting N ( v ) the open neighbourhoodof a vertex v , an open neighbourhood locating-dominating set of a graph G is a set S of vertices of G such thateach vertex v has a neighbour in S (that is, S is a total dominating set ) and for any two distinct vertices v, w of G , N ( v ) ∩ S = N ( w ) ∩ S . In other words, each vertex of G has a distinct and nonempty set of neighbourswithin S . It is not difficult to see that a graph admits an open neighbourhood locating-dominating set if andonly if it has no isolated vertices and contains no pair of open twins (open twins are vertices with the sameopen neighbourhood). We call such graphs locatable . The smallest size of an open neighbourhood locating-dominating set of a locatable graph G is its open neighbourhood location-domination number, denoted by γ OL ( G ) .The notion of open neighbourhood location-domination was introduced about a decade ago by Seo andSlater in [19], and various aspects of the problem were subsequently studied in [3, 7, 8, 10, 12, 15, 17, 20].Open neighbourhood locating-dominating sets are related to the concept of identifying codes [14] (whereopen neighbourhoods are replaced by closed neighbourhoods) and the earlier concept of locating-dominatingsets [21, 22] (where only vertices not in the solution set need to be uniquely identified). The variant oflocating-total dominating sets is also studied [11]. These concepts are special cases of more general questionsof identification in discrete structures, which are studied in various settings since the 1960s [1, 2, 16, 18]. ∗ Univ. Bordeaux, CNRS, Bordeaux INP, LaBRI, UMR5800, F-33400 Talence, France. † Univ. Orléans, INSA Centre Val de Loire, LIFO EA 4022, F-45067 Orléans Cedex 2, France. ‡ Department of Industrial Design, College of Fine Arts, University of Tehran, Tehran, Iran. § School of Mathematics, Statistics, and Computer Science, University of Tehran, Tehran, Iran. ¶ School of Mathematics, Institute for Research in Fundamental Sciences (IPM), Tehran, Iran. half-graphs ; half-graphs form an infinite family of bipartite graphs, studied and named byErdős and Hajnal (see [4]). See Figure 1 for the first five half-graphs. Half-graphs or their complements havebeen useful in previous works about locating-dominating sets [6] and identifying codes [5].We show that each half-graph H of order n satisfies γ OL ( H ) = n . Note that if a locatable graphis disconnected, its open neighbourhood location-domination number is the sum of those of its connectedcomponents, thus one obtains other examples by taking disjoint unions of half-graphs. Our main theoremproves that these are the only examples. Theorem 1.
For a connected locatable graph G of order n , γ OL ( G ) = n if and only if G is a half-graph. This characterization corrects the incomplete one from [3]. It also reveals an interesting property of openneighbourhood locating-dominating sets: indeed, there are only few concepts known to us in the area ofgraph identification problems, for which there exist infinitely many finite connected undirected graphs whichhave their whole vertex set as unique solution. (Another concept with this feature is the one of solid-locatingdominating sets, see [13].)
In a locatable graph G , some vertices have to belong to any open neighbourhood locating-dominating set:we call such vertices forced . There are two types of forced vertices: those that are forced because of thedomination condition, and those that are forced because of the location condition. Definition 2.
Let G be a locatable graph and let v be a vertex of G . Vertex v is called domination-forced if there exists a vertex w , such that v is the unique neighbour of w . Vertex v is called location-forced ifthere exist two distinct vertices x and y , such that N ( x ) ⊖ N ( y ) = { v } (where ⊖ denotes the set symmetricdifference). We can observe the following.
Proposition 3.
If there is a vertex v in a locatable graph G which is neither domination-forced nor location-forced, then V ( G ) \ { v } is an open neighbourhood locating-dominating set of G .Proof. Since v is not domination-forced, every vertex of G has a neighbour in V ( G ) \ { v } . Moreover since v is not location-forced, for every pair z, w of distinct vertices in G , there is a vertex in V ( G ) \ { v } in thesymmetric difference N ( z ) ⊖ N ( w ) , which therefore distinguishes z and w .Proposition 3 implies that in any locatable graph G of order n with γ OL ( G ) = n , every vertex isdomination-forced or location-forced (or both).We now formally define half-graphs, which were named in [4]. Definition 4.
For any integer k ≥ , the half-graph H k is the bipartite graph on vertex sets { v , . . . , v k } and { w , . . . , w k } , with an edge between v i and w j if and only if i ≤ j . 0 We now show that every half-graph H k of order n = 2 k satisfies γ OL ( H k ) = n .2 v (a) H w v w v (b) H w v w v v w (c) H w v w v v w v w (d) H w v w v v w v w v w (e) H Figure 1: The five first half-graphs.
Proposition 5.
For every k ≥ , the half-graph H k satisfies γ OL ( H k ) = 2 k .Proof. Let k ≥ . It is clear that H k is locatable, thus γ OL ( H k ) ≤ k . By Proposition 3, it suffices toprove that every vertex of H k is forced. Vertices v and w k are domination-forced, since they are the onlyneighbours of w and v k , respectively. For every integer i with ≤ i ≤ k − , w i is the only vertex inthe symmetric difference of N ( v i ) and N ( v i +1 ) . Thus, w i is location-forced. Since graph H k is symmetric,similarly v k , . . . , v are location-forced. This completes the proof.Before proving our characterization, we will use the following celebrated theorem of Bondy to upper-boundthe number of locating-forced vertices in any locatable graph. Theorem 6 (Bondy’s Theorem [2]) . Let V be an n -set, and A = {A , A , . . . , A n } be a family of n distinctsubsets of V . There is a ( n − -subset X of V such that the sets A ∩ X, A ∩ X, A ∩ X, . . . , A n ∩ X arestill distinct. Corollary 7.
Every locatable graph G of order n has at most n − location-forced vertices.Proof. Construct from graph G the set system with V ( G ) as its n -set and where the A i ’s are all the openneighbourhoods of vertices of G . Bondy’s theorem implies that there is one vertex such that removing it doesnot create two same open neighbourhoods. In other words, this vertex is not location-forced.We are now ready to prove Theorem 1 that a connected locatable graph G of order n satisfies γ OL ( G ) = n if and only if G is a half-graph. In the proof, we will say that a vertex v of a graph G is located by a set S ifthere is no vertex w = v with N ( v ) ∩ S = N ( w ) ∩ S . Also, v is total dominated by S if v has a neighbour in S . Proof of Theorem 1.
The sufficient side is proved in Proposition 5. We prove the necessary side by inductionon n . Let G be a connected locatable graph of order n with γ OL ( G ) = n . We first prove the statement for n ≤ . The graph of order is not locatable. The only locatable connected graph of order is K (which3s also the half-graph H ), and γ OL ( K ) = 2 . The only locatable connected graph of order is K , and γ OL ( K ) = 2 .Suppose G is a connected locatable graph of order n such that n ≥ and γ OL ( G ) = n . By Proposition 3,every vertex in G is either domination-forced or location-forced. By Corollary 7, there is at least one vertexthat is domination-forced, let us call it x . let y be a vertex such that x is the unique neighbour of y . Weclaim that vertex y is location-forced. If y was domination-forced, then there would exist a vertex that isdominated only by y . That vertex should be equal to x , and then G would be K , a contradiction. Therefore, y is location-forced and so, there exists a vertex z , such that N ( x ) = N ( z ) ∪ { y } .Now we remove x and y from G and call the new graph G ′ . We claim that G ′ is locatable and connected.Since every vertex that is a neighbour of x is a neighbour of z , and y has no neighbours in V ( G ′ ) , G ′ remainsconnected. This implies that G ′ has no isolated vertex (since n ≥ ). To see that G ′ is locatable, assumeby contradiction that there are two vertices s and t that are open twins in G ′ . Thus, in G , without loss ofgenerality, N ( s ) = N ( t ) ∪ { x } . But recall that N ( x ) = N ( z ) ∪ { y } . This means that z is a neighbour of s ,and not a neighbour of t . So s and t are not open twins in G ′ , a contradiction. Thus, G ′ is locatable.We now claim that γ OL ( G ′ ) = n − . By contradiction, suppose γ OL ( G ′ ) ≤ n − . Let S ′ be an openneighbourhood locating-dominating set of G ′ , with | S ′ | ≤ n − . We claim that S = S ′ ∪ { x, y } is an openneighbourhood locating-dominating set of G . First of all, S is total dominating, indeed, every vertex in G ′ hasa neighbour in S ′ , and x and y are total dominated by each other. Secondly, we have to check that S locatesall the vertices. Vertex x is located by S , since it is the only vertex that has y as its neighbour. In addition, weclaim that y is located by S . Indeed, if it was not, there would be a vertex w such that N ( w ) ∩ S = { x } . Thus N ( w ) ∩ S ′ = ∅ . Therefore, w was not total dominated by S ′ in G ′ , a contradiction. So y is located by S . Inaddition, all the vertices of G ′ are still located because they were located in G ′ by S ′ , and they remain locatedin G by the vertices in S ′ . As a result, S is an open neighbourhood locating-dominating set of G . Thus, if | S ′ | ≤ n − , then | S | ≤ n − , which is a contradiction. Therefore, γ OL ( G ′ ) = n − . By induction, we concludethat G ′ is isomorphic to the half-graph H k with k = n − and vertex set { v , v , ..., v k } ∪ { w , w , ..., w k } .Now we claim that vertex z is domination-forced in G ′ . Note that vertex z is location-forced in G (indeed z cannot be domination-forced in G , since all its neighbours are also neighbours of x ), which means that thereexist vertices v and w in G such that N ( w ) = N ( v ) ∪ { z } . Vertex v cannot be in the common neighbourhoodof x and z , and since N ( x ) = N ( z ) ∪ { y } , we have v = y . Hence, w has degree in G and N ( w ) = { x, z } .Therefore, by removing x and y from G to obtain G ′ , z is the only neighbour of w in G ′ . Thus, z isdomination-forced in G ′ , as claimed.As we know, in H k there are two domination-forced vertices: v and w k . By the symmetry of H k , withoutloss of generality, we can assume that z = w k . As a result, it follows from the fact that N ( x ) = N ( z ) ∪ { y } ,that we can label x = w k +1 and y = v k +1 and that G is isomorphic to the half-graph H k +1 . This completesthe proof. References [1] B. Bollobás and A. D. Scott. On separating systems.
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