Neighborhood structures and products of undirected graphs

Discrete Mathematics 313 (2013) 563–574

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Neighborhood structures and products of undirected graphs Martin Sonntag a,∗ , Hanns-Martin Teichert b a

Faculty of Mathematics and Computer Science, Technische Universität Bergakademie Freiberg, D–09596 Freiberg, Prüferstr. 1, Germany

b

Institute of Mathematics, University of Lübeck, D–23562 Lübeck, Ratzeburger Allee 160, Germany

article

info

Article history: Received 14 March 2012 Received in revised form 20 November 2012 Accepted 21 November 2012 Available online 23 December 2012 Keywords: Neighborhood graph Neighborhood hypergraph Graph products

abstract Let G = (V , E ) be a simple undirected graph. The neighborhood hypergraph N (G) = (V , E N ) of G has the edge set E N = {e ⊆ V | |e| ≥ 1 ∧ ∃ x ∈ V : e = NG (x)}. In a certain sense, this is a generalization of the well-known notion of the neighborhood graph N (G) = (V , E N ). For several products G1 ◦ G2 of simple undirected graphs G1 and G2 , we investigate the question how N (G1 ◦ G2 )/N (G1 ◦ G2 ) can be constructed from G1 , G2 , N (G1 ), N (G2 )/N (G1 ), N (G2 ) and vice versa. © 2012 Elsevier B.V. All rights reserved.

1. Introduction All graphs G = (V (G), E (G)) and hypergraphs H = (V (H ), E (H )) considered here may have isolated vertices but no multiple edges. In graphs, loops are forbidden. The neighborhood graph N (G) = (V , E N ) of the graph G = (V , E ) has the edge set E N = {{a, b} | a ̸= b ∧ ∃x ∈ V : {x, a} ∈ E ∧ {x, b} ∈ E }. Moreover, N (G) = (V , E N ) is the neighborhood hypergraph of the graph G = (V , E ) if and only if the edges of N (G) are the (open) neighborhoods NG (v) = {w ∈ V | {w, v} ∈ E } of the vertices v ∈ V (G), i.e. E N = {e ⊆ V | |e| ≥ 1 ∧ ∃x ∈ V : e = NG (x)}. The above definition of neighborhood graphs is the usual one (cf. the references below), where an edge {a, b} ∈ E N (G) indicates that a and b have a common neighbor v in G. In general, contrary to neighborhood hypergraphs, the edges of the neighborhood graph N (G) do not contain the full information on the sets of neighbors of the vertices of G. Our decision, to define E N (G) as the set of all nonempty neighborhoods of the vertices of G, has the consequence that vertices v ∈ V (G) of degree 1 correspond to loops in N (G), whereas they do not lead to any edge in N (G). Of course, it would also be possible to forbid loops in neigborhood hypergraphs, if we are willing to lose information about vertices of degree 1. In addition to the definition of the neighborhood NG (v), we refer to the set NG [v] = {w ∈ V | {w, v} ∈ E } ∪ {v} as the closed neighborhood of the vertex v ∈ V (G). Definitions not explicitly given here can be found in Diestel [6] and Berge [2]. Several aspects of neighborhood graphs were investigated in the last thirty years (cf. Acharya and Vartak [1], Boland, Brigham and Dutton [3,4], Brigham and Dutton [5], Exoo and Harary [7], Greenberg, Lundgren and Maybee [8], Lundgren, Merz, Maybee and Rasmussen [11], Lundgren, Merz and Rasmussen [12], Lundgren and Rasmussen [13], Lundgren, Rasmussen and Maybee [14], Miller, Brigham and Dutton [15] and Schiermeyer, Sonntag and Teichert [18]). Some of these

∗

Corresponding author. E-mail address: [email protected] (M. Sonntag).

0012-365X/$ – see front matter © 2012 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2012.11.028

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papers use the notation 2-step graph or competition graph instead of neighborhood graph. As the latter name indicates, the neighborhood graph N (G) of an undirected graph G is closely related to the competition graph C (D) of a digraph D. Surveys about competition graphs can be found in Kim [9], Lundgren [10] and Roberts [17]. In Section 2, for five products G1 ◦ G2 of graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) we investigate the construction of the neigborhood graph N (G1 ◦ G2 ) = (V , E◦N ) from G1 , G2 , N (G1 ) = (V1 , E1N ) and N (G2 ) = (V2 , E2N ). Furthermore, we give some results concerning the reconstruction of the original graphs G1 , G2 and their neighborhood graphs N (G1 ), N (G2 ), respectively, from N (G1 ◦ G2 ). In the subsequent Section 3, for the same five products G1 ◦ G2 we consider the relations between the neighborhood hypergraph N (G1 ◦ G2 ) = (V , E◦N ) on the one hand and G1 , G2 , N (G1 ) = (V1 , E1N ) as well as N (G2 ) = (V2 , E2N ) on the other hand. We place the emphasis on the question whether G1 , G2 , N (G1 ) = (V1 , E1N ) or N (G2 ) = (V2 , E2N ) can be reconstructed from N (G1 ◦ G2 ) or not. To obtain N (G1 ◦ G2 ) from G1 , G2 , N (G1 ) = (V1 , E1N ) and N (G2 ) = (V2 , E2N ) is an easier problem. In a certain sense in Sections 2 and 3 we continue our investigations in [19], where we considered competition hypergraphs of products of digraphs. All the five products G1 ◦ G2 (Cartesian sum G1 + G2 , Cartesian product G1 × G2 , normal product G1 ∗ G2 , lexicographic product G1 · G2 and disjunction G1 ∨ G2 ) considered here have the vertex set V = V1 × V2 ; using the notation  E = {{(a, b), (a′ , b′ )} | a, a′ ∈ V1 ∧ b, b′ ∈ V2 } their edge sets E◦ = E (G1 ◦ G2 ) are defined as follows: E+ = {{(a, b), (a′ , b′ )} ∈  E | ({a, a′ } ∈ E1 ∧ b = b′ ) ∨ (a = a′ ∧ {b, b′ } ∈ E2 )}, E× = {{(a, b), (a′ , b′ )} ∈  E | {a, a′ } ∈ E1 ∧ {b, b′ } ∈ E2 }, E∗ = E× ∪ E+ , E· = {{(a, b), (a′ , b′ )} ∈  E | {a, a′ } ∈ E1 ∨ (a = a′ ∧ {b, b′ } ∈ E2 )}, E∨ = {{(a, b), (a′ , b′ )} ∈  E | {a, a′ } ∈ E1 ∨ {b, b′ } ∈ E2 }. We obtain immediately E+ ⊆ E∗ ⊆ E· ⊆ E∨ and E× ⊆ E∗ . Except for the lexicographic product all these products are commutative in the sense that G1 ◦ G2 ≃ G2 ◦ G1 , where ◦ ∈ {+, ×, ∗, ∨}. In the following, let G1 = (V1 , E1 ), G2 = (V2 , E2 ) be graphs; V1 = {1, 2, . . . , r }, V2 = {1, 2, . . . , s} and the vertices of V = V1 × V2 be arranged according to the places of an (r , s)-matrix. Then, for each ◦ ∈ {+, ∗, ·, ∨}, the subgraph of G1 ◦ G2 induced by the vertices of a column and a row of this matrix scheme is isomorphic to G1 and G2 , respectively. We refer to these subgraphs as the columns and the rows of G1 ◦ G2 . Analogously, we define the columns and the rows of N (G1 ◦ G2 ) as well as of N (G1 ◦ G2 ). If G = (V , E ) and G′ = (V ′ , E ′ ) are graphs we often use the notation G ∪ G′ = (V ∪ V ′ , E ∪ E ′ ). Moreover, if e ⊆ V1 × V2 is a set of vertices of G1 ◦ G2 , then π1 (e) and π2 (e) denotes the set of the first and the second components of the vertices of e, respectively. In other words, π1 and π2 is the projection of the vertices of e onto their first and their second components, respectively. 2. Neighborhood graphs 2.1. The Cartesian sum G1 + G2 The neighborhood graph N (G1 + G2 ) can be obtained from the Cartesian sum of the neighborhood graphs of G1 and G2 and the Cartesian product of G1 and G2 . Theorem 1. For graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) it holds N (G1 + G2 ) = (N (G1 ) + N (G2 )) ∪ (G1 × G2 ). N Proof. Let (i1 , j1 ), (i2 , j2 ) ∈ V = V1 × V2 be two distinct vertices of G1 + G2 . Then {(i1 , j1 ), (i2 , j2 )} ∈ E+ if and only if there is a vertex (i, j) ∈ V1 × V2 with (i1 , j1 ), (i2 , j2 ) ∈ NG1 +G2 ((i, j)). This is equivalent to the existence of i ∈ V1 and j ∈ V2 such that one of the following cases occurs: (a): {i1 , i} ∈ E1 , {i2 , i} ∈ E1 , j1 = j2 = j or (b): i1 = i2 = i, {j1 , j} ∈ E2 , {j2 , j} ∈ E2 or (c): {i1 , i} ∈ E1 , j1 = j, i2 = i, {j2 , j} ∈ E2 or (d): i1 = i, {j1 , j} ∈ E2 , {i2 , i} ∈ E1 , j2 = j. Next we eliminate the vertices i and j in (a)–(d) and obtain (a): {i1 , i2 } ∈ E1N , j1 = j2 or

(b): i1 = i2 , {j1 , j2 } ∈ E2N or (c/d): {i1 , i2 } ∈ E1 , {j1 , j2 } ∈ E2 . Using N (G1 ) + N (G2 ) and G1 × G2 this leads to (a/b): {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) + N (G2 )) or (c/d): {(i1 , j1 ), (i2 , j2 )} ∈ E× . This is equivalent to {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) + N (G2 )) ∪ E (G1 × G2 ).



M. Sonntag, H.-M. Teichert / Discrete Mathematics 313 (2013) 563–574

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In principle, the graphs G1 , G2 , N (G1 ) and N (G2 ) can be reconstructed from N (G1 + G2 ): if E1 ̸= ∅ and E2 ̸= ∅, the edges N of Gi (i ∈ {1, 2}) can be obtained as the projection of the edges {(i1 , j1 ), (i2 , j2 )} ∈ E+ with i1 ̸= i2 and j1 ̸= j2 onto their first and second components, respectively. We can describe the reconstruction formulaically as follows. Let N E ′ = {{(i1 , j1 ), (i2 , j2 )} ∈ E+ | i1 ̸= i2 ∧ j1 ̸= j2 }, ′ π1 (E ) = {{i1 , i2 } | ∃j1 , j2 ∈ V2 : {(i1 , j1 ), (i2 , j2 )} ∈ E ′ }

and

π2 (E ′ ) = {{j1 , j2 } | ∃i1 , i2 ∈ V1 : {(i1 , j1 ), (i2 , j2 )} ∈ E ′ }. Corollary 1. If E ′ ̸= ∅, then G1 = (V1 , π1 (E ′ )) and G2 = (V2 , π2 (E ′ )). N Now let E1′ = E+ ∩ {{(i1 , 1), (i2 , 1)} | i1 , i2 ∈ V1 } and N E2′ = E+ ∩ {{(1, j1 ), (1, j2 )} | j1 , j2 ∈ V2 }.

Since the columns of N (G1 + G2 ) are isomorphic to N (G1 ) and the rows are isomorphic to N (G2 ), the neighborhood graphs of G1 and G2 can be reconstructed. Corollary 2. N (G1 ) = (V1 , π1 (E1′ )) and N (G2 ) = (V2 , π2 (E2′ )). 2.2. The Cartesian product G1 × G2 The neighborhood graph N (G1 × G2 ) is the normal product of the neighborhood graphs of G1 and G2 , if neither G1 nor G2 has isolated vertices. Theorem 2. If G1 = (V1 , E1 ) and G2 = (V2 , E2 ) do not contain isolated vertices, then N (G1 × G2 ) = N (G1 ) ∗ N (G2 ) = (N (G1 ) + N (G2 )) ∪ (N (G1 ) × N (G2 )). Proof. Again, let (i1 , j1 ), (i2 , j2 ) ∈ V = V1 × V2 be two distinct vertices. In N (G1 × G2 ) there exists the edge {(i1 , j1 ), (i2 , j2 )} ∈ N E× if and only if in G1 × G2 , both vertices have a common neighbor (i, j) ∈ V1 × V2 , i.e. (i1 , j1 ), (i2 , j2 ) ∈ NG1 ×G2 ((i, j)). The definition of the Cartesian product implies that this is equivalent to {i1 , i} ∈ E1 , {i2 , i} ∈ E1 , {j1 , j} ∈ E2 and {j2 , j} ∈ E2 . In detail, this means

{i1 , i} ∈ E1 ∧ (i1 = i2 ∨ i1 = ̸ i2 ∧ {i2 , i} ∈ E1 ) and {j1 , j} ∈ E2 ∧ (j1 = j2 ∨ j1 = ̸ j2 ∧ {j2 , j} ∈ E2 ). This is the same as i1 = i2 ∈ NG1 (i) ∨ (i1 ̸= i2 ∧ {i1 , i2 } ⊆ NG1 (i))

and

j1 = j2 ∈ NG2 (j) ∨ (j1 ̸= j2 ∧ {j1 , j2 } ⊆ NG2 (j)). Because of the non-existence of isolated vertices in G1 and G2 , we obtain equivalently i1 = i2 ∨ {i1 , i2 } ∈ E1N

and

j1 = j2 ∨ {j1 , j2 } ∈ E2N .

Now we can proceed to the neighborhood graphs N (G1 ) and N (G2 ) and the above can be expressed as i1 = i2 ∧ {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) + N (G2 ))

or

j1 = j2 ∧ {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) + N (G2 )) or

{(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) × N (G2 )). Finally, it is easy to see that this is nothing but the assertion:

{(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) + N (G2 )) ∪ E (N (G1 ) × N (G2 )).  Remark 1. Isolated vertices in G1 (in G2 ) induce rows (columns) of isolated vertices in N (G1 × G2 ), since a vertex (i, j) ∈ V with NG1 (i) = ∅ or NG2 (j) = ∅ has no neighbor in G1 × G2 and, consequently, also no neighbor in N (G1 × G2 ). Since each column (row) of N (G1 × G2 ) corresponding to a non-isolated vertex of G2 (of G1 ) is isomorphic to N (G1 ) (to N (G2 )) we easily obtain N (G1 ) and N (G2 ) from N (G1 × G2 ). N Corollary 3. Let i ∈ V1 and j ∈ V2 be non-isolated vertices in G1 and G2 , respectively. Moreover, let E1′ = E× ∩ {{(i1 , j), (i2 , j)} | ′ ′ N i1 , i2 ∈ V1 } and E2 = E× ∩ {{(i, j1 ), (i, j2 )} | j1 , j2 ∈ V2 }. Then N (G1 ) = (V1 , π1 (E1 )) and N (G2 ) = (V2 , π2 (E2′ )).

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2.3. The normal product G1 ∗ G2 Considering the normal product G1 ∗ G2 of the graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) we have a more complicated situation. For i ∈ {1, 2} let G∅i = (Vi , ∅) and G+∅ = (Vi+ , ∅), where Vi+ = {v ∈ Vi | d(v) > 0}. i Theorem 3. For the normal product of G1 and G2 , +∅ ∅ N (G1 ∗ G2 ) = ((N (G1 ) ∪ G1 ) × (N (G2 ) ∪ G2 )) ∪ (N (G1 ) + G∅2 ) ∪ (G1 + G+∅ 2 ) ∪ (G1 + N (G2 )) ∪ (G1 + G2 ).

Proof. Let (i1 , j1 ), (i2 , j2 ) ∈ V = V1 × V2 be two distinct vertices. We have to consider the following cases: (a) i1 ̸= i2 ∧ j1 ̸= j2 ,

(b) i1 ̸= i2 ∧ j1 = j2

and

(c) i1 = i2 ∧ j1 ̸= j2 .

Investigating these cases it can be useful to consider Fig. 1, where for each subcase of (a), (b) and (c) we sketch the vertices

(i1 , j1 ), (i2 , j2 ) and a third vertex (i, j) (which is used in the proof) in G1 ∗ G2 together with the relevant edges. Note that in Fig. 1 the vertices are arranged according to their places in the (r , s)-matrix scheme explained in Section 1, i.e. i, i1 , i2 denote rows and j, j1 , j2 denote columns of the matrix scheme, respectively. {(i1 , j1 ), (i2 , j2 )} ∈ E∗N is equivalent to the existence of a common neighbor (i, j) ∈ V1 × V2 in G1 ∗ G2 , i.e. (i1 , j1 ), (i2 , j2 ) ∈ NG1 ∗G2 ((i, j)). In connection with the cases (a), (b) and (c) mentioned above we have to investigate the following three situations: For(a) ({i1 , i2 } ⊆ NG1 (i) ∨ {i1 , i2 } ∈ E1 ) ∧ ({j1 , j2 } ⊆ NG2 (j) ∨ {j1 , j2 } ∈ E2 ), for(b) {i1 , i2 } ⊆ NG1 (i) ∨ ({i1 , i2 } ∈ E1 ∧ {j1 , j} ∈ E2 ), for(c) {j1 , j2 } ⊆ NG2 (j) ∨ ({i1 , i} ∈ E1 ∧ {j1 , j2 } ∈ E2 ). To demonstrate that these three cases with the described subcases include all possible configurations in N (G1 ∗ G2 ) (up to exchanging the roles of the vertices (i1 , j1 ) and (i2 , j2 )), we discuss the subcases in detail (cf. Fig. 1). Case (a) has the subcases (a1) {i1 , i2 } ⊆ NG1 (i) ∧ {j1 , j2 } ⊆ NG2 (j), (a3) {i1 , i2 } ∈ E1 ∧ {j1 , j2 } ⊆ NG2 (j) and

(a2) {i1 , i2 } ⊆ NG1 (i) ∧ {j1 , j2 } ∈ E2 , (a4) {i1 , i2 } ∈ E1 ∧ {j1 , j2 } ∈ E2 .

Owing to the fact that it suffices that there exists an i ∈ V1 and a j ∈ V2 with the required properties, it is possible to reduce the two situations in (b1) and (c1) in Fig. 1 in each case to one constellation by choosing j = j1 = j2 and i = i1 = i2 . So for (b) we have (b1): {i1 , i2 } ⊆ NG1 (i) or (b2): {i1 , i2 } ∈ E1 ∧ {j1 , j} ∈ E2 . (c) has the subcases (c1): {j1 , j2 } ⊆ NG2 (j) or (c2): {i1 , i} ∈ E1 ∧ {j1 , j2 } ∈ E2 . A final look at the cases (a)–(c) completes our proof: Case (a): i1 ̸= i2 ∧ j1 ̸= j2 . Subcase (a1) is equivalent to {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) × N (G2 )), subcase (a2) corresponds to {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) × G2 ), in subcase (a3) we have {(i1 , j1 ), (i2 , j2 )} ∈ E (G1 × N (G2 )) and in the last subcase (a4) it follows {(i1 , j1 ), (i2 , j2 )} ∈ E (G1 × G2 ). Case (b): i1 ̸= i2 ∧ j1 = j2 . Subcase (b1) is the case {(i1 , j1 ), (i2 , j1 )} ∈ E (N (G1 ) + G∅2 ) and in subcase (b2) we obtain {(i1 , j1 ), (i2 , j1 )} ∈ E (G1 + G+∅ 2 ). Case (c): i1 = i2 ∧ j1 ̸= j2 . In subcase (c1) evidently {(i1 , j1 ), (i1 , j2 )} ∈ E (G∅1 + N (G2 )) is valid and in subcase (c2) we get {(i1 , j1 ), (i1 , j2 )} ∈ E (G+∅  1 + G2 ). 2.4. The lexicographic product G1 · G2 Additionally to G∅2 = (V2 , ∅) we define G02∅ = (V20 , ∅), where V20 = {v ∈ V2 | v isolated} is the set of isolated vertices in G2 . Note that in the case V20 = ∅ we have G02∅ = (∅, ∅) as well as G1 · G02∅ = (∅, ∅), i.e. both graphs are empty. Moreover, let Ks = (V2 , EKs ), where EKs = {{a, b} ⊆ V2 | a ̸= b}. Theorem 4. For graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ), N (G1 · G2 ) = (N (G1 ) · N (G2 )) ∪ (G1 · G∅2 ) \ E (G1 · G02∅ ) ∪ G+∅ 1 · Ks .

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Proof. Let (i1 , j1 ), (i2 , j2 ) ∈ V = V1 × V2 be two distinct vertices. Then {(i1 , j1 ), (i2 , j2 )} ∈ E·N if and only if there is a common neighbor (i, j) in G1 · G2 , i.e.: (i1 , j1 ), (i2 , j2 ) ∈ NG1 ·G2 ((i, j)). Per definition, this is equivalent to

({i1 , i} ∈ E1 ∨ (i1 = i ∧ {j1 , j} ∈ E2 ))

and

({i2 , i} ∈ E1 ∨ (i2 = i ∧ {j2 , j} ∈ E2 )) .

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Fig. 1. The subcases in the proof of Theorem 3.

A simple logical transformation leads to the following four alternative cases (the notations of these cases will be explained below): (a1)/(b): (c1): (c2): (a2):

{i1 , i} ∈ E1 , {i2 , i} ∈ E1 or {i1 , i} ∈ E1 , i2 = i, {j2 , j} ∈ E2 or i1 = i, {i2 , i} ∈ E1 , {j1 , j} ∈ E2 or i1 = i2 = i, {j1 , j} ∈ E2 , {j2 , j} ∈ E2 .

Depending on whether i1 = i2 or not the first situation (a1)/(b) leads to the first two lines (a1) and (b) of the following alternative cases. The remaining lines are equivalent to their above counterparts. So we have the cases: (a1): {i1 , i2 } ∈ E (N (G1 )) or (b): i1 = i2 , dG1 (i1 ) > 0 or (c1): {i1 , i2 } ∈ E1 , dG2 (j2 ) > 0 or

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(c2): {i1 , i2 } ∈ E1 , dG2 (j1 ) > 0 or (a2): i1 = i2 , {j1 , j2 } ∈ E (N (G2 )). The notations (a1), (a2) and (c1), (c2) are chosen that way since (a1) ∨ (a2) and (c1) ∨ (c2) are equivalent to the following cases (a) and (c), respectively. (a): {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) · N (G2 )) or (b): {(i1 , j1 ), (i2 , j2 )} ∈ E G+∅ 1 · Ks or

(c): {(i1 , j1 ), (i2 , j2 )} ∈ E (G1 · G∅2 ) \ E (G1 · G02∅ ).



Thus, we make the following conclusion. Corollary 4. If neither G1 nor G2 contain isolated vertices, then N (G1 · G2 ) = (N (G1 ) · N (G2 )) ∪ G1 · G∅2 ∪ G∅1 · Ks .

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2.5. The disjunction G1 ∨ G2 Investigating the disjunction, for graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) we need the following definition:

(V1+ × V2 ) ∪∗ (V1 × V2+ ) = {{(a1 , b1 ), (a2 , b2 )} | (a1 , b1 ) ̸= (a2 , b2 ) ∧ (a1 , b1 ) ∈ V1+ × V2 ∧ (a2 , b2 ) ∈ V1 × V2+ }. Now let  G = (V1 × V2 ,  E ), where  E = (V1+ × V2 ) ∪∗ (V1 × V2+ ). Theorem 5. If G1 = (V1 , E1 ) and G2 = (V2 , E2 ), then +∅ N (G1 ∨ G2 ) = N (G1 ) ∨ G∅2 ∪ G∅1 ∨ N (G2 ) ∪ G+∅ ∪ G. 1 + Ks ∪ Kr + G2

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Proof. Let (i1 , j1 ), (i2 , j2 ) ∈ V = V1 × V2 be distinct vertices in N (G1 ∨ G2 ). Analogous to the previous proofs, {(i1 , j1 ), (i2 , j2 )} ∈ E∨N if and only if there is a common neighbor (i, j) in G1 ∨ G2 . That is, (i1 , j1 ), (i2 , j2 ) ∈ NG1 ∨G2 ((i, j)). Per definition, this is the same as {i1 , i} ∈ E1 ∨ {j1 , j} ∈ E2 and {i2 , i} ∈ E1 ∨ {j2 , j} ∈ E2 . This leads to the alternative cases: (a1)/(b1): (c1): (c2): (a2)/(b2):

{i1 , i} ∈ E1 , {i2 , i} ∈ E1 or {i1 , i} ∈ E1 , {j2 , j} ∈ E2 or {i2 , i} ∈ E1 , {j1 , j} ∈ E2 or {j1 , j} ∈ E2 , {j2 , j} ∈ E2 .

Equivalently, we obtain the alternative: (a1): (b1): (c1): (c2): (a2): (b2):

{i1 , i2 } ∈ E (N (G1 )) or i1 = i2 ∈ V1+ or i1 ∈ V1+ , j2 ∈ V2+ or i2 ∈ V1+ , j1 ∈ V2+ or {j1 , j2 } ∈ E (N (G2 )) or j1 = j2 ∈ V2+ .

(a1) or (a2) is equivalent to (a): {(i1 , j1 ), (i2 , j2 )} ∈ E (N (G1 ) ∨ G∅2 ) ∪ E (G∅1 ∨ N (G2 )); (b1) or (b2) is equivalent to +∅ (b): {(i1 , j1 ), (i2 , j2 )} ∈ E (G+∅ 1 + Ks ) ∪ E (Kr + G2 ) and

(c1) or (c2) is equivalent to (c): {(i1 , j1 ), (i2 , j2 )} ∈ (V1+ × V2 ) ∪∗ (V1 × V2+ ) =  E.



Due to the definition of  E, in case of V1+ = V1 and V2+ = V2 we have

 E = {{(a1 , b1 ), (a2 , b2 )} ⊆ V1 × V2 | (a1 , b1 ) ̸= (a2 , b2 )}. Corollary 5. If neither G1 nor G2 contain isolated vertices, then N (G1 ∨ G2 ) is a complete graph. 2.6. Additional remarks concerning the reconstruction Some (positive) results concerning the reconstruction of G1 , G2 , N (G1 ) and N (G2 ) were presented in Sections 2.1 and 2.2 (Corollaries 1–3).

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If G1 , G′1 , G2 , G′2 are graphs with G1 ̸≃ G′1 or G2 ̸≃ G′2 , where N (G1 ) = N (G′1 ) and N (G2 ) = N (G′2 ), then constructing G1 and G2 from N (G1 ◦ G2 ) can be impossible, for all of the products we consider, except the Cartesian sum. As an example, for n ≥ 5, it is well-known that the wheel Wn with n vertices has the same neighborhood graph as the complete graph Kn , i.e. N (Wn ) = N (Kn ) = Kn , but Wn ̸≃ Kn . We use these graphs in the proof of the following. Corollary 6. Let G1 = (V1 , E1 ), G2 = (V2 , E2 ) and ◦ ∈ {×, ∗, ·, ∨}. Then, in general, G1 and G2 cannot be reconstructed from N (G1 ◦ G2 ). Proof. Let r ≥ 5, s ≥ 5, {G1 , G′1 } = {Wr , Kr } and {G2 , G′2 } = {Ws , Ks }. Note that the neighborhood graphs of G1 , G′1 and of G2 , G′2 are the complete graphs with r and s vertices, respectively. Cartesian product: The assertion follows immediately from Theorem 2. Normal product: Because of

(N (G1 ) ∪ G1 ) × (N (G2 ) ∪ G2 ) = (Kr ∪ G1 ) × (Ks ∪ G2 ) = Kr × Ks , +∅ ∅ ∅ (N (G1 ) + G∅2 ) ∪ (G1 + G+∅ 2 ) = (Kr + G2 ) ∪ (G1 + G2 ) = Kr + G2 , +∅ ∅ ∅ ∅ (G∅1 + N (G2 )) ∪ (G+∅ 1 + G2 ) = (G1 + Ks ) ∪ (G1 + G2 ) = G1 + Ks , Theorem 3 leads to N (G1 ∗ G2 ) = (Kr × Ks ) ∪ (Kr + G2 ) ∪ ∅ (G1 + Ks ) = (Kr × Ks ) ∪ (Kr + Ks ). Analogously, we obtain N (G′1 ∗ G′2 ) = (Kr × Ks ) ∪ (Kr + Ks ). Lexicographic product: Due to Theorem 4 it holds N (G1 · G2 ) = Kr · Ks = N (G′1 · G′2 ).

Disjunction: The assertion results from the stronger Remark 2 below.



The next remark follows as an immediate conclusion to Corollary 5. Remark 2. Let G1 and G2 be arbitrary graphs without any isolated vertices. Then none of the graphs G1 , G2 , N (G1 ) and N (G2 ) can be reconstructed from N (G1 ∨ G2 ) ≃ Kr ·s . 3. Neighborhood hypergraphs Per definition, the edges e ∈ E N of the neighborhood hypergraph N (G) = (V , E N ) of a graph G = (V , E ) are the neighbor sets NG (v) of cardinality at least 1 of the vertices v ∈ V (G). Therefore, the definition of the graph product G1 ◦ G2 yields the edge set E◦N of the neighborhood hypergraph N (G1 ◦ G2 ) = (V , E◦N ). In the following theorem we give the edge sets E◦N of the neighborhood hypergraphs N (G1 ◦ G2 ) = (V , E◦N ) of the five graph products G1 ◦ G2 , where ◦ ∈ {+, ×, ∗, ·, ∨}. Theorem 6. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ) be graphs and V = V1 × V2 . Then: (1) (2) (3) (4) (5)

E+N E×N E∗N E·N E∨N

    = { {i} × NG2 (j) ∪ NG1 (i) × {j} | (i, j) ∈ V1 × V2 } \ {∅} = {e1 × e2 | e1 ∈ E1N ∧ e2 ∈ E2N } = {NG1 [i] × NG2[j] \ {(i, j)} | (i, j) ∈ V1 × V2 } \ {∅} = { {i} × NG2 (j) ∪ NG1 (i) × V2 | (i, j) ∈ V1 × V2 } \ {∅} = {(V1 × e2 ) ∪ (e1 × V2 ) | e1 ∈ E1N ∧ e2 ∈ E2N }.

Note that, in order to simplify the notation, we fix here that (5) is a short form for

E∨N

 {(V × e2 ) ∪ (e1 × V2 ) | e1 ∈ E1N ∧ e2 ∈ E2N },    1 {e1 × V2 | e1 ∈ E1N }, = {V1 × e2 | e2 ∈ E2N },   ∅,

if E1N if E1N if E1N if E1N

̸ ∅ and E2N ̸= ∅ = ̸= ∅ and E2N = ∅ = ∅ and E2N ̸= ∅ = ∅ = E2N .

(∗)

Proof. (1) In G1 + G2 , a vertex (i′ , j′ ) ∈ V1 × V2 is a neighbor of (i, j) if and only if (a) (i′ , j′ ) is in the same row as (i, j) (i.e. i′ = i) and j′ is a neighbor of j in G2 or (b) (i′ , j′ ) is in the same column as (i, j) (i.e. j′ = j) and i′ is a neighbor of i in G1 . (2) In the Cartesian product G1 × G2 , the neighborhood NG1 ×G2 ((i, j)) of a vertex (i, j) ∈ V (G1 × G2 ) contains all vertices (i′ , j′ ) with i′ ∈ NG1 (i) and j′ ∈ NG2 (j). Therefore, E×N = {NG1 (i) × NG2 (j) | (i, j) ∈ V1 × V2 } \ {∅}. (3) Consider a vertex (i, j) ∈ V1 × V2 in G1 ∗ G2 . Since E∗ = E× ∪ E+ , (1) and (2) lead to NG1 ∗G2 ((i, j)) = NG1 +G2 ((i, j)) ∪ NG1 ×G2 ((i, j))

= {(i′ , j′ ) | (i′ = i ∧ j′ ∈ NG2 (j)) ∨ (i′ ∈ NG1 (i) ∧ j′ = j) ∨ (i′ ∈ NG1 (i) ∧ j′ ∈ NG2 (j))} = {(i′ , j′ ) | i′ ∈ NG1 [i] ∧ j′ ∈ NG2 [j] ∧ (i′ , j′ ) ̸= (i, j)} = (NG1 [i] × NG2 [j]) \ {(i, j)}.

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(4) A vertex (i′ , j′ ) is a neighbor of the vertex (i, j) in the lexicographic product G1 · G2 if and only if (a) (i′ , j′ ) is in the same row as (i, j) (i.e. i′ = i) and j′ is a neighbor of j in G2 or (b) i′ is a neighbor of i in G1 and j′ ∈ V2 . (5) (i′ , j′ ) is a neighbor of (i, j) in the disjunction G1 ∨ G2 if and only if i′ ∈ NG1 (i) or j′ ∈ NG2 (j). Analogous to the second part in (2) we obtain (5).  Now the question arises whether or not the neighborhood hypergraph N (G1 ◦ G2 ) of the product graph G1 ◦ G2 can be obtained from N (G1 ) and N (G2 ) using suitable hypergraph products. In case of the Cartesian product G1 × G2 and the disjunction G1 ∨ G2 this is possible (see the following Remark 3). Generally, N for the other three products the same is impossible: to construct the edges in E+ , E∗N and E·N , for every i ∈ {1, 2} and all N edges e ∈ Ei it would be necessary to know the vertices v ∈ Vi with e = NGi (v). But, in general, this information cannot be obtained from the neighborhood hypergraph N (Gi ). Let H1 = (V1 , E1 ) and H2 = (V2 , E2 ) be hypergraphs and V = V1 × V2 . We define the edge sets of the Cartesian product H1 ×H H2 = (V , E× ) and the disjunction H1 ∨H H2 = (V , E∨ ) of H1 and H2 as follows:

E× = {e1 × e2 | e1 ∈ E1 ∧ e2 ∈ E2 } E∨ = {(V1 × e2 ) ∪ (e1 × V2 ) | e1 ∈ E1 ∧ e2 ∈ E2 }. For the definition of E∨ we use a convention analogous to (∗) (cf. Theorem 6). Note that the Cartesian product H1 ×H H2 is defined in [16] as the direct product of H1 and H2 . In general, if H1 and H2 are graphs, both products provide hypergraphs but not graphs, i.e. H1 ×H H2 ̸≃ H1 × H2 and H1 ∨H H2 ̸≃ H1 ∨ H2 . Using these hypergraph products, by Theorem 6, (2) and (5), we make the following conclusion. Remark 3. If G1 = (V1 , E1 ) and G2 = (V2 , E2 ) are graphs and V = V1 × V2 , then (1) N (G1 × G2 ) = N (G1 ) ×H N (G2 ) (2) N (G1 ∨ G2 ) = N (G1 ) ∨H N (G2 ). In the following subsections, we turn to the problem of reconstructing the neighborhood hypergraphs N (G1 ) = (V1 , E1N ), N (G2 ) = (V2 , E2N ) and the graphs G1 = (V1 , E1 ), G2 = (V2 , E2 ) from N (G1 ◦ G2 ) = (V , E◦N ), where ◦ ∈ {+, ×, ∗, ·, ∨}. 3.1. The Cartesian sum G1 + G2 In this subsection, we will demonstrate that (excepting some special cases) the neighborhood hypergraphs N (G1 ) = (V1 , E1N ) and N (G2 ) = (V2 , E2N ) as well as the graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) can be reconstructed from the N neighborhood hypergraph N (G1 + G2 ) = (V , E+ ). At first, we investigate the exceptional cases. Lemma 1. Let |V (N (G1 + G2 ))| = |V2 |, i.e. |V1 | = 1. Then G1 + G2 ≃ G2 and therefore N (G1 + G2 ) ≃ N (G2 ). Both, G1 and N (G1 ), have only one vertex and no edge. Apart from very special cases (e.g. E2 = ∅ or |E2 | = 1) the graph G2 is not reconstructable from N (G1 + G2 ) ≃ N (G2 ).  Because of G1 + G2 ≃ G2 + G1 analogous conclusions follow from |V (N (G1 + G2 ))| = |V1 |. N Lemma 2. Let |V1 | > 1, |V2 | > 2 and N (G1 + G2 ) be 1-uniform, i.e. E+ solely consists of loops. Then either E1 = ∅ or E2 = ∅. N Which of the sets E1 and E2 is empty can be found out if and only if |E+ | < |V1 | · |V2 |. In exactly this case N (G1 ) and N (G2 ) can be reconstructed from N (G1 + G2 ). N in contradiction to the 1-uniformity Proof. First, assume {i, i′ } ∈ E1 and {j, j′ } ∈ E2 . Then (i, j′ ), (i′ , j) ∈ NG1 +G2 ((i, j)) ∈ E+ of N (G1 + G2 ). Hence, either E1 = ∅ or E2 = ∅. N Now consider an edge {(i, j)} ∈ E+ . If i′ ∈ V1 is a neighbor of the vertex i, then for each j′ ∈ V2 we have {(i, j′ )}, {(i′ , j′ )} ∈ N ′ ′ ′ E+ , since {(i, j ), (i , j )} ∈ E+ . Consequently, N N N (a) E+ includes all singletons of the vertices of the rows i and i′ of N (G1 + G2 ), i.e. {(i, j′ )} ∈ E+ and {(i′ , j′ )} ∈ E+ , where j′ ∈ V2 . (b) E2 = ∅ as well as E2N = ∅.  (c) The set of the non-isolated vertices of G1 is V1+ = {i | i ∈ e∈E N π1 (e)}. Moreover, E1N = {{i} | i ∈ V1+ }. +

For each k ∈ V1 property (a) implies

• all singletons of the vertices of the row k are contained in E+N if and only if k ∈ V1+ and • no singleton of any vertex of the row k is contained in E+N if and only if k ∈ V1 \ V1+ . N As a consequence, in the case |E+ | < |V1 | · |V2 | we have E2 = ∅ if and only if E+N is the union of systems of singletons of vertices of entire rows of N (G1 + G2 ).

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Trivially, an analogous statement holds for E1 = ∅ and the columns of N (G1 + G2 ). N On the contrary, if |E+ | = |V1 | · |V2 | is valid, then E+N = {{(i, j)} | (i, j) ∈ V1 × V2 } is the union of the systems of the singletons of the vertices of all (entire) rows as well as the union of the systems of the singletons of the vertices of all (entire) columns of N (G1 + G2 ). In this case, both E1N = {{i} | i ∈ V1 }, E2 = ∅ and E2N = {{j} | j ∈ V2 }, E1 = ∅ is possible.  N Note that under the suppositions of Lemma 2, where E+ includes only singletons, the maximum degree of the vertices of the graphs G1 and G2 is one.

Lemma 3. Let |V1 | > 1, |V2 | > 1 and N (G1 + G2 ) be not 1-uniform. N (i) If |π1 (e)| = 1, for each e ∈ E+ , then E1 = ∅. N (ii) If |π2 (e)| = 1, for each e ∈ E+ , then E2 = ∅.

In case (i) each row of N (G1 + G2 ) is isomorphic to N (G2 ) and in case (ii) each column of N (G1 + G2 ) is isomorphic to N (G1 ). In general, G2 and G1 , respectively, are reconstructable only in special situations (cf. Lemma 1). N Proof. We consider only case (i), since (ii) can be investigated analogously. If E+ = ∅, then E1 = ∅ is trivial. Now let ′ N ′ ′ ′ (i, j), (i, j ) ∈ e ∈ E+ ̸= ∅ with j ̸= j and {j, l}, {j , l} ∈ E2 . Assume {k, k } ∈ E1 . Then (k, j), (k′ , l) ∈ e′ = NG1 +G2 ((k, l)) ∈ E+N and |π1 (e′ )| > 1 contradicting (i). Hence E1 is empty, there are no edges between different rows in G1 + G2 and the isomorphism of each row of G1 + G2 to G2 implies the isomorphism of each row of N (G1 + G2 ) to N (G2 ). 

Lemmas 1–3 include all cases where at least one of E1 and E2 is empty. The only situation where N (G1 ) or N (G2 ) cannot be reconstructed from N (G1 + G2 ) is the case where

|V1 | > 1,

|V2 | > 1 and E+N = {{(i, j)} | (i, j) ∈ V1 × V2 } (cf. Lemma 2).

In the following theorem we investigate the cases not being covered by the above Lemmas, i.e. the situation E1 ̸= ∅ and E2 ̸= ∅. N The case E1 ̸= ∅ and E2 ̸= ∅ is equivalent to the existence of an edge e ∈ E+ with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2. This becomes clear if we consider edges {x, x′ } ∈ E1 and {y, y′ } ∈ E2 and the neighborhood of the vertex (x, y) ∈ V in G1 + G2 , N since (x, y′ ), (x′ , y) ∈ NG1 +G2 ((x, y)) ∈ E+ . N Theorem 7. Let N (G1 + G2 ) = (V , E+ ) be the neighborhood hypergraph of the Cartesian sum of the graphs G1 = (V1 , E1 ) and N G2 = (V2 , E2 ) such that there exists an edge e ∈ E+ with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2. Then N (G1 ) and N (G2 ) as well as G1 and G2 can be reconstructed from N (G1 + G2 ). N Proof. By Theorem 6(1), E+ = { {i} × NG2 (j) ∪ NG1 (i) × {j} | (i, j) ∈ V1 × V2 } \ {∅}. In the following, occasionally we will make use of the existence of two edges {x, x′ } ∈ E1 and {y, y′ } ∈ E2 in G1 and G2 , respectively. Let k ∈ V1 be an arbitrary vertex. N N Because of (k, y) ∈ NG1 +G2 ((k, y′ )) ∈ E+ the vertex k appears as the first component of a vertex of an edge e ∈ E+ ; that N is k ∈ π1 (e). If for each e ∈ E+ from k ∈ π1 (e) it follows |π1 (e)| = 1, then k is isolated, since the existence of a k′ ∈ NG1 (k) N would imply (k, y′ ), (k′ , y) ∈ NG1 +G2 ((k′ , y′ )) ∈ E+ and, trivially, |π1 (NG1 +G2 ((k′ , y′ )))| ≥ 2. N On the other hand, if k ∈ π1 (e), where e ∈ E+ and |π1 (e)| ≥ 2, then k is non-isolated. In this way, the isolated vertices in G1 and the set V1+ of the non-isolated vertices of G1 (and, analogously, of G2 ) can be determined. As a consequence, for N the determination of G1 , G2 , N (G1 ) and N (G2 ) it suffices to consider the edges e ∈ E+ with |e| ≥ 2. + Now, let k ∈ V1 . N with k ∈ π1 (e) and |π1 (e)| ≥ 2. In the following, for such edges e we will write Then there exists an edge e ∈ E+ e = {(k, j1 ), . . . , (k, jp ), (i1 , j), . . . , (iq , j)}, where k, i1 , . . . , iq are pairwise distinct, j1 , . . . , jp , j are pairwise distinct and p ≥ 1, q ≥ 1. N Case 1: There is an edge e ∈ E+ such that k ∈ π1 (e), |π1 (e)| ≥ 2 and p > 1. Since each vertex (i′ , j′ ) in G1 + G2 has only neighbors in the union of row i′ and column j′ of G1 + G2 , and k appears at least twice as the first component of vertices in e = {(k, j1 ), . . . , (k, jp ), (i1 , j), . . . , (iq , j)}, it follows e = NG1 +G2 ((k, j)). Therefore, NG1 (k) = {i1 , . . . , iq } (and NG2 (j) = {j1 , . . . , jp }). N Case 2: For each edge e ∈ E+ with k ∈ π1 (e), |π1 (e)| ≥ 2 it follows that p = 1. N Let e ∈ E+ be an edge of maximum cardinality with the property k ∈ π1 (e) and |π1 (e)| ≥ 2. In detail, let e = {(k, j1 ), (i1 , j), . . . , (iq , j)}. If q > 1 then it is clear that e = NG1 +G2 ((k, j)) and, therefore, NG1 (k) = {i1 , . . . , iq } (as well as NG2 (j) = {j1 }). If q = 1, we obtain e = {(k, j1 ), (i1 , j)}. Owing to |π1 (e)| ≥ 2 we get k ̸= i1 . The assumption j1 = j would contradict our supposition that j, j1 , . . . , jp are pairwise distinct. Hence we have k ̸= i1 and j1 ̸= j. This leads to e = NG1 +G2 ((k, j)) or e = NG1 +G2 ((i1 , j1 )). In both cases {k, i1 } ∈ E1 and {j1 , j} ∈ E2 are valid. Assume, there is a vertex i2 ∈ NG1 (k)\{i1 }. Then {(k, j1 ), (i1 , j), (i2 , j)} ⊆ NG1 +G2 ((k, j)) in contradiction to the maximality of the edge e. Therefore, NG1 (k) = {i1 }. Using only the neighborhood hypergraph N (G1 + G2 ), the above considerations provide for each vertex k ∈ V1 the neighborhood NG1 (k). In this way we obtain G1 as well as N (G1 ) from N (G1 + G2 ). Analogously, G2 and N (G2 ) can be constructed from N (G1 + G2 ). 









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3.2. The Cartesian product G1 × G2 N By Theorem 6, E× = {e1 × e2 | e1 ∈ E1N ∧ e2 ∈ E2N }. Therefore, to reconstruct N (G1 ) and N (G2 ) from N (G1 × G2 ), it is N necessary to have E× ̸= ∅ which is equivalent to E1N ̸= ∅ and E2N ̸= ∅ as well as to E1 ̸= ∅ and E2 ̸= ∅. N Theorem 8. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ) be graphs and E× ̸= ∅. Then N (G1 ) and N (G2 ) can be reconstructed from N (G1 × G2 ). N Proof. Since E× ̸= ∅, we obtain EiN = {πi (e) | e ∈ E×N }, for i = 1, 2.



N Considering E× = {e1 × e2 | e1 ∈ E1N ∧ e2 ∈ E2N }, it is clear that for graphs G2 ̸= G′2 such that N (G2 ) = N (G′2 ) we know N (G1 × G2 ) = N (G1 × G′2 ).

As an example we choose graphs G2 = (V2 , E2 ) and G′2 = (V2 , E2′ ) with V2 = {1, 2, 3, 4}, E2 = {{1, 2}, {3, 4}} and E2′ = {{1, 4}, {2, 3}}. Because of E (N (G2 )) = {{1}, {2}, {3}, {4}} = E (N (G′2 )), Theorem 6(2) implies N (G1 × G2 ) = N (G1 × G′2 ) for arbitrarily chosen G1 . Therefore we make the following conclusion. Remark 4. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ). Then, in general, G1 and G2 cannot be reconstructed from N (G1 × G2 ). 3.3. The normal product G1 ∗ G2 Considering the normal product G1 ∗ G2 of the graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) we find several similarities to the Cartesian sum G1 + G2 . Corollary 7. Lemmas 1–3 remain valid for the normal product G1 ∗ G2 instead of the Cartesian sum G1 + G2 . Proof. In Lemmas 1–3 at least one of E1 and E2 is empty. Therefore, under the assumptions of the Lemmas G1 + G2 = G1 ∗ G2 is valid.  We obtain a result analogous to Theorem 7.

Theorem 9. Let N (G1 ∗ G2 ) = (V , E∗N ) be the neighborhood hypergraph of the normal product of the graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) such that there exists an edge e ∈ E∗N with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2. Then N (G1 ) and N (G2 ) as well as G1 and G2 can be reconstructed from N (G1 ∗ G2 ). Proof. By Theorem 6, E∗N = { NG1 [i] × NG2 [j] \{(i, j)} | (i, j) ∈ V1 × V2 }\{∅}. Additionally, the existence of an edge e ∈ E∗N with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2 implies E1 ̸= ∅ and E2 ̸= ∅. Thus, there are edges {x, x′ } ∈ E1 and {y, y′ } ∈ E2 . First we consider an edge e ∈ E∗N with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2, as we will see it suffices to use such edges in order to obtain G1 and G2 . Then there is a vertex (i, j) ∈ V with e = NG1 ∗G2 ((i, j)), i.e. π1 (e) = NG1 [i] and π2 (e) = NG2 [j]. For each i′ ∈ NG1 (i) = NG1 [i] \ {i} = π1 (e) \ {i} in e there are exactly n2 = |NG2 [j]| = |π2 (e)| ≥ 2 vertices having the first component i′ . On the other hand, since the vertex (i, j) have to be deleted in NG1 [i] × NG2 [j] in order to obtain e, in the edge e there are exactly n2 − 1 ≥ 1 vertices having the first component i. Analogously, for every j′ ∈ NG2 (j) = π2 (e) \ {j} in e there are exactly n1 = |NG1 [i]| = |π1 (e)| vertices with the second component j′ and exactly n1 − 1 ≥ 1 vertices having the second component j. In this way, for every e ∈ E∗N with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2 the vertices ie ∈ V1 and je ∈ V2 with e = NG1 ∗G2 ((ie , je )) are uniquely determined. The same holds for their neighborhoods NG1 (ie ) = NG1 [ie ] \ {ie } = π1 (e) \ {ie } ̸= ∅ and NG2 (je ) = NG2 [je ] \ {je } = π2 (e) \ {je } ̸= ∅. On the other hand, if i ∈ V1 is a non-isolated vertex in G1 , then there exists an edge e ∈ E∗N with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2 and a vertex j ∈ V2 such that e = NG1 ∗G2 ((i, j)), namely the vertex y from above will be a suitable vertex having the required property. Of course, the analogous result holds for an arbitrarily chosen non-isolated vertex j ∈ V2 . Hence, the investigation of the edges e ∈ E∗N with |π1 (e)| ≥ 2 and |π2 (e)| ≥ 2 including the corresponding vertices ie ∈ V1 , je ∈ V2 enables us to determine all non-isolated vertices of G1 and G2 as well as their neighborhoods: Let E 2 = {e | e ∈ E∗N ∧ |π1 (e)| ≥ 2 ∧ |π2 (e)| ≥ 2}. Then the set of the non-isolated vertices in G1 is V1+ = e∈E 2 π1 (e) = {i | ∃e ∈ E 2 : i = ie }. Moreover, for each i ∈ V1+ there exists a unique e ∈ E 2 with i = ie and we obtain NG1 (i) = π1 (e) \ {i}. Consequently, E1N = {π1 (e) \ {ie } | e ∈ E 2 }.  Analogously, in G2 we have V2+ = e∈E 2 π2 (e) = {j | ∃e ∈ E 2 : j = je } and for every j ∈ V2+ there is a unique e ∈ E 2 with j = je and NG2 (j) = π2 (e) \ {j}. Therefore, E2N = {π2 (e) \ {je } | e ∈ E 2 }. Then, V1 \ V1+ and V2 \ V2+ are the sets of the isolated vertices vertices in G1 and G2 , respectively. In this way, we obtain G1 , G2 , N (G1 ) and N (G2 ). 





Additionally, we mention that the isolated vertices in G1 are the vertices which occur in at least one edge e ∈ E∗N with |π1 (e)| = 1 (and, of course, in no edge e′ ∈ E 2 ).

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Therefore, V1 \ V1+ = {π1 (e) | e ∈ E∗N ∧ |π1 (e)| = 1}. This follows from the existence of at least one edge {y, y′ } ∈ E2 , since for any isolated vertex i ∈ V1 trivially (i, y′ ) ∈ NG1 ∗G2 ((i, y)) ∈ E∗N and NG1 ∗G2 ((i, y)) ⊆ {i} × V2 hold. The analogous result is true for the isolated vertices in G2 . In general, it is not possible to reconstruct an arbitrary graph G from its neighborhood hypergraph N (G). For the Cartesian sum and the normal product, Theorems 7 and 9 show that the neighborhood hypergraphs N (G1 + G2 ) and N (G1 ∗ G2 ) contain the complete information on G1 + G2 and G1 ∗ G2 , respectively.



Remark 5. Since the graphs Gi (i = 1, 2) can be reconstructed from N (G1 + G2 ) and N (G1 ∗ G2 ) under the assumptions of Theorems 7 and 9, trivially the same holds for G1 + G2 and G1 ∗ G2 , respectively. 3.4. The lexicographic product G1 · G2 As known from Theorem 6 we have E·N = { {i} × NG2 (j) ∪ NG1 (i) × V2 | (i, j) ∈ V1 × V2 } \ {∅}.









Theorem 10. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ). Then N (G1 ) and N (G2 ) can be reconstructed from N (G1 · G2 ). Proof. If E·N = ∅, then all vertices in G1 and all vertices in G2 are isolated, i.e. E1 = ∅ = E2 . So we assume E·N ̸= ∅ and consider the reconstruction of N (G1 ) first. (A) E1N = {{i | {i} × V2 ⊆ e} | e ∈ E·N } \ {∅}. To see this, as a first step note that for an arbitrary e ∈ E·N it holds that ∅ = {i | {i} × V2 ⊆ e} if and only if e = NG1 ·G2 ((ie , je )) with NG1 (ie ) = ∅. In this case we have e = {ie } × NG2 (je ), where je is not isolated, i.e. ∅ ̸= NG2 (je ) ⊂ V2 . (If both ie and je would be isolated we had the contradiction e = NG1 ·G2 ((ie , je )) = ∅ ̸∈ E·N ). On the other hand, suppose ∅ ̸= {i | {i} × V2 ⊆ e}. Let e = NG1 ·G2 ((ie , je )). Then ie ̸∈ {i | {i} × V2 ⊆ e}, since (ie , je ) ̸∈ NG1 ·G2 ((ie , je )). Therefore, every i ∈ V1 with {i} × V2 ⊆ e has to be a neighbor of ie and even {i | {i} × V2 ⊆ e} = NG1 (ie ) ∈ E1N . In this way we obtain all elements of E1N . N (B) E2 = {{j | ∃i ∈ V1 : (i, j) ∈ e ∧ {i} × V2 ̸⊆ e} | e ∈ E·N } \ {∅}. Let e ∈ E·N . It follows {j | ∃i ∈ V1 : (i, j) ∈ e ∧ {i} × V2 ̸⊆ e} = ∅ if and only if e is the union of nothing but entire rows of G1 · G2 . Such an entire row {i} × V2 is a subset of e = NG1 ·G2 ((ie , je )) if and only if i ∈ NG1 (ie ). Trivially, {ie } × V2 ̸⊆ e and e2 = {j | j ∈ V2 ∧ (ie , j) ∈ e} ⊂ V2 . Moreover, e2 = NG2 (je ) and e2 ∈ E2N hold if and only if je is not isolated in G2 , this is equivalent to e2 ̸= ∅. In this way the edges e ∈ E·N , which include an incomplete row of G1 · G2 , provide all edges of E2N .  Corollary 8. If there exists an edge e ∈ E·N , which includes an incomplete row of G1 · G2 , then G1 = (V1 , E1 ) can be reconstructed from N (G1 · G2 ). Proof. Using part (A) of the above proof, we obtain E1N , i.e. all non-empty neighborhoods NG1 (i′ ) of vertices i′ ∈ V1 . In the following we will describe how to find the (unique) vertex ie1 with NG1 (ie1 ) = e1 for an arbitrarily chosen e1 ∈ E1N . So let e1 ∈ E1N and e ∈ E·N such that e1 = {i | {i} × V2 ⊆ e}. Further let e′ = e \ (e1 × V2 ). Then there exists a unique ie ∈ V1 with e′ ⊆ {ie } × V2 and NG1 (ie ) = e1 . (Equivalently, ie can be also determined by {ie } = π1 (e′ ).) Each vertex i ∈ V1 , which cannot be obtained in this manner, is isolated.  Corollary 9. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ). Then, in general, G2 cannot be reconstructed from N (G1 · G2 ). Proof. We reuse the example from the end of Section 3.2. So let G2 = (V2 , E2 ) and G′2 = (V2 , E2′ ) be the graphs with V2 = {1, 2, 3, 4}, E2 = {{1, 2}, {3, 4}} and E2′ = {{1, 4}, {2, 3}}. In case of the lexicographic product it is recommended to choose special graphs for the graph G1 in order to demonstrate E (N (G1 · G2 )) = E (N (G1 · G′2 )). An appropriate family of graphs may contain all graphs G1 = (V1 , E1 ) with V1 = {1, . . . , r } and E1 = {{i, i′ } | i, i′ ∈ V1 ∧ |i − i′ | = 1}, where r ≥ 1. For i ∈ V1 we use the notation zi = {{i′ } × V2 | i′ ∈ V1 ∧ |i − i′ | = 1}. Then, for all i ∈ V1 and j ∈ V2 , the set zi consists of all vertices contained in the row i − 1 (if i > 1) and in the row i + 1 (if i < r) of G1 · G2 , respectively. Therefore, zi includes all neighbors of the vertex (i, j) in G1 · G2 being not contained in row i. It follows that

E (N (G1 · G2 )) =

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{NG1 ·G2 ((i, 1)), NG1 ·G2 ((i, 2)), NG1 ·G2 ((i, 3)), NG1 ·G2 ((i, 4))}

i∈V1

=



{zi ∪ {(i, 2)}, zi ∪ {(i, 1)}, zi ∪ {(i, 4)}, zi ∪ {(i, 3)}}

i∈V1

=



{NG1 ·G′2 ((i, 3)), NG1 ·G′2 ((i, 4)), NG1 ·G′2 ((i, 1)), NG1 ·G′2 ((i, 2))}

i∈V1

= E (N (G1 · G′2 )). Hence we have N (G1 · G2 ) = N (G1 · G′2 ), where G2 ̸= G′2 .



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3.5. The disjunction G1 ∨ G2 For given graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) Theorem 6 gives the edge set of N (G1 ∨ G2 ) as E∨N = {(V1 × e2 ) ∪ (e1 × V2 ) | e1 ∈ E1N ∧ e2 ∈ E2N }. E∨N can be completely described without using G1 and G2 since only the neighborhood hypergraphs N (G1 ) are N (G2 ) are needed. Also it is impossible to reconstruct the graphs G1 and G2 from N (G1 ∨ G2 ). This can be seen considering the same example used for Remark 4. Thus we make the following conclusion. Remark 6. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ). Then, in general, G1 and G2 cannot be reconstructed from N (G1 ∨ G2 ). On the other hand, the neighborhood hypergraphs of G1 and G2 can be reconstructed. Theorem 11. Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ). Then N (G1 ) and N (G2 ) can be reconstructed from N (G1 ∨ G2 ). Proof. If E∨N = ∅, then we get E1N = ∅ = E2N as well as E1 = ∅ = E2 , i.e. N (G1 ) and N (G2 ) as well as G1 and G2 can be reconstructed. Now let E∨N ̸= ∅ and e ∈ E∨N . Per definition, e has to be the neighborhood of a vertex in G1 ∨ G2 , i.e. there exist vertices ie ∈ V1 and je ∈ V2 such that e = NG1 ∨G2 ((ie , je )). Thus e is the union of several entire rows {i} × V2 and entire columns V1 × {j} of G1 ∨ G2 , where i ∈ {i1 , . . . , ip } ⊆ V1 (0 ≤ p < |V1 |) and j ∈ {j1 , . . . , jq } ⊆ V2 (0 ≤ q < |V2 |). From Remark 6 it follows that ie as well as je can be determined only in special cases (e.g. ie is unique if it has the degree |V1 | − 1). But we will see that the set {i1 , . . . , ip } of all numbers i with {i} × V2 ⊆ e is the neighborhood of ie in G1 , i.e. {i1 , . . . , ip } = NG1 (ie ) ∈ E1N . In this way, we can obtain the neighborhoods of all vertices in G1 , i.e. we get E1N . Therefore, we have to show {i|{i} × V2 ⊆ e} = NG1 (ie ). The first inclusion NG1 (ie ) ⊆ {i|{i} × V2 ⊆ e} follows immediately from the definition of the disjunction G1 ∨ G2 since, for every neighbor i ∈ NG1 (ie ), {i} × V2 ⊆ e is valid. Conversely, let i ∈ V1 such that {i} × V2 ⊆ e = NG1 ∨G2 ((ie , je )) is fulfilled. For each j ∈ V2 this implies (i, j) ∈ NG1 ∨G2 ((ie , je )). Hence, especially (i, je ) ∈ NG1 ∨G2 ((ie , je )) is true. From je ̸∈ NG2 (je ) the definition of the disjunction provides i ∈ NG1 (ie ). Consequently, the neighborhood hypergraph of G1 is given by

E1N = {{i | {i} × V2 ⊆ e} | e ∈ E∨N } \ {∅}. Analogously, we obtain E2N = {{j|V1 × {j} ⊆ e} | e ∈ E∨N } \ {∅}.



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