**Observation 1.**


We now turn our attention to the simultaneous strong metric dimension of graph families and look for an equivalent version of the strong resolving graph in a simultaneous version. That is, given a family of graphs G = {*<sup>G</sup>*1, *G*2, ..., *Gk*} defined over the set of vertices *V* (as described above), we say that the *simultaneous strong resolving graph* of *G*, denoted by G*SSR*, is a graph whose vertex set is *<sup>V</sup>*(G*SSR*) = *V*. In addition, two vertices *u*, *v* are adjacent in G*SSR* if the vertices *u* and *v* are mutually maximally distant in some graph *Gi* ∈ G. It is readily seen that G*SSR* can be obtained from the overlapping of the strong resolving graphs of the graphs *G*1, *G*2, ... , *Gk*. An equivalent result to that of Theorem 1 can be then derived for the simultaneous case. To this end, the next remarks make an important role.

**Remark 1.** *Let G be any connected graph. For any two vertices x*, *y* ∈ *<sup>V</sup>*(*G*)*, there are two MMD vertices <sup>x</sup>*, *y of G, such that a shortest x* − *y path contains the vertices x*, *y.*

We must recall that at least one of the vertices *x*, *y* in the result above could precisely be at least one of the vertices *<sup>x</sup>*, *y*, respectively (this could happen in case *x*, *y* are MMD or whether one of them is maximally distant from the other).

**Theorem 2.** *For any family of graphs,* G = {*<sup>G</sup>*1, *G*2, ..., *Gk*}*, Sds*(G) = *<sup>α</sup>*(G*SSR*).

**Proof.** We shall prove that any set is an SSMG for G if and only if it is a vertex cover of G*SSR*. Assume each graph of G is defined over the set of vertices *V*. Let *W* ⊂ *V* be an SSMG of G and let *uv* be an edge of G*SSR*. By the definition of G*SSR*, there is a graph *Gi* ∈ G such that *u*, *v* are MMD in *Gi*. Thus, *W* ∩ {*<sup>u</sup>*, *v*} = ∅, which means that the edge *uv* is covered by *W* in G*SSR*. Thus, *W* is a vertex cover of G*SSR*.

On the other hand, let *W* ⊂ *V* be a vertex cover of G*SSR* and let *x*, *y* be any two different vertices of *V*. If *x*, *y* are MMD in some *Gj* ∈ G, then *xy* is an edge of G*SSR*, which means that *W* ∩ {*<sup>x</sup>*, *y*} = ∅, since such edge must be covered by *W*. Assume *x* ∈ *W*. Thus, the pair of vertices *x*, *y* is strongly resolved by *x* in every *Gi* ∈ G. On the contrary, if *x*, *y* are not MMD in every *Gi* ∈ G, then the edge *xy* does not exist in G*SSR*. Moreover, by Remark 1, in every graph *Gl* ∈ G, there are two MMD vertices *xl*, *yl* such that a shortest *xl* − *yl* path of *Gl* contains *x* and *y*. Clearly, for every *Gl* ∈ G, the edge

*xlyl* belongs to *GSSR* and so, *W* ∩ {*xl*, *yl*} = ∅. Hence, for any *Gl* ∈ G, we observe that *x*, *y* are strongly resolved by *xl* or by *yl*. As a consequence, *W* is a strong resolving set for any *Gi* ∈ G and therefore *W* is a simultaneous strong resolving set of G, which completes the proof of the equality *Sds*(G) = *<sup>α</sup>*(G*SSR*).

### **3. Realization of the Simultaneous Strong Resolving Graphs with Some Consequences**

In the recent work [20], several results concerning the realization of the strong resolving graphs of graphs were presented. For instance, there was proved that there is not any graph *G* whose strong resolving graph is isomorphic to a complete bipartite graph *K*2,*r* for every *r* ≥ 2. In contrast with these facts, we shall prove that every graph *G* can represent the simultaneous strong resolving graph of some family of graphs.

**Proposition 1.** *For any graph G of order n and size m with vertex set V, there exist a family of m paths* P = {*P*1*n*, *<sup>P</sup>*2*n*,..., *Pmn* } *defined over the set of vertices V such that* P*SSR is isomorphic to G.*

**Proof.** Let *V* = {*<sup>v</sup>*1, *v*2, ... , *vn*} be the vertex set of *G*. For any edge *eij* = *vivj* of *G*, consider a path *Pij n* whose leaves are *vi* and *vj* and the remaining vertices are *V* − {*vi*, *vj*}. Since the strong resolving graph of *Pij n* is formed by a graph *K*2 on the vertices {*vi*, *vj*} and the *n* − 2 isolated vertices *V* − {*vi*, *vj*}, it is readily seen that the union (overlapping) of the *m* paths *P*2 constructed in this way, corresponding to the edges of *G*, together with the other *n* − 2 isolated vertices, is precisely the graph *G*.

Since the realization family given above is formed only by paths, one may now wonder if a given graph can be realized as the simultaneous strong resolving graph of a family of other graphs different from paths. For instance, the following two results show two other different realizations. To this end, a *multisubdivided star Sr*,*<sup>t</sup>* of order *r* + *t* + 1 is obtained from a star *S*1,*t* by subdividing some edges with some vertices until we have a graph of order *r* + *t* + 1 (clearly *r* vertices were used in this multisubdivision process). In addition, a *comet graph Cr*,*<sup>t</sup>* (where *r* ≥ 4 is an even integer and *t* ≥ 0) is a unicyclic graph of order *r* + *t* whose unique cycle has order *r*, and there is at most one vertex of degree three and at most one leaf. Note that this comet graph can be a cycle graph when *t* = 0 (in such case an even cycle indeed). In other words, a comet graph is obtained from a cycle *Cr* by attaching a path of order *t* ≥ 0 to one of its vertices.

**Proposition 2.** *For any graph G of order n with vertex set V, there exists a family of multisubdivided star graphs* F = {*<sup>G</sup>*1, *G*2,..., *Gk*} *defined over the set of vertices V such that* F*SSR is isomorphic to G.*

**Proof.** Let *V* = {*<sup>v</sup>*1, *v*2, ... , *vn*} be the vertex set of *G*. Now, for every clique *Qj* of *G* of cardinality *j*, consider a multisubdivided star graph *Sj*,*<sup>t</sup>* such that *j* + *t* + 1 = *n* whose leaves are the vertices of *Qj*, and the remaining vertices are *V* − *Qj* (taken in any order). Since the strong resolving graph of every multisubdivided star graph *Sj*,*<sup>t</sup>* is formed by a complete graph *Kj* on the vertices of *Qj* and the *n* − |*Qj*| isolated vertices in *V* − *Qj* (by using Observation 1), it is readily seen that the overlapping of the strong resolving graphs of the graphs belonging to this set of multisubdivided star graphs constructed in this way, corresponding to the cliques of *G*, together with the other corresponding isolated vertices, gives precisely the graph *G*.

In order to present our next construction, we remark (which can be easily observed) that the strong resolving graph of a comet graph is given by the disjoint union of *r* graphs *K*2 and *t* isolated vertices. We also need the following terminology. A *matching* in a graph *G* is a set of pairwise disjoint edges in the graph, and a *maximum matching* is a matching *M* such that the inclusion of any other edge of *G* to *M* leads to at least two not disjoint edges.

**Proposition 3.** *For any graph G of order n with vertex set V, there exist a family* F = {*<sup>G</sup>*1, *G*2, ... , *Gk*} *containing comet graphs and paths, defined over the set of vertices V such that* F*SSR is isomorphic to G.*

**Proof.** Let *V* = {*<sup>v</sup>*1, *v*2, ... , *vn*} be the vertex set of *G*. We consider all possible maximum matchings of *G*. If there is a maximum matching which has only one edge *e* = *uv*, then we consider a path graph *Pn*, for the family F, whose leaves are *u* and *v*. Clearly, the strong resolving graph of this path is the graph *K*2 on the vertex set {*<sup>u</sup>*, *v*} and *n* − 2 isolated vertices. Next, we take every other maximum matching *Mi* of *G* having *i* ≥ 2 edges. Now, consider a comet graph *C*2*i*,*<sup>t</sup>* with 2*i* + *t* = *n*, such that any two vertices being an edge of *Mi* are diametral in the cycle *C*2*i* of *C*2*i*,*t*, and the remaining vertices are *V* − *Mi* (taken in any order) forming the path of *C*2*i*,*<sup>t</sup>* of order *t* that are attached to one vertex of *C*2*i*. Note that the strong resolving graph of any comet graph is given by the disjoint union of *i* graphs *K*2 and *t* isolated vertices. Thus, by using all the graphs constructed as mentioned above for all the maximum matchings of *G*, it is readily seen that the overlapping of the strong resolving graphs of the graphs belonging to this set constructed in this way, corresponding to the maximum matchings of *G*, together with the other corresponding isolated vertices, gives precisely the graph *G*.

Based on the constructions above, it looks like several different families of graphs can produce the same simultaneous strong resolving graph. In this sense, it is natural to raise the following question, which roughly speaking, seems to be very challenging.

**Open question:** Given a graph *G*, is it possible to characterize all the possible families of graphs F = {*<sup>G</sup>*1, *G*2,..., *Gk*} such that F*SSR* is isomorphic to *G*?

It was proved in [15] that computing the simultaneous strong metric dimension of graph families is NP-hard, even when restricted to families of trees. An interesting consequence of Proposition 1 shows that such problem, which next appears, remains NP-hard, even when restricted to a couple of simpler families.


By using Proposition 1, we next present a reduction of the problem of computing the vertex cover number of graph to the problem of computing the simultaneous strong metric dimension of families of paths.

### **Theorem 3.** *The SSD problem is NP-complete for families of paths or multisubdivided star graphs.*

**Proof.** The problem is clearly in NP since verifying that a given set of vertices is indeed an SSMG for a graph family can be done in polynomial time. Now, let *G* be any graph with vertex set *V* of order *n* and size *m*. From Proposition 1 (resp. Proposition 2), we know there is a family of *m* paths P = {*P*1*n*, *<sup>P</sup>*2*n*, ... , *Pmn* } (resp. of multisubdivided star graphs) defined over the set of vertices *V* such that P*SSR* is isomorphic to *G*. Therefore, from Theorem 2, we have that *Sds*(P) = *<sup>α</sup>*(P*SSR*) = *<sup>α</sup>*(*G*), which completes the NP-completeness reduction based on the fact that the decision problem concerning the vertex cover number of graphs is an NP-complete problem (see [21]).

Another interesting consequence of Theorem 2 concerns the approximation of computing the simultaneous strong metric dimension of graphs families. We first note that finding the simultaneous strong resolving graph of a graph family can be polynomially done. This fact, together with the fact that computing the vertex cover number of graphs admits a polynomial-time 2-approximation, allows for claiming that computing the simultaneous strong metric dimension of graphs families also admits a polynomial-time 2-approximation.

### **4. Applications of the Simultaneous Strong Resolving Graph**

Since computing the simultaneous strong metric dimension of graph families is NP-hard even when restricted to very specific families, it is then desirable to describe as many families as possible for which its simultaneous strong metric dimension can be computed. In this sense, from now on in this section we are devoted to make this so, and a fundamental tool for it shall precisely be the simultaneous strong resolving graph and, connected with it, Theorem 2.

**Proposition 4.** *If* F *is a family of bipartite graphs, each of them is defined over the common bipartition sets U*, *V, then* Sd*s*(F) ≤ |*U*| + |*V*| − 2*.*

**Proof.** If any two vertices are MMD in some graph *Gi* ∈ F, then they belong to the same bipartition set of *Gi*. Thus, it must happen that F*SSR* is a subgraph of a graph with two connected components isomorphic to *<sup>K</sup>*|*U*<sup>|</sup> and *<sup>K</sup>*|*V*|. By using Theorem 2, we obtain that Sd*s*(F) = *<sup>α</sup>*(F*SSR*) ≤ *<sup>α</sup>*(*<sup>K</sup>*|*U*<sup>|</sup> ∪ *<sup>K</sup>*|*V*|) = |*U*| + |*V*| − 2.

Next, we particularize the result above and show that such bound is achieved in several situations.

**Proposition 5.** *If* F *is a family of bipartite graphs, each of them is defined over the common bipartition sets U*, *V, and such that it contains the complete bipartite graph <sup>K</sup>*|*U*|,|*V*|*, then* Sd*s*(F) = |*U*| + |*V*| − 2*.*

**Proof.** The result directly follows from the fact that, if *<sup>K</sup>*|*U*|,|*V*<sup>|</sup> ∈ F, then (F)*SR* is isomorphic to *<sup>K</sup>*|*U*<sup>|</sup> ∪ *<sup>K</sup>*|*V*|. Thus, from Theorem 2, we ge<sup>t</sup> the desired result, since *<sup>α</sup>*(*<sup>K</sup>*|*U*<sup>|</sup> ∪ *<sup>K</sup>*|*V*|) = |*U*| + |*V*| − 2.

Let *Cr* = *v*0*v*1 ... *vr*−1, with *r* ≥ 4 and even be a cycle. Then, let F*C* be a family of cycles defined on a common vertex set with *Cr* ∈ F*C*, and every other cycle *C* ∈ F*C* is obtained from *Cr*, by making a permutation of two vertices *vi*, *vj* of *Cr* such that either *i*, *j* ∈ {0, ... ,*<sup>r</sup>*/2 − 1} or *i*, *j* ∈ {*r*/2, ... ,*r* − <sup>1</sup>}.

**Proposition 6.** *If* F*C is a family of cycles obtained from a cycle Cr as described above, then* Sd*s*(F*C*) = *r*2*.*

**Proof.** We first note (by Observation 1) that the strong resolving graph of any cycle *C*(*j*) *r* ∈ F*C* is isomorphic to ∪*<sup>r</sup>*/2 *<sup>i</sup>*=1*K*2. Now, since any cycle of F*C* is obtained from *Cr*, by making a permutation of two vertices *vi*, *vj* of *Cr* such that either *i*, *j* ∈ {0, ... ,*<sup>r</sup>*/2 − 1} or *i*, *j* ∈ {*r*/2, ... ,*r* − <sup>1</sup>}, we deduce that there are no edges in (F*C*)*SSR* between any two vertices of the set {*<sup>v</sup>*0, *v*1, ... , *vr*/2−<sup>1</sup>}, in addition to no edges between any two vertices of the set {*vr*/2, *<sup>v</sup>*(*r*/2+1, ... , *vr*−<sup>1</sup>}. Moreover, for any vertex of the set *vj* ∈ {*<sup>v</sup>*0, *v*1, ... , *vr*/2−<sup>1</sup>}, there is an edge joining *vj* with a vertex *vk* ∈ {*vr*/2, *vr*/2+1, ... , *vr*−<sup>1</sup>} and vice versa. As a consequence of such facts, we obtain that (F*C*)*SSR* is a bipartite graph, which is a subgraph of the complete bipartite graph *Kr*/2,*r*/2. Since *<sup>α</sup>*(*Kr*/2,*r*/2) = *r*/2, by using Theorem 2, we have Sd*s*(F*C*) = *<sup>α</sup>*((F*C*)*SSR*) ≤ *<sup>α</sup>*(*Kr*/2,*r*/2) = *r*2 . On the other hand, since *Cr* ∈ F*C*, the edges *<sup>v</sup>*0*vr*/2, *<sup>v</sup>*1*vr*/2+1, ... , *vr*/2−1*vr*−<sup>1</sup> belong to (F*C*)*SSR*. Consequently, in order to cover such edges, it must happen *<sup>α</sup>*((F*C*)*SSR*) ≥ *r*/2. Therefore, by using again Theorem 2, we complete the proof.

Let *T* be a tree having all the vertices with a degree larger than two unless they are leaves. Hence, let F*T* be the family of trees defined on a common vertex set with *T* ∈ F*T*, and every other tree *T* ∈ F*T* is obtained from *T*, by making a permutation of two vertices *u*, *v* of *T* such that *u* is a leaf of *T* and *v* is not a leaf.

**Proposition 7.** *If* F*T is a family of trees obtained from a tree T with t leaves as described above, then t* − 1 ≤ Sd*s*(F*T*) ≤ *t. Moreover,* Sd*s*(F*T*) = *t if and only if every leaf of T has been used to make a permutation with other non leaf vertex of T in order to obtain another tree of* F*T.*

**Proof.** By Observation 1, the strong resolving graph of the tree *T* ∈ F*T* is isomorphic to the complete graph *Kt* together with *n* − *t* isolated vertices. Thus, *Kt* is a subgraph of the graph (F*T*)*SSR* and so, by using Theorem 2, we obtain Sd*s*(F*T*) = *α*(( F*T*)*SSR*) ≥ *α*(*Kt*) = *t* − 1, which is the lower bound. Now, observe that the strong resolving graph of any tree, other than *T*, is obtained from the strong resolving graph of *T* by removing all edges incident to one vertex, say *y*, corresponding to a leaf of *T*, choosing an isolated vertex *x* corresponding to a non leaf of *T*, and adding all the possible edges between *x* and the vertices corresponding to leaves of *T* other than *y*. Moreover, since there are no edges between any two of such chosen isolated vertices mentioned above, it is clear that the set of *t* vertices corresponding to the leaves of *T* represents a vertex cover set of (F*T*)*SSR*. By using again Theorem 2, we obtain Sd*s*(F*T*) = *α*(( F*T*)*SSR*) ≤ *t*, which is the upper bound.

On the other hand, in order for this set of *t* vertices to correspond to the leaves of *T*, which will represent a vertex cover set of minimum cardinality in (F*T*)*SSR*, it is required that all such vertices will have a neighbor not in this set of leaves. This means that every leaf of *T* has been used to make a permutation with another non leaf vertex of *T* in order to obtain a tree of F*T* other than *T*. The opposite direction is straightforward to observe. Therefore, the proof is complete.

Again, let *T* be a tree having all the vertices with a degree larger than two unless they are leaves. Let H*T* be the family of at least two unicyclic graphs *H* defined on a common vertex set such that every graph *H* ∈ H*T* is obtained from *T*, by adding an edge between any two vertices *u*, *v* of *T*. Before studying the simultaneous strong metric dimension of H*<sup>T</sup>*, we introduce some terminology and basic properties of the strong resolving graph of unicyclic graphs. Given a unicyclic graph *G* = ( *V*, *E*) with the unique cycle *Cr*, we denote by *<sup>c</sup>*2(*G*) the set of vertices of the cycle *Cr* having degree two. By *<sup>T</sup>*(*G*), we represent the set of vertices of degree one in *G*.

**Remark 2.** *Let G be a unicyclic graph. For every vertex x* ∈ *<sup>c</sup>*2(*G*)*, there exists at least one vertex y* ∈ *<sup>c</sup>*2(*G*) ∪ *T*(*G*) *such that x*, *y are mutually maximally distant in G.*

**Remark 3.** *Let G be a unicyclic graph. Then, two vertices x*, *y are mutually maximally distant in G if and only if x*, *y* ∈ *<sup>c</sup>*2(*G*) ∪ *<sup>T</sup>*(*G*)*.*

We note that any graph *H* ∈ H*T* satisfies that *<sup>c</sup>*2(*H*) is empty (whether the unicyclic graph *H* has been obtained from *T* by adding an edge between two non leaf vertices of *T*), or has cardinality 1 (whether *H* has been obtained from *T* by adding an edge between a leaf a non leaf vertex of *T*), or cardinality 2 (whether *H* has been obtained by an added edge between two leaves of *T*).

**Proposition 8.** *If* H*T is a family of unicyclic graphs obtained from a tree T with t leaves as described above, then* Sd*s*(H*T*) = *t* − 1*.*

**Proof.** By Remarks 2 and 3, and the fact that 0 ≤ |*<sup>c</sup>*2(*H*)| ≤ 2, we deduce that, for any graph *H* ∈ H*<sup>T</sup>*, it follows that *HSR* contains |*V*(*G*)| − *t* isolated vertices together with either


We note that these, one or two, extra vertices are precisely the leaves of *T*, for which an incident edge has been added to the tree *T* in order to obtain *H*. According to these facts, the simultaneous strong resolving graph (H*T*)*SSR* has a connected component which is a subgraph of a complete graph *Kt* whose vertex set is precisely the set of leaves of *T*. In this sense, by using Theorem 2, we obtain Sd*s*(H*T*) = *α*(( H*T*)*SSR*) ≤ *t* − 1.

On the other hand, consider *Q* is the subgraph induced by the set of vertices corresponding to the leaves of *T*. If item (i) above occurs for every *H* ∈ H*<sup>T</sup>*, then clearly (H*T*)*SSR* ∼= *Kt* and so

Sd*s*(H*T*) = *α*(( H*T*)*SSR*) = *t* − 1. Suppose now that Sd*s*(H*T*) < *t* − 1. This means that there are two vertices *x*, *y* corresponding to two leaves of *T* which are not adjacent in (H*T*)*SSR*. In consequence, there is a unicyclic graph *Gx*,*<sup>y</sup>* ∈ H*T* which was obtained from *T* by adding the edge *xy*. However, since H*T* has at least two graphs, there is at least a graph *G* ∈ H*<sup>T</sup>*, other than *Gx*,*<sup>y</sup>* in which the vertices *x*, *y* are not adjacent. Thus, *x*, *y* are MMD in *G* and so the edge *xy* exists in (H*T*)*SSR*, a contradiction with our supposition. Therefore, Sd*s*(H*T*) ≥ *t* − 1 and we have the desired equality.

A similar process as described above can be developed in order to ge<sup>t</sup> families of graphs for which the simultaneous strong metric dimension of graphs can be computed. However, one may need to use several assumptions while constructing such families. This is based on the fact that computing the simultaneous strong metric dimension is NP-complete for several "very simple" families of graphs (like families of paths, for instance). In connection with this, it would be desirable to find some "properties" satisfied by a graph family in order to decide if it is "easy" to compute its simultaneous strong metric dimension or not.

### **5. The Particular Case of Cartesian Product Graphs Families**

Given two graph families F1 = {*<sup>G</sup>*1, ... , *Gr*} and F2 = {*<sup>H</sup>*1, ... , *Ht*} defined over the common sets of vertices *V*1 and *V*2, respectively, the *Cartesian product graph family* F1 F2 is given by the family {*Gi Hj* : *Gi* ∈ F1, *Hj* ∈ F2}. In order to study the simultaneous strong metric dimension of Cartesian product graphs families, we also need the following definition. The *direct product graph family* F1 × F2 is given by the family {*Gi* × *Hj* : *Gi* ∈ F1, *Hj* ∈ F2}.

We shall next need the following definitions. Given two graphs *G* and *H*, the *Cartesian product graph* of *G* and *H* is a graph, denoted by *G H*, having vertex set *V*(*G H*) = *V*(*G*) × *<sup>V</sup>*(*H*). In addition, there is an edge between two vertices (*a*, *b*),(*<sup>c</sup>*, *d*) ∈ *V*(*G H*) if it is satisfied that either (*a* = *c* and *bd* ∈ *E*(*H*)) or (*b* = *d* and *ac* ∈ *E*(*G*)). In a similar way, the direct product of graphs can be defined. That is, the *direct product graph* of *G* and *H* is a graph, denoted by *G* × *H*, having vertex set *V*(*G* × *H*) = *V*(*G*) × *<sup>V</sup>*(*H*). Now, two vertices (*a*, *b*),(*<sup>c</sup>*, *d*) are adjacent in the direct product *G* × *H* whenever *ac* ∈ *E*(*G*) and *bd* ∈ *<sup>E</sup>*(*H*).

The first result concerning Cartesian product graphs families is a relationship between the simultaneous strong resolving graph of Cartesian product graphs families and that of its factors. The following equivalent result for the strong metric dimension of graph was given in [22].

**Theorem 4** ([22])**.** *Let G and H be two connected graphs. Then,* (*G <sup>H</sup>*)*SR* ∼= *GSR* × *HSR.*

By using the result above and the fact that the simultaneous strong resolving graph of a graph family equals the union (overlapping) of the strong resolving graph of each graph of the family, we deduce the next result.

**Theorem 5.** *Let* F1 = {*<sup>G</sup>*1, ... , *Gr*} *and* F2 = {*<sup>H</sup>*1, ... , *Ht*} *be two graph families defined over the common sets of vertices V*1 *and V*2*, respectively. Then,* (F1 F2)*SSR* ∼= (F1)*SSR* × (F2)*SSR.*

**Proof.** Since (F1 F2)*SSR* is given by *Gi*∈F1,*Hj*∈F2 (*Gi Hj*)*SR* and (*Gi Hj*)*SR* ∼= (*Gi*)*SR* × (*Hj*)*SR* (by Theorem 4), we ge<sup>t</sup> that

$$(\mathcal{F}\_1 \square \Box \mathcal{F}\_2)\_{SRR} \cong \bigcup\_{\mathcal{G}\_l \in \mathcal{F}\_1, H\_j \in \mathcal{F}\_2} (\mathcal{G}\_l \square H\_j)\_{SR} \cong \bigcup\_{\mathcal{G}\_l \in \mathcal{F}\_1, H\_j \in \mathcal{F}\_2} (\mathcal{G}\_l)\_{SR} \times (H\_j)\_{SR} \cong (\mathcal{F}\_1 \times \mathcal{F}\_2)\_{SRR}$$

which gives our claim.

We next give several results concerning the simultaneous strong metric dimension of Cartesian product graph families. In this sense, the result above plays an important role. Now, our next result, which is obtained by using Theorem 2 and Theorem 5, shall be used as an important tool to develop our exposition.

**Corollary 1.** *Let* F1 = {*<sup>G</sup>*1, ... , *Gr*} *and* F2 = {*<sup>H</sup>*1, ... , *Ht*} *be two graph families defined over the common sets of vertices V*1 *and V*2*, respectively. Then,* Sd*s*(F1 F2) = *α*(( F1)*SSR* × (F2)*SSR*)*.*

Due to the similarity of the results above (in this section) with those obtained in [22] concerning the strong metric dimension of graphs, we note that some analogous reasonings as that ones in [22] shall lead to several results concerning the simultaneous strong metric dimension of graph families. In this sense, we now close our exposition with some problems that are of interest in our point of view.

### **6. Conclusions and Open Problems**

A new approach to study the simultaneous strong metric dimension of graphs families has been presented in this work. That is, we have introduced the notion of simultaneous strong resolving graph of graphs families, and proved the computing the simultaneous strong metric dimension of a family of graphs is equivalent to compute the vertex cover number of this newly introduced simultaneous strong resolving graph. Based on this equivalence, several computational and combinatorial results have been deduced. For instance, we have proved that computing the simultaneous strong metric dimension of families of paths and families of multisubdivided star graphs is NP-hard. As a consequence of the study, a number of open questions have been raised. We next point out several of the most interesting ones:


**Funding:** This research received no external funding.

**Conflicts of Interest:** The author declares no conflict of interest.
