*4.1. Unicyclic Graphs*

Our first results shows the relationship between the strong metric dimension of a unicyclic graph and that of its branch restricted unicyclic graph.

**Lemma 2.** *Let G be a unicyclic graph and* T (*G*) *be its branch restricted unicyclic graph. Then*

$$
\dim\_\mathfrak{s}(G) = \dim\_\mathfrak{s}(\mathcal{T}(G)).
$$

**Proof.** By Lemma 1 and Theorem 1, we derive that *dims*(*G*) = *β*(*GSR*) = *β*((T (*G*))*SR*) = *dims*(T (*G*)) and the proof is complete.

**Theorem 4.** *Let G be a unicyclic graph with unique cycle Cr. Then*

$$\max\left\{ \left\lceil \frac{r}{2} \right\rceil, \left| T(G) \right| - 1 \right\} \le \dim\_{\mathbb{S}}(G) \le \left| T(G) \right| + \left\lfloor \frac{|c\_2(G)|}{2} \right\rfloor.$$

**Proof.** From Remark 1 we have that every strong resolving basis must contain at least |*T*(*G*)| − 1 vertices of degree one. So, *dims*(*G*) ≥ |*T*(*G*)| − 1. On the other hand, for every vertex *i* ∈ {1, ... ,*<sup>r</sup>*} there exists at least a vertex *wi* ∈ *<sup>t</sup>*(*vi*) ∪ {*vi*} such that *wi* ∈ *∂*(*G*) (notice that it could happen *wi* = *vi*). Thus we have that *dims*(*G*) = *β*(*GSR*) ≥ |*∂*(*G*)| 2≥ ! *r*2".

 On the other side, since *T*(*G*) forms a clique in *GSR* and for every *u* ∈ *<sup>c</sup>*2(*G*) there exists at least one vertex *v* ∈ *∂*(*G*) such that they are MMD, according to the description of *GSR* presented in the previous section, we have *dims*(*G*) = *β*(*GSR*) ≤ |*T*(*G*)| + ) |*<sup>c</sup>*2(*G*)| 2 \*. Therefore the proof is complete.

As we can see in the following, the bounds above are tight. In particular, we characterize all the unicyclic graphs having a unique cycle of even order that are attaining the upper bound.

**Theorem 5.** *Let G be a unicyclic graph with a unique cycle Cr of even order. Then dims*(*G*) = |*T*(*G*)| + ) |*<sup>c</sup>*2(*G*)| 2 \* *if and only if* |*<sup>c</sup>*2(*G*)| = *r* − 1*.*

**Proof.** (⇐) We assume |*<sup>c</sup>*2(*G*)| = *r* − 1. Let *v* be the only vertex of *Cr* with degree greater than two, and let *u* be the diametral vertex with *v* in *Cr*. So, every two vertices in *<sup>t</sup>*(*v*) ∪ {*u*} are MMD. Also, every two diametral vertices in *<sup>c</sup>*2(*G*) − {*u*} are MMD. Thus, *GSR* is formed by *<sup>r</sup>*−22 = |*<sup>c</sup>*2(*G*)|−<sup>1</sup> 2 connected components isomorphic to *K*2 and one component isomorphic to *<sup>K</sup>*|*t*(*v*)|+1. Since *T*(*G*) = *<sup>t</sup>*(*v*), we have that

$$\dim\_{\mathbb{S}}(G) = \beta(G\_{\mathcal{SR}}) = \beta(K\_{|T(G)|+1}) + \beta\left(\bigcup\_{i=1}^{\frac{|c\_2(G)|-1}{2}} K\_2\right) = |T(G)| + \frac{|c\_2(G)|-1}{2} = |T(G)| + \left\lfloor \frac{|c\_2(G)|}{2} \right\rfloor.$$

(⇒) We assume now that *dims*(*G*) = |*T*(*G*)| + ) |*<sup>c</sup>*2(*G*)| 2 \*is satisfied. If |*<sup>c</sup>*2(*G*)| < *r* − 1, then there are at least two vertices *x*, *y* such that *<sup>t</sup>*(*x*) ≥ 1 and *<sup>t</sup>*(*y*) ≥ 1. We consider two cases.

Case 1: *x*, *y* are diametral in *Cr*. Hence, *<sup>t</sup>*(*x*) ∪ *<sup>t</sup>*(*y*) forms a clique in *GSR* of cardinality |*t*(*x*)| + |*t*(*y*)|. Also, the vertices in *<sup>t</sup>*(*x*) ∪ *<sup>t</sup>*(*y*) have no neighbor from *<sup>c</sup>*2(*G*) in *GSR*. Note that, there could be some other vertices in *T*(*G*) having neighbors from *<sup>c</sup>*2(*G*) in *GSR*, and if there is one of such vertices, say *z*, then |*t*(*z*)| ≥ 1 and *<sup>t</sup>*(*x*) ∪ *<sup>t</sup>*(*y*) ∪ *<sup>t</sup>*(*z*) is also a clique in *GSR*. However, this will not influence on the fact that, in order to cover the edges of *GSR*, one can leave one vertex *w* of *<sup>t</sup>*(*x*) ∪ *<sup>t</sup>*(*y*) outside of the vertex cover set, by simply taking *T*(*G*) \ {*w*} as a part of such vertex cover set. Thus, we have that *β*(*GSR*) ≤ |*T*(*G*)| − 1 + ) |*<sup>c</sup>*2(*G*)| 2\*, a contradiction.

Case 2: *x*, *y* are not diametral in *Cr*. Let *<sup>x</sup>*, *y* ∈ *<sup>c</sup>*2(*G*) being diametral vertices with *x*, *y*, respectively. Hence, *<sup>t</sup>*(*x*) ∪ *<sup>t</sup>*(*y*), *<sup>t</sup>*(*x*) ∪ {*x*} and *<sup>t</sup>*(*y*) ∪ {*y*} form cliques in *GSR*. Also, *<sup>x</sup>*, *y* have no neighbor in *GSR* other than that ones in *<sup>t</sup>*(*x*), *<sup>t</sup>*(*y*), respectively. Thus, in order to cover the edges of *GSR*, we can leave outside of the vertex cover set both vertices *<sup>x</sup>*, *y*, by simply taking *T*(*G*) in such vertex cover set. On the other hand, to cover the remaining vertices in *<sup>c</sup>*2(*G*) \ {*x*, *y*} we will need at most ) |*<sup>c</sup>*2(*G*)|−<sup>2</sup> 2 \*. We then deduce that *β*(*GSR*) ≤ |*T*(*G*)| + |*<sup>c</sup>*2(*G*)|−<sup>2</sup> 2 = |*T*(*G*)| − 1 + |*<sup>c</sup>*2(*G*)|/2 − 1, a contradiction again.

Since we have contradiction on both cases above, it must happen that |*<sup>c</sup>*2(*G*)| = *r* − 1, and the proof is completed.

Note that the upper bound of Theorem 4 is also tight when the unique cycle of *G* is odd, but the characterization of the limit case seems to be a hard working task. For instance, if *G* has a unique cycle of odd order and |*<sup>c</sup>*2(*G*)| = *r* − 1, then a "relatively" similar argumen<sup>t</sup> to the first part of the proof of Theorem 5 leads to conclude that *dims*(*G*) = |*T*(*G*)| + ) |*<sup>c</sup>*2(*G*)| 2 \*. Other cases, when |*<sup>c</sup>*2(*G*)| < *r* − 1 can be hand computed, and we leave this to the reader.

**Proposition 3.** *Let G be a unicyclic graph with a unique cycle Cr of even order. Then dims*(*G*) = *r*2 *if and only if the following items hold.*

(i) |*t*(*x*)| ≤ 1 *for every x of Cr.*

(ii) *There is at most one pair of diametral vertices in Cr each one having one terminal vertex.*

**Proof.** (⇒) Assume *dims*(*G*) = *r*2 . If |*t*(*x*)| > 1 for some *x* of *Cr*, then let *x* be the vertex of *Cr* being diametral with *x* in *Cr*. Hence, *<sup>t</sup>*(*x*) ∪ *<sup>t</sup>*(*x*) (or *<sup>t</sup>*(*x*) ∪ {*x*} if *x* ∈ *<sup>c</sup>*2(*G*)) is a clique in *GSR*, and so, in order to cover the edges of *GSR*, we need at most *r*2 in connection with pairs of diametral vertices in *Cr* together with at least one extra vertex from *<sup>t</sup>*(*x*), since |*t*(*x*)| > 1 (there are at least two MMD vertices in *<sup>t</sup>*(*x*)). Thus, (i) follows.

Now, let *a* be the number of pairs of diametral vertices in *Cr* each one having one terminal vertex. Suppose that *a* ≥ 2. Also, let *b* be the number of pairs of diametral vertices in *Cr*, in which one of them has one terminal vertex and the other one belongs to *<sup>c</sup>*2(*G*), and let *c* be the number of pairs of diametral vertices in *Cr*, each one belonging to *<sup>c</sup>*2(*G*). Note that the *a* + *b* + *c* = *r*2 and that |*T*(*G*)| = 2*a* + *b*. Also, the 2*a* vertices and the *b* vertices of *<sup>T</sup>*(*G*), corresponding to that pairs mentioned above, form a clique in *GSR* such that the 2*a* vertices has no neighbors other than that ones in such clique, and such that each of the *b* vertices has exactly one other neighbor from *<sup>c</sup>*2(*G*) in *GSR*. Moreover, the *c* pairs of vertices also mentioned above, form *c* components of *GSR* isomorphic to *K*2. In consequence, we observe that *β*(*GSR*) = 2*a* − 1 + *b* + *c* = *r*2 − 1 + *a* ≥ *r*2 + 1. This is a contradiction, and the proof of (ii) is complete.

(⇐) Assume on the other hand that *G* satisfies (i) and (ii). We shall use the same notation of *a*, *b* and *c* from the implication above. By (ii), 0 ≤ *a* ≤ 1. If *a* = 1, then *dims*(*G*) = *β*(*GSR*) = 2*a* − 1 + *b* + *c* = *r*2 − 1 + *a* = *r*2 (note that the equality *β*(*GSR*) = 2*a* − 1 + *b* + *c* follows by using (i)). Also, if *a* = 0, then *dims*(*G*) = *β*(*GSR*) = *b* + *c* = *r*2(we again use (i) as explained before).

To conclude this section, we next show that the differences between the lower (partially) and upper bounds of Theorem 4, and the real value of the strong metric dimension of some unicyclic graphs can be as large as possible.

We consider the unicyclic graph *G<sup>k</sup>* with a cycle *C*2*n* = *v*1*v*2 ··· *v*2*nv*1 and 1 ≤ *k* ≤ *n* such that the vertices *vk*+1, *vk*+2, ... , *v*2*n* form the set *<sup>c</sup>*2(*G*), and each vertex *vi* for *i* ∈ {1, ... , *k*} has one terminal vertex denoted by *xi*. Since 2*n* is an even number, and according to the description of the strong resolving graph of a unicyclic graph, it clearly follows that (*G<sup>k</sup>*)*SR* consists of a graph isomorphic to *<sup>K</sup>*|*T*(*G*)<sup>|</sup> *K*1 and ) |*<sup>c</sup>*2(*G*)|−*k* 2 \* graphs isomorphic to *K*2. Thus, *dims*(*G<sup>k</sup>*) = |*T*(*G*)| + ) |*<sup>c</sup>*2(*G*)|−*k* 2 \*. Since 1 ≤ *k* ≤ |*<sup>c</sup>*2(*G*)|, we can easily observe that |*T*(*G*)| + ) |*<sup>c</sup>*2(*G*)| 2 \* − *dims*(*G<sup>k</sup>*) and *dims*(*G<sup>k</sup>*) − (|*T*(*G*)| − 1) can be arbitrarily large.

### *4.2. Bouquet of Cycles*

For the results of this subsection, we use the terminology and notations given in Section 3.2.

**Theorem 6.** *For any bouquet of cycles B* ∈ B*<sup>a</sup>*,*b*,*c,*

$$\dim\_s(B) = a + \sum\_{i=1}^{a} \frac{r\_i - 2}{2} + 2b + \sum\_{j=1}^{b} \frac{s\_j - 3}{2} + 2c - 1.$$

**Proof.** According to the description of *BSR* presented before, it follows that *BSR* consist of a graph isomorphic to *Ka*+2*b*+2*c*(*<sup>s</sup>*<sup>1</sup> − 3,*<sup>s</sup>*2 − 3, ... ,*sb* − 3, 0, 0, ... , 0) and ∑*ai*=<sup>1</sup> *ri*−2 2 graphs isomorphic to *K*2. First we consider the subgraph *H* induced by *NBSR* [*Va* ∪ *V*2*b* ∪ *<sup>V</sup>*2*c*]. Notice that *β*(*H*) = *a* + 2*b* + 2*c* − 1. In order to compute *β*(*Ka*+2*b*+2*c*(*<sup>s</sup>*<sup>1</sup> − 3,*<sup>s</sup>*2 − 3, ... ,*sb* − 3, 0, 0, ... , 0)) we need to cover the remaining edges in *BSR* corresponding to edges of the odd cycles in *B*. Since for each odd cycle *Csi* for *i* ∈ {1, . . . , *b*}, two edges of it are already considered in *H*, it remains to cover *si* − 4 edges which are inducing a path of order *si* − 3. Thus, to cover each cycle *Csi* we need *si*−3 2 vertices.

On the other hand, to cover the ∑*ai*=<sup>1</sup> *ri*−2 2 graphs isomorphic to *K*2, ∑*ai*=<sup>1</sup> *ri*−2 2 extra vertices are needed. The sum of these three quantities above gives the vertex cover number of *BSR*, and also the strong metric dimension of *B*, by using Corollary 1, which completes the proof.

### *4.3. Chains of Even Cycles*

In order to give a formula for the strong metric dimension of chains of even cycles, we need to first compute the value of the vertex cover number of a bipartite graph *Jr* as described in Section 3.3.

### **Lemma 3.** *For any bipartite graph Jr, β*(*Jr*) = *r* + 1*.*

**Proof.** We first note that if *r* is an even integer, then the set of edges *Er* = {*<sup>a</sup>*0*br*, *<sup>a</sup>*1*br*−1, ... , *ar*/2*br*/2} ∪ {*b*0*ar*, *b*1*ar*−1,..., *br*/2−1*ar*/2+<sup>1</sup>} is a maximum matching in *Jr* of cardinality *r*/2 + 1 + *r*/2 = *r* + 1.

On the other hand, if *r* is odd, then the set of edges *Er* = {*<sup>a</sup>*0*br*, *<sup>a</sup>*1*br*−1, ... , *<sup>a</sup>*(*<sup>r</sup>*−<sup>1</sup>)/2*b*(*r*+<sup>1</sup>)/2} ∪ {*b*0*ar*, *b*1*ar*−1, ... , *<sup>b</sup>*(*<sup>r</sup>*−<sup>1</sup>)/2*<sup>a</sup>*(*r*+<sup>1</sup>)/2} is a maximum matching in *Jr* of cardinality (*r* − 1)/2 + 1 + (*r* − 1)/2 + 1 = *r* + 1.

Thus, since *Jr* is bipartite, by using the famous K˝onig's Theorem, we obtain the required result.

**Theorem 7.** *For any chain of even cycles F* ∈ F*k of order n with c cut vertices,*

$$\dim\_{\mathfrak{s}}(F) = \frac{n-c}{2}.$$

**Proof.** According to the description of *FSR* presented before, the vertices *ai* and *bi*, with *i* ∈ {0, ... , *k*}, forms a component of the graph *FSR* isomorphic to a bipartite graph *Jk* of order 2*k* + 2. For completing the graph *FSR*, we need to add *<sup>n</sup>*−*c*−2*k*−2 2 graphs isomorphic to *K*2. Hence, by using Theorem 1, Lemma 3 and Observation 1, we have *dims*(*F*) = *β*(*Jk*) + *<sup>n</sup>*−*c*−2*k*−2 2 *β*(*<sup>K</sup>*2) = *n*−*c* 2 .
