In Tn,κ, if we let νκj=ν1j+1=νj (a joint/cut vertex between jth and (j+1)th polygons/cycles) for 1≤j≤n−1, then cut vertices in Tn,κ are adjacent, and this type of chain polygonal cactus is denoted by Hn,κ. In fact, cut-vertices in Hn,κ, lying in the same non-pendent polygon, are adjacent. See Fig. 2.

Lemma 2.2. For κ,n>3, it is not possible that two consecutive blocks do not contribute to form a metric generator for Hn,κ.

Proof. Contrarily, suppose that two consecutive blocks Bi and Bi+1 do not contribute to form a metric generator S for Hn,κ. Then, there are BDS x in Bi and y in Bi+1 and both x and y are neighbors of the joint νi, such that no vertex s∈S identified x and y. So, S is not a metric generator for Hn,κ, a contradiction.

Theorem 2.1. For n≥3, dim(Hn,3)=2.

Proof. By Theorem 1.1, only path graph has metric dimension equal to 1, so dim(Hn,3)≥2. Let W={ν11,ν3n} be a set of vertices of Hn,3, then metric vectors of all the vertices in Hn,3 with respect to W are:

mW(ν11)=(0,n−1), mW(ν3n)=(n−1,0),

mW(νj)=(j,n−j) for 1≤j≤n−1,

mW(ν2j+1)=(1+j,n−j) for 0≤j≤n−1.

We can see that all the metric vectors are distinct. So, W is a metric generator for Hn,3, and therefore dim(Hn,3)=2.

Theorem 2.2. For odd κ≥5, dim(H3,κ)=2.

Proof. By Theorem 1.1, only path graph has the metric dimension equals to 1. So, dim(H3,κ)≥2. Further, consider W={νκ−121,νκ+123}, and accordingly metric vectors of the vertices are: mW(νi1)={(κ−12−i,κ+12+i),1≤i≤κ−32,(i+1−κ2,3κ+12−i),κ−12<i≤κ−1,(κ−12,κ+12),i=κ. mW(νi2)={(κ−32+i,κ−12+i),1≤i≤κ−12,(i+κ−32,3κ−12−i),i=κ+12,(3κ+12−i,3κ−12−i),κ+12<i≤κ. mW(νi3)={(κ−12+i,κ+12−i),1≤i≤κ+12,(3κ+32−i,i−κ+12),κ+12<i≤κ.

Obviously for every two vertices x, y of H3,κ with x≠y, mW(x)≠mW(y). Thus, W is a metric generator for H3,κ and dim(H3,κ)≥2.

Lemma 2.3. For even κ≥4, if S is a minimum metric generator for H4,κ, then |S|≥4.

Proof. We contrarily assume that |S|=3. By Lemma 2.1, S must contain one vertex from each end block. Let a vertex u be taken from the block B1 and a vertex w be taken from the block B4. Then S does not contain any vertex from one of the remaining two blocks. Without loss of generality, we suppose that S does not contain any vertex from block B3, then we have two possibilities:

Whenever d(u,ν1)≠κ2≠d(w,ν3), then there are two vertices x and y, both are the neighbors of the joint ν3, such that they are BDS in H4,κ and mS(x)=mS(y), a contradiction.

Hence, our supposition is wrong and |S|≥4.

Theorem 2.3. For even κ≥4, dim(H4,κ)=4.

Proof. Lemma 2.3 provides the lower bound for dim(H4,κ).

Now, we prove that dim(H4,κ)≤4. For this, let S={νκ2+11,νκ2+12,νκ2+13,νκ4}, and we have to show that for any pair (x, y) of vertices in H4,κ, there is always a vertex in S which identifies the pair (x, y). For this, we consider three cases:

Case-I Whenever x, y belong to the same block Bi of H4,κ, then there are two possibilities:

If x and y are BDS, then a vertex in S, chosen from the block Bi, will identifies the pair (x, y).

Case-II If x, y do not belong to the same block Bi, then there are two possibilities:

When x belongs to the block Bi and y belongs to the block Bi+1 (or Bi−1). If x and y are BDS, then there is a vertex of S lying either in the block Bi or Bi−1 or Bi+1, which identifies the pair (x, y). Otherwise, there is always a vertex s in S lying in the block containing x or y such that d(x,s)≠d(y,s).

Case-III Whenever x or y or both x and y is (are) a joint (s), then there are two possibilities: 1.

If x and y are adjacent, then there is a vertex u∈S, such that d(x,u)=1+d(y,u) where u and y lie on a same cycle, or d(y,u)=d(x,u)+1 where u and x lie on a same cycle. Accordingly, u identifies the pair (x, y).

d(x,s2)=d(y,x)+d(y,s2) andd(y,s1)=d(x,y)+d(x,s1).

According to all these cases, it can be concluded that S is a metric generator of H4,κ.

Lemma 2.4. For even κ≥4 and any n≥3 with n≠4, if S is a minimum metric generator for Hn,κ, then |S|≥[n2]+2.

Proof. Let [n2]+2=m and S has two vertices x and y (say) from end blocks B1 and Bn respectively, by Lemma 2.1. Next, we show that S must contain at least m−2 more vertices from Hn,κ. Contrarily, assume that S contains m−3 more vertices. There are two claims to discuss:

Claim-1 Whenever d(x,ν1)=κ2 (ord(y,νn−1)=κ2), then S must contain one more vertex from B1 (or Bn).

Neighbors u and ν of x (or y) satisfy mS(u)=mS(ν), so S is not a metric generator. In this way, we get two consecutive blocks among them no vertex will contribute in S, because S contains m−3 more vertices from (n−2) blocks. It yields a contradiction of Lemma 2.2.

Claim-II Whenever d(x,ν1)≠κ2≠d(x,νn−1), then S must have at least one vertex from both the blocks B2 and Bn−1.

We suppose that S does not contain a vertex from the block B2 (say). Then there are two vertices, u1 in the block B1 and w1 in the block B2, such that mS(u1)=mS(w1), where u1∼ν1∼w1. So, S is not a metric generator. Thus our claim is true. Now, S must have at least one vertex from both the block B2 and Bn−1, and S must contain m−3 vertices from (n−2) blocks. So, there always exist two consecutive blocks from each and among them no vertex will contribute to form the set S, which is contradiction of Lemma 2.2.

Both the claims provide that our assumption is wrong. Hence S must contain at least m−2 more vertices other than x and y, which implies that |S|≥m.

Theorem 2.4. For even κ≥4 and any n≥3 with n≠4, dim(Hn,κ)=[n2]+2.

Proof. An establishment of upper and lower bounds for dim(Hn,κ) will complete the proof.

Lower bound: Lemma 2.4 provides the minimum metric generator for Hn,κ of cardinality [n2]+2, which yields the lower bound.

Upper bound: We discuss two cases according to the parity of n.

When n≥6 is even. Let W={νκ2+11,νκ2+12,νκ2+12i+1,νκn;1≤i≤n−22}. Then, with the similar reasoning given in the proof of Theorem 2.3 for the upper bound, the set W is a metric generator for Hn,κ.

Possibility 1: When r=m, then we always have a vertex s∈S such that s∈Br+1 or s∈Br−1 and s identifies the pair p.

Possibility 2: When r∈{m,m+1,m−1}, then there is a vertex s in S from the block Br (or Bm) such that s identifies the pair p.

Possibility 3: When r∉{m−1,m,m+1}, then at least one of the following two observations must true:

S contains an element s from the block Br (or Bm) such that s identifies the pair p.

Hence, S is a metric generator for Hn,κ.

Lemma 2.5. For odd κ≥5 and any n≥4, if S is a minimum metric generator for Hn,κ, then |S|≥[n2]+1.

Proof. Let [n2]+1=l. By Lemma 2.1, S must have two vertices from both the end blocks of Hn,κ. Now, we have to show that S contains at least l−2 more vertices. Contrarily, assume that S contains l−3 more vertices. Then, with the similar reasoning given in the proof of Lemma 2.4, we get two consecutive blocks such that none of them contributes in the set S, which is a contradiction of Lemma 2.2. So, our supposition is wrong. Hence, S must contain at least l−2 more vertices, which implies that |S|≥l.

Theorem 2.5. For odd κ≥5 and any n≥4, dim(Hn,κ)=[n2]+1.

Proof. dim(H4,κ)≥[n2]+1, by Lemma 2.5. Moreover, with the similar justification proposed for the proof of upper bound in Theorem 2.4, Hn,κ has a metric generator W={νκ2+12i−1,νκn;1≤i≤n2} when n is even, and has a metric generator W={νκ22i−1;1≤i≤n+12} when n is odd. It follows that dim(Hn,κ)≤[n2]+1.

Rn,κ denotes a chain cactus Tn,κ such that the cut-vertices, lying in the same non-pendent polygon of Tn,κ, are not adjacent, see Fig. 3. We further classify Rn,κ into three types:

Whenever κ is even and the distance between cut vertices is κ2, then we let νκ2+1j=ν1j+1=νj (a joint/cut vertex between jth and (j+1)th polygons/cycles) for 1≤j≤n−1.

With the similar justification proposed for the proof of Lemma 2.2, we have the following result:

Lemma 2.6. For κ>5 and n≥3, it is not possible that two consecutive blocks do not contribute to form any metric generator for Rn,κ.

Theorem 2.6. For odd κ≥3, dim(R2,κ)=2.

Proof. Since only a path graph has the metric dimension equals to 1, by Theorem 1.1, dim(R2,κ)≥2. Now, we have to prove that dim(R2,κ)≤2 by investigating a metric generator of cardinality 2. Let us consider the set of vertices W={ν11,νκ+122}. Then, metric vectors of all the vertices in R2,κ with respect to W are: mW(νi1)={(i−1,i+κ−12),1≤i<κ−12,(κ−32,κ−1),i=κ−12,(k−i+1,3κ−12−i),κ−12<i≤κ. mW(νi2)={(i,κ+12−i),1≤i≤κ+12,(κ−i+2,i−κ+12),κ+12<i≤κ.

It can be easily verified that for every pair x, y of distinct vertices, we have mW(x)≠mW(y). So, W is a metric generator for R2,κ, and dim(R2,κ)≤2.

Theorem 2.7. For even κ≥4, dim(R2,κ)=3.

Proof. The proof follows from the following two claims:

Claim I: (dim(R2,κ)≥3)

Suppose contrarily that dim(R2,κ)<3. Since any metric generator for R2,κ must contain a vertex from both end blocks, by Lemma 2.1, dim(R2,κ)≥2. Let S={x,y} be a minimum metric generator for R2,κ, where x∈B1 and y∈B2. Then, there are two possibilities:

Whenever d(x,νκ1)=κ2 or d(y,ν12)=κ2, then there are two vertices u1 and u2, lying in the same block and both are neighbors of the joint, such that ms(u1)=ms(u2), a contradiction.

Thus, according to these possibilities, S is not a metric generator. So, our supposition is wrong and dim(R2,κ)≥3.

Claim II: (dim(R2,κ)≤3)

Let us consider a set S={ν11,ν22,νκ−22} of vertices. Then, metric vectors of the vertices of R2,κ with respect to S are: mS(νi1)={(i−1,1+i,i+1),1≤i≤κ2−2,(κ2−2,κ2,κ2−1),i=κ2−1,(κ2−1,κ2+1,κ2−2),i=κ2,(κ2,κ2,κ2−2),i=κ2+1,(κ−i+1,κ−i+1,κ−i+2),κ2+1<i≤κ−2,(2,2,1),i=κ−1,(1,1,2),i=κ. mS(νi2)={(1,1,2),i=1,(κ−i+2,i−2,1+i),2≤i≤κ2+1,(κ2+2,κ2,κ2+3),i=κ2+2,(κ2+2,κ−i+2,κ−i+1),κ2+2<i≤κ.

It can be seen that all the metric vectors are different, which implies that S is a metric generator for R2,κ. Hence dim(R2,κ)≥3.

Let u,ν∈V(G) be any two vertices. Then, u,ν are called twins if either N[u]=N[ν] or N(u)=N(ν). The relation of twins between vertices of G is an equivalence relation, which partitioned V(G) into classes each of which is called a twin class. A twin class may be singleton [6]. The following results are useful tools to identify twins in a graph G.

Lemma 2.7. [6] If u and ν are twins in a connected graph G, then no vertex, except u and ν, of G identifies the vertices u and ν.

Accordingly, we have the following remark:

Remark 2.1. If U is twin class in a connected graph G with |U|=l≥2, then every metric generator for G contains at least l−1 vertices from U.

Theorem 2.8. For n≥3, dim(Rn,4)=n.

Proof. We prove the result with two cases providing lower and upper bounds for the metric dimension of Rn,4.

Case-I (Lower bound)

In Rn,4, we obtain n twin classes each of them has cardinality 2. Now, if S is a minimum metric generator for Rn,4, then S must contain at least one vertex from each twin class, by Remark 2.1. This implies that dim(Rn,4)=|S|≥n.

Case-II (Upper bound)

Let S={ν31,ν2t:2≤t≤n}⊂V(Rn,4). Then, S is a metric generator for Rn,4, because all the vertices have distinct metric vectors with respect to S as listed below:

for fixed 1≤j≤n,

mS(ν11)=(a1j,a2j,…,anj), where the lth coordinate is alj={2l−1,for1≤l≤n.

mS(ν2j)=(a1j,a2j,…,anj), where the lth coordinate is alj={2(j−l),whenever1≤l<j,0,wheneverl=j,2(l−j),wheneverj<l≤n.

mS(ν3j)=mS(ν1j+1)=(a1j,a2j,…,anj), where the lth coordinate is alj={2j−2l+1,whenever1≤l<j,1,wheneverl=j,2l−2j−1,wheneverj<l≤n.

mS(ν4j)=(a1j,a2j,…,anj), where the lth coordinate is alj={2(j−l),whenever1≤l<j,2,wheneverl=j,2(l−2j),wheneverj<l≤n.

This implies that dim(Rn,4)≤n.

Lemma 2.8. For even κ≥6, if S is a minimum metric generator for R3,κ, then |S|≥3.

Proof. By Lemma 2.1, S must contain a vertex from both the end blocks of R3,κ. Suppose that a 2-element set S={x,y} is a metric generator for R3,κ, where x lies in the block B1 and y lies in the block B3. We will discuss two possibilities:

Whenever d(x,ν1)=κ2, then there are vertices u1, w1 in the block B1 such that u1 and w1 are BDS and u1∼ν1∼w1. It follows that mS(u1)=mS(w1), a contradiction to the fact that S is a metric generator. Similarly, if d(y,ν2)=κ2, then again we get a contradiction.

It follows that our supposition is wrong, and no 2-element set is a metric generator for R3,κ. Thus |S|≥3.

Theorem 2.9. For even κ≥6, dim(R3,κ)=3.

Proof. By Lemma 2.8, dim(R3,κ)≥3. Moreover, dim(R3,κ)≤3, because the set S={νκ1,ν23,ν22} is a metric generator for R3,κ due to the following distinct metric vectors of the vertices with respect to S: mS(νl1)={(l,l+4,l+2),1≤l≤κ2−1,(κ−l,κ−l+2,κ−l),κ2−1<l≤κ−1,(0,4,2),l=κ. mS(νl2)={(1,3,1),l=1,(l,l+2,l−2),2≤l≤κ2−1,(l,κ−l,l−2),κ2−1<l≤κ2+2,(l,κ−l,κ−l+2),κ2+2<l≤κ−2,(κ−l+2,κ−l,κ−l+2),l=κ−1,(κ−l+2,2,κ−l+2),l=κ. mS(νl3)={(3,1,3),l=1,(l+2,l−2,l+2),2≤l≤κ2+1,(κ−l+4,κ−l+2,κ−l+4),κ2+1≤l≤κ.

Theorem 2.10. For even κ≥6, dim(R4,κ)=4.

Proof. Let S={ν11,ν22,ν23,ν24}⊂V(R4,κ). Then the metric vector of νlj with respect to S is given below: mS(νl1)={(l−1,l+2,l+4,l+6),1≤l≤κ2−1,(κ−l+1,κ−l+2,κ−l+4),κ2−1<l≤κ−1,(1,2,4,6),l=κ. mS(νl2)={(2,1,3,5),l=1,(l+1,l−2,l+2,l+4),2≤l≤κ2−1,(κ2+1,κ2+1−2,κ2,κ2+1+2),l=κ2,(κ−l+3,κ−l+2,κ−l,κ−l+2),κ2+1≤l≤κ−1,(3,2,2,4),l=κ. mS(νl3)={(4,3,1,3),l=1,(l+3,l+2,l−3,l+2),2≤l≤κ2−1,(κ2+3,κ2+2,κ2−2),l=κ2,(κ−l+5,κ−l+4,κ−l+2,κ−l),κ2+1≤l≤κ−1,(5,4,2,2),l=κ. mS(νl4)={(6,5,3,1),l=1,(5+l,4+l,2+l,2−l),2≤lκ2+1,(κ−l+7,κ−l+6,κ−l+4,κ−l+2),κ2+1≤l≤κ.

It can easily verify that all metric vectors are distinct. Thus, S is a metric generator for R4,κ and dim(R4,κ)≤4.

Now, we claim that if S is a minimum metric generator for R4,κ, then |S|≥4. Suppose contrarily that |S|=3. By Lemma 2.1, S must contain one vertex from both the end blocks of R4,κ. Let S={x,y,z}, where x lies in the first block B1 and y lies in the last block B4. There are two cases to discuss:

If z lies in the block B1 (or B4), then there exist BDS, u1 lies in the block B2 and w1 lies in the block B3 with u1∼ν2∼w1, such that mS(u1)=mS(w1), a contradiction.

All these possibilities conclude that our supposition is wrong and |S|≥4. Hence, dim(R4,κ)≥4.

Theorem 2.11. For n≥3, dim(Rn,5)=2.

Proof. By Theorem 1.1, only a path graph has the metric dimension equals to 1. Therefore, dim(Rn,5)≥2. Now, consider a set W={ν11,ν3n} of vertices of Rn,5. Then, metric vectors of the vertices with respect to W are: mW(νl1)={(1−l,3−l+2(n−1)),l=1,(l−1,3−l+2(n−1)),1<l≤3,(6−l,l+2n−5),3<l≤5.

For 2≤j≤n, mW(νlj)={(l+2(j−2),3−l+2(n−j)),1≤l≤3,(3−l+2j,2(n−j)+l−3),3<l≤5.

It can easily verify that for each pair of distinct vertices (x, y) in R3,κ, we have mW(x)≠mW(y). Thus, W is a metric generator for Rn,5 and dim(Rn,5)≤2. It completes the proof.

According to the similar reasoning of the proofs of Theorems 2.4 and 2.5 we have the following two results for Rn,κ.

Theorem 2.12. For even κ≥6 and any n≥5, dim(Rn,κ)=[n2]+2.

Theorem 2.13. For odd κ≥7 and any n≥3, dim(Rn,κ)=[n2]+1.

Theorem 2.14. For even κ≥6 and any n≥3, whenever d(νi,νi+1)=κ2 in Rn,κ for each 1≤i≤n−1, then dim(Rn,κ)=n.

Proof. Let S be a minimum metric generator and assume that |S|=n−1. This implies that S does not have any vertex from at least one block Bt (say), then we have two vertices u,w∈Bt such that u and w are BDS, u∼νt−1 (orνt), w∼νt−1 (orνt) and mS(u)=mS(w). This is a contradiction to the fact that S is a metric generator for Rn,κ. Hence dim(Rn,κ)=|S|≥n.

Now, let S={νκ2−1i:1≤i≤n}⊂V(Rn,κ). Then, with the similar reasoning as given for the proof of upper bound in Theorem 2.3, S is a metric generator for Rn,κ. It follows that dim(Rn,κ)≤n.

Theorem 2.15. For odd κ≥7 and any n≥3, whenever d(νi,νi+1)=κ−12 in Rn,κ for each 1≤i≤n−1, then dim(Rn,κ)=2.

Proof. By Theorem 1.1, dim(Rn,κ)≥2, because Rn,κ is not a path graph. Now, let W={νκ−321,νκ+12n} and corresponding metric vectors of the vertices are: mW(νi1)={(κ−52−i+1,κ+12−i+κ−12(n−1)−κ−12(j−1)),1≤i≤κ−32,(i−κ−32,κ+12−i+κ−12(n−1)−κ−12(j−1)),κ−32<i≤κ+12,(i−κ−32,κ−32+i+κ−12(n−3)−κ−12(j−1)),κ+12<i≤κ−2,(3κ−32−i,κ−32+i+κ−12(n−3)−κ−12(j−1)),κ−2<i≤κ.