Metric Dimension of Corona Product and Join Graph of Zero Divisor Graphs of Direct Product of Finite Fields

Subhash Mallinath Gaded1, Nithya Sai Narayana2
1R K Talreja College of Arts, Science and Commerce, Ulhasnagar-03, Maharashtra, India
2Department of Mathematics, University of Mumbai, Mumbai, Maharashtra, India

Abstract

The metric dimension of a graph is the smallest number of vertices such that all vertices are uniquely determined by their distances to the chosen vertices. The corona product of graphs G and H is the graph GH obtained by taking one copy of G, called the center graph, |V(G)| copies of H, called the outer graph, and making the jth vertex of G adjacent to every vertex of the jth copy of H, where 1j|V(G)|. The Join graph G+H of two graphs G and H is the graph with vertex set V(G+H)=V(G)V(H) and edge set E(G+H)=E(G)E(H){uv:uV(G),vV(H)}. In this paper, we determine the Metric dimension of Corona product and Join graph of zero divisor graphs of direct product of finite fields.

Keywords: Zero divisor graph, Metric dimension, Corona product, Join graph, finite fields

1. Introduction and Preliminaries

The concept of graph of a commutative ring R, was introduced by Beck [1] in 1988. Beck considered all the elements of the ring R as the vertices of the graph and two distinct vertices x and y are adjacent in the graph if and only if xy=0, the additive identity of the ring R. This definition of graph given by Beck was modified by Anderson and Livingston [2] in 1999. Anderson and Livingston considered only the non-zero zero divisors of commutative ring R to be the vertex set of the graph known as zero divisor graph of R denoted by Γ(R) in which two distinct vertices x and y are adjacent in Γ(R) if and only if xy=0. In this paper we consider the zero divisor graph of the reduced ring, R=F1×F2××Fn, of direct product of finite fields. Let Z(R) be the set of non-zero zero-divisors of the ring R=F1×F2××Fn and Γ(R) denote the graph with vertex set as Z(R) and edge set as {rs:rs=0,r,sZ(R)}.

The metric dimension of a graph is the smallest number of vertices such that all vertices are uniquely determined by their distances to the chosen vertices. For an ordered subset W={w1,w2,,wk} of vertices in a connected graph G and a vertex v of G, the metric representation of v with respect to W is the kvector r(v|W)=(d(v,w1),d(v,w2),,d(v,wk)). The set W is a resolving set for G if r(u|W)=r(v|W) implies that u=v for all pairs u,v of vertices of G. In other words, a set WV(G) is called a resolving set, if for each two distinct vertices u,vV(G) there exists wW such that d(u,w)d(v,w), where d(x,y) is the distance between the vertices x and y. The minimum cardinality of a resolving set for G is called the metric dimension of G, and denoted by β(G).

The problem of Metric dimension was introduced by Slater [3] and also independently introduced by Harary and Melter [4]. In [5-7] graphs of order n with metric dimension 1,n3,n2 and n1 have been characterized. In [8], metric dimensions for many particular classes of graphs have been determined.

The corona product of graphs G and H is the graph GH obtained by taking one copy of G, called the center graph, |V(G)| copies of H, called the outer graph, and making the jth vertex of G adjacent to every vertex of the jth copy of H, where 1j|V(G)|.

The Join graph G+H of two graphs G and H is the graph with vertex set V(G+H)=V(G)V(H) and edge set E(G+H)=E(G)E(H){uv:uV(G),vV(H)}.

In this paper, we determine the Metric dimension of Corona product and Join graph of zero divisor graphs of direct product of finite fields.

2. Results on Metric Dimension of Zero Divisor Graph of Direct Product of Finite Fields

In [9, Proposition 6.2], Raja, Pirzada & Redmond, proved that for any k2,loc(Γ(i=1kZ2))k. They also proved that [9, Theorem 6.3], loc(Γ(i=15Z2))=5. In [10], it is proved that if n=2,3,4, and order of each field is two, then β(Γ(R))=n1.

Theorem 1. [10, Theorem 2.4] If R=F1××Fn,n2 with |Fi|3,(1in), then the metric dimension β(Γ(R))=|V(Γ(R))|(2n2).

Consider the reduced rings R1=F1××Fn,(n2) and R2=J1××Jm,(m2) where Fi,(1in),Jk,(1km) are finite fields with |Fi|2,|Jk|2.

In [11], the following result is proved.

Theorem 2. [11] Let R1=F1××Fn,(n2) and R2=J1××Jm,(m2) where Fi,(1in),Jk,(1km) are finite fields with |Fi|2,|Jk|2. Then,

  1. the diameter of Corona product of Γ(R1) and Γ(R2) is 3 if n=2,m=2,|F1|=|F2|=|J1|=|J2|=2,

  2. the diameter of Corona product of Γ(R1) and Γ(R2) is 4 if n=2,m=2,|F1|3 or |F2|3 and |J1|3 or |J2|3,

  3. the diameter of Corona product of Γ(R1) and Γ(R2) is 5 if n3,m3,

  4. the girth of Corona product of Γ(R1) and Γ(R2) is 3 if n=2,m=2,|F1|,|F2|3,|J1|,|J2|3,

  5. the girth of Corona product of Γ(R1) and Γ(R2) is 3 if n3 or m3.

3. Metric Dimension of Corona Product of Zero Divisor Graphs of Direct Product of Finite Fields

In this section, we determine the Metric dimension of Corona product of zero divisor graphs of direct product of finite fields.

Theorem 3. If R1=F1××Fn,(n2) and R2=J1××Jm,(m2) where Fi,(1in),Jk,(1km) are finite fields with |Fi|=2,|Jk|=2, then the Metric dimension of Corona product of Γ(R1) and Γ(R2) is β(Γ(R1)Γ(R2))n+(2n2)m.

Proof. According to [9, Proposition 6.2], the metric dimension β(Γ(R1))n and β(Γ(R2))m. Let W0={viV(Γ(R1)):vicontains0inithposition and1in remainingpositions,(1in)}. We first prove that W0 is resolving set for the center graph Γ(R1).

Let x,yV(Γ(R1)). Let r(x|W0)=r(y|W0). Suppose x contains r1 number of 0s and y contains r2 number of 0s,(1r1,r2n1).
Case 1. r1r2.

Let r1<r2. Then there exists a position i such that y contains 0 in the ith position and x contains 1 in the ith position. Then both y and viW0 are adjacent to a vertex z with a 1 in the ith position and 0 everywhere else. Also y is not adjacent to vid(y,vi)=2. Suppose x contains 0 in the jth position (ji). Then x is adjacent to a vertex w with a 1 in the jth position and 0 everywhere else. Also w is adjacent to z and x is not adjacent to z and vi. Therefore, d(x,vi)=3r(x|W0)r(y|W0). This is a contradiction.
Case 2. r1=r2 but x and y differ in atleast one position of 0. Suppose y contains 0 in the ith position and x contains 1 in the ith position. Then from case (i), r(x|W0)r(y|W0). This is a contradiction. Thus, r(x|W0)=r(y|W0)x=yW0 is a resolving set for Γ(R1). Let Wk={viV(Γ(R1)):vicontains0inithposition and1inremaining positions,(1im)}, with 1k|V(Γ(R1))|. As proved above, similarly we can prove that Wk is resolving set for the outer graph of kth copy of Γ(R2). Let W=W0[j=1|V(Γ(R1))|Wj]. Then, W is resolving set for the Corona product of Γ(R1)Γ(R2) and for xV(Γ(R1)Γ(R2)), the metric representation of x with respect to W is the vector of length n+|V(Γ(R1))|m. Since, V(Γ(R1))=2n2, therefore, the Metric dimension of Corona product of Γ(R1) and Γ(R2) is β(Γ(R1)Γ(R2))n+(2n2)m◻

Remark 1.

  • If n=2,3,4 and m=2,3,4 and order of each field is two, then according to [10], β(Γ(R1)Γ(R2))=(n1)+(2n2)(m1).

  • If n=5 and m=5, and order of each field is two, then according to [9, Theorem 6.3], β(Γ(R1)Γ(R2))=5+(252)5.

Theorem 4. If R1=F1××Fn,(n2) and R2=J1××Jm,(m2) where Fi,(1in),Jk,(1km) are finite fields with |Fi|3,|Jk|3, then the Metric dimension of Corona product of Γ(R1) and Γ(R2) is β(Γ(R1)Γ(R2))=(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2)).

Proof. Consider the reduced rings R1=F1××Fn,(n2) and R2=J1××Jr,(r2) where Fi,(1in),Jr,(1rm) are finite fields with |Fi|3,|Jr|3. Let S0={(a1,,an)V(Γ(R1)):ai{0,1}with not allai=0and not allai=1,1in}. Then |S0|=2n2. Let W0=V(Γ(R1))S0. According to [10, Theorem 2.4], W0 is minimum resolving set for Γ(R1) and |W0|=|V(Γ(R))|(2n2). Now consider the Corona product of Γ(R1) and Γ(R2). There are |V(Γ(R1))| copies of Γ(R2) in the Corona product graph Γ(R1)Γ(R2) and the kth vertex of Γ(R1) is adjacent to each and every vertex of kth copy of Γ(R2). Let Sj={(a1,,am)V(Γ(R2)):ai{0,1}with not allai=0and not allai=1,1im} and 1j|V(Γ(R1))|. Then |Sj|=2m2. Let Wj=V(Γ(R2))Sj. According to [10, Theorem 2.4], Wj is minimum resolving set for Γ(R2) and |Wj|=|V(Γ(R2))|(2m2). Let W=W0[j=1|V(Γ(R1))|Wj], and S=S0[j=1|V(Γ(R1))|Sj]. Then W=V(Γ(R1)Γ(R2))S and |W|=(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2)) and |S|=(2n2)+|V(Γ(R1))|(2m2). We claim that W is minimum resolving set for the corona product graph Γ(R1)Γ(R2).

For xV(Γ(R1)Γ(R2)), the metric representation of x with respect to W is the vector of length (|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2)).

Clearly, r(x|W) is distinct for each xW, as the vector r(x|W) contains exactly one 0 entry with the position of 0 in distinct co-ordinate for each metric representation of xW. It is enough to show that for each yS, the metric representation r(y|W) is distinct.

For y1,y2S=S0[j=1|V(Γ(R1))|Sj], if y1Sj1,y2Sj2,j1j2, then clearly d(y1,w)d(y2,w),wWj, since Wj1 is in the j1th copy of Γ(R2) and Wj2 is in the j2th copy of Γ(R2) in corona product Γ(R1)Γ(R2). Now suppose there exists distinct y1,y2Sj(0j|V(Γ(R1))|) such that r(y1|W)=r(y2|W). In other words, d(y1,w)=d(y2,w),wW=W0[j=1|V(Γ(R1))|Wj].

Since the entries in y1 and y2 are only 0s and 1s and y1,y2Sj, this implies y1 and y2 will differ in at least one position containing 0. Suppose y1 contains 0 in the rth co-ordinate position and y2 contains 1 in the rth co-ordinate position. Then y1 is adjacent to a vertex w1Wj, with a non-zero entry other than 1 in the rth position and 0 in the remaining positions (as each field contains at least three elements). This implies d(y1,w1)=1=d(y2,w1) as r(y1|W)=r(y2|W). But since y2 and w1 both contain non-zero entry in the kth position, this implies d(y2,w1)=2or3. This is a contradiction to d(y1,w2)=1=d(y2,w1). Therefore, the metric representation r(y|W) is distinct for each yS. Thus, W is a resolving set for the corona product graph Γ(R1)Γ(R2).

Now consider W=W{x},xW. Suppose x contains r,0s in positions i1,i2,,ir and non-zero entries in remaining positions with at least one non-zero entry other than 1, and let yS=S0[j=1|V(Γ(R1))|Sj] with r,0s in positions i1,i2,,ir and 1 in remaining positions. This implies xy and x,y contains r,0s in same co-ordinate positions i1,i2,,ir. This implies the vertices adjacent(non-adjacent) to x are also adjacent(non-adjacent) to y, implies d(x,u)=d(y,u),uV(Γ(R1)Γ(R2)){x,y} implies d(x,w)=d(y,w),wW implies r(x|W)=r(y|W). Thus, W=W{x} is not a resolving set for each xW. This implies W is a minimal resolving set. Thus, the metric dimension, β(Γ(R1)Γ(R2))|W|=(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2)).

Now there are exactly C(n,r) vertices in Sj(0j|V(Γ(R1))|), such that any two vertices with r zero entries will differ in at least one position containing 0. Since |Sj|=2n2 if j=0 and |Sj|=2m2 if (1j|V(Γ(R1))|), any two vertices in Sj will either differ in the number of 0 entries or if the two vertices contain same number of 0s, then the two vertices will differ in at least one position containing 0. Thus, if SjV(Γ(R1)) with |Sj|=(2n2)+1 when j=0 or SjV(Γ(R2)) with |Sj|=(2m2)+1 where 1j|V(Γ(R1))|, then there exists at least two vertices, say, x,ySj such that both x,y contains same number of 0s and position of 0s is also same in both x and y. This implies positions of non-zero entries is also same, but since x and y are distinct, they will differ in at least one position containing non-zero entry. Since x,y both contains 0s in same co-ordinate positions, by a similar argument above, we get, d(x,u)=d(y,u),uV(Γ(R1)Γ(R2)){x,y}.

Let S be the set obtained from S=S0[j=1|V(Γ(R1))|Sj], by replacing the set Sj with Sj. Then |S|=|S|+1. Therefore, if T=V(Γ(R1)Γ(R2))S with |T|=(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2))1, then there exists at least two vertices, say, x,yV(Γ(R1)Γ(R2))T, such that, d(x,v)=d(y,v),vT, implies r(z|T)=r(w|T). This implies T cannot be a resolving set with |T|=(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2))1 implies β(Γ(R1)Γ(R2))>|T| implies β(Γ(R1)Γ(R2))>(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2))1 implies β(Γ(R1)Γ(R2))(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2)).

Hence, the metric dimension, β(Γ(R1)Γ(R2))=(|V(Γ(R1))|(2n2))+|V(Γ(R1))|(|V(Γ(R2))|(2m2)). ◻

4. Metric Dimension of Join Graph of Zero Divisor Graphs of Direct Product of Finite Fields

Theorem 5. If R1=F1××Fn,(n2) and R2=J1××Jm,(m2) where Fi,(1in),Jk,(1km) are finite fields with |Fi|=2,|Jk|=2, then the Metric dimension of Join graph of Γ(R1) and Γ(R2) is β(Γ(R1)+Γ(R2))n+m.

Proof. According to [9, Proposition 6.2], the metric dimension β(Γ(R1))n and β(Γ(R2))m. Let W1={viV(Γ(R1)):vicontains;0inithposition and1inremaining positions,(1in)}. In the proof of Theorem 3, we proved that, W1 is resolving set for Γ(R1). Let W2={viV(Γ(R1)):vicontains0inithpositionand1in remainingpositions,(1im)}. Similarly, W2 is resolving set for Γ(R2).

Let W=W1W2. Then, clearly W is resolving set for the Join graph of Γ(R1) and Γ(R2) and for xV(Γ(R1)+Γ(R2)), the metric representation of x with respect to W is the vector of length n+m. Hence, the Metric dimension of Join graph of Γ(R1) and Γ(R2) is
β(Γ(R1)+Γ(R2))n+m◻

Remark 2.

  • If n=2,3,4 and m=2,3,4 and order of each field is two, then according to [10], β(Γ(R1)+Γ(R2))=(n1)+(m1).

  • If n=5 and m=5, and order of each field is two, then according to [9, Theorem 6.3], β(Γ(R1)+Γ(R2))=5+5=10.

Theorem 6. If R1=F1××Fn,(n2) and R2=J1××Jm,(m2) where Fi,(1in),Jk,(1km) are finite fields with |Fi|3,|Jk|3, then the Metric dimension of Join graph of Γ(R1) and Γ(R2) is β(Γ(R1)+Γ(R2))=(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2)).

Proof. Let S1={(a1,,an)V(Γ(R1)):ai{0,1}with not allai=0and not allai=1}. Then |S1|=2n2. Let W1=V(Γ(R1))S1. According to [10, Theorem 2.4], W1 is minimum resolving set for Γ(R1) and |W1|=|V(Γ(R))|(2n2). Let S2={(a1,,am)V(Γ(R2)):ai{0,1}with not allai=0and not allai=1}. Then |S2|=2m2. Let W2=V(Γ(R2))S2. Then by [10, Theorem 2.4], W2 is minimum resolving set for Γ(R2) and |W2|=|V(Γ(R))|(2m2).

Consider the Join graph of Γ(R1) and Γ(R2) and let W=W1W2. For each xV(Γ(R1)),r(x|W1) is distinct, and d(x,z)=1 for all zW2 in the Join graph of Γ(R1) and Γ(R2). Similarly, for each yV(Γ(R2)),r(x|W2) is distinct, and d(y,z)=1 for all zW1 in the Join graph of Γ(R1) and Γ(R2). Thus, W=W1W2 is resolving set for the Join graph of Γ(R1) and Γ(R2) and for wV(Γ(R1)+Γ(R2)), the metric representation of w with respect to W is the vector of length (|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2)).

Now consider W=W{x},wW. Either xW1 or xW2. Without loss of generality suppose xW1. Suppose x contains r,0s in positions i1,i2,,ir and non-zero entries in remaining positions with at least one non-zero entry other than 1, and let yS1 with r,0s in positions i1,i2,,ir and 1 in remaining positions. This implies xy and x,y contains r,0s in same co-ordinate positions i1,i2,,ir. This implies the vertices adjacent(non-adjacent) to x are also adjacent(non-adjacent) to y, implies d(x,u)=d(y,u),uV(Γ(R1)+Γ(R2)){x,y} implies d(x,w)=d(y,w),wW implies r(x|W)=r(y|W). Similarly, if xW2 then r(x|W)=r(y|W). Thus, W=W{x} is not a resolving set for each xW. This implies W is a minimal resolving set. Thus, the metric dimension is β(Γ(R1)+Γ(R2))|W|=(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2)).

Now there are exactly C(n,r) vertices in Sj(j=1,2), such that any two vertices with r zero entries will differ in at least one position containing 0. Since |Sj|=2n2 if j=1 and |Sj|=2m2 if j=2, any two vertices in Sj will either differ in the number of 0 entries or if the two vertices contain same number of 0s, then the two vertices will differ in at least one position containing 0. Thus, if SjV(Γ(R1)) with |Sj|=(2n2)+1 where j=1 or SjV(Γ(R2)) with |Sj|=(2m2)+1 where j=2, then there exists at least two vertices, say, x,ySj such that both x,y contains same number of 0s and position of 0s is also same in both x and y. This implies positions of non-zero entries is also same, but since x and y are distinct, they will differ in at least one position containing non-zero entry. Since x,y both contains 0s in same co-ordinate positions, by a similar argument above, we get, d(x,u)=d(y,u),uV(Γ(R1)+Γ(R2)){x,y}.

Let S be the set obtained from S=S1S2, by replacing the set Sj with Sj. Then |S|=|S|+1. Therefore, if T=V(Γ(R1)+Γ(R2))S with |T|=(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2))1, then there exists at least two vertices, say, x,yV(Γ(R1)+Γ(R2))T, such that, d(x,v)=d(y,v),vT, implies r(z|T)=r(w|T). This implies T cannot be a resolving set with |T|=(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2))1 implies β(Γ(R1)+Γ(R2))>|T| implies β(Γ(R1)+Γ(R2))>(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2))1 implies β(Γ(R1)+Γ(R2))(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2)).

Hence, the metric dimension is β(Γ(R1)+Γ(R2))=(|V(Γ(R1))|(2n2))+(|V(Γ(R2))|(2m2))◻

5. Conclusion

We have determined the Metric dimension of Corona product and Join graph of zero divisor graphs of direct product of finite fields.

Author Contributions

Both authors Dr. Nithya Sai Narayana and Subhash Mallinath Gaded wrote the main manuscript text and reviewed the manuscript. Both authors have contributed equally to the work. Both authors read and approved the final manuscript.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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