Independent Fixed Connected Geodetic Number of a Graph

P. Titus1, S. Antin Mary2
1Department of Mathematics, University College of Engineering Nagercoil, Anna University, Tirunelveli Region.
2Department of Mathematics,Holy Cross College (Autonomous), Nagercoil, India.

Abstract

In this paper we introduce the concept of independent fixed connected geodetic number and investigate its behaviours on some standard graphs. Lower and upper bounds are found for the above number and we characterize the suitable graphs achieving these bounds. We also define two new parameters connected geo-independent number and upper connected geo-independent number of a graph. Few characterization and realization results are formulated for the new parameters. Finally an open problem is posed.

Keywords: Independent fixed connected geodetic set, Independent fixed connected geodetic number, Connected Geo-independent number, Upper connected Geo-independent number

1. Introduction

The introduction of Graph Theory is a revolution in the field of Mathematics. Various concepts were made easily understandable by its simple expression through graphical models. By a graph G we mean V, the set of vertices; E, the set of edges together with a binary operation of association. We refer to [1,2, 3 ] for basic graph theoretic terms. In G, a shortest xy path is also known as xy geodesic. The distance d(x,y) is defined as the number of edges of an xy geodesic in G. For any two vertices x and y in G, the closed interval I[x,y] is the collection of vertices on an xy geodesic. The closed interval I[S,S], where S,SV(G), is defined as the union of subintervals I[x,y] for some xS and yS. i.e., I[S,S]=xS,ySI[x,y]. A vertex v in G is called an extreme vertex or simplicial vertex if the subgraph induced by its adjacent vertices is complete.

A set SV(G) is called a geodetic set or geodomination set if every vertex of G is on some xy geodesic where x,yS. The minimum cardinality of a geodetic set of G is called as the geodetic number of G, denoted by g(G) [4, 5, 6, 7, 8]. If S is a geodetic set of G and S is connected, then S is called the connected geodetic set of G. Its minimum order is named as the connected geodetic number of G, denoted by cg(G). A connected geodetic set of cardinality cg(G) is called a cg-set of G [9]. Again parameters upper connected geodetic number and forcing connected geodetic number were defined and investigated in [10]. Santhakumaran and Titus first introduced the vertex geodomination number in [11] and further studied in [12, 13]. For any vertex x in G, a set SV(G) is called an x-geodominating set of G if every vertex v in G is on an xy geodesic for some y in S. The minimum cardinality of an x-geodominating set of G is defined as the x-geodomination number of G, denoted by gx(G). An x-geodominating set of cardinality gx(G) is called a gx-set of G. A connected x-geodominating set of G is an x-geodominating set S such that S is connected. The minimum cardinality of a connected x-geodominating set of G is the connected x-geodomination number of G and is denoted by cgx(G). A connected x-geodominating set of cardinality cgx(G) is called a cgx-set of G [14].

Let e=xy be any edge of a connected graph G of order atleast 3. A set S of vertices of G is an e-geodominating set of G if every vertex of G is either on an xu geodesic or on an yu geodesic for some element u in S. The minimum cardinality of an e-geodominating set of G is defined as the e-geodomination number of G and is denoted by ge(G) or gxy(G). An e-geodominating set of cardinality ge(G) is called a ge-set of G [15].

It is clear that x-geodominating set of G is obtained by fixing a vertex x in G and e-geodominating set of G is obtained by fixing an edge e=xy in G. Based on these concepts we defined a new parameter called independent fixed geodomination number (or independent fixed geodetic number) in [16]. Let S be an independent set of a connected graph G of order atleast 2. Let S be a subset of V(G). If each vertex v in G is on an xy geodesic for some xS and yS, then S is an S-fixed geodetic set of G . The Sfixed geodetic number gs(G) of G is the minimum cardinality of an S-fixed geodetic set of G. The independent fixed geodetic number of G is gif(G)=min{gs(G)}, where the minimum is taken over all independent sets S in G. An independent fixed geodetic set of cardinality gif(G) is described as gif-set of G. We too further proceed to infer about connectedness of an S-fixed geodetic set of G, where S is an independent set in a connected graph G. In the computation of independent fixed connected geodetic number, the succeeding theorems will be employed.

Theorem 1. [16] For any connected graph G, 1gif(G)p1.

Theorem 2. [16] Let G be a connected graph. Then gif(G)=1 if and only if there is an independent set S and its eccentric vertex y such that every vertex of G is on an xy geodesic for some xS.

Theorem 3. [16] For any complete graph Kp, gif(Kp)=p1.

2. Independent fixed connected geodetic number

Definition 1. Let S be an independent set of a connected graph G of order atleast 2. An S-fixed connected geodetic set of G is an S-fixed geodetic set S such that S is connected. The S-fixed connected geodetic number cgs(G) of G is the minimum cardinality of an S-fixed connected geodetic set of G. The independent fixed connected geodetic number cgif(G) of G is defined as cgif(G)=min{cgs(G)}, where the minimum is taken over all independent sets S in G. An independent fixed connected geodetic set of cardinality cgif(G) is called a cgif-set of G.

Example 1. Consider a graph G as shown in Figure 1. The Table 1 gives the independent sets S, their corresponding minimum S-fixed geodetic sets and minimum S-fixed connected geodetic sets, the S-fixed geodetic numbers gs(G) and the S-fixed connected geodetic numbers cgs(G). Then the independent fixed geodetic number of G is gif(G)=min{gs(G)}=2 and the independent fixed connected geodetic number of G is cgif(G)=min{cgs(G)}=3.

Table 2.1
Independent Minimum S-fixed S-fixed Minimum S-fixed S-fixed
set S geodetic set geodetic connected geodetic set connected
number gsG geodetic
number cgsG
{x1} {x2,x4,x5,x6} 4 {x2,x3,x4,x5,x6} 5
{x2} {x1,x4,x5,x6} 4 {x1,x3,x4,x5,x6} 5
{x3} {x1,x2,x4,x5,x6} 5 {x1,x2,x3,x4,x5,x6} 6
{x4} {x1,x2,x6} 3 {x1,x2,x3,x6} 4
{x5} {x1,x2,x4,x6} 4 {x1,x2,x3,x4,x6} 5
{x6} {x1,x2,x4} 3 {x1,x2,x3,x4} 4
{x1,x4} {x2,x6} 2 {x2,x3,x6} 3
{x1,x5} {x2,x4,x6} 3 {x2,x3,x4,x6} 4
{x1,x6} {x2,x4} 2 {x2,x3,x4} 3
{x2,x4} {x1,x6} 2 {x1,x3,x6} 3
{x2,x5} {x1,x4,x6} 3 {x1,x3,x4,x6} 4
{x2,x6} {x1,x4} 2 {x1,x3,x4} 3
{x4,x6} {x1,x2,x5} 3 {x1,x2,x3,x5} 4
{x1,x4,x6} {x2,x5} 2 {x2,x3,x5} 3
{x2,x4,x6} {x1,x5} 2 {x1,x3,x5} 3

Theorem 4. In a connected graph G, 1gif(G)cgif(G)p1.

Proof. For any independent set S in a connected graph G, every S-fixed connected geodetic set is an S-fixed geodetic set of G, we have gif(G)cgif(G). Then by Theorem 1, we have 1gif(G)cgif(G)p1

We intend to characterize the graphs that realize the bounds in Theorem 4. For that we use the following definition.

Definition 2. [16] Let G be a connected graph of order atleast 2. Let SV(G) and yV(G)S. The distance between the set S and the vertex y is d(S,y)=min{d(x,y):xS}. The eccentricity of the set S is e(S)=max{d(S,y):yV(G)S}. An eccentric vertex of S is a vertex v of G with d(S,v)=e(S).

Theorem 5. Let G be a connected graph of order atleast 2. Then cgif(G)=1 if and only if there is an independent set S and its eccentric vertex y such that every vertex of G is on an xy geodesic for some xS.

Proof. The result follows from Theorems 2 and 4

The following theorem gives cgif(G) for some standard graphs.

Theorem 6.

  1. If G=T or Km,n, then cgif(G)=1.

  2. If G=Cp, then cgif(G) is 1 or 2 according as p is even or odd.

  3. If G=K1+mjKj with G is neither a complete graph nor a star, then cgif(G)=j1 or mj(j1)+1 according as j2mj=1 or j2mj2.

In view of Theorem 4, we proceed to characterize graphs G with cgif(G)=p1.

Theorem 7. Let G be a connected graph of order p2. Then cgif(G)=p1 if and only if G=Kp.

Proof. Let cgif(G)=p1. Claim that G=Kp. If not, then G has a diametral path P:u0,u1,,ud of length d2. It is clear that u0 and ud are non-cut vertices of G. If u0 or ud is an end vertex of G, then S={u0,ud} is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. If both u0 and ud are non-end vertices of G, then there exist atleast two u0ud paths in G. If atleast one vertex other than u0 and ud is common for any two u0ud paths in G, then S={u0,ud} is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. Suppose that there exist two u0ud paths, say P1 and P2, has no common vertices other than u0 and ud. Let z be an adjacent vertex of u0 in P1. If z is a cut vertex of G, then let H be a component of Gz with u0 not in H. Let z1 be a vertex farthest away from z in H. Then it is clear that S={z1,ud} is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. If z is not a cut vertex and {uo,z} is not a vertex-cut of G, then S={u0} is an independent set of G and S=V(G){u0,z} is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. If z is not a cut vertex and {u0,z} is a vertex-cut of G, then there exists another u0z path, say Q, of length atleast 2. Let w be an adjacent vertex of u0 on Q. If w is a cut vertex of G, then let H1 be a component of Gw with u0 not in H1. Let w1 be a vertex farthest away from w in H1. Then it is clear that S={w1,ud} is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. If w is not a cut vertex and {w,ud} is not a vertex-cut of G, then S={w,ud} is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. If w is not a cut vertex and {w,ud} is a vertex-cut of G, then S={w} is an independent set of G and S=V(G){w,u0} is an S-fixed connected geodetic set of G. Hence cgif(G)p2, but this leads to a contradiction. Converse is clear from Theorems 3 and 4

Now we proceed to characterize graphs G with cgif(G)=p2. For that we require the definition of a special graph KmPrKn.

Definition 3. The graph G=KmPrKn is obtained from two complete graphs Km,Kn and a path Pr by joining every vertex in Km with an end vertex of Pr and joining every vertex in Kn with the other end vertex of Pr.

The graph G=KmPrKn is shown in Figure 2.

Theorem 8. Let G be a connected graph of order p3. Then cgif(G)=p2 if and only if G is either P3 or KmPrKn (m,n2 and r1).

Proof. Let cgif(G)=p2. If p=3, then by Theorems 7 and 6(i) we conclude that G=P3. If p=4, then G is either K4,K2+K2¯,K1+(K1K2),K1,3,P4 or C4. If G=K4, then by Theorem 7, cgif(G)=3=p1, but this leads to a contradiction. If G=K2+K2¯, then G has two simplicial vertices, say x and y. It is obvious that S={x} is an independent set of G and S={y} is the minimum S-fixed connected geodetic set of G. Hence cgif(G)=1=p3, but this leads to a contradiction. If G=K1+(K1K2),K1,3,P4 or C4, then by Theorem 6, cgif(G)=1=p3, but this leads to a contradiction.
Now, let p5. Since cgif(G)=p2, by Theorem 7, we have GKp. First, we show that every vertex of G is either a cut vertex or a simplicial vertex. Incase there exists a vertex, say x, in G which is neither a cut vertex nor a simplicial vertex, then x lies on a cycle in G. Let C be a largest chordless cycle containing the vertex x in G. We consider three cases.
Case (1) Length of the cycle C is more than 3 and degree of x is 2.
Let y and z be the adjacent vertices of x on C. Since C is a chordless cycle, y and z are non-simplicial vertices of G.
Subcase (1) deg y=deg z=2. Let S={x} be an independent set of G. If C is an even cycle, then y and z is on an xx1 geodesic, where x1 is an eccentric vertex of x on C; and if C is an odd cycle, then y is on an xx1 geodesic and z is on an xx2 geodesic, where x1 and x2 are the eccentric vertices of x on C. Then it is clear that S=V(G){x,y,z} is an S-fixed geodetic set of G. Since deg y=deg z=deg x=2, S is connected. Hence S is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction.
Subcase (2) deg y3 and deg z3. Since deg x=2, Gx is a connected graph. If y and z are cut vertices of G, then let G1 and G2 be the components of Gy and Gz, respectively, with the vertex x not in both G1 and G2. Let y1 be a vertex farthest away from y in G1 and let z1 be a vertex farthest away from z in G2. Then it is vivid that S={x,y1,z1} is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction.
If y and z are non-cut vertices of G and {y,z} is a vertex-cut of Gx, then y,x and z are the consecutive vertices of another chordless cycle, say C, in G. Let zx be an adjacent vertex of z on C. If z is a non-cut vertex of G, then S={x,z} is an independent set of G and S=V(G){x,z,z} is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction. If z is a cut vertex of G, then let G3 be a component of Gz with the vertex z not in G3. Let z be a vertex farthest away from z in G3. Then S1={x,z} is an independent set of G and S1=V(G){x,z,z} is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction.
If y and z are non-cut vertices of G and {y,z} is not a vertex-cut of Gx. It is clear that S={x} is an independent set of G and S=V(G){x,y,z} is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction. If either y or z is a cut vertex of G, then by the arguments similar to the above, we get a contradiction.
Subcase (3) deg y=2 and deg z3 ((or) deg y3 and deg z=2). If z is a cut vertex of G, then let G1 be a component of Gz with the vertex x not in G1. Let z be a vertex farthest away from z in G1. Then S={x,z} is an independent set of G and S=V(G){x,y,z} is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction.
If z is not a cut vertex of G, then clearly S={x} is an independent set of G and S=V(G){x,y,z} is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction.
Case (2) Length of the cycle C is more than 3 and degree of x is more than 2.
Subcase (1) {x,y} and {x,z} are non vertex-cuts of G. Then it is clear that S={x} is an independent set of G and S=V(G){x,y,z} is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction.
Subcase (2) {x,y} and {x,z} are vertex-cuts of G. Then G has two more cycles, say C1 and C2, with xy an edge of C1 and xz an edge of C2. Let xy be an adjacent vertex of x on C1 and let xz be an adjacent vertex of x on C2. If x is a cut vertex of G, then take a=x1, where x1 is a vertex farthest away from x in a component H of Gx with x not a vertex of H. Otherwise, take a=x. Similarly, if x is a cut vertex of G, then take b=x1, where x1 is a vertex farthest away from x in a component H1 of Gx with x not a vertex of H1. Otherwise, take b=x. It is clear that S={a,b} is an independent set of G and S=V(G){x,a,b} is an S-fixed connected geodetic set of G. It implies cgif(G)p3, but this leads to a contradiction.
Subcase (3) {x,y} is a vertex-cut and {x,z} is not a vertex-cut of G ((or) {x,y} is not a vertex-cut and {x,z} is a vertex-cut of G). Then G has one more chordless cycle, say C1, with xy an edge of C1. Let xy be an adjacent vertex of x on C1. If x is adjacent to y in G and x is not a cut vertex of G, then S={x} is an independent set of G and S=V(G){x,x,z} is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction. If x is adjacent to y in G and x is a cut vertex of G, then by an argument similar to Subcase (3) of Case (1), S={x,x} is an independent set of G and S=V(G){x,x,z} is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction.
Similarly, if x is a non-adjacent vertex of y on C1 and x is a non-cut vertex of G, then S={x} and S=V(G){x,x,z}. If x is a non-adjacent vertex of y and x is a cut vertex of G, then S={x,x} and S=V(G){x,x,z}, where x is a vertex farthest away from x in a component H of Gx with x not a vertex of H. In both cases, S is an independent set of G and S is an S-fixed connected geodetic set of G and hence cgif(G)p3, but this leads to a contradiction.
Case (3) Length of the cycle C is 3.

Then the graph H given in Figure 2.3 is a subgraph of G. If not, every block of G is either K2 or K3 and so every vertex of G is either a cut vertex or a simplicial vertex, which is a contradiction to our assumption. Here we take the vertex u as x and the cycle u,v,w,z,u in H as C and continue the procedure exactly similar to Case (1) and Case (2), we get cgif(G)p3, but this leads to a contradiction.

Hence in all the three cases we get a contradiction and so every vertex in G is either a cut vertex or a simplicial vertex. Since cgif(G)=p2, by Theorem 7, G is a non-complete graph. Hence G has atleast one cut vertex. Let Q={u1,u2,,ua} be the set of all cut vertices of G. We consider two cases.
Case (1) G has exactly one cut vertex, say u1.

Let G1,G2,,Gt(t2) be the components of Gu1. If t3, then S={x1,x2,,xt}, where xiGi(1it), is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction. Hence t=2. Now claim that each component Gi(1i2) has atleast two vertices. If G1 has exactly one vertex, say v, then S={v,z}, where zG2, is an independent set of G. It is clear that S=V(G){v,z,u1} is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction. Hence G has exactly one cut vertex and two components with each component having atleast two vertices. Hence G=KmPrKn (m,n2 and r=1).
Case (2) G has two or more cut vertices.

Let R={z1,z2,,zl}(l0) be the set of all cut vertices of degree 3 in G. If l=0, then G is a path and so by Theorem 6(i), cgif(G)=1<p3, but this leads to a contradiction. Now, let l1 and let S={x1,x2,,xb} (bmax{2,l}), where xi is a simplicial vertex of G in the ith component of GR. If l3, then clearly S is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction. If l=1, then clearly S is an independent set of G and S=V(G){S(QR)} is an S-fixed connected geodetic set of G. Hence cgif(G)p3, but this leads to a contradiction. Hence l=2. If GR has 3 or more components, then S is an independent set of G and S=V(G)S is an S-fixed connected geodetic set of G and so cgif(G)p3, but this leads to a contradiction. Hence GR has exactly 2 components. Since every vertex of G is either a cut vertex or a simplicial vertex, the components of GR are complete graphs. Since l=2, the two components of GR are complete graphs with atleast two vertices and R is a path. Hence G=KmPrKn (m,n2 and r1).

Conversely, let G=P3 or KmPrKn (m,n2 and r1). If G=P3, then by Theorem 6(i), cgif(G)=1=p2. If G=KmPrKn, then every vertex in KmKn is a simplicial vertex of G. It is clear that any independent set S of G contains atmost one vertex from each Km and Kn. Also, every S-fixed connected geodetic set of G contains atleast m1 vertices from Km and atleast n1 vertices from Kn. Since m2, n2 and r1, every vertex of Pr is an element of any S-fixed connected geodetic set of G. Hence S=V(G)S is an S-fixed connected geodetic set of G and so cgif(G)p2. Let S1={x,y}, where xV(Km) and yV(Kn), and let S1=V(G)S1. It is clear that S1 is an independent set of G and S1 is an S1-fixed connected geodetic set of G and so cgif(G)=p2

Based on Theorem 4, we have the following realization result.

Theorem 9. For any three positive integers a,b and p with 2abp3, it is possible to identify a connected graph G of order p with gif(G)=a and cgif(G)=b.

Proof. We consider two cases.
Case (1) 2a=bp3. Let Ppa1:v1,v2,,vpa1 be a path of order pa1 and let Ka+1 be the complete graph of order a+1. Let G be the graph obtained from Ppa1 and Ka+1 by joining an end vertex vpa1 of Ppa1 with every vertex of Ka+1. The resultant graph G is shown in Figure 4 and its order is p.

It is clear that any independent set of G contains atmost one vertex from the complete graph Ka+1. Also, atleast a vertices in Ka+1 lie in every S-fixed geodetic set of G, where S is an independent set of G. Hence gif(G)a. Let S={v1,x} and S=V(Ka+1){x}, where xV(Ka+1). Since ap3, S is an independent set of G and it is clear that every vertex of V(G)S is on a v1y geodesic for any yS. Hence S is an S-fixed geodesic set of G and so gif(G)=|S|=a. Also, since S is connected, we have cgif(G)=gif(G)=a.
Case (2) 2a<bp3. Let Ppa2:v1,v2,,vpb2,,vpa2 be a path of order pa2. Let K2 and Ka be two complete graphs of orders 2 and a, respectively. Let G be the graph obtained from Ppa2, K2 and Ka, by joining the vertex vpb1 of Ppa2 with every vertex of K2 and joining the end vertex vpa2 of Ppa2 with every vertex of Ka. The resultant graph G is shown in Figure 5 and its order is p.

It is clear that any independent set of G contains atmost one vertex from each complete graphs K2 and Ka. Also, atleast one vertex in K2 and atleast a1 vertices in Ka lie in every S-fixed geodetic set of G, where S is any independent set of G. Hence gif(G)a. Let S={v1,x,y} and let S=(V(K2){x})(V(Ka){y}), where xV(K2) and yV(Ka). Since a<bp3, S is an independent set of G and it is clear that every vertex of V(G)S lies on a uv geodesic for some u in S and v in S. Hence S is an S-fixed geodetic set of G and so gif(G)=|S|=a. But S is not connected and so cgif(G)>a. Since every S-fixed geodetic set of G contains atleast one vertex from each complete graphs K2 and Ka, every S-fixed connected geodetic set of G contains the cut vertices {vpb1,vpb,,vpa2}. Let S=S{vpb1,vpb,,vpa2}. It is clear that S is a minimum S-fixed connected geodetic set of G and so cgif(G)=|S|=b

We know that the diameter of any connected graph lies between its radius and two times of its radius. For that Ostrand [17] has given a realization result. Ostrand’s theorem can be extended so that cgif(G) can also be prescribed.

Theorem 10. For any three positive integers r,d and n with rd2r, a connected graph G can be identified with radius r, diameter d and the independent fixed connected geodetic number n.

Proof. If r=1, then d=1 or 2. If d=1, then by Theorem 7, G=Kn+1 has the desired property. Now, let d=2. Let G be the graph obtained from the complete graphs K2 and Kn+2 by merging a vertex of K2, say y, and a vertex of Kn+2. Then G has radius 1, diameter 2 and is shown in Figure 6.

Let T=V(Kn+2){y}. If S is any independent set of G, then atleast n vertices in T lie in every S-fixed connected geodetic set of G and so cgif(G)n. Let S={x,z}, where zy in Kn+2, and let S=T{z}. Clearly, S is an independent set of G and S is the minimum S-fixed connected geodetic set of G and so cgif(G)=n.

Now, let r2. We construct a graph G which meets our requirement.
Case (1) r=d. Let Kn+2 be the complete graph with vertex set V(Kn+2)={u1,u2,,un+2} and let C2r be the even cycle with vertex set V(C2r)={v1,v2,,v2r}. Let G be the graph obtained from Kn+2 and C2r by merging the edge u1u2 in Kn+2 and vrvr+1 in C2r. The resultant graph G is shown in Figure 7.

It can be easily verified that e(v)=r for any vertex vG and so rad G=diam G=r. Also, T=V(Kn+2){u1,u2} is the set of all simplicial vertices of G and any independent set of G contains atmost one element in T. If S is any independent set of G, then atleast n1 vertices in T lie in every S-fixed connected geodetic set of G and so cgif(G)n1. Also, it is clear that all the vertices of C2r are not on any xy geodesic for some xV(C2r) and yT. Hence cgif(G)>n1. Let S={v1,u3} and let S=(T{u3}){vr+1}. Clearly, every vertex of C2r is on a v1vr+1 geodesic and so S is an S-fixed geodetic set of G. Also, S is connected and so cgif(G)=n.
Case (2) r<d2r. Let Kn+1 be the complete graph with the vertex set V(Kn+1)={u1,u2,,un+1}, let Pdr be a path with the vertex set V(Pdr)={v1,v2,,vdr} and let C2r be the cycle with the vertex set V(C2r)={w1,w2,,w2r}. Let G be the graph obtained from Kn+1,Pdr and C2r by joining every vertex of Kn+1 with the vertex v1 in Pdr, and joining the vertex vdr of Pdr with the vertex w1 in C2r. The graph G is shown in Figure 8.

It can be easily verified that re(x)d, e(w1)=r and e(wr+1)=d. Hence rad G=r and diam G=d. It is clear that T=V(Kn+1) is the set of all simplicial vertices of G. Then by an argument similar to Case (1) of Theorem 2.10, we have S={u1,wr+1} is an independent set of G and S=T{u1} is a minimum S-fixed connected geodetic set of G. Hence cgif(G)=n

3. Connected Geo-independent Number

Definition 4. The minimum (maximum) independent set required to form a cgif-set of G is called a connected geo-independent set (upper connected geo-independent set) of G. The cardinality of a connected geo-independent set (upper connected geo-independent set) of G is called the connected geo-independent number (upper connected geo-independent number) of G and is denoted by cgin(G) (cgin+(G)).

Example 2. For the graph G given in Figure 2.1, we have cgif(G)=3. From Table 1, it can be easily seen that S1={x1,x4},S2={x1,x6}, S3={x2,x4} and S4={x2,x6} are the connected geo-independent sets of G, S5={x1,x4,x6} and S6={x2,x4,x6} are the upper connected geo-independent sets of G. Hence cgin(G)=2 and cgin+(G)=3.

The following result gives the connected geo-independent numbers and the upper connected geo-independent numbers of certain special classes of graphs.

Result 11.

  1. If G=Pp, then cgin(G)=1 and cgin+(G)=p2.

  2. If G=K1,p1(p3), then cgin(G)=p2 and cgin+(G)=p1.

  3. If G=Kp, then cgin(G)=cgin+(G)=1.

  4. If G=Cp, then cgin(G)=1 and cgin+(G)=p2 or p21 according as p2 is odd or even.

  5. If G=Km,n (2mn), then cgin(G)=m1 and cgin+(G)=n1.

  6. If G=Qn (n3), then cgin(G)=1 and cgin+(G)=2n1.

The following observation is an easy consequence of some of the previous results.

Observation 12. For any connected graph G of order p2,

  1. 1cgin(G)cgin+(G)β(G)p1, where β(G) is the independence number of G.

  2. 2cgin(G)+cgif(G)p and 2cgin+(G)+cgif(G)p.

Theorem 13. Let G be a connected graph of order p2. Then

  1. cgin(G)=1 if and only if cgx(G)=cgif(G) for some vertex x in G.

  2. cgin(G)=p1 if and only if G=K2.

  3. cgin+(G)=p1 if and only if G=K1,p1.

Proof. (i) Let cgin(G)=1. Then there exists a vertex, say x, in G with S={x} and S, a minimum S-fixed connected geodetic set of G. Hence every vertex of G is on an xy geodesic for some yS and S is connected, and so S is a minimum connected x-geodominating set of G. Hence cgx(G)=|S|=cgif(G). Converse is clear from the respective definitions.
(ii) Let cgin(G)=p1. If p=2, then we get the required result. So, let p3. Since the connected graph G has p1 independent vertices, all the independent vertices are end vertices of G. Hence G is a star. Then by Result 11(ii), cgin(G)=p2, but this leads to a contradiction. Conversely, let G=K2. Then by Result 11(iii), cgin(G)=1=p1.
(iii) The result follows from the proof of (ii) and Result 1(ii). 

Problem 14. Characterize graphs G for which cgin+(G)=1.

The following theorem gives a realization result.

Theorem 15. For any four positive integers a,b,c and n with 2abc, a connected graph G can be identified with cgin(G)=a, cgin+(G)=b, β(G)=c and cgif(G)=n.

Proof. Case (1) a=b. Let H1,H2,H3 and H4 be the complete graphs with vertex sets V(H1)={x,y},V(H2)={u1,u2,,uca+1},V(H3)={v1,v2,,va2} and V(H4)={w1,w2,,wn+1} respectively. Let the graph G be constructed using H1¯,H2¯,H3¯ and H4 by (i) joining the vertices x and y in H1¯ with every vertex of H2¯ and (ii) joining the vertex y in H1¯ with every vertex of H3¯H4. The resultant graph G is shown in Figure 9.

Every independent set of G contains atmost one vertex from H4. Then atleast n vertices in the complete graph H4 lie on every S-fixed geodetic set of G, where S is an independent set of G. Hence cgif(G)n. Let S={x,v1,v2,,va2,wi} and let S=V(H4){wi} for any i(1in+1). Clearly S is an independent set of G and every vertex of V(G)S is on an xs geodesic for any sS and S is connected. Hence S is an S-fixed connected geodetic set of G and so cgif(G)=|S|=n. Also, it is clear that, for any independent set S of G, every minimum S-fixed connected geodetic set contains n vertices from H4.

Let T be any independent set of G with T, a T-fixed connected geodetic set of G and |T|=cgif(G)=n. Hence TV(H4). Now claim that V(H3¯)T. If not, let zV(H3¯) and zT. Since z is an end vertex of G, z is not an internal vertex of any geodesic and so zT, which is a contradiction to TV(H4). Hence V(H3¯)T. Also, since cgif(G)=n and TV(H4), exactly one vertex in H4, say w1, belongs to T. Since w1 is adjacent to y, yT. Then either x or ui(1ica+1) but not both belongs to T.
Subcase (1) xT. Then every vertex of GH3¯ is on an xw geodesic for any wT and so T=V(H3¯){x,w1} is an independent set of G with T, a T-fixed connected geodetic set of G and |T|=n.
Subcase (2) uiT(1ica+1). If all the vertices uiT, then x is not an internal vertex of any st geodesic for any sT and tT. Hence xT, which is a contradiction to TV(H4). If atleast one vertex uiT, then ui is not an internal vertex of any st geodesic for any sT and tT. Hence uiT, which is a contradiction to TV(H4). Thus Ti=V(H3¯){x,wi}, where wiV(H4) (1in+1), is the independent set of G with Ti=V(H4){wi}, the Ti-fixed connected geodetic set of G and |Ti|=n. Hence cgin(G)=cgin+(G)=a.
Claim β(G)=c. It can be easily verified that Wi=V(H2¯H3¯){wi}, where 1in+1, is a maximum independent set of G and so β(G)=c.
Case (2) a<b. Let H1,H2,H3,H4 and H5 be the complete graphs with vertex sets V(H1)={x,y,z},V(H2)={u1,u2,,ucb+1},V(H3)={v1,v2,,va2},V(H4)={w1,w2,,wba} and V(H5)={t1,t2,,tn+1}, respectively. Let G be the graph obtained from H1¯,H2¯,H3¯,H4¯ and H5 by (i) joining the vertices x and y in H1¯ with every vertex of H2¯, (ii) joining the vertex y in H1¯ with every vertex of H3¯, (iii) joining the vertices y and z in H1¯ with every vertex of H4¯, and (iv) joining the vertex z in H1¯ with every vertex of H5. The resultant graph G is shown in Figure 10.

By an argument similar to Case (1), for any independent set S of G, every minimum S-fixed connected geodetic set contains atleast n vertices from H5. Let S=V(H3¯){x,t1} and let S=V(H5){t1}. Then clearly S is an independent set of G and S is an S-fixed connected geodetic set of G and so cgif(G)=|S|=n.

Next, prove that cgin(G)=a. By an argument similar to Case (1), to form an S-fixed connected geodetic set S with |S|=n, we need all the vertices in H3¯, the vertex x in H1¯ and exactly one vertex in H5 for S. Hence cgin(G)a. As in the above paragraph, S=V(H3¯){x,t1} is an independent set of G and S=V(H5){t1} is the S-fixed connected geodetic set with |S|=n. Hence S is a connected geo-independent set of G and so cgin(G)=|S|=a.

Claim that cgin+(G)=b. Let T be any independent set of G with T, a T-fixed connected geodetic set of G and |T|=cgif(G)=n. Then by an argument similar to Case (1), V(H3¯){x,ti}Ti,V(H5){ti}=Ti for any i (1in+1), and sTi, where sV(H2¯){y,z}. Let Si=V(H3¯H4¯){x,ti}. Then clearly Si is a maximum independent set of G with Si=V(H5){ti}, the Si-fixed connected geodetic set of G and |Si|=n. Hence cgin+(G)=|Si|=b.

It can be easily verified that Wi=V(H2¯H3¯H4¯){ti}, where 1in+1, is a maximum independent set of G and so β(G)=c

Conflict of Interest

The authors declare no conflict of interests.

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