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.
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 we mean , the set of vertices; , the set of edges together with a
binary operation of association. We refer to [1,2, 3 ] for basic graph theoretic terms. In , a shortest path is also known as geodesic. The distance is defined as the number of edges
of an geodesic in . For any two vertices and in , the closed interval is the collection of vertices on
an geodesic. The closed
interval, where
, is defined
as the union of subintervals for some and
. i.e., . A vertex in is called an extreme vertex or
simplicial vertex if the subgraph induced by its adjacent
vertices is complete.
A set is called
a geodetic set or geodomination set if every vertex of
is on some geodesic where . The minimum cardinality of a
geodetic set of is called as the
geodetic number of ,
denoted by [4, 5, 6, 7, 8]. If is a geodetic set of and is
connected, then is called the
connected geodetic set of . Its minimum order is named as the
connected geodetic number of , denoted by . A connected geodetic set of
cardinality is called a -set of [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 in , a set is called an -geodominating set of if every vertex in is on an geodesic for some in . The minimum cardinality of an -geodominating set of is defined as the -geodomination number of , denoted by . An -geodominating set of cardinality is called a -set of . A connected -geodominating set of is an -geodominating set such that is
connected. The minimum cardinality of a connected -geodominating set of is the connected -geodomination number of and is denoted by . A connected -geodominating set of cardinality is called a -set of [14].
Let be any edge of a
connected graph of order atleast
3. A set of vertices of is an -geodominating set of if every vertex of is either on an geodesic or on an geodesic for some element in . The minimum cardinality of an -geodominating set of is defined as the -geodomination number of and is denoted by or . An -geodominating set of cardinality is called a -set of [15].
It is clear that -geodominating
set of is obtained by fixing a
vertex in and -geodominating set of is obtained by fixing an edge in . Based on these concepts we defined a
new parameter called independent fixed geodomination number (or
independent fixed geodetic number) in [16]. Let be
an independent set of a connected graph of order atleast . Let be a subset of . If each vertex in is on an geodesic for some and , then is an -fixed geodetic set of . The –fixed geodetic number of is the minimum cardinality of an -fixed geodetic set of . The independent fixed geodetic
number of is ,
where the minimum is taken over all independent sets in . An independent fixed geodetic set of
cardinality is described
as -set of . We too further proceed to infer about
connectedness of an -fixed
geodetic set of , where is an independent set in a connected
graph . In the computation of
independent fixed connected geodetic number, the succeeding theorems
will be employed.
Theorem 2. [16] Let be a
connected graph. Then
if and only if there is an independent set and its eccentric vertex such that every vertex of is on an geodesic for some .
Definition 1. Let be an independent set of a connected
graph of order atleast 2. An
-fixed connected geodetic
set of is an -fixed geodetic set such that is
connected. The -fixed
connected geodetic number of is the minimum cardinality of an -fixed connected geodetic set of . The independent fixed connected
geodetic number of
is defined as ,
where the minimum is taken over all independent sets in . An independent fixed connected
geodetic set of cardinality is called a -set of G.
Example 1. Consider a graph as shown in Figure 1. The Table 1 gives
the independent sets , their
corresponding minimum -fixed
geodetic sets and minimum -fixed
connected geodetic sets, the -fixed geodetic numbers and the -fixed connected geodetic numbers . Then the independent fixed
geodetic number of is
and the independent fixed connected geodetic number of is .
Figure 1. GS.
Table 2.1
Independent
Minimum -fixed
-fixed
Minimum -fixed
-fixed
set
geodetic set
geodetic
connected geodetic set
connected
number
geodetic
number
Theorem 4. In a connected graph , .
Proof. For any independent set in a connected graph , every -fixed connected geodetic set is an
-fixed geodetic set of , we have . Then by Theorem
1, we
have .
We intend to characterize the graphs that realize the bounds in
Theorem 4. For that we use the following
definition.
Definition 2. [16] Let be a
connected graph of order atleast 2. Let and . The distance between the set
and the vertex is . The eccentricity of the set is . An eccentric vertex of is a vertex of with .
Theorem 5. Let be a connected graph of order atleast
2. Then if and only
if there is an independent set
and its eccentric vertex such
that every vertex of is on an
geodesic for some .
The following theorem gives for some standard graphs.
Theorem 6.
If or , then .
If , then is 1 or 2 according as is even or odd.
If
with is neither a complete graph
nor a star, then
or according as
or .
In view of Theorem 4, we proceed to characterize graphs with .
Theorem 7. Let be a connected graph of order . Then if and only if .
Proof. Let . Claim that .
If not, then has a diametral path
of length
. It is clear that and are non-cut vertices of . If or is an end vertex of , then is an independent set of
and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. If both and
are non-end vertices of , then there exist atleast two paths in . If atleast one vertex other than and is common for any two paths in , then is an independent set of
and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. Suppose that there exist two paths, say and , has no common vertices other than
and . Let be an adjacent vertex of in . If is a cut vertex of , then let be a component of with not in . Let be a vertex farthest away from in . Then it is clear that is an independent set of
and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. If is not a cut
vertex and is not a
vertex-cut of , then is an independent set of and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. If is not a cut
vertex and is a
vertex-cut of , then there exists
another path, say , of length atleast 2. Let be an adjacent vertex of on . If is a cut vertex of , then let be a component of with not in . Let be a vertex farthest away from in . Then it is clear that is an independent set of
and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. If is not a cut
vertex and is not a
vertex-cut of , then is an independent set of
and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. If is not a cut
vertex and is a
vertex-cut of , then is an independent set of and is an -fixed connected geodetic set of . Hence , but this leads to
a contradiction. Converse is clear from Theorems 3 and 4.
Now we proceed to characterize graphs with . For that we require the
definition of a special graph .
Definition 3. The graph is
obtained from two complete graphs and a path by
joining every vertex in with an
end vertex of and joining every
vertex in with the other end
vertex of .
Theorem 8. Let be a connected graph of order . Then if and only if is either or ( and ).
Proof. Let . If , then by
Theorems 7 and 6 we
conclude that . If , then is either or . If , then by Theorem 7, , but this leads to a
contradiction. If , then has
two simplicial vertices, say and
. It is obvious that is an independent set of and is the minimum -fixed connected geodetic set of . Hence , but this leads to a
contradiction. If or , then
by Theorem 6, , but this leads to a contradiction.
Now, let . Since , by Theorem 7, we
have . First, we show
that every vertex of is either a
cut vertex or a simplicial vertex. Incase there exists a vertex, say
, in which is neither a cut vertex nor a
simplicial vertex, then lies on a
cycle in . Let be a largest chordless cycle containing
the vertex in . We consider three cases. Case (1) Length of the cycle is more than 3 and degree of is 2.
Let and be the adjacent vertices of on . Since is a chordless cycle, and are non-simplicial vertices of . Subcase (1). Let be an independent set of . If is an even cycle, then and is on an geodesic, where is an eccentric vertex of on ; and if is an odd cycle, then is on an geodesic and is on an geodesic, where and are the eccentric vertices of on . Then it is clear that is an -fixed geodetic set of . Since , is
connected. Hence is an -fixed connected geodetic set of and so , but this leads to a
contradiction. Subcase (2) and . Since
, is a connected graph. If and are cut vertices of , then let and be the components of and , respectively, with the vertex not in both and . Let be a vertex farthest away from in and let be a vertex farthest away from in . Then it is vivid that is an independent set
of and is an -fixed connected geodetic set of . Hence , but this leads to a
contradiction.
If and are non-cut vertices of and is a vertex-cut of , then and are the consecutive vertices of another
chordless cycle, say , in
. Let be an adjacent vertex of
on . If is a non-cut vertex of , then is an independent set of and is an -fixed connected geodetic set of . Hence , but this leads to a
contradiction. If is a cut
vertex of , then let be a component of with the vertex not in . Let be a vertex farthest away
from in . Then is an independent set of and is
an -fixed connected geodetic set
of . Hence , but this leads to a
contradiction.
If and are non-cut vertices of and is not a vertex-cut of . It is clear that is an independent set of and is an -fixed connected geodetic set of and so , but this leads to a
contradiction. If either or is a cut vertex of , then by the arguments similar to the
above, we get a contradiction. Subcase (3) and ((or) and ). If is a cut vertex of , then let be a component of with the vertex not in . Let be a vertex farthest away from
in . Then is an independent set of
and is an -fixed connected geodetic set of and so , but this leads to a
contradiction.
If is not a cut vertex of , then clearly is an independent set of and is an -fixed connected geodetic set of . Hence , but this leads to a
contradiction. Case (2) Length of the cycle is more than 3 and degree of is more than 2. Subcase (1) and are non vertex-cuts of . Then it is clear that is an independent set of and is an -fixed connected geodetic set of . Hence , but this leads to a
contradiction. Subcase (2) and are vertex-cuts of . Then has two more cycles, say and , with an edge of and an edge of . Let be an adjacent vertex of
on and let be an adjacent vertex
of on . If is a cut vertex of , then take , where is a vertex farthest away from
in a component of with not a vertex of . Otherwise, take . Similarly, if is a cut vertex of , then take , where is a vertex farthest away
from in a component
of with not a vertex of . Otherwise, take . It is clear that is an independent set of and is an -fixed connected geodetic set of . It implies , but this leads to a
contradiction. Subcase (3) is a vertex-cut and is not a vertex-cut of ((or) is not a vertex-cut and is a vertex-cut of ). Then has one more chordless cycle, say , with an edge of . Let be an adjacent vertex of
on . If is adjacent to in and is not a cut vertex of , then is an independent set of
and is an -fixed connected geodetic set of and so , but this leads to a
contradiction. If is
adjacent to in and is a cut vertex of , then by an argument similar to Subcase
(3) of Case (1), is an independent set
of and is an
-fixed connected geodetic set of
. Hence , but this leads to a
contradiction.
Similarly, if is a
non-adjacent vertex of on and is a non-cut vertex of , then and . If is a non-adjacent vertex of and is a cut vertex of , then and , where
is a vertex farthest
away from in a component
of with not a vertex of . In both cases, is an independent set of and is an -fixed connected geodetic set of and hence , but this leads to a
contradiction. Case (3) Length of the cycle is 3.
Figure 3. G.
Then the graph given in Figure
2.3 is a subgraph of . If not,
every block of is either or and so every vertex of is either a cut vertex or a simplicial
vertex, which is a contradiction to our assumption. Here we take the
vertex as and the cycle in as and continue the procedure exactly
similar to Case (1) and Case (2), we get , but this leads to a
contradiction.
Hence in all the three cases we get a contradiction and so every
vertex in is either a cut vertex
or a simplicial vertex. Since , by Theorem 7, is a non-complete graph. Hence has atleast one cut vertex. Let be the set of
all cut vertices of . We consider
two cases. Case (1) has
exactly one cut vertex, say .
Let
be the components of . If
, then , where , is an
independent set of and is an -fixed connected geodetic set of and so , but this leads to a
contradiction. Hence . Now claim
that each component has atleast two vertices. If has exactly one vertex, say , then , where , is an independent set of . It is clear that is an -fixed connected geodetic set of and so , but this leads to a
contradiction. Hence has exactly
one cut vertex and two components with each component having atleast two
vertices. Hence ( and ). Case (2) has two
or more cut vertices.
Let be the set of all cut vertices of degree in . If , then is a path and so by Theorem 6(i),
, but this
leads to a contradiction. Now, let and let , where is a simplicial vertex of in the component of . If , then clearly is an
independent set of and is an -fixed connected geodetic set of and so , but this leads to a
contradiction. If , then clearly
is an independent set of and is an -fixed connected geodetic set of . Hence , but this leads to a
contradiction. Hence . If has 3 or more components, then is an independent set of and is an -fixed connected geodetic set of and so , but this leads to a
contradiction. Hence has
exactly 2 components. Since every vertex of is either a cut vertex or a simplicial
vertex, the components of are
complete graphs. Since , the two
components of are complete
graphs with atleast two vertices and is a path.
Hence ( and ).
Conversely, let or and . If , then by Theorem 6(i), . If , then
every vertex in is a
simplicial vertex of . It is clear
that any independent set of contains atmost one vertex from each
and . Also, every -fixed connected geodetic set of contains atleast vertices from and atleast vertices from . Since , and , every vertex of is an element of any -fixed connected geodetic set of . Hence is an -fixed connected geodetic set of and so . Let , where and , and let . It is clear that is an independent set of and is an -fixed connected geodetic set of and so .
Based on Theorem 4, we have the following realization
result.
Theorem 9. For any three positive integers and with , it is possible to identify a connected graph
of order with and .
Proof. We consider two cases. Case (1). Let be a path of order and let be the complete graph of order
. Let be the graph obtained from and by joining an end vertex of with every vertex of . The resultant graph is shown in Figure 4 and its order is
.
Figure 4. G.
It is clear that any independent set of contains atmost one vertex from the
complete graph . Also,
atleast vertices in lie in every -fixed geodetic set of , where is an independent set of . Hence . Let and , where . Since , is an independent set of and it is clear that every vertex of
is on a geodesic for any . Hence is an -fixed geodesic set of and so . Also,
since is connected, we have . Case (2). Let be a path of order
. Let and be two complete graphs of orders
and , respectively. Let be the graph obtained from , and , by joining the vertex of with every vertex of and joining the end vertex of with every vertex of . The resultant graph is shown in Figure 5 and its order is
.
Figure 5. G.
It is clear that any independent set of contains atmost one vertex from each
complete graphs and . Also, atleast one vertex in and atleast vertices in lie in every -fixed geodetic set of , where is any independent set of . Hence . Let and let , where and .
Since , is an independent set of and it is clear that every vertex of
lies on a geodesic for some in and in . Hence is an -fixed geodetic set of and so . But
is
not connected and so . Since every -fixed
geodetic set of contains atleast
one vertex from each complete graphs and , every -fixed connected geodetic set of contains the cut vertices .
Let . It is clear that is a minimum -fixed connected geodetic set of and so .
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 can also be prescribed.
Theorem 10. For any three positive integers and with , a connected graph can be identified with radius , diameter and the independent fixed connected
geodetic number .
Proof. If , then
or . If , then by Theorem 7, has the desired property. Now,
let . Let be the graph obtained from the complete
graphs and by merging a vertex of , say , and a vertex of . Then has radius 1, diameter 2 and is shown
in Figure 6.
Figure 6. G.
Let . If is any independent set of , then atleast vertices in lie in every -fixed connected geodetic set of and so . Let , where in , and let . Clearly, is an independent set of and is the minimum -fixed connected geodetic set of and so .
Now, let . We construct a
graph which meets our
requirement. Case (1). Let
be the complete graph with
vertex set and let
be the even cycle with vertex set . Let
be the graph obtained from and by merging the edge in and in . The resultant graph is shown in Figure 7.
Figure 7. G.
It can be easily verified that for any vertex and so . Also, is the set of
all simplicial vertices of and
any independent set of contains
atmost one element in . If is any independent set of , then atleast vertices in lie in every -fixed connected geodetic set of and so . Also, it is clear
that all the vertices of are
not on any geodesic for some
and . Hence . Let and let .
Clearly, every vertex of is
on a geodesic and so
is an -fixed geodetic set of . Also, is
connected and so . Case (2). Let be the
complete graph with the vertex set ,
let be a path with the
vertex set and
let be the cycle with the
vertex set . Let
be the graph obtained from and by joining every vertex of with the vertex in , and joining the vertex of with the vertex in . The graph is shown in Figure 8.
Figure 8. G.
It can be easily verified that , and
. Hence and . It is clear that is the set of all simplicial
vertices of . Then by an argument
similar to Case (1) of Theorem 2.10, we have is an independent set
of and is a minimum -fixed connected geodetic set of . Hence .
3. Connected Geo-independent
Number
Definition 4. The minimum (maximum) independent
set required to form a -set
of is called a connected
geo-independent set (upper connected geo-independent set)
of . The cardinality of a
connected geo-independent set (upper connected geo-independent set) of
is called the connected
geo-independent number (upper connected geo-independent
number) of and is denoted by
().
Example 2. For the graph given in Figure 2.1, we have . From Table 1, it can be easily
seen that , and are the connected
geo-independent sets of , and are the upper
connected geo-independent sets of . Hence and .
The following result gives the connected geo-independent numbers and
the upper connected geo-independent numbers of certain special classes
of graphs.
Result 11.
If , then and .
If ,
then and .
If , then .
If , then and or
according as is odd or even.
If , then and .
If , then and .
The following observation is an easy consequence of some of the
previous results.
Observation 12. For any connected graph of order ,
, where is the independence number of
.
and .
Theorem 13. Let be a connected graph of order . Then
if and only if
for some vertex
in .
if and
only if .
if and
only if .
Proof. (i) Let . Then there exists a vertex,
say , in with and , a minimum -fixed connected geodetic set of . Hence every vertex of is on an geodesic for some and is
connected, and so is a
minimum connected -geodominating
set of . Hence .
Converse is clear from the respective definitions.
(ii) Let . If , then we get the required result. So,
let . Since the connected
graph has independent vertices, all the
independent vertices are end vertices of . Hence is a star. Then by Result 11(ii),
, but this leads to a
contradiction. Conversely, let . Then by Result 11(iii), .
(iii) The result follows from the proof of (ii) and Result 1(ii).
Problem 14. Characterize graphs for which .
The following theorem gives a realization result.
Theorem 15. For any four positive integers and with , a connected graph can be identified with , , and .
Proof.Case (1). Let and be the
complete graphs with vertex sets and
respectively. Let the graph be
constructed using and by (i) joining the vertices and in with every vertex of and (ii) joining the vertex
in with every vertex of . The resultant graph
is shown in Figure 9.
Figure 9. G.
Every independent set of
contains atmost one vertex from . Then atleast vertices in the complete graph lie on every -fixed geodetic set of , where is an independent set of . Hence . Let and
let for any
. Clearly
is an independent set of and every vertex of is on an geodesic for any and is
connected. Hence is an -fixed connected geodetic set of and so . Also,
it is clear that, for any independent set of , every minimum -fixed connected geodetic set contains
vertices from .
Let be any independent set of
with , a -fixed connected geodetic set of and . Hence
. Now claim
that . If
not, let and
. Since is an end vertex of ,
is not an internal vertex of any geodesic and so , which is a contradiction to
. Hence . Also, since
and , exactly one vertex
in , say , belongs to . Since is adjacent to , . Then either or but not both
belongs to . Subcase (1). Then every vertex of is on an geodesic for any and so is an
independent set of with , a -fixed connected geodetic set of and . Subcase (2). If all the vertices , then is not an internal vertex of any geodesic for any and . Hence , which is a contradiction to
. If atleast
one vertex , then is not an internal vertex of any
geodesic for any and . Hence , which is a contradiction
to . Thus , where
, is the independent
set of with , the -fixed connected geodetic set of and . Hence .
Claim . It can be easily
verified that , where , is a maximum independent set of and so . Case (2).
Let and be the complete graphs with vertex
sets and ,
respectively. Let be the graph
obtained from
and by (i) joining the vertices
and in with every vertex of , (ii) joining the vertex in with every vertex of , (iii) joining the vertices
and in with every vertex of , and (iv) joining the vertex
in with every vertex of . The resultant graph is shown in Figure 10.
Figure 10. G.
By an argument similar to Case (1), for any independent set of , every minimum -fixed connected geodetic set contains
atleast vertices from . Let and let
. Then clearly
is an independent set of and is an -fixed connected geodetic set of and so .
Next, prove that . By
an argument similar to Case (1), to form an -fixed connected geodetic set with , we need all the
vertices in , the vertex
in and exactly one vertex in for . Hence . As in the above paragraph,
is an
independent set of and is the -fixed connected geodetic set with . Hence is a connected geo-independent set of
and so .
Claim that . Let
be any independent set of with , a -fixed connected geodetic set of and . Then by
an argument similar to Case (1), for any , and , where . Let .
Then clearly is a maximum
independent set of with , the -fixed connected geodetic set of and . Hence .
It can be easily verified that , where , is a maximum independent set of and so .
Conflict of
Interest
The authors declare no conflict of interests.
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