1. Introduction
A pebbling move is defined as the removal of two pebbles from one
vertex, followed by the placement of one pebble on an adjacent vertex,
while the other pebble is eliminated. The pebbling number of a vertex
in a graph is the smallest positive integer such that, for every distribution
of pebbles on the vertices
of , one pebble can be moved to
through a sequence of pebbling
moves. The pebbling number is
the maximum of over all
vertices of . For more information
on graph pebbling, interested readers can refer to [5].
The pebbling problem focuses on ensuring that one pebble can reach
any specified target vertex. This problem can be interpreted as the
transmission of information from one specific location to a target
location. However, a natural question arises: how can information reach
all locations of a graph simultaneously within a reasonable time? In
pebbling terminology, this requires placing one pebble on all the
vertices of the graph
simultaneously. For example, in an Adhoc Network, if there is congestion
in the network, the information must reach neighboring locations
immediately. This question is addressed by a variation of pebbling,
called the cover pebbling number.
The cover pebbling number of a graph is the minimum number of
pebbles required such that, for any configuration of pebbles on the vertices of
, each vertex will have at least
one pebble after a series of pebbling moves. At the end of this process,
no vertex is left uncovered. Crull et al. [3] introduced the concept of cover pebbling,
determining the cover pebbling number for complete graphs, paths, and
trees. In [6], Hulbert and
Munyan determined the cover pebbling number of the -cube. Later, in 2010, Subido and
Aniversario [10] expanded upon
the work of Crull et al. by determining the cover pebbling number of
graphs via a key vertex. Furthermore, Vuong and Wyckoff [11] introduced the stacking
conjecture, where all pebbles for cover pebbling are placed on a single
vertex. For any connected graph ,
by the stacking theorem, . Any vertex
satisfying this equation is a key
vertex of the graph. Selecting a suitable key vertex is particularly
challenging for random graphs, graphs with a large number of vertices,
and graphs with varying diameters. Specifically, the intuition of
obtaining the cover pebbling number of graphs by distributing pebbles
over the vertices of is
equivalent to finding the cover pebbling number by stacking all the
pebbles at a suitable key vertex.
Santhakumaran et al. [9]
introduced the concept of monophonic distance in graphs. The monophonic
distance between and , denoted as , is the length of the longest
– monophonic path in . For any two vertices and in a connected graph , a –
path is a monophonic path if it contains no chords [9]. (A chord is a line segment
connecting two points on a curve.) Lourdusamy et al. [7] defined the
monophonic pebbling number using monophonic paths. The monophonic
pebbling number, , of a
connected graph , is the smallest
positive integer such that any
distribution of pebbles on allows one pebble to be moved to any
specified vertex using monophonic paths via a sequence of pebbling
moves.
Building on these definitions, we introduce the concept of a
monophonic cover pebbling number. The monophonic cover pebbling number,
, is the minimum
number of pebbles required to cover all the vertices of with at least one pebble on each vertex
after a series of pebbling transformations using monophonic paths. While
the pebbling number of a graph is determined using the shortest distance
in a graph , the monophonic
pebbling number is determined using monophonic distance. These two
concepts play a crucial role in applications such as the supply of goods
and transportation problems. When geodesic paths are unavailable,
monophonic paths can serve as alternatives. The choice of paths directly
impacts the cost of goods. Similarly, these concepts are applicable in
network information transmission from one node to another. The
monophonic cover pebbling number ensures the equitable distribution of
goods across all customers using monophonic paths. For basic
terminologies in graph theory, readers can refer to [1] and [2].
In this paper, we determine the monophonic cover pebbling number for
various graphs, including the gear graph, sunflower planar graph, sun
graph, closed sun graph, tadpole graph, lollipop graph, double star-path
graph, and a class of fuses.
Notation 1.1.
The notation stands for the
gear graph, which is taken from [4]. The notation stands for the sunflower planar
graph, also taken from [4]. The notation stands for the sun graph, and stands for the closed sun
graph, both referenced in [4]. The notation stands for the tadpole graph, which is taken from
[12].
Theorem 1.2.
[8] For the path , is .
Definition 1.3.
[10] Let . Then is called a key or source vertex if
is maximum.
Notation 1.4.
Throughout this article, we denote:
as the source
vertex.
as the monophonic path,
and contains the vertices
that are not on .
as the monophonic cover
pebbling number.
as the monophonic
distance.
as the neighborhood
of .
2. of families of network graphs
Theorem 2.1.
For the gear graph , is .
Proof. Let and , where and .
Without loss of generality, let
be odd. Let . To cover the vertices
it
will cost pebbles; to cover it
will cost pebbles; to cover
it will cost pebbles; to cover it will cost pebbles, and there is no pebble left to
cover . Thus,
Now we prove .
Case 1: Let .
From Table 1, to cover the
vertices , we require pebbles; to cover we
require pebbles; to cover we need pebbles; to cover we need pebbles, and to cover we need pebble. Thus, in this case, we use
pebbles.
Table 1 Monophonic distance from , , to
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Case 2: Let .
From Table 1, to cover the
vertices , we need pebbles; to cover we require
pebbles; to cover we need pebbles; to cover we need pebble. Thus, in this case, we use
pebbles.
Case 3: Let .
From Table 1, to cover the
vertices , we
need pebbles; to cover , we need pebbles; to cover , we need pebble. Thus, here we use
pebbles. 
Theorem 2.2.
The of the sunflower planar
graph is:
Proof. Let and where and . The center vertex
has degree , vertices have degree 5, and
vertices have
degree 2.
Case 1: When is even.
Table 2 Monophonic distance from , , to (even )
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Subcase 1.1: .
From Table 2,
to cover it will cost
pebbles; to cover it will cost pebbles; to cover it will cost pebbles; to cover it will cost pebbles; to cover it will cost pebble. Thus, in this case to cover all
the vertices we used
Subcase 1.2: .
From Table 2,
to cover and it requires pebbles; to cover it requires pebbles; to cover , it requires pebbles; to cover
and , it requires pebbles; to cover , it requires pebbles; to cover , it requires pebble. Thus, in this case we used
Subcase 1.3: .
From Table 2,
to cover it will cost pebbles; to cover it will cost pebbles; to cover it will cost pebble. Thus, in this case we used
Case 2: When is odd.
Table 3 Monophonic distance from to (odd n)
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Subcase 2.1: .
From Table 3,
to cover it will cost
pebbles; to cover it will cost pebbles; to cover it will cost pebbles;
to cover it will cost pebbles; to cover it will cost pebble. Thus, in this case we used
pebbles.
Subcase 2.2: Let .
From Table 3,
to cover it will cost
pebbles; to cover it
will cost pebbles; to cover
it will cost
pebbles; to cover it will
cost pebbles; to cover it will cost pebbles; to cover it will cost pebble. Thus, in this case we used
pebbles.
Subcase 2.3: Let .
From Table 3,
to cover it will cost pebbles; to cover it will cost
pebbles; to cover it will cost pebble. Thus, in this case we used
pebbles. 
Proof. Let and where and . The degree of is
and is . Let To cover the vertices , we require pebbles; to cover , we require pebbles; to cover , we require pebbles. Now there is no pebble to
cover . Hence,
Now we show
Table 4 Monophonic distance from , to
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Case 1: Let where .
Fix . From Table 4, to cover the vertices , which are at the
monophonic distance , we require
pebbles; to cover , which are at the
monophonic distance , we require
pebbles; to cover , we require pebbles; and to cover , we require pebble. Thus, in this case, we used
pebbles.
Case 2: Let where .
Fix . From Table 4, to cover the vertices , which are at
the monophonic distance , we
require pebbles; to cover
,
which are adjacent to , we
require pebbles; to cover
, we require pebble. Thus, in this case, we used

Proof. Let and , where and . The
degree of is , and the degree of is .
Case 1: When is even.
Let . To cover the
vertices , it requires
pebbles; to cover the vertices , it requires pebbles, and there is no pebble left to cover . Thus,
Now we show .
Table 5 Monophonic distance from and to
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Subcase 1.1: Let .
From Table 5, to
cover , it costs pebbles; to cover , which are at
monophonic distances , it costs pebbles; to cover , it costs pebble. Thus, in this subcase, we used
pebbles.
Subcase 1.2: Let .
From Table 5, to
cover , it costs pebbles; to cover , it costs pebbles; to cover , which are at
monophonic distances , it costs pebbles; to cover , it costs pebble. Thus, in this subcase, we used
Case 2: When is odd.
Let To cover the vertices , it
requires pebbles; to cover
the vertices , it requires pebbles,
and there is no pebble to cover . Thus,
Now we prove
Table 6 Monophonic distance from and to
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Subcase 2.1: Let .
From Table 6, to
cover , it will cost pebbles; to cover , which are at
the monophonic distances , it will cost pebbles;
to cover , it will cost pebble. Thus, in this case, we used
pebbles.
Subcase 2.2: Let .
From Table 6, to
cover , it will cost pebbles; to cover , it will cost pebbles; to cover , which are at
the monophonic distances , it will cost pebbles; to cover , it will cost pebble. Thus, in this case, we used
pebbles. 
Proof. Let and where and . Consider a bridge between and .
Case 1: When is even.
Let To cover the vertices we need
pebbles; to cover we need pebbles; to cover we need pebbles; to cover we need pebbles but we have
pebbles.
Thus, there must be a vertex without cover. Hence, Now we prove
Subcase 1.1: Let .
To cover the vertices , it will cost pebbles; to cover it will cost pebbles; to cover it will cost pebbles; to cover we need pebbles. Thus, we
used pebbles.
Subcase 1.2: Let .
To cover the vertices , it will cost pebbles; to cover the vertices
it will cost pebbles; to cover the vertices it will cost pebbles; to cover it will cost pebbles. Thus, we used
pebbles.
Case 2: When is odd.
Let To cover the vertices , we need pebbles; to cover , we need pebbles; to cover , we need Thus, in this case: The equality holds when or . 
Proof. Let where vertices form a complete graph and
vertices form a path of order
. Without loss of generality,
consider a bridge between and
. Let To
cover the vertices , it will cost pebbles; to cover the
vertices , it will cost pebbles, and there is no pebble left to cover . Thus,
Now we prove
Case 1: Let .
Let the monophonic path of length . By Theorem 1.2, to cover , we require pebbles; to cover the
vertices ,
which are at a monophonic distance of , we need pebbles. Thus, to cover all
the vertices, it will cost pebbles.
Case 2: Let .
Consider the monophonic path of length . By Theorem 1.2, to cover , we require pebbles; to cover the
vertices ,
which are adjacent to , it will
cost pebbles. Thus, to cover
all the vertices from , it will
cost pebbles.
By symmetry, the same can be proven for the vertices .
We observe that if the source vertices are , then
the monophonic cover pebbling number of these vertices will lie between
Case 1 and Case 2. 
Theorem 2.7.
For the double star-path graph ,
Proof. Let and where , , and . Let To cover the vertices , it will cost pebbles; to cover , it will cost pebbles; to cover , it will cost pebbles, but we have pebbles. Thus, there will be a
vertex without cover. Hence,
Now we prove
Case 1: Let .
Let us consider the monophonic path of length . By Theorem 1.2, to cover , we require pebbles; to cover , we require pebbles; to cover , which are at
monophonic distance , we need
pebbles. Thus, in this case,
we used pebbles. By symmetry, the
same can be proven for the vertices .
Case 2: Let .
Let us consider the monophonic path of length . By Theorem 1.2, to cover , we require pebbles; to cover , we require pebbles; to cover , which are at
monophonic distance , we need
pebbles. Thus, in this case,
we used pebbles. Since .
Case 3: Let , where .
There exist two monophonic paths. Let of length and
of length . By Theorem 1.2, to cover , we need pebbles, and to cover , we need pebbles. Now to cover , we need pebbles; to cover , we need pebbles. Thus, to cover all
the vertices, we require pebbles. Similarly, we can prove for
and . 
Theorem 2.8.
For the class of fuses ,
Proof. Let and
where , , and . Consider the monophonic
path of length . Let
To cover the vertices , we
require pebbles. By
Theorem 1.2, to cover the
vertices of , we require pebbles, but we have only
pebbles. Thus, there
will be a vertex without cover. Hence,
Now we prove
Case 1: Let .
Consider the path . By
Theorem 1.2, to cover , we require pebbles. To cover the vertices , which
are at the monophonic distance ,
we require pebbles.
Thus, with a configuration of pebbles, we can cover all the vertices. Similarly,
we can prove for the vertices .
Case 2: Let , where .
There exist two different monophonic paths: of length
and of length . By
Theorem 1.2, to cover , we require pebbles; to cover , we require pebbles; and to cover the
vertices , we require pebbles. Thus, to cover
all the vertices, we require 