Let be any graph. If there exists an injection , such that is prime for every , then we say is a prime distance graph (PDG). The problem of characterizing the family of all prime distance graphs (PDGs) with chromatic number 3 or 4 is challenging. In the fourth part of this series of articles, we determined which fans are PDGs and which wheels are PDGs. In addition, we showed: (1) a chain of mutually isomorphic PDGs is a PDG, and (2) the Cartesian product of a PDG and a path is a PDG. In this part of the series, we improve (1) by showing that there exists a chain of arbitrary PDGs which is a PDG. We also show that the following graphs are PDGs: (a) any graph with at most three cycles, (b) the one-point union of cycles, and (c) a family of graphs consisting of paths with common end vertices.
Keywords: Chromatic number, Distance graph, Prime distance graph, Prime distance labeling
1. Introduction
By a graph we mean a pair
, where is any arbitrary set and
is a subset of the set of all
2-element subsets of . The
elements of and are called, respectively, the vertices
and edges of . If then is said to be finite. The graphs
considered here are simple with finite or countably infinite vertex
sets. By a distance graph , we
mean a graph, whose vertex set is the set of integers and two vertices and form an edge if and only if If , the set of primes, then we call
, the prime distance graph.
Given a graph with vertex set
and edge set and a set of colors, an ordinary coloring is an
assignment of elements of to the
elements of in any order. A
coloring is
called proper if for all Such a coloring is called a
proper -coloring if . The least for which has a proper -coloring is called the chromatic
number of and is denoted by
. One can consult
[1-15] for more on
this topic.
Eggleton, Erdős, and Skilton attempted the task of finding the
chromatic number of the prime distance graph where . They found that [9]. The
problem of computing
where is still open.
Voigt and Walther in [16],
Kemnitz and Kolberg in [2],
and Yegnanarayanan et al. [17-21] have contributed more on this
topic.
Observe that where is a 4-coloring of for any . This is because if , then
and . Thus and . As is a monotonic
function, if
then . So, if then the tough task is to classify as per its chromatic number.
We say a graph is of Class
if and only if . It is obvious that is the only distance graph
that is Class 1, and is Class
2 whenever is a singleton set. It
is already known from the results available in the literature that if
2 then is Class 2 if and only if 2 . Similarly, if 2 and 3 , then is Class 3. If with then is Class 4 [6]. Also, if with
, then is Class 3. If or is contained in a subset , then or 4. It is an unsolved
problem to classify these sets
whether they belong to Class 3 or 4 (see [6,9,22,23]).
Let be any graph. If
there exists an injection such that
is prime for every then
we say a prime distance
labeling of and is a prime distance graph. The
value of is referred to
as the induced label of the edge For the sake of brevity, if this
induced label is prime, we say the edge is prime.
It is known that a bipartite graph is a prime distance graph [24]. So every graph with is a prime distance graph. In
particular, trees are prime distance graphs. It is still open to find a
characterization for graphs with or . In this paper we demonstrate the
existence of several new collections of families of prime distance
graphs with or .
Laison, Starr, and Walker showed that not only is every bipartite
graph a prime distance graph, but
a prime distance labeling can be
found such that for any
particular vertex and
for all [24]. They showed the same is true for
cycles.
Theorem 1 (Laison, et al., [24]). Let be a bipartite graph or a cycle and
be any vertex. There
exists a prime distance labeling such that .
Next we share some observations that are used in the constructions
throughout this paper. The following three observations are from [25].
Observation 2. Let be any graph. An injection is a prime distance
labeling of if and only if is a prime distance labeling of
Observation 3. If is a prime distance graph and is a subgraph of , then is also a prime distance
graph.
Observation 4. Let be any graph, an injection, and
any real number. The function
is a prime distance labeling of
if and only if is a prime distance labeling of .
We call the operation from the previous observation lifting by
Corollary 1. If and are both prime distance graphs, then
the disjoint union is
also a prime distance graph.
Proof. Let and be prime distance labelings of and respectively and be the maximum label of . Label the vertices by Since
is injective, it is a prime distance labeling of by Observation 4.
Let be
the path on vertices and be the cycle
on vertices.
2. New Results
Lemma 1. Let be a prime distance graph. Then a graph
obtained from by adding a pendant edge to is also a prime distance
graph.
Proof. Let and be the
pendant edge of where Suppose is a prime distance labeling of and is the smallest prime such that Then a
prime distance labeling of is
since
is clearly an injection and the induced edge labels are prime.
Notice that Lemma 1 can be repeatedly applied to make
larger graphs from a given prime distance graph Let be such a graph and and such that . Then it is possible to
construct the prime distance labeling from repeatedly applying Lemma 1 such that for every This will be important in the
proof of Theorem 6.
2.1 Graphs Containing two or
Less Cycles
A graph which contains no cycles is necessarily bipartite, and thus a
prime distance graph by Theorem 1. A graph is unicyclic if it contains
exactly one cycle. If the cycle contained by is even, then belongs to Class . Otherwise, belongs to Class . Our next result shows all unicyclic
graphs are prime distance graphs.
Theorem 5. Every unicyclic graph is a prime distance graph. Furthermore,
for any vertex lying on the cycle
of , there exists a prime distance
labeling such that
Proof. Let be a
unicyclic graph that contains
connected components. If , the
vertices of on the cycle may be
labeled by Theorem 1. The result now
follows by inductively applying Lemma 1.
On the other hand, if
label the connected component containing the cycle as in the previous
case. The remaining components constitute a bipartite subgraph of Therefore, admits a distance prime labeling which may be suitably lifted
by Observation 4 to obtain the result.
A graph is bicyclic
if it contains exactly two cycles. If both cycles contained in are even, then belongs to Class . Otherwise, belongs to Class .
Theorem 6. Every bicyclic graph is a prime
distance graph.
Proof. Let be a
bicyclic graph containing cycles and . We establish two cases based on the
structure of . and are not in the same connected
component.
In this case, is the disjoint
union of two unicyclic graphs. Therefore, the prime distance labeling
exists by Theorem 5 and Corollary 1.
and belong to the same connected
component.
We may assume is the -point intersection of two unicyclic
subgraphs and with common vertex such that is contained in , is contained in , , and is contained in . By Theorem 5 and observation 4 a nonnegative prime distance
labeling of exists such that for all Lifting by gives a prime distance labeling of such that and for all
Let be a nonnegative prime
distance labeling of such that
Such a labeling exists by
Theorem 5. We combine the labelings and to produce a labeling as follows. Let It is clear that is injective since for all , for all , and and Further, every edge is
prime since and are prime distance labelings.
2.2 Graphs Containing more
than Two Cycles
Figure 1 The One Point Union of , and
The graph of edge-disjoint
cycles glued
together at a common vertex is called a one point union of
cycles. See Figure 1 for
an example. Before investigating this family of graphs, we state a
conjecture of de Polignac.
Conjecture 1 (de Polignac, 1849). For every
positive integer , there are
infinitely many primes that differ by .
Notice that the famous Twin Prime Conjecture corresponds to the case
in de Polignac’s Conjecture.
Laison, Starr, and Walker showed that the one point union of -cycles is a prime distance graph if and
only if the Twin Prime Conjecture is true [24]. We prove some analogous but broader
results using de Polignac’s Conjecture next.
Theorem 7. If de Polignac’s Conjecture is true,
then the one point union of cycles is a prime
distance graph for any
and
Proof. Let and
be mutually distinct
primes for such that
. If de
Polignac’s Conjecture is true, we may find such primes. Label the
vertices of each cycle by
consecutively around the cycle such that the vertex labeled is common to all cycles.
The lengths of the edges in form the set Since every
member of this set is prime and the labeling described is injective, we
have completed the proof.
A graph is tricyclic if it contains exactly cycles. If a graph is tricyclic, then the Class of is if all three of the cycles are even,
and is otherwise. For any , a -chorded -cycle is a tricyclic graph with
vertex set
and edge set for some The edge
is called the chord.
For practical reasons, the next theorem only requires the truth of a
much weaker version of Polignac’s conjecture: the existence of
a prime gap of size for
sufficiently large primes.
Theorem 8. If de Polignac’s Conjecture is true,
then every -chorded -cycle is a prime distance
graph.
Proof. Label the cycle using any prime
distance labeling of such that and (see Theorem 3 in [24] for example). Let and
label the cycle
in order by for any
such that both and are both prime. Such a pair can
be found if Conjecture 1 is true. Since is a prime distance labeling, we need
only check that the edges on the cycle are all prime. Since
these labels all belong to the set we have proven the
theorem.
Theorem 9. Every disconnected tricyclic graph is
a prime distance graph.
Proof. If is a
disconnected tricyclic graph, then it is either the disjoint union of
three unicyclic graphs or a bicyclic graph with a unicyclic graph.
Therefore, the result follows from Theorems 5 and 6 and Corollary 1.
Theorem 10. If de Polignac’s Conjecture is true,
then every connected tricyclic graph is a prime distance graph.
Proof. Let be a
connected tricyclic graph. Then
either contains a -chorded cycle
or the one point union of three cycles. Therefore, the labeling can be
carried out by Theorems 8 or 7,
respectively. Labeling the remaining vertices in by Lemma 1 concludes
the proof.
2.2.1 Graphs containing more
than three cycles
For any set of graphs and integer , we define an -chain graph, as any edge-disjoint union of such that for every
and otherwise.
Notice that two -chains of the
same set of graphs are not
necessarily isomorphic. The authors showed that if for all where is a prime distance graph, then there
exists a corresponding -chain of
that is a prime distance graph
[25]. Here, we prove a much
stronger result.
Theorem 11. Let be any
set of prime distance graphs. Then there exists an -chain that is a prime distance
graph.
Proof. Let be a prime distance labeling of and be a vertex such that . We can make this assumption
by Observation 4. Denote by the vertex with the
largest label, Form an
-chain by gluing to for Then it is easy to see that the labeling is a prime distance labeling of
One possible generalization of the -chorded graphs can be defined as
follows. Let
with be any set of
non-decreasing integers. We define the rainbow graph as the union of
paths of the form for Since we are only
considering simple graphs, we require Simply put, is isomorphic to the
graph obtained by gluing together the left endpoints and gluing together
the right endpoints of paths. The
lengths of these paths may differ. We call the vertices and the rainbow ends. Notice that
is isomorphic to a
-chorded See Figure 2 for an example of
Figure 2 The Rainbow Graph
Theorem 12. If de Polignac’s Conjecture is true,
then every rainbow graph is a prime distance
graph.
Proof. For , let be a prime such
that is also prime. If
de Polignac’s Conjecture is true, not only does such a set of primes
exist, but we may further assume that the primes are all distinct ( may be chosen so that ). For , label the path with in
order. It is easy to check that the induced edge labels comprise the set
Since every member of this set is prime, the proof is
complete.
Let be a
set of rainbow graphs. We define
the rainbow range as the graph
obtained by chaining together the rainbows by gluing the vertex of to the vertex of for Notice that the next
result does not necessarily follow from Theorem 11.
Theorem 13. If de Polignac’s Conjecture is true,
then every rainbow range is a prime distance
graph for any .
Proof. Let and be a prime
distance labeling of as given
in the proof of Theorem 12 with and maximum label
Apply in the natural way to the subgraph
isomorphic to in
Again, use the proof of Theorem 12 to find a
prime distance labeling of
such that and for all vertices Then apply the
labeling to the subgraph
in . Continue in this way to label the
subgraphs . Since
lifting each labeling by 2 preserves the induced edge labels on each
subgraph isomorphic to for
, and introduces no
conflicts in the labels of the ends of the rainbows, we see that is a prime distance graph.
3. Conclusion
We have provided many new families of graphs with chromatic number at
least which are prime distance
graphs. One possible direction forward would be to prove Theorems 7,8,10,12 and 13 independently of de Polignac’s
Conjecture (or show that the conjecture is necessary).
Acknowledgment
This research was funded by National Board of Higher Mathematics
Department of Atomic Energy, Government of India, Mumbai, by their grant
no. 02011/10/21NBHM- (R.P)/R&D/8007/Date: 13-07-2021. J George
Barnabas and Yegnanarayanan Venkatraman gratefully acknowledges the
same.
Funding
This work was supported by National Board of Higher Mathematics,
Government of India.
Role of Funding
Support
The National Board of Higher Mathematics-NBHM, Department of Atomic
Energy, Government of India has provided scholarship support for George
Barnabas to complete his doctoral thesis work.
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