International Journal of Trend in Scientific Research and Development (IJTSRD)
Volume 4 Issue 5, August 2020 Available Online: www.ijtsrd.com e-ISSN: 2456 โ 6470
@ IJTSRD | Unique Paper ID โ IJTSRD31865 | Volume โ 4 | Issue โ 5 | July-August 2020 Page 430
On the Radio Number of Certain Classes of Circulant Graphs Kins Yenoke
PG & Research, Department of Mathematics, Loyola College, Chennai, Tamil Nadu, India
ABSTRACT Radio labelling problem is a special type of assignment problem which maximizes the number of channels in a specified bandwidth. A radio labelling of a connected graph ๐บ = (๐, ๐ธ) is an injection โ: ๐(๐บ) โ ๐ such that ๐(๐ฅ, ๐ฆ) + |๐(๐ฅ) โ ๐(๐ฆ)| โฅ 1 + ๐(๐บ)โ ๐ฅ, ๐ฆ โ ๐(๐บ), where ๐(๐บ) is the diameter of the graph ๐บ. The radio number of ๐ denoted ๐๐(โ), is the maximum number assigned to any vertex of ๐บ. The radio number of ๐ฎ, denoted ๐๐(๐บ), is the minimum value of ๐๐(โ) taken over all labellingโs โ of ๐บ. In this paper we have obtained the radio number certain classes of
circulant graphs, namely ๐บ (๐; {1,2 โฆ โ๐
2โ โ
1}) , ๐บ (๐; {1,๐
2}) , ๐บ (๐; {1,
๐
3}) and ๐บ (๐; {1,
๐
5}).
KEYWORDS: Labelling, Radio labelling, Radio number, Circulant graphs
How to cite this paper: Kins Yenoke "On the Radio Number of Certain Classes of Circulant Graphs" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-4 | Issue-5, August 2020, pp.430-434, URL: www.ijtsrd.com/papers/ijtsrd31865.pdf Copyright ยฉ 2020 by author(s) and International Journal of Trend in Scientific Research and Development Journal. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (CC BY 4.0) (http://creativecommons.org/licenses/by/4.0)
1. INTRODUCTION In the modern world, all the communications are based on the wireless mode. In the end of the 20th century most of the communication were developed based on the wireless networks. These communications are working based on the specified allocation of the electromagnetic spectrum. One of the main applications is based on the low frequency long wavelength waves called radio waves. Among this, a fixed bandwidth ranges from 88.1 MHZ to 108 MHZ was allotted for the channel assignment of FM radio stations. 88.1 MHz and finished at 108MHz This real-life minimax problem motivated Chartrand et al. [6] in the year 2001 to introduced a new labelling called the radio labelling. He defined the formal graph theoretical definition for radio labelling problem as follows: Let ๐๐๐๐(๐บ) be the diameter of a connected graph G. An injection h from the vertex set of G to N such that ๐(๐ข, ๐ฃ) +|โ(๐ข) โ โ(๐ฃ)| โฅ 1 + ๐๐๐๐(๐บ)) for every pair of vertices in G. The radio number of h, denoted by ๐๐(โ), is the maximum number assigned to any vertex of G. The radio number of G, denoted by ๐๐(๐บ), is the minimum value of ๐๐(โ) taken over all labellingโs โ of ๐บ. The radio number problem is NP-hard [9], even for graphs with diameter 2. In the past two decades plenty of research articles are published in this area and also developed new labelling problems based on the radio number. 2. An Overview of the Paper For the past 20 years, several authors studied the radio labelling problem and its variations in various networks
and graphs. Radio labelling problem is a particular case of radio K- Chromatic number [7]. In the recent years few new labellingโs were introduced by different authors based on the k value, namely, radio mean labelling, radio multiplicative labelling, radial radio labelling etc. The radio number of square cycles was determined by Liu et.al [12]. Bharati et.al [2,3] obtained the bounds for the hexagonal mesh as 3๐2 โ 3๐ + 2 + 12 โ ๐(๐ โ ๐ โ 1) โค๐โ2
๐=0
๐๐(๐บ) โค ๐(3๐2 โ 4๐ โ 1) + 3 and completely determined the radio number of graphs with small diameters. Kchikech et.al [10] studied the radio k-labelling of graphs. Fernandez et al. [8] computed the radio number for gear graph. Kins et. al. [11] investigated the radio number for mesh derived architectures and wheel extended graphs. In this paper we have investigated the radio labelling of certain classes of circulant graphs. 3. Circulant Graphs Circulant graphs have been used for several decades in the design of telecommunication networks because of their optimal fault-tolerance and routing capabilities [5]. For designing certain data alignment networks, the circulant graphs are being used. for complex memory systems [13]. Most of the earlier research concentrated on using the circulant graphs to build interconnection networks for distributed and parallel systems [2]. By using circulant graph we can adapt the performance of the network to user needs. Itโs a regular graph which includes standard such as the complete graph and the cycle.
IJTSRD31865
International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470
@ IJTSRD | Unique Paper ID โ IJTSRD31865 | Volume โ 4 | Issue โ 5 | July-August 2020 Page 431
Definition 3.1: An undirected circulant graph denoted by
๐บ(๐; ยฑ{1,2 โฆ ๐}), 1 โค ๐ โค โ๐
2โ , ๐ โฅ 3, is defined as a graph
with vertex set ๐ = {0,1 โฆ ๐ โ 1} and the edge set ๐ธ ={(๐, ๐): |๐ โ ๐| โก ๐(๐๐๐ ๐), ๐ โ {1,2 โฆ ๐}. Remark 1: In this paper for our convenience we take the vertex set ๐ as {๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐}, which is named in clockwise order.
Remark 2: It is clear that, when ๐ =๐
2, the circulant graph
๐บ (๐; ยฑ {1,2 โฆ โ๐
2โ}), become a complete graph.
Lemma 3.1: The diameter of the circulant graph
๐บ (๐; {1,2 โฆ โ๐
2โ โ 1}) is 2.
Proof: As we know that, the circulant graph
๐บ (๐; {1,2 โฆ โ๐
2โ โ 1}) is obtained from the circulant graph
๐บ (๐; ยฑ {1,2 โฆ โ๐
2โ}) by the removal of an unique one edge
from each vertex with maximum distance on the outer
cycle. Since the diameter of ๐บ (๐; ยฑ {1,2 โฆ โ๐
2โ}) is 1, it is
obvious that the diameter of the circulant graph
๐บ (๐; {1,2 โฆ โ๐
2โ โ 1}) is 2.
Theorem 3.1: The radio number of the circulant graph
๐บ (๐; {1,2 โฆ โ๐
2โ โ 1}) is given by ๐บ (๐; {1,2 โฆ โ
๐
2โ โ 1}) =
๐.
Proof: Let ๐ (๐บ (๐; {1,2 โฆ โ๐
2โ โ 1})) = {๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐}. Define
an injection โ: {๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐} โ ๐ as follows: โ(๐ฃ๐) = 2๐ โ
1, ๐ = 1,2 โฆ โ๐
2โ, โ (๐ฃ
โ๐
2โ+๐
) = 2๐, ๐ = 1,2 โฆ โ๐
2โ.
Next, we claim that the defined injection h is a valid radio labeling. Using Lemma 3.1, we must verify the radio labelling condition ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 3 โ ๐ฅ, ๐ฆ โ
๐ (๐บ (๐; {1,2 โฆ โ๐
2โ โ 1})).
Case 1: If ๐ฅ = ๐ฃ๐ and ๐ฆ = ๐ฃ๐, 1 โค ๐ โ ๐ โค โ๐
2โ, then
๐(๐ฅ, ๐ฆ) โฅ 1, โ(๐ฅ) = 2๐ โ 1 and โ(๐ฆ) = 2๐ โ 1. Hence ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + 2(๐ โ ๐) โฅ 3, since ๐ โ ๐.
Case 2: If ๐ฅ = ๐ฃโ๐
2โ+๐
and ๐ฆ = ๐ฃโ๐
2โ+๐
, 1 โค ๐ โ ๐ โค โ๐
2โ,
then ๐(๐ฅ, ๐ฆ) โฅ 1, โ(๐ฅ) = 2๐ and โ(๐ฆ) = 2๐. Hence ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + 2(๐ โ ๐) โฅ 3, since ๐ โ ๐.
Case 3: If ๐ฅ = ๐ฃ๐ and ๐ฆ = ๐ฃโ๐
2โ+๐
, 1 โค ๐ โค โ๐
2โ, 1 โค ๐ โค โ
๐
2โ,
then either ๐(๐ฅ, ๐ฆ) โฅ 1 and |โ(๐ฅ) โ โ(๐ฆ)| โฅ 2 or ๐(๐ฅ, ๐ฆ) =2 |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1. In both the cases we have ๐(๐ฅ, ๐ฆ) +|โ(๐ฅ) โ โ(๐ฆ)| โฅ 3. Thus, ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 3 for all ๐ฅ, ๐ฆ โ
๐ (๐บ (๐; {1,2 โฆ โ๐
2โ โ 1})).
Therefore, โ is a radio labelling and ๐๐(๐บ) โค ๐. Since the mapping is an injection, all the ๐ vertices of
๐บ (๐; {1,2 โฆ โ๐
2โ โ 1}) received different radio labelling.
Hence, we conclude that the radio number of
๐บ (๐; {1,2 โฆ โ๐
2โ โ 1}) is exactly ๐.
Lemma 3.2: The diameter of ๐บ (๐; {1,๐
3}) is โ
๐
6โ +1,
whenever ๐ โก 0(๐๐๐ 3). Proof: As in the proof of Lemma 3.2, with a common
difference of length ๐
3โ 1 in the outer cycle, we construct
the graph ๐บ (๐; {1,๐
3}) by joining the vertices ๐ฃ1 to ๐ฃ๐
3, ๐ฃ2 to
๐ฃ๐
2+1 โฆ. Therefore, the diameter of ๐บ (๐; {1,
๐
3}) is โ
๐
6โ +1.
Theorem 3.2: The radio number of ๐บ (๐; {1,๐
3}) , ๐ โก
0(๐๐๐3), satisfies ๐๐ (๐บ (๐; {1,๐
2})) โค
{(
๐
6+ 1) (
๐
2+ 1) + โ
๐
12โ + 1, if ๐ is even
โ๐
6โ (
๐โ1
2) +, if ๐ is odd.
Proof: We partition the vertex set ๐ = {๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐} into
four disjoint set ๐1, and ๐2, where ๐1 = {๐ฃ1, ๐ฃ2 โฆ ๐ฃโ๐
2โ} , ๐2 =
{๐ฃโ
๐
2โ+1
, ๐ฃ๐ฃโ๐2โ+2
โฆ ๐ฃ๐}. We discuss the proof for n even and
odd case separately.
Figure 1: Radio labelling of circulant graphs ๐ฎ(๐; {๐,
๐
๐})
with ๐ = ๐๐ ๐๐๐ ๐๐. Case 1: n is even
International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470
@ IJTSRD | Unique Paper ID โ IJTSRD31865 | Volume โ 4 | Issue โ 5 | July-August 2020 Page 432
Define a mapping โ: ๐ (๐บ (๐; {1,๐
3})) โ ๐ as follows:
โ(๐ฃ๐) = (๐
6+ 1) (๐ โ 1) + 1, ๐ = 1,2 โฆ
๐
2.
โ (๐ฃ๐
2+๐
) = (๐
6+ 1) (๐ โ 1) + โ
๐
12โ + 1, ๐ = 1,2 โฆ
๐
2. See
Figure 1.
Case 1.1: Suppose ๐ฅ = ๐ฃ๐ and ๐ฆ = ๐ฃ๐, 1 โค ๐ โ ๐ โค๐
2,
then ๐(๐ฅ, ๐ฆ) โฅ 1. Also, โ(๐ฅ) = (๐
6+ 1) (๐ โ 1) + 1, and
โ(๐ฆ) = (๐
6+ 1) (๐ โ 1) + 1,. Hence ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ
โ(๐ฆ)| โฅ (๐
6+ 1) (๐ โ ๐) โฅ 2 + โ
๐
6โ, since ๐ โ ๐.
Case 1.2: If ๐ฅ = ๐ฃ๐
2+๐ and ๐ฆ = ๐ฃ๐
2+๐, 1 โค ๐ โ ๐ โค
๐
2, then
|โ(๐ฅ) โ โ(๐ฆ)| = |(๐
6+ 1) (๐ โ 1) + โ
๐
12โ + 1 โ
((๐
6+ 1) (๐ โ 1) + โ
๐
12โ + 1)| โฅ (
๐
6+ 1) (๐ โ ๐) and
๐(๐ฅ, ๐ฆ) โฅ 1. Hence ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ โ๐
6โ +
2, since ๐ โ ๐.
Case 1.3: If ๐ฅ = ๐ฅ = ๐ฃ๐ and ๐ฆ = ๐ฆ = ๐ฃ๐
2+๐ , 1 โค ๐, ๐ โค
๐
2
then ๐(๐ฅ, ๐ฆ) โฅ 1 and |โ(๐ฅ) โ โ(๐ฆ)| โฅ |((๐
6+ 1) (๐ โ 1) +
1) โ ((๐ โ 1) (๐
4โ 1) + 2)| โฅ โ
๐
6โ + 1.
Therefore ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ โ๐
6โ + 2 .
Thus, the radio labelling condition is true for the case when n is even. Case 2: n is odd
Define an injection โ: ๐ (๐บ (๐; {1,๐
3})) โ ๐ as follows:
โ(๐ฃ๐) = โ๐
6โ (๐ โ 1) + 1, ๐ = 1,2 โฆ
๐+1
2.
โ (๐ฃ๐
2+๐
) = โ๐
6โ (๐ โ 1) + โ
๐
12โ + 1, ๐ = 1,2 โฆ
๐โ1
2.
Proceeding as in previous case, we can show h is a radio
labelling and that ๐๐ (๐บ (๐; {1,๐
2})) โค
{(
๐
6+ 1) (
๐
2+ 1) + โ
๐
12โ + 1, if ๐ is even
โ๐
6โ (
๐โ1
2) +, if ๐ is odd.
.
Lemma 3.3: For ๐ โก 0(๐๐๐ 4), the diameter of
๐บ (๐; {1,๐
2}) is โ
๐
4โ.
Proof: Let {๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐} be the vertices in the outer circle.
We construct the graph ๐บ (๐; {1,๐
2}) by joining the vertices
๐ฃ1 to ๐ฃ๐
2, ๐ฃ2 to ๐ฃ๐
2+1 โฆ with a common difference of length
๐
2โ 1 till the process of joining gets over. Therefore, the
maximum distance from a vertex to another vertex is at
least half of ๐
2โ 1, which is equal to โ
๐
4โ, since ๐ โก
0(๐๐๐ 4).
Theorem 3.3: The radio number of ๐บ (๐; {1,๐
2}) , ๐ โก
0(๐๐๐ 4), ๐ > 16, satisfies ๐๐ (๐บ (๐; {1,๐
2})) โค
(๐
2โ 1) (
๐
4โ 1) + 4.
Proof: We partition the vertex set ๐ = {๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐} into four disjoint set ๐1, ๐2 , ๐3 and ๐4. Let ๐1 =
{๐ฃ1, ๐ฃ2 โฆ ๐ฃ๐
4} , ๐2 = {๐ฃ๐
4+1, ๐ฃ๐
4+2 โฆ ๐ฃ๐
2}, ๐3 =
{๐ฃ๐
2+1, ๐ฃ๐
2+2 โฆ ๐ฃ3๐
2
} and ๐3 = {๐ฃ3๐
2+1, ๐ฃ3
๐
2+2 โฆ ๐ฃ๐}.
Define a mapping โ: ๐ (๐บ (๐; {1,๐
2})) โ ๐ as follows:
โ(๐ฃ2๐โ1) = (๐ โ 1) (๐
4โ 1) , ๐ = 1,2 โฆ โ
๐
8โ.
โ(๐ฃ2๐) = (โ๐
8โ + (๐ โ 1)) (
๐
4โ 1) + 1, ๐ = 1,2 โฆ โ
๐
8โ.
โ (๐ฃ๐
4+2๐โ1
) = (๐ โ 1) (๐
4โ 1) + 2, ๐ = 1,2 โฆ โ
๐
8โ.
โ (๐ฃ๐4
+2๐) = (โ
๐
8โ + (๐ โ 1)) (
๐
4โ 1) + 2, ๐ = 1,2 โฆ โ
๐
8โ.
โ (๐ฃ๐โ(2โ
๐8
โ+2๐โ3) = (
๐
4+ ๐ โ 1)(
๐
4โ 1)) + 4, ๐ = 1,2 โฆ โ
๐
8โ
โ (๐ฃ๐โ(2โ
๐8
โ+2๐โ2)) = (โ
๐
8โ + ๐ +
๐
4โ 1)(
๐
4โ 1)) + 4, ๐
= 1,2 โฆ โ๐
8โ.
โ(๐ฃ๐โ2(๐โ1)) = (๐
4+ (๐ โ 1)) (
๐
4โ 1) + 3, ๐ = 1,2 โฆ โ
๐
8โ.
โ(๐ฃ๐โ(2๐โ1)) = (โ๐
8โ +
๐
4+ (๐ โ 1)) (
๐
4โ 1) + 3, ๐ =
1,2 โฆ โ๐
8โ. See Figure 2.
Next, we verify that ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + โ๐
4โ for
all ๐ฅ, ๐ฆ โ ๐ (๐บ (๐; {1,๐
2}))
Figure 2: A circulant graph graph ๐(๐๐, {๐, ๐๐}) and its
radio labelling Case 1: Suppose ๐ฅ = ๐ฃ2๐โ1 and ๐ฆ = ๐ฃ2๐โ1๐ , 1 โค ๐ โ ๐ โค
โ๐
8โ, then the distance between them is at least 1.
Also, โ(๐ฅ) = (๐ โ 1) (๐
4โ 1) and โ(๐ฆ) = (๐ โ 1) (
๐
4โ 1).
Hence ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 +๐
4(๐ โ ๐) โฅ 1 +
โ๐
4โ , since ๐ โ ๐.
International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470
@ IJTSRD | Unique Paper ID โ IJTSRD31865 | Volume โ 4 | Issue โ 5 | July-August 2020 Page 433
Case 2: If ๐ฅ = ๐ฃ2๐ and ๐ฆ = ๐ฃ2๐, 1 โค ๐ โ ๐ โค โ๐
8โ, then
โ(๐ฅ) = (โ๐
8โ + (๐ โ 1)) (
๐
4โ 1) and โ(๐ฆ) = (โ
๐
8โ +
(๐ โ 1)) (๐
4โ 1) and ๐(๐ฅ, ๐ฆ) โฅ 1. Hence ๐(๐ฅ, ๐ฆ) +
|โ(๐ฅ) โ โ(๐ฆ)| โฅ๐
4(๐ โ ๐) โฅ 1 + โ
๐
4โ , since ๐ โ ๐.
Case 3: If ๐ฅ = ๐ฃ๐
4+2๐โ1 and ๐ฆ = ๐ฃ๐
4+2๐โ1, 1 โค ๐ โ ๐ โค โ
๐
8โ,
then ๐(๐ฅ, ๐ฆ) โฅ 1 and |โ(๐ฅ) โ โ(๐ฆ)| โฅ |(๐ โ 1) (๐
4โ 1) +
2 โ ((๐ โ 1) (๐
4โ 1) + 2)| โฅ โ
๐
4โ , since ๐ โ ๐.
Therefore ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + โ๐
4โ .
Case 4: If ๐ฅ = ๐ฃ๐
4+2๐ and ๐ฆ = ๐ฃ๐
4+2๐, 1 โค ๐ โ ๐ โค โ
๐
8โ, then
๐(๐ฅ, ๐ฆ) โฅ 1 and the modulus difference of
โ(๐ฅ)and โ(๐ฆ) is altleast โ๐
4โ.
Therefore ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + โ๐
4โ.
|โ(๐ฅ) โ โ(๐ฆ)| โฅ |(๐ โ 1) (๐
4โ 1) + 2
โ ((๐ โ 1) (๐
4โ 1) + 2)| โฅ โ
๐
4โ , since ๐
โ ๐.
Therefore ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + โ๐
4โ .
Case 5: Suppose x and y are of the form ๐ฃ
๐โ(2โ๐
8โ+2๐โ3
and
๐ฃ๐โ(2โ
๐
8โ+2๐โ3
, 1 โค ๐ โ ๐ โค โ๐
8โ, then the distance between
them is at least 1 and |โ(๐ฅ) โ โ(๐ฆ)| โฅ |((๐
4+ ๐ โ 1)(
๐
4โ
1)) + 4) โ (๐
4+ ๐ โ 1)(
๐
4โ 1)) + 4| โฅ |(
๐
16
2+ ๐
๐
4) โ
(๐
16
2+ ๐
๐
4)|.
Hence ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 1 + โ๐
4โ, since ๐ โ ๐.
Case 6: If ๐ฅ = ๐ฃ2๐โ1 and ๐ฆ = ๐ฃ๐
4+2๐, 1 โค ๐ โค โ
๐
8โ, 1 โค ๐ โค
โ๐
8โ, then ๐(๐ฅ, ๐ฆ) โฅ 3 and โ(๐ฅ) = (๐ โ 1) (
๐
4โ 1) , โ(๐ฆ) =
(โ๐
8โ + (๐ โ 1)) (
๐
4โ 1) + 2.
Therefore ๐(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ 3 + |(โ๐
8โ +
(๐ โ 1)) (๐
4โ 1) + 2 โ ((๐ โ 1) (
๐
4โ 1))| โฅ 3 +
โ๐
8โ (
๐
4โ 1) > โ
๐
4โ + 1.
Case 7: Suppose ๐ฅ = ๐ฃ
๐โ(2โ๐
8โ+2๐โ2)
and ๐ฃ๐โ(2๐โ1), 1 โค
๐, ๐ โค โ๐
8โ, then โ(๐ฅ) = (โ
๐
8โ + ๐ +
๐
4โ 1)(
๐
4โ 1)) + 4 and
โ(๐ฆ) = (โ๐
8โ +
๐
4+ (๐ โ 1)) (
๐
4โ 1) + 3.
Also ๐(๐ฅ, ๐ฆ) โฅ 2. Hence, we get(๐ฅ, ๐ฆ) + |โ(๐ฅ) โ โ(๐ฆ)| โฅ
2 +๐
4+ 1 > โ
๐
4โ + 1.
Similarly, we can prove the rest of the cases. Thus, h is a
valid radio labelling and satisfies ๐๐ (๐บ (๐; {1,๐
2})) โค
(๐
2โ 1) (
๐
4โ 1) + 4.
Lemma 3.4: If ๐ โก 0(๐๐๐ 10), then the diameter of
๐บ (๐; {1,๐
5}) is
๐
10+ 2.
Figure 3: Radio labelling of a circulant graph
๐(๐ง; {๐,๐ง
๐}) with ๐ง = ๐๐.
Proof: As the proof is similar to Lemma 3.2, we omit the proof.
Theorem 3.4: The radio number of ๐บ (๐; {1,๐
5}) , ๐ โก
0(๐๐๐ 10), satisfies ๐๐ (๐บ (๐; {1,๐
2})) โค {
๐
20(๐ + 18) +
โ๐
20โ.
Proof: We omit the proof. Figure 3, illustrates the proof of the theorem 4. Conclusion In this article we have obtained the radio number of certain classes of circulant graphs. Further the work is extended to other extensions of channel assignment problems such as radial radio number, antipodal radio mean labelling etc. References [1] K. F. Benson - M. Porter, M. Tomova โThe radio
numbers of all graphs of order ๐ and diameter ๐ โ 2โ vol.68, pp. 167-190, 2013.
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[3] Bharati Rajan, Kins Yenoke, โOn the radio number of hexagonal meshโ, Journal of Combinatorial Mathematics and Combinatorial Computing, Vol. 79, pp. 235-244, 2011.
International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470
@ IJTSRD | Unique Paper ID โ IJTSRD31865 | Volume โ 4 | Issue โ 5 | July-August 2020 Page 434
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[6] G. Chartrand, D. Erwin, P. Zhang, โA radio labeling problem suggested by FM channel restrictionsโ, Bull. Inst. Combin. Appl. Vol 33, pp.77-85, 2001.
[7] G. Chartrand, L. Nebeskยทy and P. Zhang, Radio k-colorings of paths, Discussiones Mathematicae Graph Theory, Vol. 24, pp. 5-21, 2004.
[8] C. Fernandez, A. Flores, M. Tomova and Cindy Wyels, The Radio Number of Gear Graphs, http: //arxiv.org/P Scache/arxiv/pdf /0809/0809.2623v1.pdf.
[9] D. Fotakis, G. Pantziou, G. Pentaris, and P. Spirakis, Assignment in mobile and radio networks, DIMACS Series in Discrete Mathematics and Theoretical Computer Science, Vol. 45, pp. 7390, 1999.
[10] M. Kchikech, R. Khennoufa and O. TogniLinear and cyclic radio k labelings of trees. Discussiones Mathematicae Graph Theory, 27 (1): 105-123, (2007).
[11] Kins Yenoke, P. Selvagopal, K. M. Baby Smitha, Ronald Cranston, โBounds for the Radio Number of Mesh derived Architectureโ International Journal of Innovative Science, Engineering and Technology, Vol 9, pp. 4806-4810, 2020.
[12] D. Liu and M. Xie, โRadio Number for Square Cyclesโ, Congr. Numerantium, Vol 169 pp. 105 - 125, 2004.
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