Cmjv05i04p0358

JOURNAL OF COMPUTER AND MATHEMATICAL SCIENCES An International Open Free Access, Peer Reviewed Research Journal www.compmath-journal.org

ISSN 0976-5727 (Print) ISSN 2319-8133 (Online) Abbr:J.Comp.&Math.Sci. 2014, Vol.5(4): Pg.358-366

On the Radio Number of Extended Mesh Kins Yenoke Department of Mathematics, Loyola College, Chennai, INDIA. (Received on: July 18, 2014) ABSTRACT A radio labeling of a connected graph is an injection from the vertices of to the natural numbers such that ( , ) + | − ( )| ≥ 1 + ( ) for every pair of distinct vertices and of . The radio number of denoted , is the maximum number assigned to any vertex of . The radio number of , denoted ( ), is the minimum value of ( ) taken over all labelings of . In this paper we determine bounds for the radio number of the enhanced mesh.

Keywords: Labeling, Radio labeling, Radio number and Extended mesh.

1. INTRODUCTION Interest in graph labeling problems began in the mid-1960’s with a conjecture of Ringel16 and a paper by Rosa17. In the intervening years dozens of graph labelings techniques have been studied in over several papers. Despite the large number of papers, there are relatively few general results or methods on graph labeling. Indeed most of the results focus on particular classes of graphs and utilize adhoc methods.

Frequently, the same classes have been done by several authors. Labeled graphs serve as useful models for a broad range of applications such as coding theory, x-ray,

crystallography, radar, astronomy, circuit design, channel assignments of FM radio stations and communication network addressing applications5, 6.

2. AN OVERVIEW OF THE PAPER The general problem that inspired radio labeling is what has been known as the channel assignment problem: the goal is to assign radio channels in a way so as to avoid interference between radio transmitters that are geometrically close. The use of graph theory to study the Channel Assignment Problem and related problems dates back at least to 1970 (see [15). The problem was

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first put into a graph theoretic context by Hale10, 9 in 1980. Since then, a number of graph colorings have been inspired by the Channel Assignment Problem.

In 2001 Chartrand et al.7 were motivated by regulations for channel assignments of FM radio stations to introduce radio labelings of graphs. Two stations that share the same channel must be separated by at least 115 kilometers; however, the actual required separation depends on the classes of the two stations.

Two channels are considered to be firstadjacent (or simply adjacent) if their frequencies differ by 200 kHz, that is, if they are consecutive on the FM dial. For example, two FM stations on channels 92.4 MHz and 92.6 MHz are adjacent. The distance between two radio stations on adjacent channels must be at least 72 kilometers. Again, the actual restriction depends on the classes of the stations. The distance between two radio stations whose channels differ by 400 or 600 kHz (secondor third-adjacent channels) must be at least 31 kilometers. Once again, the actual required separation depends on the classes of the stations.

For a connected graph G of diameter d and an integer k with 1≤ k ≤ d; a −radio coloring (sometimes called a radio −coloring) of G is an assignment f of colors (positive integers) to the vertices of G such that ( , ) + | − ( )| ≼ 1 + for every two distinct vertices u and v of G. It is important to note that −radio labeling is actually a generalization of the classical idea of vertex coloring. Vertex coloring corresponds to − radio labeling with = 1.

A radio labeling of a connected graph G is an injection f from the vertices of G to the natural numbers such that ( , ) + | − ( )| ≼ 1 ( )) for every two distinct vertices u and v of G. The radio number of f, denoted by ( ), is the maximum number assigned to any vertex of G. The radio number of G, denoted by ( ), is the minimum value of ( ) taken over all labelings of .

The radio numbers for different families of graphs have been studied by several authors. For paths and cycles, the radio number problem has been studied by Chartrand et al.8 and by Zhang18 and completely determined by Liu and Zhu14.

The radio number for square of paths and of cycles was investigated by Liu et al.13.

Bharati et al.2 completely determined the radio number of wheels, fans, double fans, Dutch-t-mill graph, star graph, uniform ( + − 1) −cyclic split graph and uniform −cyclic star split graphs.

Khennoufa et al.12 have investigated the radio number of hypercube. Bharati et al.3, 4 obtained the bounds for the radio number hexagonal mesh and uniform theta graphs.

Katherine et al.11 completely determined the radio number of all graphs of order and diameter − 2.

In this paper we have determined the upper and lower bounds for the radio number of extended mesh.

3. EXTENDED MESH In this section we obtain bounds for the radio number of extended mesh ( , ). First we recall the definition of the extended mesh.

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The × mesh denoted , is defined as the Cartesian product ௠× ௥ of paths on m and n vertices respectively.

The number of vertices in , is and diameter is + − 2.

The architecture obtained by making each 4-cycle in ( , ) into a complete graph is called an extended mesh [1]. It is denoted by ( , ). The number of vertices of vertices in ( , ) is and diameter is , − 1.

In this paper we name the vertices of ( , ) as ŕŻĄáˆşŕŻ?ŕŹżŕŹľáˆťŕŹžŕŻž , = 1,2 ‌ , = 1,2 ‌ . See Figure 3.1.

Define a mapping : ( , ) → as ௥ follows: ŕŻœ = ଶ + 1 + − 1 − 1 , = 1,2 ‌

௡(௡ିଵ) ଶ

.

௥

ଵ

ŕł™áˆşŕł™ŕ°ˇŕ°­áˆťŕŹžŕŻœ = + ଷ − 2 ଶ − + 4 + ଶ

ŕ°Ž

ଶ ௡

− 1 , = 1, 2 ‌ . ೙ఎ = 1. ŕś„ŕ°Žŕśˆ

೙ఎ

ŕś„ ŕ°Ž ŕśˆŕŹžŕŻœ

ଶ

ଵ

= ೙ఎ + ଷ − 2 ଶ − + 4 + ଶ

௥

− 1 , = 1, 2 ‌ . ଶ

ŕł™(೙జభ)ŕŹžŕŻœ = + 1 + − 1 − 1 , = ŕ°Ž

1,2 ‌

೙(೙డభ) ఎ

.

Claim: The mapping f is a valid radio labeling . We must show that the radio condition , + | − | ≼ 1 +

= 1 + − 1 = , holds for all pair of vertices , ∈ ( , .

Figure 3.1: A general naming of ॹࢄ(࢓, ࢔).

3.1 Upper Bounds Theorem 3.1 Let n be odd. Then the radio number of extended mesh ( , ) satisfies ௥ , ≤ 2 − 1 + ଵ ଷ ଶ

ଶ

− 2 ଶ − + 4 .

Proof: We partition the vertex set ( , ) into 3 disjoint sets ଵ , ଶ and ଷ . Let ଵ = ଵ , ଶ ‌ ŕł™(೙డభ) , ŕ°Ž

ଶ = ŕł™(೙డభ)ାଵ, ŕł™(೙డభ)ାଶ ‌ ŕł™(೙జభ) and ŕ°Ž

ŕ°Ž

ŕ°Ž

ŕ°Ž

ŕ°Ž

ଷ = ŕł™(೙జభ)ାଵ, ŕł™(೙జభ)ାଶ ‌ ௥ఎ .

Case 1: Suppose u and w are any two vertices in ଵ . Then = ௞ and = ŕŻ&#x; , ௥(௡ିଵ) ௥ 1 ≤ ≠≤ ଶ . Therefore ( )= ଶ ! + ௥

1 + − 1 − 1 , ( )= ଶ ! + 1 + − 1 − 1 and ( , ) ≼ 1. Hence , + | − | ≼ 1 + | − 1 ( − )| ≼ .

Case 2: Suppose u and w are any two vertices in ଶ . Then = ௞ and = ŕŻ&#x; for ௥ some and , where ଶ ( − 1) + 1 ≤ ≠௥

≤ ଶ ( + 1).

Case 2.1: If = ŕł™(௡ିଵ)ା௾ and = ŕł™(௡ିଵ)ାŕŻ&#x; , ŕ°Ž

௡ ଵ ! + ଶ ଷ ଶ

ŕ°Ž

௥

1 ≤ ≠≤ ଶ,

then

=

− 2 ଶ − + 4 + − 1 ,

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Kins Yenoke, J. Comp. & Math. Sci. Vol.5 (4), 358-366 (2014) ௡

ଵ

= ଶ ! + ଶ ଷ − 2 ଶ − + 4 + − 1 and ( , ) ≥ 1. Therefore , + | − | ≥ 1 + | − 1 ( − )| ≥ .

Case 2.2: If = ೙మ

ቒ మ ቓା௞

௡

1 ≤ ≠≤ , then ଶ ଵ

and = ೙మ

ቒ à°® ቓାà¯&#x;

,

, ≥ 1 and

= "೙మ# + ଶ ଷ − 2 ଶ − + 4 + ଵ "೙మ# + ଶ ଷ

− 1 , = − 2 ଶ − + 4 + − 1 . Therefore , + | − | ≥ 1 + | − 1 ( − )| ≥ .

Case

2.3:

If

= = ೙(௡ିଵ)ା௞

and

à°®

௡

௡ !+ ଶ

= ೙మ , 1 ≤ ≤ ଶ , then = ଵ ଷ ଶ

ቒమቓ

− 2 ଶ − + 4 + − 1 , = 1 and , ≥ 1. Therefore , + ௡ ଵ | − | ≥ 1 + $ ! + ଷ − 2 ଶ − ଶ ଶ

+ 4 + − 1 − 1$ ≥ $ ଵ ଷ ଶ

Case

− 2 ଶ + $ > .

2.4:

If

௡ !+ ଶ

= = ೙(௡ିଵ)ା௞

and

à°®

=

− 1 and , ≥ 2. Hence , + ௡ | − | ≥ 1 + | | ≥ $ "೙మ# − ! + − 1 ( − )$ ≥ 2 + − 2 = .

ଶ

Case 2.5: If = ೙మ and = ೙మ ௡

1 ≤ ≤ ଶ ,

ቒమቓ

then

, ≥ 1

ଵ ଷ ଶ

− 2 ଶ − + 4 + − 1 − 1$ > .

Case 3: Suppose u and w are any two vertices in ଷ . Then = ೙(௡ାଵ)ା௞ and

=

à°®

à³™ (௡ାଵ)ାà¯&#x; à°®

1≤ ≠≤

,

ቒ మ ቓା௞

,

and

௡(௡ିଵ) . ଶ

Therefore ( , ) ≥ 1 and ( )= + 1 + − 1 − 1 , ( )= + 1 + − 1 − 1 .

Hence , + | − | ≥ 1 + | − 1 ( − )| ≥ .

Case 4: ∈ ଵ and ∈ ଶ .

Case 4.1: If = ௞ and = ೙ሺ௡ିଵሻାà¯&#x; ,

1≤ ≤

௡ሺ௡ିଵሻ , ଶ

௡

1 ≤ ≤ ଶ , then

, ≥ 1 and =

− 1 − 1 , =

௡ , 1 ≤ , ≤ ଶ , then ቒ à°® ቓାà¯&#x; ௡ ଵ ! + ଶ ଷ − 2 ଶ − + 4 + − 1 , ଶ ଵ = "೙మ# + ଶ ଷ − 2 ଶ − + 4 +

= ೙మ

ଵ

= 1, = "೙మ# + ଷ − 2 ଶ − ଶ + 4 + − 1 . Therefore , + | − | ≥ 1 + $ "೙మ# +

ଵ ଷ ଶ

à°®

௡ !+1+ ଶ ௡ !+ ଶ

− 2 ଶ − + 4 + − 1 . Therefore , + | − | ≥ 1 + | − 1 ( − − 1)| ≥ .

Case 4.2: If = ௞ and = ೙మ , 1 ≤ ≤ ௡ሺ௡ିଵሻ , ଶ

ቒమቓ

then either , ≥ 1 and

| − | ≥ ௡ ଶ

௡ ! + ଶ

− 1

, = and | − | ≥ , + | − | ≥ .

or ௡ . ଶ

Hence

Case 4.3: If = ௞ and = ೙మ 1≤ ≤

௡ሺ௡ିଵሻ , ଶ

௡

1 ≤ ≤ ଶ ,then

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ቒ à°® ቓାà¯&#x;

,

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Kins Yenoke, J. Comp. & Math. Sci. Vol.5 (4), 358-366 (2014)

, ≥ 1 and =

ଵ ଷ ଶ

௡ !+1+ ଶ

− 2 ଶ − + 4 + − 1 . Therefore

, + | − | ≥ 1 ೙మ + 2 > .

− 1 − 1 , = "೙మ# +

Figure 3.2: The extended mesh ࡱࢄ(ૠ, ૠ) with a radio labeling.

Case 5: Suppose ∈ ଵ and ∈ ଷ then = ௞ and = à³™(௡ାଵ)ାà¯&#x; , 1 ≤ , ≤ ௡(௡ିଵ) . ଶ

à°®

Therefore

in

this

case

either

( , ) ≥ 2 and | − | ≥ ଷ(௡ିଵ) ଶ

௡

௡ !+ ଶ

or , ≥ ଶ + 1 and | − ௡

| ≥ ଶ + 1.

Hence | − | ≥ .

, +

Case 6.1: If = ೙ሺ௡ିଵሻା௞ and = ೙ሺ௡ାଵሻାà¯&#x; , à°®

1≤ ≤

à°®

௡ሺ௡ିଵሻ , ଶ

1≤ ≤

௡ሺ௡ିଵሻ , ଶ

then either , ≥ 1 and | − | ≥ + 2 or , = − 1 and | − | ≥ 3. Hence in both the possibilities, we have , + | − | ≥ .

ቒమቓ

à°®

1≤ ≤ then either , ≥ 1 and ( ) = 1|, = + 1 + − 1 − 1 | . Therefore , + | − | ≥ 1 + > .

Case 4.3: If = ೙మ 1≤ ≤

Case 5: ∈ ଶ and ∈ ଷ . ௡ ଶ ,

Case 6.2: If = ೙మ and = ೙ሺ௡ାଵሻା௞ ,

௡ ଶ ,

1≤

and = ೙ሺ௡ାଵሻାà¯&#x; ,

ቒ మ ቓା௞ ௡ሺ௡ିଵሻ ≤ ଶ ,

à°®

then

, ≥ 1 and = " # + ଵ ଷ − 2 ଶ − + 4 + − 1 , = ଶ + 1 + − 1 − 1 . Therefore , + | − | ≥ 1 + % "೙మ# − 1% > . ೙ మ

Thus , + | − | ≥ for all , ∈ , , is odd. Since the vertex ೙మ ೙ receives the maximum label, ቒ మ ቓାඃ మ ඇ

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Kins Yenoke, J. Comp. & Math. Sci. Vol.5 (4), 358-366 (2014)

the radio number of extended mesh . satisfies ௥

, ≤ 2 − 1 + ଶ

ଵ ଷ ଶ

− 2 ଶ − + 4 , n odd.

The radio labeling of 7,7 is depicted in Figure 3.2.

Theorem 3.2: Let n be even. Then the radio number of extended mesh , satisfies ௥ satisfies , ≤ ଶ ଶ − + 1 + 1.

Figure 3.3: The extended mesh ॹࢄ(ૠ, ૠ) with a radio labeling.

As in theorem 3.1, we can prove that f is a valid radio labeling. Since the vertex ೙ఎ ೙ receives the maximum label we ఎ

ା ାଵ మ

Proof: Let us partition the vertex set ( , into 3 disjoint sets ଵ , ଶ and

have the required conclusion.

We now proceed to determine the lower bounds for the radio number of extended mesh.

ଷ , where ଵ = & ଵ , ଶ ‌ ೙ఎ ',

3.2 Lower Bounds

ŕ°Ž

In3, Bharati Rajan and Kins Yenoke have proved the following result which we will use to find the lower bound of extended mesh.

ଶ = & ŕł™ŕ°ŽŕŹžŕŹľ, ŕł™ŕ°ŽŕŹžŕŹś ‌ ೙ఎజ೙ ' and ŕ°Ž

ŕ°Ž

ŕ°Ž

ଷ = & ŕł™ŕ°ŽŕŹžŕŻĄŕŹžŕŹľ, ŕł™ŕ°ŽŕŹžŕŻĄŕŹžŕŹś ‌ ௥ఎ '. ŕ°Ž

ŕ°Ž

Define a mapping : ( , ) → as follows: ŕŻĄáˆşŕŻ?ŕŹżŕŹľáˆťŕŹžŕŻœ = − 1 − 1 + − + ,

= 1,2 ‌ . , = 1,2 ‌ 2 ௥ ଶ ೙ఎ = ଶ − 1 + 1 + − 1 ŕ°Ž ŕŹžŕŻœ

− 1 + 1, = 1,2 ‌ ௥

௡ ଶ

+ 1.

ŕł™áˆşŕŻĄŕŹžŕŹľáˆťŕŹžŕŹľŕŹžŕŻœ = + − 1 − 1 , ௥

= 1,2 ‌ − 1. ŕ°Ž

೙ఎ

ଶ

ŕ°Ž ା௡ŕŻ?ŕŹžŕŻœ

=

ଶ

௡మ ଶ

− 2 + − 1 − 1 + ௥

− + + 2, = 1,2 ‌ , = 1,2 ‌ − 1. ଶ

Theorem 3.3: (As Theorem 2 in3): Let G be a simple connected graph of order n. Let ଴ , ଵ ‌ ௞ be the number of vertices having eccentricities (଴ , (ଵ ‌ (௞ , where

= (଴ > (ଵ > â‹Ż > (௞ =

( ).Then ( ) ≼ ‍ۓ‏ ۖ

ÝŠ − 2áˆşÝ€ − Ý ŕŻž áˆť + ŕˇ? 2áˆşÝ€ − Ý ŕŻœ áˆťÝŠŕŻœ , ௞

݂݅ ݊௞ > 1

௞ ‍۔‏ Ű–ÝŠ − áˆşÝ€ − Ý ŕŻž áˆť − (Ý€ − Ý ŕŻžŕŹżŕŹľ ) + ŕˇ? 2áˆşÝ€ − Ý ŕŻœ áˆťÝŠŕŻœ , ‍ە‏ ŕŻœŕ­€ŕŹľ ŕŻœŕ­€ŕŹľ

݂݅ ݊௞ = 1

Theorem 3.4: Let n be odd. Let (଴ , (ଵ ‌ (ቔ೙ቕ be the eccentricities of the ŕ°Ž

vertices of the extended mesh , with = (଴ > (ଵ > â‹Ż > (ቔ೙ቕ = ). Then ŕ°Ž

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364

Kins Yenoke, J. Comp. & Math. Sci. Vol.5 (4), 358-366 (2014) â€ŤÝŠÝŽâ€ŹŕľŤâ€ŤÜşÜ§â€ŹáˆşÝŠ, ÝŠáˆťŕľŻ ≼

ݩଶ

Theorem 3.5: Let , be an extended mesh of order Ă— . If n is odd, then

௥ ቔ á‰•ŕŹżŕŹľ ଶ

+ 1 + 8 ŕˇ? Ý…(ÝŠ − (2Ý… + 1).

ŕł™

ቔ á‰•ŕŹżŕŹľ

ŕŻœŕ­€ŕŹľ

Proof: When n is odd, the eccentricities (଴ , (ଵ ‌ (ቔ೙ቕ of , are given by ŕ°Ž

(଴ = − 1, , (ଵ = − 2 ‌

௥ (ቔ೙ቕ = ଶ . ŕ°Ž ௥ ଶ .

That

is (ŕŻœ = − + 1 , 0 ≤ ≤ The number of vertices having eccentricities (଴ is 4 − 1 , (ଵ is 4 − 3 ‌ (á‰”ŕł™á‰•ŕŹżŕŹľ is ŕ°Ž

4 Ă— 2 and (ቔ೙ቕ is 1. That is, ଴ = 4 − 1 , ŕ°Ž

ଵ = 4 − 3 ‌ á‰”ŕł™á‰•ŕŹżŕŹľ 4 Ă— 2, ቔ೙ቕ = 1. ŕ°Ž

ŕ°Ž

In

other words ŕŻœ = 4( − 2 + 1 , 0 ≤ ≤ ௥ ௥ ଶ − 1, ଶ = 1. See Figure 3.4. Now

ŕ°Ž ( − (2 + 1) ≤ ଶ + 1 + 8 âˆ‘ŕŻœŕ­€ŕŹľ

௥

, ≤ 2 − 1 + ଶ

ଵ ଷ ଶ

− 2 ଶ − + 4 .

Theorem 3.6 Let n be even. Let (଴ , (ଵ ‌ (ŕł™ŕŹżŕŹľ be the eccentricities of the ŕ°Ž

vertices of the extended mesh , with = (଴ > (ଵ > â‹Ż > (ŕł™ŕŹżŕŹľ = ). ŕ°Ž

Then , ≼ ଶ − + 2 ௥ ିଵ ଶ

+ 8 + ( − (2 + 1). ŕŻœŕ­€ŕŹľ

௥

= (଴ = − 1 and = ଶ and hence by

Figure 3.4: The subgraphs of ŕĄąŕ˘„áˆşŕŤ˘, ŕŤ˘áˆť induced by vertices with eccentricities ࢋŕ˘? , ŕŤ™ ≤ ŕ˘? ≤ ŕŤ?.

Theorem 3.3, we have, , ≼ − − !௞ − − !௾ିଵ + âˆ‘ŕŻžŕŻœŕ­€ŕŹľ 2 − !ŕŻœ ŕŻœ ௥

ଶ

௥

= − − 1 − − ( − 1 − + 1 + ଶ

ŕł™ ቔ á‰•ŕŹżŕŹľ ŕ°Ž

ଶ

௥

âˆ‘ŕŻœŕ­€ŕŹľ 2 . 4( − 2 + 1 + 2( − 1 − ) ௥

ቔ á‰•ŕŹżŕŹľ

ଶ

ŕ°Ž = ଶ − 2 − 2 − 3 + 8 âˆ‘ŕŻœŕ­€ŕŹľ ( −

ଶ

ŕł™

௥

2 + 1 + (2 − 2 − 2 ) ŕł™ ቔ á‰•ŕŹżŕŹľ ŕ°Ž

ଶ

= ଶ + 1 + 8 âˆ‘ŕŻœŕ­€ŕŹľ ( − 2 + 1 ).

Figure 3.5: The subgraphs of ŕĄąŕ˘„áˆşŕŤĄ, ŕŤĄáˆť induced by vertices with eccentricities ࢋŕ˘? , ŕŤ™ ≤ ŕ˘? ≤ ŕŤœ.

Proof: When n is even, the eccentricities (଴ , (ଵ ‌ (ŕł™ŕŹżŕŹľ of , are given by ŕ°Ž

௥

(଴ = − 1, (ଵ , = − 2 ‌ (ŕł™ŕŹżŕŹľ=ଶ . ŕ°Ž

The

number of vertices having eccentricities (଴ is 4 − 1 , (ଵ is 4 − 3 ‌ (ŕł™ŕŹżŕŹľ is 4 Ă— 2 and

(ቔ೙ቕ ఎ

is

4.

That is

ŕ°Ž

௥

(ŕŻœ = −

+ 1 , 0 ≤ ≤ − 1, ŕŻœ = 4( − ଶ ௥

2 + 1 , 0 ≤ ≤ − 1. See Figure 3.5. ଶ

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365

Kins Yenoke, J. Comp. & Math. Sci. Vol.5 (4), 358-366 (2014) ௥

Hence by theorem 3.3, with = − 1, ଶ , ≼ − − (௞ + âˆ‘ŕŻžŕŻœŕ­€ŕŹľ 2 − (ŕŻœ ŕŻœ = − 2( 2 + 1 ) ଶ

ŕł™

ିଵ ௥ ŕ°Ž − 1 − ) + âˆ‘ŕŻœŕ­€ŕŹľ 2 . 4( ଶ ŕł™

4.

−

ିଵ

ŕ°Ž = ଶ − + 2 + 8 âˆ‘ŕŻœŕ­€ŕŹľ ( − 2 + 1 ).

5.

Theorem 3.7: Let , be an extended mesh of order Ă— . If n is even, then

6.

ŕł™ ିଵ ŕ°Ž

ଶ − + 2 + 8 âˆ‘ŕŻœŕ­€ŕŹľ ( − 2 + 1 ) ≤ ௥ , ≤ ଶ ଶ − + 1 + 1.

4. CONCLUSION In this paper we have obtained the upper and lower bounds for the radio number of extended mesh. The radio number problems for Torus mesh, Honeycomb mesh, Silicate network, Enhanced mesh etc., are under investigation.

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