RADIO NUMBER FOR PRISM RELATED GRAPHS Dn

Sci.Int(Lahore),26(2),551-555,2014

ISSN 1013-5316; CODEN: SINTE 8

551

RADIO NUMBER FOR PRISM RELATED GRAPHS Dn‫כ‬ SaimaNazeer Department of Mathematics, Lahore College for Women University, Lahore-Pakistan E-mail: [email protected]

ImranaKousar Department of Mathematics, Lahore College for Women University, Lahore-Pakistan E-mail: [email protected] WaqasNazeer* Department of Mathematics, Lahore Leads University, Lahore-Pakistan E-mail: [email protected]

ABSTRACT: For a connected graph G , let d ( x, y ) denote the distance between any two vertices in G , which is the length of the shortest path between them. The diameter of G is the maximum distance between any pair of vertices of G denoted by diam(G ) . A radio labeling of G is an assignment of

G such that for any vertices x and y the following inequality | f ( x ) − f ( y ) |≥ diam(G ) − d ( x, y ) + 1 holds. The span of a radio labeling is the maximum integer assigned to a vertex. The radio number of a graph G is the minimum span taken over all radio labelings ∗ of G . In this paper, we determine the radio number for the prism related graphs, Dn when n = 4k + 2 . positive integer to the vertices of

AMS Subject Classification: 05C12, 05C15, 05C78. Key words: channel assignment, radio labeling, radio number and Prism graph.

INTRODUCTION Radio labeling of a graph is initiated by assigning frequencies to the radio transmitters [2] in such a way that two geographically close transmittersmust be assigned frequencies with large differences. On the other hand, thetransmitters which are far away from each other may receive channels withsmall frequencies. By considering transmitters as the vertices of the graph, we allocatepositive integers to the vertices of the graph such that our purpose is tominimize the largest integer to be used. The radio number for paths andcycles were determined in [4]. The radio number for the square of paths wereinvestigated by Daphne Der-Fen Liu and Melanie Xie in [5], they have alsodiscussed the problem for the square of a cycles in [6]. Daphne Der-Fen Liudetermined the lower bound for the radio number of spiders in terms of thelength of their legs and characterized the spiders that achieve this boundin [7]. The spiders are trees with at most one vertex of degree more thantwo. The radio number for the generalized prism graphs were presented byPaul Martinez et.al. in [8]. The radio number for Cartesian product ofcycles with itself was established by Marc Morris- Rivera et al. in [9]. Theone those of Gear graphs was studied in [3].Let G be a connected graph. The distance d ( x, y ) between any pair

span of f . The minimum span taken over all radiolabelings

G is called the radio number of G and it is denoted by rn(G ). In this paper, we determine the radio number of the ∗ graph Dn which is an extension of the prism graph defined in [8] as follows. For eachvertex bi , of the outer cycle introducing a new vertex ai , andjoin ai to bi and bi −1 , where b0 = bn . Thus i = 1, 2,3,..., n , of

V ( D n∗ ) =

n

U { a , b , c } Here {ci } , i

i

i

are inner cycle vertices

i =1

and {bi } , are outer cyclevertices and {ai }, i = 1, 2,3,..., n are adjacent vertices to outer cycle.In this paper, we determine the radio number for the prism related graphs,

Dn∗ when n = 4k + 2 . MAIN RESULTS The main theorem of this paper is: Theorem 1:For the Prism related graphs,

Dn∗ , n = 4k + 2, k ≥ 1. rn( Dn∗ ) = 6k 2 + 18k + 6. ∗

ofvertices in G is the length of the shortest path between them. Thediameter of G , denoted by diam(G ) = d , is the

Remark: Note that, the diameter of Dn is:

G. A maximum distance between anytwo vertices in function f : V (G ) → N ∪ {0} is called aradio labeling

A Lower Bound For Dn , n = 4k + 2

diam( Dn∗ ) = 2k + 2, when n = 4k + 2. ∗

∗

which is also known as multilevel distance labeling of G if is one-one and satisfying the condition f

In this section, the lower bound for Dn , where

| f ( x) − f ( y ) |≥ diam(G ) + 1 − dG ( x, y ) for any pair of vertices x, y . The largest number in f (V ) is known as

Lemma 1:Let Dn be the prism related graphs,

is determined. ∗

n = 4k + 2.

n = 4k + 2

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552

(i)

For each vertex a1 ∈ {ai :1 ≤ i ≤ n}, which is anadjacent

Therefore, if u , v, w are three vertices which are adjacent to

vertex on the outer cycle there is exactly one vertex at a

the outer cycle of Dn then

∗

(ii)

(iv)

d (u , v ) + d (v, w) + d ( w, u ) ≤ 2d + 1.

For each vertex b1 ∈ {bi :1 ≤ i ≤ n} on the outercycle (i) there is exactly one vertex on the inner cycle

Lemma 3: ∗ Let u , v, w be three vertices of Dn ,

twovertices on the outer and one vertex on the inner cycles then

vertex on the outer cycle there are two vertices on an inner ∗ (ii) cycle {ci :1 ≤ i ≤ n} at a distance d , diameter of Dn .

Let u , v, w be three vertices of Dn ,

For each vertex b1 ∈ {bi :1 ≤ i ≤ n} on the outer cycle

− 1 , diameter of

∗ n

D .

(i) For each vertex c1 ∈ {ci :1 ≤ i ≤ n} on the innercycle there is exactly one vertex at a distance d

(i)

− 1 , diameter of Dn∗ .

Proof: We show that d ( a1 , a2 k + 2 ) = d = 2k + 2 .

a1 → b1 → b2 → b3 → ... → b2 k +1 → a2 k + 2

(iii)

Now d ( a1 , c2 k + 2 ) = 2k + 2.

(iv)

a1 → b4 k + 2 → c2 k + 2 k + 2 → c4 k +1 → ... → c2 k + 2 . Similarly, we show that d ( a1 , c2 k +1 ) = 2k + 2. d (b1 , b2 k + 2 ) = 2k + 1 = d − 1. b1 → b2 → b3 → b4 → ... → b2 k + 2 . (v) d (c1 , c2 k + 2 ) = 2k + 1 = d − 1.

∗

n = 4k + 2 with

twovertices on the inner and one vertex on the outer cycles then

d (u , v ) + d (v, w) + d ( w, u ) ≤ 2d .

Proof: By Lemma 1(iv), d (b1 , b2 k + 2 ) = 2k + 1 = d − 1 . For each vertex c1 on the inner cycle there is only one vertex b2 k + 2 on the outer cycle at a distance

d . i.e

Therefore,

d (b1 , b2 k + 2 ) + d (b2 k + 2 , c1 ) + d (c1 , b1 ) = d − 1 + d + 1 = 2d Thus, if u , v, w are three vertices with two vertices on the ∗

outer and onevertex on the inner cycles of Dn , n = 4k + 2 then

Now d (b1 , c2 k + 2 ) = 2k + 2,

b1 → c1 → c2 → c3 → ... → c2( k ) + 2 .

d (u , v ) + d (v, w) + d ( w, u ) ≤ 2d .

d (c1 , b2 k + 2 ) = d ,

Since n = 4k + 2, there are equal number of vertices on the left half andright half of the adjacent vertices on the outer cycle. The path from a1 to a2 k + 2 is of length 2k + 2 as

(ii)

n = 4k + 2 with

{ci :1 ≤ i ≤ n} at a distance d , diameter of D For each vertex a1 ∈ {ai :1 ≤ i ≤ n} , which is anadjacent

there is exactly one vertex at a distance d (v)

∗

distance d , diameter of Dn .

∗ n

(iii)

Sci.Int(Lahore),26(2),551-555,2014

(ii)

c1 → c2 → c3 → c4 → ... → c2 k + 2 . Lemma 2:Let u , v, w be three vertices which are adjacent

d (u , v ) + d (v, w) + d ( w, u ) ≤ 2d . By Lemma 1 (v), d (c1 , c2 k + 2 ) = 2k + 1 = d − 1 .

For each vertex b1 on the outer cycle there is only one vertex c2 k + 2 on the inner cycle at a distance

d . i.e

d (b1 , c2 k + 2 ) = d , Therefore,

d (c1 , c2 k + 2 ) + d (c2 k + 2 , b1 ) + d (b1 , c1 ) = d − 1 + d + 1 = 2d . Thus, if u , v, w are three vertices with two vertices on the ∗

inner and one vertex on the outer cycles of Dn , n = 4k + 2 then

d (u , v ) + d (v, w) + d ( w, u ) ≤ 2d . n = 4k + 2 then Lemma 4:Let f be a radio labelling of d (u , v ) + d (v, w) + d ( w, u ) ≤ 2d + 1. Dn∗ , n = 4k + 2, k ≥ 1 . Proof: By Lemma 1(i), d ( a1 , a2 k + 2 ) = 2k + 2 = d . (i) Suppose {xi :1 ≤ i ≤ n} is the set of vertices which Now d ( a2 k + 2 , a4 k + 2 ) = 2k + 1 = d − 1 and a path of areadjacent on the outer cycle and f ( xi ) < f ( x j ) length 2k + 1 between a2 k + 2 to a4 k + 2 is whenever i < j . Then | f ( xi + 2 ) − f ( xi ) |≥ φ ( n) ,where a2 k + 2 → b2 k + 2 → b2 k +3 → ... → b2 k + 2 k +1 = b4 k +1 → a4 k + 2 φ ( n) = k + 2. (ii) Suppose { yi :1 ≤ i ≤ n} is the set of vertices on theouter and d ( a4 k + 2 , a1 ) = 2. and inner cycles and f ( yi ) < f ( y j ) whenever i < j . ∗

to outer cycle of Dn ,

Sci.Int(Lahore),26(2),551-555,2014

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Then | f ( yi + 2 ) − f ( yi ) |≥ ψ ( n) ,where

For i = 1, 2,3,...n − 1 , set d i = d ( xi , xi +1 ) and

ψ (n) = ⎢ k + ⎥ = k + 3. 2⎦ ⎣

f i = f ( xi +1 ) − f ( xi ) . Then f i ≥ d − d i + 1 for all i .By Lemma 5(i ) , the span of

Proof: Let {xi , xi +1 , xi + 2 } be any set of three vertices which

a distance labeling f of Dn for

⎢

(i)

553

5⎥

∗ n

areadjacent on the outer cycle of D , n = 4k + 2. Applying the radio condition to each pair in the vertex set {xi , xi +1 , xi + 2 } and take the sum of these inequalities.

∗

the vertices which are adjacent to the outer cycle is n −1

f ( xn ) = ∑ fi = f1 + f 2 + f 3 + .... + f n − 2 + f n −1 i =1

= [ f ( x2 ) − f ( x1 )] + [ f ( x3 ) − f ( x2 )] + ... + [ f ( xn −1 ) − f ( xn − 2 )] + [ f f ( xi +1 ) − f ( xi ) |≥ diam(G ) − d ( xi +1 , xi ) + 1 f ( xi + 2 ) − f ( xi +1 ) |≥ diam(G ) − d ( xi + 2 , xi +1 ) + 1 = ( f1 + f 2 ) + ( f3 + f 4 )) + ( f5 + f 6 ) + ... + ( f n −3 + f n − 2 ) + f n −1 f ( xi + 2 ) − f ( xi ) |≥ diam(G ) − d ( xi + 2 , xi ) + 1 n−2 f ( xi +1 ) − f ( xi ) | + | f ( xi + 2 ) − f ( xi +1 ) | + | f ( xi + 2 ) − f ( xi ) |≥ 23diam(G ) + 3 − d ( xi +1 , xi ) − d ( xi + 2 , xi +1 ) − d ( xi + 2 , xi ) = ∑ ( f 2i −1 + f 2i ) + f n −1 We drop the absolute sign because | | | |

f ( xi ) < f ( xi +1 ) < f ( xi + 2 ) and using Lemma 2, we have 2[ f ( xi + 2 ) − f ( xi )] ≥ 3 + 3d − (2d + 1) = d + 2 d + 2 2k + 4 [ f ( xi + 2 ) − f ( xi )] ≥ = = k + 2. 2 2 Thus, φ ( n ) = k + 2. (ii) Now let { yi , yi +1 , yi + 2 } be any set of threevertices on the outer and inner cycles of

Dn∗ , n = 4k + 2 .

i =1

n−2 φ (n) + 1 2 2 Thus, f ( xn ) ≥ 2k + 4 k + 1. ≥

Applying Lemma 2 and Lemma 5(ii) to the vertices xn −1 , xn , y1 such that f ( xn −1 ) < f ( xn ) < f ( y1 ) , then we obtain:

| f ( y1 ) − f ( xn −1 ) |≥ k + 2 f ( y1 ) ≥ 2k 2 + 5k + 2.

Applying radio condition to each pair in the above manner and using Lemma 4, we get

2[ f ( yi + 2 ) − f ( yi )] ≥ 3 + 3d − 2d = d + 3 d + 3 2k + 2 + 3 5 [ f ( xi + 2 ) − f ( xi )] ≥ = =k+ 2 2 2 5⎥ ⎢ Thusψ (n) = ⎢ k + ⎥ = k + 3. 2⎦ ⎣ ∗ Theorem 1:For the Prism related graphs, Dn , n = 4k + 2 , k ≥ 1. rn( Dn∗ ) ≥ 6k 2 + 18k + 6. ∗ n

3n vertices. First divide set ofvertices into three subsets {a1 , a2 , a3 ,..., an } , {b1 , b2 , b3 ,..., bn } and {c1 , c2 , c3 ,..., cn }.

Proof:A Prism related graph D has

∗

Let f be a distance labeling for Dn . We order the vertices ∗ n

of D which are adjacent to the outer cycle by

x1 , x2 , x3 ,..., xn with f ( xi ) < f ( xi +1 ) and the vertices on the outerand inner cycles by y1 , y2 , y3 ,..., y2 n with f ( yi ) < f ( yi +1 ) , diam( Dn∗ ) = d = 2k + 2.

∗

By Lemma 5(ii), the span of distance labeling f of Dn forthe vertices on the outer and inner cycles is

f ( y2 n ) − f ( y1 ) = ′ 1

∑f i =1

′ i

= ( f + f ) + ( f + f 4′ ) + ... + ( f 2′n −3 + f 2′n − 2 ) + f 2′n −1

=

2n−2 2

∑ i =1

≥

′ 2

2 n −1

′ 3

f 2′i −1 + f 2′i ) + f 2′n −1

2n − 2 ψ ( n) + 1 2

2n − 2 ( k + 3) + 1 2 f ( y2 n ) ≥ 4k 2 + 13k + 4 + f ( y1 )

i.e. f ( y2 n ) − f ( y1 ) ≥

f ( yn ) ≥ 6k 2 + 18k + 6. ∗

An Upper Bound For Dn ,

n = 4k + 2

To complete the proof of Theorem 1, it remains to find ∗

distance labeling f for Dn with span equal to the desired numbers. The labeling is generated by pair of sequences, the distance gap sequence

D =

( d1 , d 2 , d3 .... , d n−1 )

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5544

D ′ = ( d1′ , d 2′ , d 3′ ,..., d 2′ n −1 )

Sci.Int(L Lahore),26(2),5551-555,2014

In Figuure below, orddinary and optim mal radio labelling of

rn ( D6∗ ) is shown

a the color gaap sequence and

F = ( f1 , f 2 , f 3 ,...., f n −1 ) F ′ = ( f1′ , f 2′ , f 3′ ,..., f 2′n −1 ) ′

T distance gaap sequences D and D aree given by The

⎧2k + 2,, ⎪ di = ⎨ ⎪k + 2, ⎩

if

i

is i odd ;

if

i is even.

if

i is odd ;

a and

⎧2k + 2,, ⎪ d =⎨ ⎪k + 1, ⎩ ′ i

if i is even e . ′

F each i , we have d ( xi , xi +1 ) = d i , d ( yi , yi +1 ) = d i For ′

a d = d ( xn , y1 ) = k + 1. and ′

F and T color gap sequences The s a F are givven by

⎧1, ⎪ fi = ⎨ ⎪k + 1, ⎩ ⎧1, ⎪ fi = ⎨ ⎪k + 2, ⎩

if

i

is odd ;

if

i is evven.

if i is odd o ;

′

if i is even.

f ′ = k + 1. T position fuunction The

p : V ( Dn∗ ) = {ai , bi , ci :1 ≤ i ≤ n} → {x1 , x2 ,..., xn , y1 , y2 ,..., y2 n } is defined as foollows. For n = 4k + 2 , we define p (a2i −1 ) = x2i −1 fo or i = 1, 2,...,, 2k + 1,

X

p (a2i + 4 ) = x2i fo or i = 1, 2,..., 2k + 1, p (c3i − 2 ) = y2i −1 fo or i = 1, 2,3,...., n, X p (b3i +3 ) = y2i fo or i = 1, 2,3,..., n. T span of f is equal to The f1 + f 2 + f 3 +,..., f n − 2 + f n −1 + f ′ + f1′ + f 2′ + f 3′ +,...., f 2′n − 2 + f 2′n −1

12

= 6k 2 + 18k + 6. E Example:

7

y9

y

5

y

x5

y10

y 3

11

y2

y

8

x2

Open problem p

6

y

1

y6

X

4

y

+[( f1′ + f 3′ + f 5′ +,..., + f 2′n −1 )] + [( f 2′ + f 4′ + f 6′ +,..., + f 2′n − 2 )]

2n 2n − 2 n n−2 ( k + 1) + k + 1 + (1) + ( k + 2) = (1) + 2 2 2 2

y

y

3

= [( f1 + f 3 + f 5 +,..., + f n −1 )] + [( f 2 + f 4 + f 6 +,..., + f n − 2 )] + f ′

1

X4

Sci.Int(Lahore),26(2),551-555,2014

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∗

The radio number of Dn for

n = 4 k , n = 4 k + 1, n = 4 k + 3

is an open problem for

researcher. Bibliography [1] M. Ali, M. T. Rahim, G. Ali , On Two Families of Graphs With Constant Metric Dimension, Journal of Prime Research in Mathematics, 8,95-101,(2012). [2] G. Chartrand, D. Erwin, P. Zhang and F. Harary, Radio labelings of graphs, Bull. Inst.Combin. Appl., 33 77-85(2001). [3] C. Fernandez, A.Flores, M.Tomova and C. Wyles , The Radio Number of Gear Graphs, Preprint arxive:0809. 2623VI [Math. Co], 15 Sep 2008.

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[4] D.D. F. Liu and X. Zhu, Multilevel Distance Labeling For Paths And Cycles, SIAM J. Discrete Math., 19 (3):610-620(electronic), (2009), [5] D.D. F. Liu and M.XieRadio Numbers For Squares Paths, Ars Combin, 90, 307-319(2009). [6] D.D. F. Liu and M. Xie, Radio Numbers of Squares of Cycles, Congr. Numer, 169, 101-125(2001), [8] D. Liu, Radio Number For Spiders ,manuscript, 2004. [9] P. Murtinez, J. OrtiZ, M. Tomova, andC. Wyles, Radio Numbers For Generalized Prism Graphs, Kodai Math. J., 22,131-139(1999). [10] M. M. Rivera, M. Tomova, C. WyelsAnd A. Yeager The Radio Number of Cn Cn arXiv: 1007. 5344VI. [Math. Co] 29 july, 2010.