Radio Labeling of Ladder Radio Labeling of Ladder GraphsGraphsJosefina FloresJosefina FloresKathleen LewisKathleen Lewis

From: California State University Channel IslandsFrom: California State University Channel IslandsAdvisors: Dr. Tomova and Dr.WyelsAdvisors: Dr. Tomova and Dr.Wyels

Funding: NSF, NSA, and Moody’s, via the SUMMA program.

Distance: Distance: dd((u,vu,v)) Length of shortest path Length of shortest path

between two vertices between two vertices uu and and vvEx:Ex: dd((vv11,,vv66)=2)=2

Diameter: diam(Diameter: diam(GG)) Maximum distance in a graph Maximum distance in a graph

over all vertices.over all vertices.Ex: Ex: diam(diam(GG)=3)=3

Graph TerminologyGraph Terminology

V1V2

V3

V4 V5

V6

G

Radio LabelingRadio Labeling• A function A function c c thatthat assigns positive integer values to each assigns positive integer values to each

vertex so as to satisfy the vertex so as to satisfy the radio conditionradio condition

dd((uu,,vv)) + |c + |c((uu))-c-c((vv))| | ≥ ≥ diam(diam(GG)) + + 1.1.

diam (diam (GG) - diameter of graph ) - diameter of graph dd((uu,,vv) - distance between vertices u and v) - distance between vertices u and v

cc((uu),c(),c(vv) – label assigned to vertices) – label assigned to vertices

1 + |1 – 1 + |1 – cc((vv)| ≥ 4)| ≥ 4

11 ++ c c((vv) ) – – 11 ≥≥ 44

cc((vv)) ≥≥ 44

dd((uu,,vv)) ++ |c |c((uu) – ) – cc((vv))| | ≥≥ 44 141

84

27 610

13

151 + |4 – 1 + |4 – cc((vv)| ≥ 4)| ≥ 4

11 + c + c((vv) ) – – 44 ≥≥ 44

cc((vv)) ≥≥ 77 10

G

Sample LabelingSample Labeling

d(u,v) |c(u) – c(v)|

1 3

2 2

3 1

SpanSpan(c) – Maximum label value assigned to a vertex in a graph. – Maximum label value assigned to a vertex in a graph.

diam(G)=3

Can we get a lower span?Can we get a lower span?

Span(Span(cc)=10)=10

Yes we can!Yes we can!

dd((uu,,vv)) + |c + |c((uu) – ) – cc((vv))| | ≥≥ diam(G) + 1diam(G) + 1

410

1

2 6

1

13

15

7 10

4

8

What is Radio Number?What is Radio Number?The radio number of G, The radio number of G, rnrn((GG), is the minimum span, ), is the minimum span,

taken over all possible radio labelings of taken over all possible radio labelings of GG..

G

rn(G)

VV(2,5)(2,5)

VV(1,2)(1,2)VV(1,1)(1,1) VV(1,3)(1,3) VV(1,4)(1,4) VV(1,5)(1,5) VV(1,6)(1,6) VV(1,7)(1,7)

VV(2,2)(2,2)VV(2,1)(2,1) VV(2,3)(2,3) VV(2,4)(2,4) VV(2,6)(2,6) VV(2,7)(2,7)

What is the distance between VV(1,2)(1,2) and VV(2,5)(2,5)?

Odd Ladders Odd Ladders

3,7:7 knL

),( )5,2()2,1( VVd

|52||21| 4

12: knLn

Lower BoundLower Bound

.424)(Then

. of labeling radioany be Let :Theorem2

12

kkcspan

Lc k

Lower BoundLower Bound

).(),(1)diam()( 11 iiii xcxxdGxc

.1)diam()()(),( 11 Gx-cxcxxd iii i

.ifonly and if)()( jixcxc ji

Proof: List the vertices of Ln as {x1, x2, …, x2n} in increasing label order:

The radio condition implies

Rewrite this as

)(x),(1 )diam()()( 122122 nnnn cxxdGxccspan

Expansion of the InequalityExpansion of the Inequality

)( 12nxc

12

112

1

),( 11] )diam()[12()(

1)(n

iiin xxdGnxc

xc

)(x),(1 )diam( 221222 nnn cxxdG

Key IdeaKey Idea

14

112 ),( 11] )diam()[12()(

k

iiin xxdGnxc

c(x2n) is the span of the labeling c.

The smallest possible value of c(x2n) corresponds to the largest possible value of

.),(14

11

k

iii xxd

σσ--ττ Notation Notation

5)2(2)2(5,22

Vx

7L

6)3(1)3(6,13

Vx

VV(1,1)(1,1) VV(1,2)(1,2) VV(1,3)(1,3) VV(1,4)(1,4) VV(1,5)(1,5) VV(1,6)(1,6) VV(1,7)(1,7)

VV(2,2)(2,2)VV(2,1)(2,1) VV(2,3)(2,3) VV(2,4)(2,4) VV(2,5)(2,5) VV(2,6)(2,6) VV(2,7)(2,7)

)()( iii Vx

2x

3x

|)3()2(||)3()2(|

and between Distance 32

xx

Maximizing the DistanceMaximizing the Distance

)2()12()2()12(

)3()2()2()1()2()1(

nnnn

14

11),(

k

iii xxd

VV(1,(1,nn))VV(1,1)(1,1) VV(1,2)(1,2) VV(1,(1,kk)) VV(1,(1,kk+1)+1)

VV(2,2)(2,2)VV(2,1)(2,1) VV(2,(2,kk)) VV(2,(2,kk+1)+1)

VV(1,(1,nn-1)-1)

VV(2,(2,nn-1)-1) VV(2,(2,nn))

Maximizing the DistanceMaximizing the Distance

)2()12()12()22(

)4()3()3()2()2()1(

nnnn

).12( is sum theofpart first for the max valueThe n

)2()12()12()22(

)4()3()3()2()2()1(

nnnn

Maximizing the DistanceMaximizing the Distance

)2()12()12()22(

)4()3()3()2()2()1(

nnnn

.12,...,2each for twiceappears )( nii

.2,1each for once appears )( nii

terms.negative

and positive ofnumber equalan are There

Maximizing the DistanceMaximizing the Distance

)2()12()12()22(

)4()3()3()2()2()1(

nnnn

of sum in the twiceup showonly will4

4)2( and4)1(Let

)2()1(:1Case

nττ

nττ

each times3appear will7 and 2

7)2( and2)1(Let

)2()1( :2 Case

nττ

nττ

Positive Negative

Maximizing Distance of Maximizing Distance of LL7

3,7 kn

1,1,1,1

2,2,2,2

3,3,3,3

7,7,7,7

5,5,5,56,6,6,6

4 4

Using the best caseUsing the best case

Maximizing Distance of Maximizing Distance of LL2k+1

Positive Negative4)12( k

4)2(

4)2(

k

k

kkkk ,,,

2,2,2,2

1,1,1,1

1k1k

][][ )1(4)1(4|)()1(|1

12

2

12

1

kikiiik

i

k

ki

k

i

424)( 2 kkcspan

Lower Bound for LLower Bound for L22kk+1+1

224)( 2 kkcspan

14)1(4)1(4 ][][1

12

2

kkikik

i

k

ki

14

112 ),( 11] )diam()[12()(

k

iiin xxdGnxc

Upper BoundUpper Bound

.424)( :Theorem 212 kkLrn k

Labeling AlgorithmLabeling Algorithm

xx33 xx1212 xx66 xx88 xx1010 xx11 xx44 xx1313 xx77 xx99 xx1111 xx22 xx55

xx1414

xx1515

xx1616

xx1717

xx1818

xx1919

xx2020

xx2121

xx2222

xx2323

xx2424

xx2525

xx2626

13P

13L

The Upper BoundThe Upper Bound

Radio condition:Radio condition:

The upper bound:The upper bound:

1)(diam)()(),( 11 Gxcxcxxd iiii

424)( 22 kkxc n

ConclusionConclusion

424)(424 212

2 kkLrnkk k

424)( 212 kkLrn k

Even LaddersEven Ladders

724)(224 22

2 kkLrnkk k

ReferencesReferences

D. Liu and X. Zhu, D. Liu and X. Zhu, Multilevel Distance labelings Multilevel Distance labelings for paths and cycles, for paths and cycles, SIAM J. Discrete Math. SIAM J. Discrete Math. 1919 (2005), No. 3, 610-621. (2005), No. 3, 610-621.