radio labeling cartesian products of path graphs eduardo calles and henry gómez advisors: drs....

Radio Labeling Cartesian Radio Labeling Cartesian Products of Path GraphsProducts of Path Graphs

Eduardo Calles and Henry GEduardo Calles and Henry Góómezmez

Advisors: Drs. Maggy Tomova and Cindy WyelsAdvisors: Drs. Maggy Tomova and Cindy Wyels

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

Grid GraphsGrid GraphsThe path graph The path graph PPnn contains contains nn consecutive vertices consecutive vertices

connected along a sequence of connected along a sequence of n n - 1 edges.- 1 edges.

When we “multiply” two path graphs (take the When we “multiply” two path graphs (take the Cartesian product), the graph product is a grid graph.Cartesian product), the graph product is a grid graph.

Ex: P5

Ex: P3 □ P3

Transition to TablesTransition to Tables

v1 v2

v3

v7v9

v4 v5

v8

v6

Vertices are now represented by boxes.

Two boxes represent adjacent vertices if they share an edge.

v1 v2 v3

v4 v5 v6

v7 v8 v9

Strategy for Establishing an Upper Strategy for Establishing an Upper Bound for Bound for rnrn((PPnn □□ PPnn))

Specify the order in which we’ll label vertices.Specify the order in which we’ll label vertices.

Give the vertices the minimum label values Give the vertices the minimum label values required so as to satisfy the radio condition.required so as to satisfy the radio condition.

Calculate the span of this labeling.Calculate the span of this labeling.

This span is an upper bound for This span is an upper bound for rnrn((PPnn □□ PPnn).).

2

12

1239

1041

16 36 64 73 50 26 10 24

34 62 75 71 52 28 1214

60 77 69 54 3032

79 67 5658

6581

51 63 7472

70 53 61 76 4442

40 68 55 59 78 46 2220

18 38 66 57 80 48 24 86

Order of Vertex LabelsOrder of Vertex Labels

4

6 8

4 2

6 8

4 7 2

9 1 5

6 3 811

13

14

15

16

17

18

19

20

21

22

24

23

25

1043

46

42

16

35

14

18

27

28

31

30

33

38

40

24

26

20 22

44

29

32

36

37

34 45

32&

13

311

12

210 47

48

49

General Ordering (for Odds)General Ordering (for Odds)

1,1 1,2k+11,k+11,k

2k+1,2k+1

k+1,1

k,1

2k+1,1 2k+1,k 2k+1,k+1

k+1,k+1

Stage 0: (k+1, k+1)

Stage 1: top right side,

top left side,

bottom left side,

bottom right side,

Stage 2: repeat the cycle

k = number of stages

Any stages greater than stage 0 complete the diamond cycle pattern

Distance traveled to start new stage is 2k;

Distance traveled within stages is 2k+1 each time.

Calculating the Span for Odd GraphsCalculating the Span for Odd Graphs

StageStage Number of Number of vertices to labelvertices to label

Label Values addedLabel Values added

00 11 11

11 8 = 8(1)8 = 8(1) (2(2kk+1) + (8-1)(2+1) + (8-1)(2kk))

22 16 = 8(2)16 = 8(2) (2(2kk+1) + (16-1)(2+1) + (16-1)(2kk))

33 24 = 8(3)24 = 8(3) (2(2kk+1) + (24-1)(2+1) + (24-1)(2kk))

ii 8(8(ii)) (2(2kk+1) +[8(+1) +[8(ii)-1](2)-1](2kk))

kk 8(8(kk)) (2(2kk+1) +[8(+1) +[8(kk)-1](2)-1](2kk))

.188

)]12()18(2[)(

23

1

kkk

kikPPrnk

inn

Strategy to Establish a Lower Bound Strategy to Establish a Lower Bound for for rnrn((PPnn □□ PPnn))

Let Let cc be any radio labeling of be any radio labeling of PPnn □ □ PPnn..

1.1. Develop equation relating span(Develop equation relating span(cc) to the sum of ) to the sum of distances between consecutively-labeled vertices.distances between consecutively-labeled vertices.

2.2. Minimize the span by maximizing this sum of Minimize the span by maximizing this sum of distances.distances.

This minimum span is a lower bound for This minimum span is a lower bound for rnrn((PPnn □□ PPnn).).

General Equation for span(General Equation for span(cc))List the vertices of List the vertices of PPnn □ □ PPnn as {as {xx11,…,,…,xxnn22} in } in

increasing label order:increasing label order:

Radio Condition gives a necessary condition:Radio Condition gives a necessary condition:

Rewrite:Rewrite: )(),(1)diam()( 11 iiii xcxxdGxc

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

jixcxc ji ifonly and if)()(

)(),(1 )diam()(11 2222

nnnn

xcxxdGxc

Expansion of the InequalityExpansion of the Inequality

)(12n

xc

1

11

2

1

2

), 11] )diam()[1()(

)(n

iii xd(xGncspan

xc

)(),(1 )diam(212 222

nnn

xcxxdG

To minimize the span, maximize the sum of the distances.

Calculating Distance Calculating Distance dd((xxii,,xxii+1+1))

)()( :Convert to iii vx

What is the distance between x3 and x4?

3

1321

)4()3()4()3(

),( )4()4()3()3(43

vvxxd

xx11 = vv(1,1)(1,1) xx33 = = vv(1,3)(1,3)

xx44 = = vv(2,1)(2,1)

xx22 = = vv(1,2)(1,2)

xx55 = = vv(2,2)(2,2)

xx66 = = vv(2,3)(2,3)

xx77 = = vv(3,1)(3,1)xx88 = = vv(3,2)(3,2)

xx99 = = vv(3,3)(3,3)i

i

xi

xi

ofindex column thegives )( and

ofindex rowthegives)(where

Maximizing the Sum of DistancesMaximizing the Sum of Distances

,)1()()1()(),( Recall 1 iiiixxd ii

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

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

)2()1()2()1(), So1

11

2

kkkk

xd(xn

iii

),(),(), 22

2

121

1

11 nn

n

iii xxdxxdxd(x


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

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

)2()1()2()1( Examine

kkkk

By examining the sum of distances, σ’s and τ’s appear

Positive Negative

2

]4)12(4[)1(

kk


times)12(4 taken )2(

times)12(4 taken)2(

kk

kk

times)12(4 taken

times)12(4 taken 2

kk

k

times)12(4taken1 k

]4)12(4)[(1( kk

n = 2k+1

2

]4)12(4[)1(

kk

Same amount need to be added and subtracted.

times)12(4 taken )12( kk

Generalized Lower Bound for Generalized Lower Bound for Odds where Odds where n n = 2= 2kk + 1 + 1

][1

12

2

)12(4

k

i

k

ki

iik

188)( 231212 kkPPrn kk

1)12(

11

21212

2

), 11] )diam()[1)12(()(k

iiikk xd(xGkPPrn

Evens ( n = 2k)

Odds ( n = 2k + 1)

Our ResultsOur Results

188)(188 231212

23 kkkPPrnkk kk

148)(4448 2322

23 kkPPrnkkk kk

radio labeling cartesian products of path graphs eduardo calles and henry gómez advisors: drs....

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