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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
Maximizing the Sum of DistancesMaximizing the Sum of Distances
)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
Maximizing the Sum of DistancesMaximizing the Sum of Distances
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