radio labeling of ladder graphs
DESCRIPTION
Radio Labeling of Ladder Graphs. Josefina Flores Kathleen Lewis From: California State University Channel Islands Advisors: Dr. Tomova and Dr.Wyels. Funding: NSF, NSA, and Moody’s, via the SUMMA program. Distance: d ( u,v ) Length of shortest path between two vertices u and v Ex: - PowerPoint PPT PresentationTRANSCRIPT
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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.
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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
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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
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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
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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)
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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
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Lower BoundLower Bound
.424)(Then
. of labeling radioany be Let :Theorem2
12
kkcspan
Lc k
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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
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)(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
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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
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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
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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))
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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
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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
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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
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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
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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
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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
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Upper BoundUpper Bound
.424)( :Theorem 212 kkLrn k
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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
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The Upper BoundThe Upper Bound
Radio condition:Radio condition:
The upper bound:The upper bound:
1)(diam)()(),( 11 Gxcxcxxd iiii
424)( 22 kkxc n
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ConclusionConclusion
424)(424 212
2 kkLrnkk k
424)( 212 kkLrn k
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Even LaddersEven Ladders
724)(224 22
2 kkLrnkk k
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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.