Undergraduate Research Cindy Wyels CSU Channel Islands Research with Undergraduates Session MathFest 2009 Overview Distance labeling schemes Radio labeling Research with undergrads context ID: 728398
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Slide1
Graph Labeling Problems Appropriate for
Undergraduate Research
Cindy WyelsCSU Channel Islands
Research with Undergraduates Session
MathFest, 2009Slide2
Overview
Distance labeling schemes
Radio labeling
Research with undergrads: context
Problems for undergraduate research
Radio numbers of graph families
Radio numbers and graph properties
Properties of radio numbers
Radio numbers and graph operations
Achievable radio numbersSlide3
Distance Labeling
Motivating Context: the Channel Assignment Problem
General Idea:
geographically close
transmitters must be assigned channels with
large frequency differences
;
distant
transmitters may be assigned channels with relatively
close frequencies
.Slide4
Channel Assignment via Graphs
The diameter of the graph G, diam(G
), is the longest distance in the graph.Model: vertices correspond to transmitters.
The
distance
between vertices
u
and
v
,
d
(
u
,v), is the length of the shortest path between u and v.
u
v
w
d
(
u
,
v
) = 3
d
(
w,v
) = 4
diam(
G
) = 4Slide5
Defining Distance Labeling
All graph labeling starts with a function f : V
(G) → N
that satisfies some conditions.
f
(
v
) = 3
f
(
w
) = 1
2
1
3
1
3
1
5
3
w
vSlide6
Some distance labeling schemes
f : V(G
) → N satisfies ______________
k
-labeling:
Antipodal: (same)
Radio: (same)
L
d
(2,1):
Slide7
Radio:
4
1
6
3
1
4
7
2
The
radio number
of a graph
G
, rn(
G
), is the smallest integer
m
such that
G
has a radio labeling
f
with
m
= max{
f
(
v
) |
v
in
V
(
G)}.
rn
(
P
4
) = 6Slide8Slide9
Radio Numbers of Graph Families
Standard problem: find rn
(G) for all graphs G
belonging to some family of graphs.
“… determining the radio number seems a difficult problem even for some basic families of graphs.”
(Liu and Zhu)
Complete graphs, wheels, stars (generally known)
S
5
4
1
4
5
3
6
diam(
S
n
) = 2
rn(
S
n
) =
n
+ 1Slide10
Radio Numbers of Graph Families
Complete
k-partite graphs (Chartrand, Erwin, Harary, Zhang)Paths and cycles (Liu, Zhu)
Squares of paths and cycles (Liu, Xie)
Spiders (Liu)Slide11
Radio Numbers of Graph Families
Gears (REU ’06)
Products of cycles (REU ’06)Generalized prisms (REU ’06)
Grids* (REU ’08)
Ladders (REU ’08)
Generalized gears* (REU ’09)
Generalized wheels* (REU ’09)
Unnamed families (REU ’09)Slide12
Radio Numbers & Graph Properties
Diameter
GirthConnectivity
(your favorite set of graph properties)
Question:
What can be said about the radio numbers of graphs with these properties?Slide13
E.g. products of graphs
The (box) product of graphs G and H,
G □ H, is the graph with vertex set V(
G
)
×
V
(
H
)
,
where (
g1, h1) is adjacent to (
g2,
h2) if and only if
g1 = g
2 and h1 is adjacent to
h2 (in H), and
h1 =
h2 and g
1 is adjacent to g2
(in G).
a
1
3
5
b
(
a
,
1
)
(
b
,
3
)
(
a
,
5
)
(
b
,
5
)
Radio Numbers & Graph OperationsSlide14
Graph Numbers and Box Products
Coloring: χ
(G□H) = max{χ
(
G
),
χ
(
H
)}
Graham’s Conjecture:
π
(
G□
H) ≤ π
(G) ∙ π
(H)
Optimal pebbling: g
(G□
H) ≤ g(
G) ∙ g
(H)
Question:
Can rn(
G □ H
) be determined by rn(G) and rn(
H)? If not, what else is needed?Slide15
REU ’07 students at JMM
Bounds on radio numbers of products of graphsSlide16
REU ‘07 Results – Lower Bounds
Radio Numbers: rn(
G □ H) ≥ rn(
G
) ∙ rn(
H
) - 2
Number of Vertices:
rn(
G
□
H
) ≥ |V
(G)| ∙ |V
(H)|
Gaps: rn(
G □
H) ≥ (½(|V(
G)|∙|V
(H)| - 1)(
φ(G
) - φ(
H) – 2) Slide17
Analysis of Lower Bounds
Product
Radio No.
Vertices
Gap
C
4
□
P
2
5
8
–
C
n
□
P
2
n
2
/8
2
n
–
C
4
□
C
4
8
16
30
C
n
□
C
n
n
2
/4
n
3
/8
n
2
P
4
□
P
4
10
16
30
P
100
□
P
100
9,800
10,000
499,902
P
n
□
P
n
n
2
n
2
n
3
/4
Pete
□
Pete
18
100
100Slide18
Theorem (REU ’07):
Assume G and H
are graphs satisfying diam(G) - diam(H) ≥ 2 as well as
rn(
G
) =
n
and rn(
H
) =
m
. Then
rn(
G
□
H) ≤ diam(G)(
n+m
-2) + 2mn - 4n
- 2m + 8.
REU ’07 proved two other theorems providing upper bounds under different hypotheses.
REU ‘07 Results – Upper BoundsSlide19
Need lemma giving M = max{
d(u,v)+d
(v,w)+d(w
,
v
)}.
Assume
f
(
u
) <
f
(
v) < f(w).
Summing the radio condition
d(u,v
) + |f(u) -
f(v)| ≥ diam(
G) + 1
for each pair of vertices in {u, v,
w} gives M + 2
f(w) – 2
f(u)
≥ 3 diam(G) + 3
i.e.f
(w) – f
(u) ≥ ½(3 diam(
G) + 3 – M
).Using Gaps
gapSlide20
Have f(
w) – f(u) ≥ ½(3 diam(
G) + 3 – M) = gap.
If |
V
(
G
)| =
n
, this yields
Using Gaps, cont.
gap
+ 1
gap
+ 2
gap
2
gap
+ 2
2
gap
+ 1
gap
1
2
gapSlide21
Using Gaps to Determine a Lower Bound for the Radio Number of Prisms
Y
6
Choose any three vertices
u
,
v
, and
w
.
d
(
u
,
v
) +
d
(
u
,
w
) +
d
(
v
,
w
)
≤ 2∙diam(
Y
n
) (
n
even)
u
v
wSlide22
Assume we have a radio labeling
f of Yn, and f(
u) < f(v) < f(w). Then Slide23
Strategies for establishing an upper bound for rn
(G)Define a labeling, prove it’s a radio labeling, determine the maximum label.
Might use an intermediate labeling that orders the vertices {x1, x2
, …
x
s
} so that
f
(
x
i
) >
f
(xj) iff i > j.
Using patterns, iteration, symmetry, etc. to define a labeling makes it easier to prove it’s a radio labeling.