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Graph Labeling Problems Appropriate for
Undergraduate Research
Cindy WyelsCSU Channel Islands
Research with Undergraduates Session
MathFest, 2009
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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 numbers
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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.
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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
vw
d(u,v) = 3
d(w,v) = 4 diam(G) = 4
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Defining Distance LabelingAll graph labeling starts with a function
f : V(G) → N
that satisfies some conditions.
f(v) = 3
f(w) = 12
1
3
1
3
1 5
3
w
v
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Some distance labeling schemesf : V(G) → N satisfies ______________
k-labeling:
Antipodal: (same)
Radio: (same)1)(diam)()(),( Gvfufvud
)(diam)()(),( Gvfufvud
)(,1)()(),( GVvukvfufvud
Ld(2,1):
2),(when1),(when2
)()(vuddvudd
vfuf
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Radio: 1)(diam)()(),( Gvfufvud
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(P4) = 6
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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)
S5
4
14
5
3
6
3)()(),( vfufvuddiam(Sn ) = 2
rn(Sn) = n + 1
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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)
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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)
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Radio Numbers & Graph Properties
Diameter Girth Connectivity (your favorite set of graph properties)
Question: What can be said about the radio numbers of graphs with these properties?
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E.g. products of graphsThe (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 = g2 and h1 is adjacent to h2 (in H), andh1 = h2 and g1 is adjacent to g2 (in G).
a1
3
5
b
(a, 1)
(b, 3)
(a, 5)
(b, 5)
Radio Numbers & Graph Operations
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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?
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REU ’07 students at JMM
Bounds on radio numbers of products of graphs
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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)
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Analysis of Lower Bounds
Product Radio No. Vertices GapC4 □ P2 5 8 – Cn □ P2 n2/8 2n –C4 □ C4 8 16 30Cn □ Cn n2/4 n3/8 n2
P4 □ P4 10 16 30P100 □ P100 9,800 10,000 499,902Pn □ Pn n2 n2 n3/4
Pete □ Pete 18 100 100
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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. Thenrn(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 Bounds
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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 + 2f(w) – 2f(u) ≥ 3 diam(G) + 3
i.e.f(w) – f(u) ≥ ½(3 diam(G) + 3 – M).
Using Gaps
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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
2gap + 22gap +
1gap
1 2
gap
even.212
odd,12
1
)(nngap
nngapGrn
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Using Gaps to Determine a Lower Bound for the Radio Number of Prisms
Y6
Choose any three vertices u, v, and w.
21)(diam nYn
d(u,v) + d(u,w) + d(v,w) ≤ 2∙diam(Yn) (n even)
u v
w
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Assume we have a radio labeling f of Yn, and f(u) < f(v) < f(w). Then
1)(diam)()(),( nYufwfwud1)(diam)()(),( nYufvfvud
1)(diam)()(),( nYvfwfwvd
3)(diam3)(2)(2),(3 nYufwfwud
23)()()(
nYdiamufwf
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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, … xs} so that f(xi) > f(xj) iff i > j.Using patterns, iteration, symmetry, etc. to define a labeling makes it easier to prove it’s a radio labeling.