Graph Labeling Problems Appropriate for Undergraduate Research Cindy Wyels CSU Channel Islands Research with Undergraduates Session MathFest, 2009
Feb 25, 2016
Graph Labeling Problems Appropriate for
Undergraduate Research
Cindy WyelsCSU Channel Islands
Research with Undergraduates Session
MathFest, 2009
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
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.
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
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
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
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
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
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)
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)
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?
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
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?
REU ’07 students at JMM
Bounds on radio numbers of products of graphs
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)
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
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
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
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
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
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
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.