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.