1. Graph Labeling Problems Appropriate for Undergraduate Research Cindy Wyels CSU Channel Islands Research with Undergraduates Session MathFest, 2009

2. Overview Distance labeling schemes Distance labeling schemes Radio labeling Radio labeling Research with undergrads: context Research with undergrads: context Problems for undergraduate research 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

3. 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.

4. 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) = 4

5. 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 15 3 w v

6. Some distance labeling schemes f : V(G) → N satisfies ______________ k-labeling: Antipodal:(same) Radio: (same) L d (2,1):

7. Radio: 41 6 3 1472 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 ) = 6

9. 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) S54S54 1 4 5 3 6 diam(S n ) = 2 rn(S n ) = n + 1

10. 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)

11. 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)

12. 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?

13. 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 (g 1, h 1 ) is adjacent to (g 2, h 2 ) if and only if g 1 = g 2 and h 1 is adjacent to h 2 (in H), and h 1 = h 2 and g 1 is adjacent to g 2 (in G). a 1 3 5 b (a, 1) (b, 3) (a, 5) (b, 5) Radio Numbers & Graph Operations

14. 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?

15. REU ’07 students at JMM Bounds on radio numbers of products of graphs

16. 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)

17. Analysis of Lower Bounds Product Radio No. VerticesGap C 4 □ P 2 58– C n □ P 2 C n □ P 2 n 2 /8 2n2n2n2n– C 4 □ C 4 81630 C n □ C n n 2 /4 n 3 /8 n2n2n2n2 P 4 □ P 4 101630 P 100 □ P 100 9,80010,000499,902 P n □ P n n2n2n2n2 n2n2n2n2 n 3 /4 Pete □ Pete 18100100

18. 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 Bounds

19. 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

20. 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 + 2 2gap + 1 gap 12

21. Using Gaps to Determine a Lower Bound for the Radio Number of Prisms Y6Y6 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 w

22. Assume we have a radio labeling f of Y n, and f(u) < f(v) < f(w). Then

23. 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 {x 1, x 2, … x s } so that f(x i ) > f(x j ) iff i > j. Using patterns, iteration, symmetry, etc. to define a labeling makes it easier to prove it’s a radio labeling.