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Graph Labeling Problems Appropriate for Undergraduate Research. Cindy Wyels CSU Channel Islands. Research with Undergraduates Session MathFest, 2009. Overview. Distance labeling schemes Radio labeling Research with undergrads: context Problems for undergraduate research
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Graph Labeling Problems Appropriate for Undergraduate Research Cindy Wyels CSU 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 u v w 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. d(u,v) = 3 d(w,v) = 4 diam(G) = 4 The diameter of the graph G, diam(G), is the longest distance in the graph.
Defining Distance Labeling All graph labeling starts with a function f : V(G) → N that satisfies some conditions. 1 f(v) = 3 f(w) = 1 3 3 3 2 v 1 1 5 w
Some distance labeling schemes f : V(G) → N satisfies ______________ Ld(2,1): k-labeling: Antipodal: (same) Radio: (same)
Radio: 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)}. 4 1 6 3 rn(P4) = 6
Radio Numbers of Graph Families 3 4 S5 4 1 6 5 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) diam(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 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 = g2 and h1 is adjacent to h2 (in H), and h1 = h2 and g1 is adjacent to g2 (in G). (a, 1) 1 a b 3 (b, 3) 5 (b, 5) (a, 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)
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
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 gap
Have f(w) – f(u) ≥ ½(3 diam(G) + 3 – M) = gap. If |V(G)| = n, this yields Using Gaps, cont. gap + 2 2gap + 2 gap + 1 2gap + 1 1 2 gap gap gap
Using Gaps to Determine a Lower Bound for the Radio Number of Prisms Y6 u v Choose any three vertices u, v, and w. w d(u,v) + d(u,w) + d(v,w) ≤ 2∙diam(Yn) (n even)
Assume we have a radio labeling f of Yn, and f(u) < f(v) < f(w). Then
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.