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

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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

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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 v w d(u,v) = 3 d(w,v) = 4 diam(G) = 4

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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

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

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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

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

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Graph Numbers and Box Products Coloring: (GH) = max{(G), (H)} Grahams Conjecture: (GH) (G) (H) Optimal pebbling: g(GH) 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) ((|V(G)||V(H)| - 1)((G) - (H) 2)

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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

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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. 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

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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) (3 diam(G) + 3 M). Using Gaps

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Have f(w) f(u) (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

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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) 2diam(Y n ) (n even) u v w

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

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Strategies for establishing an upper bound for rn(G) Define a labeling, prove its 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 its a radio labeling.