Radio Labeling Cartesian Products of Path Graphs Eduardo Calles and Henry Gómez Advisors: Drs. Maggy Tomova and Cindy Wyels Funding: NSF, NSA, and Moody’s, via the SUMMA program.

Grid Graphs The path graph P n contains n consecutive vertices connected along a sequence of n - 1 edges. When we “multiply” two path graphs (take the Cartesian product), the graph product is a grid graph. Ex: P 5 Ex: P 3 □ P 3

Transition to Tables v1v1 v2v2 v3v3 v7v7 v9v9 v4v4 v5v5 v8v8 v6v6 Vertices are now represented by boxes. Two boxes represent adjacent vertices if they share an edge. v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v7v7 v8v8 v9v9

Strategy for Establishing an Upper Bound for rn(P n □ P n ) Specify the order in which we’ll label vertices. Give the vertices the minimum label values required so as to satisfy the radio condition. Calculate the span of this labeling. This span is an upper bound for rn(P n □ P n ).

Order of Vertex Labels

General Ordering (for Odds) 1,11,2k+11,k+11,k 2k+1,2k+1 k+1,1 k,1 2k+1,1 2k+1,k2k+1,k+1 k+1,k+1 Stage 0: (k+1, k+1) Stage 1: top right side, top left side, bottom left side, bottom right side, Stage 2: repeat the cycle k = number of stages Any stages greater than stage 0 complete the diamond cycle pattern Distance traveled to start new stage is 2k; Distance traveled within stages is 2k+1 each time.

Calculating the Span for Odd Graphs Stage Number of vertices to label Label Values added = 8(1) (2k+1) + (8-1)(2k) 2 16 = 8(2) (2k+1) + (16-1)(2k) 3 24 = 8(3) (2k+1) + (24-1)(2k) i 8(i) (2k+1) +[8(i)-1](2k) k 8(k) (2k+1) +[8(k)-1](2k)

Strategy to Establish a Lower Bound for rn(P n □ P n ) Let c be any radio labeling of P n □ P n. 1.Develop equation relating span(c) to the sum of distances between consecutively-labeled vertices. 2.Minimize the span by maximizing this sum of distances. This minimum span is a lower bound for rn(P n □ P n ).

General Equation for span(c) List the vertices of P n □ P n as {x 1,…,x n 2 } in increasing label order: Radio Condition gives a necessary condition: Rewrite:

Expansion of the Inequality To minimize the span, maximize the sum of the distances.

Calculating Distance d(x i,x i+1 ) What is the distance between x 3 and x 4 ? x 1 v (1,1) x 1 = v (1,1) x 3 = v (1,3) x 4 = v (2,1) x 2 = v (1,2) x 5 = v (2,2) x 6 = v (2,3) x 7 = v (3,1) x 8 = v (3,2) x 9 = v (3,3)

Maximizing the Sum of Distances

By examining the sum of distances, σ’s and τ’s appear

PositiveNegative Maximizing the Sum of Distances n = 2k+1 Same amount need to be added and subtracted.

Generalized Lower Bound for Odds where n = 2k + 1

 Evens ( n = 2k)  Odds ( n = 2k + 1) Our Results