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The Growth Degree of Vertex Replacement Rules

This presentation by Nicholas Ross, supervised by Michelle Previte at Penn State Erie, delves into the growth degree of vertex replacement rules applied to finite graphs. It examines the function ( f(m,x,X) ), which counts vertices within a specified distance from a vertex ( x ). The paper provides a mathematical framework for understanding the diameter of ( Rn(G) ) graphs and presents conjectures relating growth degrees to fractal dimensions. The study highlights the independence of the growth degree from initial graph structures and their implications in graph theory.

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The Growth Degree of Vertex Replacement Rules

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  1. The Growth Degree of Vertex Replacement Rules Presenter: Nicholas Ross Advisor: Michelle Previte Penn State Erie, The Behrend College Date: April 2006

  2. Growth Degree For each nonnegative integer m, let f(m,x,X) denote the number of vertices at distance at most m from some arbitrary vertex, x. Distance Vertices 0 1 1 5 x 2 13 3 25 mf(m,x) f(m, x) = 2m2 + 2m + 1 X

  3. Vertex Transitive Graphs X X f(m,x,X) = f(m,X) = 2m2 + 2m + 1

  4. A Vertex Replacement Rule, Ris a finite set of finite graphs called replacement graphs. Every replacement graph in the replacement rule has a designated set of boundary vertices. H1 H2

  5. A Replacement In Action v1 v1 H1 w1 w2 w1 w2 v3 v2 w3 G v2 v3 w3 H2 R(G)

  6. The Sequence {Rn(G)} G R(G) R2(G) R3(G)

  7. The Limit Graph, X

  8. Another Example H G R(G)

  9. The Sequence {Rn(G)} G R(G) R2(G) R3(G)

  10. Not Vertex Transitive X X

  11. Growth Degree Since our limit graphs are not usually vertex transitive, we need to pick a point from which to measure. x f(m, x, X) = the number of vertices at distance at most mfrom a pointx in space X. X

  12. To compute f(m,x,X) we approximate X by Rn(G) X x x Rn(G)

  13. f(m,x,X) = f(diam(Rn(G)), x, X) ≈the total number of vertices in Rn(G) Now we need a way to find the diameter of Rn(G) and the number of vertices in Rn(G).

  14. N(۰) =Total # of vertices in ۰ N(Rn(G)) = 3 + 3n H G N(R3(G)) = 30 R3(G)

  15. N(Rn(G)) For replacement rules with exactly one replacement graph the general formula is N(Rn(G)) = nrep(G) + nrep(G) * nrep(H) * + rep(G) * rep(H)n rep(H)n - 1 rep(H) - 1 rep(H) ≠ 1 Example: N(Rn(G)) = 3 + 3 * 0 * + 1 * 3n = 3 + 3n 3n – 1 3 – 1

  16. The diameter of Rn(G) depends on a simple boundary connecting path in H. L(σ) = the length of a path, σ, in H. rep(σ) = the number of replaceable vertices on σ in H. Example: L(σ) = 1 rep(σ) = 2 H

  17. Diameter of Rn(G) 2n – 1 2 – 1 diam(Rn(G)) = 2 + (1) = 2n + 1 H G R3(G) R3(G) = 23 + 1 = 9

  18. Diameter of Rn(G) For replacement rules with exactly one replacement graph the general formula is rep(σ)n – 1 rep(σ) – 1 diam(Rn(G)) = diam(G) + L(σ) , rep(σ) ≠ 1 Example: diam(Rn(G)) = 2 + 1 * = 2n + 1 2n – 1 2 – 1

  19. Putting It Together f(diam(Rn(G)), x, X) ≈ total number of vertices in Rn(G) Example: f(2n + 1, x, X) ≈ 3n+ 3 f(2n + 1 x, X) ≈ 3n f(m, x, X) ≈ mln3/ln2

  20. Conjecture 1 Let G be a finite initial graph with at least 1 replaceable vertex, and let R be a replacement rule with only 1 graph. The growth degree of the limit graph X of {Rn(G)} is independent of the vertex x in X and the initial graph G.

  21. How Does This Relate to Fractals? Let (Rn(G),1) = Rn(G) scaled to have diameter 1. Then {(Rn(G),1)} usually converges to a fractal, Y. (G,1) (R(G),1) (R2(G),1) Y = lim (Rn(G),1)

  22. Conjecture 2 The growth degree of the limit X of Rn(G) is the same as the fractal dimension (i.e. Hausdorff dimension) of the limit Y of the sequence {(Rn(G),1)} of scaled vertex replacements.

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