The Growth Degree of Vertex Replacement Rules

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.