The Growth Degree of Vertex Replacement Rules

Presenter: Nicholas Ross

Advisor: Michelle Previte

Penn State Erie, The Behrend College

Date: April 2006

Growth Degree

f(m, x) = 2m2 + 2m + 1

Distance Vertices

0 11 52 133 25m f(m,x)

X

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.

x

Vertex Transitive Graphs

X X

f(m,x,X) = f(m,X) = 2m2 + 2m + 1

Every replacement graph in the replacement rule has a

designated set of boundary vertices.

H1 H2

A Vertex Replacement Rule, R is a finite set of finite graphs called

replacement graphs.

A Replacement In Action

R (G)

H1

H2

G

w3

w2w1

v3 v2

v1

w3

w1 w2

v3v2

v1

The Sequence {Rn(G)}

G R(G) R2(G) R3(G)

The Limit Graph, X

Another Example

H

G

R(G)

The Sequence {Rn(G)}

R3(G)R(G)G R2(G)

Not Vertex Transitive

X X

Growth Degree

f(m, x, X) = the number of vertices at distance at most m from a point x in space X.

Since our limit graphs are not usually vertex transitive, we need to pick a point from which to measure.

X

x

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

X Rn(G)

xx

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).

N(۰) =Total # of vertices in ۰

R3(G)

N(Rn(G)) = 3 + 3n

H G

N(R3(G)) = 30

N(Rn(G))

N(Rn(G)) = nrep(G) + nrep(G) * nrep(H) * + rep(G) * rep(H)n

rep(H)n - 1 rep(H) - 1

For replacement rules with exactly one replacement graph the general formula is

Example:

N(Rn(G)) = 3 + 3 * 0 * + 1 * 3n = 3 + 3n

3n – 1 3 – 1

rep(H) ≠ 1

The diameter of Rn(G) depends on a simple boundary connecting path in H.

L(σ) = the length of a path, σ, in H.

H

rep(σ) = the number of replaceable vertices on σ in H.

Example: L(σ) = 1rep(σ) = 2

Diameter of Rn(G)

diam(Rn(G)) = 2 + (1)

R3(G)

2n – 1 2 – 1

= 2n + 1

H G

R3(G) = 23 + 1 = 9

Diameter of Rn(G)

For replacement rules with exactly onereplacement graph the general formula is

diam(Rn(G)) = diam(G) + L(σ) , rep(σ) ≠ 1rep(σ)n – 1 rep(σ) – 1

Example:

diam(Rn(G)) = 2 + 1 * = 2n + 12n – 1 2 – 1

Putting It Together

f(diam(Rn(G)), x, X) ≈ total number of vertices in Rn(G)

f(m, x, X) ≈ mln3/ln2

f(2n + 1, x, X) ≈ 3n + 3

f(2n + 1 x, X) ≈ 3n

Example:

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.

How Does This Relate to Fractals?

(G,1) (R(G),1) (R2(G),1) Y = lim (Rn(G),1)

Let (Rn(G),1) = Rn(G) scaled to have diameter 1.

Then {(Rn(G),1)} usually converges to a fractal, Y.

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.