The Growth Degree of Vertex Replacement Rules Presenter: Nicholas Ross Advisor: Michelle Previte Penn State Erie, The Behrend College Date: April 2006
Jan 12, 2016
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.