transition trees and their growth...
TRANSCRIPT
TRANSITION TREES AND THEIR GROWTH RATES
JOSEPH PREVITE, MICHELLE PREVITE, AND MARY VANDERSCHOOT
Abstract. In this paper we define a class of infinite rooted trees calledtransition trees for which the growth function can be calculated usingpowers of a nonnegative matrix. We show that if this matrix is primitive,has spectral radius greater than 1, and does not have 1 as an eigenvalue,then the tree has exponential growth and the growth rate is the spectralradius of the matrix. We also describe a copy-and-paste method that canbe used to construct any self similar infinite tree G from a finite generatortree H . We show that each generator tree H has an associated matrixwhose spectral radius is the exponential growth rate of the infinite treegenerated by H .
1. Introduction.
Our primary interest will be a class of trees (called transition trees) forwhich there exists a way to count the vertices at any distance to the rootusing powers of a nonnegative square matrix. We first review some terminol-ogy and definitions related to trees, as well as some results on nonnegativematrices from Perron-Frobenius theory.
An undirected (directed) graph G consists of a nonempty set of verticesV (G) and a set of undirected (directed) edges E(G) (D(G)) where eachedge is associated to one or more vertices (an ordered pair of vertices) calledits endpoints. The directed edge from v to w will be denoted (v,w). In anundirected graph, the edges have no orientation so (v,w) and (w, v) representthe same edge.
A path from a vertex v to a vertex w in a directed (undirected) graph is afinite sequence of vertices v0, . . . , vℓ with v0 = v and vℓ = w for which thereexists a directed (undirected) edge from vi to vi+1. The number of edges ina path is called the length of the path. A cycle is a path that starts andends at the same vertex and has length greater than zero. A path or cycleis simple if it does not contain the same edge more than once.
A tree is an undirected graph that has no simple cycles. A tree is rooted
if one vertex is designated as the root. Let G be a rooted tree with rootr, and suppose each edge of G has a positive integer length. The distance
between vertices v and w, denoted d(v,w), is the sum of the lengths of theedges along the unique simple path from v to w. Define the height of G by
h(G) = maxv∈V (G)
d(v, r).
For i ≥ 0, let ci(G) be the number of vertices in G of distance i from theroot, that is, let
ci(G) = |{v ∈ V (G) : d(v, r) = i}|.(If only one graph is under consideration, we will write ci instead of ci(G).)An infinite rooted tree is a rooted tree with unbounded height.
Date: May 30, 2013.
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The level of a vertex v of G is the number of edges along the uniquesimple path from v to r. (If all edges have length one, then d(v, r) is thelevel of v.) Given vertices v and w of G, if v lies on the unique simple pathfrom w to the root, then v is an ancestor of w, and w is a descendant ofv. If v is an ancestor of w and w is k levels farther away from the root thanv, then w is a k-th descendant of v. A vertex of G is called a leaf if ithas no descendants.
w
rG
v
Figure 1. Edge (v,w) of G has length 2; all other edgeshave length 1.
In Figure 1, w is a level three vertex, a leaf, and a first descendant ofv. Also note that v is a level two vertex, and a second descendant of r.Moreover, h(G) = d(w, r) = 4 and c3(G) = 7.
We now recall the definitions of growth function and exponential growthof an infinite rooted tree. The growth function describes how fast G expandsfrom the root.
Definition 1.1. Let G be an infinite rooted tree. The growth function of
G is given by f(G,n) =
n∑
i=0
ci(G).
Definition 1.2. We say an infinite rooted tree G has exponential growth
if there exists λ > 0 and positive constants α1 and α2 such that
α1λn ≤ f(G,n) ≤ α2λ
n
for all n. The constant λ is called the growth rate of G. Observe that Ghas exponential growth with growth rate λ if and only if
limn→∞
ln f(G,n)
n= ln λ.
The following notation, definitions and results on nonnegative matricesfrom Perron-Frobenius theory can be found in [1], [2] and [4]. We will denotethe (i, j) entry of A by ai,j, and the (i, j) entry of An by (An)i,j. A realmatrix A is nonnegative (positive) if ai,j ≥ 0 (ai,j > 0) for every (i, j). Wewrite A ≥ 0 if A is nonnegative, and A > 0 if A is positive. A nonnegativesquare matrix A is called irreducible if for any (i, j) there exists k suchthat (Ak)i,j > 0. A nonnegative square matrix A is called primitive if
Ak > 0 for some k.
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The spectrum of a square matrix A, denoted σ(A), is the set of alleigenvalues of A. The spectral radius of A is defined by ρ(A) = max
λ∈σ(A)|λ|.
Theorem 1.3. Perron-Frobenius Let A be an irreducible matrix.
(a) A has a positive real eigenvalue r = ρ(A) such that r ≥ |λi| for anyeigenvalue λi of A.
(b) Furthermore, r has algebraic and geometric multiplicity one, and hasa positive eigenvector x.
(c) Any non-negative eigenvector of A is a multiple of x.
(d) mini
∑
j
ai,j ≤ r ≤ maxi
∑
j
ai,j.
(e) If A is primitive, then r > |λi| for any eigenvalue λi of A, λi 6= r.(f) If A is primitive, then the unique vector p defined by
p > 0, Ap = rp, ||p||1 = 1,
is called the Perron vector.
Theorem 1.4. Let A be a primitive matrix, and let r = ρ(A). Then
limk→∞
(A
r
)k
= L =pqT
qT p> 0
where p and q are the respective right- and left-hand Perron vectors of A.
The associated directed graph, G(A) of an n × n nonnegative matrix Aconsists of n vertices v1, . . . , vn, where there exists a directed edge from vi tovj, denoted (vi, vj) if and only if ai,j 6= 0. A graph is strongly connected
if for any two vertices vi and vj there is a directed path from vi to vj. Thefollowing theorems give necessary and sufficient conditions for a matrix tobe irreducible or primitive.
Theorem 1.5. Let A be a nonnegative matrix. Then A is irreducible if andonly if its associated directed graph is strongly connected.
Theorem 1.6. (Theorem 9.2.1, [4]) Let A be an irreducible matrix. ThenA is primitive if and only if the associated directed graph has two cycles ofrelatively prime lengths.
We now define transition trees and find sufficient conditions for such treesto have exponential growth.
Definition 1.7. An infinite rooted tree G is called a transition tree ifthere exists a square matrix A of size N × N for some N ∈ Z
+ such thatfor all n ≥ 0,
An
c1
c2...
cN
=
c1+n
c2+n...
cN+n
. (1)
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That is, c(n) = Anc(0), where c(n) =[
c1+n c2+n · · · cN+n
]T, and
ci(G) = |{v ∈ V (G) : d(v, r) = i}|. The matrix A is called a transition
matrix of G.
The transition matrix for a transition tree is not unique. For example,
A1 = [2] and A2 =
[0 14 0
]are transition matrices for the full infinite
binary tree. (In a full binary tree, each vertex that is not a leaf has exactlytwo first descendants.)
Theorem 1.8. Let G be a transition tree with transition matrix A. If Ais primitive, ρ(A) > 1, and 1 6∈ σ(A), then G has exponential growth withgrowth rate r = ρ(A).
Proof. Let G be a transition tree with N ×N transition matrix A such that
c(n) = Anc(0) where c(n) =[c1+n c2+n · · · cN+n
]T. Assume that A is
primitive, 1 6∈ σ(A), and ρ(A) > 1. Let r = ρ(A) and let L be the positive
matrix (as described in Theorem 1.4) given by L = limk→∞
(A
r
)k
.
We now derive a formula for the growth function f(G, kN) in terms of
the matrix A. Let e be the vector in RN given by e =
[1 1 · · · 1
]T.
Observe that eT Anc(0) = eTc(n) =∑N
i=1 ci+n. Note that c0 = 1, and the
matrix AN − I is invertible since 1 6∈ σ(A). Therefore
f(G, kN) =
kN∑
i=0
ci
= 1 + eT (I + AN + A2N + · · · + A(k−1)N )c(0) (2)
= 1 + eT (AN − I)−1(AkN − I)c(0)
= 1 + eT (AN − I)−1AkNc(0) − eT (AN − I)−1c(0). (3)
Since the definitions for exponential growth and exponential growth rate
involve f(G,n) not f(G, kN) we next find bounds forf(G,n)
λnthat allow us
to make use of Equation (3). For every n ≥ 0 there exists an integer kn ≥ 0such that
knN ≤ n ≤ (kn + 1)N. (4)
Observe that kn → ∞ as n → ∞. Since f(G,n) gives the number of verticesin G of distance at most n to the root, it immediately follows from Equation(4) that
0 < f(G, knN) ≤ f(G,n) ≤ f(G, (kn + 1)N). (5)
Using Equations (4) and (5), and the fact that r > 1 we have
f(G, knN)
rknNrN≤ f(G,n)
rn≤ f(G, (kn + 1)N)
rknN. (6)
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From Equation (6), Definition 1.2, and convergent sequence results itfollows that G has exponential growth with growth rate r if there exists realnumbers ℓ1 and ℓ2, 0 < ℓ1 ≤ ℓ2, such that
limk→∞
f(G, kN)
rkNrN= ℓ1, and lim
k→∞
f(G, (k + 1)N)
rkN= ℓ2.
We will first show that ℓ1 exists. Dividing through Equation (3) by rkNrN
yields
f(G, kN)
rkNrN=
1
rkNrN+
eT (AN − I)−1AkNc(0)
rkNrN− eT (AN − I)−1c(0)
rkNrN. (7)
Since r > 1, the first and last terms of the right-hand side of Equation (7)limit to 0 as k → ∞, that is,
limk→∞
1
rkNrN= 0 and lim
k→∞
eT (AN − I)−1c(0)
rkNrN= 0.
And, since limk→∞
AkN
rkN= L, we have
ℓ1 = limk→∞
f(G, kN)
rkNrN=
eT (AN − I)−1Lc(0)
rN.
We next show that ℓ1 > 0. From Equation (2) we have that
f(G, kN)
rkNrN=
1 + eT (I + AN + · · · + A(k−1)N )c(0)
rkNrN. (8)
Since A is primitive, r > 1, and the entries of e and c(0) are positive, itimmediately follows from Equation (8) that
f(G, kN)
rkNrN≥ eT A(k−1)Nc(0)
rkNrN. (9)
Since limk→∞
eT A(k−1)Nc(0)
rkNrN= lim
k→∞
eT A(k−1)Nc(0)
r(k−1)Nr2N=
eT Lc(0)
r2N> 0, it follows
that ℓ1 > 0.Arguing as before, ℓ2 also exists and, by Equation (6), ℓ2 ≥ ℓ1. �
2. Copy-Paste Trees
In this section we describe a class of trees (called copy-paste trees) whichwill be shown to be transition trees in Theorem 2.13. Copy-paste treesare constructed by gluing together copies of a given tree according to rulesdescribed in Definition 2.1. This method is a variation of the method ofconstructing infinite graphs using replacement rules (see [3]).
Definition 2.1. G is a copy-paste tree if it is obtained by the followingprocedure: Start with a single copy of a finite rooted tree H with a set ofleaves of cardinality at least 2 designated as boundary leaves, denoted VB(H).(Let VN (H) denote the set of non-boundary leaves of H.) We let the root ofthis first copy of H be the root of G. Then to each boundary vertex in this
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first copy of H, attach another copy of H at its root. Proceed by attachingto each boundary vertex of these copies of H yet another copy of H at theroot. Following this pattern indefinitely yields an infinite rooted tree G. Wecall H the generator of G.
Before stating our main results, we provide some examples of copy-pastetrees.
Example 2.2. Let H1 be the generator in Figure 2 with VB(H1) = {v1, v2}.Suppose all edges of H1 have length one. Figure 2 shows the first three stepsin the construction of the copy-paste tree G generated by H1. Observe thatG is the infinite full binary tree.
The generator of a copy-paste tree is not unique. Let H2 be the genera-tor in Figure 3. Suppose all edges of H2 have length one, and VB(H2) ={w1, w2, w3, w4}. Then H2 also generates the infinite full binary tree.
Step 1 Step 3H1
v2v1
Step 2
Figure 2. The first steps in the construction of a copy-paste tree.
w3 w4
H2
w1 w2
Figure 3. H2 is also a generator of the infinite full binary tree.
Example 2.3. Let H be the generator in Figure 4. Suppose all edges of Hhave length one and VB(H) = {v1, v2}. The copy-paste tree generated by His also shown in Figure 4.
Example 2.4. Let H be the generator in Figure 5 with VB(H) = {v1, v2}.Suppose that edge (r, v2) has length two, and edge (r, v1) has length one. Thecorresponding copy-paste tree G is also shown in Figure 5.
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v2
H G
v1
Figure 4. The copy-paste tree generated by H.
v1
r
v2
H G
Figure 5. A generator H and its corresponding copy-pastetree G.
Example 2.5. Let H be the generator shown in Figure 6 with VB(H) ={v1, v3} and VN (H) = {v2}. Suppose all edges of H have length one. Thesequence {cn} in the resulting copy-paste tree G satisfies the Lucas sequence:
cn+1 = cn + cn−1, c0 = 1, c1 = 3.
8
v3
H G
v1 v2
Figure 6. The tree H is a generator for the Lucas tree G.
Suppose G is a copy-paste tree with generator H of height N . For j =1, . . . , N , define
bi(H) = |{v ∈ VB(H) : d(v, r) = i}|.So bi(H) gives the number of boundary vertices of H of distance i to the root.(For the generator H in Figure 5 of Example 2.4, N = 2, b1(H) = b2(H) = 1,c1(G) = 1, and c2(G) = 2.) As before, if only one generator graph H andits corresponding copy-paste tree G are under consideration, we will oftenwrite bi for bi(H), and ci for ci(G).
In Theorem 2.13, we will show that the following matrix is a transitionmatrix for G:
A =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . . 00 0 0 · · · 1bN bN−1 bN−2 · · · b1
.
The proof of Theorem 2.13 relies on partitioning the vertices of G in away that leads to a recursive formula for ci(G) (Lemma 2.9). We first makethe following definitions:
Definition 2.6. Define the closed ball of radius n about a vertex v ∈ V (G),denoted Bn(G, v), by
Bn(G, v) = {z ∈ |G| : d(z, v) ≤ n}where |G| is the realization of G, each edge of length ℓ is isometric to [0, ℓ],and d is the associated path metric.
Definition 2.7. For i ≥ 1, define the i−shell, denoted Si, by
Si = Bi(G, r) \ Bi−1(G, r).
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Figure 7 illustrates Definition 2.7. It includes the S1, . . . , S6 shells (withcircles denoting points not in Si) for the copy-paste tree of Example 2.3,depicted in Figure 4.
S6
S1
S3
S2
S4
S5
Figure 7. Shells of the copy-paste tree with generator H ofExample 2.3.
The following observation immediately follows from Definitions 2.1 and2.7.
Observation 2.8. Let b be a boundary vertex in a copy-paste tree G, andlet D(G, b) be the subtree of G with root b. Then for all i ≥ 1,
(Bi(G, b) \ Bi−1(G, b)) ∩ D(G, b)
is isometric to Si.
In Figure 8, boundary vertices b and c are indicated in the copy-pastetree G of Example 2.3, depicted in Figure 4. Since H has height 4, andG is constructed by repeatedly pasting copies of H at its root to boundaryvertices, we see the same shell pattern S1, S2, S3, . . . descending from b andc as descends from the root. By Observation 2.8, the following two subsetsof S4 and S6, respectively, are isometric to S2:
(B2(G, b) \ B1(G, b)) ∩ D(G, b) and (B2(G, c) \ B1(G, c)) ∩ D(G, c).
In the next lemma we show that if H is a generator tree of height N , thenevery Sn shell can be partitioned into isometric copies of S1, S2, . . . , SN .This is a technical but important lemma. The reader may find it helpful torefer to Figure 10 in Example 2.10 for an illustration of the definitions andformulas in the proof of Lemma 2.9. Recall that
ci = ci(G) = |{v ∈ V (G) : d(v, r) = i}|
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(B2(G
, c)\ B
1(G
, c))∩
D(G
, c)
c
b
S2
(B3(G, b) \ B2(G, b)) ∩ D(G, b)
(B2(G, b) \ B1(G, b)) ∩ D(G, b)
(B5(G, b) \ B4(G, b)) ∩ D(G, b)
r
Figure 8. Shell patterns.
and
bi = bi(H) = |{v ∈ VB(H) : d(v, r) = i}|.
Lemma 2.9. Let G be a copy-paste tree with a generator H of height N .For all n > N , there exists a partition Pn of Sn such that each element ofPn is isometric to one of the shells S1, S2, . . . , SN .
Moreover, if ti(Sn) is the number of isometric copies of Si in the partitionPn of Sn, then the following formulas hold:
(a) cn = t1(Sn)c1 + t2(Sn)c2 + · · · + tN (Sn)cN .(b) ti(SN+1) = bN+1−i for i = 1, . . . , N .(c) For n > N , t1(Sn+1) = tN (Sn)t1(SN+1).(d) For n > N , ti(Sn+1) = ti−1(Sn) + tN (Sn)ti(SN+1), for 2 ≤ i ≤ N .
Proof. Let n > N . For each x ∈ |G| such that d(x, r) = n, let a(x) ∈ |G|such that a(x) lies on the unique simple path from x to r and such thatd(a(x), r) = n − N . Since H has height N , the unique simple path from xto a(x) must contain a boundary vertex. If a(x) is a boundary vertex, letk(x) = a(x). If a(x) is not a boundary vertex, then a(x) lies in a uniquecopy of H that was used in the construction of G, which we will denoteH(a(x)). In this case, let k(x) be the unique boundary vertex of H(a(x))that lies on the unique simple path from x to a(x).
Figure 9 illustrates the definitions of both a(x) and k(x) for four verticesof S7 for the copy-paste tree in Figure 4 of Example 2.3 for which N = 4.
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Since v ∈ S7 and N = 4, it follows that d(a(v), r) = 7 − 4 = 3. Note thata(v) is a boundary vertex, and hence a(v) = k(v).
a(v) = a(x) = k(v) = k(x)
k(y) = k(z)
a(y) = a(z)
GH(a(y)) = H(a(z))
v x y z
Figure 9. Some examples of a(x) and k(x) for copy-pastetree of Example 2.3.
Define an equivalence relation on Vn = {x ∈ |G| : d(x, r) = n} by xRy ifa(x) = a(y) and k(x) = k(y). (In Figure 9, vRx and yRz.) Let Pn denotethe partition of Sn formed by the equivalence classes of R, and let Pn(x)denote the partition element of Pn containing x.
Observe that
Pn(x) = (Bi(G, k(x)) \ Bi−1(G, k(x))) ∩ D(G, k(x))
for some boundary vertex k(x) and 1 ≤ i ≤ N . By Observation 2.8, Pn(x)is isometric to Si. Thus each element of the partition Pn of Sn is isometricto one of the shells S1, S2, . . . , SN . The formula in Part (a) immediatelyfollows.
We now use the partition to find formulas for counting the number ofisometric copies of S1, S2, . . . , SN in any Sn shell, where n ≥ N . We firstconsider the shell SN+1. Let x ∈ |G| such that d(x, r) = N + 1, and letPN+1(x) be the partition element of SN+1 containing x. Then
PN+1(x) = (Bi(G, k(x)) \ Bi−1(G, k(x))) ∩ D(G, k(x)),
for some 1 ≤ i ≤ N and boundary vertex k(x). Note that k(x) is in the initialcopy of H used in the construction of G since d(a(x), r) = N + 1 − N = 1.Moreover, d(k(x), r) = N +1−i. It now follows that the number of isometriccopies of Si in SN+1 is at most the number of boundary vertices in H ofdistance N +1−i to the root, i.e., ti(SN+1) ≤ bN+1−i. To establish equality,note that if v is an arbitrary boundary vertex of H of distance N + 1− i tothe root for some 1 ≤ i ≤ N , then
(Bi(G, v) \ Bi−1(G, v)) ∩ D(G, v) ⊂ SN+1
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is a partition element of PN+1 that is isometric to Si. Thus the formula inPart (b) holds.
We now establish the formulas in in Parts (c) and (d). Let n > N . Letx ∈ |G| such that d(x, r) = n, and let Pn(x) be the partition element of Sn
containing x. If k(x) = a(x) (i.e., a(x) is a boundary vertex), then
Pn(x) = (BN (G, k(x)) \ BN−1(G, k(x))) ∩ D(G, k(x))
is isometric to SN . Note that
(BN+1(G, k(x)) \ BN (G, k(x))) ∩ D(G, k(x)) ⊂ Sn+1
is isometric to SN+1, which is itself partitioned into copies of S1, . . . , SN . Inparticular, each isometric copy of SN in the Sn shell contributes the follow-ing partition elements to Pn+1: t1(SN+1) isometric copies of S1, t2(SN+1)isometric copies of S2, . . . , and tN (SN+1) copies of SN . Thus ti(Sn+1) ≥tN (Sn)ti(SN+1) for 1 ≤ i ≤ N .
If k(x) 6= a(x) (i.e., a(x) is not a boundary vertex), then
Pn(x) = (Bi−1(G, k(x)) \ Bi−2(G, k(x))) ∩ D(G, k(x)),
for some i where 2 ≤ i ≤ N . In this case, Pn(x) is isometric to Si−1, and
(Bi(G, k(x)) \ Bi−1(G, k(x))) ∩ D(G, k(x)) ⊂ Sn+1
is a partition element of Pn+1 that is isometric to Si. Thus each isometriccopy of Si−1 in the Sn shell (where 2 ≤ i ≤ N) contributes exactly oneisometric copy of Si to the Sn+1 shell. Therefore ti(Sn+1) ≥ ti−1(Sn) for2 ≤ i ≤ N . The formulas in Parts (c) and (d) now follow.
�
Example 2.10. Figure 10 shows the copy-paste tree depicted in Figure 4of Example 2.3. The generator H has two boundary vertices, v1 and v2
(indicated in Figure 10) with d(v1, r) = 1 and d(v2, r) = 4. Thus b1 = 1,b2 = b3 = 0, and b4 = 1. Since H has height 4, S5 is the first shell thatcan be partitioned into isometric copies of S1, S2, S3, and S4. (The partitionelements of S5, S6 and S7 are circled.)
Consider the vertices w and y of S5 indicated in Figure 10. Note that a(w)and a(y) (the ancestors of w and y, respectively, that are four levels closerto the root) are in the original H. Note that a(w) = k(w) = v1 and hence wis in an isometric copy of S4. Whereas a(y) 6= v1 and hence k(y) = v2 andso y is in an isometric copy of S1. That is,
t1(S5) = b4 = 1, t2(S5) = b3 = 0, t3(S5) = b2 = 0, and t4(S5) = b1 = 1,
which verifies the formula in Part (b) of Lemma 2.9.As described in the proof of Lemma 2.9, each copy of S1 in S5 contributes
one copy of S2 to S6. Similarly, each S2 or S3 in S5 contributes one copyof S3 or S4 to S6. But each S4 contributes one S5, which itself consists oft1(S5) = 1 copy of S1 and t4(S5) = 1 copy of S4. Hence, as shown in Figure10, t1(S6) = 1, t2(S6) = 0, t3(S6) = 1, and t4(S6) = 1.
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S7
a(y)
k(y) = v2
y
a(w) = k(w) = v1
w
S6
S5
Figure 10. Partition elements of S5, S6 and S7 for the copy-paste tree of Example 2.3.
The following calculations illustrate the formulas in Parts (a), (c) and(d) in Lemma 2.9 for the shell S7.
t1(S7) = t4(S6)t1(S5) = 1 · 1 = 1
t2(S7) = t1(S6) + t4(S6)t2(S5) = 1 + 1 · 0 = 1
t3(S7) = t2(S6) + t4(S6)t3(S5) = 1 + 1 · 0 = 1
t4(S7) = t3(S6) + t4(S6)t4(S5) = 0 + 1 · 1 = 1
c7 = c1t1(S7) + c2t2(S7) + c3t3(S7) + c4t4(S7)
= 2 · 1 + 3 · 1 + 4 · 1 + 5 · 1= 14
The recursive formulas given in Lemma 2.9 can be easily implementedusing matrices. Define the copy-paste matrix A associated to a generator
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H of height N to be the N × N matrix
A =
t1(S2) t2(S2) t3(S2) · · · tN (S2)t1(S3) t2(S3) t3(S3) · · · tN (S3)
......
......
...t1(SN+1) t2(SN+1) t3(SN+1) · · · tN (SN+1)
. (10)
The matrix in (10) can be simplified by noting that for i = 1, . . . N ,ti(Si) = 1 and ti(Sn) = 0 for i 6= n. Thus the matrix in (10) can berewritten as
A =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . ....
0 0 0 · · · 1t1(SN+1) t2(SN+1) t3(SN+1) · · · tN (SN+1)
. (11)
By Lemma 2.9 Part (b), we can also write the matrix in (11) as
A =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . ....
0 0 0 · · · 1bN bN−1 bN−2 · · · b1
. (12)
Remark 2.11. Let A be the copy-paste matrix associated to H and let ai,j
be the (i, j) entry of A. Then bi = bi(H) = aN,N−i+1.
The copy-paste matrix for the copy-paste tree of Example 2.4 depicted in
Figure 5 is
[0 11 1
]; the copy-paste matrix for the copy-paste tree of Example
2.3 depicted in Figure 4 is
0 1 0 00 0 1 00 0 0 11 0 0 1
.
The next lemma gives several properties of copy-paste matrices.
Lemma 2.12. Let A be copy-paste matrix of the form (12).
(a) If λ is an eigenvalue of A, then[1 λ λ2 · · · λN−1
]Tis an eigen-
vector of A corresponding to λ.(b) For all N ≥ 2, the determinant of A − I is given by
|A − I| = (−1)N
(1 −
N∑
i=1
bi
).
(c) 1 6∈ σ(A).
15
(d) If bN 6= 0, then A is irreducible.(e) If bN 6= 0, then ρ(A) > 1.
Proof. Part (a): Let v =[v1 v2 · · · vN
]Tbe an eigenvector of A associ-
ated to λ. Since Av = λv we have that
vi = λvi−1 for i = 2, . . . , N and (13)
bNv1 + bN−1v2 + · · · + b1vN = λvN (14)
Since v is an eigenvector, v1 6= 0. (If v1 = 0, then v2 = · · · = vN = 0by Equation 13.) Without loss of generality, v1 = 1. From Equation (13) itfollows that
vi = λi−1 for i = 2, . . . , N. (15)
Substituting Equation (15) into Equation (14) yields
bN + bN−1λ + bN−2λ2 + · · · + b1λ
N−1 = λN . (16)
Using Equation (16), it is now straightforward to verify
A[1 λ λ2 · · · λN−1
]T= λ
[1 λ λ2 · · · λN−1
]T.
Part (b): We will argue by induction on N . When N = 2, we have that
|A − I| =
∣∣∣∣−1 1b2 b1 − 1
∣∣∣∣ = 1 − b1 − b2.
So the base case holds. Next assume the determinant formula is valid forany (k − 1) × (k − 1) matrix of this form, where k ≥ 3. We must show thedeterminant formula is valid for a k × k matrix of this form. Let
A =
0 1 0 0 · · · 00 0 1 0 · · · 00 0 0 1 · · · 0...
......
.... . .
...0 0 0 0 · · · 1bk bk−1 bk−2 bk−3 · · · b1
.
Using a cofactor expansion along the first column of A − I, we have that
|A−I| = −1
∣∣∣∣∣∣∣∣∣∣∣
−1 1 0 · · · 00 −1 1 · · · 0...
......
......
0 0 0 · · · 1bk−1 bk−2 bk−3 · · · b1 − 1
∣∣∣∣∣∣∣∣∣∣∣
+(−1)k+1bk
∣∣∣∣∣∣∣∣∣∣∣
1 0 0 · · · 0−1 1 0 · · · 0
0 −1 1 · · · 0...
......
......
0 0 · · · −1 1
∣∣∣∣∣∣∣∣∣∣∣
.
By applying the induction hypothesis and properties of triangular matrices,we see
|A − I| = (−1)(−1)k−1
(1 −
k−1∑
i=1
bi
)+ (−1)k+1bk = (−1)k
(1 −
k∑
i=1
bi
),
16
which is the desired result.Part(c): By Definition 2.1, H has at least two boundary vertices. Thus
the sum of the entries in the last row of A is at least 2. By Part (b) it followsthat |A − I| 6= 0, and thus 1 6∈ σ(A).
Part(d): By Theorem 1.5, it suffices to show that G(A), the associateddirected graph of A, is strongly connected. Let x1, . . . , xN denote the verticesG(A). The matrix A has the property that for i = 1, . . . , N − 1,
ai,i+1 = 1 and ai,j = 0 for j 6= i + 1.
Thus for i = 1, . . . , N − 1, there is a directed edge in G(A) from xi toxi+1. Since aN,1 = bN 6= 0 there is a directed edge in G(A) from xN to x1.Therefore G(A) has a cycle of length N , namely xN , x1, x2, . . . , xN−1, xN ,that includes every vertex of G(A). Hence G(A) is strongly connected.
Part (e): By Part (d), A is irreducible. It follows by Theorem 1.3 that
ρ(A) is a positive eigenvalue of A, and mini
∑
j
ai,j ≤ ρ(A) ≤ maxi
∑
j
ai,j.
The result now follows from the fact that mini
∑
j
ai,j = 1 and 1 6∈ σ(A).
�
We now show that every copy-paste matrix is a transition matrix, andtherefore every copy-paste tree is a transition tree.
Theorem 2.13. Let A be the copy-paste matrix associated to a generatorH of height N (as given in (11)), and let G be the corresponding copy-pastetree. For any n ≥ 1,
(An)i,j = tj(Sn+i).
Furthermore, A is a transition matrix, and G is a transition tree.
Proof. We proceed by induction on n. When n = 1 the result clearly holds.Next we assume the result holds for n = k, k ≥ 1. That is, assume
Ak =
t1(Sk+1) t2(Sk+1) · · · tN (Sk+1)t1(Sk+2) t2(Sk+2) · · · tN (Sk+2)
...... · · · ...
t1(Sk+N ) t2(Sk+N ) · · · tN (Sk+N )
.
We must show that Ak+1 =
t1(Sk+2) t2(Sk+2) · · · tN (Sk+2)t1(Sk+3) t2(Sk+3) · · · tN (Sk+3)
...... · · · ...
t1(Sk+1+N ) t2(Sk+1+N ) · · · tN (Sk+1+N )
.
Because Ak+1 = AAk we have that
17
Ak+1 =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . . 00 0 0 · · · 1
t1(SN+1) t2(SN+1) t3(SN+1) · · · tN(SN+1)
t1(Sk+1) t2(Sk+1) · · · tN (Sk+1)t1(Sk+2) t2(Sk+2) · · · tN (Sk+2)t1(Sk+3) t2(Sk+3) · · · tN (Sk+3)t1(Sk+4) t2(Sk+4) · · · tN (Sk+4)
...... · · ·
...t1(Sk+N) t2(Sk+N) · · · tN(Sk+N)
.
Note that multiplication in this order results in a matrix whose first N −1rows are just the rows of Ak shifted up by one row. Thus the entries in thefirst N − 1 rows of Ak+1 are the desired result.
To show the last row of Ak+1 is the desired result we write Ak+1 as theproduct Ak+1 = AkA.
Ak+1 =
t1(Sk+1) t2(Sk+1) · · · tN(Sk+1)t1(Sk+2) t2(Sk+2) · · · tN(Sk+2)t1(Sk+3) t2(Sk+3) · · · tN(Sk+3)t1(Sk+4) t2(Sk+4) · · · tN(Sk+4)
...... · · ·
...t1(Sk+N) t2(Sk+N) · · · tN (Sk+N)
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . . 00 0 0 · · · 1
t1(SN+1) t2(SN+1) t3(SN+1) · · · tN (SN+1)
.
The last row of the matrix product AkA is the vector
tN (Sk+N)t1(SN+1)t1(Sk+N) + tN (Sk+N )t2(SN+1)t2(Sk+N) + tN (Sk+N )t3(SN+1)
...tN−1(Sk+N) + tN (Sk+N )tN (SN+1)
T
.
By Parts (c) and (d) of Lemma 2.9 we have the desired result for the lastrow of Ak+1.
The following calculation uses Part (a) of Lemma 2.9 to verify that c(n) =Anc(0):
An
c1...
cN
=
t1(Sn+1) · · · tN (Sn+1)...
......
t1(Sn+N ) · · · tN (Sn+N )
c1...
cN
=
c1t1(Sn+1) + · · · + cN tN (Sn+1)...
c1t1(Sn+N ) + · · · + cN tN (Sn+N )
=
cn+1...
cN+n
.
�
18
3. Copy-Paste Trees have Exponential Growth
Throughout the remainder of this paper, GH and AH will denote thecopy-paste tree generated by H and the copy-paste matrix associated to H,respectively. Define a path from the root of a tree to a leaf to be non-
branching if every vertex along the path that is not an endpoint of thepath has degree two. Define a tree to be non-branching if every simplepath from the root to a leaf is non-branching.
Recall that both H1 and H2 (Figures 2 and 3, respectively) of Example2.2 generate the infinite full binary tree. The associated copy-paste matrices
are AH1= [2] and AH2
=
[0 14 0
]. Note that AH1
is primitive, 1 6∈ σ(AH1),
and ρ(AH1) = 2 > 1. Thus by Theorem 1.8, the infinite full binary tree has
exponential growth with growth rate ρ(AH1) = 2. Although ρ(AH2
) = 2,Theorem 1.8 does not apply to H2 since AH2
is not primitive.The purpose of this section is to show, using a bounding argument, that
every copy-paste tree GH has exponential growth. Moreover, the growthrate of GH is the spectral radius of a certain copy-paste matrix AH′ , whichis determined by the number of boundary vertices of H and their distancesto the root.
Let H be a generator tree of height N for a copy-paste tree GH and letM be the maximum distance over all boundary vertices in H to the root.
From H, we will construct two non-branching trees H ′ and H of height Msuch that the growth function f(G,n) satisfies
f(GH′ , n) ≤ f(GH , n) ≤ f(GH
, n)
and the copy-paste matrices are given by
AH′ = AH =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
.... . .
...0 0 0 · · · 1
bM (H) bM−1(H) bM−2(H) · · · b1(H)
.
In Lemmas 3.8 and 3.9, it will be shown that GH′ and GH
have exponen-tial growth with growth rate ρ(AH′). Consequently, GH also has exponentialgrowth with growth rate ρ(AH′).
We first give an informal description of the construction of H ′ and H. Thetree H ′ will be constructed by replacing each path in H of length ℓ from theroot to a boundary vertex b with a single edge of length ℓ. Moreover, H ′
will have no non-boundary leaves. So if H has k boundary vertices, then H ′
is a non-branching tree having exactly k edges and k boundary vertices. Inparticular, H ′ will have exactly k + 1 vertices consisting of the k boundaryvertices and the root.
19
The tree H will be constructed so that it has the same number of boundaryleaves as H and at least the same number of non-boundary leaves as H.
Additionally, H will be non-branching and all edges will have length one.
To construct H, each path of length ℓ in H from the root to a boundaryvertex will be replaced with a non-branching path of length ℓ. Each pathfrom the root to a non-boundary leaf will be replaced with non-branchingpaths of length M from the root to non-boundary leaves. The number ofsuch paths will be be chosen to ensure that f(GH , n) > f(GH , n). Forinstance, if M = 4 and N = 100, then replacing the path in H of length100 from the root to a non-boundary leaf by a single path of length 4 toa non-boundary leaf will not guarantee that f(G
H, n) provides an upper
bound for f(GH , n). In our construction, we will replace the path of length100 by N − M = 96 such paths of length 4, which certainly provides anupper bound.
We now give a formal description of the constructions of H ′ and H. Asbefore, denote the boundary leaves of H by VB(H) = {v1, . . . , vn}, and thenon-boundary leaves of H by VN (H) = {w1, . . . , wp}. Let M and qi bedefined as follows:
M = maxv∈VB(H)
d(v, r),
and, for 1 ≤ i ≤ p,
qi =
{0 if d(wi, r) ≤ Md(wi, r) − M if d(wi, r) > M
Definition 3.1. Define H ′ to be the non-branching rooted tree with root r′,having exactly n+1 vertices r′, v′1, . . . , v
′
n and such that VB(H ′) = {v′1, . . . , v′n}and d(v′i, r
′) = d(vi, r) for all i = 1, . . . , n.
Definition 3.2. Define H to be the non-branching rooted tree with root rsuch that all edges have length one, and with the following properties for itsleaves:
(a) VB(H) = {v1, . . . , vn} and d(vi, r) = d(vi, r) for all i = 1, . . . , n.
(b) VN (H) = {w1, . . . , wj} where j = p +
p∑
i=1
qi, and d(wi, r) = M for
all i = 1, . . . , j.
Let m = gcd(d(v1, r), . . . , d(vn, r)). Define H to be unscaleable if m = 1,and H to be scaleable if m > 1. Note that H1 in Figure 2 of Example 2.2is unscaleable whereas H2 in Figure 3 is scaleable since m = 2. In Lemma3.8, it will be shown that AH′ is primitive if H is unscaleable. Moreover, ifH is unscaleable, then GH′ and G
Hhave exponential growth with growth
rate ρ(AH′).
If m > 1, then two additional trees SH ′ and SH will be constructed from
H ′ and H, respectively, by scaling by a factor of 1m the distances of each leaf
20
to the root. The effects of such scaling on the spectral radius of the copy-paste matrix and the growth rate of the copy-paste tree will be examined inLemma 3.9. In particular, it will be shown that if m > 1, then GH′ and GH
have exponential growth with growth rate m√
ρ(ASH′) = ρ(AH′).
Definition 3.3. If m > 1, define SH ′ be the non-branching rooted tree withroot r′ having exactly n + 1 vertices r′, v′1, . . . , v
′
n, and such that VB(SH ′) =
{v′1, . . . , v′n} and d(v′i, r′) =
d(vi, r)
mfor all i = 1, . . . , n.
Definition 3.4. If m > 1, define SH be the non-branching rooted tree withroot r such that all edges have length one, and with the following propertiesfor its leaves:
(a) VB(SH) = {v1, . . . , vn} and d(vi, r) =d(vi, r)
mfor all i = 1, . . . , n.
(b) VN (SH) = {w1, . . . , wj} where j = p +
p∑
i=1
qi, and d(wi, r) = Mm for
all i = 1, . . . , j.
Example 3.5. For the generator H2 of Example 2.2 depicted in Figure 3,
m = 2. The trees H2, H ′
2, H2 and SH ′
2 = SH2 are shown below.
H2 H ′
2 H2 SH ′
2 = SH2
Figure 11. Trees H ′
2, SH ′
2, H2, and SH2 corresponding to H2.
Example 3.6. Let H be the generator tree shown below in Figure 12. Sup-pose edges (x, y) and (z,w3) have length 2, and all other edges have length1. Suppose also that the set of boundary vertices is VB(H) = {v1, v2, v3},and the set on non-boundary leaves is VN (H) = {w1, w2, w3, w4}. Note thatthe height of H is N = d(w4, r) = 6, and the distances of the boundaryvertices to the root are d(v1, r) = d(v2, r) = 4 and d(v3, r) = 2. Thusm =gcd(2, 4) = 2. The maximum distance over all boundary vertices to theroot is M = 4, and the values of qi are q1 = q2 = q3 = 0, and q4 = 2.
Figure 12 shows the non-branching trees H ′, SH ′, H and SH associatedto H along with the copy-paste tree GH′ . Note that edges (r′, v′1) and (r′, v′2)have length 4 in H ′ but length 2 in SH ′, and edge (r′, v′3) has length 2
in H ′ but length 1 in SH ′. All edges of H and SH have length 1, and
VN (H) = {w1, . . . , w6}.Note that the copy-paste matrices of H,H ′, SH ′, H, and SH of Example
3.6 are:
21
r
w2z
H
y
v3
w1 v1
w4
w3v2
r
v1w1
w1 w2
w2
w6
w4 w6
w4 v1
GH′
v′
1 v′
2
v′
3
H ′
r′
v′
2
v′
3
SH ′
r′
x
r
SH
v3
w5
H
v3
w3 w5
w3
v2
v2
v′
1
Figure 12. Trees H ′, GH′ , SH ′, H, and SH correspondingto H.
AH =
0 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 10 0 2 0 1 0
, AH′ = AH =
0 1 0 00 0 1 00 0 0 12 0 1 0
, and
ASH′ = ASH
=
[0 12 1
].
The following observation lists several useful formulas that relate bi(H),
the number of boundary vertices of H of distance i to the root, to bi(H′), bi(H),
bi(SH ′) and bi(SH). It also describes the relationships between the copy-paste matrices and ci, the number of vertices in a tree of distance i to theroot.
Observation 3.7. It follows immediately from Definitions 3.1–3.4 that
22
(a) For i ≥ 0, bi(H′) = bi(H) and ci(H
′) ≤ ci(H). Hence f(GH′ , n) ≤f(GH , n).
(b) For i ≥ 0, bi(H) = bi(H) and ci(H) ≥ ci(H). Hence f(GH , n) ≤f(GH , n).
(c) For i ≥ 0, bi(H′) = bi(H) = bi(H), and both H ′ and H have height
M . Hence AH′ = AH . (If N = M , then AH = AH′ .)
(d) If m > 1, then bi(SH ′) = bi(SH) for all i ≥ 0, and both SH ′ and
SH have height Mm . Hence ASH′ = ASH .
(e) If m > 1, then bim(H ′) = bi(SH ′) and cim(H ′) = ci(SH ′). Henceci(GSH′) = cim(GH′) for i ≥ 0.
(f) If m > 1, then bim(H) = bi(SH) and cim(H) = ci(SH). Henceci(GSH
) = cim(GH
) for i ≥ 0.
By Part (c) of Observation 3.7,
AH′ = AH =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
.... . .
...0 0 0 · · · 1
bM (H) bM−1(H) bM−2(H) · · · b1(H)
. (17)
Parts (a) and (b) of Observation 3.7 imply
f(GH′ , n) ≤ f(GH , n) ≤ f(GH
, n).
Lemma 3.8. Let GH be a copy-paste tree with generator H of height N > 1and let AH′ be the copy-paste matrix associated to H ′. If m = 1, then AH′ isprimitive and GH′ and GH have exponential growth with growth rate ρ(AH′).
Proof. Let AH′ be the M × M copy-paste matrix of H ′ given by Equation(17). Let G(AH′) be the associated directed graph of AH′ , and let x1, . . . , xM
denote the vertices of G(AH′). Note that bM (H) 6= 0 by definition of M . Inthe proof of Lemma 2.12 Part (d), it was shown that bM (H) 6= 0 impliesthat G(AH′) is strongly connected. In particular, it has a cycle of length Mthat contains every vertex, namely,
xM , x1, x2, . . . , xM−1, xM .
Hence, by Theorem 1.5, AH′ is irreducible.By Remark 2.11, bi(H) = aM,M−i+1 for i = 1, . . . ,M , where ai,j denotes
the i, j entry of AH′ . Therefore, if bi(H) 6= 0, then (xM , xM−i+1) is a directededge in G(AH′). Using the cycle described in the previous paragraph, itfollows that
xM , xM−i+1, xM−i+2, . . . , x(M−i)+(i−1), xM
is a cycle of length i in G(AH′).If m = 1, then there exist boundary vertices v and w of H such that
gcd(d(v, r), d(w, r)) = 1. Let j = d(v, r) and k = d(w, r). Then bj(H) 6= 0,
23
bk(H) 6= 0 and gcd(j, k) = 1. Thus G(AH′) has a cycle of length j and acycle of length k. Since k and j are relatively prime, Theorem 1.6 impliesAH′ is primitive.
By Lemma 2.12, ρ(AH′) > 1 and 1 6∈ σ(AH′). Thus Theorem 1.8 impliesGH′ has exponential growth rate with growth rate ρ(AH′). Since AH′ is also
the copy-paste matrix associated to H, the result also holds for GH
.�
Lemma 3.9. Let GH be a copy-paste tree with generator H of height N > 1.
Suppose m > 1. Let H ′, SH ′, H and SH be the trees described in Definitions3.1–3.4.
(a) If GSH′ has exponential growth with growth rate λ, then GH′ has
exponential growth with growth rate m√
λ. If GSH
has exponentialgrowth with growth rate λ, then GH has exponential growth with
growth rate m√
λ.(b) If λ is an eigenvalue of AH′ , then λm is an eigenvalue of ASH′ .
Conversely, if s is an eigenvalue of ASH′ then m√
s is an eigenvalue
of AH′ . Moreover, ρ(AH′) = m√
ρ(ASH′).(c) The copy-paste trees GH′ and G
Hhave exponential growth with growth
rate ρ(AH′).
Proof. Part(a): Since the proof for GSH and GH is analogous to the prooffor GSH′ and GH′ , we will include only the proof of the latter. Because GSH′
has exponential growth with growth rate λ, there exist positive constantsα1 and α2 such that
α1λn ≤ f(GSH′ , n) ≤ α2λ
n.
Thus, using Part (e) of Observation 3.7,
f(GSH′ , n) =n∑
i=0
ci(GSH′) =n∑
i=0
cim(GH′) ≤nm∑
i=0
ci(GH′) = f(GH′ , nm).
By Observation 3.7 part (e), it also follows that
f(GH′ , nm) = c0(GH′)
+(c1(GH′) + · · · + cm(GH′))
+(cm+1(GH′) + · · · c2m(GH′))
...
+(c(n−1)m+1(GH′) + · · · + cnm(GH′))
≤ m(c0(GH′) + cm(GH′) + c2m(GH′) + · · · cnm(GH′))
= m(c0(GSH′) + c1(GSH′) + · · · cn(GSH′))
= mf(GSH′ , n). (18)
Therefore
f(GSH′ , n) ≤ f(GH′ , nm) ≤ mf(GSH′ , n).
24
Hence
α1λn ≤ f(GH′ , nm) ≤ (mα2)λ
n.
And so
α1(λ1/m)nm ≤ f(GH′ , nm) ≤ (mα2)(λ
1/m)nm.
It now follows that GH′ has exponential growth with growth rate m√
λ.Part (b): Recall that the copy-paste matrix of H ′ is
AH′ =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . ....
0 0 0 · · · 1bM (H) bM−1(H) bM−2(H) · · · b1(H)
.
By Lemma 2.12 Part (a), v =[1 λ λ2 · · · λM−1
]Tis an eigenvector
of AH′ associated to λ. Note that the last equation obtained from AH′v =λv is
bM (H) + λbM−1(H) + λ2bM−2(H) + · · · + λM−1b1(H) = λM . (19)
Since m|M , there exists k ∈ Z+ such that M = km. Since m > 1, the
only possible nonzero entries in the last row of AH′ have the form bmi(H)for some i ∈ Z
+. Thus Equation (19) can be written as
λmk = bmk(H) + λmbm(k−1)(H) + · · · + λm(k−1)bm(H). (20)
Since bi(SH ′) = bmi(H′) = bmi(H) and SH ′ has height k, the copy-paste
matrix of SH ′ is given by
ASH′ =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . ....
0 0 0 · · · 1bmk(H) bm(k−1)(H) bm(k−2)(H) · · · bm(H)
.
25
Using Equation (20), the following calculation verifies that λm is an eigen-value of ASH′ :
ASH′
1λm
λ2m
...
λ(k−1)m
=
λm
λ2m
...
λ(k−1)m
bmk(H) + bm(k−1)(H)λm + · · · + bm(H)λ(k−1)m
=
λm
λ2m
...
λ(k−1)m
λmk
= λm
1λm
...
λ(k−2)m
λ(k−1)m
.
For the converse, suppose s is an eigenvalue of ASH′ . By Lemma 2.12 Part
(a), w =[1 s s2 · · · sk−1
]Tis an eigenvector associated to s. The last
equation obtained from ASH′w = sw, is
bmk(H) + bm(k−1)(H)s + · · · bm(k−j)sj + · · · + bm(H)sk−1 = sk. (21)
Since M = mk, Equation (21) can be rewritten as
bM (H) + bM−m(H)s + · · · bM−jmsj + · · · + bm(H)sk−1 = sk. (22)
Since H is scaleable, the only possible nonzero entries in the last row ofAH′ have the form bmi(H), that is, bM−j(H) = bmk−j(H) = 0 for j 6= im.Using that bM−j(H) = 0 for j 6= im and Equation (22) we have
bM (H)+bM−1(H)s1/m+bM−2(H)s2/m+· · ·+bM−m(H)sm/m+· · ·+b1(H)s(M−1)/m = sk
A straightforward calculation verifies that[1 s1/m s2/m · · · s(M−1)/m
]T
is an eigenvector of AH′ with corresponding eigenvalue s1/m.Since bM (H) > 0, it follows from Lemma 2.12 Part (d) that AH′ and ASH′
are irreducible. Thus by Theorem 1.3, ρ(AH′) and ρ(ASH′) are positive
eigenvalues of A′ and ASH′ , respectively. Therefore ρ(AH′) = m
√ρ(ASH′).
Part (c): By construction, the trees SH ′ and SH are unscaleable. ByLemma 3.8, GSH′ and GSH have exponential growth with growth rateρ(ASH′). By Part (a), the copy-paste trees GH′ and GH have exponential
growth with growth rate m√
ρ(ASH′). By Part (b), ρ(AH′) = m√
ρ(ASH′).Thus GH′ and G
Hhave exponential growth with growth rate ρ(AH′).
�
Theorem 3.10. Let G be a copy-paste tree with generator H having bound-ary vertices VB(H) = {v1, . . . , vn}. Let bi(H) = |{v ∈ VB(H) : d(v, r) = i}|,
26
M = maxv∈VB(H)
d(v, r), and
AH′ =
0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...
......
. . ....
0 0 0 · · · 1bM (H) bM−1(H) bM−2(H) · · · b1(H)
.
Then G has exponential growth with growth rate ρ(AH′).
Proof. Construct trees H ′ and H as described in Definitions 3.1 and 3.2,respectively. Observe that AH′ is the copy-paste matrix associated to H ′ as
well as to H. Since
f(GH′ , n)
n≤ f(G,n)
n≤
f(GH , n)
n,
it suffices to show that GH′ and GH
have exponential growth with growthrate ρ(AH′). Let m = gcd(d(v1, r), . . . , d(vn, r)). There are two cases toconsider: m = 1 or m > 1.
If m = 1, then by Lemma 3.8, GH′ and GH
have exponential growth
rate with growth rate ρ(AH′). If m > 1, then construct SH ′ and SH asdescribed in Definitions 3.3 and 3.4, respectively. By Part (c) of Lemma 3.9,GH′ and G
Hhave exponential growth with growth rate ρ(AH′).
�
We conclude by using Theorem 3.10 to calculate the growth rates of thecopy-paste trees considered throughout the paper.
Example 3.11. Let H be the generator of Example 2.3 as depicted in Fig-ure 4. Note that H has height N = 4 and the maximum distance to theroot over all boundary vertices is M = 4. Also note that b1(H) = 1,b2(H) = 0, b3(H) = 0, b4(H) = 1. Since the g.c.d. of distances fromthe boundary vertices to the root is m = 1, H is unscaleable. Hence AH =
AH′ =
0 1 0 00 0 1 00 0 0 11 0 0 1
. Thus by Theorem ??, GH has exponential growth
with growth rate ρ(AH′) ≈ 1.3803.
Example 3.12. Let H be the generator of Example 2.4 as depicted in Figure5. Here b1(H) = 1, b2(H) = 1, N = M = 2, and m = 1. Thus AH′ =
AH =
[0 11 1
]. Thus GH has exponential growth with growth rate ρ(AH′) =
(1 +√
5)/2.
Example 3.13. Let H be the generator of Example 2.5 as depicted in Figure6. As in Example 3.12, b1(H) = 1, b2(H) = 1, N = M = 2, m = 1, and
27
AH′ = AH =
[0 11 1
]. Thus the Lucas tree generated by H has exponential
growth with growth rate ρ(AH′) = (1 +√
5)/2.
Example 3.14. Let H be the generator tree of Example 3.6 as depicted inFigure 12. Recall that N = 6, M = 4, b4(H) = 2, b3(H) = 0, b2(H) = 1,
b1(H) = 0, m = 2, and AH′ =
0 1 0 00 0 1 00 0 0 12 0 1 0
. Thus the exponential growth
rate of GH is ρ(AH′) =√
2.
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