Growth Series for Rooted Trees Katholieke Universiteit Leuven 2 June 2014

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1 Growth Series for Rooted Trees Katholieke Universiteit Leuven 2 June 2014 recent research results by Ian Dalton and Eric Freden

2 Motivation 1: Growth series for groups are usually computed by finding normal forms (geodesic words) and counting them. Such normal forms can represent a spanning tree in the Cayley graph. Thus the growth series of the group is the growth series of a rooted tree, with all edges having weight 1, and vertex valences usually variable (but uniformly bounded). The figure shows such a partial spanning tree for Z Z.

3 Partial abstract spanning tree for Baumslag-Solitar BS(1,3) (not embedded in Cayley graph) showing color coded cone types

4 Motivation 2: The asymptotic growth rate ω for any Baumslag-Solitar group can be found by computing the growth series of the corresponding Bass-Serre tree. This tree has constant valence and non-constant (but uniformly bounded) edge weights that correspond to edge projections from Cayley graph to tree. Below is such a tree for BS(1,3). It has 3 cone types and ω=2 (exercise).

5 The tree for BS(2,4) has a much more complicated weighting scheme, with infinitely many cone types. Every cone contains a copy of every other cone. Our algorithm for assigning edge weights to this tree requires exponential space (essentially storing a weight scheme for each cone type). The computation of the growth series appears to rely on the edge weight schema.

6 Motivation 3: As an administrator in a small undergraduate university, I have limited time for research. My colleagues tend to be students who have not yet learned anything about cohomology, measure theory, representation theory, hyperbolic groups and non-positive curvature, varieties, etc. They have learned combinatorics, basic group theory, theory of computation, algorithm design, and some complex analysis.

7 Define a valence scheme as an algorithm that builds a rooted tree by recursively assigning leaf nodes/edges to existing leaves. For example, the standard valence scheme for a binary tree assigns two new leaves to each existing leaf. Define a weight scheme as an algorithm that recursively assigns edge weights to an already existing rooted tree. For example, our complicated algorithm for BS(2,4) assigns edge weights to a regular 5-ary tree (OK, the root has 6 children ). The growth series for a rooted tree is the generating function T (z)= r 0 τ r z r where τ r counts the nodes at weighted distance r from the root.

8 All the trees that we are interested in are infinite (indeed, have exponential growth) thus the full application of a valence scheme, weight scheme, or growth series will require infinite time. Define a valence, weight, or growth algorithm as Wilfian if its application to height h [or radius r] takes at most poly(h) [or poly(r)] many computational steps where poly(x) is a polynomial that depends on the specific instance. Some Wilfian examples include: a valence scheme for the regular rooted n-ary tree a weight scheme for the BS(1,3) Bass-Serre tree a valence/weight scheme for a tree with finitely many cone types the Taylor series for any D-finite function the Taylor series for the secant function

9 Every valence/weight scheme determines a unique growth series. What is the computational relationship here? Folklore: If a valence/weight scheme is Wilfian, so is the associated growth series. The idea is that nodes/edges must be partitioned into polynomially many equivalence classes upon which valences/weights are uniform. Show that these classes can be easily counted and correlated with radius. What about a converse? Given a Wilfian growth series is there a Wilfian valence and/or weight scheme for an associated rooted tree? Obviously, the series must have positive integer coefficients, be increasing, have 1 as constant coefficient, etc.

10 All edges have weight 1 These trees both have growth series 1+z+z 2 +4z 3 +4z 4 +4z 5 +16z 6 +16z There are many non-isomorphic trees with the same growth series.

11 An admissible growth series is trivially associated to a variable valence tree with constant edge weight 1 as per the previous slide, so If an admissible growth series is Wilfian, there exists an associated tree with valence/root scheme that is Wilfian. Open question: given an admissible Wilfian growth series is every associated valence/weight scheme on a rooted tree (equivalent to one that is) Wilfian? We have some partial results...

12 Proposition: Suppose T is a rooted tree with a finite set of edge weights {w 1, w 2,., w n } and finite set of vertex valences {v 1, v 2,.,v m }. Suppose that edge weights and vertex valences are constant on each height. Then the valence/weight scheme is Wilfian if and only if the growth series is Wilfian. Proof: ( ) In this situation, the r th nonzero growth series term is v 1 v 2 v r z w 1+w 2 + +w r where v i (respectively, w i ) is the vertex valence (respectively edge weight) at height i, for 1 i r. Formally divide by the (r-1) st term to recover v r and w r. This is accomplished in polynomial time by hypothesis. ( ) Is the folklore result mentioned earlier.

13 Unfortunately, constant valence and weight along heights is a strong assumption that rarely occurs. Rephrase: Given an admissible Wilfian growth series is every associated valence/weight scheme on a rooted tree Wilfian? as Can a computationally nice growth series have a computationally ugly tree? We address the latter question by a series of examples.

14 Perhaps a cleverly constructed non-wilfian weight scheme can produce a Wilfian growth series? Consider this basic example: The right side edges have weight sequence w 1, w 2, w 3, w 4,. All left trending edges are weighted 1. Color scheme shows self-similarity. There are infinitely many cone types, each containing every other.

15 Let R(z)=z w 1 + z w 1 + w 2+z w 1 + w 2 + w 3+ be the OGF for the 1+ R(z) variably weighted side.then T (z)= 1 zr( z).

16 Let R(z)=z w 1 + z w 1 + w 2+z w 1 + w 2 + w 3+ be the OGF for the 1+ R(z) variably weighted side.then T (z)= 1 zr( z). This leads to the same situation as the special case Proposition. If T(z) is Wilfian, so is R(z), and the weights are easily recoverable. The rational form for T(z) insinuates the self-similarity of cones, etc. Subexamples include: w k =1 for all k (standard binary tree) w k =1+k th bit in the binary expansion of any non-computable real number (R, hence T, is not computable) w k =2 if k is a triangular number, else 1 (R is lacunary, so T is Wilfian but not D-finite)

17 Eye Candy! Julia-type set for the sub example lacunary R(z) showing the structure of singularities on the unit circle.

18 How about starting with a rational growth series and trying to create an ugly weight scheme? Can we jam T(z) = 1/(1-2z) onto a regular ternary tree using edge weights 1 and 2? The root has the vertex at distance 0. We need two vertices at distance 1,

19 How about starting with a rational grow series and trying to create an ugly weight scheme? Can we jam T(z) = 1/(1-2z) onto a regular ternary tree using edge weights 1 and 2? The root has the vertex at distance 0. We need two vertices at distance 1, and four more at distance 2,

20 How about starting with a rational grow series and trying to create an ugly weight scheme? Can we jam T(z) = 1/(1-2z) onto a regular ternary tree using edge weights 1 and 2? The root has the vertex at distance 0. We need two vertices at distance 1, and four more at distance 2, and eight more at distance 3,

21 How about starting with a rational grow series and trying to create an ugly weight scheme? Can we jam T(z) = 1/(1-2z) onto a regular ternary tree using edge weights 1 and 2? The root has the vertex at distance 0. We need two vertices at distance 1, and four more at distance 2, and eight more at distance 3, and sixteen more at distance 4,...

22 This left-to-right deterministic weight scheme (the Dalton algorithm) can be continued indefinitely, produces infinitely many different cone types, and appears to require exponential time. However, it runs in linear time and space (in terms of weighted distance d) assuming that the growth coefficients are read from file and the tree is ternary. The Dalton algorithm can be applied to other growth series, in particular the transcendental Taylor series whose n th coefficient is floor(exp(n)). This has exponent ω = e.

23 Storage table for Dalton algorithm applied to tree growing like e

24 Does the Dalton algorithm work for the BS(2,4) tree? Given the first 36 growth series terms, weight set {1,2,3}, and a 5-ary tree, does the algorithm reconstruct the tree weightings correctly? No! Even with the addition of a finite set of initial conditions, non-deterministic choices are eventually required in order to create (an isomorphic copy of) the correct subtree of radius r. Making these choices deterministic appears to require using the original (exponential time/space) weight scheme! We conclude with some conjectures.

25 Conjecture: both the growth series and weight scheme for the BS(2,4) tree is non-wilfian. Conjecture: the same is true for BS(p,q) when 1<p<q. Conjecture: if there is no Wilfian method to produce a particular weight scheme for a rooted tree, then the induced growth series is not Wilfian. The ternary tree with constant weight 1 has ω=3. With constant weight 2, it grows as ω= 3. Conjecture Lemma: the Dalton algorithm can be applied to a ternary tree to get any exponential growth rate between 3 and 3.

26 Thank You! (especially the organizers)