Pattern_avoidance_in_ternary_trees

arXiv:1110.2225v2[math.CO]27Dec2011
Pattern Avoidance in Ternary Trees
Nathan Gabriel1
Department of Mathematics
Rice University
Houston, TX 77251, USA
Katherine Peske1
Department of Mathematics and Computer Science
Concordia College
Moorhead, MN 56562, USA
Lara Pudwell1
Department of Mathematics and Computer Science
Valparaiso University
Valparaiso, IN 46383, USA
Lara.Pudwell@valpo.edu
Samuel Tay1
Department of Mathematics
Kenyon College
Gambier, OH 43022, USA
Abstract
This paper considers the enumeration of ternary trees (i.e., rooted ordered trees in
which each vertex has 0 or 3 children) avoiding a contiguous ternary tree pattern. We
begin by ﬁnding recurrence relations for several simple tree patterns; then, for more
complex trees, we compute generating functions by extending a known algorithm for
pattern-avoiding binary trees. Next, we present an alternate one-dimensional notation
for trees which we use to ﬁnd bijections that explain why certain pairs of tree patterns
yield the same avoidance generating function. Finally, we compare our bijections to
known “replacement rules” for binary trees and generalize these bijections to a larger
class of trees.
1
Partially supported by NSF grant DMS-0851721
1

1 Introduction
The notion of one object avoiding another has been studied in permutations, words, parti-
tions, and graphs. Although pattern avoidance has proven to be a useful language to describe
connections between various combinatorial objects, it has also attracted broad interest as
a pure enumerative topic. One combinatorial problem that has received much attention in
recent years is to count the number of permutations of length n avoiding a certain smaller
permutation. Here, permutation π avoiding permutation ρ means that π has no subsequence
that is order-isomorphic to ρ. Although the classical case of the permutation pattern problem
allows ρ to be given as any subsequence of π, a special case that can be attacked successfully
via a variety of techniques is studying when π contains ρ as a consecutive subpermutation.
This latter question can be answered by a variety of techniques including inclusion-exclusion.
There also exist algorithmic techniques, such as the Goulden-Jackson cluster method [4, 6]
to approach this question using generating functions. Two natural questions arise: “Given a
permutation ρ, how many permutations of length n avoid ρ?” and “When do two forbidden
permutations ρ1 and ρ2 have the same avoidance generating function?”
In this paper we consider the analogous questions for plane trees. All trees in the paper
are rooted and ordered. We will focus on ternary trees, that is, trees in which each vertex has
0 or 3 (ordered) children. A vertex with no children is a leaf and a vertex with 3 children is
an internal vertex. A ternary tree with k internal vertices has 2k + 1 leaves, and the number
of such trees is
(3k
k )
2k+1
(OEIS A001764). It is clear then that there only exist ternary trees
with an odd number of leaves. The first few ternary trees are depicted in Figure 2.
Conceptually, a plane tree T avoids a tree pattern t if there is no instance of t anywhere
inside T. Pattern avoidance in vertex-labeled trees has been studied in various contexts by
Steyaert and Flajolet [10], Flajolet, Sipala, and Steyaert [3], Flajolet and Sedgewick [2], and
Dotsenko [1]. Recently, Khoroshkin and Piontkovski [5] considered generating functions for
general unlabeled trees but in a different context.
In 2010, Rowland [7] explored contiguous pattern avoidance in binary trees (that is,
rooted ordered trees in which each vertex has 0 or 2 children). He had two motivations for
choosing these particular trees; first, there is a clear and natural definition of what it means
for a rooted ordered tree to contain a contiguous ordered pattern that is unclear for general
trees, and second, there is natural bijection between n-leaf binary trees and n-vertex trees.
His study had two main objectives. First, he developed an algorithm to find the generating
function for the number of n-leaf binary trees avoiding a given tree pattern; he adapted this
to count the number of occurrences of the given pattern. Second, he determined equivalence
classes for binary tree patterns, classifying two trees s and t as equivalent if the same number
of n-leaf binary trees avoid s as avoid t for n ≥ 1. He completed the classification for all
binary trees with at most eight leaves, using these classes to develop replacement bijections
between equivalent binary trees.
In this paper, we extend Rowland’s work by exploring pattern avoidance in ternary
trees, and in some cases to general m-ary trees (that is, trees where each vertex has 0 or
m children). We first compute recurrence relations to count trees that avoid ternary tree
patterns with at most seven leaves. Next, we adapt Rowland’s algorithm to find functional
equations for the avoidance generating functions of arbitrary ternary tree patterns. Finally,
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we give bijections between trees avoiding several pairs of equivalent tree patterns, and begin
generalizing this process to fit the more general case of m-ary trees. The appendix contains
all the equivalence classes of ternary tree patterns with at most nine leaves found using the
Maple package TERNARYTREES. The Maple package itself is given at the third author’s
website (http://faculty.valpo.edu/lpudwell/maple.html).
1.1 Definitions and Notation
Following Rowland’s definition of avoidance, a ternary tree T contains t as a tree pattern
if t is a contiguous, rooted, and ordered subtree of T. Conversely, T avoids t if there is no
such subtree of T. For example, consider the three trees shown in Figure 1. T contains
t because this pattern occurs beginning at the center child of the root of T (see bolded
subtree). However, T avoids s because no vertex in T has children extending from both its
left and center children.
T = t = s =
Figure 1: Three ternary trees
We define Avt(n) to be the set of n-leaf ternary trees that avoid the pattern t, and
avt(n) = |Avt(n)|. We will be particularly interested in determining the generating function
gt(x) =
∞
n=0
avt(n)xn
for various patterns t.
Before we explore particular ternary tree patterns, we list all of the 3, 5, and 7-leaf ternary
trees. Note, however, if tr
is the left–right reflection of t, then avt(n) = avtr (n) by symmetry,
so left–right reflections are omitted. We label trees with a double subscript notation. The
first subscript gives the number of leaves of the tree, and the second subscript distinguishes
between distinct tree patterns of the same depth. We will use these labels throughout the
remainder of the paper.
2 Recurrences for Simple Tree Patterns
In this section, we find recurrence relations for the number of trees avoiding several of the
trees in Figure 2. For each tree, we discuss the structure of trees that avoid the given tree
pattern, how a recurrence and generating function can be found from this structure, and we
list any other equivalent tree patterns. If t is clear from context, we will simply write Av(n)
and av(n) in lieu of Avt(n) and avt(n).
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t31 = t51 = t52 =
t71 = t72 = t73 = t74 =
t75 = t76 = t77 =
Figure 2: 3, 5, and 7-leaf ternary trees
2.1 Avoiding t51
and t52
To find avt51
(n), let us look at how an n-leaf tree T must be structured in order to avoid
t51 . Consider any internal vertex v of T. v’s left child can have no descendants, thus it must
be a leaf. v’s center child can be the root of a subtree of any number of leaves k, where
1 ≤ k ≤ n − 2. Finally, v’s right child can also be the root of a subtree, but because there
are n total leaves, this subtree must have precisely n − k − 1 leaves. Thus, there are av(k)
possible subtrees beginning at v’s center child, and av(n − k − 1) possible subtrees at v’s
right child that also avoid t51 . Taking the summation of these over the possible values of k
gives the recurrence relation
av(n) =
n−2
k=1
av(k)av(n − k − 1) where n ≥ 3.
Our initial conditions for this recurrence are av(0) = 0, because there are no ternary
trees with 0 leaves; av(1) = 1, because there is one tree with one leaf, and it avoids any tree
pattern with more than one leaf; and av(2) = 0, because there are no trees with 2 leaves.
We can now compute gt1 (x) = ∞
k=0 av(k)xk
using standard techniques to obtain
gt1 (x) =
1 −
√
1 − 4x2
2x
.
The first few terms of this sequence are (for n ≥ 0)
0, 1, 0, 1, 0, 2, 0, 5, 0, 14, . . .
Two things are worth noting about this avoidance sequence. First, the non-zero terms
are the Catalan numbers (OEIS A000108). Second, the sequence is interspersed by zeros
because there are no ternary trees with an even number of leaves. This second observation
will be true for the avoidance sequence of any ternary tree pattern.
For trees avoiding t52 , we only need to make one alteration; namely, that it is the center
child, instead of the leftmost child, of each vertex that cannot have any children. Therefore,
we find that
gt51
(x) = gt52
(x) =
1 −
√
1 − 4x2
2x
.
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2.2 Avoiding t71
and t72
Next, we find the number of n-leaf trees that avoid t71 such that n ≥ 3. As before, we consider
any internal vertex v of a tree T that avoids t71 . There are two nonexclusive possibilities for
which of v’s children are internal vertices. First, v’s leftmost child has no children, but both
its center and right children can. Otherwise, v’s center child has no children, but both its left
and right children can. These two cases are equivalent to avoiding t51 and t52 , respectively.
However, this double-counts one instance: that is, when both the left and the center child of
v have no children. There are exactly av(n − 2) trees counted by both of the first two cases.
Subtracting this from the recurrence relation, we are left with
av(n) = 2
n−2
k=1
av(k)av(n − k − 1) − av(n − 2).
Our initial conditions for this recurrence relation are again av(0) = 0, av(1) = 1, and
av(2) = 0. Using standard techniques, we obtain
gt71
(x) =
x2
+ 1 −
√
x4 − 6x2 + 1
4x
,
which gives the Little Schröder numbers (OEIS A001003), interspersed by zeros: 0, 1, 0, 1,
0, 3, 0, 11, 0, 45, 0, 197, 0,. . .
This is also the avoidance sequence for t72 . As before, two cases exist for avoiding t72
(either the left and center or the right and center children of v have descendants), as well as
a term that needs to be subtracted to avoid double-counting (when neither the left nor the
right children of v have their own children). Thus, we have
gt71
(x) = gt72
(x) =
x2
+ 1 −
√
x4 − 6x2 + 1
4x
.
Before considering other tree patterns, we further examine the connection between trees
avoiding t72 and the Little Schröder numbers (sn). To do this, we look at one well-known
combinatorial interpretation of the Little Schröder numbers: sn is the number of binary trees
with n vertices and with each right edge “colored” to be either solid or dashed [8]. We note
that elsewhere in this paper, we have concerned ourselves with strictly binary trees (each
internal vertex has precisely 2 children) or strictly ternary trees (each internal vertex has
precisely 3 children). In the current interpretation of the Little Schröder numbers, however,
these binary trees are not strict; that is, an internal vertex may have 1 or 2 children. Consider
the following map from this set of colored (non-strict) binary trees to the set of the (strict)
ternary trees avoiding t72 :
1. For each vertex v in the binary tree, draw a vertex v∗
.
2. Consider each parent-child pair in the binary tree:
• If w is the left child of v, then w∗
is the center child of v∗
.
• If w is the right child of v via a solid edge, then w∗
is the left child of v∗
.
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• If w is the right child of v via a dashed edge, then w∗
is the right child of v∗
.
3. For each vertex v∗
created in step 1, if v∗
has 0, 1, or 2 children, add children until v∗
has exactly 3 children.
For example, the colored binary tree in Figure 3 is mapped to the ternary tree in the
same Figure.
→
Figure 3: A colored binary tree and its corresponding t72 -avoiding ternary tree
Note that any ternary tree that is produced by this algorithm certainly avoids t72 since
a vertex will not have both a solid right edge and a dashed right edge at the same time, and
accordingly a vertex in the resulting ternary tree will never have both a right child and a
left child with children of their own. This process has an obvious inverse.
As an example, look at s3 = 11. There are eleven 3-vertex colored binary trees, and
eleven 5-leaf trees avoiding t72 . Each colored binary tree is shown with its image under our
map in Figure 4.
→ → →
→ → →
→ → →
→ →
Figure 4: Mapping colored binary trees to t72 -avoiding ternary trees
2.3 Avoiding t73
and t77
To ﬁnd the number of n-leaf trees that avoid t73 , n ≥ 3, we consider two cases for any internal
vertex v of a tree T that avoids t73 . First, v’s left child has no children, while the center
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and right children are roots of subtrees with k leaves and n − k − 1 leaves respectively with
1 ≤ k ≤ n − 2. The second case is when v’s left child has three children; to avoid t73 , a left-
vertex child cannot have another consecutive left-vertex child. The four other vertices (the
center and right children of v and the center and right children of v’s left child) are the roots
of subtrees with ℓ, m, k, and n − ℓ − m − k − 1 leaves respectively with 1 ≤ ℓ, m, k ≤ n − 4.
Therefore, avt73
(n) is given by the sum of these two cases
av(n) =
n−2
k=1
av(k)av(n − k − 1) +
n−4
ℓ=1
n−ℓ−3
m=1
n−ℓ−m−2
k=1
av(ℓ)av(m)av(k)av(n − ℓ − m − k − 1)
To find the recurrence relation for trees avoiding t77 , we see that instead of avoiding
two consecutive left-children vertices, we avoid two consecutive middle-children vertices.
Therefore, avt73
(n) = avt77
(n) for n ≥ 1. From this recurrence we compute the avoidance
sequence: 0, 1, 0, 1, 0, 3, 0, 11, 0, 46, 0, 207, 0, . . . (OEIS A006605 interspersed by zeros).
Clearly, it would be extremely difficult to solve this recurrence directly for the generating
function gt73
(x). It turns out that tree patterns t74 , t75 , and t76 , have the same avoidance
generating function as t71 and t72 , but we have not been able to find their recurrence relations
by hand with an argument parallel to those above because complex problems arise with
overcounting and undercounting. Instead, we now adapt Rowland’s generating function
algorithm for trees avoiding binary tree patterns to deal with ternary tree patterns.
3 A Generating Function Algorithm
As we saw in the previous section, it is straightforward to compute a given ternary tree
pattern’s avoidance generating function by hand for a few small tree patterns. However, this
type of computation quickly becomes impractical for increasingly complex tree patterns.
For this reason, we develop an algorithm to find a functional equation satisfied by the
avoidance generating function gt(x) for any ternary tree pattern t. First, however, we make
one notational adjustment. Now, let g(t;p)(x) be the generating function for the number of
n-leaf ternary trees that avoid t and contain the tree pattern p at their root. Recall, we have
already specified that tree T contains t as a pattern if T contains t as a contiguous, rooted,
ordered subgraph. T contains pattern p at the root if T contains a copy of p where the root
of p coincides with the root of T. Therefore, with our new notation, the generating function
for all trees avoiding t is given by gt(x) = g(t; )(x), because all ternary trees begin with the
single vertex root.
The algorithm we use to find g(t; )(x) is very similar to Rowland’s algorithm for binary
trees [7], but it accounts for an additional child at each internal vertex. The algorithm
produces a sequence of generating functions using a recursive method. Initially, g(t; )(x), the
generating function we are interested in, is written in terms of another generating function.
Then, for each new generating function g(t;p)(x) introduced in the recursive step, we deduce
another recurrence in terms of other generating functions. If t is a tree pattern, we will use tℓ,
tc, and tr to denote the left, center, and right subtrees of t respectively. When appropriate we
may write t = (tℓtctr). We also require one more operation on tree patterns; we will use s ∩t
7

to denote the intersection of tree patterns s and t. Conceptually, s ∩ t is the tree pattern
produced by drawing s and t so that their roots coincide. More formally, if v is a single
vertex, then t ∩ v = t and recursively s ∩ t = (sℓscsr) ∩ (tℓtctr) = ((sℓ ∩ tℓ) (sc ∩ tc) (sr ∩ tr)).
We note that although considering s and t as trees makes the intersection notation seem to
be a misnomer, the set of trees with tree pattern s at their root intersected with the set of
trees with tree pattern t at their root is indeed the set of trees with tree pattern s∩t at their
root. With this new notation, we are now prepared to give an algorithm to ﬁnd g(t; )(x)
(for t not equal to the single vertex tree).
First notice that g(t; )(x) = x + g(t; )(x). This is because, unless t is the single vertex
tree, the generating function for trees with a single vertex at the root will always account
for the one tree with one leaf. Then, the rest of the trees avoiding t have a pattern at the
root, so gt(x) = g(t; )(x) = x + g(t; )(x).
Next, we have introduced a new generating function g(t; )(x), and we need to derive
a new recurrence for this function. To do this, we recognize that a t-avoiding tree with
pattern p at its root is made up of three t-avoiding subtrees with pℓ, pc, and pr at their
respective roots. However, g(t;pℓ)(x), g(t;pc)(x), and g(t;pr)(x) each account for trees avoiding t
individually, which overcounts when the trees have root pattern of pℓ ∩ tℓ, pc ∩ tc, and pr ∩ tr
respectively. Therefore, we have
g(t;p)(x) = g(t;pℓ)(x) · g(t;pc)(x) · g(t;pr)(x) − g(t;pℓ∩tℓ)(x) · g(t;pc∩tc)(x) · g(t;pr∩tr)(x).
This observation holds not only for the tree pattern , but for any non-trivial tree pattern
p. We repeatedly use this observation to derive a new recurrence for each generating function
g(t;p)(x) that arises in our computation until we have a complete system of equations. We
are guaranteed that such a system will eventually be complete since if t has depth d, each
pattern p introduced in this process has depth at most d and there are ﬁnitely many tree
patterns of depth at most d. Once we have a complete system of equations, we eliminate
all unwanted variables until we have a functional equation for g(t; )(x). Here, then, is the
algorithm to compute gt(x) for any non-trivial ternary tree pattern t.
1. Initialize Eq = {g(t; )(x) = x + g(t; )(x)}, V ar = { }, P = { }, and P1 = ∅.
2. For p ∈ P, do:
• Let Eq = Eq ∪ {g(t;p)(x) = g(t;pℓ)(x) · g(t;pc)(x) · g(t;pr)(x) − g(t;pℓ∩tℓ)(x) · g(t;pc∩tc)(x) ·
g(t;pr∩tr)(x)}.
• Let Var = Var ∪ {p}.
• Let P1 = (P1 ∪ {pℓ, pc, pr, pℓ ∩ tℓ, pc ∩ tc, pr ∩ tr}) Var.
3. If P1 = ∅ then let P = P1, P1 = ∅, and go to Step 2.
If P1 = ∅ then eliminate all variables in Var {g(t; )(x)} from the system of equations
Eq to compute a functional equation for g(t; )(x).
We illustrate this algorithm by using it to compute the avoidance generating function for
t73 .
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In step 1, we initialize
Eq = {g(t73 ; )(x) = x + g(t73 ; )(x)}
Var = { }
P = { }
P1 = ∅
In step 2, we consider ∈ P to obtain
Eq = Eq ∪ {g(t73 ; )(x) = (g(t73 ; )(x))3
− g
(t73 ; )
(x) · (g(t73 ; )(x))2
}
Var = Var ∪ { }
P1 = { }
Since P1 = ∅, relabel P1 as P, and consider ∈ P to obtain
Eq = Eq ∪ {g
(t73 ; )
(x) = g(t73 ; )(x) · (g(t73 ; )(x))2
− g
(t73 ; )
(x) · (g(t73 ; )(x))2
}
Var = Var ∪ { }
P1 = ∅
Since P1 = ∅, we consider the three equations in Eq. Let a = g(t73 ; )(x), b = g(t73 ; )(x),
and c = g
(t73 ; )
(x). Then,
a = x + b
b = a3
− ca2
c = ba2
− ca2
Eliminating b and c gives the equation xa4
+ xa2
− a + x = 0.
For very simple trees, we can usually solve the resulting functional equation directly for
g(t; )(x); however, this quartic functional equation is a more characteristic result for complex
tree patterns. Although this functional equation does not have a simple explicit solution,
we can use it to compute arbitrarily many coefficients of the generating function gt73
(x) by
making the substitution a = k
n=0 av(n)xn
, isolating the coefficients of each power of x, and
setting them each equal to zero. From the functional equation xa4
+xa2
−a+x = 0, we find
the sequence av(n), 0 ≤ n ≤ 25, to be 0, 1, 0, 1, 0, 3, 0, 11, 0, 46, 0, 207, 0, 979, 0, 4797,
9

0, 24138, 0, 123998, 0, 647615, 0, 3428493, 0, 18356714,. . . (OEIS A006605 interspersed by
zeros).
A complete classification of ternary trees t with up to 9 leaves is given in the Appendix,
along with functional equations for gt(x) and 20 terms of the corresponding avoidance se-
quences.
Note that given a tree pattern t, the algorithm given in this section generates a system
of equations each of which has maximum total degree 3. Auxiliary variables in this system
will be eliminated to produce a polynomial functional equation for gt(x). This guarantees
that gt(x) is always algebraic. Khoroshkin and Piontkovski [5] independently showed that
generating functions are algebraic in the case of general pattern-avoiding trees; however,
their work is done in a different context.
4 Bijections on Ternary Trees
Now that we have discussed two methods for enumerating pattern-avoiding trees, we look
for connections between specific sets of those trees. Recall that several of the ternary trees
in this paper have had the same pattern avoidance sequence as one or more other trees.
That is, for some distinct i and j, we found that gtki
(x) = gtkj
(x). Such patterns are said
to be Wilf equivalent. We will now explain why certain pairs of tree patterns {tki
, tkj
} are
Wilf equivalent. As Rowland did, we accomplish this through finding bijections between the
members of Avtki
(n) and those of Avtkj
(n). In order to present these bijections in a clear and
concise way, we first present an alternate notation for ternary trees. We use this notation
to describe bijections that explain all Wilf equivalences between 5 and 7-leaf ternary tree
patterns. We then generalize these maps.
4.1 Word Notation for Trees
In this subsection we represent trees as sets of integer words. This notation easily extends
to m-ary trees (i.e., trees where each vertex has 0 or m children). At the foundation of this
word representation are m-leaf parents.
Definition 1. An m-leaf parent is an internal vertex, v, of an m-ary tree such that v has
exactly m children, all of which are leaves.
For example, t74 has one 3-leaf parent and t71 has two 3-leaf parents.
Word notation represents an m-ary tree with a set of words, where each word follows the
path from the root to an m-leaf parent. We construct such a set from the following Tree-Set
Algorithm:
1. Label the children of each internal vertex of an m-ary tree from left to right, 1 through
m. (In ternary trees, then, a vertex’s left child is labeled 1, its center child 2, and its
right child 3.)
2. Denote a path from the root of a tree to an m-leaf parent by an integer word x1 · · · xk,
where k is the length of the path from the root to the m-leaf parent, such that xi ∈ Z
10

and 1 ≤ xi ≤ m for 1 ≤ i ≤ k. The first number x1 in the word represents the child
of the root labeled x1; xi then refers to the child of the vertex given by xi−1 that is
labeled with the value of xi.
As an example, let us look again look at t74 and t71 . In t74 , the only 3-leaf parent is
reached by a path beginning at the root, going to the root’s left child, then to this vertex’s
center child; in word notation, t74 is denoted by the set {12}. For t71 , we reach one of its
3-leaf parents by going to the root’s left child, and the other by the root’s center child; this
becomes a set {1, 2}. Note that an ordered m-ary tree T is uniquely defined by the set
of paths from its root to each m-leaf parent. The single vertex tree is represented by the
empty set, {}, and the 3-leaf tree (t31 ) by the set containing the empty word, {ǫ}. The set
{21, 23, 321} denotes the tree T given in Figure 1 and the set {13, 223} denotes the ternary
tree given in Figure 3. Also, note that the Tree-Set Algorithm above is clearly reversible. In
particular:
1. Create an m-ary tree from each word by the following procedure:
(a) Create a root.
(b) Give the root m children, labeled left to right from 1 to m.
(c) For the word x1x2 · · · xk give the x1-st child of the root m children. Label these
children 1 through m as before, repeating the process at each level where xi
denotes giving m children to the xi-th child of the vertex that was given children
by xi−1.
2. Take the intersection of all trees obtained from step 1 to find the final m-ary tree.
We note that representing trees as words or as sets of words is not a new idea. For
example, Stanley [8] shows how to represent plane trees as certain integer sequences and as
parenthesizations of words subject to certain constraints.
Now that we have this alternate representation of m-ary trees, we consider specific prop-
erties of the word notation that are relevant to our questions of pattern avoidance. First
note that any set of words where one word is a prefix of another is redundant for the rep-
resentation of trees. Since each word specifies a path within a tree, if word w1 is a prefix
of word w2, the path specified by w1 is necessarily a part of the path specified by w2, and
including w1 is superfluous.
Definition 2. Let S be a set of words on the alphabet {1, . . . , m}, such that no word is a
prefix of another word in S. Namely, S is an arbitrary set of words describing an m-ary
tree. We write A(m) for the set of all such sets S.
Consider a set of words {Li}l
i=1 in A(m) corresponding to tree pattern t. Note that tree
T, given by {Mh}p
h=1, contains t if there exist {Mhi
}l
i=1 where each Mhi
begins with the
same (possibly empty) prefix as all other Mhi
’s, followed by exactly the ordered sequence of
elements of Li; this may or may not be then followed by an additional sequence of numbers.
For example, the tree pattern t = {1323, 1223}, is contained by
T1 = {3231323, 11322, 3231223112}.
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Notice that, after the prefix 323, the first and third words of T1 have exactly the sequence of
each word of t. However, T2 = {31323, 1223} avoids t. Even though it contains each sequence
of numbers from the words of t, T2’s words do not begin with the same prefix before the
sequences begin.
In the following subsections, Btki
,tkj
: A(m) → A(m) will denote a bijection from trees
avoiding tki
to trees avoiding tkj
. We find such maps by analyzing the word notation for our
pattern-avoiding trees. We are now ready to present bijections between Wilf equivalent tree
patterns.
4.2 The patterns t51
and t52
A tree avoids t51 if none of its left vertices have children, and avoids t52 if none of its center
vertices have children. In order to map a tree T avoiding t51 to a tree avoiding t52 , we define
our bijection Bt51 ,t52
(T) to “switch” the center subtree of every vertex with the left subtree
of the same vertex. In terms of word notation, a tree avoids t51 if it has no 1’s in its words,
and avoids t52 if it has no 2’s in its words. Thus, we define Bt51 ,t52
(T) to replace every 1 in
T’s words with 2, and every 2 with a 1. For example, Bt51 ,t52
({233, 32}) = {133, 31}. It is
clear that this map is one-to-one, onto, and preserves number of leaves.
and t77
A tree avoids t73 if no two consecutive left vertices have children, and avoids t77 if no two
consecutive center vertices have children. In word notation, then, a tree avoids t73 if it has
no pairs of consecutive 1’s, and it avoids t77 if it has no pairs of consecutive 2’s. Thus, we
define Bt73 ,t77
(T) = Bt51 ,t52
(T). That is, for a tree T avoiding t73 , Bt73 ,t77
(T) replaces each 1
in T’s set of words with a 2, and each 2 with a 1. Because we originally defined Bt51 ,t52
(T)
on A(3) defining Bt73 ,t77
(T) in this way is reasonable. For example, Bt73 ,t77
({13, 22, 322}) =
{23, 11, 311}. As stated before, this map is a bijection.
and t72
For a tree T to avoid t71 , no vertex v can have children descending from both its left and
center children; to avoid t72 , no vertex can have children descending from both its left and
right children. Therefore, we define a bijection Bt71 ,t72
(T) to switch the right and center
subtrees of each vertex. Using word notation, this is equivalent to defining Bt71 ,t72
(T) to
replace every 2 with a 3 and every 3 with a 2. For example, Bt71 ,t72
({121, 1232, 322, 331}) =
{131, 1323, 221, 233}. Again, it is clear that Bt71 ,t72
is one-to-one, onto, and preserves number
of leaves.
and t75
Using word notation, a tree avoids t74 if none of its words have a 1 followed by a 2; it
avoids t75 if none of its words have a 1 followed by a 3. Therefore, we define Bt74 ,t75
(T) =
Bt71 ,t72
(T). That is, Bt74 ,t75
(T) replaces each 2 with a 3, and each 3 with a 2. For example,
12

Bt74 ,t75
({1313, 3213, 323}) = {1212, 2312, 232}. As in all previous examples, it is clear that
Bt74 ,t75
is a bijection that preserves number of leaves.
and t76
Using word notation, a tree avoids t75 if none of its words have a 1 followed by a 3; it avoids
t76 if none of its words have a 2 followed by a 1. Therefore, we define Bt75 ,t76
(T) to replace
every 1 with a 2, every 2 with a 3, and every 3 with a 1. Bt75 ,t76
clearly maps words that
do not contain the sequence 13 to words that do not contain the sequence 21. For example,
Bt75 ,t76
({1, 21, 3212}) = {2, 32, 1323}. As in all previous examples, it is again clear that
Bt75 ,t76
is a bijection that preserves number of leaves.
and t74
This bijection is a bit more complicated than the previous maps. Consider the function
Bt71 ,t74
(T) defined by the following procedure applied each word L ∈ T:
1. If L contains no 1 followed by a 2, do nothing to L.
2. Otherwise, locate the first copy of 1 · · ·12 in L = x1 · · ·x|L|. In particular, suppose that
x1 · · · xi−1 has no copy of 1 · · ·12, xi−1 = 1, xi = xi+1 = · · · = xj = 1 and xj+1 = 2.
We map L to the pair of words x1 · · · xj, x1 · · · xi−1xj+1 · · ·x|L|. So x1 · · ·xj contains no
occurrence of 12.
3. Iterate step 2 for each new word created until we have produced L1, L2, . . . , Lp, none
of which contain a 1 followed by a 2.
For example, if T1 = {1232311121}, our first iteration maps T1 to {1, 232311121}. Our
second iteration gives {1, 2323111, 232321}. All of these words are 1 · · ·12-free, so we have
Bt71 ,t74
(T1) = {1, 2323111, 232321}.
In order to prove that Bt71 ,t74
is a bijection, we first construct an inverse function,
B−1
t71 ,t74
(T). We will start at the root of tree T and work downward to consider all oc-
currences of t71 ; that is, word pairs of the form p01s1, p02s2, where p0 denotes a common
prefix in the two words and s1, s2 are suffixes. Note that for each occurrence of t71 it is
possible for there to be multiple words of the form p02si. Our inverse map is given by the
following process:
1. On each level of T beginning at the root, replace each occurrence of p02si that is part
of an occurrence of t71 with p012si. If there are multiple occurrences of t71 on the same
level, then applying step 2 to one occurrence of t71 does not affect the words denoting
any other occurrence of t71 on that level. Therefore, the order with which we apply
this step to each occurrence of t71 at the ith level is irrelevant.
2. Iterate step 1 at each successive level, beginning with the root.
13

3. Discard any words that are a prefix of another in T’s set of words.
For example,
• For T2 = {1, 2323111, 232321}, at the first level we have an occurrence of t3 given by
1 ∩ 2323111 and from 1 ∩ 232321. So, from step 1, we replace {2323111, 232321} with
{12323111, 1232321}.
• We now have the set {1, 12323111, 1232321}; step 2 requires that we check the next
levels in order from lowest to highest and we see that t3 does not occur until the sixth
level, and is given by the words 12323111, 1232321. Thus, we replace 1232321 with
12323121 to obtain {1, 12323111, 12323121}.
• With the third iteration of step 1, we replace 12323121 with 123231121. The fourth
iteration replaces 123231121 with 1232311121, and we are left with the set {1, 12323111,
1232311121}. After applying step 3, we see that B−1
t71 ,t74
(T2) = {1232311121}. We note
that B−1
t71 ,t74
(T2) = T1 as expected from our earlier example.
It is easy to see that Bt71 t74
(T) does in fact map trees avoiding t71 to trees avoiding t74 . It
is also clear that B−1
t71 ,t74
(Bt71 ,t74
(T)) = T, so Bt71 ,t74
(T) has a clear inverse. Thus Bt71 ,t74
(T)
is a bijection.
We further claim that Bt71 ,t74
(T) preserves the numbers of leaves of T. Step 1 in our
bijection is the only step that changes the structure of T. Consider an arbitrary occurrence
of t74 such that the path from the root of T to the root of the occurrence is given by the
prefix p0 and such that there is no occurrence of t74 in p0. Then step 1 in our bijection will
map all words with the prefix p012 to the word p01 and words with the prefix p02. As a
result of no longer having any words with the prefix p012, we see that the vertex given by
the path p01 has one more leaf in Bt71 ,t74
(T) than it did in T (namely its second child is a
leaf, but was not a leaf in T). However we also see that the vertex given by the path p0 has
one less child that is a leaf as a result of having words with p02 as a prefix. There could
not have already been a word in T with p02 as a prefix since this would entail having an
occurrence of t71 at the vertex given by the path p0. Having the word p01 does not add to
or subtract from the number of leaves we have since it is a prefix of the word it replaces
and thus creates no new vertices. With each iteration of step 2 on an occurrence of t74 this
same reasoning holds. So we see that the number of leaves in Bt71 ,t74
(T) is the same as the
number of leaves in T.
We have now accounted for all Wilf equivalences between tree patterns with 5 or 7 leaves.
4.8 General Approaches to Bijections
In this section, we generalize the previous six bijections to bijections for a larger class of
tree patterns. In particular, we present a bijection between certain tree patterns that have
the same avoidance generating function and have only one m-leaf parent. These are tree
patterns that are represented by a single word in word notation.
Consider two m-ary tree patterns {L1} and {L2} in word notation. If there exists a
bijection b : {1, . . . , m} → {1, . . . , m} that maps L1 to L2, then there is a bijection BL1,L2 (T)
14

from the set of trees avoiding {L1} to the set of trees avoiding {L2}. In particular, BL1,L2 (T)
is the map that replaces each letter i in the lists of an L1-avoiding tree with b(i).
For example, if {L1} = {11213}, {L2} = {22321}, the map b sends 1 → 2, 2 → 3, and
3 → 1. If we consider a tree T that avoids {L1} given in list notation, then BL1,L2 (T) replaces
each integer i with b(i). For example BL1,L2 ({13231, 22321}) = {21312, 33132}.
In general, if T is a tree denoted by a set of words that avoids {L1}, when we apply
the bijection b to the letters of T, we have BL1,L2 (T) to be a set of words in which no word
contains any instance of {L2}, that is, BL1,L2 (T) avoids {L2}.
Theorem 1. The map BL1,L2 (T) is one-to-one, onto, and preserves the number of leaves of
T.
Proof. The fact that BL1,L2 (T) is one-to-one and onto follows from the fact that BL1,L2 (T)
is induced by the map b, which is just a permutation of {1, . . . , m}.
To show that BL1,L2 (T) preserves the number of leaves of T, it is enough to show that
for each internal vertex v1 in a given tree T there is a unique internal vertex v2 in BL1,L2 (T)
that has the same number of children that are leaves as does v1. Consider the word p1
describing the path to v1 in T. Let x be the number of distinct letters in {1, . . . , m} that
follow p as a preﬁx in a word of T. Then v1 has m−x children that are leaves. Now consider
BL1,L2 ({p1}) = {p2}. Since x distinct letters followed p1 in T, there must be x distinct letters
that follow p2 as a preﬁx in BL1,L2 (T), meaning the vertex whose path from the root is given
by p2 has m − x children that are leaves. This holds true for every vertex of T, proving that
the number of leaves remains the same.
These generalized bijections for trees have one more interesting property. Consider the
word notation for an m-ary tree. This can also be read as the word notation for an M-ary
tree for any M ≥ m. This says that once we have a bijection BL1,L2 (T) for a pair of m-ary
trees, we have necessarily discovered a Wilf equivalence for corresponding pairs of M-ary
trees for each M ≥ m.
4.9 Final notes on bijections
In this section we have given bijections explaining all non-trivial Wilf equivalences between
ternary tree patterns with 5 or 7 leaves. We note that all but one of the bijections given in
Section 4 are guaranteed to exist by the results of Section 4.8. The bijection between t71 -
and t74 -avoiding trees is the lone exception to this method, since it involved “cutting” the
integer words representing trees, rather than just relabeling them. In the appendix, we also
give computational data for equivalent ternary tree patterns with 9 leaves; it turns out that
all 9-leaf ternary tree patterns fall into just three distinct Wilf classes. Many, but not all, of
these equivalences can also be explained with the generalized bijections of Section 4.8.
Moreover, all bijections except the one for t71 and t74 may clearly be seen as replacement
bijections in the sense of Rowland’s binary tree patterns paper [7]. In fact, we can see
that this last bijection is not a replacement bijection by considering a well-chosen example.
Consider the tree whose word representation is {1232311121}, as shown in Figure 5. This
tree contains two copies of the tree pattern t74 , or {12}. One of those copies is at the root,
15

and the other is near the bottom of the tree, with one copy of t31 appended to its deepest
left leaf. On the other hand, the image of this tree is represented by {1, 2323111, 232321}
(also shown in Figure 5). This tree contains two copies of the tree t71 , or {1, 2}, one at the
root, and one deeper in the tree. However, this lower copy has a copy of t51 and a copy of t31
appended to its deepest leaves. In a true replacement bijection, the copies of t74 should be
transformed to copies of t71 , and the subtrees descending from leaves of a copy of t74 should
be moved in entirety to be subtrees descending from leaves of a copy of t71 . This is clearly
not the case in this bijection. It remains an open question to determine whether there exists
a replacement bijection between ternary trees avoiding t71 and ternary trees avoiding t74 .
Figure 5: The ternary trees with word representations {1232311121} and {1, 2323111,
232321}
Many of our bijections overlap with Rowland’s idea of replacement bijections; however,
we propose that considering trees as sets of integer words provides more insight into the
process of developing Wilf equivalent tree patterns.
5 Conclusion
Throughout this paper, we have investigated pattern avoidance in ternary trees, extending
previous work for binary trees. We began by finding recurrence relations and generating
functions by hand for several simple ternary tree patterns. To make the computation of
avoidance sequences easier, however, we developed an algorithm, based on Rowland’s al-
gorithm for binary trees, to find the generating function for trees avoiding any given tree
pattern. Next we classified the tree patterns, grouping together those with the same avoid-
ance sequence. From here, we were able to find bijections between the sets of trees avoiding
specific pairs of two equivalent tree patterns, tki
and tkj
; for these pairs of trees, we trans-
formed any tree avoiding tki
into one that avoids tkj
. After stating several bijections between
specific pairs of tree patterns, we then generalized this to bijections between trees avoiding
patterns in the same equivalence class of trees under permutations of {1, . . . , m} .
16

6 Acknowledgement
The authors thank Eric Rowland for a number of presentation suggestions and for support
with generating the many tree graphics required for this paper.
Appendix
This appendix lists ternary trees with at most nine leaves, classifying them by their avoidance
generating function and sequence. For each class, we give a functional equation satisfied by
a = gt(x), and we list the first 20 terms (including zeros) of the corresponding avoidance
sequence. If gt(x) has a simple explicit form it is included, and if the avoidance sequence for
a class is listed in the Online Encyclopedia of Integer Sequences (without interspersed zeros)
[9], we include the appropriate reference. For brevity, left–right reflections are omitted.
Class 5
• xa2
− a + x = 0
• gt(x) = 1−
√
1−4x2
2x
• 0, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, 0, 132, 0, 429, 0, 1430, 0, 4862, . . .
• OEIS A000108: Catalan numbers
Class 7.1
• 2xa2
− x2
a − a + x = 0
• gt(x) =
(x2+1)−
√
(x2+1)2−8x2
4x
• 0, 1, 0, 1, 0, 3, 0, 11, 0, 45, 0, 197, 0, 903, 0, 4279, 0, 20793, 0, 103049, . . .
• OEIS A001003: Little Schröder numbers
Class 7.2
17

• xa4
+ xa2
− a + x = 0
• 0, 1, 0, 1, 0, 3, 0, 11, 0, 46, 0, 207, 0, 979, 0, 4797, 0, 24138, 0, 123998, . . .
• OEIS A006605: number of modes of connections of 2n points
Class 9.1
• 3xa2
− 3x2
a − a + x3
+ x = 0
• gt(x) = (3x2+1)−
√
1−6x2−3x4
6x
• 0, 1, 0, 1, 0, 3, 0, 12, 0, 54, 0, 261, 0, 1323, 0, 6939, 0, 37341, 0, 205011, . . .
• OEIS A107264: Expansion of −(3x−1)−
√
1−6x−3x2
6x2 .
Class 9.2
18

• xa4
− x2
a3
+ 2xa2
− x2
a − a + x = 0
• 0, 1, 0, 1, 0, 3, 0, 12, 0, 54, 0, 261, 0, 1324, 0, 6954, 0, 37493, 0, 206316, . . .
• OEIS A200740: Generating function satisfies x2
A4
(x) − x2
A3
(x) + 2xA2
(x) − xA(x) −
A(x) + 1 = 0
Class 9.3
• xa6
+ xa4
+ xa2
− a + x = 0
• 0, 1, 0, 1, 0, 3, 0, 12, 0, 54, 0, 262, 0, 1337, 0, 7072, 0, 38426, 0, 213197, . . .
• OEIS A186241: Generating function given by x3
A6
(x)+x2
A4
(x)+xA2
(x)−A(x)+1 = 0
References
[1] V. Dotsenko, Pattern avoidance in labelled trees, preprint,
http://arxiv.org/abs/1110.0844.
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[3] P. Flajolet, P. Sipala, and J. M. Steyaert, Analytic variations on the common subexpres-
sion problem, Automata, Languages, and Programming: Proc. of ICALP 1990, Lecture
Notes in Computer Science, Vol. 443, Springer, 1990, pp. 220–234.
[4] I. Goulden and D. Jackson, An inversion theorem for cluster decompositions of sequences
with distinguished subsequences, J. London Math. Soc. 20 (1979), 567–576.
[5] A. Khoroshkin and D. Piontkovski, On generating series of finitely presented operads, in
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[6] J. Noonan and D. Zeilberger, The Goulden-Jackson cluster method: extensions, applica-
tions, and implementations, J. Difference Equations and Applications 5 (1999), 355–377.
[7] E. S. Rowland, Pattern avoidance in binary trees, J. Combin. Theory, Ser. A 117 (2010),
741–758.
[8] R. P. Stanley, Enumerative Combinatorics, Cambridge University Press, 1999.
[9] N. Sloane, The Encyclopedia of Integer Sequences. Available at http://oeis.org, 2011.
[10] J. M. Steyaert and P. Flajolet, Patterns and pattern-matching in trees: an analysis,
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