Extremal properties of the Colless balance index for rooted binary trees

TL;DR

This paper investigates the extremal properties of the Colless index in rooted binary trees, deriving formulas for its minimal value, characterizing trees that achieve this minimum, and revealing connections to fractal curves.

Contribution

It provides a recursive and explicit formula for the minimal Colless index, characterizes trees with minimal index, and links these findings to fractal geometry and other tree balance indices.

Findings

Derived a recursive formula for the minimal Colless index.

Characterized trees with minimal Colless index as maximally balanced and 'greedy from the bottom' trees.

Established a connection between the minimal Colless index and the Blancmange (Takagi) curve.

Abstract

Measures of tree balance play an important role in various research areas, for example in phylogenetics. There they are for instance used to test whether an observed phylogenetic tree differs significantly from a tree generated by the Yule model of speciation. One of the most popular indices in this regard is the Colless index, which measures the degree of balance for rooted binary trees. While many statistical properties of the Colless index (e.g. asymptotic results for its mean and variance under different models of speciation) have already been discussed in different contexts, we focus on its extremal properties. While it is relatively straightforward to characterize trees with maximal Colless index, the analysis of the minimal value of the Colless index and the characterization of trees that achieve it, are much more involved. In this note, we therefore focus on the minimal value of the Colless index for any given number of leaves. We derive both a recursive formula for this minimal value, as well as an explicit expression, which shows a surprising connection between the Colless index and the so-called Blancmange curve, a fractal curve that is also known as the Takagi curve. Moreover, we characterize two classes of trees that have minimal Colless index, consisting of the set of so-called \emph{maximally balanced trees} and a class of trees that we call \emph{greedy from the bottom trees}. Furthermore, we derive an upper bound for the number of trees with minimal Colless index by relating these trees with trees with minimal Sackin index (another well-studied index of tree balance).

Equations

$$\mathcal{C}(T)$$

$$=\sum _{v\in \mathring{V}(T)}|{\kappa }_T(v_1)-{\kappa }_T(v_2)|,$$

$$\mathcal{C}(T)=\mathcal{C}(T_a)+\mathcal{C}(T_b)+n_a-n_b.$$

$$\mathcal{C}(T)$$

$$=\sum _{v\in \mathring{V}(T_a)}|{\kappa }_{T_a}(v_1)-{\kappa }_{T_a}(v_2)|+\sum _{v\in \mathring{V}(T_b)}|{\kappa }_{T_b}(v_1)-{\kappa }_{T_b}(v_2)|+|{\kappa }_T(a)-{\kappa }_T(b)|$$

$$=\mathcal{C}(T_a)+\mathcal{C}(T_b)+|{\kappa }_T(a)-{\kappa }_T(b)|$$

$$=\mathcal{C}(T_a)+\mathcal{C}(T_b)+n_a-n_b.$$

$$\mathcal{C}(T)$$

$$\mathcal{C}(\tilde{T})=\mathcal{C}(\hat{T})+\mathcal{C}(T_b)+n_a-n_b<\mathcal{C}(T_a)+\mathcal{C}(T_b)+n_a-n_b=\mathcal{C}(T).$$

$$c_{2n}=2c_n,$$

$$c_{2n+1}=c_{n+1}+c_n+1.$$

$$(i)\text{leavevmode}{c}_{2n}\le 2c_n,$$

$$(ii)\text{leavevmode}{c}_{2n+1}\le c_{n+1}+c_n+1.$$

$$c_{2n}=\min \{$$

$$c_{2n-4}+c_4+2n-8,\dots ,c_{n+1}+c_{n-1}+2,2c_n\}.$$

$$c_{2n+1}=\min \{$$

$$c_{2n-3}+c_4+2n-7,\dots ,c_{n+2}+c_{n-1}+3,c_{n+1}+c_n+1\}.$$

$$c_{2n}\ge 2c_n\text{ and }$$

$$c_{2n+1}\ge c_{n+1}+c_n+1.$$

$$c_{2n+2}=\min \{$$

$$c_{n+2}+c_n+2,2c_{n+1}\}.$$

$$c_{2n+2}\ge \min \{$$

$$c_n+c_{n-1}+1+c_2+c_1+1+2n-4,\dots ,2c_{\frac{n}{2}+1}+2c_{\frac{n}{2}}+2,2c_{n+1}\}$$

$$=\min \{$$

$$c_n+c_1+n-1+c_{n-1}+c_2+n-3+2,\dots ,$$

$$2(c_{\frac{n}{2}+1}+c_{\frac{n}{2}}+1),2c_{n+1}\}.$$

$$c_{n+1}=\min \{$$

$$\dots ,c_{\frac{n}{2}+1}+c_{\frac{n}{2}}+1\},$$

$$c_{2n+2}\ge \min \{$$

$$=2c_{n+1}.$$

$$c_{2n+2}\ge \min \{$$

$$c_{\frac{n+1}{2}+1}+c_{\frac{n+1}{2}}+1+c_{\frac{n+1}{2}}+c_{\frac{n-1}{2}}+1+2,2c_{n+1}\}$$

$$=\min \{$$

$$c_n+c_1+n-1+c_{n-1}+c_2+n-3+2,\dots ,$$

$$c_{\frac{n+1}{2}+1}+c_{\frac{n-1}{2}}+2+2c_{\frac{n+1}{2}}+2,2c_{n+1}\}.$$

$$c_{n+1}=\min \{$$

$$\dots ,2c_{\frac{n+1}{2}}\},$$

$$c_{2n+2}\ge \min \{$$

$$=2c_{n+1}.$$

$$c_n=\sum _{i=0}^{k_n-2}\frac{s(2^{i-k_n+1}\cdot n)}{2^{i-k_n+1}},$$

$$T(x)=\sum _{i=0}^{\infty }\frac{s(2^i\cdot x)}{2^i},$$

Full text

Measures of tree balance play an important role in various research areas, for example in phylogenetics. There they are for instance used to test whether an observed phylogenetic tree differs significantly from a tree generated by the Yule model of speciation. One of the most popular indices in this regard is the Colless index, which measures the degree of balance for rooted binary trees. While many statistical properties of the Colless index (e.g. asymptotic results for its mean and variance under different models of speciation) have already been discussed in different contexts, we focus on its extremal properties. While it is relatively straightforward to characterize trees with maximal Colless index, the analysis of the minimal value of the Colless index and the characterization of trees that achieve it, are much more involved. In this note, we therefore focus on the minimal value of the Colless index for any given number of leaves. We derive both a recursive formula for this minimal value, as well as an explicit expression, which shows a surprising connection between the Colless index and the so-called Blancmange curve, a fractal curve that is also known as the Takagi curve. Moreover, we characterize two classes of trees that have minimal Colless index, consisting of the set of so-called maximally balanced trees and a class of trees that we call greedy from the bottom trees. Furthermore, we derive an upper bound for the number of trees with minimal Colless index by relating these trees with trees with minimal Sackin index (another well-studied index of tree balance).

1 Introduction

Rooted trees are used in different research areas, ranging from computer science (for an overview see Chapter 2.3 in Knuth [1]) to evolutionary biology. In particular, they are often used to represent the evolutionary relationships among different species, where the leaves of the tree represent a set of extant species and the root represents their most recent common ancestor. Often it is of interest to study the structure and shape of a phylogenetic tree, in particular its degree of balance. It is for example well known that the Yule model in phylogenetics (a speciation model, which assumes a pure birth process and a constant rate of speciation among species) leads to rather imbalanced trees (e.g. Mooers and Heard [2], Steel [3]). Thus, information on the balance of a tree can be used to test whether an observed phylogenetic tree is consistent with this model (see for example Mooers and Heard [2], Blum and François [4], Bartoszek [5]). Note, however, that tree balance does not only play an important role in phylogenetics, but for example also in computer science (see for example Nievergelt and Reingold [6], Walker and Wood [7], Chang and Iyengar [8], Andersson [9], Pushpa and Vinod [10]).

In order to measure the degree of balance for a tree, several balance indices have been introduced, e.g. the Sackin index (cf. Sackin [11]), the Colless index (cf. Colless [12]) and, more recently, the total cophenetic index (cf. [13]). While these indices differ in their definitions and computations, they all assign a single number to a tree that captures its balance.

Here, we focus on the Colless index and explore its extremal properties (for statistical properties of the Colless index see for example Blum et al. [14]). It is relatively straightforward to analyze the maximal Colless index a tree on $n$ leaves can have. In fact, it can be shown that the so-called caterpillar tree on $n$ leaves (i.e. the unique rooted binary tree with only one “cherry”, i.e. one pair of leaves adjacent to the same node) is the unique tree with maximal Colless index (cf. Lemma 1 in Mir et al. [15]) and this maximal value turns out to be $\frac{(n-1)(n-2)}{2}$.

In contrast, the analysis of the minimal value of the Colless index is much more involved and is one of the main aims of this manuscript. On the one hand, for many years, no explicit expression for the minimal Colless index for a tree on $n$ leaves had been known. At the time when this manuscript was about to be submitted, an independent study for the first time presented such an expression (cf. Coronado and Rosselló [16]). However, their expression differs from the one we will derive in this manuscript. Ours is based on a fractal curve, namely the so-called Blancmange curve, and therefore shows the fractal structure and the symmetry of the minimum Colless index.111We will discuss more differences between the present manuscript and the one by Coronado and Rosselló [16] in the discussion section. On the other hand, this minimum may be achieved by several trees, which raises the question of how many such trees there are. This question has so far not been directly addressed anywhere in the literature. In the following, we will present an upper bound for the number of trees with minimal Colless index. Thus, our manuscript is the first study which both analyzes the minimal Colless index itself and gives an upper bound on the number of trees with minimal Colless index. Moreover, we characterize some classes of trees that achieve the minimum value. Last, in the discussion section we also show how the results of Coronado and Rosselló [16] can be used both to improve our bound as well as to give a recursive formula for the number of trees with minimal Colless value.

To be precise, we first prove a recursive formula for the minimal Colless index, before deriving an explicit expression. Surprisingly, the latter is strongly related to a fractal curve, the so-called Blancmange curve. This curve is also known as the Takagi curve (cf. Takagi [17]), which for example plays a role in number theory, combinatorics and analysis (cf. Allaart and Kawamura [18]).

While the recursive formula directly gives rise to a class of trees with minimal Colless index (the class of so-called maximally balanced trees (cf. Mir et al. [13]), we additionally introduce another class of trees with minimal Colless index, namely the class of greedy from the bottom trees. We also show that the leaf partitioning induced by the root of these two classes of trees are extremal – i.e. no tree with minimum Colless index can have a smaller difference between the number of leaves in the left and right subtrees than the maximally balanced tree or a larger difference between these numbers than the greedy from the bottom tree.

We then turn to the combinatorial task of determining the number of trees with minimal Colless index for a given number of leaves. By showing that all trees with minimal Colless index also have minimal Sackin index, we derive an upper bound for this number, since the number of trees with minimal Sackin index is known in the literature (cf. Theorem 8 in Fischer [19]). Using some knowledge on the leaf partitioning induced by the root of trees with minimum Colless index, we are able to improve this bound even further. However, note that the connection between the Sackin index and the Colless index is intriguing also in its own right – it shows that two of the most frequently used tree balance indices are actually closely related, even though their definition is very different and even though proving properties of the Colless index is mathematically more involved.

We end our manuscript by linking our results with the results of Coronado and Rosselló [16] and by pointing out some directions for future research.

2 Basic definitions and preliminary results

Before we can present our results, we need to introduce some definitions and notations. Throughout this manuscript a rooted tree is a tree $T=(V(T),E(T))$ with node set $V(T)$ and edge set $E(T)$, where one node is designated as the root (called $\rho$). We use $V_{L}(T)\subseteq V(T)$ (with $|V_{L}(T)|=n$) to denote the leaf set of $T$ (i.e. $V_{L}(T)=\{v\in V:deg(v)\leq 1\}$) and by $\mathring{V}(T)$ we denote the set of internal nodes, i.e. $\mathring{V}(T)=V(T)\setminus V_{L}(T)$. If $|V_{L}(T)|=1$, $T$ consists of only one node and no edge and for technical reasons this node is at the same time defined to be the root and the only leaf in the tree. Whenever there is no ambiguity we simply denote $E(T)$, $V(T)$, $\mathring{V}(T)$ and $V_{L}(T)$ as $E$, $V$, $\mathring{V}$ and $V_{L}$.

Now, for $n\geq 2$, a rooted binary tree is a rooted tree where the root has degree $2$ and all other internal nodes have degree $3$. For $n=1$, we consider the unique tree consisting of only one node as rooted and binary.

Furthermore, we implicitly assume that all edges in $T$ are directed away from the root and whenever there exists a path from $u$ to $v$ in $T$, we call $u$ an ancestor of $v$ and $v$ a descendant of $u$. In addition, the direct descendants of a node are called children of this node, and in a binary tree with $n\geq 2$ leaves each interior node has exactly two children. Two leaves $x$ and $y$ are said to form a cherry, denoted by $[x,y]$, if they have the same direct ancestor.

Given a node $v$ of $T$, we call the set $C_{T}(v)$ of its descendant leaves the cluster of $v$ and use $\kappa_{T}(v)$ to denote its cardinality. If $v$ itself is a leaf, we set $C_{T}(v)=\{v\}$, i.e. $\kappa_{T}(v)=1$. Moreover, we denote the subtree of $T$ rooted at $v$ by $T_{v}$.

Recall that a rooted binary tree $T$ can be decomposed into its two maximal pending subtrees $T_{a}$ and $T_{b}$ rooted at the direct descendants $a$ and $b$ of $\rho$, and we denote this by $T=(T_{a},T_{b})$. We use $n_{a}$ and $n_{b}$ to denote the number of leaves of $T_{a}$ and $T_{b}$, respectively, and without loss of generality assume throughout this manuscript that $n_{a}\geq n_{b}$.

The depth $\delta_{T}(v)$ of a node $v$ is the number of edges on the unique shortest path from the root to $v$. Additionally, the height of a tree is defined as $h(T)=\max\limits_{v\,\in\,V_{L}}\delta_{v}$, i.e. it is the maximum depth of all leaves.

Given a rooted binary tree $T$ and an internal node $v\in\mathring{V}$ with children $v_{1}$ and $v_{2}$, the balance value of $v$ is defined as $bal_{T}(v)=|\kappa_{T}(v_{1})-\kappa_{T}(v_{2})|$. We call an internal node $v$ balanced if $bal_{T}(v)\leq 1$. Based on this we call a tree on $n$ leaves maximally balanced if all its internal nodes are balanced. Recursively, a rooted binary tree is maximally balanced if its root is balanced and both its maximal pending subtrees are maximally balanced. Note that for all $n\in\mathbb{N}$, there exists a unique maximally balanced tree on $n$ leaves (cf. Mir et al. [13]), which we denote by $T^{mb}_{n}$ (cf. Figure 1).

Two other particular trees, which will be needed in the following, are the so-called caterpillar tree $T^{cat}_{n}$ and the so-called fully balanced tree $T^{fb}_{k}$, where the latter denotes the unique tree with $n=2^{k}$ leaves in which all leaves have depth precisely $k$ (cf. Figure 1). Note that we often call $T^{fb}_{k}$ the fully balanced tree of height $k$ and that we have $T^{fb}_{k}=(T_{a},T_{b})=(T^{fb}_{k-1},T^{fb}_{k-1})$, where both $T_{a}$ and $T_{b}$ are fully balanced trees of height $k-1$. Moreover, note that $T^{fb}_{k}=T^{mb}_{2^{k}}$, because in the special case of $n=2^{k}$, $T_{k}^{fb}$ is the unique tree with $bal_{T_{k}^{fb}}(v)=0$ for all $v\in\mathring{V}$.

The caterpillar tree on the other hand denotes the unique rooted binary tree with $n$ leaves that has only one cherry (cf. Figure 1).

We are now in a position to define the most important concept of this manuscript, namely the Colless index.

Definition 1** (Colless [12]).**

The Colless index of a rooted binary tree $T$ is defined as

[TABLE]

where $v_{1}$ and $v_{2}$ denote the children of $v$.

Note that $\mathcal{C}(T)\geq 0$ as it is defined as a sum of absolute values. As an example consider the three trees depicted in Figure 1. Here, we have: $\mathcal{C}(T^{cat}_{7})=15$, $\mathcal{C}(T^{mb}_{7})=2$ and $\mathcal{C}(T^{fb}_{3})=0$.

Note that the smaller the Colless index of a tree, the more balanced we consider it, i.e. whenever we have for two trees $T_{1}$ and $T_{2}$ on $n$ leaves that $\mathcal{C}(T_{1})<\mathcal{C}(T_{2})$, then $T_{1}$ is called more balanced than $T_{2}$. For example, in Figure 1, $T_{7}^{mb}$ is more balanced than $T_{7}^{cat}$.

Also note that the Colless index of a rooted binary tree can be calculated recursively by considering the standard decomposition of a tree.

Lemma 1**.**

Let $T=(T_{a},T_{b})$ be a rooted binary tree. Let $n_{a}$ and $n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$, respectively, where $n_{a}\geq n_{b}$. Then, we have

[TABLE]

Proof.

By Definition 1 we have

[TABLE]

This completes the proof. ∎

This lemma has a direct consequence, which will be useful when analyzing the minimal Colless index and trees with minimal Colless index throughout this manuscript.

Lemma 2**.**

Let $T=(T_{a},T_{b})$ be a rooted binary tree on $n$ leaves. Then, if $T$ has minimal Colless index for $n$, i.e. for all other trees $\widetilde{T}$ with $n$ leaves we have $\mathcal{C}(T)\leq\mathcal{C}(\widetilde{T})$, both $\mathcal{C}(T_{a})$ and $\mathcal{C}(T_{b})$ are minimal for $n_{a}$ and $n_{b}$, respectively.

Proof.

Assume $\mathcal{C}(T)$ is minimal. By Lemma 1 we have

[TABLE]

Now assume that $\mathcal{C}(T_{a})$ is not minimal, i.e. there is a tree $\widehat{T}$ on $n_{a}$ leaves such that $\mathcal{C}(\widehat{T})<\mathcal{C}(T_{a})$. Then we can construct a tree $\widetilde{T}$ on $n$ leaves such that $\widetilde{T}=(\widehat{T},T_{b})$, i.e. we replace $T_{a}$ in $T$ by $\widehat{T}$ to derive $\widetilde{T}$. Now, for $\widetilde{T}$ we have by Lemma 1

[TABLE]

This contradicts the minimality of $\mathcal{C}(T)$, which implies that the assumption was wrong. So $\mathcal{C}(T_{a})$ has to be minimal, and analogously, $\mathcal{C}(T_{b})$ has to be minimal, too. ∎

Note that Lemma 2 analogously holds for trees with maximal Colless index.

3 Results

The main aim of this section is to thoroughly analyze the minimal Colless index of rooted binary trees and trees that achieve it. We derive both a recursive as well as an explicit formula for the minimal Colless index and characterize two classes of trees with minimal Colless index. We end by providing an upper bound for the number of trees with minimal Colless index for any number of leaves.

3.1 Recursive formula for the minimal Colless index

In the following we establish a recursive formula for the minimal Colless index. Therefore, let $c_{n}$ denote the minimal Colless index for rooted binary trees with $n$ leaves. Then, we have the following statement:

Theorem 1**.**

Let $c_{n}$ be the minimal Colless index for a rooted binary tree with $n$ leaves. Then, $c_{1}=c_{2}=0$, and for all $n\in\mathbb{N}_{\geq 1}$ we have

[TABLE]

Note that the sequence of integers obtained from Theorem 1 corresponds to sequence A296062 in the On-Line Encyclopedia of Integer Sequences (Sloane [20]), linking the minimal Colless index to the base-2 logarithm of the number of isomorphic maximally balanced trees with $n$ leaves (note that up to isomorphism, the maximally balanced tree is unique (Mir et al. [13]), but there may be several isomorphic maximally balanced trees with $n$ leaves. For example, for $n=3$, there are 2 isomorphic maximally balanced trees).

Proof of Theorem 1.

Let $n\in\mathbb{N}_{\geq 1}$. For $n=1$ there is just one tree, which consists of only one leaf and has Colless index 0. For $n=2$ again there is just one tree, which consists of one cherry and has Colless index 0 as well. Thus, $c_{1}=c_{2}=0$ as claimed.

We now show

[TABLE]

Ranging over all $n_{a},n_{b}$ with $n_{a}+n_{b}=n$, by Lemma 1 and Lemma 2 we have

[TABLE]

Thus, it follows immediately that $c_{2n}\leq 2c_{n}$. Analogously by Lemma 1 and Lemma 2, we have

[TABLE]

Therefore, we derive $c_{2n+1}\leq c_{n+1}+c_{n}+1$. Thus, $(i)$ and $(ii)$ hold.

We now show by induction on $n$ that

[TABLE]

For $n=1$ we have $c_{2}\geq 2c_{1}$, since $0\geq 2\cdot 0$, and $c_{3}\geq c_{2}+c_{1}+1$, since $c_{3}\geq 1$ (for 3 leaves there is just one tree which has Colless index 1). This completes the base case of the induction.

Now we assume that (3) and (4) hold for all natural numbers up to $n$ and show that they also hold for $n+1$.

We start by proving (3). The proof of (4) is given in the Appendix.

By Lemma 1 and Lemma 2 we have that

[TABLE]

In the following we consider two cases: $n$ even and $n$ odd. First, let $n$ be even. Thus by the inductive hypothesis, we can rewrite (5) as

[TABLE]

Again by Lemma 1 and Lemma 2, we have that

[TABLE]

and thus we have for example $c_{n+1}\leq c_{n}+c_{1}+n-1$. Then by using (7), (6) becomes the following

[TABLE]

This completes the proof of Equation (3) for $n$ even.

Now, let $n$ be odd. Similarly, we can rewrite (5) by using the inductive hypothesis as

[TABLE]

Again by Lemma 1 and Lemma 2, we have that

[TABLE]

and thus we have for example $c_{n+1}\leq c_{n}+c_{1}+n-1$. Then by using (10), (9) becomes the following

[TABLE]

This completes the proof of Equation (3) for $n$ odd. So in both cases we obtain $c_{2n+2}\geq 2c_{n+1}$ and thus (3) holds for all $n$.

Similarly by induction, we can show that (4) holds for all $n$. The detailed proof is given in the Appendix.

Together with $(i)$ and $(ii)$ this completes the proof. ∎

3.2 Explicit expression for the minimal Colless index

Theorem 1 implies that we can recursively calculate the minimal Colless index for all $n$. However, we can also directly calculate it by applying the following theorem. Note that the recursions stated in Theorem 1 are needed to prove Theorem 2.

Theorem 2**.**

Let $c_{n}$ denote the minimal Colless index for a rooted binary tree with $n$ leaves. Let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. Then,

[TABLE]

where $s(x)=\min\limits_{z\in\mathbb{Z}}|x-z|$, i.e. $s(x)$ is the distance from $x$ to the nearest integer.

Remark 1**.**

Note that the expression above shows a surprising connection to the so-called Blancmange curve, a fractal curve. This curve is also known as the Takagi curve (cf. Takagi [17]) and plays a role in different areas such as combinatorics, number theory and analysis (cf. Allaart and Kawamura [18]) and is defined as $T:[0,1]\rightarrow\mathbb{R}$ with

[TABLE]

where $s(x)$ is defined as in Theorem 2 (adapted from Allaart and Kawamura [18]). In contrast to $T:[0,1]\rightarrow\mathbb{R}$, for $c_{n}$ we have $c_{n}:\mathbb{N}\rightarrow\mathbb{N}$, the sum of $c_{n}$ runs from [math] to $k_{n}-2$ (and not up to infinity), and the index is shifted. So the two functions are not identical, but it is intriguing that this explicit formula for $c_{n}$ links the Colless index from phylogenetics to other research areas such as number theory.

Figure 2 shows the minimal Colless index $c_{n}$ for $n=1,\ldots,128$ and illustrates the fractal property of it.

Before we give the proof of Theorem 2, we shortly have a look at some interesting properties of $s(x)=\min\limits_{z\in\mathbb{Z}}|x-z|$ and for example show that subadditivity holds for $s(x)$. These properties will be of relevance various times in the remainder of this manuscript.

Lemma 3**.**

Let $s(x)=\min\limits_{z\in\mathbb{Z}}|x-z|$, i.e. $s(x)$ is the distance from $x$ to the nearest integer. Let $a\in\mathbb{R}$. Then, we have:

For $b\in\mathbb{R}$: $s(a+b)\leq s(a)+s(b).$ 2. 2.

For $b\in\mathbb{Z}$: $s(a)=s(a+b)$.

Proof.

Let $a,b\in\mathbb{R}$. Then,

[TABLE] 2. 2.

Now let $a\in\mathbb{R}$ and $b\in\mathbb{Z}$. By 1. we immediately have $s(a+b)\leq s(a)+s(b)$. The fact that $b\in\mathbb{Z}$ leads to $s(b)=0$, which results in $s(a+b)\leq s(a)$. Additionally, again by 1., we have for $b\in\mathbb{Z}$ that $s(a)=s(a+b-b)\leq s(a+b)+s(-b)=s(a+b)$. Combining both arguments results in $s(a)=s(a+b)$ for all $b\in\mathbb{Z}$.

∎

The proof of Theorem 2 requires one more lemma, the proof of which is given in the Appendix.

Lemma 4**.**

Let $n\in(2^{k_{n}-1},2^{k_{n}})$ be odd, where $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$, and let $n-1>2^{k_{n}-1}$. Moreover, let $s(x)=\min\limits_{z\in\mathbb{Z}}|x-z|$ and let

[TABLE]

Then for $0\leq i\leq k_{n}-3$,

[TABLE]

We are now in the position to prove Theorem 2.

Proof of Theorem 2.

Let $c_{n}$ denote the minimal Colless index for a rooted binary tree $T_{n}$ with $n$ leaves and let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. Moreover, let

[TABLE]

Then, the statement in Theorem 2 becomes $c_{n}=\sum\limits_{i=0}^{k_{n}-2}f_{i}(n)$. We prove this statement by induction on $n$. If $n=1$, there is only one rooted binary tree $T_{1}$ consisting of only one leaf. Thus, $\mathcal{C}(T_{1})=0$, which is minimal (since there is only one tree for $n=1$) and thus $\mathcal{C}(T_{1})=c_{1}$. On the other hand, we have $k_{1}=\lceil\log_{2}(1)\rceil=0$ and thus $\sum\limits_{i=0}^{k_{n}-2}f_{i}(n)=\sum\limits_{i=0}^{-2}f_{i}(n)$ is the empty sum, which by convention is [math]. This completes the base case of the induction.

Now, suppose that the claim holds for all rooted binary trees with up to $n-1$ leaves and consider a rooted binary tree $T_{n}$ with $n$ leaves. We now distinguish two cases: $n$ even and $n$ odd.

If $n$ is even, we have:

[TABLE]

This completes the proof for $n$ even.

If $n$ is odd, we already have by Theorem 1 that $c_{n}=c_{\frac{n+1}{2}}+c_{\frac{n-1}{2}}+1$. Moreover, as $k_{n}=\lceil\log_{2}(n)\rceil$ and by the fact that $n$ is odd, we have $n\in(2^{k_{n}-1},2^{k_{n}})$, and therefore $n+1\in(2^{k_{n}-1},2^{k_{n}}]$, which gives us $\lceil\log_{2}(n+1)\rceil=k_{n}$. This leads to $k_{\frac{n+1}{2}}=\lceil\log_{2}(\frac{n+1}{2})\rceil=\lceil\log_{2}(n+1)-\log_{2}(2)\rceil=k_{n}-1$.

Again by the fact that $n$ is odd, we have $n-1\in[2^{k_{n}-1},2^{k_{n}})$.

So first, if $n-1=2^{k_{n}-1}$ we have $\lceil\log_{2}(n-1)\rceil=k_{n}-1$. Moreover, we have $k_{\frac{n-1}{2}}=\lceil\log_{2}(\frac{n-1}{2})\rceil=\lceil\log_{2}(n-1)-\log_{2}(2)\rceil=k_{n}-1-1=k_{n}-2$. As $n-1=2^{k_{n}-1}$, we have $n=2^{k_{n}-1}+1$. This leads to

[TABLE]

Note that $2^{i-k_{n}+2}\leq 2^{(k_{n}-3)-k_{n}+2}=\frac{1}{2}$ results in $s(2^{i-k_{n}+2})=2^{i-k_{n}+2}-0=2^{i-k_{n}+2}$. Therefore,

[TABLE]

This gives us

[TABLE]

We now show that this indeed equals

[TABLE]

Similarly, we have $2^{i-k_{n}+1}\leq 2^{(k_{n}-2)-k_{n}+1}=\frac{1}{2}$ and thus $s(2^{i-k_{n}+1})=2^{i-k_{n}+1}$, which leads to

[TABLE]

This completes the proof for $n-1=2^{k_{n}-1}$.

The last case we need to consider is the case that $n$ is odd and $n-1\in(2^{k_{n}-1},2^{k_{n}})$. In this case, we have $\lceil\log_{2}(n-1)\rceil=k_{n}$ (since $n-1>2^{k_{n}-1}$). Furthermore, $k_{\frac{n-1}{2}}=\lceil\log_{2}(\frac{n-1}{2})\rceil=\lceil\log_{2}(n-1)-\log_{2}(2)\rceil=k_{n}-1$.

Therefore,

[TABLE]

So the claim holds for all cases, which completes the proof. ∎

The following proposition states some properties of the minimal Colless index $c_{n}$. We use the explicit expression for the minimal Colless index stated in Theorem 2 to show the third part of this proposition.

Proposition 1**.**

Let $n\in\mathbb{N}_{\geq 1}$ and let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. Then, we have for the minimal Colless index $c_{n}$:

If $n=2^{k_{n}}+1$, then $c_{n}=k_{n}$. 2. 2.

If $n=2^{k_{n}}-1$, then $c_{n}=k_{n}-1$. 3. 3.

For $n\in(2^{k_{n}-1},2^{k_{n}})$ and $j\in\{1,\dots,2^{k_{n}-1}-1\}$ we have $c_{2^{k_{n}-1}+j}=c_{2^{k_{n}}-j}.$

The proof of Proposition 1 is given in the Appendix. Note, however, that these properties are also reflected in Figure 2.

3.3 Classes of trees with minimal Colless index

Now that we have analyzed the minimal Colless index $c_{n}$, we turn our attention to trees that achieve it. Before considering the class of maximally balanced trees and introducing the class of greedy from the bottom trees for arbitrary $n$, we start with analyzing the special case where $n$ is a power of two. In particular, we consider the fully balanced tree $T_{k}^{fb}$ of height $k$ and show that its name is indeed justified in the sense that it is the unique tree with minimal Colless index for $n=2^{k}$ leaves. This observation has been stated in the literature several times without formal proof (e.g. Heard [21], Rogers [22], Mir et al. [13, 15]), which is why we provide a formal proof in the following.

Theorem 3**.**

*Let $T$ be a rooted binary tree with $n=2^{k}$ leaves. Then, we have:

$\mathcal{C}(T)=0$ if and only if $T=T_{k}^{fb}$.*

Proof.

First, suppose that $T=T^{fb}_{k}$. Then by definition of $T^{fb}_{k}$, we have $bal_{T_{k}^{fb}}(v)=0$ for all $v\in\mathring{V}$. Thus, $\mathcal{C}(T_{k}^{fb})=\sum\limits_{v\in\mathring{V}}bal_{T_{k}^{fb}}(v)=0$.

Now, let $T$ be a rooted binary tree with $n=2^{k}$ leaves and suppose that $\mathcal{C}(T)=0$. We prove by induction on $k$ that $\mathcal{C}(T)=0$ implies that $T$ equals $T_{k}^{fb}$. For $k=0$, we have $2^{0}=1=n$, and there is only one rooted binary tree $T$, which is by definition a fully balanced tree. We now assume that the statement holds up to $k-1$ and consider a rooted binary tree $T=(T_{a},T_{b})$ with $n=2^{k}$ leaves. Note that without loss of generality, $n\geq 2$ (else we consider the base case of the induction again).

By Lemma 1 we have

[TABLE]

which implies $\mathcal{C}(T_{a})=\mathcal{C}(T_{b})=(n_{a}-n_{b})=0$. In particular, $n_{a}=n_{b}=2^{k-1}$. Therefore, by the inductive hypothesis, we have that $T_{a}$ and $T_{b}$ are both fully balanced trees of height $k-1$. This concludes that $T$ is the fully balanced tree of height $k$. This completes the proof. ∎

3.3.1 Maximally balanced trees

While we have already seen that for $n=2^{k}$ there is exactly one tree with minimal Colless index, for arbitrary $n$ there might be several ones.

In Theorem 1 we have already seen how to calculate the minimal Colless index for a rooted binary tree with $n$ leaves recursively. This theorem directly yields a construction principle for trees with minimal Colless index. Given a number $n$ of leaves, we construct a tree $T=(T_{a},T_{b})$ with $\mathcal{C}(T)=c_{n}$, where

$n_{a}=n_{b}=\frac{n}{2}$, if $n$ is even;

• 2.

$n_{a}=\frac{n+1}{2}$ and $n_{b}=\frac{n-1}{2}$, if $n$ is odd,

and where $T_{a}$ and $T_{b}$ are constructed recursively by the same principle. In particular, this implies that for every internal node $v$ of $T$ we have: $bal_{T}(v)\leq 1$, which in turn implies that $T$ is the maximally balanced tree.

Note that this approach can be seen as a “greedy from the top strategy”222In Section 3.3.2 we will additionally consider a “greedy from the bottom strategy” for building trees with minimal Colless index. of bipartitioning the leaf set of each subtree (starting at the root and going towards the leaves) into two sets such that the difference of their cardinalities is minimized.

By using Theorem 1, we will now formally show that trees constructed according to this principle, i.e. maximally balanced trees, indeed have minimal Colless index.

Theorem 4**.**

Let $T_{n}^{mb}$ be the maximally balanced tree with $n$ leaves. Then, $T_{n}^{mb}$ has minimal Colless index, that is $\mathcal{C}(T_{n}^{mb})=c_{n}$.

Proof.

The proof is by induction on $n$. If $n=1$, $T_{1}^{mb}$ consists only of a leaf and thus $\mathcal{C}(T_{1}^{mb})=0=c_{1}$, which completes the base case of the induction.

Now we assume that for all maximally balanced trees with up to $n$ leaves the claim holds and consider the maximally balanced tree $T_{n+1}^{mb}$ with $n+1$ leaves. By definition all internal nodes of $T_{n+1}^{mb}$ are balanced. Thus, it remains to show that $\mathcal{C}(T_{n+1}^{mb})=c_{n+1}$.

If $n+1$ is even, then $n_{a}=n_{b}=\frac{n+1}{2}$ and we have

[TABLE]

If $n+1$ is odd, then $n_{a}=\frac{n+2}{2}$ and $n_{b}=\frac{n}{2}$. Thus,

[TABLE]

Therefore, in both cases $T_{n+1}^{mb}$ has minimal Colless index, which completes the proof. ∎

Note, however, that the maximally balanced tree on $n$ leaves is not necessarily the only tree with minimal Colless index. For example, both trees depicted in Figure 3 have minimal Colless index; one of them is the maximally balanced tree on $6$ leaves, the other one is a so-called greedy from the bottom tree, which we will introduce in the following section.

3.3.2 Greedy from the bottom trees

We now introduce another class of trees with minimal Colless index, which we call greedy from the bottom trees, or GFB trees for short. These trees can be constructed according to the following algorithm:

Definition 2** (GFB tree).**

Let $T$ be a rooted binary tree on $n$ leaves that results from Algorithm 1. Then, $T$ is called greedy from the bottom tree or GFB tree for short and is denoted by $T^{gfb}_{n}$.

Note that Algorithm 1 greedily clusters trees of minimal size starting with single nodes and proceeding until only one tree is left. Thus, in principle it goes from the leaves towards the root (in contrast to the greedy strategy presented in Section 3.3.1, which goes from the root towards the leaves), which is why we call the resulting trees GFB trees.

Remark 2**.**

Note that for $n=1$ Algorithm 1 returns one single node. Moreover, note that the GFB tree is unique for all positive integers $n$ (as it results from Algorithm 1, which is deterministic).

The following lemma shows that both maximal pending subtrees of a GFB tree are also GFB trees, which is an intuitive property and follows directly from Algorithm 1.

Lemma 5**.**

Let $T$ be a GFB tree with $n\geq 2$ leaves and standard decomposition $T=(T_{a},T_{b})$. Then, $T_{a}$ and $T_{b}$ are also GFB trees.

Proof.

Let $T=(T_{a},T_{b})$ be a GFB tree and let $n_{a}$ and $n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$, respectively. This means that Algorithm 1 induces a bipartition of the $n$ leaves into two disjoint sets of sizes $n_{a}$ and $n_{b}$, respectively. Note that up to the last iteration of the while-loop in Algorithm 1 these two sets are independent of each other.

Now, applying Algorithm 1 to $n_{a}$ and $n_{b}$ leaves, respectively, results in two unique GFB trees $T^{gfb}_{n_{a}}$ and $T^{gfb}_{n_{b}}$ with $n_{a}$ and $n_{b}$ leaves. As the leaf sets of $T_{a}$ and $T_{b}$ of sizes $n_{a}$ and $n_{b}$, respectively, are independent and do not influence each other, this implies $T_{a}=T^{gfb}_{n_{a}}$ and $T_{b}=T^{gfb}_{n_{b}}$. This completes the proof. ∎

In the following we will see that GFB trees always have minimal Colless index, even though in general the GFB tree $T_{n}^{gfb}$ is different from the maximally balanced tree $T_{n}^{mb}$ on $n$ leaves (cf. Figure 3). We can, however, characterize GFB trees in terms of the sizes of their maximal pending subtrees. This information will be very useful in subsequent analyses, in particular when showing that GFB trees have minimal Colless index (cf. Theorem 5).

Proposition 2**.**

Let $T_{n}^{gfb}$ be a GFB tree with $n\geq 2$ leaves and standard decomposition $T_{n}^{gfb}=(T_{a},T_{b})$. Let $n_{a}$ and $n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$, respectively, such that without loss of generality, $n_{a}\geq n_{b}$. Let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$, i.e. $n\in(2^{k_{n}-1},2^{k_{n}}]$. Then, we have:

If $n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$, we have $n_{a}=n-2^{k_{n}-2}$ and $n_{b}=2^{k_{n}-2}$. In particular, $T_{b}$ is the fully balanced tree of height $k_{n}-2$ and we have $k_{n_{a}}\coloneqq\lceil\log_{2}(n_{a})\rceil=k_{n}-1$. 2. 2.

If $n=3\cdot 2^{k_{n}-2}$, we have $n_{a}=2^{k_{n}-1}$ and $n_{b}=2^{k_{n}-2}$. In particular, $T_{a}$ is the fully balanced tree of height $k_{n}-1$ and $T_{b}$ is the fully balanced tree of height $k_{n}-2$. 3. 3.

If $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}}]$, we have $n_{a}=2^{k_{n}-1}$ and $n_{b}=n-2^{k_{n}-1}$. In particular, $T_{a}$ is the fully balanced tree of height $k_{n}-1$ and we have $k_{n_{b}}\coloneqq\lceil\log_{2}(n_{b})\rceil=k_{n}-1$.

The proof of Proposition 2 is given in the appendix. However, it has an interesting consequence:

Corollary 1**.**

Let $n\in\mathbb{N}_{\geq 2}$ and $n\neq 3\cdot 2^{k_{n}-2}$, where $k_{n}=\lceil\log_{2}(n)\rceil$. Then, $T_{n-1}^{gfb},T_{n}^{gfb}$ and $T_{n+1}^{gfb}$ have a common maximal pending subtree, which is fully balanced:

If $n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$, this common subtree is a fully balanced tree of height $k_{n}-2$. 2. 2.

If $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}}]$, this common subtree is a fully balanced tree of height $k_{n}-1$.

Proof.

The claimed statements are a direct consequence of Proposition 2:

If $n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$, we distinguish between two cases:

• (a)

If $n-1>2^{k_{n}-1}$, i.e. if $n-1$, $n$ and $n+1$ are all in $(2^{k_{n}-1},3\cdot 2^{k_{n}-2}]$, by Proposition 2, Parts 1 and 2, $T_{n-1}^{gfb},T_{n}^{gfb}$ and $T_{n+1}^{gfb}$ all contain a fully balanced tree of height $k_{n}-2$ as a maximal pending subtree.

• (b)

If $n-1=2^{k_{n}-1}$, by Proposition 2, Part 3, $T_{n-1}^{gfb}$ contains a fully balanced tree of height $(k_{n}-1)-1=k_{n}-2$ as a maximal pending subtree (since $k_{n-1}=\lceil\log_{2}(n-1)\rceil=\lceil\log_{2}(2^{k_{n}-1})\rceil=k_{n}-1$). Moreover, by Proposition 2, Part 1, $T_{n}^{gfb}$ and $T_{n+1}^{gfb}$ also contain a fully balanced tree of height $k_{n}-2$ as a maximal pending subtree. 2. 2.

If $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}}]$, we again distinguish between two cases:

• (a)

If $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}})$, then we have $n-1$ and $n+1\in[3\cdot 2^{k_{n}-2},2^{k_{n}}]$ and by Proposition 2, Parts 2 and 3, $T_{n-1}^{gfb},T_{n}^{gfb}$ and $T_{n+1}^{gfb}$ all contain a fully balanced tree of height $k_{n}-1$ as a maximal pending subtree.

• (b)

If $n=2^{k_{n}}$, then $n-1=2^{k_{n}}-1$ and $n+1=2^{k_{n}}+1$. By Proposition 2, Part 2, $T_{n}^{gfb}$ and $T_{n-1}^{gfb}$ contain a fully balanced tree of height $k_{n}-1$ as a maximal pending subtree. Now, as $k_{n+1}=\lceil\log_{2}(n+1)\rceil=\lceil\log_{2}(2^{k_{n}}+1)\rceil=k_{n}+1$, by Proposition 2, Part 1, $T_{n+1}^{gfb}$ contains a fully balanced tree of height $(k_{n}+1)-2=k_{n}-1$ as a maximal pending subtree. This completes the proof.

∎

We will now show that GFB trees always have minimal Colless index, i.e. we will prove the following theorem:

Theorem 5**.**

Let $T_{n}^{gfb}$ be a GFB tree with $n$ leaves. Then, $T_{n}^{gfb}$ has minimal Colless index, i.e. $\mathcal{C}(T_{n}^{gfb})=c_{n}$.

In order to prove Theorem 5, we require the following technical lemma, the proof of which can be found in the Appendix.

Lemma 6**.**

Let $s(x)=\min\limits_{z\in\mathbb{Z}}|x-z|$, i.e. $s(x)$ is the distance from $x$ to the nearest integer. Let $n\in\mathbb{N}$ and let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. Then,

For $n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2}]$, we have

[TABLE] 2. 2.

For $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}}]$, we have

[TABLE]

We can now prove Theorem 5.

Proof of Theorem 5.

Let $T_{n}^{gfb}$ be a GFB tree with $n$ leaves. Let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. In order to show that $T_{n}^{gfb}$ has minimal Colless index, we show that $\mathcal{C}(T_{n}^{gfb})=c_{n}=\sum\limits_{i=0}^{k_{n}-2}\frac{s(2^{i-k_{n}+1}\cdot n)}{2^{i-k_{n}+1}}$. We do this by induction on $n$.

For $n=1$ we have $\mathcal{C}(T_{1}^{gfb})=0=c_{1}$, which gives the base case of the induction.

Now, we assume that the statement given in Theorem 5 holds for all GFB trees with up to $n-1$ leaves and we show that it also holds for the GFB tree with $n$ leaves. We now distinguish between 3 cases:

$n=3\cdot 2^{k_{n}-2}$:

By Proposition 2 we know that $T_{3\cdot 2^{k_{n}-2}}^{gfb}$ has the following standard decomposition: $T_{3\cdot 2^{k_{n}-2}}^{gfb}=(T_{k_{n}-1}^{fb},T_{k_{n}-2}^{fb})$. In particular, both maximal pending subtrees of $T_{3\cdot 2^{k_{n}-2}}^{gfb}$ are fully balanced trees. By Theorem 3, we have $\mathcal{C}(T_{k_{n}-1}^{fb})=0$ and $\mathcal{C}(T_{k_{n}-2}^{fb})=0$. Thus, using Lemma 1, we have

[TABLE]

On the other hand,

[TABLE]

Thus, $T_{3\cdot 2^{k_{n}-2}}^{gfb}$ has minimal Colless index.

• 2.

$n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$:

By Proposition 2 we know that $T_{n}^{gfb}$ has the following standard decomposition: $T_{n}^{gfb}=(T_{n-2^{k_{n}-2}},T_{k_{n}-2}^{fb})$. In particular, $T_{k_{n}-2}^{fb}$ is the fully balanced tree of height $k_{n}-2$, and thus by Theorem 3, $\mathcal{C}(T_{k_{n}-2}^{fb})=0$. Moreover, for $T_{n-2^{k_{n}-2}}$ we have: $\lceil\log_{2}(n-2^{k_{n}-2})\rceil=k_{n}-1$. Thus, using Lemma 1 and Lemma 2, we have

[TABLE]

Thus, $T_{n}^{gfb}$ has minimal Colless index.

• 3.

$n\in(3\cdot 2^{k_{n}-2},2^{k_{n}}]$:

This case follows analogously to the previous case, but instead of using Part 1 of Lemma 6, we use Part 2. For completeness the proof can be found in the Appendix.

Thus, in all cases, $\mathcal{C}(T_{n}^{gfb})=c_{n}$, which completes the proof. ∎

Note that for $n=2^{k_{n}}$ Theorem 5 together with Theorem 3 implies that $T^{gfb}_{2^{k_{n}}}=T^{fb}_{k_{n}}$.

3.3.3 Further characterizing and counting trees with minimal Colless index

So far, we have seen that there are two classes of trees with minimal Colless index, namely maximally balanced trees and GFB trees. However, there are trees that are neither maximally balanced nor GFB, but still have minimal Colless index (e.g. tree $T_{1}$ depicted in Figure 4). In the following, we will thus try to further characterize and count trees with minimal Colless index.

In particular, we will show that the leaf partitioning of $n$ leaves into $n_{a}$ and $n_{b}$ as induced by Algorithm 1 is the most extreme one a tree with minimal Colless index can have. This means that given a tree $T$ with minimal Colless index, the difference in the number of leaves of its two maximal pending subtrees $n_{a}-n_{b}$ cannot be larger than it is in a GFB tree. To be precise, we have the following theorem:

Theorem 6**.**

Let $T^{gfb}_{n}=(T_{a},T_{b})$ be a GFB tree on $n$ leaves with $n\in(2^{k_{n}-1},2^{k_{n}})$ (where $k_{n}=\lceil\log_{2}(n)\rceil$), i.e.

[TABLE]

where we have $n_{a}-n_{b}=n-2^{k_{n}-1}$ in the first case, $n_{a}-n_{b}=2^{k_{n}-2}$ in the second case and $n_{a}-n_{b}=2^{k_{n}}-n$ in the last case. Now, suppose that $\widehat{T}=(\widehat{T}_{a},\widehat{T}_{b})$ is a tree with a more extreme leaf partitioning, i.e. $\widehat{n}_{a}-\widehat{n}_{b}>n_{a}-n_{b}$ or, to be more precise,

[TABLE]

Then, we have

[TABLE]

i.e. $\widehat{T}$ does not have minimal Colless index (where the first equality follows from Theorem 5).

The proof of this theorem requires the following lemma, which provides an upper bound on the maximal minimal Colless index for any given $n\in(2^{k_{n}-1},2^{k_{n}})$. This upper bound is also depicted in Figure 2.

Lemma 7**.**

Let $n\in(2^{k_{n}-1},2^{k_{n}})$ with $k_{n}=\lceil\log_{2}(n)\rceil$. Let $\bar{c}_{n}$ denote the maximal minimal Colless index for $n\in(2^{k_{n}-1},2^{k_{n}})$, i.e.

[TABLE]

Then, we have

[TABLE]

Proof.

By Theorem 2, we have for the minimal Colless index

[TABLE]

This means that $c_{n}$ is bounded from above by $2^{k_{n}-1}$. In particular, $\bar{c}_{n}=\max\limits_{n\in(2^{k_{n}-1},2^{k_{n}})}c_{n}<2^{k_{n}-1}$, which completes the proof. ∎

We are now in a position to prove Theorem 6, which is divided into three subcases. For $n=3\cdot 2^{k_{n}-2}$ the proof is straightforward, while the cases $n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$ and $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}})$ are more technical and require Lemma 3 and Lemma 7.

Proof of Theorem 6.

Let $n\in(2^{k_{n}-1},2^{k_{n}})$ with $k_{n}=\lceil\log_{2}(n)\rceil$. Let $T^{gfb}_{n}=(T_{a},T_{b})$ be the GFB tree on $n$ leaves. We now distinguish between three cases (only case 1. is given here; the other two cases are shown in the Appendix):

$n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$:

From Proposition 2, we have for $T^{gfb}_{n}=(T_{a},T_{b})$ that $n_{a}=n-2^{k_{n}-2}$ and $n_{b}=2^{k_{n}-2}$. In particular, $T_{b}$ is the fully balanced tree of height $k_{n}-2$ and we have $\mathcal{C}(T_{b})=c_{n_{b}}=0$ by Theorem 3. For $T_{a}$ we have $k_{n_{a}}=\lceil\log_{2}(n_{a})\rceil=k_{n}-1$. Now, consider $\mathcal{C}(T^{gfb}_{n})$. By Lemmas 1 and 2 and by Theorem 5 we have

[TABLE]

Now, suppose that $\widehat{T}=(\widehat{T}_{a},\widehat{T}_{b})$ with $\widehat{n}_{a}=n_{a}+j$ and $\widehat{n}_{b}=n_{b}-j$ (and $j\in\{1,\ldots,2^{k_{n}-2}-1\}$) is also a tree with minimal Colless index, i.e. $\mathcal{C}(\widehat{T})=\mathcal{C}(T^{gfb}_{n})=c_{n}$. Consider $\mathcal{C}(\widehat{T})$. Again, by Lemmas 1 and 2, we have

[TABLE]

Now, comparing $\mathcal{C}(T^{gfb}_{n})$ (see Equation (14)) and $\mathcal{C}(\widehat{T})$ (see Equation (16)) it immediately follows that we have $\mathcal{C}(T^{gfb}_{n})<\mathcal{C}(\widehat{T})$ if $c_{n_{a}}<2j$, which would contradict the minimality of $\mathcal{C}(\widehat{T})$. Thus, we have $c_{n_{a}}\geq 2j$. We now claim that $c_{n_{a}}\geq 2j$ implies $k_{\widehat{n}_{b}}=\lceil\log_{2}(\widehat{n}_{b})\rceil=k_{n_{b}}=k_{n}-2$. To see this, consider the following:

• (a)

As $\widehat{n}_{b}<n_{b}$, we have $k_{\widehat{n}_{b}}\leq k_{n_{b}}=k_{n}-2$.

• (b)

Suppose $k_{\widehat{n}_{b}}\leq k_{n}-3$. As $\widehat{n}_{b}=n_{b}-j=2^{k_{n}-2}-j$, this implies $j\geq 2^{k_{n}-3}$. In particular, $2j\geq 2^{k_{n}-2}$.

Now, consider $c_{n_{a}}$. As $k_{n_{a}}=k_{n}-1$ it follows from Lemma 7 that

[TABLE]

However, as $2j\geq 2^{k_{n}-2}$, this implies $c_{n_{a}}<2j$, which contradicts the assumption that $c_{n_{a}}\geq 2j$. Thus, $k_{\widehat{n}_{b}}>k_{n}-3$.

Thus, in total we have $k_{\widehat{n_{b}}}=k_{n}-2$. This, however, implies that

[TABLE]

In particular, $j\in(0\cdot 2^{k_{n}-2},\frac{1}{2}\cdot 2^{k_{n}-2})$ and $2^{2-k_{n}}\cdot j\in(0,\frac{1}{2})$. This, in turn, implies

[TABLE]

a fact that will be used later on.

Moreover, we have for $\widehat{T}_{a}$: $k_{\widehat{n}_{a}}=\lceil\log_{2}(\widehat{n}_{a})\rceil\in\{k_{n}-1,k_{n}\}$:

• (a)

As $\widehat{n}_{a}>n_{a}$ and $k_{n_{a}}=k_{n}-1$, it follows that $k_{\widehat{n}_{a}}\geq k_{n}-1$.

• (b)

However, as $\widehat{n}_{a}<n$, it also follows that $k_{\widehat{n}_{a}}\leq k_{n}$.

In total $k_{\widehat{n}_{a}}\in\{k_{n}-1,k_{n}\}$.

Thus, for $\widehat{T}$ we now distinguish between two cases:

1. (a)

$k_{\widehat{n}_{a}}=k_{n}-1$ and $k_{\widehat{n}_{b}}=k_{n}-2$:

Continuing from Equation (16) and using Theorem 2, we have

[TABLE]

Now, as by assumption both $T^{gfb}_{n}$ and $\widehat{T}$ are trees with minimal Colless index, we have (using Equations (15) and (18))

[TABLE]

This, however, is a contradiction. Thus, $\mathcal{C}(\widehat{T})$ is not minimal, which completes the proof for this subcase. 2. (b)

$k_{\widehat{n}_{a}}=k_{n}$ and $k_{\widehat{n}_{b}}=k_{n}-2$:

Again, continuing from Equation (16) and using Theorem 2, we have

[TABLE]

Again, as by assumption both $T^{gfb}_{n}$ and $\widehat{T}$ are trees with minimal Colless index, we have (using Equations (15) and (19))

[TABLE]

This, however, is a contradiction. Thus, $\mathcal{C}(\widehat{T})$ is not minimal, which completes the proof for this subcase.

For $n=3\cdot 2^{k_{n}-2}$ the proof is straightforward and for $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}})$ the proof is similar to the case shown above. Thus, both cases are given in the Appendix.

In all cases, we have $\mathcal{C}(\widehat{T})>\mathcal{C}(T^{gfb}_{n})$, which completes the proof. ∎

To summarize, we have seen that both maximally balanced trees and GFB trees have minimal Colless index (cf. Theorems 4 and 5). Moreover, for a tree $T=(T_{a},T_{b})$ with minimal Colless index we always have:

Corollary 2**.**

Let $T=(T_{a},T_{b})$ be a tree on $n$ leaves with minimal Colless index, i.e. $\mathcal{C}(T)=c_{n}$. Let $n_{a},n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$, respectively, where $n_{a}\geq n_{b}$. Then

[TABLE]

where $n_{a}^{gfb}$ and $n_{b}^{gfb}$ denote the number of leaves of $T^{gfb}_{a}$ and $T^{gfb}_{b}$ in $T_{n}^{gfb}=(T_{a}^{gfb},T_{b}^{gfb})$ and $n_{a}^{mb}$ and $n_{b}^{mb}$ denote the number of leaves of $T_{a}^{mb}$ and $T_{b}^{mb}$ in $T_{n}^{mb}=(T_{a}^{mb},T_{b}^{mb})$.

Proof.

Let $T=(T_{a},T_{b})$ be a tree on $n$ leaves with minimal Colless index, i.e. $\mathcal{C}(T)=c_{n}$. Let $n_{a},n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$, respectively, where $n_{a}\geq n_{b}$.

Now, recall that both maximally balanced trees and GFB trees have minimal Colless index (cf. Theorems 4 and 5).

Assume $n_{a}<n_{a}^{mb}$ and $n_{b}>n_{b}^{mb}$. As $n_{a}^{mb}=n_{b}^{mb}=\frac{n}{2}$ for $n$ even and $n_{a}^{mb}=\frac{n+1}{2}$ and $n_{b}^{mb}=\frac{n-1}{2}$ for $n$ odd, this assumption contradicts $n_{a}\geq n_{b}$. Thus, $n_{a}\geq n_{a}^{mb}$ and $n_{b}\leq n_{b}^{mb}$.

Additionally, $n_{a}\leq n_{a}^{gfb}$ and $n_{b}\geq n_{b}^{gfb}$ is a direct consequence of Theorem 6. This completes the proof. ∎

Note that Corollary 2 only gives a necessary and not a sufficient condition. Consider for example tree $T_{2}$ in Figure 4 on $12$ leaves. Here, $n_{a}=7$ and $n_{b}=5$, i.e.

[TABLE]

but $T_{2}$ does not have minimal Colless index. This is due to the fact that if $n_{a}\neq n_{b}$ and $n_{b},n_{b}$ odd, the resulting tree will not have minimal Colless index:

Theorem 7**.**

Let $T=(T_{a},T_{b})$ be a tree on $n$ leaves and let $n_{a}$ and $n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$. If $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ odd, then $C(T)>c_{n}$, i.e. $T$ does not have minimal Colless index.

Proof.

Let $T=(T_{a},T_{b})$ be a tree on $n$ leaves. Let $n_{a}$ and $n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$ with $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ odd. Without loss of generality let $n_{a}>n_{b}$. The fact that $n_{a}$ and $n_{b}$ are both odd, results in

[TABLE]

By (20) we also have

[TABLE]

We prove the statement by contradiction and assume that $C(T)=c_{n}$, i.e. $T$ has minimal Colless index. Then,

[TABLE]

Additionally, by Lemma 1, Lemma 2 and (21)

[TABLE]

which results in

[TABLE]

Similarly, by Lemma 1, Lemma 2 and (20)

[TABLE]

which results in

[TABLE]

By (23) and (24) and the fact that $n$ is even we can rewrite (22) in the following way

[TABLE]

This results in $c_{n}\geq c_{n}+2$ which is a contradiction, and therefore completes the proof. ∎

Thus, if $n_{a}\neq n_{b}$, and $n_{a},n_{b}$ odd, the resulting tree does not have minimal Colless index. However, it can easily be verified that if $n_{a}\neq n_{b}$, but both $n_{a}$ and $n_{b}$ are even, the resulting tree may or may not have minimal Colless index. Consider for example $n=24$: In this case $\lceil\log_{2}(24)\rceil=5$ and thus by Theorem 2 we have:

[TABLE]

Thus, while $n_{a}=16,n_{b}=8$ yields a Colless minimum, $n_{a}=14,n_{b}=10$ does not.

We now turn to the number of trees on $n$ leaves with minimal Colless index and state an upper bound for it. This bound is implied by the fact that we can relate trees with minimal Colless index with trees with minimal Sackin index, another index of tree balance which is defined as follows:

Definition 3** (Sackin [11]).**

Let $T$ be a rooted binary tree. Then, its Sackin index is defined as

[TABLE]

where $n_{u}$ denotes the number of leaves of the subtree of $T$ rooted at $u$.

For the Sackin index, the number of trees with minimal Sackin index is known (cf. Theorem 8 in Fischer [19]).

Theorem 8** (adapted from Theorem 8 in Fischer [19]).**

Let $\widetilde{s}(n)$ denote the number of binary rooted trees with $n$ leaves and with minimal Sackin index and let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. For any partition of $n$ into two integers $n_{a}$, $n_{b}$, i.e. $n=n_{a}+n_{b}$, we use $k_{n_{a}}$ and $k_{n_{b}}$ to denote $\lceil\log_{2}(n_{a})\rceil$ and $\lceil\log_{2}(n_{b})\rceil=\lceil\log_{2}(n-n_{a})\rceil$, respectively. Moreover, let

[TABLE]

Then, the following recursion holds:

$\widetilde{s}(1)=1$**

• 2.

$\widetilde{s}(n)=\sum\limits_{\begin{subarray}{c}(n_{a},n_{b}):\\ n_{a}+n_{b}=n,\\ n_{a}\geq\frac{n}{2},\\ k_{n_{a}}=k-1,\\ k_{n_{b}}=k-1,\\ n_{a}\neq n_{b}\end{subarray}}\left(\widetilde{s}(n_{a})\cdot\widetilde{s}(n_{b})\right)+\widetilde{f}(n)+\widetilde{s}(n-2^{k-2}).$**

Note that above recursion yields sequence A299037 in the On-Line Encyclopedia of Integer Sequences (Sloane [20]).

In the following we will show that every tree with minimal Colless index also has minimal Sackin index. Note, however, that the converse is not true: tree $T_{2}$ depicted in Figure 4 has minimal Sackin index, but does not have minimal Colless index. Nevertheless, as we will show next, the number of trees with minimal Sackin index on $n$ leaves provides an upper bound for the number of leaves with minimal Colless index.

Proposition 3**.**

Let $T$ be a rooted binary tree on $n$ leaves that has minimal Colless index, i.e. $\mathcal{C}(T)=c_{n}$. Then, $T$ has minimal Sackin index.

The proof of Proposition 3 requires the following corollary from Fischer [19]:

Corollary 3** (Corollary 4 in Fischer [19]).**

Let $T$ be a rooted binary tree with $n\in\mathbb{N}_{\geq 2}$ leaves. Moreover, let $T=(T_{a},T_{b})$ be the standard decomposition of $T$ into its two maximal pending subtrees, let $n_{i}$ denote the number of leaves in $T_{i}$ for $i\in\{a,b\}$, respectively, such that $n_{a}\geq n_{b}$. Let $k_{n}\coloneqq\lceil\log_{2}(n)\rceil$. Then, the following equivalence holds: $T$ has minimal Sackin index if and only if $T_{a}$ and $T_{b}$ have minimal Sackin index and $n_{a}-n_{b}\leq\min\{n-2^{k_{n}-1},2^{k_{n}}-n\}$.

Proof of Proposition 3.

We show the statement by induction on $n$. For $n=1$, there is only one tree. This tree trivially has both minimal Colless index as well as minimal Sackin index, which completes the base case of the induction.

Suppose that the claim holds for all trees with fewer than $n$ leaves. Now, let $T=(T_{a},T_{b})$ be a tree with $n$ leaves and minimal Colless index, i.e. $\mathcal{C}(T)=c_{n}$. Let $n_{a}$ and $n_{b}$ denote the number of leaves of $T_{a}$ and $T_{b}$, respectively, and without loss of generality let $n_{a}\geq n_{b}$. Then from Theorem 6 we have that:

[TABLE]

First, let $T$ have $n=3\cdot 2^{k_{n}-2}$ leaves. Then by (25), $n_{a}-n_{b}\leq 2^{k_{n}-2}=\min\{n-2^{k_{n}-1},2^{k_{n}}-n\}=\min\{2^{k_{n}-2},2^{k_{n}-2}\}$. Moreover, by Lemma 2 both $T_{a}$ and $T_{b}$ have minimal Colless index and as $n_{a},n_{b}<n$ they also have minimal Sackin index by the inductive hypothesis. In total, by Corollary 3 this implies that $T$ has minimal Sackin index.

Now, let $T$ have $n\in(2^{k_{n}-1},3\cdot 2^{k_{n}-2})$ leaves. Then by (25), $n_{a}-n_{b}\leq n-2^{k_{n}-1}=\min\{n-2^{k_{n}-1},2^{k_{n}}-n\}$. Again, using Lemma 2 and the inductive hypothesis, by Corollary 3 this implies that $T$ has minimal Sackin index.

Last, let $T$ have $n\in(3\cdot 2^{k_{n}-2},2^{k_{n}}]$ leaves. Then by (25), $n_{a}-n_{b}\leq 2^{k_{n}}-n=\min\{n-2^{k_{n}-1},2^{k_{n}}-n\}$, i.e. using Lemma 2 and the inductive hypothesis, by Corollary 3, $T$ has minimal Sackin index. This completes the proof. ∎

As every tree with minimal Colless index has minimal Sackin index (while the converse is not true), a direct consequence of Proposition 3 is the following corollary:

Corollary 4**.**

Let $\widetilde{c}(n)$ denote the number of binary rooted trees with $n$ leaves and minimal Colless index and let $\widetilde{s}(n)$ denote the number of binary rooted trees with $n$ leaves and minimal Sackin index. Then, we have $\widetilde{c}(n)\leq\widetilde{s}(n)$.

Remark 3**.**

Recall that not every tree with minimal Sackin index also has minimal Colless index. Thus, $\widetilde{s}(n)$ is not a sharp bound for $\widetilde{c}(n)$. While for a tree $T=(T_{a},T_{b})$ with $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ odd, it is totally possible to have minimal Sackin index (as long as $n_{a}$ and $n_{b}$ satisfy the conditions of Corollary 3), for the Colless index this is never possible (cf. Theorem 7). Thus, we can tighten the upper bound for $\widetilde{c}(n)$ by excluding all pairs of $n_{a}$ and $n_{b}$, where $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ odd from the recursion given in Theorem 8333In the last section of this manuscript, we will discuss how to improve this bound even further, based on the results by Coronado and Rosselló [16], and even derive a recursive formula for the number of trees with minimal Colless index. We denote the resulting tighter upper bound by $\widetilde{b}(n)$. In particular, we have for all $n\in\mathbb{N}$: $\widetilde{c}(n)\leq\widetilde{b}(n)\leq\widetilde{s}(n)$. This relation is depicted in Figure 5.

Note that starting at $n=1$ and continuing up to $n=32$, the sequences $\widetilde{c}(n)$ and $\widetilde{b}(n)$ are $(1,1,1,1,1,2,1,1,1,3,3,4,3,3,1,1,1,4,6,10,16,21,13,11,13,21,16,10,6,4,1,1)$ and $(1,1,1,1,1,2,1,1,1,3,3,4,3,3,1,1,1,4,6,10,16,21,25,20,25,21,16,10,6,4,1,1)$, respectively (where $\widetilde{c}(n)$ results from an exhaustive enumeration of trees with minimal Colless index444See the discussion section for a recursive formula to calculate $\widetilde{c}(n)$ based on the results by Coronado and Rosselló [16]. and $\widetilde{b}(n)$ is obtained from $\widetilde{s}(n)$ (cf. Theorem 8) by excluding all pairs of $n_{a}$ and $n_{b}$, where $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ odd). The sequence for $\widetilde{c}$ has been submitted to the On-Line Encyclopedia of Integer Sequences OEIS (Sloane [20]) as it so far had not been contained in it. It is currently under review there and will shortly be published as sequence A307689.

To summarize, in this section we have further characterized trees with minimal Colless index. Additionally, we have given two upper bounds for the number of trees with minimal Colless index by first relating the Colless index to the Sackin index and then improving the obtained bound by using Theorem 7.

4 Discussion

In this manuscript, we have thoroughly analyzed extremal properties of the Colless balance index. We have focused on the minimal Colless index of a tree with $n$ leaves and have both given a recursive formula as well as an explicit expression for this value, where the latter shows a surprising connection of the minimal Colless index to the Blancmange/Takagi curve, a fractal curve. While the recursive formula555Note that this formula was independently also discovered by Coronado and Rosselló [16]. directly yields a class of trees with minimal Colless index, namely the class of maximally balanced trees, we have subsequently introduced another class of trees with minimal Colless index, namely the class of GFB trees. Note that this class of trees might somehow be related to the explicit formula for the Colless value stated by Coronado and Rosselló [16]. On the other hand, our own explicit formula, as stated above, is more suitable to express the fractal structure of the minimal Colless index by relating it to the famous Blancmange curve.

Anyway, while the two mentioned classes of trees, i.e. maximally balanced trees and GFB trees, as well as their corresponding leaf partitionings yield trees with minimal Colless index, we have additionally shown that a tree $T=(T_{a},T_{b})$ with $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ odd, cannot have minimal Colless index, while for $n_{a}\neq n_{b}$ and $n_{a},n_{b}$ even it may or may not have minimal Colless index.

However, an independent full characterization of trees with minimal Colless index has recently been achieved by Coronado and Rosselló [16], and the authors also characterize valid leaf partionings $n_{a}$ and $n_{b}$ for a tree with $n=n_{a}+n_{b}$ leaves and minimal Colless index. This characterization, which can be found in Proposition 11 of Coronado and Rosselló [16], can be used to improve the bound $\widetilde{b}(n)$ as presented in Figure 5 of our manuscript by summing only over those pairs $(n_{a},n_{b})$ that are valid due to this proposition. However, note that this improved bound $\widehat{b}(n)$ is still not sharp – this can be seen e.g. by considering the tree depicted in Figure 6. This tree is a tree on $n=23$ leaves with minimal Sackin index and with $n_{a}=12$ and $n_{b}=11$, which is a combination of $n_{a}$ and $n_{b}$ that is explicitly allowed by Proposition 11 of Coronado and Rosselló [16] (which is correct, because there are in fact trees with 23 leaves and minimal Colless index and leaf partitioning $(n_{a},n_{b})=(12,11)$). However, as subtree $T_{a}$ with $n_{a}=12$ leaves consists of two maximal pending subtrees with 7 and 5 leaves, respectively (and thus a combination of two different odd numbers), by Theorem 7 in our manuscript, $T_{a}$ is not a tree with minimal Colless value, and thus by Lemma 2, $T$ is not a tree with minimal Colless value, either. In fact, the minimal Colless value for $n=23$ is $c_{23}=10$ by Theorem 2, but the tree depicted in Figure 6 has Colless value $\mathcal{C}(T)=12$.

By relating trees with minimal Colless index with trees with minimal Sackin index, we have shown that the two classic and most frequently used tree balance indices are actually closely related, and we have used this insight to present an upper bound for the number of Colless minimal trees.

However, by denoting the set of valid $(n_{a},n_{b})$ pairs by $QB(n)$, i.e. QB(n)=\{(n_{a},n_{b}):n_{a}+n_{b}=n\mbox{ and \exists$a tree$T$on$n$leaves:$\mathcal{C}(T)=c_{n}} \mbox{ and with leaf partitioning (n_{a},n_{b})}\}, a set which was characterized by Coronado and Rosselló [16], one can quite easily derive a recursive formula for the number of Colless minima (in the same way as Theorem 8 works for the number of Sackin minima):

$\widetilde{c}(1)=1$

• 2.

$\widetilde{c}(n)=\sum\limits_{\begin{subarray}{c}(n_{a},n_{b}):\\ n_{a}+n_{b}=n,\\ n_{a}\neq n_{b},\\ (n_{a},n_{b})\in QB(n)\end{subarray}}\left(\widetilde{c}(n_{a})\cdot\widetilde{c}(n_{b})\right)+\widetilde{f}(n),$

where \widetilde{f}(n)=\begin{cases}0&\mbox{if n is odd}\\ {\widetilde{c}(\frac{n}{2})+1\choose 2}&\mbox{else. }\end{cases}.

Note that the binomial coefficient in $\widetilde{f}$ prevents counting symmetries twice in the case where $n_{a}=n_{b}=\frac{n}{2}$. The correctness of this formula is a direct consequence of Lemma 2, which implies that each Colless minimal tree has two maximal pending subtrees which are also Colless minimal, combined with the definition of the set $QB(n)$, which ensures that we only sum over pairs $(n_{a},n_{b})$ which indeed imply Colless minimal trees on $n$ leaves.

Note that this is the first formula in the literature which enables us to calculate $\widetilde{c}(n)$, and we have submitted the resulting sequence to the Online Encyclopedia of Integer Sequences Sloane [20] as it was not previously listed there. It is currently under review there and will shortly be published as sequence A307689. However, it would definitely be of interest to find an explicit formula for $\widetilde{c}(n)$ and to analyze if the fractal structure of the sequence of the minimal Colless index induced by the Blancmange curve is reflected in the sequence of the number of trees that achieve it (as is suggested by Figure 5). These are topics for future research.