**Open Journal of Discrete Mathematics**

Vol.04 No.04(2014), Article ID:50292,11 pages

10.4236/ojdm.2014.44013

Balance in Random Trees

Azer Akhmedov, Warren Shreve

Department of Mathematics, North Dakota State University, Fargo, ND, USA

Email: azer.akhmedov@ndsu.edu, warren.shreve@ndsu.edu

Copyright © 2014 by authors and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

Received 30 July 2014; revised 25 August 2014; accepted 20 September 2014

ABSTRACT

We prove that a random labeled (unlabeled) tree is balanced. We also prove that random labeled and unlabeled trees are strongly -balanced for any. Definition: Color the vertices of graph with two colors. Color an edge with the color of its endpoints if they are colored with the same color. Edges with different colored endpoints are left uncolored. is said to be balanced if neither the number of vertices nor and the number of edges of the two different colors differs by more than one.

**Keywords:**

Random Trees, Balance, Equicolorable Graphs

1. Introduction

The notion of a balanced graph is defined [1] as follows:

Definition 1.1 Let be a finite simple graph, be an integer, be a map.

For all, we write. We also write . The map is called a coloring.

The case of is especially interesting. In this case, the sets are called the sets of black vertices, white vertices, black edges, and white edges respectively. If the coloring is fixed we may drop it in the notation.

Definition 1.2 A finite simple graph is called balanced if there exists a coloring

such that and. A map satisfying this condition is called a balanced coloring.

The graph in Figure 1 is balanced since we have shown a balanced coloring of it.

It is not difficult to see that:

Figure 1. With the given coloring, the graph has 4 black and 3 white vertices; it also has 2 white edges (labeled with a “W”) and 1 black edge (labeled with a “B”).

1) The complete graph is balanced if and only if or is even.

2) The star is balanced if and only if; see Figure 2 for a balanced coloring of.

3) The double star is balanced if and only if. (The double star is a connected graph where some two adjacent vertices have degree and, and all other vertices have degree 1).

In [2] , the author introduces a somewhat similar notion of a cordial graph, a generalization of both graceful and harmonious graphs. It has been conjectured by A. Rosa, G. Ringel and A. Kotzig that every tree is graceful (Graceful Tree Conjecture, [3] ), and it has been conjectured by R. Graham and N. Sloane that every tree is harmonious (see [4] ). While these conjectures are still open, in [5] it is proved that every tree is cordial.

Not every tree is balanced; in this paper, we will be interested in the property of being balanced for a random labeled and unlabeled tree, as well as for random labeled graphs.

The main results of the paper are Theorem A and Theorem B, stated below.

Theorem A A random labeled (unlabeled) tree is balanced; more precisely, if denotes the number of all labeled (unlabeled) trees on vertices, and denotes the number of all balanced labeled (unla-

beled) trees on vertices, then and.

Remark 1.3 In this paper, for simplicity, we consider only uniform models of random graphs and random trees. The results can be extended to a large class of non-uniform models as well. Note that (see [6] or [7] ) and for some positive constants and (see [8] ).

We also would like to introduce the notion of -balanced graphs.

Definition 1.4 Let. A finite simple graph is called -balanced if there exists a coloring such that and for all distinct. The map will be called a -balanced coloring.

Definition 1.5 Let. A finite simple graph is called strongly -balanced if there exists a coloring such that, and for all distinct. The map will be called a strongly -balanced coloring.

In more popular terms, a finite simple graph is strongly -balanced if and only if it is -equitably colorable. In Section 5 we study some basic properties of -balanced graphs. We prove the following theorem.

Theorem B. For all, a random (labeled) tree is strongly -balanced.

Remark 1.6 Let us emphasize that Theorem B is originally due to B. Bollobás and R. Guy (see [9] ). Our proof in this paper is very different with some ingredients which might be interesting independently.

Remark 1.7 It has been proved by I. Ben-Eliezer and M. Krivelevich (see [10] ) that a random graph is balanced. For, it seems quite plausible that a random graph is indeed -balanced. However, notice that the clique number of a random graph on vertices is at least (see [11] ) thus a random graph is not strongly -balanced.

Figure 2. A balanced coloring of.

Acknowledgement: We would like to thank to B. Gittenberger for the discussion and for bringing the reference [12] to our attention; to I. Pak for his comments; to M. Krivelevich for bringing [9] to our attention. We also would like to extend our gratitude to the anonymous referee for many helpful remarks, and for pointing out a flaw in the proof of Lemma 5.3.

Notes:

1) For any finite simple graph, we will denote the maximum degree of by.

2) A vertex of degree one will be called a leaf vertex or simply a leaf. A non-leaf vertex is called a pre- leaf vertex if it is adjacent exactly to leaves where. A pre-leaf vertex of degree two is called special.

3) For, there exists a unique tree up to isomorphism with vertices and maximum degree at most two; we will call this tree a path on vertices, and denote it with.

4) For a finite simple graph and for a subset, the induced subgraph will be called the full subgraph of on.

5) For a tree and a non-leaf vertex, a subset will be called a branch of with respect to if is a maximal subset such that the full subgraph is connected and where denotes the distance in the tree.

2. Characterization of Balanced Graphs

In this section we observe some basic facts on balanced and -balanced graphs. Let us first prove a very simple lemma which provides a necessary and sufficient condition for a graph to be balanced.

Lemma 2.1 Let be a finite simple graph with vertices, and degrees. is balanced if and only if there exists a partition such that

1)

2)

Proof. Let.

Assume is balanced with a balanced coloring.

Let.

Since is a balanced coloring, we get so condition 1) is satisfied.

For every, we denote

and for every, we denote

Then. On the other hand, since is balanced, we have .

Then. Thus condition 2) is also satisfied.

To prove the converse, assume conditions 1) and 2) are satisfied. We define the coloring as follows: for every we set and for every we set.

Then we have.

On the other hand,

Then by condition 2), we get.

Corollary 2.2 It is proved in [1] that an -regular finite simple graph with vertices is balanced if and only if is even or. This fact also follows immediately from Lemma 2.1. In [13] , the authors deduce the same fact from their characterization of balanced graphs.

Lemma 2.1 shows that the balancedness of a graph completely depends on the degree sequence of it. This is no longer the case for -balanced graphs for. In fact, the trees and in Figure 3 have the same degree sequence, and it is not difficult to see that is 3-balanced while is not.

The fact that, for, the -balancedness is not determined by the degree sequence causes difficulties in proving that random graphs are -balanced. It also seems plausible that, generically, -balancedness is a weaker condition than balancedness, although it does not seem easy to describe (with a good sufficient condition) when exactly is this true. It is useful to point out the following simple fact.

Proposition 2.3 For all distinct there exists a finite simple graph which is -balanced but not -balanced.

Proof. Let be a prime number such that.

Let us first assume that. If divides, then the graph is -balanced but not -balanced. If does not divide then the graph is -balanced but not -balanced.

Now assume that. Then the graph is -balanced but not -balanced.

3. Combinatorial Lemmas

Let. The elements of consist of sequences of positive integers of length such that no term is bigger than. We denote.

Now we introduce the notion of balanced sequences:

Definition 3 A sequence (an element) is called balanced if and only if there exists a partition such that

1)

2)

The partition will be called a balanced partition.

In these new terms, Lemma 2.1 states that a graph is balanced if and only if its degree sequence is balanced.

When the sequence is not balanced, we would like to measure how far it is from being balanced.

Figure 3. The trees and have the same degree sequence; is 3-balanced while is not.

Definition 3.1 Let be any finite sequence of non-negative integers. The quantity

will be called the balance of.

Remark 3.2 By Lemma 2.1, a sequence is balanced if and only if. The quantity, somewhat roughly, measures how far the sequence is from being balanced. For an example, let

and be a sequence of length 8. It is easy to see .

The following easy lemma will be useful:

Lemma 3.3 Let be any finite sequence of non-negative integers. Then.

Proof. We will present a constructive proof.

Without loss of generality, we may assume that. First, let us assume that is even, so let. We will build two subsets of inductively such that, and

.

We divide the sequence into pairs, and we will abide by the rule that exactly one element of each pair belongs to and the other element belongs to. We start by letting . Assume now we have built the subsets such that and for all.

Let. If then we let but if then we let , and we proceed by induction. Then we let. Clearly, we have.

If is odd, then we may replace by and apply the previous argument.

We will need the following notations

Definition 3.4 Let. We will denote

Lemma 3.5 Let such that and. Then is balanced.

Proof. Let. Without loss of generality we may assume that

. If then is clearly balanced so let and let

.

By Lemma 3.3, hence there exists a partition such that and. Then there exists a partition such that and. By letting we obtain that , , and.

4. Proof of Theorem A

First, we will discuss the case of labeled trees. The following theorem of J. W. Moon will play a crucial role.

Theorem 4.1 (see [14] ) If is a fixed positive constant, then in a random labeled tree with vertices, the maximum degree satisfies the following inequality

Remark 4.2 By choosing we obtain that

in a random tree with vertices.

We will use only the upper bound in the inequality of Remark 4.2. Besides the upper bound on the maximum degree in random trees, we also need a lower bound on the number of vertices with degree 1, and with degree 2. Notice that, since the sum of degrees of a tree with vertices is exactly, at least half of the vertices have degree either 1 or 2. However, we need a linear lower bound for the number of vertices of degree 1 and for the number of vertices of degree 2 separately.

Let be the random variable which denotes the number of vertices of degree in a labeled tree with vertices. Also let. It has been proved by A. Rényi (see [15] ) that the asymptotic distribution of random variable is normal with mean and variance. A similar result has been proved for the random variable, by A. Meir and J. W. Moon (see [16] ), namely, that the asymptotic distribution of the random variable is normal with mean and variance. Combining these two results we can state the following theorem (due to A. Rényi and A. Meir-J.W. Moon)

Theorem 4.3 Let be fixed real numbers,; and for, let denotes the probability that. Then

We need the following immediate corollary of this theorem

Corollary 4.4 In a random labeled tree with vertices, for all,.

Now, in the case of random labeled trees, Theorem A immediately follows from Theorem 4.1, Corollary 4.4, and Lemma 3.5.

The case of unlabeled trees: We will use the results analogous to Theorem 4.1 and Theorem 4.3. The analogue of Theorem 4.1 is proved by W. Goh and E. Schmutz:

Theorem 4.5 (see [12] ) There exists positive constants such that in a random unlabeled tree with vertices, the maximum degree satisfies the inequality.

Now, for any let the random variable denotes the number of vertices of degree in a random unlabeled tree with vertices. The following theorem is due to M. Drmota and B. Gittenberger; in the case of

, as a special case, it provides an analogue of Theorem 4.3.

Theorem 4.6 (see [17] ) For arbitrary fixed natural, there exist positive constants and such that the limiting distribution of is normal with mean and variance.

Corollary 4.7 For all and, in a random unlabeled tree with vertices.

Now, in the case of unlabeled trees, the claim of Theorem A follows from Theorem 4.5, Lemma 3.5, and Corollary 4.7.

5. -Balanced Trees: Proof of Theorem B

In this section we will assume that. The fact that the -balancedness is not determined by the degree sequence causes significant difficulties in proving that random graphs are balanced. We nevertheless prove that random trees are strongly -balanced by more careful study of -balancedness.

First, we need to prove the following technical lemma.

Lemma 5.1 Let be a tree and be distinct vertices of with degree at least. If are distinct pre-leaf vertices of then there exists a strongly 3-balanced coloring of such that and.

Proof. The proof is by induction on. For the claim is obvious (since, in this case, will be isomorphic either to a path or to the double star), so we will assume that and the claim holds for all trees of order less than.

Assume that at least one of the following two conditions holds:

(c1) there exists such that;

(c2) there exists a leaf vertex not adjacent to any of the vertices.

Then there exists a leaf such that if is a full subgraph on, then, in the tree, we have, and are still pre-leaf vertices.

By inductive hypothesis, there exists a strongly 3-balanced coloring of such that and. Let be the unique vertex of adjacent to. Without loss of generality, we may assume that and.

If then we let thus extending to a strongly 3-balanced coloring of such that and.

If, however, then there exists such that; also, since,

there exists a branch of with respect to which is disjoint from. Let be a leaf vertex in. Then and. We define as follows:

Notice that because of the inequality, we have and. Then the map is a strongly 3-balanced coloring.

Now, suppose that neither of the conditions (c1) and (c2) hold. Let be the path in starting at and ending at (it may possibly consist of just the vertices and). Then the tree satisfies the following conditions: there exists two vertices in and paths starting at respectively such that any vertex of either belongs to one of the paths or it is a leaf vertex adjacent to one of the vertices. Then it is straightforward to build a strongly 3-balanced coloring satisfying the conditions and.

The following proposition is interesting in itself; it will also play a key role in proving Theorem B.

Proposition 5.2 If is a tree with then is strongly -balanced. Moreover, for

any two distinct pre-leaf vertices and of there exists a strongly 3-balanced coloring such that.

Proof. The proof will be by induction on. For we have hence is isomorphic to a path. Thus, the claim is obvious. Let us now assume that, and the claim holds for all trees of

order less than with.

Let and. We will consider the following three cases separately:

Case 1..

Let be a leaf of, , and let be the full subgraph of on. Then we have

By inductive hypothesis, there exists a strongly 3-balanced coloring of.

On the other hand, is adjacent to exactly one vertex in; let be this vertex. Let be any element of. We extend the coloring of to a strongly 3-balanced coloring by defining.

Case 2..

Let be distinct leaves and be the only vertices of adjacent to respectively (and are not necessarily distinct). Let also be the full subgraph of on the set. Then we still have the inequality. Hence, by inductive assumption, there exists a strongly 3-balanced coloring of.

Then there exist distinct such that and. Thus we can extend to a strongly 3-balanced coloring of by defining and.

Case 3..

The major difference in this case compared with the previous two cases is that when we obtain by deleting some arbitrary three leaves from, we may lose the inequality. (Notice that possesses three leaf vertices unless it is isomorphic to a path). Suppose are the vertices adjacent to respectively. Note that are not necessarily distinct. If we have the inequality then by inductive assumption we would have a strongly 3-balanced coloring. However, if then it becomes problematic to extend to a strongly 3-balanced coloring. Thus we need to employ different and more careful tactics.

We will prove the following lemma which suffices for the proof of Proposition 5.2 in the case.

Lemma 5.3 Let be a tree with vertices where. If are distinct pre-leaf vertices of then there exists a strongly 3-balanced coloring such that.

Proof. The proof of the lemma will be again by induction on. The “part” of the claim will be needed to make the step of the induction. For, the graph is isomorphic to a path thus the claim is obvious. For it can be seen by a direct checking. (We leave this to a reader as a simple exercise.) Thus let us assume that.

Let. Let also. We will consider the following cases (the notations in each case will be independent of the notations of other cases):

Case A: The vertices and are the only pre-leaf vertices of.

The claim is obvious when, so we may assume that. We will consider two sub- cases:

Sub-case 1:.

Then there exist leaves such that is adjacent to and are adjacent to. Let be the full subgraph of on. Notice that and are still pre-leaf vertices of. By the inductive hypothesis, has a strongly 3-balanced coloring such that. We may assume that. Then we extend to a strongly 3-balanced coloring by letting.

Sub-case 2:.

In this case there exists a path in where is a leaf, are vertices of degree two, and

where are the leaves adjacent to. Then we have the inequalities and. We will construct the required strongly 3-balanced coloring explicitly as follows.

First, for all we let when is odd, and when is even. We also let and.

Now we need to define the coloring on the remaining set

Let where for all. (Thus we are reorder the elements of the set from closest to the farthest from the leaf.) Then, for all, we let when is odd, and when is even.

For the rest of the proof we will assume that has more than two pre-leaf vertices.

Case B: and is not special.

Let be distinct leaves such that is adjacent to, is adjacent to, and is adjacent to a vertex distinct from and. We let be the full subgraph on. Then and we have. By inductive hypothesis, there exists a strongly 3-balanced coloring such that. Without loss of generality we may assume that. Then we extend to a strongly 3-balanced coloring as follows: if then we let; and if then we let.

Case C: and is special.

Let be the only leaf adjacent to, be the unique non-leaf vertex adjacent to, be a leaf vertex not adjacent to, and be the unique vertex adjacent to. We let be the full subgraph on. Then and. By inductive hypothesis, there exists a strongly 3-balanced coloring. Then we extend to a strongly 3-balanced coloring as follows: we let such that is distinct from and. Then we define such that is distinct from and. Finally we let such that is distinct from and. Notice also that we obtain.

Case D: and there exists a leaf vertex adjacent to.

This case is similar to Case B. Since and, we have. If, we let be leaves adjacent to respectively; and if, we let be leaves adjacent to respectively, and be a leaf not adjacent to either of the vertices. We define be the full subgraph on. Then hence admits a strongly 3-balanced coloring such that. We extend to a strongly 3-balanced coloring to as in Case B.

Case E:, is special and there exists a leaf vertex adjacent to.

This case is similar to Case C. Let be the only leaf adjacent to, be the unique non-leaf vertex adjacent to, be a leaf vertex adjacent to. We let be the full subgraph on. Then and. By inductive hypothesis, there exists a strongly 3-balanced coloring. Then we extend to a strongly 3-balanced coloring as follows: we let such that is distinct from and. Then we define such that is distinct from and. Finally we let such that is distinct from and.

Case F:, and there is no leaf vertex adjacent to.

Since, there exists a special vertex adjacent to. Let be the unique leaf adjacent to. Let also be a leaf not adjacent to any of the vertices (such a leaf exists because), and let be the unique vertex adjacent to.

We define to be the full subgraph on. By inductive assumption, there exists a strongly 3-balanced coloring, moreover, if then.

If, then we let be any element of distinct from. Then we let be any element of distinct from and. Finally, we let be any element of distinct from and. Thus we have extended to a strongly 3-balanced coloring such that.

If then and we may assume that. Then we let

be any element of distinct from and; then we let be any element of distinct from and; finally we let be any element of distinct from and.

Case G:.

In this case the claim follows immediately from Lemma 5.1.

Now we can prove an analogous result for -balanced graphs.

Proposition 5.4 Let be a tree with vertices where and. Then is strongly -balanced.

Proof. The proof is by induction on. For, the claim is true by Proposition 5.2.

Assume now. The tree has vertices such that. Moreover, for all distinct, the vertices and are not connected by an edge. Let also, and be a full subgraph on the subset. Then is a forest with vertices but with. This implies that is a subgraph of a tree with vertices where.

Then. By inductive hypothesis, we obtain that

is strongly -balanced, hence is strongly -balanced. Since no two elements of are adjacent, we obtain that is strongly -balanced.

Now, for random labeled trees, Theorem B follows immediately from Theorem 4.1 and Proposition 5.4; and for random unlabeled trees, it follows immediately from Theorem 4.5 and Proposition 5.4.

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