The Number of Subtrees of Trees with Given Degree Sequence

Transcription

1 arxiv: v1 [math.co] 3 Sep 2012 The Number of Subtrees of Trees with Given Degree Sequence Xiu-Mei Zhang 1,2, Xiao-Dong Zhang 1,, Daniel Gray 3 Hua Wang 4 1 Department of Mathematics, Shanghai Jiao Tong University 800 Dongchuan road, Shanghai, , P. R. China 2 Department of Mathematics,Shanghai sau University 2727 jinhai road, Shanghai, , P. R. China 3 Department of Mathematics, University of Florida Gainesville, FL 32611, United States 4 Department of Mathematical Sciences, Georgia Southern University Stateboro, GA 30460, United States Abstract This paper investigates some properties of the number of subtrees of a tree with given degree sequence. These results are used to characterize trees with the given degree sequence that have the largest number of subtrees, which generalizes the recent results of Kirk Wang. These trees coincide with those which were proven by Wang independently Zhang et al. to minimize the Wiener index. We also provide a partial ordering of the extremal trees with different degree sequences, some extremal results follow as corrollaries. Key words: Tree; subtree; degree sequence; majorization; AMS Classifications: 05C05, 05C30 This work is supported by the National Natural Science Foundation of China (No: ), the National Basic Research Program (973) of China (No.2006CB805900), a grant of Science Technology Commission of Shanghai Municipality (STCSM, No: 09XD ). Corresponding author: Xiao-Dong Zhang ( xiaodong@sjtu.edu.cn) 1

2 1 Introduction All graphs in this paper will be finite, simple undirected. A tree T= (V, E) is a connected, acyclic graph where V(T) E(T) denote the vertex set edge set respectively. We refer to vertices of degree 1 of T as leaves. The unique path connecting two vertices u, v in T will be denoted by P T (u, v). The number of edges on P(u, v) is called distance dist T (u, v), or for short dist(u, v) between them. We call a tree (T, r) rooted at the vertex r (or just by T if it is clear what the root is) by specifying a vertex r V(T). The height of a vertex v of a rooted tree T with root r is h T (v)=dist T (r, v). For any two different vertices u, v in a rooted tree (T, r), we say that v is a successor of u u is an ancestor of v if P T (r, u) P T (r, v). Furthermore, if u v are adjacent to each other dist T (r, u)=dist T (r, v) 1, we say that u is the parent of v v is a child of u. Two vertices u, v are siblings of each other if they share the same parent. A subtree of a tree will often be described by its vertex set. The number of subtrees of a tree has received much attention. It is well known that the path P n the star K 1,n 1 have the most least subtrees among all trees of order n, respectively. The binary trees that maximize or minimize the number of subtrees are characterized in [5, 7]. Formulas are given to calculate the number of subtrees of these extremal binary trees. These formulas use a new representation of integers as a sum of powers of 2. Number theorists have already started investigating this new binary representation [1]. Also, the sequence of the number of subtrees of these extremal binary trees (with 2l leaves, l = 1, 2, ) appears to be new [4]. Later, a linear-time algorithm to count the subtrees of a tree is provided in [11]. In a related paper [6], the number of leaf-containing subtrees are studied for binary trees. The results turn out to be useful in bounding the number of acceptable residue configurations. See [3] for details. An interesting fact is that among binary trees of the same size, the extremal one that minimizes the number of subtrees is exactly the one that maximizes some chemical indices such as the well known Wiener index, vice versa. In [2], subtrees of trees with given order maximum vertex degree are studied. The extremal trees coincide with the ones for the Wiener index as well. Such correlations between different topological indices of trees are studied in [8]. Recently, in [13] [9] respectively, extremal trees are characterized regarding the Wiener index with a given degree sequence. Then it is natural to consider the following 2

3 question. Problem 1.1 Given the degree sequence the number of vertices of a tree, find the upper bound for the number of subtrees, characterize all extremal trees that attain this bound. It will not be a surprise to see that such extremal trees coincide with the ones that attain the minimum Wiener index. Along this line, we also provide an ordering of the degree sequences according to the largest number of subtrees. With our main results, Theorems , one can deduce extremal graphs with the largest number of subtrees in some classes of graphs. This generalizes the results of [5], [2], etc. The rest of this paper is organized as follows: In Section 2, some notations the main theorems are stated. In Section 3, we present some observations regarding the structure of the extremal trees. In Section 4, we present the proofs of the main theorems. In Section 5, we show, as corollaries, characterizations of the extremal trees in different categories of trees including previously known results. 2 Preliminaries For a nonincreasing sequence of positive integersπ=(d 0,, d n 1 ) with n 3, lett π denote the set of all trees withπas its degree sequence. We can construct a special tree T π T π by using breadth-first search method as follows. Firstly, label the vertex with the largest degree d 0 as v 01 (the root). Secondly, label the neighbors of v 0 as v 11, v 12,..., v 1d0 from left to right let d(v 1i ) = d i for i = 1,, d 0. Then repeat the second step for all newly labeled vertices until all degrees are assigned. For example, ifπ = (4, 4, 3, 3, 3, 3, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), Tπ is shown in Fig. 1. There is a vertex v 01 (the root) in layer 0 with the largest degree 4; its four neighbors are labeled as v 11, v 12, v 13, v 14 in layer 1, with degrees 4, 3, 3, 3 from left to right; nine vertices v 21, v 22,, v 29 in layer 2; five vertices v 31, v 32, v 33, v 34, v 35 in layer 3. The number of vertices in each layer i, denoted by s i can be easily calculated as s 0 = 1, s 1 = d 0 = 4, s 2 = d 1 + d 2 + d 3 + d 4 s 1 = =9, s 3 = d 5 + +d 13 s 2 = 5. 3

4 v01 v 11 v 12 v 13 v 14 v 21 v 22 v 23 v 24 v 25 v 26 v 27 v 28 v 29 v 31 v 32 v 33 v 34 v 35 Figure 1 To explain the structure properties of Tπ, we need the following notation from [12]. Definition 2.1 ([12]) Let T= (V, E) be a tree with root v 0. A well-ordering of the vertices is called a BFS-ordering if satisfies the following properties. (1) If u, v V, u v, then h(u) h(v) d(u) d(v); (2) If there are two edges uu 1 E(T) vv 1 E(T) such that u v, h(u)=h(u 1 ) 1 h(v)=h(v 1 ) 1, then u 1 v 1. We call trees that have a BFS-ordering of its vertices a BFS-tree. It is easy to see that Tπ has a BFS-ordering any two BFS-trees with degree sequenceπ are isomorphic (for example, see [12]). And the BFS-trees are extremal with respect to the Laplacian spectral radius. Letπ = (d 0,, d n 1 ) π = (d 0,, d n 1 ) be two nonincreasing sequences. If k i=0 d i k i=0 d i for k=0,, n 2 n 1 i=0 d i= n 1 i=0 d i, then the sequenceπ is said to major the sequenceπ denoted byπ π. It is known that the following holds (for example, see [10] or [12]). Proposition 2.2 (Wei [10]) Letπ=(d 0, d n 1 ) π = (d 0,, d n 1 ) be two nonincreasing graphic degree sequences. Ifπ π, then there exists a series of graphic degree sequencesπ 1,,π k such thatπ π 1 π k π, whereπ i π i+1 differ at exactly two entries, say d j (d j ) d k (d k ) ofπ i (π i+1 ), with d j = d j+ 1, d k = d k 1 j<k. The main results of this paper can be stated as follows. 4

5 Theorem 2.3 With a given degree sequenceπ, Tπ is the unique tree with the largest number of subtrees int π. Theorem 2.4 Given two different degree sequencesππ 1. Ifπ π 1, then the number of subtrees of Tπ is less than the number of subtrees of T π1. 3 Some Observations In order to prove Theorems , we need to introduce some more terminologies. For a vertex v of a rooted tree (T, r), let T(v), the subtree induced by v, denote the subtree of T (rooted at v) that is induced by v all its successors. For a tree T vertices v 1,v 2,...,v m 1,v m of T, let f T (v 1,v 2,...,v m 1,v m ) denote the number of subtrees of T that contain the vertices v 1,v 2,...,v m 1,v m. In particular, f T (v) denotes the number of subtrees of T that contain v. Letϕ(T) denote the number of non-empty subtrees of T. Let W be a tree x, y be two vertices of W. The path P W (x, y) from x to y can be denoted by x m x m 1... x 2 x 1 y 1 y 2...y m 1 y m for odd dist(x, y) or x m x m 1... x 2 x 1 zy 1 y 2...y m 1 y m for even dist(x, y), where x m x, y m y. Let G 1 be the graph resulted from W by deleting all edges in P W (x, y). The connected components (in G 1 ) containing x i, y i z are denoted by X i, Y i Z, respectively, for i=1, 2,..., m. We also let X k be the connected component of W containing x k after deleting the edge x k 1 x k Y k be the connected component of W containing y k after deleting the edge y k 1 y k, for k=1,, m. Figure 2 shows such a labelling according to a path of odd length (without z). X k 1... X 2 X 1 Y 1 Y 2... Y k 1 X k x k x k 1 x 2 x 1 y 1 y 2 y k 1 y k Y k Figure 2: Labelling of a path the components We need the next two lemmas from [2] to proceed. Lemma 3.1 ([2]) Let W be a tree with a path P W (x m, y m )= x m x m 1... x 2 x 1 (z)y 1 y 2...y m 1 y m from x m to y m. If f Xi (x i ) f Yi (y i ) for i=1, 2,..., m, then f W (x m ) f W (y m ). Furthermore, if this inequality holds, then f W (x m )= f W (y m ) if only if f Xi (x i )= f Yi (y i ) for i=1, 2,..., m. 5

6 Now let X Y be two rooted trees with roots x y. Let T be a tree containing vertices x y. Then we can build T by identifying the root x of X with x of T the root y of Y with y of T, T by identifying the root x of X with y of T the root y of Y with x of T. X x T y Y Y x T y X Figure 3: Constructing T (left) T (right) Lemma 3.2 ([2]) Let T,T,T be as in Figure 2. If f(x) f W (y) f X (x) f Y (y), then ϕ(t ) ϕ(t ) with equality if only if f T (x)= f T (y) or f X (x )= f Y (y ). From Lemmas , we immediately achieve the following observation. We leave the proof to the reader. Lemma 3.3 Let T be a tree int π P(x m, y m )= x m x m 1... x 2 x 1 (z)y 1 y 2... y m 1 y m be a path of T. Let T be the tree from T by deleting the two edges x k x k+1 y k y k+1 adding two edges x k+1 y k y k+1 x k. If f Xi (x i ) f Yi (y i ) for i=1,, k 1 k m 1, f X k+1 (x k+1 ) f Y k+1 (y k+1 ), then ϕ(t) ϕ(t ) with equality if only if f X k+1 (x k+1 )= f Y k+1 (y k+1 ) or f Xi (x i )= f Yi (y i ) for i=1,, k. For convenience, we refer to trees that maximize the number of subtrees as optimal. In terms of the structure of the optimal tree, we have the following version of Lemma 3.3. Corollary 3.4 Let T be an optimal tree int π P(x m, y m )= x m x m 1... x 2 x 1 (z)y 1 y 2...y m 1 y m be a path of T. If f Xi (x i ) f Yi (y i ) for i=1,, k with at least one strict inequality 1 k m 1, then f X k+1 (x k+1 ) f Y k+1 (y k+1 ). Lemma 3.5 Let T be an optimal tree int π P(x m, y m )= x m x m 1... x 2 x 1 (z)y 1 y 2...y m 1 y m be a path of T. If f Xi (x i ) f Yi (y i ) for i=1,, k with at least one strict inequality 1 k m 1, then f Xk+1 (x k+1 ) f Yk+1 (y k+1 ). 6

7 Proof. If k=m 1, then by Corollary 3.4, the assertion holds since f X m (x m )= f Xm (x m ) f Y m (y m )= f Ym (y m ). Hence we assume that 1 k m 2. Suppose that f Xk+1 (x k+1 )< f Yk+1 (y k+1 ). Denote by M the number of subtrees of T not containing vertices x k y k. Let W be the connected component of T by deleting the two edges x k x k+1 y k y k+1 containing vertices x k y k. Then ϕ(t) = { 1+ f Xk+1 (x k+1 )[1+ f X k+2 (x k+2 )] } [ f W (x k ) f W (x k, y k )]+ { 1+ fyk+1 (y k+1 )[1+ f Y k+2 (y k+2 )] } [ f W (y k ) f W (x k, y k )]+ { 1+ fxk+1 (x k+1 )[1+ f X k+2 (x k+2 )] }{ 1+ f Yk+1 (y k+1 )[1+ f Y k+2 (y k+2 )] } f W (x k, y k )+ M. Oh the other h, let T be the tree from T by deleting four edges x k x k+1, x k+1 x k+2, y k y k+1 y k+1 y k+2 adding four edges x k y k+1, y k+1 x k+2, y k x k+1 x k+1 y k+2. Clearly, T T π ϕ(t ) = { 1+ f Yk+1 (y k+1 )[1+ f X k+2 (x k+2 )] } [ f W (x k ) f W (x k, y k )]+ { 1+ fxk+1 (x k+1 )[1+ f Y k+2 (y k+2 )] } [ f W (y k ) f W (x k, y k )]+ { (1+ fyk+1 (y k+1 )[1+ f X k+2 (x k+2 )] }{ 1+ f Xk+1 (x k+1 )[1+ f Y k+2 (y k+2 )] } f W (x k, y k )+ M. Hence ϕ(t ) ϕ(t) = ( f Yk+1 (y k+1 ) f Xk+1 (x k+1 )){[1+ f X k+2 (x k+2 )][ f W (x k ) f W (x k, y k )] [1+ f Y k+2 (y k+2 )][ f W (y k ) f W (x k, y k )]+ f w (x k, y k )( f X k+2 (x k+2 ) f Y k+2 (y k+2 ))}. Obviously, we have f W (y k ) > f W (x k, y k ) f W (x k ) > f W (x k, y k ). By Lemma 3.1, we have f W (x k )> f W (y k ). Further by Corollary 3.4, we have f X k+1 (x k+1 ) f Y k+1 (y k+1 ). Since f X k+1 (x k+1 ) = f Xk+1 (x k+1 )(1+ f X k+2 (x k+2 )) f Y k+1 (y k+1 ) = f Yk+1 (y k+1 )(1+ f Y k+2 (y k+2 )), we have f X k+2 (x k+2 ) f Y k+2 (y k+2 ) since we assumed f Xk+1 (x k+1 )< f Yk+1 (y k+1 ). Therefore, ϕ(t )>ϕ(t)>0, contradicting to the optimality of T. So the assertion holds. Lemma 3.6 Let P be a path of an optimal T int π whose end vertices are leaves. (i) If the length of P is odd (2m 1), then the vertices of P can be labeled as x m x m 1 x 1 y 1 y 2 y m such that f X1 (x 1 ) f Y1 (y 1 ) f X2 (x 2 ) f Y2 (y 2 ) f Xm (x m )= f Ym (y m )=1. (ii) If the length of P is even (2m), then the vertices of P can be labeled as x m+1 x m x m 1 x 1 y 1 y 2 y m such that f X1 (x 1 ) f Y1 (y 1 ) f X2 (x 2 ) f Y2 (y 2 ) f Xm (x m ) f Ym (y m )= f Xm+1 (x m+1 )=1. 7

8 Proof. We provide the proof of part (i), part (ii) can be shown in a similar manner. Obviously, the vertices of P may be labeled as x r x r 1 x 1 y 1 y 2 y s such that f X1 the maximum among f Xi f X j for i=1, 2,, r j=1, 2,, s, where r+s=2m. Therefore, there is only one of the following three cases: Case 1: If the number of the maximum components is one, then there exists a 1 k m such that f X1 (x 1 )> f Y1 (y 1 ), f Y1 (y 1 )= f X2 (x 2 ),, f Yk 1 (y k 1 )= f Xk (x k ), f Yk (y k )> f Xk+1 (x k+1 ) (1) Next we will prove (1). It is divided into three subcases. Case 1.1: If f Y1 (y 1 )> f X2 (x 2 ), then we have k=1 (1)holds. Case 1.2: If f Y1 (y 1 )< f X2 (x 2 ), then the vertices of P may be relabeled such thaty i is instead by x i+1 for i=1,, s X i is instead by y i 1 for i=2,, r. Hence it is the same as the subcase 1.1. Case 1.3: If f Y1 (y 1 )= f X2 (x 2 ). Then we must have f Y2 (y 2 )> f X3 (x 3 ) or f Y2 (y 2 )< f X3 (x 3 ) or f Y2 (y 2 )= f X3 (x 3 ). Case 1.3.1: If f Y2 (y 2 )> f X3 (x 3 ), then we have k=2 (1)holds. Case 1.3.2: If f Y2 (y 2 )< f X3 (x 3 ), then the vertices of P may be relabeled such thaty i is instead by x i+1 for i=1,, s X i is instead by y i 1 for i=2,, r. Hance, the case is the same as the subcase Case 1.3.3: If f Y2 (y 2 )= f X3 (x 3 ), we can continue to analyze like f Y1 (y 1 )= f X2 (x 2 ). Then we have k 3 (1)holds. Next we will prove that if (1) holds, then we must have r=s=m. Otherwise, if r < s, then by Lemma 3.5, f Xi (x i ) f Yi (y i ) for i = 1,, r. Hence by Corollary 3.4 we have f X r (x r ) f Y r (y r ). On the other h, it is clear that f X r (x r )=1 f Y r (y r ) 2, contradiction. If r>s, then r s+2 since r+s=2m. Now we consider the path from vertex x s+1 to y s. By Lemma 3.5, we have f Yi (y i ) f Xi+1 (x i+1 ) for i=1,, s. Further, by Corollary 3.4, we have f Y s (y s ) f X s+1 (x s+1 ). Similarly, since f Y s (y s )=1 f X s+1 (x s+1 ) 2, contradiction. Therefore r= s=m. Now by Lemma 3.5 applied to the path from x m to y m, we have f Xi (x i ) f Yi (y i ) for i=1,, m. On the other h, by Lemma 3.5 applied to the path from y m 1 to x m, we have f Yi (y i ) f Xi+1 (x i+1 ) for i=1, 2,..., m 1. Hence the assertion holds. Case 2: If the number of the maximum components is 2k 2. Then the path P can be 8

9 labeled as x m x m 1 x 1 y 1 y 2 y m such that f X1 (x 1 )= f Y1 (y 1 )= = f Xk (x k )= f Yk (y k )> f Xk+1 (x k+1 ) f Yk+1 (y k+1 ), (2) the vertices x 1, x 2,, x k, y 1, y 2,, y k are in the maximum components respectively. That is to say all the maximum components are adjoining. Otherwise, there must be two pair vertices satisfying the first inequality in (1). Hence either of them, the vertices of P may be labeled as x r1 x r1 1 x 1 y 1 y 2 y s1 or x r2 x r1 1 x 1 y 1 y 2 y s2. By the case 1, we can have r 1 = r 2 = s 1 = s 2 = m. But it is impossible.therefore, if there are more than one component with the most subtrees containing the vertex on the path P, then all of them must adjoin. Case 3: If the number of the maximum components is 2k+1>2. Then the path P can be labeled as x m x m 1 x 1 y 1 y 2 y m such that f X1 (x 1 )= f Y1 (y 1 )= = f Xk (x k )= f Yk (y k )= f Xk+1 (x k+1 )> f Yk+1 (y k+1 ), (3) We omits the details. Then Cases (2) or (3) can be hled in the same manner, we omit the details here. Following the conditions in Lemma 3.6, we have the following. Lemma 3.7 (i) If case (i) of Lemma 3.6 holds, then f T (x 1 ) f T (y 1 )> f T (x 2 ) f T (y 2 )> > f T (x m ) f T (y m ). Moreover, if f T (x k )= f T (y k ) for some 1 k m, then f T (x i )= f T (y i ) for i=k,, m. (ii) If case (ii) of Lemma 3.6 holds, then f T (x 1 )> f T (y 1 ) f T (x 2 )> f T (y 2 ) f T (x m )> f T (y m ) f T (x m+1 ). Moreover, if f T (y k )= f T (x k+1 ) for some 1 k m, then f T (y i )= f T (x i+1 ) for i=k, m. Proof. We only prove part (i), part (ii) is similar. For any 2 k m, let W k 1 be the connected component of T containing vertices x k 1 y k 1 after removing the edges x k 1 x k y k 1 y k. For k=1 k=m, it is easy to see f T (x 1 ) f T (y 1 )= f X1 (x 1 )(1+ f X 2 (x 2 )) f Y1 (y 1 )(1+ f Y 2 (y 2 )) f T (x m ) f T (y m )= f Xm (x m )(1+ f Wm 1 (x m 1 )) f Ym (y m )(1+ f Wm 1 (y m 1 )). 9

10 Moreover, f T (x k )= f Xk (x k )(1+ f X k+1 (x k+1 ))(1+ f Wk 1 (x k 1 )+ f Wk 1 (x k 1,, y k 1 ) f Yk (y k )(1+ f Y k+1 (y k+1 ))) (4) f T (y k )= f Yk (y k )(1+ f Y k+1 (y k+1 ))(1+ f Wk 1 (y k 1 )+ f Wk 1 (y k 1,, x k 1 ) f Xk (x k )(1+ f X k+1 (x k+1 ))). (5) By equations (4) (5), we have f T (x k ) f T (y k ) = f Xk (x k )(1+ f Wk 1 (x k 1 ))(1+ f X k+1 (x k+1 )) f Yk (y k )(1+ f Wk 1 (y k 1 ))(1+ f Y k+1 (y k+1 )). (6) Now we claim that for 1 k m 1, f X k+1 (x k+1 ) f Y k+1 (y k+1 ), (7) If there is at least one strict inequality in f Xi (x i ) f Yi (y i ) for i=1,, k, then by Lemma 3.5, (7) holds. If f Xi (x i )= f Yi (y i ) for i=1,, k there exists a k<l<msuch that f Xi (x i )= f Yi (y i ) for i = 1,, l 1 f Xl (x l ) > f Yl (y l ). Then by Lemma 3.5, we have f X l+1 (x l+1 ) f Y l+1 (y l+1 ). Moreover, f X k+1 (x k+1 )= l j l f Xi (x i )+ f X l+1 (x l+1 ) f Xi (x i ) (8) j=k+1 i=k+1 i=k+1 l j l f Y k+1 (y k+1 )= f Yi (y i )+ f Y l+1 (y l+1 ) f Yi (y i ). (9) j=k+1 i=k+1 i=k+1 By equations (8) (9), the claim holds. If f Xi (x i )= f Yi (y i ) for i=1,, m, then by equations (8) (9), we have f X k+1 (x k+1 )= f Y k+1 (y k+1 ) the claim holds. Hence (7) is proved. On the other h, by Lemma 3.1, we have f Wk 1 (x k 1 ) f Wk 1 (y k 1 ). Together with (7), we see that (6) 0. Then f T (x k ) f T (y k ). Now we prove f T (y k ) f T (x k+1 ) for any 1 k m 1. Let U k be the connected component of T containing vertex x k after removing the edges y k 1 y k (if k=1, let y 0 = x 1 ) 10

11 x k x k+1. Then f T (y k )= f Yk (y k )(1+ f Y k+1 (y k+1 ))(1+ f Uk (y k 1 )+ f Uk (y k 1,, x k ) f Xk+1 (x k+1 )(1+ f X k+2 (y k+2 ))) (10) f T (x k+1 )= f Xk+1 (x k+1 )(1+ f X k+2 (x k+2 ))(1+ f Uk (x k )+ f Uk (x k,,y k 1 ) f Yk (y k )(1+ f Y k+1 (y k+1 ))). (11) Similar to (7), we can show that f Y k+1 (y k+1 ) f X k+2 (x k+2 ). By Lemma 3.1, we have f Uk (y k 1 ) f Uk (x k ). Hence (10) (11) imply that f T (y k ) f T (x k+1 ) = f Yk (y k )(1+ f Uk (y k 1 ))(1+ f Y k+1 (y k+1 )) f Xk+1 (x k+1 )(1+ f Uk (x k ))(1+ f X k+2 (x k+2 )) = f Y k (y k )(1+ f Uk (y k 1 )) f X k+1 (x k+1 )(1+ f Uk (x k )) 0. (12) Moreover, if f T (x k )= f T (y k ) for some 1 k m, then by (6), we have f Xk (x k )= f Yk (y k ), f X k (x k )= f Y k (y k ), f Wk 1 (x k 1 )= f Wk 1 (y k 1 ). (13) Since f X k = f Xk (x k )(1+ f X k+1 (x k+1 )) f Y k = f Yk (y k )(1+ f Y k+1 (y k+1 )), we have f X k+1 (x k+1 )= f Y k+1 (y k+1 ) by (13). On the other h, since f Wk (x k )= f Xk (x k )(1+ f Wk 1 (x k 1 )+ f Wk 1 (x k 1,, y k 1 ) f Yk (y k )) f Wk (y k )= f Yk (y k )(1+ f Wk 1 (y k 1 )+ f Wk 1 (y k 1,, x k 1 ) f Xk (x k )), we have f Wk (x k )= f Wk (y k ) by (13). Hence f T (x k+1 ) = f X k+1 (x k+1 )(1+ f Wk (x k )+ f Wk (x k,, y k ) f Y k+1 (y k+1 )) = f Y k+1 (y k+1 )(1+ f Wk (y k )+ f Wk (x k,, y k ) f X k+1 (x k+1 ))= f T (y k+1 ). (14) Therefore we have f T (x i )= f T (y i ) for i=k,, m. Finally, we prove that f T (y i )> f T (x i+1 ) for i=1,, m 1. Suppose that f T (y k )= f T (x k+1 ) for some 1 k m. Then by equation (12), we have f Yk (y k )= f Xk+1 (x k+1 ) f Y k (y k )= f X k+1 (x k+1 ). Moreover, f Yk (y k )(1+ f Y k+1 (y k+1 ))= f Y k (y k )= f X k+1 (x k+1 )= f Xk+1 (x k+1 )(1+ f X k+2 (x k+2 )). 11

12 Hence f Y k+1 (y k+1 )= f X k+2 (x k+2 ). Continuing this way in an inductive manner, we have f Y m 1 (y m 1 )= f X m (x m ). But f Y m 1 (y m 1 ) 2 f X m (x m )=1, contradiction. Combining the above results, we have proved part (i). The next Lemma relates the number of subtrees to the structure of the tree. Lemma 3.8 For a path P(x m, y m )= x m x m 1... x 2 x 1 (z)y 1 y 2...y m 1 y m in an optimal tree T, if f Xi (x i ) f Yi (y i ) for i=1,, k, 1 k m 1, then d(x k ) d(y k ). Moreover, if f Xi (x i )= f Yi (y i ) for i=1,, k, 1 k m 1, then d(x k )=d(y k ). Proof. Suppose that d(x k )<d(y k ), let r=d(y k ) d(x k ) 1 y k u i Y k for i=1,, r. Further let W be the connected component of T containing vertices x k y k after removing the r edges y k u 1,, y k u r. Let X be the single vertex x k let Y be the connected component of T containing vertex y k after removing all edges incident to y k except for the r edges y k u 1,, y k u r. Since f Xi (x i ) f Yi (y i ) for i=1,, k, it is easy to see that f W (x k )> f W (y k ) f X (x k )=1<2 f Y (y k ). By Lemma 3.2, there exists another tree T T π such thatϕ(t)<ϕ(t ), contradicting to the optimality of T. Therefore the assertion holds. The case of equality is similar. From Lemmas 3.6, we have the following Lemma that decides the center of the optimal tree. Lemma 3.9 Let T be an optimal tree int π. If f T (v 0 ) = max{ f T (v), v V(T)}, then d(v 0 )=max{d(v), v V(T)}. Proof. The assertion clearly holds for small trees, so we assume that V(T) 4. Suppose that d(v 0 )<max{d(v), v V(T)}. Then there exists a vertex w such that d(v 0 )<d(w). By Theorem 9.1 in [5], f T (v) is maximized at one or two adjacent vertices of T. Thus we have f T (v 0 )> f T (v) for v V(T)\{v 0 }, or f T (v 0 )= f T (v 1 )> f T (v) for v V(T)\{v 0, v 1 } v 0 v 1 E(T). Case 1: f T (v 0 )> f T (v) for v V(T)\{v 0 }. Hence, f T (v 0 )> f T (w). It is easy to see that v 0 is not a leaf (otherwise, let u be a neighbor of v 0 we have f T (u)> f T (v 0 )). Let P be a path containing vertex v 0 w whose end vertices are leaves. Let the length of P be 2m 1 (the even length case is similar). Then by Lemma 3.6, the vertices of P can be labeled as P= x m x 1 y 1 y m such that f X1 (x 1 ) f Y1 (y 1 ) f X2 (x 2 ) f Y2 (y 2 ) f Xm (x m )= f Ym (y m )=1. 12

13 Hence by Lemma 3.7, we have f T (x 1 ) f T (y 1 ) f T (x 2 ) f T (y 2 )... f T (x m ) f T (y m ). Therefore x 1 must be v 0 w must be x k for 2 k m or y j for 1 j m. By Lemma 3.8, we have d(v 0 )=d(x 1 ) d(x k )=d(w) or d(v 0 )=d(x 1 ) d(y j )=d(w), contradiction. Hence the assertion holds. Case 2: f T (v 0 )= f T (v 1 )> f T (v) for v V(T)\{v 0, v 1 } v 0 v 1 E(T). If w=v 1, then by Lemma 3.8, we have d(w)=d(v 1 )=d(v 0 )<d(w), contradiction. Hence we assume that w v 1. First note that v 0 v 1 are not leaves. Let P be a path containing vertices v 0, v 1 w whose end vertices are leaves. Let the length of P be 2m 1 (the even case is similar), then by Lemma 3.6, the vertices of P can be labeled as P= x m x 1 y 1 y m such that f X1 (x 1 ) f Y1 (y 1 ) f X2 (x 2 ) f Y2 (y 2 ) f Xm (x m ) f Ym (y m )=1. Hence by Lemma 3.7, we have f T (x 1 ) f T (y 1 ) f T (x 2 ) f T (y 2 )... f T (x m ) f T (y m ). Therefore{x 1, y 1 } = {v 0, v 1 } w must be x k or y k for 1 < k m. By Lemma 3.8, d(v 0 ) d(w) d(v 1 ) d(w), contradiction. Combining cases (1) (2), the assertion is proved. Lemma 3.10 Let T be an optimal tree int π. If there is a path P=u l u l 1 u 1 v 0 v 1 v k with f T (v 0 )=max{ f T (v) : v V(P)}, f T (u 1 ) f T (v 1 ), l=k(or l=k+1), then f T (u 1 ) f T (v 1 ) f T (u 2 ) f T (u k ) f T (v k ) (or f T (u k+1 )) d(u 1 ) d(v 1 ) d(u 2 ) d(u k ) d(v k ) (or d(u k+1 )). Proof. Clearly, there exists a path Q that contains the path P its end vertices are leaves. We assume l=k(the l=k+1case is similar). Let the length of Q be 2m 1 (the even length case is similar). By Lemmas , The vertices of Q can be labeled as Q= x m x m 1 x 1 y 1 y m such that f T (x 1 ) f T (y 1 )> f T (x 2 ) f T (y 2 )>...> f T (x m ) f T (y m ) 13

14 d(x 1 ) d(y 1 ) d(x 2 ) d(y 2 ) d(x m )=d(y m )=1. Case 1: v 0 = x 1. We must have u 1 = y 1 v 1 = x 2. Then u i = y i v i = x i+1 for i=1,, k. Hence the assertion holds. Case 2: v 0 = x i for i>1. Then f T (v 0 ) f T (x 1 ) f T (y 1 ) f T (x i )= f T (v 0 ), which implies f T (x 1 )= f T (y 1 )= f T (v 0 ) contradicts to Theorem 9.1 in [5]. Case 3: v 0 = y i. Then i=1 f T (x 1 )= f T (y 1 )= f T (v 0 ). We must have u 1 = x 1 v 1 = y 2. Then u i = x i v i = y i+1 for i=1,, k. So the assertion holds. Now for an optimal tree T int π, let v 0 V(T) be the root of T with f T (v 0 )=max{ f T (v) : v V(T)} d(v 0 )=max{d(v) : v V(T)}. Corollary 3.11 If there is a path P=u k u 1 wv 1 v 2 v k with dist(u k, v 0 )=dist(v k, v 0 )= dist(w, v 0 )+k f T (u 1 ) f T (v 1 ), then f T (u 1 ) f T (v 1 ) f T (u 2 ) f T (u k ) f T (v k ) d(u 1 ) d(v 1 ) d(u 2 ) d(u k ) d(v k ). If there is a path P = u k+1 u 1 wv 1 v 2 v k with dist(u k+1, v 0 ) = dist(v k, v 0 )+1 = dist(w, v 0 )+k+1 f T (u 1 ) f T (v 1 ), then f T (u 1 ) f T (v 1 ) f T (u 2 ) f T (u k ) f T (v k ) f T (u k+1 ) d(u 1 ) d(v 1 ) d(u 2 ) d(u k ) d(v k ) d(u k+1 ). Proof. If w=v 0, then the assertion follows from Lemma If w v 0, then there exists a path Q containing vertices u k,, u 1, w, v 0 whose end vertices are leaves. By Lemma 3.10, we have f T (w) f T (u 1 ) f T (u k ). Similarly, there exists a path R containing vertices v k,, v 1, w, v 0 whose end vertices are leaves we have f T (w) f T (v 1 ) f T (v k ). Therefore f T (w)=max{ f T (v) : v V(P)}, the assertion follows from Lemma Proofs of Theorems Now we are ready to prove Theorems

15 Proof. of Theorem 2.3. Let T be an optimal tree int π. By Lemma 3.9, there exists a vertex v 0 such that f T (v 0 )=max{ f T (v) : v V(T)} d(v 0 )=max{d(v) : v V(T)}. Let v 0 be the root of T put V i ={v : dist(v, v 0 )=i} for i=0,, p+1 with V(T)= p+1 i=0 V i. Denote by V i = s i for i=1,, p+1. We now can relabel the vertices of V(T) by the recursion method. For V 0, relabel v 0 by v 01 as the root of tree T. The vertices of V 1 (consisting of all neighbors v 01 ) are relabeled as v 11,, v 1,s1, satisfying: f T (v 11 ) f T (v 12 ) f T (v 1,s1 ) f T (v 1i )= f T (v 1 j ) implies d(v 1i ) d(v 1 j ) for 1 i< j s 1. Generally, we assume that all vertices of V i are relabeled as{v i1,, v i,si } for i=1,, t. Now consider all vertices in V t+1. Since T is tree, it is easy to see that s 1 = d(v 01 ) s t+1 = V t+1 =d(v t1 )+ +d(v t,st ) s t. Hence for 1 r s t, all neighbors in V t+1 of v tr are relabeled as satisfy the conditions: v t+1,d(vt1 )+ +d(v t,r 1 ) (r 1)+1,, v t+1,d(vt1 )+ +d(v t,r ) r f T (v t+1,i ) f T (v t+1, j ) (15) f T (v t+1,i )= f T (v t+1, j ) implies d(v t+1,i ) d(v t+1, j ) (16) for d(v t1 )+ +d(v t,r 1 ) (r 1)+1 i< j d(v t1 )+ +d(v t,r ) r. In this way, we have relabeled all vertices of V(T)= p+1 i=0 V i. Therefore, we are able to define a well-ordering of vertices in V(T) as follows: v ik v jl, if 0 i< j p+1 or i= j 1 k<l s i. (17) We need to prove that this well-ordering is a BFS-ordering of T. In other words, T is isomorphic to T π. We first prove, for t = 0,, p + 1, the following inequalities. f T (v t1 ) f T (v t2 ) f T (v t,st ) f T (v t+1,1 ) (18) 15

16 d(v t1 ) d(v t2 ) d(v t,st ) d(v t+1,1 ). (19) For any two vertices v ti v t j with 1 i< j s t, there exists a path P=v ti v k+1,l w k v k+1,r v t j with l<r, where dist(v ti, v 01 )=dist ( v t j, v 01 )=dist(w k, v 01 )+t k. Then we have f T (v k+1,l ) f T (v k+1,r ), f T (v ti ) f T (v t j ) d(v ti ) d(v t j ) by Corollary On the other h, we consider the path Q=v t+1,1 v t1 v 11 v 01 v 1s1 v t,st. Then f T (v t,st ) f T (v t+1,1 ) d(v t,st ) d(v t+1,1 ) by Corollary Therefore (18) (19) hold for t=0,, p+1. That is f T (v 01 ) f T (v 11 ) f T (v 1,s1 ) f T (v 21 ) f T (v 2,s2 ) f T (v p+1,sp+1 ) (20) d(v 01 ) d(v 11 ) d(v 1,s1 ) d(v 21 ) d(v 2,s2 ) d(v p+1,1 ) d(v p+1,sp+1 ). (21) By (17), (20) (21), it is easy to see that this well ordering satisfies all conditions in Definition 2.1. Hence T has a BFS-ordering. Further, by Proposition 2.2 in [12], T is isomorphic to T π. So T π is the unique optimal tree int π having the largest number of subtrees. Proof. of Theorem 2.4. By proposition 2.2, without loss of generality, we assume that π=(d 0, d 1,, d i,, d j,, d n 1 ) π 1 = (d 0, d 1,, d i + 1,, d j 1,, d n 1 ) with i< j, then we haveπ π 1. Let T π be the optimal tree int π. By the proof of Theorem 2.3, the vertices of T π can be labeled as the V={v 0,, v n 1 } such that f T π (v 0 ) f T π (v 1 ) f T π (v n 1 ) d(v 0 ) d(v 1 ) d(v n 1 ), where d(v l )=d l for l=0,, n 1. Moreover, v 0 is the root of Tπ. There exists a vertex v k such that v j v k E(Tπ) with k> j. Let W be the tree achieved from Tπ by removing the subtree induced by v k. Moreover, let X be the single vertex v i Y be the subtree induced by v k with the edge v j v k added, respectively. Clearly, f T (v i )= f W (v i )+ f W (v i, v j )( f Y (v j ) 1) f T (v j )= f W (v j )+ f W (v j )( f Y (v j ) 1). Hence by f W (v i, v j )< f W (v j ) f T (v i ) f T (v j ), we have f W (v i )> f W (v j ). On the other h, let T 1 be the tree from T by deleting the edge v j v k adding the edge v i v k. Then the degree sequence of T 1 isπ 1. By Lemma 3.2, we haveϕ(t π)<ϕ(t 1 ). Henceϕ(T π)<ϕ(t 1 ) ϕ(t π 1 ). The assertion is then proved. 16

17 5 Applications of the Main Theorems In the end we use Theorems to achieve extremal graphs with the largest number of subtrees in some classes of graphs. As corollaries, we provide proofs to some results in [5], [2], etc. LetT (1) (2) n, be the set of all trees of order n with the largest degree,t n,s be the set of all trees of order n with s leaves,t n,α (3) be the set of all trees of order n with the independence numberαt (4) n,β be the set of all trees of order n with the matching numberβ. Corollary 5.1 ([5]) Let T be any tree of order n. Then n+1 2 ϕ(t) 2n 1 + n 1 with left equality if only if T is a path of order n the right equality if only if T is the star K 1,n 1. Proof. Let T be a tree of order n with degree sequenceτ. Letπ 1 = (2,, 2, 1, 1) π 2 = (n 1, 1,, 1) with n terms. Clearly the path P of order n is the only tree with the degree sequenceπ 1 the star K 1,n 1 of order n is the only tree with degree sequenceπ 2. Furthermore,π 1 τ π 2. Hence by Theorems , the assertion holds. Corollary 5.2 ([2]) There is only one optimal tree T (1) int n, with 3, where T is T π n( 2)+2 with degree sequenceπas follows: Denote p= log ( 1) 1 n ( 1)p 2 = ( 2 1)r+q for 0 q< 1. If q=0, putπ=(,,, 1,, 1) with the number ( 1)p 1 2 of degree. If q 1, putπ=(,,, q, 1,, 1) with the number ( 1)p r +r of degree Proof. For any tree T of order n with the largest degree, letπ 1 = (d 0,, d n 1 ) be the nonincreasing degree sequence of T. Assume that T has p+2 layers. Then there is a vertex in layer 0 (i.e. the root), there are vertices in layer 1, there are ( 1) vertices in layer 2,, there are ( 1) p 1 vertices in layer p, there are at most ( 1) p vertices in layer p+1. Hence 1+ + ( 1)+ + ( 1) p 1 < n 1+ + ( 1)+ + ( 1) p. Thus ( 1) p 2 2 < n ( 1)p

18 Hence n( 2)+2 p= log ( 1) 1 there exist integers r 0 q< 1such that n ( 1)p 2 = ( 1)r+q. 2 Therefore the degrees of all vertices from layer 0 to layer p 1 are there are r vertices in layer p with degree. Denote by m= ( 1)p r 1. Then there are m+1 vertices 2 with degree in T. Hence the degree sequence of T T n, isπ=(d 0,, d n 1 ) with d 0 = =d m =, d m+1 = =d n 1 = 1 for q=0; isπ=(d 0,, d n 1 ) with d 0 = =d m =, d m+1 = q, d m+2 = =d n 1 = 1 for q=1. It follows from d i that k i=0 d i k i=0 d i for k=0,, m. Further by d i= 1 d i for k=m+2,, n 1, we have k d i = 2(n 1) i=0 n 1 i=k+1 d i 2(n 1) for k=m+1, n 1. Thusπ 1 π. Hence by Theorems ,ϕ(T) ϕ(t ) with equality if only if T= T. n 1 i=k+1 d i= Remark If =3 in Corollary 5.2, then the result is precisely Theorem 2.1 in [7]. Corollary 5.3 There is only one optimal tree Ts (2) int n,s where Ts is obtained from t paths of order q+2 s t paths of order q+1 by identifying one end of the s paths. Here n 1= sq+t, 0 t< s. In other words, for any tree of order n with s leaves, with equality if only if T is T s. ϕ(t) (q+2) t + (q+1) s t+2 Proof. Let T be any tree int (2) n,s with the nonincreasing degree sequenceπ 1 = (d 0,, d n 1 ). Thus d n s 1 > 1 d n s = =d n 1 = 1. Let T π be a BFS-tree with degree sequence π=(s, 2,, 2, 1,, 1), where there are the number s of 1 s inπ. It is easy to see that π 1 π. By Theorem 2.4, the assertion holds. Corollary 5.4 There is only one optimal tree Tα (3) int n,α, where Tα is T π with degree sequenceπ=(α, 2,, 2, 1,, 1) with numbers n α 1 of 2 s αof 1 s, i.e., Tπ is obtained from the star K 1,α by adding n α 1 pendent edges to n α 1 leaves of K 1,α. In other words, for any tree of order n with the independence numberα, with equality if only if T is T α. ϕ(t) 2 2α n+1 3 n α 1 + 2n α 2 18 k i=0 d i

19 Proof. For any tree T of order n with the independence numberα, let I be an independent set of T with sizeατ=(d 0,, d n 1 ) be the degree sequence of T. If there exists a leaf u with u I, then there exists a vertex v Iwith (u, v) E(T). Hence I {u}\{v} is an independent set of T with size α. Therefore, one can always construct an independent set of T with sizeαthat contains all leaves of T. Hence there are at mostαleaves. Then d n α 1 2 τ π. By Theorems , the assertion holds. Corollary 5.5 There is only one optimal tree Tβ (4) int n,β, where T β is T π with degree sequenceπ = (n β, 2,, 2, 1,, 1). Here the number of 1 s is n β. That is, Tπ is obtained from the star K 1,n β by addingβ 1 pendent edges toβ 1 leaves of K 1,n β. In other words, for any T T (4) n,β, ϕ(t) 2 n 2β+1 3 β 1 + n β 2 with equality if only if T is Tβ. Proof. For any tree T of order n with matching numberβ, letτ=(d 0,, d n 1 ) be the degree sequence of T. Let M be a matching of T with sizeβ. Since T is connected, there are at leastβvertices in T such that their degrees are at least two. Hence d β 1 2 τ π. By Theorems , the assertion holds. References [1] C. Heuberger H. Prodinger, On α-greedy expansions of numbers, Adv. in Appl. Math., 38(2007), [2] R. Kirk H. Wang, Largest number of subtrees of trees with a given maximum degree, SIAM J. Discrete Math., 22(2008) [3] B. Knudsen, Optimal multiple parsimony alignment with affine gap cost using a phylogenetic tree, in Lecture Notes in Bioinformatics, Vol.2812, Springer-Verlag, 2003, pp [4] The On-Line Encyclopedia of Integer Sequences, A092781, com/ njas/sequences. [5] L. A. Székely H. Wang, On subtrees of trees, Advances.in Applied Mathematics, 34(2005)

20 [6] L. A. Székely H. Wang, Binary trees with the largest number of subtrees with at least one leaf, Congr. Numer., 177(2005) [7] L. A. Székely H. Wang, Binary trees with the largest number of subtrees, Discrete Applied Mathematics, 155(2007) [8] S. G. Wagner, Correlation of graph-theoretical indeces, SIAM J. Discrete Math. 21(2007) [9] H. Wang, The extremal values of the Wiener index of a tree with given degree sequence, Discrete Applied Mathematics, 156(2008) [10] W.-D.Wei, The class U(R, S ) of (0,1) matrices, Discrete Mathematics 39(1982) [11] W.-G. Yan Y.-N. Yeh, Enumeration of subtrees of trees, Theoretical Comuter Science, 369(2006) [12] X.-D. Zhang, The Laplacian spectral radii of trees with degree sequences Discrete Mathematics 308(2008) [13] X.-D. Zhang, Q.-Y. Xiang, L.-Q. Xu, R.-Y. Pan, The Wiener Index of Trees with Given Degree Sequences, MATCH Commun.Math.Comput.Chem., 60(2008) [14] X.-D. Zhang, Y. Liu M.-X. Han, Maximum Wiener Index of Trees with Given Degree Sequence, MATCH Commun.Math.Comput.Chem., 64(2010)