ARTICLE IN PRESS Theoretical Computer Science ( ) – Contents lists available at ScienceDirect Theoretical Computer Science journal homepage: www.elsevier.com/locate/tcs Separation numbers of trees Tao Jiang ∗ , Zevi Miller, Dan Pritikin Miami University, Oxford, OH 45056, USA article info Article history: Received 10 September 2008 Received in revised form 3 April 2009 Accepted 4 May 2009 Communicated by E. Pergola Keywords: Antibandwidth Bandwidth Separation Labeling abstract Let G be a graph on n vertices. Given a bijection f : V (G) → {1, 2, . . . , n}, let |f | = min{|f (u) − f (v)| : uv ∈ E (G)}. The separation number s(G) (also known as antibandwidth [T. Calamoneri, A. Massini, L. Török, I. Vrt’o, Antibandwidth of Complete kary trees, Electronic Notes in Discrete Mathematics 24 (2006), 259–266; A. Raspaud, H. Schroder, O. Sykora, L. Török, I. Vrt’o, Antibandwidth and cyclic antibandwidth of meshes and hypercubes, Discrete Mathematics 309 (2009) 3541–3552] of G is then max{|f |} over all such bijections f of G. We study the case when G is a forest, obtaining the following results. 1. Let F be a forest in which each component is a star. Then s(F ) = minimum value of %X | − |Y % over all bipartitions (X , Y ) of F . 2. Let d be the maximum √ degree of a tree T on n vertices. Then (a) s(T ) ≥ 2n − c1 nd, and (b) s(T ) ≥ 2n − c2 d2 logd n, where c1 and c2 are absolute constants. n−µ , 2 where µ is the We give constructions showing that the bound (a) is asymptotically tight when d is in 1 the range n 3 < d ≤ 1 3 n , where 0 < q < Thd 1 3 n , 12 while (b) is asymptotically tight when d is in the range nq ≤ d ≤ is any fixed constant, and when d ≥ 4 is an absolute constant. We also show that for h ≥ 3 and odd d ≥ 3, we have s(Thd ) = n 2 − Θ (d2 + dh), where is the symmetric d-ary tree of height h, improving the estimates obtained in the first of the above-mentioned references. © 2009 Elsevier B.V. All rights reserved. 1. Introduction We let [a, b] denote the set of integers x with a ≤ x ≤ b. By a labeling for a graph G on n vertices we mean a bijection f : V (G) → [1, n]. Let |f | denote min{|f (x) − f (y)| : xy ∈ E (G)}, and let s(G) = max{|f |} over all labelings. We call s(G) the separation number of G. In this paper we seek tight bounds on this parameter when G is a forest, in terms of n and the maximum degree d. The separation of G is sometimes called the antibandwidth of G since it can be viewed as dual to the well-known bandwidth B(G) of a graph G (defined as the minimum of max{|f (x) − f (y)| : xy ∈ E (G)} over all labelings f of G). Thus the study of B(G) concerns minimizing the longest ‘‘stretch’’ |f (x) − f (y)| of any edge xy under f , while the study of s(G) concerns maximizing the shortest such ‘‘stretch’’. Also s(G) is not to be confused with the vertex separation v s(G) defined as follows. For a labeling f of G let fi (G) be the number of vertices u in G for which f (u) ≤ i and there is a vertex v such that f (v) > i and uv ∈ E (G). ∗ Corresponding address: Miami University, Mathematics and Statistics, 301 S. Patterson Avenue, 45056 Oxford, OH, United States. Tel.: +1 513 529 3422; fax: +1 513 529 1493. E-mail addresses: jiangt@muohio.edu (T. Jiang), millerz@muohio.edu (Z. Miller), pritikd@muohio.edu (D. Pritikin). 0304-3975/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.tcs.2009.05.011 Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS 2 T. Jiang et al. / Theoretical Computer Science ( ) – Letting M (f ) be the maximum of fi (G), 1 ≤ i ≤ n − 1, then v s(G) is the minimum of M (f ) over all labelings f of G. Results on v s(G) and on other labeling (or ‘‘layout’’) problems for graphs can be found in the survey [2] . The separation problem was first studied in [4], where the primary concern was to study the complexity of this problem and its variants. There it was observed that the corresponding decision problem ‘‘given a graph G, is s(G) > k?’’ is NPcomplete, even for the case k = 1 (by a simple reduction from the hamiltonian path problem). The main results gave reductions of certain multiprocessor job scheduling problems to variants of the separation problem. Given in [7] are bounds for the separation of grids, (where in [8], using the term ‘‘antibandwidth’’ for separation, one of the bounds was shown to be exact) and an asymptotically optimal lower bound for !n−2 " mthe# separation of the n-dimensional hypercube Qn (refined further in [8]). Finally the exact formula s(Qn ) = 2n−1 - m=0 ⌊ m ⌋ was derived in [10]. 2 In [5] a generalization was considered, where we map a graph G into a graph H, and let s(G, H ) be the maximum, over all injections f : V (G) → V (H ), of the minimum of distH (f (x), f (y)), over all edges xy of G (where distH refers to distance in H). There the parameter s(G, H ) was studied in the case where G = Kp and H is a tree, and also where G = Kp,q and H is a hypercube. Bounds for s(G, H ) in terms of eigenvalues for certain pairs G, H were developed in [6]. In [3] s(G, H ) was studied for the case when G is a path or a power of a path and H is a two-dimensional grid, with applications to data storage. In this paper, we study s(T ) for arbitrary trees T and obtain asymptotically tight estimates of s(T ) in terms of the order n and the maximum degree d of T . Note the trivial upper bound s(G) ≤ ⌊ 2n ⌋, when G has no isolated vertices, since the vertex mapped to ⌊ 2n ⌋ + 1 has a neighbor. Thus we will derive asymptotically tight lower bounds in the form s(T ) ≥ 2n − r (n, d), for some function r of n and d. Earlier and independent of our work, Calamoneri et al. [1] studied the special case of T = Thd , where Thd is a symmetric d-ary tree of height h. They proved that s(Thd ) = n+21−d when d is even and that 2n − O(d2 h) ≤ s(Thd ) ≤ 2n − O(h) when d is odd. At the end of the last section, we will improve these estimates to show that 2n − O(d2 + dh) ≤ s(Thd ) ≤ 2n − O(d2 + dh) when d is odd. We consider only simple graphs without isolated vertices. For finite sets X , Y , we refer to ||X | − |Y || as the discrepancy of (X , Y ). Given a bipartite graph G, let the discrepancy of G, denoted by µ(G), be the minimum discrepancy value over all bipartitions (X , Y ) of G. We say that G is balanced if µ(G) = 0. A subset S of V (G) is called a balancing set for G if µ(G − S ) = 0 or 1. For a vertex v of a graph G let NG (v) denote the set of neighbors of v , and let d(v) = |NG (v)|, called the degree of v. For a subset W of V (G) we let NG (W ) denote ∪v∈W NG (v). When the context is clear, we will drop the subscript G. For graph theoretic notations not defined here see [9]. 2. Basic results and star forests We first prove a simple but useful lemma, already implicit in [7], including the proof here for completeness. Observe that in a forest F with bipartition (X , Y ) where |X | ≥ |Y |, X has a vertex of degree at most one in F . This is because the average |E (F )| |X |+|Y |−1 degree |X | among vertices in X is at most ≤ 2|X|X|−| 1 < 2. |X | Lemma 2.1. Let T be a forest with a bipartition (X , Y ) where p = |X | ≥ |Y | = q. Then one can order the vertices in X as x1 , x2 , . . . , xp and the vertices in Y as y1 , y2 , . . . , yq such that if xi yj ∈ E (T ) then j ≤ i. Proof. Suppose the claim fails for some bipartition (X , Y ) and T , and consider a failing case with q as small as possible. Clearly q > 0. By earlier discussion, some vertex x1 in X has at most one neighbor. Let y1 denote that neighbor if N (x1 ) ,= ∅, else letting y1 be any vertex in Y1 . Then T − {x1 , y1 } is a forest with bipartition (X − x1 , Y − y1 ) with |X − x1 | ≥ |Y − y1 |, so by minimality the claim holds for this bipartition. Thus one can order the vertices in X − x1 , Y − y1 as x2 , x3 , . . . , xp and y2 , y3 , . . . , yq respectively such that if xi yj ∈ E (T ) then j ≤ i. Since x1 has no neighbor other than y1 , the orderings x1 , x2 , . . . , xp and y1 , y2 , . . . , yq verify the claim for (X , Y ), completing the proof. ! For completeness, we reprove the resulting lower bound on s(T ) for forests T . Lemma 2.2 ([7]). Let T be a forest on n vertices with discrepancy µ = µ(T ). Let (X , Y ) be an arbitrary bipartition of T where |X | ≥ |Y |. Then (a) s(T ) ≥ |Y |, and n−µ (b) s(T ) ≥ 2 . Proof. (a) Suppose |X | = p, |Y | = q. By Lemma 2.1, we can name the vertices in X as x1 , . . . , xp and the vertices in Y as y1 , . . . , yq such that j ≤ i for each edge xi yj ∈ E (T ). Define labeling f : V (T ) → [1, p + q] as follows. For each i = 1, 2, . . . , q, let f (yi ) = i and f (xi ) = q + i. If p > q, assign labels in [2q + 1, p + q] to xq+1 , . . . , xp in an arbitrary way. Consider any edge uv ∈ E (T ) where u ∈ X and v ∈ Y . If u ∈ {xq+1 , . . . , xp }, then f (u) − f (v) ≥ 2q + 1 − q = q + 1. If u = xi for some i ∈ {1, . . . , q} then v = yj for some j ≤ i and f (u) − f (v) = (q + i) − j ≥ q. Thus we have |f | ≥ q, so s(T ) ≥ |Y |. (b) Let (X , Y ) be a bipartition of T with |X | − |Y | = µ. Since |X | + |Y | = n, we have |Y | = (n − µ)/2. So s(T ) ≥ |Y | = (n − µ)/2. ! Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS T. Jiang et al. / Theoretical Computer Science ( ) 3 – a b Fig. 1. (a) 2 ≤ s(T ) ≤ 3 by Corollary 2.5, bipartition indicated by shading (b) A star forest T and a bipartition realizing µ(T ) = 1. So s(T ) = 7 by Theorem 2.7. Let f be a labeling of G. Define an orientation Df (which we abbreviate by D when f is fixed) of G by orienting each edge uv ∈ E (G) from u to v if f (u) < f (v) or from v to u if f (v) < f (u). Call a vertex with in-degree 0 in D a source, a vertex with out-degree 0 in D a sink, and a vertex with both in-degree and out-degree at least one in D a level vertex. Let A = A(f ), B = B(f ), C = C (f ) denote the sets of sources, sinks, and level vertices, respectively in Df . Let dC = dC (f ) = max{d(x) : x ∈ C }. We will drop the reference to f when the context is clear. Lemma 2.3. Let G be a bipartite graph with no isolated vertex. Let f be a labeling of G. Then |f | ≤ min{|A(f )|, |B(f )|}. If C (f ) ,= ∅, n−dC (f )+1 . then |f | ≤ 2 Proof. Let A = A(f ), B = B(f ), C = C (f ). Since G has no isolated vertex, some vertex has a positive out-degree in Df ; let x be one with largest f -label. Then the n − f (x) vertices whose f -labels are larger than f (x) are sinks. Thus, n − f (x) ≤ |B|. Let y be an out-neighbor of x. Then f (x) < f (y) ≤ n. We have |f | ≤ f (y) − f (x) ≤ n − f (x) ≤ |B|. Similarly, by considering the vertex with the smallest f -label that has a positive in-degree, we have |f | ≤ |A|. It follows that |f | ≤ min{|A|, |B|}. Suppose that C ,= ∅. Let x be a level vertex with d(x) = dc (f ). Let u be an in-neighbor of x with largest f -label and v an outneighbor of x with smallest f -label. Then f (u) < f (x) < f (v). By our choice of u and v , the f (v) − f (u) − 1 vertices receiving labels in the open interval (f (u), f (v)) are non-neighbors of x. Hence, f (v)− f (u)− 1 ≤ n − d(x). So, f (v)− f (u) < n − d(x)+ 1. n−d(x)+1 Note that (f (v) − f (x)) + (f (x) − f (u)) = f (v) − f (u). Thus, we have |f | ≤ min{f (v) − f (x), f (x) − f (u)} ≤ = 2 n−dC +1 . 2 ! We now derive a general upper bound on s(G) that allows us to determine the exact value of s(G) in some cases. Let γ (G) denote the minimum cardinality of a balancing set of G. If G is already balanced, then we let γ (G) = 0. Clearly γ (G) ≤ µ(G) for any bipartite graph G. The penult degree d∗ (G) of G is defined as follows. If G has no vertex of degree larger than 1 then d∗ (G) = 1; otherwise d∗ (G) is the least vertex degree in G that is larger than 1. See Fig. 1 below for illustrations of γ (T ) and µ(T ) and how these parameters are used in bounding s(T ) in the theorems which follow. Theorem 2.4. Let G be a bipartite graph on n vertices with no isolated vertices. Then (a) s(G) ≤ (n − γ (G))/2, and n−d∗ (G)+1 n−µ(G) (b) s(G) ≤ max{ , 2 }. 2 Proof. Let f be a labeling of G with |f | = s(G), and consider the orientation D = Df . Let A = A(f ), B = B(f ), C = C (f ). Let a = |A|, b = |B|. Note that each of A and B is independent in G, so by Lemma 2.3 we have |f | ≤ min{a, b}. By symmetry, we may assume that a ≥ b. By removing the c level vertices and a − b sources, we can split the remaining vertices into two independent sets of equal sizes. Hence, γ (G) ≤ c + a − b. Since a + b + c = n, we have b ≤ (n − γ (G))/2. So, s(G) = |f | ≤ b ≤ (n − γ (G))/2, proving the first statement. For the second statement, suppose first that D has no level vertices. Then (A, B) is a bipartition of G. We have |f | ≤ n−µ(G) n−µ(G) min{a, b}. Since a + b = n and |a − b| ≥ µ(G), we have min{a, b} ≤ . Hence s(G) = |f | ≤ as desired. Suppose 2 2 ∗ n−dC +1 ≤ n−d 2(G)+1 , as desired. ! instead that C ,= ∅. Note that dC ≥ d∗ . By Lemma 2.3, s(G) = |f | ≤ 2 Lemma 2.2 and Theorem 2.4 immediately yield the following. Corollary 2.5. Let T be a forest on n vertices. Then n−µ(G) 2 ≤ s(T ) ≤ n−γ (G) . 2 Corollary 2.6. Let T be a forest on n vertices. If d∗ (T ) ≥ µ(T ) + 1, then s(T ) = n−µ(T ) . 2 n−µ(T ) . In general, however, µ(T ) and γ (T ) can differ If T is a forest with µ(T ) = γ (T ) then Corollary 2.5 yields s(T ) = 2 drastically. In such cases, Corollary 2.6 could be useful. For instance, if T is a star with m leaves, then µ(T ) = m − 1 while T) = 1. It is natural to γ (T ) ≤ 2. Also, d∗ (T ) = m = µ(T ) + 1. So by Corollary 2.6 and n = m + 1, we have s(T ) = n−µ( 2 n−µ(T ) ask whether s(T ) = is still valid when T is a star forest, i.e., a vertex-disjoint union of stars. Neither Corollary 2.5 nor 2 Corollary 2.6 gives a definite answer. In the next theorem we prove that this equality indeed holds for star forests. Theorem 2.7. Let T be a star forest. Let µ = µ(T ). Then s(T ) = Proof. By Lemma 2.2, it remains to show that s(T ) ≤ n−µ . 2 n−µ . 2 n−µ . 2 Let f be a labeling of T with |f | = s(T ). We need to show that Consider D = Df . If there is a level vertex with degree at least µ + 1, then by Lemma 2.3, |f | ≤ |f | ≤ and we are done. n−dC +1 2 ≤ Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 n−µ 2 ARTICLE IN PRESS 4 T. Jiang et al. / Theoretical Computer Science ( ) – a b Fig. 2. (a) The labeling of A′ ∪ B′ given in Theorem 3.1 (b) The labeling of T using the labeling of A′ ∪ B′ . Thus we may assume that every level vertex has degree at most µ. Note that each level vertex has degree at least 2 and is the center of a star component of T . Let T0 be the subforest of T obtained by removing each star component that has a level vertex at the center. Let F1 , F2 , . . . , Fp denote the star components removed. For each i, let li denote the number of leaves in p Fi ; we have łi ≤ µ by our earlier assumption. Let m = Σi=1 li − p. Claim 1. µ(T0 ) ≥ µ + m. Proof of Claim 1. Suppose that µ0 = µ(T0 ) ≤ µ − 1 + m. We derive a contradiction by showing that we can obtain a bipartition of T with discrepancy at most µ− 1. Let (X , Y ) be a bipartition of T0 with discrepancy µ0 , where |X | ≥ |Y |. Let q be q the largest integer such that |X |+ q ≥ |Y |+ Σi=1 li . Let X ′ be the set containing X and the centers of F1 , . . . , Fq and Y ′ be the set containing Y and the leaves of F1 , . . . , Fq . Let T ′ = T0 ∪ F1 ∪· · ·∪ Fq . Then (X ′ , Y ′ ) is a bipartition of T ′ , and by the definition of p q, |X ′ | ≥ |Y ′ |. Suppose first that q = p. In this case, T ′ = T . We have |X ′ |−|Y ′ | = (|X |+ p)−(|Y |+ Σi=1 li ) = |X |−|Y |− m = ′ ′ µ0 − m ≤ µ − 1. So (X , Y ) is a bipartition of T with discrepancy at most µ − 1, a contradiction. Hence, we may assume that q+1 q < p. By our choice of q, |X |+ q + 1 < |Y |+ Σi=1 li . That is, |X ′ |+ 1 < |Y ′ |+ lq+1 ≤ |Y ′ |+µ. Thus, (X ′ , Y ′ ) is a bipartition of T ′ with discrepancy at most µ − 1. Now, one by one we add Fq+1 , Fq+2 , . . . , Fp to T ′ , always placing the center of an added star component in the larger part and leaves in the smaller part of the current bipartition. It is easy to see that in the end we obtain a bipartition of T with discrepancy at most µ− 1, a contradiction. Thus µ0 = µ(T0 ) ≥ µ+ m, completing the proof of Claim 1. Let A = A(f ), B = B(f ). Claim 2. min{|A|, |B|} ≤ n−µ . 2 Proof of Claim 2. By our assumption, each vertex in T0 is either a source or sink in D. Let A0 = V (T0 )∩ A and B0 = V (T0 )∩ B. Then (A0 , B0 ) is a bipartition of T0 . By Claim 1, ||A0 | − |B0 || ≥ µ + m. For each i ∈ {1, . . . , p}, the center of Fi is a level vertex while its li leaves are in A ∪ B with at least one in each of A and B. Hence |V (Fi ) ∩ A| and |V (Fi ) ∩ B| differ by at most li − 1. p n−µ Hence, ||A| − |B|| ≥ ||A0 | − |B0 || − Σi=1 (li − 1) ≥ (µ + m) − m = µ. Since |A| + |B| ≤ n, we have min{|A|, |B|} ≤ 2 , completing the proof of Claim 2. By Lemma 2.3, s(G) = |f | ≤ min{|A|, |B|} ≤ n−µ , 2 as required. ! We note that Theorem 2.7 implies that the separation problem restricted to star forests is already NP-hard, by a reduction from the PARTITION problem. 3. A good measure of separation in trees In this short section, we establish a connection between the separation number of a tree and a parameter involving independent sets of T . Theorem 3.1. Let W be an independent set in a tree T . Let (A, B) be a bipartition of T − W with |A| ≤ |B|. Then s(T ) ≥ |A| − |N (W )|. Proof. Let A′ = A − N (W ) and B′ = B − N (W ). The forest F induced in T by A′ ∪ B′ has (A′ , B′ ) as a bipartition. Let m = min{|A′ |, |B′ |}. In particular, m ≥ |A′ | ≥ |A| − |N (W )|. By Lemma 2.2 and its proof, F has a labeling g with |g | ≥ m in which without loss of generality all of A′ appears before B′ . Take the linear ordering of V (F ) associated with g, insert W between A′ and B′ , insert A ∩ N (W ) before A′ , and B ∩ N (W ) after B′ , where within each of A ∩ N (W ), W and B ∩ N (W ) the ordering is arbitrary. Let f be the resulting labeling of T . Since m ≥ |A| − |N (W )|, it suffices to show that |f | ≥ m. See Fig. 2 below for an illustration of f . Let uv be an edge in T with f (v) − f (u) = |f |. We are done if either all of A′ or all of B′ lies between u and v in f , since then f (v) − f (u) ≥ min{|A′ |, |B′ |} = m. It is easy to see, from the definition of f and the independence of A, W , B, that the only remaining case is when u ∈ A′ and v ∈ B′ . But then f (v) − f (u) ≥ g (v) − g (u) ≥ m, completing the proof. ! For each independent set W in T , let ϕ(W ) = 21 µ(T − W )+ 12 |W |+|N (W )|. Let ϕ(T ) = min{ϕ(W ) : W is an independent set of T }. The next two theorems show that ϕ(T ) provides a good measure of how far s(T ) is from the trivial upper bound 2n . As a result, we can get good bounds on s(T ) by finding good bounds on ϕ(T ). Fig. 3 below illustrates the function ϕ and its use in the theorem which follows. Theorem 3.2. Let T be an n-vertex tree. Then s(T ) ≥ n 2 − ϕ(T ). Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS T. Jiang et al. / Theoretical Computer Science ( ) 5 – Fig. 3. ϕ(T ) = 2, so s(T ) ≥ 5 − 2 = 3 by Theorem 3.2. Proof. Let W be independent in T with ϕ(W ) = ϕ(T ). Let (A, B) be a bipartition of T − W , with |A| ≤ |B| and |B| − |A| = µ(T − W ) = µ. We have |A| ≥ n−|W2 |−µ . By Theorem 3.1, s(T ) ≥ |A| − |N (W )| ≥ 2n − µ2 − |W2 | − |N (W )| = n − ϕ(W ) = 2n − ϕ(T ). ! 2 Theorem 3.3. Let T be an n-vertex tree. Then s(T ) ≤ ⌊ 2n ⌋ + 1 − ⌊ ϕ(T ) ϕ(T ) 5 ⌋≤ n 2 +2− ϕ(T ) 5 . n 2 Proof. Let m = ⌊ 5 ⌋. Suppose s(T ) ≥ ⌊ ⌋ + 2 − m. We derive a contradiction by finding an independent set W with ϕ(W ) < 5m ≤ ϕ(T ). Let f be an optimal labeling of T , so that |f | ≥ ⌊ 2n ⌋ + 2 − m. Let A denote the set of vertices receiving the first ⌊ 2n ⌋ − m labels, W the set of vertices receiving the next 2m labels, and C the set of vertices receiving the last ⌈ 2n ⌉ − m labels. Since |f | ≥ ⌊ 2n ⌋ + 2 − m, each of A, W , B induces an independent set. This also implies that µ(T − W ) = 0 or 1. Since a vertex in W has f -label at most ⌊ 2n ⌋ + m and |f | ≥ ⌊ 2n ⌋ + 2 − m, vertices in N (W ) ∩ A must receive labels in the interval [1, 2m − 2]. So, |N (W ) ∩ A| ≤ 2m − 1. Similarly, one can show that |N (W ) ∩ B| ≤ 2m − 1. Now, ϕ(W ) = 21 µ(T − W ) + 12 |W | + |N (W )| ≤ 21 + 12 (2m) + 4m − 2 < 5m, a contradiction. This completes the proof. ! For the rest of the paper, we develop bounds on s(T ) by bounding ϕ(T ). For the most part, we will be focusing on finding the correct order of magnitude of ϕ(T ) in terms of the order n of T and the maximum degree d of T . 4. Separation for trees of maximum degree d; lower bounds In this section, we derive lower bounds on the separation for trees T with maximum degree d, and in the next section n we show that these bounds are asymptotically tight when d is an absolute constant and when nq < d < 12 for any fixed constant q ∈ (0, 1), where n = |V (T )|. By Theorem 3.2, to find a good lower bound on s(T ), it suffices to find a good upper bound on ϕ(T ). We accomplish this in two stages. In the first stage we use a variant of the usual inorder numbering of trees to first find a set M for which µ(T − M ) and |M | are small. In the second stage, we use this set M to carefully construct our independent set W with small ϕ(W ). Before we introduce our numbering algorithm, we need some notation. Let T be a tree rooted at r. For each vertex v in T let Tv denote the subtree of T rooted at v and let n(v) denote |V (Tv )|. For v ∈ V (T ) − {r } let v − denote the parent of v , i.e., the neighbor of v on the r , v -path in T . A neighbor of v other than v − is a child of v . For any child x of v , we call Tx a branch below v. Order the children v1 , v2 , . . . , vc of each vertex v so that n(v1 ) ≤ n(v2 ) ≤ · · · ≤ n(vc ). We now number the vertices of T from 1 to n as follows: we proceed recursively by traversing the lightest branch below r, then r, then the remaining branches below r in nondecreasing order of size, provided there are at least two branches below r. If there is only one such branch below r, then r is traversed first, and then the branch below r. As the tree is traversed, the labels 1 through n = |V (T )| are assigned to the vertices in the order visited. Below is the formal algorithm. See Fig. 4 for an illustration of the labeling. Procedure Inorder(T , v): Input: a tree T rooted at v Output: a vertex labeling l : V (T ) → [1, n] (similar to the usual inorder numbering), n = |V (T )| 1. Let v1 , v2 , . . . , vc be the children of v in nondecreasing order of branch size n(v1 ) ≤ n(v2 ) ≤ · · · ≤ n(vc ) 2. If c = 0 (i.e., v is a leaf) then l(v) ← least integer from [1, n] not already assigned as a label If c = 1 then l(v) ← least integer from [1, n] not already assigned as a label Apply Inorder(Tv1 , v1 ) If c ≥ 2 then Apply Inorder(Tv1 , v1 ) l(v) ← least integer from [1, n] not already assigned as a label For i = 2 to c, apply Inorder(Tvi , vi ) 3. If all labels in [1, n] have been assigned, then halt. To analyze this labeling, we use the following notation. Let the two partite sets of T be R and B (red and blue). For each i with 1 ≤ i ≤ n(T ), let C (i) be the set of vertices labeled 1 through i. Let C (0) = ∅. Let U (i) = V (T ) − C (i). Let M (i) consist of those vertices in C (i) having at least one neighbor in U (i), and let L(i) = C (i) − M (i). We call C (i), L(i), M (i) and U (i) the Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS 6 T. Jiang et al. / Theoretical Computer Science ( ) – Fig. 4. Inorder labeling of a tree; (a) order in which objects in Tx are numbered in procedure Inorder(Tx , x), (b) fully labeled(shaded), mixed (dark), unlabeled(open) vertices after 7 vertices are labeled. labeled, fully labeled, mixed, and unlabeled vertices of T (respectively) at the ith step of the procedure. We drop the index i when the context is clear. See Fig. 4 for an illustration. For each subset S of V (T ), let µ∗ (S ) = |S ∩ R| − |S ∩ B|. For each i = 1, . . . , n, we have |µ∗ (C (i)) − µ∗ (C (i − 1))| = 1. µ∗ (V (T )) ⌋. For the rest of the paper we fix i to be this value, and Since µ∗ (C (0)) = 0, there exists an i such that µ∗ (C (i)) = ⌊ 2 let L = L(i), U = U (i), M = M (i) and still refer to these sets as fully labeled, unlabeled, and mixed vertices. We will now analyze the structure of L, M , and U. Let M1 be the set of mixed vertices having at least one unlabeled child and M2 the set of mixed vertices v for which the parent v − of v is the only unlabeled neighbor of v . Then M = M1 ∪ M2 . Let x1 , x2 , . . . , xp be the mixed vertices, in nondecreasing order of distance from the root r. See Fig. 5a below for an illustration of the mixed vertices and the sets M1 and M2 . Lemma 4.1. (a) There is no edge xy in T − M with x ∈ L and y ∈ U. (b) Let y be any vertex in T . Then at most one branch under y can contain a mixed vertex. (c) All the mixed vertices lie on a path P from the root r to xp . (d) For i = 1, . . . , p − 1, if xi ∈ M2 then xi+1 ∈ M1 . Proof. (a) This is clear from the definitions of L and U. (b) Suppose otherwise that y1 , y2 are children of y such that Ty1 contains a mixed vertex xr and Ty2 contains a mixed vertex xs . Assume also that y1 appears before y2 in the ordering of the children of y. In our algorithm Inorder, by the time we label vertices in Ty2 , all of V (Ty1 ) and y should have been labeled. So, xr is already fully labeled, a contradiction. (c) This follows from part (b) immediately. (d) Let P be the path from r to xp given in (c). For each vertex v ∈ V (P ) − xp , let v + denote its child on P. Suppose that − + − xi , xi+1 ∈ M2 . Then x+ i ∈ L and xi+1 ∈ U. As we move along P from xi to xi+1 we must encounter a vertex z such that z ∈ L + and z ∈ U. Such z would be a mixed vertex, contradicting xi and xi+1 being consecutive mixed vertices on P. ! We have completed the first stage. We now go to the second stage of constructing our independent set W . As we traverse the path P of Lemma 4.1(c) from r to xp , we encounter the mixed vertices in the order x1 , . . . , xp . Select a distinguished integer q, 1 ≤ q ≤ p, based on some criteria to be described later. We build W as a disjoint union of independent sets W1 and W2 , where W1 is derived from the segment x1 , x2 , . . . , xq of mixed vertices while W2 is derived from the segment xq+1 , xq+2 , . . . , xp . To get W1 , start with the initialization W1 = {xq }. Now we treat the xi , i ≤ q, in decreasing order of i, by greedily including xi in W1 if doing so preserves the independence of the current W1 . Otherwise, we instead include in W1 the unlabeled neighbors of xi . Recall that for any vertex v on the path P, v − denotes the parent of v in T , and as in the proof of Lemma 4.1(d), we let v + denote the child of v on P. Then from this description we see that if we do not include xi in W1 , then it must be due to x+ i already being included in W1 when we processed xi+1 earlier. Below is the formal algorithm. 1. (Initialization) W1 = {xq }, i ← q − 1. 2. If i ≥ 1, then (a) If x+ / W1 , then W1 ← W1 ∪ {xi }. i ∈ (b) If x+ i ∈ W1 , then W1 ← W1 ∪ {unlabeled neighbors of xi }. (c) i ← i − 1. 3. If i = 0, halt. Otherwise, go to step 2. We construct W2 similarly, except that we treat the xi in increasing order of i. 1. (Initialization) W2 = ∅, i ← q + 1. 2. If i ≤ p, then (a) If x− / W2 , then W2 ← W2 ∪ {xi }. i ∈ (b) If x− ∈ W2 , then W2 ← W2 ∪ {unlabeled neighbors of xi }. i (c) i ← i + 1. 3. If i = p + 1, halt. Otherwise, go to step 2. Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS T. Jiang et al. / Theoretical Computer Science ( a ) 7 – b Fig. 5. (a) The path P containing the mixed vertices xi and surrounding branches, labeled vertices shaded (b) A segment of the main path near xq and construction of W1 and W2 , dark rings indicating vertices added to W1 ∪ W2 . Finally let W = W1 ∪ W2 . Thus, W is obtained from M by retaining certain (possibly all) vertices in M, and replacing the remaining vertices in M by their unlabeled neighbors. Fig. 5b below illustrates the sets W1 , W2 , and W . Lemma 4.2. (a) There is no edge xy in T − W with x ∈ L ∪ (M − W ) and y ∈ U − W . (b) We have µ(T − W ) ≤ |W | + 1 and s(T ) ≥ 2n − ϕ(W ) ≥ 2n − 12 − |W ∪ N (W )|. Proof. (a) Suppose such an edge xy exists. By Lemma 4.1 (a), there is no edge between L and U. So we must have x ∈ M − W . But by our construction of W , all unlabeled neighbors of x are included in W , and so x has no neighbor in U − W , contradicting y ∈ U − W. (b) Recall that for each subset S of V (T ), µ∗ (S ) = |S ∩ R| − |S ∩ B|. By our choice of i, we have µ∗ (C (i)) = ⌊µ∗ (V (T ))/2⌋, where C (i) = L ∪ M. Since V (T ) = L ∪ M ∪ U, it follows that µ∗ (L ∪ M ) and µ∗ (U ) differ by at most 1. Thus µ∗ ((L ∪ M ) − W ) and µ∗ (U − W ) differ by at most |W | + 1. By (a) there is no edge in T − W between (L ∪ M ) − W and U − W . Thus, by switching the red vertices with the blue vertices in U − W if necessary, we obtain a bipartition of T − W with discrepancy at most |W | + 1. Now, ϕ(T ) ≤ ϕ(W ) ≤ 21 (|W | + 1) + 21 |W | + |N (W )| ≤ 21 + |W ∪ N (W )|. By Theorem 3.2, we have s(T ) ≥ n 2 − ϕ(T ) ≥ n 2 − 1 2 − |W ∪ N (W )|. ! Next we establish an upper bound on |W ∪ N (W )|, which together with Lemma 4.2 will give us lower bounds on s(T ). For each xi ∈ M, let di be the number of unlabeled children of xi , and let fi = di + 1. Note that if xi ∈ M2 , then di = 0, and so fi = 1. Lemma 4.3. (a) For each 1 ≤ i ≤ p − 1, we have n(xi ) ≥ fi · n(xi+1 ). Thus, n ≥ f1 · f2 · · · fp . (b) n ≥ (f1 · f2 · · · fq )(|W2 ∪ N (W2 )| − 1). Proof. (a) If xi ∈ M2 then fi = 1, so the claim follows trivially from Txi+1 ⊆ Txi . Suppose then that xi ∈ M1 . We consider two cases. Case 1: x+ i is labeled. Recall that the procedure Inorder (T , xi ) labels the branches below xi in the order of nondecreasing size, and all vertices of a branch must be labeled before the labeling of the next branch can begin. Since the branch Tx+ contains mixed vertices (e.g. i xi+1 ), the labeling of Tx+ has not been completed. Hence for each of the di unlabeled children y of xi we have |Tx+ | ≤ |Ty |. It i i follows that n(xi ) ≥ (di + 1)n(x+ i ) ≥ (di + 1)n(xi+1 ) = fi · n(xi+1 ). Case 2: x+ i is unlabeled. + Note that some branch B under x+ i contains mixed vertices (e.g. xi+1 ). Since some vertex in B has been labeled but xi has + + ′ not, B must be the lightest branch under xi , and xi has at least one other branch B in which the labeling has not started. By ′ our design, |B′ | ≥ |B|. Hence, n(x+ i ) ≥ |B| + |B | ≥ 2|B| ≥ 2n(xi+1 ). Similarly, the fact that Tx+ contains mixed vertices says i that we have not finished labeling Tx+ , which implies that each of the branches under the other di − 1 unlabeled children of xi is at least as heavy as Tx+ . So, i i n(xi ) ≥ di n(x+ i ) ≥ 2n(xi+1 )di ≥ n(xi+1 )(di + 1) = n(xi+1 )fi . It follows by induction that n ≥ f1 · f2 · · · fp . (b) Note Txq+1 ⊇ (W2 ∪ N (W2 )) − x− q+1 . The claim follows from (a) by induction. ! Next we develop an optimization lemma which we will use in conjunction with Lemma 4.3 to bound |W ∪ N (W )|. For simplicity, we will be somewhat generous in our estimates. Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS 8 T. Jiang et al. / Theoretical Computer Science ( ) – Lemma 4.4. Let D, N be real numbers D ≤ N. Let t be an arbitrary positive integer. Let y1 , y2 , . . . , yt be real $t such that 2 ≤ ! t numbers satisfying 2 ≤ yi ≤ D and i=1 yi ≤ N. Then i=1 yi ≤ 2D logD N. !m Proof. Among all multisets satisfying the constraints, pick {y1 , . . . , ym } such that i yi is maximum and m is a minimum subject to this condition. It suffices to prove the claims for this multiset. Suppose for some i ,= j that 2 < yi ≤ yj < D. If yy yy yi yj ≤ 2D then replace yi , yj by 2, i2 j , and otherwise replace yi , yj by D, Di j . Either way, one can check that the new multiset ! satisfies the constraints, but yi will be larger than before, a contradiction. So, we may assume that among the yi values at most one is strictly between 2 and D. ! $m yi ≤ D log N ≤ 2D logD N. Suppose first that D < 4. Since each yi ≥ 2 we have 2m ≤ i yi ≤ N. So, m ≤ log N. Thus, Next, suppose D ≥ 4. If there are two 2’s among the yi ’s, then replacing two by a single 4 yields a smaller multiset with the same product and sum as before, contradicting our choice of {y1 , . . . , ym }$ . So at most one yi equals 2. By our earlier m discussion, at most one yi is strictly between 2 and D. Thus, we have 4Dm−2 ≤ i yi ≤ N. Thus, m − 2 ≤ logD N4 ≤ logD N ! 2 and m ≤ logD N + 2. If N ≥ D or m ≤ 2, then we have m ≤ 2 logD N and yi ≤ 2D logD N, and we are done. So, we $m−1 may assume N < D2 and m ≥ 3. By our discussion, at least one yi , say ym , is D. Now, we have i=1 yi ≤ ND < D and !m−1 i=1 yi ≤ $m−1 !m yi < D. Thus, i=1 i=1 yi ≤ D + D ≤ 2D logD N. ! Lemma 4.3 yields the following bound on |W | independent of our choice of q. Lemma 4.5. Suppose ∆(T ) = d, where d ≥ 1. Then |W | ≤ 4(d + 1) logd+1 n. Proof. For each xi ∈ M1 , we! have fi = di + 1 ≥ 2 and fi ≤ d + 1. Also, by Lemma 4.3, Πxi ∈M1 fi ≤ n. By Lemma 4.4, with D = d + 1, N = n, we have xi ∈M1 fi ≤ 2(d + 1) logd+1 n. By Lemma 4.1 (d), |M2 | ≤ |M1 |+ 1. Since for each xi ∈ M2 , fi = 1, we have ! xi ∈M2 fi = |M2 | ≤ |M1 |+ 1 ≤ 1 + 12 ! xi ∈M1 fi . So, xi ∈M fi ≤ 32 ( xi ∈M1 fi ) + 1 ≤ 4(d + 1) logd+1 n. Finally, note that in forming W we include for each i either xi or its ! unlabeled neighbors, so |W | ≤ xi ∈M fi ≤ 4(d + 1) logd+1 n. ! ! ! We now define the distinguished integer q on which the above construction of W1 and W2 was based. Recall that di is the number of unlabeled children of the mixed vertex xi , and fi = 1 if xi ∈ M2 while fi = di + 1 if xi ∈ M1 . Setting k = nd , we let ' & √ fi ≥ k or j = p . q = min j : dj ≥ ⌊ k⌋ or % i≤j We now estimate |W ∪ N (W )|. Lemma 4.6. Suppose ∆(T ) = d, where d ≥ 1 and k = √ (a) |W1 | ≤ 1 + f1 + f2 + · · · + fq−1 ≤ 8 k + 1. n d ≥ 4. We have √ (b) |W2 ∪ N (W2 )| ≤ (1 + o(1)) √n = (1 + o(1)) nd. k (c) |W ∪ N (W )| ≤ (9 + o(1)) √n k √ = (17 + o(1)) nd. (d) |W ∪ N (W )| ≤ 4(d + 1)2 logd+1 n. Proof. (a) In the first inequality the term 1 accounts for xq , while the summand fi is at least as large as the number of unlabeled neighbors of xi . Hence the first inequality follows.√ Consider now the second inequality. Let X = {i ≤ q − 1 : xi ∈ $q−1 M1 } and Y = {i ≤ q − 1 : xi ∈ M2 }. By our choice of q, di ≤ ⌊ k⌋ − 1 for each 1 ≤ i ≤ q − 1 and i=1 fi ≤ k. Hence for each √ √ √ ! ≤ k. By Lemma 4.4, with D = k and N = k, we have i∈X fi ≤ 2 k log√k k = 4 k. Now √ √ ! !q−1 since |Y | ≤ |X |+ 1, by Lemma 4.1(d) and fi = 1 for i ∈ Y , we get i∈Y fi = |Y | ≤ 4 k + 1. It follows that i=1 fi ≤ 8 k + 1, i ∈ X , 2 ≤ fi ≤ proving (a). √ k and $ i∈X fi (b) We consider cases, based on the defining property of q. Case 1: q = p. Since |W2 | = 0 in this case, (b) follows trivially. √ Case 2: dq ≥ k. Recalling again that the branches of xq are labeled in nondecreasing order of size and that the branch Tx+ contains mixed q vertices (e.g., xq+1 ), we find as before that |Tx+ | ≤ |Ty | for any unlabeled child y of xq . As there are dq unlabeled such children, q we get √ n ≥ (dq + 1) · |Tx+ | ≥ (⌊ k⌋ + 1) · |Tx+ | ≥ q q √ k · |Tx+ |. q Since all but at most one vertex of W2 ∪ N (W2 ) are contained in Tx+ , this vertex being xq in case xq+1 = x+ q , we have |W2 ∪ N (W2 )| ≤ |Tx+q | + 1 ≤ √n k q + 1 and the claim follows. Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS T. Jiang et al. / Theoretical Computer Science ( ) ≥ k. Applying Lemma 4.3 (b) we get n ≥ (|W2 ∪ N (W2 )| − 1) |W2 ∪ N (W2 )| ≤ 1 + nk ≤ (1 + o(1)) √nk , as required. Case 3: 9 – $q i=1 fi $q i=1 f i ≥ k(|W2 ∪ N (W2 )| − 1). Thus (c) Applying (a) and (b) of this lemma, we have |W ∪ N (W )| ≤ |W1 ∪ N (W1 )|+|W2 ∪ N (W2 )| ≤ |W1 |+ d|W1 |+(1 + o(1)) √n ≤ √ k √ 8 k + 1 + nk (8 k + 1) + (1 + o(1)) √n ≤ (17 + o(1)) √n . k k (d) By Lemma 4.5, |W | ≤ 4(d + 1) logd+1 n. Thus, we have |N (W ) ∪ W | ≤ (d + 1)|W | ≤ 4(d + 1)2 logd+1 n. ! Lemmas 4.6 and 4.2 together imply our main result below. Theorem 4.7. Let T be an n-vertex tree with maximum degree d, where 1 ≤ d ≤ (a) s(T ) ≥ (b) s(T ) ≥ n 2 n 2 n . 4 We have √ √ − (17 + o(1))( nd) = 2n − Ω ( nd), and − 4(d + 1)2 logd+1 n = 2n − Ω (d2 logd n). 5. Extremal tree constructions with maximum degree d √ In this section we show that the lower bounds of Theorem 4.7 are best possible, up to constant factors in the Ω ( nd) and√Ω (d2 logd n) terms, for suitable ranges on the maximum degree d. Toward that goal, we construct trees T with ϕ(T ) = Ω ( nd) or Ω (d2 logd n). Using Theorem 3.2, we then get upper bounds for s(T ) which asymptotically match the lower bounds of Theorem 4.7. Since we are only interested in asymptotics, we will be very generous with constant factors. We start with a lemma that will be useful when we analyze discrepancy. Lemma 5.1. Let T be a rooted tree. Let F be a subgraph of T and v a vertex in F . Let F1 , F2 , . . . , Fp be the components of F − v !p that contain a child of v in T . Then µ(F ) − µ(F − v) ≤ 1 + i=1 2µ(Fi ). Also, µ(F ) − µ(F − v) ≤ 2|V (Tv )|, where Tv is the subtree of T rooted at v . Proof. Each Fi , being a tree, has a unique bipartition. Take a bipartition (A, B) of F − v with discrepancy µ(F − v) and color vertices in A red and vertices in B blue. If the parent of v is in F we may assume that it is colored blue. Now, color v red. Note that (A ∪ v, B) may not be a bipartition of F since v may have red neighbors (which can only be children of v ). To amend this, we switch red vertices with blue vertices in each Fi that contains a red neighbor of v ; such a switch changes the overall #red vertices − #blue vertices count in an Fi by at most 2µ(Fi ). The final red and blue sets form a bipartition of F with !such p discrepancy at most µ(F − v) + 1 + i=1 2µ(Fi ). !p For the second statement, note that 1 + i=1 2µ(Fi ) ≤ 2|V (Tv )|. ! The next lemma will be used to extend constructions that work only for specific values of n and d to all values of n, and d in a certain range in terms of n. Lemma 5.2. Let T ∗ be a tree and T be a tree obtained from T ∗ by attaching a path P to T ∗ at a vertex v ∈ V (T ∗ ) . Then ϕ(T ) ≥ ϕ(T ∗ ) − 21 . Proof. Let W be an independent set of T with ϕ(W ) = ϕ(T ), where ϕ(W ) is defined relative to T . Let W1 = W ∩ V (T ∗ ) and W2 = W ∩ (V (P − v)). Then W1 and W2 partition W . Observe that in any bipartition (A, B) of P − v − W2 , we have ||A| − |B|| ≤ |W2 | + 1. Hence, µ(T − W ) ≥ µ(T ∗ − W1 ) − |W2 | − 1. Now, we have ϕ(T ) = ϕ(W ) = + ≥ 1 2 1 2 1 2 1 1 2 2 µ(T − W ) + |W | + |NT (W )| ≥ (|W1 | + |W2 |) + |NT (W )| = ∗ 1 2 [µ(T ∗ − W1 ) − |W2 | − 1] 1 µ(T ∗ − W1 ) + |W1 | + |NT (W )| − 2 1 1 2 2 µ(T − W1 ) + |W1 | + |NT ∗ (W1 )| − ∗ 1 2 1 ≥ ϕ(T ) − . ! 2 Now we are ready for our constructions. We will first construct trees for specific pairs n and d. Then we will use Lemma 5.2 to extend our constructions. We need some further notation. Let T be a tree with root r. For each integer i ≥ 0, let Li denote the set of vertices at distance i from r. Given a list of positive integers d1 , d2 , . . . , dh . Let T (d1 , d2 , . . . , dh ) be the (h + 1)-level tree rooted at r in which for each i = 1, . . . , h each vertex in Li−1 has di children in Li . When d1 = d2 = · · · = dh = d, we use Thd to denote T (d1 , . . . , dh ), commonly known as a symmetric d-ary tree of height h. Let n = |V (Thd )|. A calculation 2 ) when h is odd and µ(Thd ) = (n − 1)(1 − d+2 1 ) + 1 when h is even. So, always n(1 − d+2 1 ) ≤ shows that µ(Thd ) = n(1 − d+ 1 µ(Thd ) ≤ n(1 − 2 d+1 ) + 1. The following lemma, in combination with Theorem 3.3, will be used to show that the lower bound in Theorem 4.7(a) is √ 1 best possible, up to the constant factor implied in the Ω ( nd) term, when d is in the range n 3 ≤ d ≤ n . 12 Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS 10 T. Jiang et al. / Theoretical Computer Science ( ) – Lemma 5.3. Let m, d ≥ 3 be odd positive integers. Let T = T (m, d, m). (a) Let W be an independent set in T . If µ(T − W ) ≤ (b) ϕ(T ) ≥ 1 md. 12 1 md, 3 then |W ( 1 md. 6 N (W )| ≥ Proof. (a) Let r denote the root. Suppose the vertices of W are x1 , . . . , xp in nondecreasing order of level. Initially let F = T . Then we remove x1 , . . . , xp in order from F ; updating F and µ(F ) at each step. We consider two cases. Case 1: x1 = r. Note that after one step, F = T − x1 = T − r, which consists of m copies of T (d, m). Since m is odd, µ(F ) = µ(T (d, m)) = md + 1 − d ≥ 32 md. Since W is independent, the other vertices in W − x1 lie in L2 ∪ L3 . Suppose W has a vertices in L2 and b vertices in L3 . When we remove a vertex xi in W ∩ L2 , by Lemma 5.1, we can decrease µ(F ) by at most 2m + 1. When we remove a vertex xi in W ∩ L3 , we can decrease µ(F ) by at most one. Since µ(T − x1 ) ≥ 32 md while µ(T − W ) ≤ 13 md, we must have (2m + 1)a + b ≥ 2 md 3 Case 2: x1 ,= r. − 13 md = 31 md. Now, we have |W ∪ N (W )| = (m + 2)a + 2b ≥ 16 md. If |W ∩ L1 | ≥ 31 m then already |W ∩ N (W )| ≥ 13 md ≥ 16 md and we are done. So suppose otherwise. For each xi ∈ W ∩ L1 the removal of xi decreases µ(F ) by at most 2µ(T (d, m)) ≤ 2md. Note however that µ(T ) = m + m2 d − 1 − md. Hence, µ(T − W ∩ L1 ) ≥ m2 d + m − 1 − md − ( 31 m)2md > 32 md (with a lot of room to spare). As in Case 1 (with T − W ∩ L1 playing the role of F = T − x1 ), the contribution to |W ∪ N (W )| from vertices in W ∩ (L2 ∪ L3 ) must be at least 61 md. (b) It suffices to prove that for each independent set W , ϕ(W ) ≥ 1 md 6 1 2 1 md. So, 12 1 md. 12 suppose µ(T − W ) ≤ µ(T − W ) ≥ ≥ ϕ(W ) ≥ 21 |W | + |N (W )| ≥ 1 md. 3 ! 1 md. 12 If µ(T − W ) ≥ 1 with maximum degree d such that s(T ) ≤ then certainly ϕ(W ) ≥ Then by part (a), we have |W | + |N (W )| ≥ Proposition 5.4. Let n ≥ 48 be an integer. Let d ≥ 4 be an integer such that n 3 < d ≤ n 2 1 md, 3 √ 1 n. 12 1 md. 6 So, There exists a tree on n vertices − c nd for some absolute constant c. 2 Proof. First assume d is even so that d−1 is odd. Let m be the largest ) odd integer such that M = 1+m+m(d−1)+m (d−1) ≤ 1 n. It is easy to check since d > n 3 that m ≤ d − 1. Also, m > 1 2 ∗ n , d and since d ≤ n 12 ∗ we have m ≥ 3. Let T be obtained from T = T (m, d − 1, m) by attaching a path P of length n − M to a leaf of T . Note that T ∗ has M vertices while T has n vertices and max degree d. √ √ 1 1 1 nd − 21 . By Theorem 3.3, s(T ) ≤ 21 n − 120 nd + 25 . md − 21 ≥ 24 By Lemmas 5.2 and 5.3, ϕ(T ) ≥ ϕ(T ∗ ) − 12 ≥ 12 ∗ If d is odd, we let T = T (m, d − 2, m), where m is the largest odd integer such that |V (T (m, d − 2, m)| < n. We obtain T by attaching path of length n − |V (T ∗ )| to T ∗ at a vertex of degree d − 1. ! The next lemma and the proposition following it will show that the lower bound in Theorem 4.7(b) is best possible, up to the constant factor implied in the Ω (d2 logd n) term, when d is in the range nq ≤ d ≤ n1/3 , where q > 0 is any fixed constant. Lemma 5.5. Let d ≥ 3 be an odd integer. Let T be a symmetric d-ary tree of height h ≥ 3 and order n = |V (T )|. (a) Let W be an independent set in T . If µ(T − W ) ≤ (b) ϕ(T ) ≥ 1 2 d . 24 Proof. (a) Clearly, it suffices to prove that |W | ≥ 1 6d d . 12 n, then |W ∪ N (W )| ≥ 1 2 d . 12 Let r denote the root of T . Suppose first that r ∈ W . Note that T − r n 1 (1 − d+2 1 ) ≥ 3d . Since there are an odd each of which has the same discrepancy µ(Thd−1 ) ≥ n− d n d number of these copies, we get µ(T − r ) = µ(Th−1 ) ≥ 3d . Since W is an independent set, if x ∈ W − r, then x ∈ Lj for consists of d copies of Thd−1 , some j ≥ 2. By Lemma 5.1, for any subgraph F of T containing x, since |V (Tx )| ≤ n , 6d n 3d d 12 n , we d2 d . 12 have µ(F − x) ≥ µ(F ) − 2n . d2 Since while µ(T − W ) ≤ there must be at least such x’s. So |W | ≥ Suppose instead that r ∈ / W . Then for each x ∈ W , x ∈ Lj for some j ≥ 1. By Lemma 5.1, for any subgraph F of T 2 n that contains x, we have µ(F − x) ≥ µ(F ) − 2n . Since µ(T ) ≥ (1 − d+ )n while µ(T − W ) ≤ 6d , we must have |W | ≥ d 1 µ(T − r ) ≥ [(1 − 2 d+1 )n − n 6d ]/ 2n ≥ d d . 12 (b) It suffices to prove that ϕ(W ) ≥ ≥ 1 2 n , 6d then ϕ(W ) ≥ |W | + |N (W )| ≥ 1 2 d2 . 24 d2 24 µ(T − W ) ≥ ! for each independent set W of T . Note that since h ≥ 3, we have n ≥ d3 . If µ(T − W ) n 12d ≥ d2 . 24 Otherwise, µ(T − W ) ≤ n . 6d By (a), |W | + |N (W )| ≥ d2 . 12 Hence, ϕ(W ) ≥ 1 Proposition 5.6. Let n ≥ 64 be an integer. Let d ≥ 4 be an integer such that nq < d ≤ n 3 , where q < 31 is a fixed positive number. Then there exists a tree on n vertices with maximum degree d such that s(T ) ≤ 2n − c ′ d2 logd n for some absolute constant c ′ . Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS T. Jiang et al. / Theoretical Computer Science ( ) – 11 Proof. First assume d is even so that d − 1 is odd. Let h be the largest integer such that D = 1 + (d − 1) + (d − 1)2 + · · · + 1 (d − 1)h ≤ n. It is easy to check that h ≥ ⌊logd n⌋. Since d ≤ n 3 , this yields h ≥ 3. Let T ∗ = Thd−1 and T be obtained from T ∗ by attaching a path P of length n − D to a leaf of T ∗ . Then T is an n-vertex tree with maximum degree d. 1 2 1 2 By Lemmas 5.2 and 5.5, ϕ(T ) ≥ 16 d − 12 . Thus, by Theorem 3.3 we have s(T ) ≤ 21 n − 80 d + 21 . Since logd n ≤ 1q , we 10 q 21 n 2 have s(T ) ≤ 2 − ( 80 )d logd n + 10 . If d is odd, we let h be the largest integer such that |V (Thd−2 )| < n. Let T ∗ = Thd−2 . We obtain T by adding a path of length n − |V (T ∗ )| to T ∗ at a vertex of degree d − 1. ! Next, we show that Theorem 4.7(b) is best possible, up to the constant factor implied in the Ω (d2 logd n) term, when d is an absolute constant. In what follows, n(H ) denotes the number of vertices in H. Lemma 5.7. Let d ≥ 3 be an odd integer. Let T be a symmetric d-ary tree of height h ≥ 3 and order n = |V (T )|. (a) Let W be an independent set of T . If µ(T − W ) ≤ (b) ϕ(T ) ≥ 1 d 4 (h − 1) ≥ 1 d 4 1 2 √ (logd n − 2). n, then |W | ≥ ⌊ 2h ⌋ ≥ 1 2 logd n − 1. 2 . For any p ≥ 1, by earlier discussions, we have β · n(Tpd ) ≤ µ(Tpd ) ≤ β · n(Tpd ) + 1. Let r denote Proof. (a) Let β = 1 − d+ 1 the root of T . Suppose the vertices of W are x1 , x2 , . . . , xp in nondecreasing order of level. We use induction to prove for each i = 1, . . . , ⌊ 2h ⌋ − 1 that µ(T − {x1 , . . . , xi }) ≥ βn di − 2di − 1. For the basis step, let i = 1. If x1 = r, then T − x1 consists of d copies of Thd−1 each having discrepancy µ(Thd−1 ) ≥ β · n(Thd−1 ) = β n−d 1 . Since d is odd, we have µ(T − x1 ) = µ(Thd−1 ) ≥ β( n−d 1 ) = βdn − βd ≥ βdn − 2d − 1. So the claim holds. If x1 ,= r, then x1 ∈ Lj for some j ≥ 1. Each of the d subtrees under x1 has discrepancy at most β djn+1 + 1 ≤ β dn2 + 1. By Lemma 5.1, µ(T − x1 ) ≥ µ(T ) − 2d(β dn2 + 1) − 1. Also, µ(T ) ≥ β n. Thus, we have µ(T − x1 ) ≥ β n − 2d(β dn2 + 1) − 1 ≥ (1 − 2d )β n − 2d − 1 ≥ βdn − (2d + 1), since d ≥ 3. So the claim holds. For the induction step, let i > 1. Suppose first that xi ∈ L0 ∪ L1 . . . ∪ Li−1 . Note that T ′ = T − L0 ∪ L1 ∪ · · · ∪ Li−1 consists of di copies of Thd−i each with the same discrepancy µ(Thd−i ). Since d is odd, we have µ(T ′ ) = µ(Thd−i ). Note that n−2di−1 di − 1. Hence, µ(T ′ ) = µ(Thd−i ) ≥ β · n(Thd−i ) ≥ β( dni − 1) ≥ βdni − 1. Note that T − {x1 , . . . , xi } can be obtained from T ′ by adding at most |L0 ∪ L1 ∪ · · · ∪ Li−1 | ≤ 2di−1 vertices. It follows that µ(T − {x1 , . . . , xi }) ≥ β n − 1 − 2di−1 ≥ β n − 2di − 1. So the claim holds. Suppose next that xi ∈ / L0 ∪ L1 ∪ · · · ∪ Li−1 . Then xi ∈ Lj for some j ≥ i. In this case, each branch below xi has discrepancy at most β djn+1 + 1 ≤ β din+1 + 1. By Lemma 5.1, for any subgraph F of G containing xi , we have µ(F − xi ) ≥ µ(F ) − [2d(β din+1 + 1) + 1] = µ(F ) − 2dβin − 2d − 1. By induction hypothesis, µ(T − {x1 , . . . , xi−1 }) ≥ dβi−n1 − 2di−1 − 1. So µ(T − {x1 , . . . , xi }) ≥ ( dβi−n1 − 2di−1 − 1) − 2dβin − 2d − 1 ≥ βdni − 2di − 1. This completes the induction step, so the claim n(Thd−i ) ≥ ≥ n di holds. 2 Note that β = 1 − d+ ≥ 12 and n ≥ 9 since h ≥ 3. We have for each i ≤ ⌊ 2h ⌋ − 1, µ(T − {x1 , . . . , xi }) ≥ βdni − 2di − 1 ≥ √ √ √ 1 1√ n n − n ≥ 2√n/d − n ≥ 2 n. Since µ(T − W ) < 12 n, we must have |W | ≥ ⌊ 2h ⌋ ≥ 21 logd n − 1. 2di √ h h µ(T − W ) ≥ 41 n ≥ 41 d 2 ≥ 14 d(3) 2 −1 ≥ 41 dh ≥ 14 d(logd n − 1). If µ(T − W ) < 2 n, then by (a), |W | + |N (W )| ≥ d⌊ 2h ⌋ ≥ 12 d(h − 1). Hence ϕ(W ) ≥ 1 |W | + |N (W )| ≥ 14 d(h − 1). Since this holds for all W , ϕ(T ) ≥ 41 d(h − 1) ≥ 41 d(logd n − 2). ! 2 (b) Let W be any independent set of T . If µ(T − W ) ≥ √ 1 1 2 √ n, then ϕ(W ) ≥ 1 2 Proposition 5.8. Let d ≥ 4 be a fixed positive integer. Let n ≥ 1 + (d − 1) + (d − 1)2 + (d − 1)3 be an integer. There exists a tree on n vertices with maximum degree d such that s(T ) ≤ 2n − c ′′ d logd n, for some absolute constant c ′′ . Proof. First assume d is even so that d − 1 is odd. Let h be the largest integer such that D = 1 +(d − 1)+(d − 1)2 +· · ·+(d − 1)h ≤ n. Note that h ≥ ⌊logd n⌋ ≥ logd n − 1. Let T ∗ be the complete (d − 1)-ary tree of height h and let T be obtained from T ∗ by attaching path P of length n − D to a leaf of T ∗ . Then T has n vertices and maximum degree d. Then by Lemmas 5.2 and 21 d 1 5.7, we have ϕ(T ) ≥ ϕ(T ∗ ) − 21 ≥ 41 d(h − 1) − 12 ≥ 41 d(logd n − 2) − 12 . By Theorem 3.3, s(T ) ≤ 21 n + 10 + 10 − 20 d logd n. If d is odd, let h be the largest integer such that |V (Thd−2 )| < n. Let T ∗ = Thd−2 . We obtain T by adding a path of length n − |V (T ∗ )| to T ∗ at a vertex of degree d − 1. ! We summarize our results below. Proposition 5.9. Let n be an integer tending to infinity. Let d ≥ 4 be an integer. Let g (n, d) = min{s(T ) : T is a tree with n vertices and maximum degree d}. There exist absolute constants c1 , c2 , c3 , c4 , c5 , c6 such that 1 (a) If n 3 < d ≤ n , 12 1 then n 2 √ − c1 nd ≤ g (n, d) ≤ n 2 then 2n √ − c2 nd. (b) If nq ≤ d ≤ n 3 , where 0 < q < 31 is fixed, − c3 d2 logd n ≤ g (n, d) ≤ 2n − c4 d2 logd n. n (c) If d ≥ 4 is an absolute constant, then 2 − c5 logd n ≤ g (n, d) ≤ 2n − c6 logd n. Please cite this article in press as: T. Jiang, et al., Separation numbers of trees, Theoretical Computer Science (2009), doi:10.1016/j.tcs.2009.05.011 ARTICLE IN PRESS 12 T. Jiang et al. / Theoretical Computer Science ( ) – Fig. 6. The symmetric d-ary tree of height h. Finally, we focus on the symmetric d-ary tree Thd . By Propositions 5.5 and 5.7 ϕ(Thd ) ≥ Thus, in particular, we have ϕ( Thd )≥ 1 48 Thd 2 n 2 Thd 1 2 d 24 and ϕ(Thd ) ≥ 1 240 2 1 d 4 (h − 1) ≥ 81 dh. (d + dh). By Theorem 3.3, we have s( ) ≤ + 2 − (d + dh). We show next n n 2 − O ( d h ) ≤ s ( − O ( h) obtained in [1]. ) ≤ 2 2 that this is asymptotically tight. Our bounds improve the estimates Proposition 5.10. For all integers h ≥ 3 and odd integers d ≥ 3, we have n 2 − 3(d2 + dh) ≤ s(Thd ) ≤ n 2 +2− 1 240 (d2 + dh). Proof. It remains to prove the lower bound. By Theorem 3.2, it suffices to find an independent set S with ϕ(S ) ≤ 3(d2 + dh). We draw T = Thd in the plane in the natural noncrossing fashion where the root r is at the top. Let x1 , . . . , xd denote r’s children from left to right. As before, for each i, let Li be the set of vertices in T at distance i from r. Let (X , Y ) denote the unique bipartition of T , where |Y | ≥ |X |. Observe that Y contains Lh , the set of leaves of T . Suppose d = 2k + 1. Let S0 = {x1 , . . . , xk+1 }. Note that there are d2 copies of Thd−2 rooted in L2 . After the deletion 2 d(d+1) of these become components by themselves. If we flip the first (from the left) d 2−1 of these of S0 , (k + 1)d = 2 d Th−2 -components (interchanging X -vertices with Y -vertices in them), we obtain a bipartition (X ′ , Y ′ ) of T − S0 with µ(Thd−2 ) − d − 1 ≤ µ(T − S0 ) = |Y ′ | − |X ′ | ≤ µ(Thd−2 ) + d + 1. Now, there are d branches under xd , each being a copy of Thd−2 . Let A and B denote two of these branches. Note that the roots of A and B are in L2 . Let P be a path of length h − 2 in A that starts at A’s root and moves down the levels such that each vertex is the rightmost child of the previous vertex. Define the path Q similarly for B. Let y2 be the root of A, which is the only element in V (P ) ∩ L2 . Let y3 be the only vertex in V (Q ) ∩ L3 . Let y4 be the only vertex in V (P ) ∩ L4 . Let y5 be the only vertex in V (Q ) ∩ L5 . We continue like this, alternating between P and Q as we move from one level to the next, obtaining y2 , . . . , yh−1 in that order. Here, we start our indices at 2 to be consistent with the level number and we stop at level h − 1. Note that the subtree rooted at any yi is a copy of Thd−i (See Fig. 6). Sequentially, delete y2 , . . . , yh−1 , increasing |Y ′ | − |X ′ | by at most h − 2. When we delete yi , the copy of Thd−i rooted at yi 1 of these copies, interchanging X ′ -vertices with Y ′ -vertices. breaks into d copies of Thd−i−1 , at which point we flip the first d− 2 1 · 2µ(Thd−i−1 ) = (d − 1)µ(Thd−1−i ). Such a flip reduces |Y ′ | − |X ′ | by d− 2 2 Tqd Tqd 2 Tqd Tqd 2 !h−1 1 Thd−1−i . i=2 d dq+1 −1 d . So, Tqd d−1 h − i 2 d −1 1 d+ h−1−i 1 d−1 2 d from 1 n T h−2 , it is d+1 Hence, after doing the flipping for each i = 2, . . . , h − 1, |Y ′ | − |X ′ | is further reduced by p = q ( − )µ( Recall that (1 − d+1 )n( ) ≤ µ( ) ≤ (1 − d+1 )n( ) + 1. Also, n( ) = 1 + d + d + · · · + = !h−1 !h−1 2 dq+1 −1 d (1 − d+1 ) d−1 + ǫq , for some 0 ≤ ǫq ≤ 1. Hence, p = i=2 (d − 1)µ(Th−1−i ) = i=2 (d − 1)[( − ) ! (1 − d+2 1 )(n(Thd−2 ) − 1) − (1 − d+2 1 )(h − 2) + (d − 1) hi=−21 ǫh−1−i . Since µ(Thd−2 ) is within 1 ( − ) µ( ) = +ǫ ) ( ) ]= easy to see that |p − µ(Thd−2 )| ≤ dh. Recall that before removing yi ’s, µ(Thd−2 ) − d − 1 ≤ |Y ′ | − |X ′ | ≤ µ(Thd−2 ) + d + 1, that the removals change |Y ′ | − |X ′ | by at most h − 2, and the flips reduce |Y ′ | − |X ′ | by p. For the new X ′ , Y ′ , we have ||Y ′ | − |X ′ || ≤ d + 1 + h − 2 + dh ≤ 2dh. Let S = S0 ∪ {y2 , . . . , yh−1 }. (In Fig. 2, vertices in S are circled.) We have argued that µ(T − S ) ≤ 2dh. Observe also that S is an independent set in T with |S | ≤ d + h. We have ϕ(S ) = 12 |S |+ 21 µ(T − S )+|N (S )| ≤ 12 (d + h)+ 12 · 2dh + d(d + h) ≤ 3(d2 + dh), completing the proof. ! Acknowledgement The first author’s research was partially supported by National Security Agency under grant number H98230-07-1-027. Please cite this article in press as: T. 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