Maximizing the number of independent subsets over trees with maximum degree 3. Clemens Heuberger and Stephan G. Wagner

Transcription

1 FoSP Algorithmen & mathematische Modellierung FoSP Forschungsschwerpunkt Algorithmen und mathematische Modellierung Maximizing the number of independent subsets over trees with maximum degree 3 Clemens Heuberger and Stephan G. Wagner Project Area(s): Analysis of Digital Expansions with Applications in Cryptography Konkrete Mathematik: Fraktale, Ziffernfunktionen und Punktverteilungen The Hardy-Littlewood Method in the Analysis of Digit Problems and Enumerative Combinatorics Institut für Optimierung und Diskrete Mathematik (Math B) Report , April 2007

2 MAXIMIZING THE NUMBER OF INDEPENDENT SUBSETS OVER TREES WITH MAXIMUM DEGREE 3 CLEMENS HEUBERGER AND STEPHAN G. WAGNER Abstract. The number of independent vertex subsets is a graph parameter that is, apart from its purely mathematical importance, of interest in mathematical chemistry. In particular, the problem of maximizing or minimizing the number of independent vertex subsets within a given class of graphs has been investigated in many papers. In view of the applications of this graph parameter, trees of restricted degree are of particular interest. In this paper, we give a characterization of the trees with maximal degree 3 which maximize the number of independent subsets, and show that these trees also minimize the number of independent edge subsets. The structure these trees show is quite interesting and unexpected it can be described by means of a novel binary digit system using the digits and 4. The proof mainly depends on an exchange lemma for branches of a tree.. Introduction The parameter number of independent subsets of a graph first appears in the mathematical literature in a paper of Prodinger and Tichy [4] in 982. They call it the Fibonacci number of a graph in view of the fact that the number of independent subsets of a path on n vertices is exactly the Fibonacci number F n+2. Among other results, they show that the number of independent vertex subsets is maximal among all trees on n vertices for the star graph S n and minimal for the path P n. In subsequent papers, Kirschenhofer, Prodinger and Tichy study the asymptotic behavior of this parameter for complete t-ary trees and for families of simply generated trees [8, 9]. Independently, Merrifield and Simmons [3] introduced the number of independent vertex subsets (which they call the σ-index) to the chemical literature in 989, showing connections between the σ-index of a molecular graph and physicochemical properties such as boiling points. Meanwhile, the number of independent subsets of a graph is called the Merrifield-Simmons index in mathematical chemistry, and there is already a substantial amount of literature on chemical applications as well as on graph-theoretical properties of this index (see [3, 8] and the references therein). It belongs to the group of so-called topological indices, which are used to give quantitative descriptions for the structure of a (molecular) graph. Another representative of this group is the so-called Hosoya index [5], defined as the number of independent edge subsets (which are also referred to as matchings). One of the major mathematical problems in connection with these indices is the question which graphs from a given class (typically, trees) maximize or minimize the index value [6,, 2, 9]. It is known that the extremal graphs for the Merrifield-Simmons index and the Hosoya index usually coincide, where an extremal graph maximizes one of these indices while it minimizes the other, compare [9]. It is also conjectured that the trees which minimize the Hosoya index also minimize the energy (the sum of the absolute values of a graph s eigenvalues), see []. In view of the chemical applications, it is quite natural to consider trees of restricted degrees. For instance, trees of maximal degree 4 which minimize the Hosoya index have been determined Date: April 4, Mathematics Subject Classification. 05C69; 05C05 A63. Key words and phrases. Independent sets, binary trees, maximal degree, digit expansion. This paper was written while C. Heuberger was a visitor at the Center of Experimental Mathematics at the University of Stellenbosch. He thanks the center for its hospitality. This work was also supported by the Austrian Science Foundation FWF, projects S9606 and S96, that are part of the Austrian National Research Network Analytic Combinatorics and Probabilistic Number Theory.

3 2 CLEMENS HEUBERGER AND STEPHAN G. WAGNER by means of an extensive computer search in []; [7] provides a faster algorithm for this purpose. However, we believe that this question is also very interesting from a purely graph-theoretical point of view. Quite a lot of similar results are known in the graph-theoretic literature as well: for instance, Hedman [4] studies the (essentially equivalent) problem of maximizing the number of cliques in graphs with a given maximal clique size. In this paper, we will restrict ourselves to the study of trees with maximal degree 3; if not stated otherwise, all trees in the following are assumed to have maximum degree 3. In spite of the simple problem statement, the problem of determining the trees from this set which maximize the number of independent subsets is quite intricate and reveals an interesting and unexpected structure, which can be described by means of a novel binary digit system. For other graph parameters, namely the Wiener index (sum of all distances between pairs of vertices) and the number of subtrees, the extremal trees of maximal degree 3 are already known (see [2, 7, 5, 6]) basically, the solution is given by the complete binary trees. Thus it was quite a surprise to find that the complete binary tree is not maximal with respect to the number of independent subsets. Yet, complete trees play a major role in the description of the extremal trees our main results read as follows: Theorem. Let n be a positive integer. Then there is a unique tree T with n vertices and maximal degree 3 that simultaneously maximizes the number of independent vertex subsets and minimizes the number of independent edge subsets. It can be decomposed as M l M 0 M M 2 M 3 M l 2 M l with M k {B k, B k+2 } for 0 k < l and M l {B l, B l+, B l+2 }. Here, B h denotes the complete binary tree of height h. Furthermore, there are unique d 0,..., d m such that d k {, 4} for 0 k < m, d m {, 2, 4}, and m n + = d k 2 k, k=0 and they are given by m = l and d k = 2 δ k for 0 k m, where δ k {0,, 2} is defined by M k = B k+δk. The paper is organized as follows: we start with basic definitions and a recursive characterization of the number of independent vertex subsets. Section 3 provides the necessary preliminary results. In Section 4, the main exchange lemma is formulated and used to prove the main result of this paper. In the final section, we show how the proof can be adapted to the case of independent edge subsets. 2. Notations Definition 2.. () Let G be a graph. Then σ(g) is defined to be the number of independent subsets of G. (2) For a rooted tree T with root v, we also define σ 0 (T) to be the number of independent subsets of G not containing the root v and σ (T) to be the number of independent subsets of G containing the root v. The ratio σ 0 (T)/σ(T) is abbreviated as ρ(t). Note that we do not mention the root v in the notations σ 0 (T), σ (T), and ρ(t) since the roots will usually be anonymous. The empty set is always an independent subset of G, even if G is the empty graph. Therefore, σ(g) is always positive.

4 MAXIMIZING THE NUMBER OF INDEPENDENT SUBSETS 3 v v v 2 v k T T 2... T k Figure. Rooted tree with branches For a rooted tree T with root v, the connected components T,..., T k of T v are called the branches of v, cf. Figure. Taking the neighbor v j of v contained in T j as root of T j, T j is again a rooted tree. The following recursive formulæ will be used throughout the paper without further reference. Lemma 2.2. Let T be a rooted tree with root v and branches T,..., T k. Then k () σ 0 (T) = σ(t j ), (2) (3) σ (T) = ρ(t) = j= k σ 0 (T j ), j= + k j= ρ(t j). Proof. The recursive formulæ () and (2) are easy to see and well known, cf. for instance [3, 0]. Equation (3) is a consequence of () and (2). 3. Rooted trees A rooted binary tree is a rooted tree where every vertex has 0 or 2 children. The (rooted) complete binary tree of height n is denoted by B n, i.e., B is a single vertex and B n has branches B n and B n, cf. Figure 2. It is convenient to define B 0 to be the empty graph with σ 0 (B 0 ) = and σ (B 0 ) = 0. Note that equations () and (2) remain valid with this setting. Obviously, the number of vertices of B n equals B n for n 0. (a) B (b) B 2 (c) B 3 Figure 2. Complete binary trees The sequence ρ(b n ), n 0, has been studied by Kirschenhofer, Prodinger, and Tichy in [8]. By (3), we have ρ(b n ) = + ρ(b n ) 2 for n, ρ(b 0 ) =. We list the first few values of ρ(b n ) since they occur frequently: ρ(b 0 ) =, ρ(b ) = 2, ρ(b 2) = 4 5.

5 4 CLEMENS HEUBERGER AND STEPHAN G. WAGNER Lemma 3. ([8]). The sequence ρ(b n ), n 0, is convergent. Its limit c is the unique real solution of the equation c 3 + c =. The subsequence with odd indices is strictly increasing, 2 = ρ(b ) < ρ(b 3 ) < < ρ(b 2k ) < ρ(b 2k+ ) < < c, whereas the subsequence with even indices is strictly decreasing, = ρ(b 0 ) > ρ(b 2 ) > > ρ(b 2k ) > ρ(b 2k+2 ) > > c. We note a first rough estimate of ρ(t) for rooted trees T. Lemma 3.2. Let T be a rooted tree. Then 2 ρ(t) 4 5, unless T is empty, where ρ(t) =. Proof. Since ρ(b 0 ) =, we can assume that T has at least one vertex and proceed by induction on the number of vertices. Denote the branches of T by T and T 2 (allowing T j to be the empty graph) where each T j satisfies /2 ρ(t j ) by the induction hypothesis. We obtain 2 = + ρ(t) = + ρ(t )ρ(t 2 ) Definition 3.3. Let T be a possibly rooted tree. The outline graph of T is defined to be T with all maximal subtrees isomorphic to some B k, k 0, replaced by a special leaf B k. In this process, we attach a leaf B 0 to internal nodes (non-leaves and non-root) of degree 2. This ensures that the outline graph of a rooted tree is a rooted binary tree. An example is shown in Figure 3. The outline of a rooted tree B k is just the rooted tree consisting of the single leaf B k. = 4 5. B 2 B B 0 Figure 3. Reduction to the outline graph If enough information on the outline of a rooted tree is available, we can determine it from its ρ-value. Lemma 3.4. Let j 0 be an integer and T be a rooted tree whose outline does not contain any B k for 0 k j. If j is even and ρ(b j+2 ) ρ(t) or j is odd and ρ(t) ρ(b j+2 ), then T {B j, B j+2 }. Proof. We first assume that T = B l for some l. Since the outline of T does not contain a B k for k j, we conclude that l j. The monotonicity properties of ρ(b n ) in Lemma 3. imply that l {j, j + 2} and we are done. Thus we may assume that the outline of T is a binary rooted tree with more than one vertex. We proceed by induction on j. The case of j = 0 has been shown in Lemma 3.2. Assume that j. By assumption, the outline of T does not contain a B 0, or equivalently, T is a binary

6 MAXIMIZING THE NUMBER OF INDEPENDENT SUBSETS 5 rooted tree. Thus T has two non-empty branches T and T 2, respectively. By our assumption, the outline of both T and T 2 does not contain a B k for k j. Assume that j is odd. Then + ρ(t )ρ(t 2 ) = ρ(t) ρ(b j+2) = + ρ(b j+ ) 2, which is equivalent to (4) ρ(b j+ ) 2 ρ(t )ρ(t 2 ). Without loss of generality, we may assume ρ(t ) ρ(t 2 ), thus ρ(b j+ ) ρ(t 2 ). By the induction hypothesis, T 2 {B j, B j+ }. Since the outline of T 2 does not contain any B l for l j, we clearly have T 2 B j and therefore T 2 = B j+. From (4) we see that ρ(b j+ ) ρ(t ) and the same argument shows that T = B j+, too. Thus T = B j+2, which has been handled earlier. The proof for even j follows by exchanging odd and even and reversing all inequality signs in the previous paragraph. 4. Optimal Trees A tree T is called maximal, if σ(t) σ(t ) for all trees T with the same number of vertices as T. Our key tool is the following local exchange lemma. Lemma 4.. Let T be a maximal tree. If there are (possibly empty) rooted trees T, T 2, T 3, T 4 and a tree T 0 such that T can be decomposed as T T 0 T 3 v... w T 2 T 4 and such that ρ(t ) < ρ(t 3 ), then max{ρ(t ), ρ(t 2 )} min{ρ(t 3 ), ρ(t 4 )}. Proof. We need the following auxiliary quantities: σ 00 (T 0 ): number of independent subsets of T 0 containing neither v nor w. σ 0 (T 0 ): number of independent subsets of T 0 containing v, but not w. σ 0 (T 0 ): number of independent subsets of T 0 containing w, but not v. σ (T 0 ): number of independent subsets of T 0 containing both v and w. Define F(T, T 2, T 3, T 4 ) := σ 00 (T 0 ) + σ 0 (T 0 )ρ(t )ρ(t 2 ) Then it is easily seen that Since T is maximal, we must have + σ 0 (T 0 )ρ(t 3 )ρ(t 4 ) + σ (T 0 )ρ(t )ρ(t 2 )ρ(t 3 )ρ(t 4 ). σ(t) = F(T, T 2, T 3, T 4 )σ(t )σ(t 2 )σ(t 3 )σ(t 4 ). F(T, T 2, T 3, T 4 ) F(T π(), T π(2), T π(3), T π(4) ) for all permutations π of {, 2, 3, 4}. The first and the fourth summand of F(T π(), T π(2), T π(3), T π(4) ) are independent of the permutation π, so the sum of the second and the third summand is maximal. By the rearrangement inequality (note that all quantities are positive), we conclude that ρ(t 2 ) ρ(t 4 ) and σ 0 (T 0 ) σ 0 (T 0 ). Exchanging T 3 and T 4 does not change anything, however, applying the rearrangement inequality in that situation, we obtain ρ(t 2 ) ρ(t 3 ) and ρ(t ) ρ(t 4 ).

7 6 CLEMENS HEUBERGER AND STEPHAN G. WAGNER Remark 4.2. It is a consequence of Lemma 4. that a maximal tree T contains at most vertex of degree 2: Set T 2 = T 3 = B 0 and T B 0 in Lemma 4., which is applicable since ρ(t ) < ρ(t 3 ) = ρ(b 0 ) = by Lemma 3.2. We obtain = ρ(b 0 ) = ρ(t 2 ) ρ(t 4 ), i.e., T 4 = B 0 by Lemma 3.2 and w has degree. This result has already been proved in [7]. For our purposes, we need a stronger version of Remark 4.2. Here, we use the outline graph of the maximal tree T. Note that since we do not assume it to be a rooted tree, a B 0 is attached to all vertices of degree 2. For instance, if T = B l for some l 2, its outline graph is: B 0 B l Lemma 4.3. Let T be a maximal tree and let j be the least nonnegative integer such that the outline graph of T contains a B j. Then the outline graph of T contains B j only once. Proof. If B j is contained twice in the outline graph of T, the two copies of B j are not branches of the same vertex, since they would have been merged to B j+ in that case. Therefore, the outline of T has the shape T T 3... B j B j for some rooted trees T and T 3 with T B j and T 3 B j. By our assumption, the outlines of these rooted trees do not contain a B l for l < j. Assume that j is odd. We claim that ρ(t k ) ρ(b j+2 ) for k {, 3}. Otherwise we have ρ(t k ) < ρ(b j+2 ) for some k, which implies T k {B j, B j+2 } by Lemma 3.4. The first possibility has already been ruled out and the second is impossible because of the strict inequality. Thus the claim is proved. We conclude that ρ(b j ) < ρ(t ), which implies that ρ(t 3 ) ρ(b j ) by Lemma 4., a contradiction. Once again, the proof for even j follows by exchanging odd and even and reversing all inequality signs in the previous paragraph. The following lemma performs one step of the decomposition of a maximal tree T. Lemma 4.4. Let T be a maximal tree, k be a nonnegative integer and assume that the outline graph of T can be decomposed as L k R k for some rooted trees L k (possibly empty) and R k with (5a) k is odd and ρ(b k ) < ρ(l k ) < ρ(b k+2 ) or (5b) k is even and ρ(b k+2 ) < ρ(l k ) < ρ(b k ) or L k = B k. Assume that R k is non-empty and the outline of R k does not contain any B l with l < k. Then we have R k {B k, B k+, B k+2 } or the outline of R k can be decomposed as

8 MAXIMIZING THE NUMBER OF INDEPENDENT SUBSETS 7 v R k+ M k for M k {B k, B k+2 } and a non-empty rooted tree R k+ whose outline does not contain any B l for l k. In the latter case, we have (6a) k is odd and ρ(b k+3 ) < ρ(l k+ ) < ρ(b k+ ) or (6b) k is even and ρ(b k+ ) < ρ(l k+ ) < ρ(b k+3 ) where L k+ is defined as follows: L k v M k Furthermore, M k B k if L k = B k. Before we prove the lemma it is worth pointing out that its assumptions are satisfied for k = j and L k = B j by Lemma 4.3. Proof of Lemma 4.4. If R k {B k, B k+, B k+2 }, there is nothing to prove. Thus we assume that this is not the case. We consider the case of odd k, the other case follows by reversing the signs of the inequalities. Before constructing the required decomposition of R k, we prove the estimates (6a) in order to smooth the structure of the remainder of the proof. Consider first the case of M k = B k+2. Then ρ(b k+3 ) = + ρ(b k+2 ) 2 < ρ(l k+) = + ρ(l k )ρ(b k+2 ) < + ρ(b k ) 2 = ρ(b k+) by Lemma 3.. If M k = B k, we cannot have L k = B k by Lemma 4.3 and the assumption R k B k+, thus we again have ρ(b k+3 ) = + ρ(b k+2 ) 2 < ρ(l k+) = + ρ(l k )ρ(b k ) < + ρ(b k ) 2 = ρ(b k+). By assumption, R k is not empty. We claim that R k has two non-empty branches T and T 2. For k 2, the outline of R k contains neither a B 0 nor a B, so the root of R k has degree 2. For k =, we have R k B by assumption and the outline of R k does not contain a B 0 by assumption, too. For k = 0, we again have R k B. Additionally, we must have L 0 = B 0 by (5b) and Lemma 3.2. Since the outline of T contains at most one copy of B 0 by Lemma 4.3, none of the branches of R 0 can be empty. So this claim is proved. We claim that the outlines of T and T 2 do not contain any B l with l < k. That situation could only occur if T = T 2 = B k and R k = B k, and this has already been ruled out. Next, we claim that if the outline of T m contains B k for some m {, 2}, then T m = B k. Consider first the case that L k = B k. In that case, the outline graph of R k does not contain a B k by Lemma 4.3. Thus the outline graph of T m contains B k if and only if T = T 2 = B k and therefore R k = B k+, and this is not possible anymore. So we may assume L k B k and m = for the remainder of the proof of this claim. We can decompose the outline of T such that we get a decomposition of T as

9 8 CLEMENS HEUBERGER AND STEPHAN G. WAGNER L k T 0... T 3 T 2 B k for a suitable rooted tree T 3. Since the outline of T 3 is a subtree of the outline of T, it does not contain a B l for l < k. We have ρ(b k ) < ρ(l k ) by (5a) and therefore, by Lemma 4., ρ(t 3 ) ρ(l k ) < ρ(b k+2 ). From Lemma 3.4, we conclude that T 3 = B k (the other possibility cannot occur due to the strict inequality). But this is a contradiction to the fact that T 3 is contained in the outline of T, since it can be merged to a B k+. Our claim is proved. If both T = T 2 = B k, we were working with R k = B k+, which had been excluded. If exactly one of the T m equals B k, say T = B k, then we set M k = B k and R k = T 2 and we are done. We therefore only have to deal with the case that the outlines of both T and T 2 do not contain any B l for l k. We claim that if T 2 contains a B k+2 as a proper subtree, then T = B k+2 and we are done. For clarity we emphasize that B k+2 is just assumed to be a subtree of T 2, we neither require it to be contained in the outline of T 2 nor to be a branch of T 2. We can decompose T as L k T 0... T 4 T B k+2 for a suitable rooted tree T 4. By (5a), we have ρ(l k ) < ρ(b k+2 ) and we apply Lemma 4. to deduce ρ(t ) ρ(b k+2 ). Since the outline of T is known to contain no B l for l k, we obtain T = B k+2 from Lemma 3.4, as we claimed. If T 2 = B k+2, we are done anyway, so we are only left with the case that neither T nor T 2 contains B k+2 as a subtree. Consider the outline graph of T m for some m {, 2}. A leaf of the outline of T m is of the form B l for some l by the definition of an outline graph. However, the range for l is already very restricted: We have l k + since the outline graph of T m has been proved to contain no smaller B l. On the other hand, we have l k +, since T m does not contain B k+2 as a subtree. In short, all leaves of T m equal B k+. Consider a leaf B k+ of maximum height (distance from the root). Assume that this leaf B k+ has a parent w, then w has two children (we know that the outline graph of T m does not contain a B 0 since k 0) and the second child of w has to be a leaf since our original leaf was assumed to be of maximum height. Thus we found the subgraph w B k+ B k+ of the outline graph of T m, this would have been contracted to a B k+2, contradiction. So the only possibility is that T m = B k+ for both m {, 2}. But then, we have R k = B k+2, and this concludes the proof. Theorem. Let n be a positive integer. Then there is a unique maximal tree T with n vertices and maximal degree 3. It can be decomposed as

10 MAXIMIZING THE NUMBER OF INDEPENDENT SUBSETS 9 M l M 0 M M 2 M 3 M l 2 M l with M k {B k, B k+2 } for 0 k < l and M l {B l, B l+, B l+2 }. Furthermore, there are unique d 0,..., d m such that d k {, 4} for 0 k < m, d m {, 2, 4}, and m (7) n + = d k 2 k, k=0 and they are given by m = l and d k = 2 δ k for 0 k m, where δ k {0,, 2} is defined by M k = B k+δk. Thus the unique maximal tree with n vertices can be described in terms of the binary digit expansion (7) of n+ with digits {, 2, 4}, where the digit 2 is an exceptional digit which can only be used as most significant digit. Proof of the Theorem. Let T be a maximal tree with n vertices. Let j be the least integer such that the outline graph of T contains B j. Then applying Lemma 4.4 inductively for k = j,..., l where l j is the least integer such that R l {B l, B l+, B l+2 }, we get a decomposition of T as B j M l M j M j+ M l 2 M l with M k {B k, B k+2 } for j k < l and M l {B l, B l+, B l+2 }. Lemma 4.4 gives the extra information that M j B j, but we deliberately omit this result, since we prove uniqueness of the representation by other means. For k j, we further decompose L k = B k into its two branches L k = B k and M k = B k. This yields the structure described in the theorem, where L 0 = B 0 is discarded, but M 0 = B 0 may be retained. We count the number of vertices in the representation given in the theorem. We have l l (8) n = + ( + V (M k ) ) = + 2 k+δ k, k=0 i.e., the digits d k = 2 δ k indeed satisfy (7). Assume that there is another expansion (9) n + = m d k2 k with d k {, 4} for k < m and d m {, 2, 4}. Let h be the least integer such that d h d h. Then (8) and (9) yield l m (0) d k 2 k h = d k 2k h. k=h k=0 If h < min{l, m}, we consider (0) modulo 2 and see that d h d h (mod 2). Since d h, d h {, 4}, this implies d h = d h, contradiction. Therefore, we have h = min{l, m}, say h = l. If m = l, then k=h k=0

11 0 CLEMENS HEUBERGER AND STEPHAN G. WAGNER (0) simply states that d l = d l, contradiction. Thus m > l and d l {, 4}. We have d l d l (mod 2), which can only hold if d l = 2 and d l = 4. But then the left hand side of (0) is smaller than its right hand side, contradiction. Thus the uniqueness of the digit expansion is shown. Therefore, a maximal tree T is uniquely characterized, hence it is unique. 5. Independent edge subsets In this section, we will give a proof of an analogous theorem for independent edge subsets that will basically follow the same lines. We start with the appropriate definitions: Definition 5.. () Let G be a graph. Then z(g) is defined to be the number of independent edge subsets of G. (2) For a rooted tree T with root v, we also define z 0 (T) to be the number of independent edge subsets of G not containing an edge incident with the root v and z (T) to be the number of independent edge subsets of G containing an edge incident with the root v. The ratio z 0 (T)/z(T) is abbreviated as τ(t). Again, it is possible to give simple recursive formulas for z, z 0 and z : Lemma 5.2. Let T be a rooted tree with root v and branches T,..., T k. Then k () z 0 (T) = z(t j ), (2) (3) z (T) = τ(t) = j= k j= z 0 (T j ) z(t j ) k z(t j ), i= + k j= τ(t j). Proof. The formulæ for z 0 and z are easy to prove and can be found in [3, 0] again, and the identity for τ(t) follows at once. From formula (3), we obtain a simple recursion for τ(b n ). We define z 0 (B 0 ) = 0, z (B 0 ) = (so that () and (2) remain valid) and have, by (3), τ(b n ) = The first values are given by + 2τ(B n ) for n, τ(b 0 ) = 0. τ(b 0 ) = 0, τ(b ) =, τ(b 2 ) = 3. It is an easy exercise to prove the following lemma: Lemma 5.3. The sequence τ(b n ), n 0, is convergent with limit 2. The subsequence with odd indices is strictly decreasing, = τ(b ) > τ(b 3 ) > > τ(b 2k ) > τ(b 2k+ ) > > 2, whereas the subsequence with even indices is strictly increasing, 0 = τ(b 0 ) < τ(b 2 ) < < τ(b 2k ) < τ(b 2k+2 ) < < 2. Furthermore, we have the following analogue of Lemma 3.2 (the proof is basically the same): Lemma 5.4. Let T be a rooted tree. Then τ(t), 3 unless T is empty, where τ(t) = 0.

12 MAXIMIZING THE NUMBER OF INDEPENDENT SUBSETS Note that it doesn t make a difference in the proof of Lemma 3.4 whether we have a product or a sum in the denominator of formula (3). Hence we obtain the following analogue of Lemma 3.4 (we also remark that the roles of odd and even are interchanged in view of Lemma 5.3): Lemma 5.5. Let j 0 be an integer and T be a rooted tree whose outline does not contain any B k for 0 k j. If j is odd and τ(b j+2 ) τ(t) or then T {B j, B j+2 }. j is even and τ(t) τ(b j+2 ), Now, it only remains to prove an analogue of the key step in our proof: Lemma 5.6. Let T be a minimal tree in the sense that it minimizes z(t) among all trees of maximal degree 3 on n vertices. If there are (possibly empty) rooted trees T, T 2, T 3, T 4 and a tree T 0 such that T can be decomposed as T T 0 T 3 v... w T 2 T 4 and such that τ(t ) < τ(t 3 ), then Proof. Again, we need four auxiliary quantities: Define max{τ(t ), τ(t 2 )} min{τ(t 3 ), τ(t 4 )}. z 00 (T 0 ): number of independent edge subsets of T 0 not containing an edge that is incident with either v or w. z 0 (T 0 ): number of independent edge subsets of T 0 containing an edge incident with v, but no edge incident with w. z 0 (T 0 ): number of independent edge subsets of T 0 containing an edge incident with w, but no edge incident with v. z (T 0 ): number of independent edge subsets of T 0 containing an edge incident with v and containing an edge incident with w. G(T, T 2, T 3, T 4 ) := z 00 (T 0 )( + τ(t ) + τ(t 2 ))( + τ(t 3 ) + τ(t 4 )) Then it is easily seen that In view of the minimality of z(t), we must have + z 0 (T 0 )( + τ(t 3 ) + τ(t 4 )) + z 0 (T 0 )( + τ(t ) + τ(t 2 )) + z (T 0 ). z(t) = G(T, T 2, T 3, T 4 )z(t )z(t 2 )z(t 3 )z(t 4 ). G(T, T 2, T 3, T 4 ) G(T π(), T π(2), T π(3), T π(4) ) for all permutations π of {, 2, 3, 4}. Ignoring the assumption ρ(t ) < ρ(t 3 ) for the moment, we see that the minimum of the first summand among all possible permutations is attained if (4) max{τ(t ), τ(t 2 )} min{τ(t 3 ), τ(t 4 )} or min{τ(t ), τ(t 2 )} max{τ(t 3 ), τ(t 4 )} by standard arguments (note that the sum of the two factors does not depend on the permutation). The sum of the second and third summand is minimized if max{τ(t ), τ(t 2 )} min{τ(t 3 ), τ(t 4 )} in the case z 0 (T 0 ) z 0 (T 0 ) and is minimized if min{τ(t ), τ(t 2 )} max{τ(t 3 ), τ(t 4 )} in the case that z 0 (T 0 ) z 0 (T 0 ). Therefore, the minimality of G yields (4). The assumption ρ(t ) < ρ(t 3 ) implies the first possibility.

13 2 CLEMENS HEUBERGER AND STEPHAN G. WAGNER From this point on, the remaining steps are literally the same as in the case of independent vertex subsets (note again that the only slight difference lies in the fact that we have a sum in the denominator of formula (3), as opposed to the product in formula (3), but this doesn t alter the argument). Hence we just state the resulting theorem as it has been stated in the introduction, the trees which maximize the number of independent vertex subsets among all trees of maximal degree 3 are also the ones which minimize the number of independent edge subsets. Theorem 2. Let n be a positive integer. Then there is a unique tree T with n vertices and maximal degree 3 that minimizes z(t). It can be decomposed as M l M 0 M M 2 M 3 with the same M k as in Theorem. References M l 2 M l [] M. Fischermann, I. Gutman, A. Hoffmann, D. Rautenbach, D. Vidović, and L. Volkmann. Extremal Chemical Trees. Z. Naturforsch., 57a:49 52, [2] M. Fischermann, A. Hoffmann, D. Rautenbach, L. Székely, and L. Volkmann. Wiener index versus maximum degree in trees. Discrete Appl. Math., 22(-3):27 37, [3] I. Gutman and O. E. Polansky. Mathematical concepts in organic chemistry. Springer-Verlag, Berlin, 986. [4] B. Hedman. Another extremal problem for Turán graphs. Discrete Math., 65(2):73 76, 987. [5] H. Hosoya. Topological index as a common tool for quantum chemistry, statistical mechanics, and graph theory. In Mathematical and computational concepts in chemistry (Dubrovnik, 985), Ellis Horwood Ser. Math. Appl., pages Horwood, Chichester, 986. [6] Y. Hou. On acyclic systems with minimal Hosoya index. Discrete Appl. Math., 9(3):25 257, [7] F. Jelen and E. Triesch. Superdominance order and distance of trees with bounded maximum degree. Discrete Appl. Math., 25(2-3): , [8] P. Kirschenhofer, H. Prodinger, and R. F. Tichy. Fibonacci numbers of graphs. II. Fibonacci Quart., 2(3):29 229, 983. [9] P. Kirschenhofer, H. Prodinger, and R. F. Tichy. Fibonacci numbers of graphs. III. Planted plane trees. In Fibonacci numbers and their applications (Patras, 984), volume 28 of Math. Appl., pages Reidel, Dordrecht, 986. [0] X. Li, Z. Li, and L. Wang. The inverse problems for some topological indices in combinatorial chemistry. J. Comput. Biol., 0:47 55, [] X. Li, H. Zhao, and I. Gutman. On the Merrifield-Simmons index of trees. MATCH Commun. Math. Comput. Chem., 54(2): , [2] S. B. Lin and C. Lin. Trees and forests with large and small independent indices. Chinese J. Math., 23(3):99 20, 995. [3] R. E. Merrifield and H. E. Simmons. Topological Methods in Chemistry. Wiley, New York, 989. [4] H. Prodinger and R. F. Tichy. Fibonacci numbers of graphs. Fibonacci Quart., 20():6 2, 982. [5] L. A. Székely and H. Wang. On subtrees of trees. Adv. in Appl. Math., 34():38 55, [6] L. A. Székely and H. Wang. Binary trees with the largest number of subtrees. Discrete Appl. Math., 55(3): , [7] R. F. Tichy and S. G. Wagner. Algorithmic generation of molecular graphs with large Merrifield-Simmons index. Preprint available at [8] N. Trinajstić. Chemical graph theory. CRC Press, Boca Raton, FL., 992. [9] S. Wagner. Extremal trees with respect to Hosoya Index and Merrifield-Simmons Index. MATCH Commun. Math. Comput. Chem., 57():22 233, Institut für Mathematik B, Technische Universität Graz, Austria address: clemens.heuberger@tugraz.at Department of Mathematical Sciences, University of Stellenbosch, South Africa address: swagner@sun.ac.za