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Pesquisa Operacional
versão impressa ISSN 01017438
Pesqui. Oper. vol.33 no.1 Rio de Janeiro jan./abr. 2013
https://doi.org/10.1590/S010174382013000100008
Some results about the connectivity of trees
Lilian Markenzon^{I}; Nair Maria Maia de Abreu^{II, }^{*}; Luciana Lee^{III}
^{I}NCE  Universidade Federal do Rio de Janeiro, Brazil. Email: markenzon@nce.ufrj.br
^{II}Programa de Engenharia de Produção/COPPE, Universidade Federal do Rio de Janeiro. Email: nair@pep.ufrj.br
^{III}Universidade Federal do Mato Grosso. Email: lucianalee@ufmt.br
ABSTRACT
The second smallest Laplacian eigenvalue of a graph G is called algebraic connectivity, denoted a(G). The ordering of trees via this graph invariant is frequently studied in the literature. In this paper, we present a new invariant, the Internal Degree Sequence (IDS), that also supports an accurate evaluation of the connectivity of trees. We compare the IDS with a(G) for all elements in six classes of trees known to have the largest algebraic connectivity and we show that the IDS provides a strict total ordering of the elements of these classes. This result is also proved for a subclass of trees of diameter 4.
Keywords: trees; internal degree sequence, algebraic connectivity.
1 PRELIMINARIES
Let G = (V,E) be a simple graph, n = V its order and m = E its size. The degree of a vertex v is denoted by d(v). A graph is connected if for any two vertices u,v ∈ V there is a path from u to v. The graph is said to be disconnected if it has two or more connected components. The cutvertex (or cutpoint) is a vertex whose removal increases the number of componentsof G.
The Laplacian matrix L(G) = [ℓ_{ij}] is defined as follows: ℓ_{ij} = d_{i}, for i = j; ℓ_{ij} = 1 if (v_{i},v_{j}) is an edge of G and ℓ_{ij} = 0 otherwise. The spectrum of the Laplacian of G is the sequence of their eigenvalues (µ_{1}, ..., µ_{n1}, µ_{n}), given in nonincreasing order, µ_{1} > ... > µ_{n1} > µ_{n}. Since L(G) is a semidefinite positive singular matrix, µ_{n} = 0. It is well known that µ_{n1} = 0 if and only if G is a disconnected graph. Due to that, Fiedler [3] called µ_{n1} the algebraic connectivity of G, denoted a(G). For a survey about a(G) see Abreu [1]. The relation between the algebraic connectivity and the cutpoints of a graph has been addressed by Kirkland [6].
In this paper, we introduce a new concept for the connectivity of trees, the Internal Degree Sequence (IDS), taking into account the number of connected components obtained after the removal of each cutvertex. An ordering for trees with n vertices according to their IDS is defined. In Section 3, we compare the IDS with the algebraic connectivity a(G) for all elements in six classes of trees known to have the largest algebraic connectivity, presented by Yuan et al. [8], and we show that the IDS provides a strict total ordering of the elements of these classes. This result is also proved for a subclass of trees of diamenter 4 in Section 4.
2 THE NEW INVARIANT
In this section, we present a new invariant, the internal degree sequence (IDS), taking into account the number of connected components obtained after the removal of each cutvertex of the tree. The IDS allows us to differentiate the connectivity of trees, providing an extra resource to compare them. Another important feature to highlight is the simplicity of the determination of the invariant.
The degree sequence of a graph is the nonincreasing sequence of its vertex degrees. Let T = (V,E) be a tree and [d_{1}, d_{2}, ..., d_{n}] its degree sequence. Let V = {v_{1}, v_{2}, ..., v_{n}} be the ordered set of vertices of T, being d_{i} = d(v_{i}).
A vertex v of T such that d(v) > 1 is called an internal vertex of T. It is a cutvertex of T. A vertex v such that d(v) = 1 is called a leaf or an external vertex of T.
Let T = (V,E) be a tree and let L ⊂ V be the set of leaves of T. The removal of a vertex v ∈ V  L results in a disconnected tree; observe that there are d(v) components in the resulting tree. It is not difficult to see that the degree of an internal vertex is directly related with its relevance in the analysis of the connectivity. The next definition takes this fact into consideration.
Definition 1. The nonincreasing sequence of degrees of the internal vertices of a tree T is called the internal degree sequence (IDS) of a tree T. It is denoted
IDS(G) = [d_{1}, d_{2}, ..., d_{p}].
Definition 2. Let s = [d_{1}, d_{2}, ..., d_{p}] be a nonincreasing sequence of integers; s is a valid IDS for a tree T if T has n = Σ_{i = }_{1}_{,p} d_{i}  p + 2 vertices, p internal vertices and d_{i} ≠ 1, 1 < i < p.
It is known that a string t is preceded by a string s in lexicographic order if
 s is a prefix of t, or
 if c and d are respectively the first characters of s and t in which s and t differ, then c precedes d in character order.
It is possible to perform a lexicographic comparison of the IDS values of two trees. For instance, Figure 1 shows all trees with 7 vertices with their respective IDS in lexicographic order.
This example shows some coincidences with the algebraic connectivity that must be highlighted. Among all trees, the maximum and the minimum IDS occur with the star and the path, respectively; the same happens with the algebraic connectivity (Grone & Merris [5]). This observation leads us to compare the two approaches, as we will see in the next section.
It is possible to generate all valid IDS for a tree with n vertices. The sequences can be build recursively, computing each element of the sequence from the first one. Proposition provides some necessary results to build the sequences.
Proposition 1. Let T be a tree with order n and IDS(T) = [d_{1}, ...,d_{p}] its internal degree sequence. Let S = . Then
1. the maximum value for d_{i} is
2. the minimum value for d_{i} is
Proof. Let T be a tree of order n and let [d_{1}, d_{2}, ...,d_{p}] be the IDS of T. It is known that the sum of the degrees of the vertices of V is 2(n  1). The sum of the degrees of the p internal vertices of T is S = = n + p  2.
1. The maximum degree of the first vertex of the sequence occurs when all the others have the minimum degree for an internal vertex (i.e. 2). Since S = n + p 2, the first element d_{1} has value S  2(p  1) = n  p.
When computing d_{i}, 1 < i < p, the previous elements d_{1}, ..., d_{i}_{1} are already determined. It is possible to consider a sequence [d_{i}, ..., d_{p}] with sum of degrees S  . So, applying the previous reasoning and considering that, by definition, d_{i} < d_{i}_{1}, the result is obtained.
2. By Definition 1, d_{1} > ... > d_{p}. In order to force d_{1} to be minimum, the remaining degrees of S must be scattered in other positions of the sequence, obtaining the value S/p. If the remainder is zero then d_{1} = S/p. Otherwise, let r be the remainder. The positions d_{1}, d_{2}, ..., d_{r} will have the value S/P + 1. The minimum value of d_{1} is ⌈S/p⌉.
As in item(1), we consider [d_{1}, ..., d_{i}_{1}] already determined, and the sum of the degrees S  . So, the result is obtained.
The next proposition shows how any sequence realizes a tree that obeys a specific condition.
Proposition 2. Given a nonincreasing sequence of integers s = [d_{1}, d_{2}, ..., d_{p}], d_{i} ≠ 1, there exists always a tree T with n =  p +2 vertices such that IDS(T) = s.
Proof. The proof is constructive. By the definition of a valid sequence it is known that T has n =  p + 2 vertices and p internal vertices. The following steps build iteratively the tree T:

Step 1: Let T be a tree with a central vertex v and d_{1} vertices,
w_{1}, ..., w_{d}_{1}, all adjacent to v.

Step 2: For i = 2, ..., p do
Add d_{i}  1 new vertices, all adjacent to a leaf w_{j} of T.
The graph T, resulting from Step 1, is a tree; as so, it has at least two leaves. Step 2 adds, at each iteration, d_{i } 1 new vertices, all adjacent to the same vertex w_{j}. At this point, vertex w_{j} is no more a leaf, but at least one new leaf is created, maintaining the graph T as a tree. So, tree T has d_{1}+ Σ_{i = 2,p} (d_{i }  1) leaves and p internal vertices.
It is interesting to note that if the leaf w_{j} chosen in Step 2 is the most recently added leaf, the resulting tree is a caterpillar (a tree such that if all leaves and their incident edges are removed, the remainder of the graph forms a path).
Since the lexicographic order provides a total ordering for all the valid sequences of graphs with a given n, the proof of the following proposition is immediate.
Proposition 3. The invariant IDS provides a total ordering for trees with n vertices.
3 IDS AND ALGEBRAIC CONNECTIVITY
In 1987, Grone & Merris [4] started an investigation about connectivity of trees through a(G), proving in [5] that it is possible to present a strict total ordering of trees with diameter 3. After that, a large number of results comparing trees via a(G) have appeared in the literature as, for instance, Fallat & Kirkland [2] that extend the results of [5]. In this section, we compare some of the known results with those obtained using IDS.
Proposition 4. [2] Let P_{d}_{1} be a path with length d  2 labeled from 1 to d  1. Let T_{d}(k,ℓ) be a tree with n vertices and diameter d, obtained from P_{d}_{1} by adding k pendant vertices adjacent to vertex 1 and ℓ pendant vertices adjacent to vertex d  1. Then
The internal vertices of T_{d}(k,ℓ) are the vertices of the path. So, the IDS of T_{d}(k, ℓ) has d  1 elements and can be easily determined:
For this class, it is easy to prove that the ordering provided by IDS is a strict total ordering. Moreover, the orders given by a(G) and IDS are isomorphic. However, this behavior of the two measures does not occur in general.
Yuan et al. [8] introduced six classes of trees and show that, for n > 15, a(T_{i}) > a(T_{j}) if 1 < i < j < 6 and T_{i} is any tree in the class C_{i} and T_{j} is any tree in the class C_{j}. These classes are defined below and we show the IDS of their elements.
Definition 3. Let n, k, p and q be nonnegative integers with 3k + 2p + q = n  1. Let T (k,p,q) be the tree of order n which contains a vertex v such that T (k,p,q)  v = kK_{1,2} ∪ pK_{1,1} ∪ qK_{1}.
Let be the class of all trees T(k,p,d). The following subclasses of were defined in [8]:
 C_{1} = {T(0, 0, n  1)};
 C_{2} = {T(0, 1, n  3)};
 C_{3} = {T(0, p, q)  p > 2};
 C_{4} = {T(1, 0, n  4)};
 C_{5} = {T(1, p, q)  p > 1};
 C_{6} = {T(k, p, q)  k > 2}.
Notice that {T(k,p,q)  n = 3k + 2p + q + 1} = ∪_{i}_{ = 1,6} C_{i}.
For each n, classes C_{1}, C_{2} and C_{4} consist of only one tree. The trees belonging to C_{3} and C_{6} have the same algebraic connectivity, proved by Zhang [9] and Yuan et al. [8], respectively. Two important results about the ordering of all these trees are also presented by Yuan et al. [8] in Theorem 1.
Proposition 5. If T(k,p,q) is any tree in class C_{3}, then a(T) = (3 )/2.
Proposition 6. If T(k,p,q), k > 2, is any tree in class C_{6}, then a(T) = 2  .
Theorem 1. Let T_{i} be a tree in class C_{i} for i = 1, ...,6. Then:
1. For n > 8, a(T_{1}) > a(T_{2}) > a(T_{3}) > a(T_{4}) > a(T_{5}) > a(T_{6});
2. For n > 15 and T ∉ C_{i}, a(T) < a(T_{6}) = 2  .
For the six classes it is possible to establish the IDS of their elements. And more, since several trees have the same algebraic connectivity, the IDS provides a good insight to distinguish accurately between them.
Let T_{i} be a tree such that T_{i} ∈ C_{i}, n > 8. So, we have,
 IDS(T_{1}) = [n  1];
 IDS(T_{2}) = [n  2, 2] ;
 IDS(T_{3}) = [n  p  1, ], p > 2 ;
 IDS(T_{4}) = [n  3,3];
 IDS(T_{5}) = [n  p  3,3, ], p > 1 ;
 IDS(T_{6}) = [n  2k  p  1, , ], k > 2 .
The next proposition presents the comparison of the trees belonging to the six classes taking into account their IDS.
Proposition 7. Let T_{i} be any tree of order n, n > 8, such that T_{i} ∈ C_{i}. Then
1. IDS(T_{1}) > IDS(T_{2}) > IDS(T_{4});
2. IDS(T_{4}) > IDS(T_{3});
3. IDS(T_{4}) > IDS(T_{5});
4. IDS(T_{4}) > IDS(T_{6}).
Proof.
1. Each class C_{1}, C_{2} and C_{4} has just one element. It is immediate to state that
IDS(T_{1}) > IDS(T_{2}) > IDS(T_{4}).
2. The tree T ∈ C_{3} with the greatest IDS has p = 2 and IDS(T) = [n  4,2,2]. As IDS(T_{4}) = [n  3,3],
IDS(T_{4}) > IDS(T_{3}).
3. Similar reasoning can be made for classes C_{5} and C_{6}.
The tree T' ∈ C_{5} with the greatest IDS has p = 1. The tree T" ∈ C_{6} with the greatest IDS has k = 2 and p = 0. So, IDS(T') = [n  4,3,2] and IDS(T") = [n  5,3,3]. Comparing with the IDS of T_{4}
IDS(T_{4}) > IDS(T_{5}) and IDS(T_{4}) > IDS(T_{6}).
It is impossible to compare the trees belonging to classes C_{3}, C_{5} and C_{6} as a whole. Figure 2 shows how the IDS of the elements of these classes depend only on the types of their subtrees.
However a very interesting result can be stated, showing an advantage of the invariant when the trees have the same algebraic connectivity.
Theorem 2. Given n, the IDS of the elements of the classes C_{i}, i = 1,..., 6, determine a strict total ordering of the trees T(k,p,d).
Proof. We must prove that the trees belonging to C_{3}, C_{5} and C_{6} have different IDS values, both within and among classes.
Let T_{i} = T(k_{i},p_{i},q_{i}), T_{j} = T(k_{j},p_{j},q_{j}) ∈ C_{3}. We know that, for T_{3} ∈ C_{3}, IDS(T_{3}) = [n  p  1, ], p > 2 . So,
IDS(T_{i}) = IDS(T_{j}) ⇒ p_{i} = p_{j}.
The order n of T_{i} and T_{j} is given and p_{i} = p_{j}. As n = 2p + q + 1, q_{i} = q_{j} and the trees are the same and all trees in C_{3} have different IDS's.
The proof is analogous when T_{i},T_{j} ∈ C_{5} and T_{i},T_{j} ∈ C_{6}.
We must now show that trees belonging to different classes have different IDS. It is known that
IDS(T_{3}) = [n  p_{i } 1, ],
IDS(T_{5}) = [n  p_{j } 3, 3, ] and
IDS(T_{6}) = [n  2k  p  1, , ], k > 2.
Let T ∈ C_{3}. The tree T cannot have the same IDS as a tree belonging to C_{5} or C_{6} because the tree T has no internal vertex of degree 3 which is mandatory in the two other classes.
Let T' ∈ C_{5}. The tree T' cannot have the same IDS as a tree belonging to C_{6} because T' has exactly one internal vertex of degree 3 while the trees of C_{6} have at least two vertices with such condition.
So, the invariant IDS provides a strict total ordering for trees belonging to the classes C_{i}, i = 1,...,6.
4 TREES WITH DIAMETER 4
An important observation about the trees belonging to T, presented in the previous section, is that all of them have diameter less or equal 4. The approach of looking at trees with small diameter was taken up again by Wang & Tan [7]. They give continuity to the work of Yuan et al. [8], showing all trees of order n > 45 with algebraic connectivity in the interval [(5)/2 , 2).
The subclass of trees of diameter 4 is defined as follows by Zhang [10].
Definition 4. Let p_{1}, p_{2}, ...,p_{k} be integers such that
p_{1} > p_{2} > ... > p_{k} > 0, p_{1} > p_{2} > 0, k > 2, k + 1 + p_{1 }+ ... + p_{k} = n.
Let T(n,k,p_{1},p_{2},...,p_{k}) be a tree with diameter 4, N(c) = {v_{1}, ...,v_{k}}, being c the unique center of the tree, d(v_{1}) = p_{1 }+ 1, ..., d(v_{k}) = p_{k} + 1. Let T be the class of all trees T(n,k,p_{1},p_{2},...,p_{k}).
Wang & Tan[7] defined new subclasses of trees as follows:
 C'_{1} = {S(3, n  5)}, the tree of order n obtained by joining the center of K_{1,3} to the center of K_{1, n5} with an edge;
 C'_{2} = {T(n, k, 3, 1, p_{3}, ..., p_{k})};
 C'_{3} = {T(n, k, 3, 2, p_{3}, ..., p_{k})};
 C'_{4} = {T(n, k, 3, 3, p_{3}, ..., p_{k})};
Notice that C'_{1} is not a subclass of T.
Wang & Tan [7] proved interesting results about these classes:
Proposition 8. Let T be a tree of order n > 45. If T ∉ T ∪(C'_{i} ), then a(T) < (5  )/2.
Theorem 3. If T is a tree of order n > 45, then (5  )/2 < a(T) < 2  if and only if T ∈ C'_{i}.
The authors were able to provide the three trees with the first, second and third largest algebraic connectivity in T_{n} \T. However few results are presented about the ordering of all trees with diameter 4.
Using the IDS, an interesting result can be shown. We define a subclass of T. Let T' be the class of all T(n, k, p_{1}, p_{2}, ..., p_{k}) such that k > p_{1}, i.e., the set of all trees of diameter 4 such that the center of the tree has the greatest degree. It is possible to determine the IDS of a tree T ∈ T' directly from its notation.
Definition 5. Let T(n, k, p_{1}, ..., p_{z}, ..., p_{k}) with k > p_{1}. Let p_{z}, 2 < z < k, be the last nonzero element of the notation. Then
IDS(T) = [k, p_{1}+1, p_{2}+1, ..., p_{z}+1].
Theorem 4. Given n, the IDS of the elements of the class T' determine a strict total ordering of the trees T(n, k, p_{1}, p_{2}, ..., p_{k}) with k > p_{1}.
Proof. Let T(n, k, p_{1}, p_{2}, ..., p_{k}) ∈ T'. Let S be the sequence [n, k, p_{1}, p_{2}, ..., p_{z}, ...,p_{k}]. It is immediate to conclude that S is a nonincreasing sequence. The sequence IDS(T) is build from S and its elements obey exactly the original ordering. So, given n, there is a unique tree with this sequence as IDS.
Analogous result can be shown for all trees of diameter 4 such that the central vertex has the smallest degree.
ACKNOWLEDGMENTS
This work was supported by grants 305372/20092, 305016/20067, CNPq, Brazil and project 027/CAP/2011, Universidade Federal do Mato Grosso, Brazil.
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Received October 8, 2012
Accepted March 15, 2013
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