Linear Algebra and its Applications 439 (2013) 2240–2249
Contents lists available at SciVerse ScienceDirect
Linear Algebra and its Applications
www.elsevier.com/locate/laa
Distance spectral radius of trees with fixed
number of pendent vertices ✩
Wenjie Ning a,∗ , Liangqi Ouyang b , Mei Lu a
a
b
Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China
Department of Mechanical Engineering, Tsinghua University, Beijing 100084, China
a r t i c l e
i n f o
Article history:
Received 16 October 2012
Accepted 25 June 2013
Available online 12 July 2013
Submitted by R. Brualdi
a b s t r a c t
The distance spectral radius ρ (G ) of a graph G is the largest eigenvalue of the distance matrix D (G ). In this paper, we characterize
the graph with minimum distance spectral radius among trees
with fixed number of pendent vertices.
© 2013 Elsevier Inc. All rights reserved.
MSC:
05C50
15A18
Keywords:
Tree
Distance matrix
Distance spectral radius
Pendent vertex
1. Introduction
In this paper, a graph means a simple connected undirected graph. Let G = ( V , E ) be a graph
with n vertices. For two vertices v i and v j (i = j) in G, the distance between v i and v j , denoted
by di j or d( v i , v j ), is the length of a shortest path joining v i and v j . The distance matrix of G is the
matrix D (G ) = (di j ). It is obvious that D (G ) is a real symmetric matrix. Thus its eigenvalues are real
numbers. We denote by ρ (G ) the largest eigenvalue of D (G ), and call it the distance spectral radius
of G. By the Perron–Frobenius theorem, we know that ρ (G ) > 0 and there exists a unique positive
✩
This work is partially supported by National Natural Science Foundation of China (Nos. 10971114, 10990011, 11171097).
Corresponding author.
E-mail addresses: [email protected] (W. Ning), [email protected] (L. Ouyang), [email protected]
(M. Lu).
*
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved.
http://dx.doi.org/10.1016/j.laa.2013.06.030
W. Ning et al. / Linear Algebra and its Applications 439 (2013) 2240–2249
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Fig. 1. T min (8, 4).
unit vector x = (x1 , x2 , . . . , xn ) T called Perron eigenvector such that D (G )x = ρ (G )x, where xi is the
coordinate for the vertex v i .
The distance matrix of a graph came up in many areas, including communication network design
[6], graph embedding theory [4,7,8], molecular stability [11,16] and network flow algorithms [3,5].
Balaban et al. [1] proposed the use of ρ (G ) as a molecular descriptor, while in [10] it was used to
infer the extent of branching and model boiling points of alkanes. So it is of interest to study the
distance matrix and its eigenvalues.
Estimating bounds for ρ (G ) is of great interest, and many results have been obtained. Subhi and
Powers in [17] proved that for n 3 the path P n has the maximum distance spectral radius among
trees on n vertices. Stevanović and Ilić in [18] generalized this result, and proved that among trees
with fixed maximum degree , the broom graph has maximal distance spectral radius. Denote by
A (n, m) (n 2m) the tree obtained from the star S n−m+1 by attaching a pendent edge to each of
certain m − 1 non-central vertices of S n−m+1 . Ilić in [12] proved that the tree A (n, m) minimizes the
distance spectral radius among n-vertex trees with matching number m. In [2], Bose et al. showed
that among graphs with r pendent vertices, K nr is the unique graph with minimal distance spectral
radius for 0 r n − 1 but r = n − 2 and the double star S (n − 3, 1) is the unique graph with
minimal distance spectral radius for r = n − 2, where K nr is a graph obtained by joining r independent
vertices to one vertex of the complete graph K n−r . For other results, the readers may refer to [9–22].
In this paper, we characterize the graph with minimum distance spectral radius among trees with
fixed number of pendent vertices.
Suppose G is a graph. Let N ( v ) be the set of vertices adjacent to v in G. The degree of v, denoted
by d( v ), is equal to | N ( v )|. A vertex of degree one is called a pendent vertex. The edge incident with
a pendent vertex is known as a pendent edge. Denote by K(G ) the number of pendent vertices in G.
For k 1, we say a path P = w v 1 v 2 · · · v k is a pendent path if d( w ) > 2, d( v 1 ) = d( v 2 ) = · · · =
d( v k−1 ) = 2 (if they exist) and d( v k ) = 1. In this case, we say P is adjacent to w. Two pendent
paths P = w v 1 v 2 · · · v p (d( v p ) = 1) and Q = vu 1 u 2 · · · u q (d(u q ) = 1) are adjacent if w and v are the
same. The lengths of P and Q are almost the same if | p − q| 1. Set n j = |{ v ∈ V (G ) | d( v ) j }|.
For n 2, m 2, j 1, let T (n, m) = { T | T is a tree of order n and K( T ) = m}, T (n, m, j ) = { T ∈
T (n, m) | n3 = j } and ρmin (T (n, m)) = min{ρ ( T ) | T ∈ T (n, m)}. It is easy to check that when n =
2, 3, 4, 5, there is exactly one tree in T (n, m), for each possible value of m. The unique tree in T (n, 2)
is P n . So we only consider the case n 6 and m 3 in the next sections. For n 6, m 3, denote
by T min (n, m) the tree T ∈ T (n, m, 1) such that the lengths of any two pendent paths in T are almost
the same. Fig. 1 illustrates the graph T min (n, m) with n = 8, m = 4.
This paper is organized as follows. In Section 2, we introduce some lemmas that will be used later.
In Section 3, we give a graph transformation that decreases the distance spectral radius. Special cases
are also studied here. In Section 4, we find the extremal tree that uniquely minimizes the distance
spectral radius among trees with fixed number of pendent vertices.
2. Lemmas
Let G be a graph and v 0 be a vertex of G. We denote by G ( v 0 , k) the graph obtained from G by attaching at the vertex v 0 a pendent path P = v 0 v 1 · · · v k of length k. Let ρ be the distance spectral radius of G ( v 0 , k) and x0 , x1 , . . . , xk be the coordinates of the Perron vector x of G ( v 0 , k) for vertices v 0 ,
v 1 , . . . , v k , respectively. Let S be the sum of all coordinates of x. Then we have the following lemma.
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Lemma 2.1. (See [18].)
k
xi = x0 f (k) +
i =1
S
ρ
g (k),
where
f (x) =
t (t 2x − 1)
(1 + t 2x+1 )(t − 1)
and
t = 1 + 1 /ρ +
g (x) =
,
t (t x − 1)(t x+1 − 1)
(1 + t 2x+1 )(t − 1)2
2 ρ + 1 /ρ .
Furthermore, f (x) and g (x) are monotonically increasing functions.
Lemma 2.2. For t > 1, we have 2 + t − t 2 − 2t 3 + t 5 > 0 and 2 − t 2 + t 3 − 2t 5 + t 7 > 0.
Proof. Let f (t ) := 2 + t − t 2 − 2t 3 + t 5 and g (t ) := 2 − t 2 + t 3 − 2t 5 + t 7 . By a simple calculation,
we have f (t ) = 5t 4 − 6t 2 − 2t + 1, f (t ) = 20t 3 − 12t − 2 and f (3) (t ) = 60t 2 − 12. It is easy to see
that f (3) (t ) > 0 when t 1. Thus, we get f (t ) f (1) = 6 for t > 1, which indicates that f (t ) is
strictly monotone increasing when t 1. Note that f (1) = −2 < 0 and f (t ) f (1.2) = 0.328 > 0
for t 1.2. Thus, for t > 1, f (t ) gets its minimum value at some point, say b ∈ (1, 1.2).
On the other hand, we have f (b) = f (1) + f (ξ )(b − 1) by the Lagrange theorem, where ξ ∈
(1, b). Because of f (t ) being strictly monotone increasing when t 1, we get −2 = f (1) < f (ξ ) <
f (1.2) = 0.328, which indicates that f (ξ )(b − 1) > −2 ×(1.2 − 1) = −0.4 and f (b) = f (1)+ f (ξ )(b −
1) > 1 − 0.4 = 0.6. Note that when t > 1, f (t ) gets its minimum value at b. Thus, 2 + t − t 2 − 2t 3 +
t 5 > 0 holds for t > 1.
For the inequality g (t ) = 2 − t 2 + t 3 − 2t 5 + t 7 > 0 when t > 1, we can also prove it with the same
method we used above, so we omit it here. 2
Lemma 2.3. Suppose n is an integer. Let f (x) and g (x) be defined as in Lemma 2.1. Then 3 f (1) > f (2),
3g (1) > g (2), 2 f (n) > f (n + 1) for n 1 and 2g (n) > g (n + 1) for n 2.
ρ
Proof. It had been showed in [12] that f (1) = g (1) = ρ +2 ,
f (2) =
2ρ (ρ + 1)
ρ + 6ρ + 4
2
and 3g (1) > g (2). Since
3 f (1) − f (2) =
,
g (2) =
ρ (3ρ + 2)
ρ 2 + 6ρ + 4
ρ > 0, we obtain
ρ 3 + 12ρ 2 + 8ρ
> 0.
(ρ + 2)(ρ 2 + 6ρ + 4)
Furthermore, for any integer n 1,
2 f (n) − f (n + 1) =
t(
2(−1+t 2n )
1+t 1+2n
+
1−t 2+2n
)
1+t 3+2n
−1 + t
and
2(−1+t n )
2+n
t (−1 + t 1+n )( 1+t 1+2n + 11+−tt3+2n )
2g (n) − g (n + 1) =
.
(−1 + t )2
Since t > 1, we get −1 + t > 0 and −1 + t 1+n > 0.
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From Lemma 2.2 and the fact that t > 1, we have
2(−1 + t 2n )
1 + t 1+2n
+
1 − t 2+2n
1 + t 3+2n
−1 + t 2n (2 + t − t 2 − 2t 3 + t 2n+3 )
(1 + t 1+2n )(1 + t 3+2n )
−1 + t 2n (2 + t − t 2 − 2t 3 + t 5 )
(1 + t 1+2n )(1 + t 3+2n )
−1 + t 2 (2 + t − t 2 − 2t 3 + t 5 )
(1 + t 1+2n )(1 + t 3+2n )
(t − 1)3 (t + 1)2 (t 2 + t + 1)
=
(1 + t 1+2n )(1 + t 3+2n )
> 0.
=
Thus, we obtain 2 f (n) > f (n + 1) when n 1.
For n 2,
2(−1 + t n )
1 + t 1+2n
+
1 − t 2+n
1 + t 3+2n
−1 + t n (2 − t 2 + t n (t − 2t 3 + t n+3 ))
(1 + t 1+2n )(1 + t 3+2n )
n
−1 + t (2 − t 2 + t n (t − 2t 3 + t 5 ))
(1 + t 1+2n )(1 + t 3+2n )
−1 + t n (2 − t 2 + t 2 (t − 2t 3 + t 5 ))
(1 + t 1+2n )(1 + t 3+2n )
−1 + t n (2 − t 2 + t 3 − 2t 5 + t 7 )
=
,
(1 + t 1+2n )(1 + t 3+2n )
=
which follows from the fact that t − 2t 3 + t n+3 t − 2t 3 + t 5 > 0 when t > 1. From Lemma 2.2 and
the fact that t > 1, we have
−1 + t n 2 − t 2 + t 3 − 2t 5 + t 7 −1 + t 2 2 − t 2 + t 3 − 2t 5 + t 7
= −1 + 2t 2 − t 4 + t 5 − 2t 7 + t 9
= (t − 1)3 (t + 1)2 t 4 + t 3 + t 2 + t + 1
> 0.
Thus, we obtain 2g (n) > g (n + 1) when n 2.
2
Let G be a graph and w be a vertex of G. For p , q 1, denote by G ( w , p , q) the graph obtained
from G by attaching at w two pendent paths P = w v 1 v 2 · · · v p and Q = wu 1 u 2 · · · u q .
Lemma 2.4. (See [18].) If p q 1, then ρ (G ( w , p , q)) < ρ (G ( w , p + 1, q − 1)).
Lemma 2.5. Let T ∈ T (n, m, j ) such that ρ ( T ) = min{ρ ( T ) | T ∈ T (n, m, j )}, then the lengths of any two
adjacent pendent paths in T are almost the same.
Proof. Suppose to the contrary, there are two adjacent pendent paths P = w v 1 v 2 · · · v p and Q =
wu 1 u 2 · · · u q such that | p − q| > 1. Without loss of generality, we assume p − q > 1, then p − 1 q + 1 1. Let T = T − v p −1 v p + u q v p . Then T ∈ T (n, m, j ). By Lemma 2.4, ρ ( T ) < ρ ( T ), a contradiction. 2
Let T be an arbitrary tree and uv an arbitrary edge of T . Let x be the Perron eigenvector corresponding to the distance spectral radius ρ ( T ) = ρ and S the sum of all coordinates of x. Denote by S u
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Fig. 2. Tree T .
Fig. 3. Transformed tree T .
the sum of all coordinates of the Perron vector in the component containing vertex u, after removing
the edge uv. Then the following lemma holds.
Lemma 2.6. (See [12].) For an arbitrary edge uv of T , ρ (xu − x v ) = S v − S u = S − 2S u .
Lemma 2.7. (See [12].) Let G = ( V , E ) be a graph of order n. Let D = D (G ) and ρ = ρ (G ). Then
ρ
x T Dx
xT x
=
2
i < j xi x j d i j
2
i =1 x i
n
and the equality holds if and only if x is the Perron eigenvector corresponding to ρ (G ).
3. Transformation
In this section, we give a graph transformation which will be useful to derive our main results.
3.1. Transformed tree T Let T and T be trees showed in Fig. 2 and Fig. 3, respectively, where v is a vertex of degree
p + q + 1 and G is a subtree of T and T (p + q 2, p 0 and q 1). Let ρ = ρ ( T ), ρ = ρ ( T ),
D = D ( T ) and D = D ( T ). Let x be the Perron eigenvector corresponding to ρ and S the sum of all
coordinates of x. Denote by S the sum of Perron coordinates of vertices not in G. Without loss of
generality, denote by P 1 , . . . , P p , P p +1 , . . . , P p +q the p + q paths adjacent to w in T . The lengths of
P 1 , . . . , P p , P p +1 are equal to k + 1 and the lengths of P p +2 , . . . , P p +q are equal to k if q 2. Let xu
be the Perron coordinate of any vertex u ∈ V (G ). Let ai , j be the Perron coordinate of the vertex in P i ,
the distance between which and w is equal to j in T . Using symmetry, we can assume that
a 1 , j = a 2 , j = · · · = a p , j = a p +1 , j = a j
a p +2, j = a p +3, j = · · · = a p +q, j = a j
k+1
k
Denote by α =
j =1 a j .
j =1 a j and β =
( j = 1, 2, . . . , k + 1),
( j = 1, 2, . . . , k).
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Proposition 3.1. (1) S > α ; (2) α = x w f (k + 1) + ρS g (k + 1); (3) xu x w for any u ∈ V (G ) if S S + α .
Proof. (1) By a simple calculation, we have
S = ( p + 1)
k +1
k
a j + (q − 1)
j =1
aj = ( p + 1)α + (q − 1)β.
j =1
1, then S If p 1, q 2α > α ; if p = 0, q 2, then S α + β > α . In either case, we always
have S > α .
k+1
(2) Note that P 1 is a pendent path of length k + 1 in T . By Lemma 2.1, we have α =
j =1 a j =
x w f (k + 1) + ρS g (k + 1).
(3) For every edge uu in G with d T (u , w ) > d T (u , w ), by Lemma 2.6 and (1), we have
ρ (xu − xu ) = S u − S u = 2S u − S 2 S + x w − S + α = S + 2x w − α > 0,
which implies xu > xu . Thus for any u ∈ V (G ), xu x w if S S + α .
2
Proposition 3.2. If S > S + α , then ρ > ρ .
Proof. By Lemma 2.7, we have
ρ
=
=
x T Dx
xT x
x T D x + ( S − α )( S − S ) − ( S − α )α + ( S − α )( S − S − α )
xT x
x D x + 2( S T
− α )( S −
S
− α)
xT x
>
xT D x
xT x
= ρ.
The last inequality holds from Proposition 3.1(1) and the condition S > S + α .
2
Lemma 3.1. Let T and T be defined as above. Suppose G contains a pendent path of length at least k + 1. Then
ρ > ρ.
Proof. By Proposition 3.2, we just need to show that S > S + α .
Suppose S S + α . Then xu x w for any u ∈ V (G ) by Proposition 3.1(3). Let P = uv 1 v 2 · · · v t be
a pendent path in G, where t k + 1 and d( v t ) = 1. Denote by x1 , x2 , . . . , xt the Perron coordinates
of v 1 , v 2 , . . . , v t , respectively. By Lemma 2.1 and Proposition 3.1(3), we have
S S + xu +
t
i =1
S
xi = S + xu + xu f (t ) + g (t )
S + xu + xu f (k + 1) +
ρ
S
ρ
S + x w + x w f (k + 1) +
g (k + 1)
S
ρ
g (k + 1).
By Proposition 3.1(2), we have
S
S S + x w + x w f (k + 1) + g (k + 1) = S + x w + α > S + α ,
ρ
a contradiction.
2
Lemma 3.2. Let T and T be defined as above. Suppose G contains two pendent paths of lengths at least k. If
k 2, then ρ > ρ .
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Proof. By Lemma 3.1, we will assume that G contains two pendent paths of lengths equal to k. By
Proposition 3.2, we just need to show that S > S + α .
Suppose S S + α . Then xu x w for any u ∈ V (G ) by Proposition 3.1(3). Let P 1 = u 1 v 1 v 2 · · · v k
and P 2 = u 2 v k+1 v k+2 · · · v 2k be two pendent paths in G, where d( v k ) = d( v 2k ) = 1. Denote by
x1 , x2 , . . . , x2k the Perron coordinates of v 1 , v 2 , . . . , v 2k , respectively. By Proposition 3.1(2), Lemmas 2.1
and 2.3, we have
S S +
2k
xi
i =1
= S + xu 1 f (k) +
S
ρ
S + 2x w f (k) +
S
g (k) + xu 2 f (k) +
2S
ρ
> S + x w f (k + 1) +
ρ
g (k)
g (k)
S
ρ
g (k + 1) = S + α ,
2
a contradiction.
Lemma 3.3. Let T and T be defined as above. Suppose G contains m (m 3) pendent paths of lengths at
least k. If k 1, then ρ > ρ .
Proof. By Lemma 3.2, it suffices to prove the case k = 1. By Proposition 3.2, we just need to show
that S > S + α .
Suppose S S + α . Then xu x w for any u ∈ V (G ) by Proposition 3.1(3). According to the hypothesis, there are m pendent vertices v 1 , v 2 , . . . , v m adjacent to u 1 , u 2 , . . . , um in G, respectively. Denote
by x1 , x2 , . . . , xm the Perron coordinates of v 1 , v 2 , . . . , v m , respectively. From Lemmas 2.1 and 2.3,
Proposition 3.1(2), we have
S S +
m
xi = S +
i =1
a contradiction.
xu i f (1) +
i =1
S + 3x w f (1) +
> S + x w f (2) +
m 3S
ρ
S
ρ
S
ρ
g (1)
g (1)
g (2) = S + α ,
2
3.2. Special cases
Let T # and T ∗ be the trees depicted in Fig. 4 and Fig. 5, where n 8. Let
D # = D ( T # ) and D ∗ = D ( T ∗ ).
ρ # = ρ ( T # ), ρ ∗ = ρ ( T ∗ ),
Lemma 3.4. ρ # > ρ ∗ .
Proof. Let x be the Perron eigenvector corresponding to
k = 1, 2, . . . , n. By Lemma 2.7, we have
ρ ∗ . Using symmetry, we have xk = xn+1−k ,
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Fig. 4. T # (n 8).
Fig. 5. T ∗ (n 8).
Fig. 6. T 6 .
Fig. 7. T 6 .
ρ# xT D # x
xT x
x D ∗ x + x1 (2xn − x2 − x3 +
T
=
+
=
=
=
>
k=4 xk ) + xn (2x1
xT x
(xn − x1 )(x2 + x3 ) + (x1 + xn )
n−3
k=4 xk
− xn−2 − xn−1 +
(x1 + xn )
n−3
n−3
k=2 xk )
+ (x1 − xn )(xn−2 + xn−1 )
xT x
x T D ∗ x + x1 (2xn − x2 − x3 +
+
n−1
n−1
k=4 xk ) + xn (2x1
xT x
− xn−2 − xn−1 +
n−3
k=2 xk )
k=4 xk
xT x
x T D ∗ x + x1 (2xn +
n−3
k=4 xk ) + xn (2x1
xT x
x T D ∗ x + 4x1 xn + 2(x1 + xn )
+
n−3
k=4 xk ) + (x1
+ xn )
n−3
k=4 xk
n−3
k=4 xk
xT x
xT D ∗ x
xT x
= ρ∗.
2
By a simple calculation, we easily have the following lemma.
Lemma 3.5. Let T 6 , T 6 , T 7 , T 7 be the trees depicted in Figs. 6, 7, 8 and 9, respectively. Then ρ ( T 6 ) > ρ ( T 6 ),
ρ ( T 7 ) > ρ ( T 7 ).
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Fig. 8. T 7 .
Fig. 9. T 7 .
4. Trees with fixed number of pendent vertices
We have our main result as follows.
Theorem 4.1. Let T ∈ T (n, m) (n 6, m 3) and ρ ( T ) = ρmin (T (n, m)). Then T = T min (n, m).
Proof. Suppose T ∈ T (n, m, j ) ( j 1) and
ing fact.
ρ ( T ) = ρmin (T (n, m)). By Lemma 2.5, we have the follow-
Fact 1. The lengths of any two adjacent pendent paths in T are almost the same.
Fact 2. j = 1.
Proof of Fact 2. Suppose j 2. By the definition of T (n, m, j ), there are j vertices in T whose degrees
are at least 3. Let A = { w 1 , . . . , w j } ⊆ V ( T ) such that d( w i ) 3 for each w i ∈ A, then | A | = j 2. Let
B = w i w i ∈ A and there are d( w i ) − 1 pendent paths adjacent to w i .
2
Claim. | B | 2.
Proof of Claim. Without loss of generality, we assume that w 1 , w 2 are two vertices in A such that the
distance between them is the largest and P = w 1 v 1 · · · v k w 2 is the only path connecting w 1 and w 2 .
We claim that both w 1 and w 2 belong to B.
Suppose to the contrary that w 2 ∈
/ B. By the definition of a pendent path, there must be a
path Q = w 2 u 1 · · · ul (l 1) with u 1 , u 2 , . . . , ul not on P and d(ul ) 3. Thus, the path P ∪ Q =
w 1 v 1 · · · v k w 2 u 1 · · · ul is the only path connecting w 1 and ul in T , with w 1 and ul both in A. Note
that the length of P ∪ Q is longer than P , which is a contradiction to the selection of w 1 and w 2 .
Similarly, we can prove that w 1 ∈ B. Thus, | B | 2. 2
By the above Claim, without loss of generality, we assume w 1 , w 2 ∈ B and the length of the
longest pendent path adjacent to w 2 is more than and equal to that of w 1 .
Suppose d( w 1 ) = p + q + 1, w ∈ N ( w 1 ) and there are p pendent paths of length k + 1 and q pendent paths of length k (q 1 and k 1), each of which is adjacent to w 1 and does not contain w.
Since d( w 1 ) 3, we have p + q 2. By taking w 1 as v in Fig. 2, we construct T from T as in
Section 3.1. Then T ∈ T (n, m). Let G be the subtree containing w in T − w w 1 . Then w 2 ∈ V (G ).
By Lemma 3.1, there is no pendent path in G, whose length is at least k + 1. That is, the lengths
of pendent paths in G are at most k. Since w 2 ∈ V (G ), we have that the lengths of pendent paths
W. Ning et al. / Linear Algebra and its Applications 439 (2013) 2240–2249
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adjacent to w 1 are equal to k by assumption, i.e., p = 0 and q 2. Since w 2 ∈ A, there is at least one
pendent path of length k adjacent to w 2 by assumption. We will complete our proof by considering
the following two cases.
Case 1. k 2.
If there is only one pendent path of length k adjacent to w 2 in G, then there is at least one
pendent path of length less than k adjacent to w 2 in G because of w 2 ∈ B. By replacing w 1 with w 2 ,
we can derive a contradiction from Lemma 3.1.
So we can assume there are at least two pendent paths adjacent to w 2 in G whose lengths are
equal to k, then ρ ( T ) > ρ ( T ) by Lemma 3.2, a contradiction.
Case 2. k = 1.
In this case, the lengths of pendent paths in T adjacent to w 1 and w 2 are equal to 1.
By Lemma 3.3, we can assume there are exactly two pendents v 2 , v 2 in G. Then both of v 2 , v 2 are
adjacent to w 2 .
Also by Lemma 3.3, we assume that there are exactly two pendents adjacent to w 1 . Thus T ∈
{ T # , T 6 , T 7 }. By Lemmas 3.4 and 3.5, we have ρ ( T # ) > ρ ( T ∗ ), ρ ( T 6 ) > ρ ( T 6 ) and ρ ( T 7 ) > ρ ( T 7 ), our
final contradictions. 2
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