MATCH
Communications in Mathematical
and in Computer Chemistry
MATCH Commun. Math. Comput. Chem. 74 (2015) 81-96
ISSN 0340 - 6253
On the Maximum and Minimum
Zagreb Indices of Trees with a Given
Number of Vertices of Maximum Degree
Bojana Borovićanin1 , Tatjana Aleksić Lampert2
Department of Mathematics and Informatics, Faculty of Science,
University of Kragujevac, 34000 Kragujevac, Serbia
e-mail: 1 [email protected]
e-mail: 2 [email protected]
(Received January 26, 2015)
Abstract
For a (molecular) graph G the first Zagreb index M1 (G) is defined as the sum of the squares
of the vertex degrees, and the second Zagreb index M2 (G) is equal to the sum of the products
of the pairs of adjacent vertices’ vertex degrees. Let Tn,k be the class of trees with n vertices
of which k vertices have the maximum degree. In this paper we determine the extremal trees
of the class Tn,k , i.e., those with minimal (maximal) first Zagreb index or minimal (maximal)
second Zagreb index.
1
Introduction
All graphs considered in this paper are simple, connected graphs. Let G = (V, E) be
such a graph, where V = V (G) is its vertex set and E = E(G) is its edge set. An edge
connecting two vertices u and v in the graph G is denoted by uv. The degree dG (v) (or
d(v) for short) of a vertex v ∈ V (G) is the number of edges that are incident with v in the
graph G. A vertex v for which dG (v) = 1 is called a pendent vertex and a vertex of degree
three or more is called a branching vertex. The maximum vertex degree in the graph G is
denoted by ∆(G). A graph T that has n vertices and n − 1 edges is called a tree. For a
vertex v ∈ V (T ), such that 2 ≤ dT (v) ≤ ∆(T )−1, we say that its edge rotating capacity is
equal to dT (v) − 1. The total edge rotating capacity of a tree T is equal to the sum of the
edge rotating capacities of its vertices that satisfy the condition 2 ≤ dT (v) ≤ ∆(T ) − 1.
-82A sequence of positive integers π = (d1 , d2 , . . . , dn ) is called the degree sequence of
G if di = dG (v) (i = 1, . . . , n) holds for some v ∈ V (G). Throughout this paper, we
order the vertex degrees non-increasingly, i.e., d1 ≥ d2 ≥ · · · ≥ dn . Also, a sequence
π = (d1 , d2 , . . . , dn ) is called a tree degree sequence if there exists a tree T having π as its
degree sequence. Furthermore, it is well know that the sequence π = (d1 , d2 , . . . , dn ) is a
degree sequence of an n-vertex tree if and only if
n
X
di = 2(n − 1) .
(1)
i=1
If π = (d1 , d2 , . . . , dn ) and π 0 = (d01 , d02 , . . . , d0n ) are two different degree sequences, we
P
P
P
P
write π C π 0 if and only if ni=1 di = ni=1 d0i and ji=1 di ≤ ji=1 d0i for all j = 1, 2, . . . , n.
Such an ordering is called majorization [18, 19]. Also, we use Γ(π) to denote the class of
connected graphs that have the degree sequence π.
Molecular structure descriptors (topological indices) are used in mathematical chemistry to describe the properties of chemical compounds.
Some of the well studied molecular structure descriptors are the first and second Zagreb
indices, M1 (G) and M2 (G), respectively. They were introduced in 1972 by Gutman and
Trinajstić [8, 9], as follows
X
M1 (G) =
d2G (v)
(2)
dG (u)dG (v) .
(3)
v∈V (G)
and
M2 (G) =
X
uv∈E(G)
These indices reflect the extent of branching within the molecular carbon-atom skeleton, which allows them to be viewed as molecular structure descriptors [1, 21]. The main
properties of M1 and M2 were summarized in [10, 20] and the references therein. There
has been great interest in studing extremal graphs that minimize (or maximize) Zagreb
indices in different classes of graphs, recently [12, 13, 15]. Goubko and Gutman [4, 7]
described the trees with minimal first and second Zagreb indices among the trees with
a fixed number of pendent vertices. Lin [17] characterized trees with a fixed number
of vertices of degree two that maximize and minimize the first Zagreb index. In 2007,
Deng [3] described the graphs that have the largest and smallest Zagreb indices within the
classes of trees, unicyclic, and bicyclic graphs, by introducing some new transformations
of the graphs in these classes. One of the present authors [2] characterized the trees that
-83maximize and minimize the first and second Zagreb indices among the trees with a given
number of segments or given number of branching vertices.
Every tree contains at least two minimum degree vertices (i.e., pendent vertices) and
some maximum degree vertices. It is interesting to consider the trees with fixed number of
maximum degree vertices. Let Tn,k be the class of trees with n vertices that have exactly
k (≤ n − 2) vertices having the maximum degree. Recently, Lin [16] determined the trees
that maximize the Wiener index in the class Tn,k . In this paper we characterize the trees
that maximize (minimize) the first Zagreb index in this class, as well as the trees that
maximize (minimize) the second Zagreb index in the class Tn,k .
It is obvious that the path Pn is the unique element of Tn,n−2 . So, in the following we
consider the class Tn,k where k ≤ n − 3.
Let T be a tree from the class Tn,k with maximum vertex degree ∆. By ni (i =
1, 2, . . . , ∆) we denote the number of vertices of degree i in T . Then,
∆
X
ni = n .
(4)
ini = 2(n − 1) .
(5)
i=1
Also, from (1) it is obvious that
∆
X
i=1
By combining (4) and (5) we obtain
−n1 + n3 + 2n4 + · · · + (∆ − 2)n∆ = −2 .
This leads to the conclusion that
n1 ≥ 2 + (∆ − 2)n∆ = 2 + k(∆ − 2)
(6)
n ≥ n1 + n∆ ≥ 2 + k(∆ − 1) ,
(7)
n ≥ 2 + k(∆ − 1) .
(8)
and
i.e.,
Therefore,
k≤
n−2
.
∆−1
(9)
-84Since we assumed that k ≤ n − 3 (i.e., T 6= Pn ), then ∆ ≥ 3, which implies that
n−2
.
2
(10)
n−2
+1 .
k
(11)
k≤
Also, from (8) we see that
∆≤
Since ∆ is a positive integer number we obtain
∆≤b
n−2
c+1 .
k
(12)
By the previous considerations the following lemma holds.
Lemma 1.1. If T ∈ Tn,k is a tree with the maximum vertex degree ∆ then ∆ ≤ b n−2
c + 1.
k
2
The first Zagreb index
In this section, we first characterize the trees with maximal first Zagreb index in the
class Tn,k .
1
Lemma 2.1. If Tmax
is a tree with maximal first Zagreb index in Tn,k , then its maximum
vertex degree is equal to b n−2
c + 1.
k
1
Proof. Let ∆ be the maximum vertex degree in the tree Tmax
. Then, by Lemma 1.1,
∆ ≤ b n−2
c + 1. Let ∆max = b n−2
c + 1 and n − 2 = kb n−2
c + r, where 0 ≤ r < k.
k
k
k
Assume that ∆ < ∆max .
1
Let {v1 , . . . , vn } be the vertex set of a tree Tmax
with the degree sequence π =
(d1 , . . . , dn ). Then
∆ = d1 = · · · = dk = ∆max − t,
t>0.
By (6) we see that
n1 ≥ k(∆ − 2) + 2 .
(13)
Therefore, n1 = k(∆ − 2) + 2 + n01 , where n01 ≥ 0. Using (4) we obtain
n = k(∆ − 2) + 2 + n01 +
∆−1
X
ni + k ,
i=2
which implies
∆−1
X
i=2
ni + n01 = r + kt .
(14)
-85Also, from (5), by direct calculation, we have
∆−1
X
ini + n01 = 2(r + kt) .
(15)
i=2
By subtracting the relation in (14), from the relation in (15), we obtain
∆−1
X
(i − 1)ni = r + kt ≥ kt ≥ k .
(16)
i=2
1
Since the sum in (16) is equal to the total edge rotating capacity of Tmax
, we conclude
that the total edge rotating capacity of this tree is greater than or equal to k.
1
Let vi ∈ V (Tmax
) (k < i ≤ n) be a vertex that has positive edge rotating capacity
and let di (2 ≤ di ≤ ∆ − 1) be its degree. If we define a tree T1 with the vertex degree
sequence π1 = (d11 , . . . , d1n ) such that d11 = ∆ + 1, d1i = di − 1 and d1j = dj (j ∈ {2, . . . , n},
j 6= i) then
1
M1 (T1 ) − M1 (Tmax
) = (d1 + 1)2 − d21 + (di − 1)2 − d2i = 2(∆ − di + 1) > 0 .
1
Therefore, M1 (T1 ) > M1 (Tmax
), but T1 6∈ Tn,k . Since the total edge rotating capacity of
1
Tmax
is at least k, then we can conclude the following.
1
We can repeat the above described transformation of the tree Tmax
k times on every
vertex of degree ∆. In each step we define a tree Tl with the degree sequence πl =
(dl1 , . . . , dln ) such that dll = ∆ + 1, dli = dl−1
− 1 and dlj = dl−1
(j ∈ {1, . . . , n}, j 6= i, l),
i
j
where l = 2, . . . k and dl−1
is a degree of an arbitrary vertex vi ∈ V (Tl−1 ) (k < i ≤ n)
i
that has positive edge rotating capacity (this vertex exists since the total edge rotating
capacity of a tree Tl−1 is at least k − l + 1). It is possible that after some of the described
transformations we obtain a tree whose degrees dk+1 , . . . , dn are not in non-increasing
order. Besides, each application of this transformation strictly increases the first Zagreb
index. Finally, we obtain a tree Tk ∈ Tn,k that has the maximum vertex degree equal to
∆ + 1 = (∆max − t) + 1 and satisfies the condition M1 (Tk ) > M1 (Tk−1 ) > · · · > M1 (T1 ) >
1
1
M1 (Tmax
). This contradicts the fact that Tmax
has the maximum first Zagreb index in
the class Tn,k .
This proves that ∆ = ∆max = b n−2
c + 1.
k
Remark 2.1. The statement of Lemma 2.1 also holds for k = n − 2, i.e., T = Pn .
-86Theorem 2.1. Let T ∈ Tn,k , where 1 ≤ k ≤
2
n
2
− 1. Then
2
M1 (T ) ≤ k∆ + p(∆ − 1) + µ2 + n − k − p − 1,
and the equality holds if and only if T
has the vertex degree sequence
(∆, . . . , ∆, ∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1), where ∆ = b n−2
c + 1, p = b n−2−k(∆−1)
c and µ =
k
∆−2
| {z } |
{z
} | {z }
k
p
n−k−p−1
n − 1 − k(∆ − 1) − p(∆ − 2).
1
Proof. Let π = (d1 , . . . , dn ) be the vertex degree sequence of a tree Tmax
with maximal
first Zagreb index in the class Tn,k . By Lemma 2.1 we have that d1 = d2 = · · · = dk = ∆ =
b n−2
c+1. As in the previous lemma, let n−2 = k(∆−1)+r, where 0 ≤ r < k. Now, from
k
(6) it follows that n1 ≥ k(∆−2)+2 = n−k −r. Therefore, dn = dn−1 = · · · = dk+r+1 = 1.
Bearing in mind that n∆ = k and n1 = n01 + k(∆ − 2) + 2, where n01 ≥ 0, after using
(4) and (5) in the same manner as in (14) and (16) we obtain
∆−1
X
ni + n01 = r
(17)
(i − 1)ni = r .
(18)
i=2
and
∆−1
X
i=2
Since ni ≥ 0 (i = 2, . . . , ∆ − 1), from (18) it follows that p = n∆−1 ≤
non-negative integer number we obtain p ≤
r
.
∆−2
Since p is a
r
b ∆−2
c.
r
r
Suppose that b ∆−2
c > 0 and p < b ∆−2
c. Then, by (18),
P∆−2
i=2
(i − 1)ni ≥ ∆ − 2 and
there exist ni and nj (2 ≤ i < j ≤ ∆ − 2), where ni 6= 0 and nj 6= 0 (or there exists
ni ≥ 2 , where 2 ≤ i ≤ ∆ − 2 ) and the relation (18) is satisfied. Furthermore, since π =
(∆, . . . , ∆, dk+1 , . . . , dk+r , 1, . . . , 1), there exist numbers dk+j1 and dk+i1 (1 ≤ j1 < i1 ≤ r),
| {z }
k
such that dk+j1 = j > dk+i1 = i (or dk+j1 = dk+i1 = i, if ni ≥ 2).
Let π 0 = (d01 , . . . , d0n ) be a sequence of positive integers such that d0x = dx for x 6= k + j1
and x 6= k + i1 , d0k+j1 = dk+j1 + 1 = j + 1 and d0k+i1 = dk+i1 − 1 = i − 1. Since
Pn 0
0
0
i=1 di = 2n − 2, then π is the vertex degree sequence of a tree T , and
1
M1 (T 0 ) − M1 (Tmax
) = 2(j − i + 1) > 0 .
1
Since T 0 ∈ Tn,k and M1 (T 0 ) > M1 (Tmax
) we obtain a contradiction.
From the previous observation we conclude that p = n∆−1 = b n−2−k(∆−1)
c and the
∆−2
relation (18) now becomes
∆−2
X
(i − 1)ni = r − p(∆ − 2) ≤ ∆ − 3 .
i=2
-87In the same manner as previously, it can easily be proved that it has to be nµ = 1,
where µ = r − p(∆ − 2) + 1, i.e., µ = n − 1 − k(∆ − 1) − p(∆ − 2), and ni = 0 for i 6= µ,
2 ≤ i ≤ ∆ − 2, since in the opposite case we can again construct a tree whose M1 is
1
greater than M1 (Tmax
).
1
Therefore, the tree Tmax
with maximal first Zagreb index in the class Tn,k has the
vertex degree sequence π = (∆, . . . , ∆, ∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1).
| {z } |
{z
} | {z }
k
p
n−k−p−1
1
The first Zagreb index of the tree Tmax
can now be easily calculated.
Remark 2.2. Theorem 2.1 also holds for k = n − 2, i.e., T = Pn .
In the following theorem we will describe the trees that have the minimum first Zagreb
index in the class Tn,k , using the similar idea as in [2].
1
Lemma 2.2. Let Tmin
be a tree with minimal first Zagreb index in the class Tn,k , where
1≤k≤
n
2
− 1. Then, its maximum vertex degree ∆ equals 3.
1
Proof. Suppose that ∆ ≥ 4 and u is a vertex of maximum degree ∆ in Tmin
.
1
Fig. 1. The graphs Tmin
and T 1 in Lemma 2.2
1
Let P = v0 v1 . . . vi−1 u(= vi )vi+1 . . . vl be the longest path in Tmin
that contains u.
1
Also, let vi−1 , vi+1 , u1 , u2 , . . . , u∆−2 be the vertices adjacent to u in Tmin
, and z1 a pendent
vertex connected to u via u1 (it is possible that z1 ≡ u1 ) (Fig. 1). If we define a tree T 1
in the following way
1
T 1 = Tmin
− uu2 + u2 z1 ,
then
1
M1 (T 1 ) − M1 (Tmin
) = (∆ − 1)2 + 22 − ∆2 − 1 < 0 .
Obviously, the tree T 1 has k − 1 vertices of degree ∆.
(19)
-88In the same manner, we can apply the transformation described in (19) on every vertex
u of degree ∆. In each step from a tree T i we will obtain a tree T i+1 (1 ≤ i ≤ k−1) that has
smaller first Zagreb index than its predecessor. After k repetitions of the transformation
we arrive at the tree T k that has k vertices having the maximum degree ∆ − 1. Obviously,
T k ∈ Tn,k and
1
M1 (T k ) < M1 (Tmin
).
1
This contradicts the choice of Tmin
as the tree that minimizes M1 in the class Tn,k .
Theorem 2.2. Let T ∈ Tn,k where 1 ≤ k ≤
n
2
− 1. Then
M1 (T ) ≥ 2k + 4n − 6 ,
and the equality holds if and only if the tree T has the vertex degree sequence
(3, . . . , 3, 2, . . . , 2, 1, . . . , 1).
| {z } | {z } | {z }
k
n−2k−2
Proof. Let
1
Tmin
k+2
be a tree with minimal M1 in the class Tn,k . According to Lemma 2.2
the vertex degree sequence of this tree is π = (3, . . . , 3, 2, . . . , 2, 1, . . . , 1), when k ≤
| {z } | {z } | {z }
k
n2
n
2
− 1.
n1
Hence, using the equality (1) we obtain
n1 + 2n2 + 3k = 2(n1 + n2 + k) − 2 ,
(20)
which implies that n1 = k + 2 and n2 = n − 2k − 2. Therefore,
1
M1 (Tmin
) = k · 32 + (n − 2k − 2) · 22 + (k + 2) = 2k + 4n − 6 .
Corollary 2.1. Let T k be a tree with minimal first Zagreb index in the class Tn,k and T p
a tree with minimal first Zagreb index in the class Tn,p (1 ≤ k, p ≤
n
2
− 1). If k > p then
M1 (T k ) > M1 (T p ).
Proof. This result is a direct consequence of Theorem 2.2.
3
The second Zagreb index
Let π be a degree sequence. If G ∈ Γ(π) and M2 (G) ≥ M2 (G0 ) holds for any other
graph G0 ∈ Γ(π), then we say that the graph G has the maximumum second Zagreb index
in Γ(π). To prove our main result in this section we will use the following two lemmas.
-89Lemma 3.1. [18] Let π and π 0 be two different non-increasing tree degree sequences such
that π C π 0 . Let T and T 0 be the trees with maximal second Zagreb indices in Γ(π) and
Γ(π 0 ), respectively. Then, M2 (T ) < M2 (T 0 ).
Lemma 3.2. [22] Let T be a tree with maximal second Zagreb index, with ni vertices of
degree i and maximum vertex degree ∆, and let Ts be a subgraph induced by vertices of
degree ≥ s. Then Ts is connected.
Theorem 3.1. Let T ∈ Tn,k , where 1 ≤ k ≤
n
2
− 1. Then
M2 (T ) ≤ (k − 1)∆2 + 2p(∆ − 1)2 + µ(∆ + µ − 1) + ∆(n − k − (∆ − 1)p − µ),
where ∆ = b n−2
c + 1, p = b n−2−k(∆−1)
c and µ = n − 1 − k(∆ − 1) − p(∆ − 2). The equalk
∆−2
ity holds if and only if the following conditions are satisfied.
(i) The tree T has the vertex degree sequence (∆, . . . , ∆, ∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1).
| {z } |
{z
} | {z }
p
k
n−k−p−1
(ii) Every vertex of degree ∆−1 is adjacent to a vertex of degree ∆ and to ∆−2 pendent
vertices.
(iii) The vertex of degree µ (when µ > 1) is adjacent to a vertex of the degree ∆ and to
µ − 1 pendent vertices.
(iv) The remaining pendent vertices are attached to the vertices of degree ∆.
2
Proof. Let ∆ be the maximum vertex degree in a tree Tmax
with maximal second Zagreb
index in the class Tn,k . By Lemma 1.1 we know that ∆ ≤ b n−2
c + 1.
k
Let ∆max = b n−2
c + 1, and n − 2 = kb n−2
c + r (0 ≤ r < k). With π = (d1 , . . . , dn ) we
k
k
will denote a tree degree sequence such that
π = (∆max , . . . , ∆max , ∆max − 1, . . . , ∆max − 1, µ, 1, . . . , 1)
|
{z
} |
{z
} | {z }
k
p
n−k−p−1
r
where p = b ∆max −2 c and µ = r + 1 − p(∆max − 2).
2
Assume that ∆ < ∆max . Therefore, Tmax
has a degree sequence π 0 = (d01 , . . . , d0n ) 6= π.
Then d01 = d02 = · · · = d0k = ∆ = ∆max − t, t > 0 and
k
X
i=1
d0i = k∆ = k∆max − kt <
k
X
i=1
di .
-90Also, by (14), it holds that π 0 = (∆, . . . , ∆, d0k+1 , . . . , d0k+(r+kt) , 1, . . . , 1) and
| {z }
k
k+p
X
d0i ≤ (k + p)∆max − p − (k + p)t <
i=1
k+p
X
di = (k + p)∆max − p .
(21)
i=1
Having in mind that
k+(r+kt)
X
k+(p+1)
X
di =
i=k+1
di + (r + kt − p − 1) · 1 = p(∆max − 1) + µ + r + kt − p − 1 = 2r + kt
i=k+1
and
k+(r+kt)
X
d0i = 2(n − 1) − (n − k − r − kt) · 1 = r + k(∆max − 1) + k + r + kt = 2r + k∆max + kt
i=1
we obtain
k+(r+kt)
X
k+(r+kt)
d0i = k∆max + 2r + kt =
i=1
X
di .
(22)
i=1
Now, from relations (21) and (22) we conclude that there exists j (p ≤ j < r + kt),
s
s
l
l
X
X
X
X
such that
d0i <
di for s = 1, . . . , j and
d0i =
di for l = j + 1, . . . , n, which
i=1
i=1
i=1
i=1
implies
π0 C π .
(23)
Let T be a tree with maximal second Zagreb index among the trees with the degree
2
sequence π. Then T ∈ Tn,k and by Lemma 3.1 and (23), we have that M2 (T ) > M2 (Tmax
).
2
This contradicts the choice of Tmax
. Therefore, ∆ = ∆max = b n−2
c + 1.
k
Next, we notice that for two different vertex degree sequences
r
π = (∆, . . . , ∆, ∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1) (p = b
c and µ = r + 1 − p(∆ − 2))
| {z } |
{z
} | {z }
∆−2
p
k
n−k−p−1
and
π 0 = (∆, . . . , ∆, d0k+1 , . . . , d0k+r , 1, . . . , 1), where
| {z }
k
r
X
d0k+i = 2r ,
i=1
it holds
π0 C π .
(24)
-91P
In order to prove this, notice that the sum ri=1 d0k+i = 2r can be reduced to the
P
sum ∆−1
i=2 (i − 1)ni = r (ni is the number of vertices of degree i) in the same way as
r
was done in (18). Next, we conclude that n∆−1 ≤ p = b ∆−2
c. Also, for n∆−1 = p, let
P∆−2
µ − 1 = i=2 (i − 1)ni = r − p(∆ − 2) ≤ ∆ − 3. If µ > 1, then, by setting nµ = 1,
and the remaining ni ’s (2 ≤ i ≤ ∆ − 2) equal zero, we will get the degree sequence
(∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1) such that (d0k+1 , . . . , d0k+r ) C (∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1),
|
{z
} | {z }
|
{z
} | {z }
p
r−p−1
p
r−p−1
P
r
for any degree sequence (d0k+1 , . . . , d0k+r ), satisfying i=1 d0k+i = 2r, different from it.
2
The previous consideration leads us to the conclusion that a tree Tmax
with maximal
second Zagreb index in the class Tn,k must belong to Γ(π), i.e., it has to be (d0k+1 , . . . , d0k+r ) =
(∆ − 1, . . . , ∆ − 1, µ, 1, . . . , 1). In the opposite case, if T is a tree with maximal second
|
{z
} | {z }
p
r−p−1
Zagreb index among the trees with the degree sequence π, then T ∈ Tn,k and by Lemma
2
2
3.1 and (24), we have that M2 (T ) > M2 (Tmax
). This contradicts the choice of Tmax
.
2
Furthermore, using Lemma 3.2 we can conclude that the vertices in Tmax
of degree ∆
2
induce a connected subgraph Tk of Tmax
. The tree Tk contains k − 1 edges that connect
2
vertices of degree ∆ in Tmax
. As there are k∆ edges with at least one end vertex in V (Tk ),
we conclude that there exist exactly R = k∆ − (k − 1) = k(∆ − 2) + 2 edges with one
r
end vertex in V (Tk ). Since ∆ ≥ 3, it holds R ≥ k + 2. Also, as p = b ∆−2
c, and r < k,
it is obvious that p < k < R, i.e., the number of vertices of degree ∆ − 1 is smaller than
2
the number of vertices of degree ∆ in Tmax
and for an arbitrary vertex of degree ∆ − 1
there exists a corresponding vertex of degree ∆ that can be adjacent to it.
2
In the following, we prove that vertices in Tmax
have to be connected in a certain way.
Namely, we prove that each vertex of degree ∆ − 1 has exactly one neighbor of degree ∆
and ∆ − 2 pendent neighbors. In the opposite case, since vertices of degree ≥ ∆ − 1 are
2
connected in Tmax
(by Lemma 3.2), there exists a vertex (say u) of degree ∆−1 connected
2
to a vertex v of degree ∆ − 1. Besides, let vw1 ∈ E(Tmax
), where w1 is a vertex of degree
∆. Since R > k > p, then there exists a vertex w2 of degree ∆ (it may be w1 ≡ w2 ), with
a neighbor x (d(x) = 1 or d(x) = µ) (Fig. 2).
2
Fig. 2. The graphs Tmax
and T 0 in Theorem 3.1
-922
Suppose that w1 6≡ w2 and let T 0 = Tmax
− uv − w2 x + vx + w2 u. Then
2
M2 (T 0 ) − M2 (Tmax
) = ∆ − 1 − d(x) .
2
Since d(x) = 1 or d(x) = µ ≤ ∆ − 2, in both cases it holds M2 (T 0 ) − M2 (Tmax
) > 0,
2
2
and therefore, M2 (T 0 ) > M2 (Tmax
). This contradicts the choice of Tmax
.
The proof is analogous if w1 ≡ w2 .
Also, it is obvious that a vertex u of degree µ (when µ > 1) is adjacent to a vertex of
degree ∆ and to µ − 1 pendent vertices. Namely, having in mind Lemma 3.2, we conclude
that a vertex u has a neighbor of degree ∆ or ∆ − 1, and its remaining neighbors are
pendent vertices. If u has a neighbor v of degree ∆ − 1, and w is a vertex of degree ∆
2
with a pendent neighbor x, then for a tree T 0 , obtained from Tmax
by deleting the edges
uv and wx, and adding the edges ux and wu, it is satisfied
2
M2 (T 0 ) − M2 (Tmax
) = µ − 1 > 0,
2
2
and therefore M2 (T 0 ) > M2 (Tmax
). This contradicts the choice of Tmax
.
2
By the previous considerations we can calculate M2 (Tmax
) as follows.
2
M2 (Tmax
) = (k − 1)∆2 + p∆(∆ − 1) + ∆µ + p(∆ − 1)(∆ − 2) + µ(µ − 1)
+ ∆(n − k − p(∆ − 1) − µ)
= (k − 1)∆2 + 2p(∆ − 1)2 + µ(∆ + µ − 1) + ∆(n − k − (∆ − 1)p − µ) .
This proves the theorem.
Remark 3.1. Theorem 3.1 also holds for k = n − 2, i.e. T = Pn .
In the following we will characterize the trees with minimal second Zagreb index in
the class Tn,k , using the similar idea as in [2].
2
Lemma 3.3. Let Tmin
be a tree with minimal second Zagreb index in the class Tn,k , where
1≤k≤
n
2
− 1. Then, its maximum vertex degree ∆ equals 3.
2
Proof. Suppose that ∆ ≥ 4 and u is a vertex of maximum degree ∆ in Tmin
.
-93-
2
Fig. 3. The graphs Tmin
and T 1 in Lemma 3.3
2
Let P = v0 v1 . . . vi−1 u(= vi )vi+1 . . . vl be the longest path in Tmin
that contains u.
2
Also, let vi−1 , vi+1 , u1 , u2 , . . . , u∆−2 be the vertices adjacent to u in Tmin
, and z1 a pendent
vertex connected to u via u1 (it is possible that z1 ≡ u1 ) (Fig. 3). We define a tree T 1 as
2
T 1 = Tmin
− uu2 + u2 z1 .
(25)
If z1 ≡ u1 then
1
M2 (T ) −
2
M2 (Tmin
)
∆−2
X
= − d(vi−1 ) + d(vi+1 ) +
!
d(uj )
− (∆ − 2)(d(u2 ) − 1) < 0 .
j=3
Otherwise, let z2 be the vertex that is adjacent to z1 . Then
2
M2 (T 1 ) − M2 (Tmin
)
= − d(vi−1 ) + d(vi+1 ) + d(u1 ) +
∆−2
X
!
d(uj ) + (∆ − 2)d(u2 )
+ d(z2 )
j=3
< d(z2 ) − ∆ − d(u2 ) < 0 .
In both cases we obtained the tree T 1 that has k − 1 vertices of degree ∆.
After applying the transformation described in (25) k times, on every vertex u of
degree ∆ (as described in the proof of Lemma 2.2), we will obtain a tree T k that has k
2
vertices having the maximum degree ∆−1 and M2 (T k ) < M2 (Tmin
). Obviously, T k ∈ Tn,k
2
and this contradicts the choice of Tmin
.
The following lemma proves that the vertices in the tree that minimizes M2 in the
class Tn,k have to be connected in a certain way.
Lemma 3.4. Let T be a tree from the class Tn,k .
(i) Let u, z, v, w
∈
V (T ) be the vertices such that uz, zv, vw
∈
E(T ) and
dT (u) = 1, dT (z) = 2, dT (v) = 3, dT (w) = 2 or dT (w) = 3. If T ∈ Tn,k is a
0
-94tree defined as
T 0 = T − uz − vw + zw + uv ,
then M2 (T 0 ) < M2 (T ).
(ii) Let u, y, z, v, w, t ∈ V (T ) be the vertices such that uy, yz, zv, wt ∈ E(T ) and dT (u) =
dT (v) = dT (w) = dT (t) = 3 and dT (y) = dT (z) = 2. Let T 0 ∈ Tn,k be a tree defined
as
T 0 = T − yz − zv − wt + zw + zt + yv .
Then M2 (T 0 ) < M2 (T ).
Proof.
(i) It is easy to see that
M2 (T ) − M2 (T 0 ) = (dT (v) − 2)(dT (w) − 1) > 0 .
Therefore, M2 (T 0 ) < M2 (T ).
(ii) It holds that M2 (T ) − M2 (T 0 ) = 1, which implies M2 (T 0 ) < M2 (T ).
Lemma 3.4 implies that in the tree that minimizes M2 in the class Tn,k (1 ≤ k ≤
n
−1)
2
between any two vertices of degree 3 there has to be at least one vertex of degree 2 (if
possible). The remaining vertices of degree 2 (if they exist) can be placed arbitrarily
between two vertices of degree 2, or between a vertex of degree 2 and a vertex of degree
3.
Theorem 3.2. Let T ∈ Tn,k , where 1 ≤ k ≤ n2 − 1. Then

3k + 4n − 10, if n ≥ 3k + 1
M2 (T ) ≥
6k + 3n − 9, if n < 3k + 1 .
(26)
The equality holds if and only if the following three conditions are satisfied.
(i) The tree T has the vertex degree sequence (3, . . . , 3, 2, . . . , 2, 1, . . . , 1).
| {z } | {z } | {z }
k
n−2k−2
k+2
(ii) Between any two vertices of degree 3 in T there should be at least one vertex of
degree 2, if possible.
(iii) The remaining vertices of degree 2 (if they exist) in T are placed between two vertices
of degree 2, or between a vertex of degree 2 and a vertex of degree 3.
-952
Proof. Let Tmin
be a tree that minimizes M2 in the class Tn,k . According to Lemma 3.3
this tree has the vertex degree sequence π = (3, . . . , 3, 2, . . . , 2, 1, . . . , 1), when k ≤
| {z } | {z } | {z }
k
n2
n
2
− 1.
n1
Also, by the equality in (1) we obtain
n1 + 2n2 + 3k = 2(n1 + n2 + k) − 2,
(27)
which implies that n1 = k + 2 and n2 = n − 2k − 2. By the considerations discussed above,
2
between two vertices of degree 3 in Tmin
there should be at least one vertex of degree 2.
If this is possible for every two vertices of degree 3, then we obtain n − 2k − 2 ≥ k − 1, i.e.
n ≥ 3k + 1. Also, the condition in (iii) follows from the previous discussion. The second
2
Zagreb index of Tmin
can now be easily calculated. This completes the proof.
Acknowledgement. The research of the authors is supported by the Serbian Ministry of
Education, Science and Technological Development (Grant No. 174033).
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