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Complete characterization of graphs for direct comparing Zagreb indices
Discrete Applied Mathematics 215 (2016) 146–154
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Complete characterization of graphs for direct comparing Zagreb indices Batmend Horoldagva a,b , Kinkar Ch. Das b,∗ , Tsend-Ayush Selenge b,c a
Department of Mathematics, Mongolian National University of Education, Baga toiruu-14, Ulaanbaatar, Mongolia
b
Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea
c
Department of Mathematics, National University of Mongolia, Baga toiruu-1, Ulaanbaatar, Mongolia
article
abstract
info
The classical2 first and second Zagreb indices of a graph G are defined as M1 (G) = v∈V dG (v) and M2 (G) = uv∈E (G) dG (u) dG (v) , where dG (v) is the degree of the vertex v of graph G. Recently, Furtula et al. (2014) studied the difference between the Zagreb indices and mentioned a problem to characterize the graphs for which M1 (G) > M2 (G) or M1 (G) < M2 (G) or M1 (G) = M2 (G). In this paper we completely solve this problem. © 2016 Elsevier B.V. All rights reserved.
Article history: Received 10 May 2015 Received in revised form 30 June 2016 Accepted 4 July 2016 Available online 27 July 2016 Keywords: Graph First Zagreb index Second Zagreb index
1. Introduction Let G = (V , E ) be a simple graph with |V (G)| = n vertices and |E (G)| = m edges. For vi ∈ V (G), dG (vi ) is the degree of the vertex vi of G, i = 1, 2, . . . , n. The maximum vertex degree is denoted by ∆. The average of the degrees of the vertices adjacent to vertex vi is denoted by µG (vi ). For two vertices vi , vj ∈ V (G), the distance between vi and vj , denoted by dG (vi , vj ), is the length (i.e., the number of edges) of a shortest path connecting them in G. A pendant vertex is a vertex of degree one. In 1972, Gutman and Trinajstić derived a formula for estimating the total π -electron energy of conjugated systems. Within this study, two vertex-degree based invariants were encountered, that eventually were named the first (M1 ) and the second (M2 ) Zagreb indices [1]. Soon after that, M1 and M2 were recognized as measures of the branching of the carbon-atom molecular skeleton [11], and since then these are frequently used for structure–property modeling [18,19]. The first Zagreb index M1 and the second Zagreb M2 of graph G are defined as M1 (G) =
dG (vi )2
and
M2 (G) =
vi ∈V (G)
dG (vi ) dG (vj ).
vi vj ∈E (G)
Some recent mathematical results on the Zagreb indices are reported in [4–10,13–15], where also references to the previous mathematical research in this area can be found. Furtula et al. [10] mentioned that these two Zagreb indices were introduced simultaneously and examined almost always together, but the relations between them were not found until now. Caporossi et al. [3] conjectured the following relation between M1 (G) and M2 (G): M2 (G) M1 (G) n
∗
≤
m
.
Corresponding author. E-mail addresses: [email protected] (B. Horoldagva), [email protected] (K.Ch. Das), [email protected] (T.-A. Selenge).
http://dx.doi.org/10.1016/j.dam.2016.07.008 0166-218X/© 2016 Elsevier B.V. All rights reserved.
B. Horoldagva et al. / Discrete Applied Mathematics 215 (2016) 146–154
147
Fig. 1. Two graphs G and G′ .
In [12], it has been shown that the above conjecture is not true in general. After that several research papers appeared in the literature regarding the above conjecture. Recently, Furtula et al. [10] studied the difference between the Zagreb indices and mentioned the following problem: Problem 1 ([10]). To characterize the graphs for which M1 (G) > M2 (G) or M1 (G) < M2 (G)
or M1 (G) = M2 (G).
For a subset W of V (G), let G − W be the subgraph of G obtained by deleting the vertices of W and the edges incident with them. If W = {vi }, then the subgraph G − W will be written as G − vi for short. For any two nonadjacent vertices vi and vj in graph G, we use G + vi vj to denote the graph obtained from adding a new edge vi vj to graph G. As usual, Pn , Cn and K1, n−1 , denote, respectively, the path, the cycle and the star on n vertices. A cyclic graph is a graph containing at least one graph cycle. The cycle of a graph G is denoted by C (G). For other undefined notations and terminology from graph theory, the readers are referred to [2]. The paper is organized as follows. In Section 2, we obtain the difference of Zagreb indices for cyclic graphs. In Section 3, we find the difference of Zagreb indices for trees and graphs. Moreover, we solve Problem 1. 2. Difference of Zagreb indices for cyclic graphs In this section we give the difference of Zagreb indices for cyclic graphs. Transformation A. Suppose that G is a nontrivial connected graph and x is a given vertex in G which has at least one adjacent vertex of degree one. Also let u and v be two adjacent vertices of degrees greater than one in G. If x1 , x2 , . . . , xk (k ≥ 1) are all pendant vertices adjacent to vertex x, then we transform G into another graph G′ as follows: G′ = G − {uv, xx1 , xx2 , . . . , xxk } + {ux1 , x1 x2 , x2 x3 , . . . , xk v}.
(1)
The above referred graphs are illustrated in Fig. 1. Denote by 1M (G) = M2 (G) − M1 (G). We now give the following result which is useful for our main result in the next section. Lemma 2.1. Let G and G′ be two graphs as shown in Fig. 1. Then
1M (G) − 1M (G′ ) ≥ dG (u) − 2 dG (v) − 2
(2)
with equality holding if and only if there is exactly one non-pendant vertex of degree two adjacent to vertex x in G. Proof. Let G′ be the graph obtained from G by Transformation A. Then from (1), we have dG′ (w) = dG (w) for w ̸= x, xi whereas dG′ (x) = dG (x) − k and dG′ (xi ) = 2 for 1 ≤ i ≤ k. Thus M1 (G) − M1 (G′ ) = 2k dG (x) − k2 − 3k
(3)
and
M2 (G) − M2 (G′ ) = dG (u) − 2
dG (v) − 2 + k dG (x) + k dG (x) µG (x) − k2 − 4k.
(4)
Since G is connected, we have dG (x)µG (x) − dG (x) ≥ 1. From (3) and (4), we get
1M (G) − 1M (G′ ) = dG (u) − 2 dG (v) − 2 + k dG (x) µG (x) − dG (x) − 1 ≥ dG (u) − 2 dG (v) − 2 .
(5)
Moreover, the equality holds in (5) if and only if there is exactly one non-pendant vertex of degree two adjacent to vertex x in G. Thus we complete the proof of the lemma.
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Milošević et al. [16] proved that if a connected graph G is not a star, then 1M (G) ≥ −2. Moreover, it has been proved that if G is a connected graph that is neither a tree nor a cycle, then 1M (G) ≥ 1. The following result is obtained from [14]: Lemma 2.2 ([14]). Let G be a cyclic graph, possessing two adjacent vertices u and v , both of degree greater than or equal to 2. Also let x be a pendant vertex of G, which is adjacent to a vertex y(̸= u, v). Consider the transformation G′ = G − yx − uv + ux + xv . (i) If y ̸∈ V (C (G)), then 1M (G) ≥ 1M (G′ ). (ii) If y ∈ V (C (G)), then 1M (G) ≥ 1M (G′ ) + 1. From the above result, we obtain the following: Proposition 2.3. Let G be a connected cyclic graph. Then 1M (G) ≥ 0 with equality holding if and only if G is isomorphic to a cycle Cn . Proof. If G contains a pendant vertex, then we apply the transformation in Lemma 2.2 and we obtain 1M (G) ≥ 1M (G′ ). Continuing the above procedure several times and we obtain a graph G∗ such that dG∗ (vi ) ≥ 2 for any vi ∈ V (G∗ ). Thus we have
1M (G) ≥ 1M (G′ ) ≥ · · · ≥ 1M (G∗ ). For G = G∗ , we have
1M (G) =
dG (vi ) dG (vj ) − dG (vi ) − dG (vj )
vi vj ∈E (G)
=
dG (vi ) − 1
dG (vj ) − 1 − m
vi vj ∈E (G)
≥ 0. Hence 1M (G) ≥ 0. Moreover, the equality holds if and only if G is isomorphic to a cycle Cn .
3. Difference of Zagreb indices for trees and graphs From Proposition 2.3, it is sufficient to consider our main problem for trees. In other words we characterize all trees T satisfying −2 ≤ 1M (T ) ≤ 0. Let Pn−a be a path of order n − a with end vertices u and v . The broom B(n, a) (n ≥ a + 3) consists of the path Pn−a with a independent vertices adjacent to one pendant vertex u of the path. Then we say that the pendant vertex v is the origin of the broom B(n, a). For a vertex vi ∈ V (T ), the eccentricity of vi , denoted by εT (vi ), is the maximum distance from vi to other vertices in T . Let T1 = T (n1 , n2 , . . . , nd ) be a tree of order n with a vertex v of degree d (εT1 (v) = 2) and adjacent vertices of v have degrees dT1 (v1 ), dT1 (v2 ), . . . , dT1 (vd )(ni = dT1 (vi ), 1 ≤ n1 ≤ n2 ≤ · · · ≤ nd ) such that
d
i=1
ni + 1 = n. Thus we have
T1 − v = K1, n1 −1 ∪ K1, n2 −1 ∪ · · · ∪ K1, nd −1 . Let T2 = T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ) (see Fig. 2) be a tree with a vertex v (εT2 (v) = 3) of degree p + q such that T2 − v = K1, n1 −1 ∪ · · · ∪ K1, np −1 ∪ K1, np+1 ∪ · · · ∪ K1, np+q . If T ∼ = T2 = T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ), then the number of vertices in T is equal to 1 + q + p
M2 (T ) = 2(p + q) q + (p + q)
p
ni +
i =1
q
(ni − 1) ni + 2
i=1
q
np+i +
i =1
(np+i − 1) np+i
i=1
and M1 (T ) = (p + q)2 +
p
n2i + ni − 1 +
i =1
q
n2p+i + np+i − 1 + 4 q,
i=1
we have
1M (T ) = M2 (T ) − M1 (T )
= q2 − p2 + p − 3q + (p + q)
p i=1
ni − 2
p
ni
i=1
= q2 − p2 + p − 3q + (p + q) dT (v) µT (v) − 2q − 2 dT (v) µT (v) − 2q = dT (v)2 µT (v) − 2 dT (v) µT (v) − dT (v)2 + dT (v) as dT (v) = p + q.
p+q i=1
ni . By
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149
Fig. 2. Tree T2 .
Hence we get the following result:
1M (T ) = dT (v) dT (v) − 2 µT (v) − 1 − 1 .
(6)
Lemma 3.1. Let T be a tree with n vertices and v be its vertex of degree greater than one. If each connected component of T − v is star or broom with origin w , vw ∈ E (T ), then
1M (T ) = dT (v) dT (v) − 2 µT (v) − 1 − 1 . Proof. For εT (v) = 1, we have T T (n1 , n2 , . . . , nd ). Then
(7)
∼ = K1, n−1 and hence the equality holds in (7). For εT (v) = 2, we have T ∼ =
M2 (T ) = dT (v)2 µT (v) − dT (v) µT (v) +
dT (u)2
u: v u∈E (T )
and M1 (T ) = dT (v)2 + dT (v) µT (v) − dT (v) +
dT (u)2 .
u: v u∈E (T )
Thus, from the above two results, we get (7). For εT (v) = 3, we have T ∼ = T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ). By (6), we get the result in (7). Otherwise, εT (v) ≥ 4. Then there exists a pendant neighbor (say, x) in T such that the distance between v and x is strictly greater than two, that is, dT (v, x) ≥ 3. Let y be the unique vertex of degree two which is adjacent to x. Let T ′ be a tree obtained from T by contracting an edge xy. Then still each connected component of T ′ − v is star or broom with origin w , vw ∈ E (T ′ ) and |V (T ′ )| = |V (T )| − 1, dT ′ (v) = dT (v) and µT ′ (v) = µT (v). Hence
1M (T ′ ) = M2 (T ′ ) − M1 (T ′ ) = M2 (T ) − M1 (T ) = 1M (T ).
(8)
By the above described construction, the value of 1M (T ) remains constant. If T is the tree T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ), then from (6), the result in (7) remains valid. Otherwise, we continue the construction as follows. We choose a pendant neighbor x such that dT ′ (v, x) ≥ 3 and a vertex y of degree 2 is adjacent to x in T ′ . Repeating the above procedure sufficient number of times, we arrive at T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ) = T2 . Therefore dT2 (v) = dT (v) and µT2 (v) = µT (v). Thus we have ′
1M (T ) = 1M (T ′ ) = · · · = 1M (T2 ) = dT (v) dT (v) − 2 µT (v) − 1 − 1 . This completes the proof of the lemma.
We now obtain the difference of Zagreb indices for any tree T . Theorem 3.2. Let T be a tree with n vertices and v be its vertex of degree greater than one. Then
1M (T ) ≥ dT (v) (dT (v) − 2)(µT (v) − 1) − 1 with equality holding if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ).
(9)
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Proof. If each connected component of T − v is star or broom with origin w , vw ∈ E (T ), then the equality holds in (9) by Lemma 3.1. Otherwise, there is at least one connected component of T − v , say T3 , is different from star and broom. Then there exists a vertex x in T3 which has at least one adjacent vertex of degree one and dT (x)µT (x) − dT (x) > 1. Let now u (a vertex in T3 ) be adjacent to v in T . Then clearly dT (u) ≥ 2. We now apply Transformation A on T and obtain a new tree T 1 with dT 1 (v) = dT (v) and µT 1 (v) = µT (v). Then by Lemma 2.1, we get
1M (T ) − 1M (T 1 ) ≥ dT (u) − 2 dT (v) − 2 ≥ 0,
that is, 1M (T ) ≥ 1M (T 1 ).
(10)
Now we repeat the above procedure sufficient number of times and obtain a tree T r such that its each connected component T r −v is star or broom with origin w , wv ∈ E (T r ). Therefore dT r (v) = dT (v) and µT r (v) = µT (v). By Lemmas 2.1 and 3.1, we have
1M (T ) ≥ 1M (T 1 ) ≥ 1M (T 2 ) ≥ · · · ≥ 1M (T r −1 ) > 1M (T r ) = dT (v) (dT (v) − 2)(µT (v) − 1) − 1 . The inequality in the last step is strict, because there is exactly one non-pendant vertex x of degree greater than two such that dT r −1 (x) µT r −1 (x) ≥ dT r −1 (x) + 2. Consequently, the inequality in (10) is strict. This completes the proof. Corollary 3.3. Let T be a tree of order n with maximum degree ∆. Then
1M (T ) ≥ ∆ (∆ − 2)(µ∆ − 1) − 1 ,
(11)
where µ∆ is the average degree of the adjacent vertices of the maximum degree vertex v in T . Moreover, the equality holds in (11) if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ). Corollary 3.4. Let T be a tree of order n. Also let v be a vertex in T such that dT (v) ≥ 3. If dT (v) µT (v) − dT (v) > 3, then 1M (T ) > 0. Proof. Since dT (v)µT (v) − dT (v) > 3 and dT (v) ≥ 3, we have
(dT (v) − 2) (µT (v) − 1) > (dT (v) − 2) ·
3 dT (v)
Thus by Theorem 3.2, we get the required result.
=3−
6 dT (v)
≥ 1.
Given n and n1 , n2 , . . . , nd (d ≥ 2) are positive integers such that 1 + i=1 ni ≤ n and 1 ≤ n1 ≤ n2 ≤ · · · ≤ nd . Denote by Tn (n1 , n2 , . . . , nd ), the set of trees T of order n with a vertex v of degree d and adjacent vertices of v have degrees n1 , n2 , . . . , nd , and each connected component of T − v is star or broom with origin w , vw ∈ E (T ). Then by Lemma 3.1, we have
d
1M (T ) = d (d − 2)
d 1
d i=1
ni − 1
− 1 = const
for all T ∈ Tn (n1 , n2 , . . . , nd ). Lemma 3.5. Let n and n1 , n2 , . . . , nd (d ≥ 2) be given positive integers such that 1 + i=1 ni < n. Then Tn (n1 , n2 , . . . , nd ) = ∅ if and only if there is a tree T of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni ̸= 2 for all i, 1 ≤ i ≤ d.
d
Proof. Let Tn (n1 , n2 , . . . , nd ) = ∅. Suppose that there is a tree T of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni = 2 for at least one i (1 ≤ i ≤ d). Then there is a vertex u of degree two in T (n1 , n2 , . . . , nd ) which is adjacent to v . We now construct tree T with n vertices from T (n1 , n2 , . . . , nd ) in the following: The r pendant edges are attached to the pendant vertex, neighbor of vertex u in T (n1 , n2 , . . . , nd ) such that r + 1 + dT (v) dT (vi ) = n. Then the degrees of adjacent vertices of v in the constructed tree T are same as the degrees of adjacent i=1 vertices of v in the tree T (n1 , n2 , . . . , nd ) and dT (v) = d. Thus the constructed tree T belongs to Tn (n1 , n2 , . . . , nd ) and it contradicts that Tn (n1 , n2 , . . . , nd ) = ∅. Hence there is a tree T of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni ̸= 2 for all i, 1 ≤ i ≤ d. Conversely, let T be a tree of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni ̸= 2 for all i, 1 ≤ i ≤ d. Suppose that Tn (n1 , n2 , . . . , nd ) ̸= ∅. Then there is a tree T of order n in Tn (n1 , n2 , . . . , nd ). Thus we have d T ⊃ T (n1 , n2 , . . . , nd ) because 1 + i=1 ni < n. From this result with ni ̸= 2 for all 1 ≤ i ≤ d, there exists a connected component of T − v which is neither star nor broom with origin w , vw ∈ E (T ). Therefore T ̸∈ Tn (n1 , n2 , . . . , nd ) and it contradicts that T ∈ Tn (n1 , n2 , . . . , nd ). Hence Tn (n1 , n2 , . . . , nd ) = ∅. Let n1 , n2 , . . . , nd (d ≥ 2) be given positive integers. In view of the proof of Lemma 3.5, for a given integer n such that d 1 + i=1 ni < n, we can construct all graphs in Tn (n1 , n2 , . . . , nd ) from T (n1 , n2 , . . . , nd ). For example, if n = 17, then all the trees in T17 (1, 2, 2, 3, 5) are shown in Fig. 3. Moreover, Tn (3, 3, 4, 5) = ∅, n > 16.
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151
Fig. 3. Tree T (1, 2, 2, 3, 5) and all the trees in T17 (1, 2, 2, 3, 5).
Lemma 3.6. Let T be a tree of order n with T ⊇ T (n1 , n2 , . . . , nd ), where dT (v) = d (≥3). Then
1M (T ) ≥ 1M (T (n1 , n2 , . . . , nd )) with equality holding if and only if T ∈ Tn (n1 , n2 , . . . , nd ). Moreover, (i) for all i (1 ≤ i ≤ d), ni ̸= 2, then 1M (T ) = 1M (T (n1 , n2 , . . . , nd )) if and only if T ∼ = T (n1 , n2 , . . . , nd ), (ii) for any i (1 ≤ i ≤ d), ni = 2, then 1M (T ) = 1M (T (n1 , n2 , . . . , nd )) if and only if T ∈ Tn (n1 , n2 , . . . , nd ) ̸= ∅. Proof. Let v be a vertex of T (n1 , n2 , . . . , nd ) in which adjacent vertices of it have degrees n1 , n2 , . . . , nd . Then by Lemma 3.1, we have d 1 ni − 1 − 1 . 1M (T (n1 , n2 , . . . , nd )) = d (d − 2)
(12)
d i=1
Since T ⊇ T (n1 , n2 , . . . , nd ), we have v ∈ V (T ). Then clearly dT (v) = d and µT (v) ≥ Theorem 3.2, we have
1 d
d
i=1
ni . For dT (v) ≥ 3, by
1M (T ) ≥ dT (v) (dT (v) − 2)(µT (v) − 1) − 1
(13)
with equality holding if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ). Using d the result dT (v) = d and µT (v) ≥ 1d i=1 ni , from (12) and (13), we get
1M (T ) ≥ 1M (T (n1 , n2 , . . . , nd )).
(14)
Moreover, the equality holds in (14) if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ) and µT (v) = 1d di=1 ni , that is, T ∈ Tn (n1 , n2 , . . . , nd ). (i) Let ni ̸= 2 for all i (1 ≤ i ≤ d). If 1 +
d
i=1
ni < n, then by Lemma 3.5, Tn (n1 , n2 , . . . , nd ) = ∅ and hence the inequality
in (14) is strict. Otherwise, 1 + i=1 ni = n. Then T ∼ = T (n1 , n2 , . . . , nd ) and the equality holds in (14). (ii) Let ni = 2 for any i (1 ≤ i ≤ d). Then by Lemma 3.5, we have Tn (n1 , n2 , . . . , nd ) ̸= ∅. Then it follows that the equality holds in (14) if and only if T ∈ Tn (n1 , n2 , . . . , nd ) ̸= ∅.
d
The double broom DB(n, a, b) consists of the path Pn−a−b together with a independent vertices adjacent to one pendant vertex of the path and b independent vertices adjacent to the other pendant vertex. The graph DB(n, a, b) is said to be a balanced double broom if |a − b| ≤ 1. Let Bn be the class of trees T of order n such that T ∼ = B(n, ∆ − 1) or T ∼ = DB(n, a, b). Then by Lemma 3.1, it is easy to see that if T ∈ Bn , then 1M (T ) = −2. Also we have 1M (Pn ) = −2 and 1M (K1, n−1 ) = 1 − n. We denote by T∗n (see Fig. 4), a class of trees T of order n obtained from double broom DB(n − 2, a, b) (double broom is of order n − 2) by attaching two pendant edges to the two non-adjacent vertices of average degree two. If T ∈ T∗n , then there exist two vertices v and y such that dT (u) = 3, µT (u) = 5/3 (u = v, y) and v y ̸∈ E (T ). By Lemma 3.5, one can see easily that Tn (2, 2, 2) ̸= ∅, Tn (1, 2, 3) ̸= ∅, Tn (1, 1, 2, 2) ̸= ∅ (n ≥ 7) and Tn (1, 2, 2) ̸= ∅ (n ≥ 6). We now obtain the complete characterization of 1M (T ) ≤ 0 for trees. Theorem 3.7. Let T be a tree with n ≥ 3 vertices and different from a star. Then
(i) 1M (T ) = 0 if and only if T ∼ = T (1, 1, 1, 3) or T ∈ Tn (2, 2, 2), Tn (1, 2, 3), Tn (1, 1, 2, 2), T∗n . (ii) 1M (T ) = −1 if and only if T ∼ = T (1, 1, 3) or T ∈ Tn (1, 2, 2). (iii) 1M (T ) = −2 if and only if T ∼ = Pn or T ∈ Bn .
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Fig. 4. Tree T , T ∈ T∗n .
Proof. If T ∼ = Pn , then 1M (T ) = −2. Otherwise, there is a vertex of degree greater than or equal to 3 in T . Let S be a set of vertices v (dT (v) ≥ 3) in T such that dT (v) µT (v) − dT (v) is maximum, that is, S = {v ∈ V (T ) : dT (v) ≥ 3 and dT (v) µT (v) − dT (v) is maximum}. Let v be the maximum degree vertex of degree dT (v) (≥3) in S. Then we have dT (v) ≤ ∆. Also let A = dT (v) µT (v) − dT (v). Then one can see easily that A ≥ 0. If A ≥ 4, then 1M (T ) > 0 by Corollary 3.4. We now consider the following cases: Case (1): A = 0. In this case dT (v) µT (v) = dT (v). Therefore vertex v is adjacent to the pendant vertices in T , that is, T ∼ = K1, n−1 , a contradiction. Case (2): A = 1. In this case dT (v) µT (v) = dT (v) + 1. Therefore vertex v is adjacent to exactly one non-pendant vertex in T . Since dT (v) ≥ 3 and dT (v) µT (v) − dT (v) is maximum with maximum value 1, then all the vertices of degree greater than or equal to 3 in T must be adjacent to exactly one non-pendant vertex. Then T is isomorphic to broom or double broom. Hence T ∈ Bn . Moreover, 1M (T ) = −2 if T ∈ Bn . Case (3): A = 2. Then we have dT (v) µT (v) = dT (v) + 2. If dT (v) ≥ 5, then
(dT (v) − 2) (µT (v) − 1) = 2 −
4 dT (v)
> 1.
By Theorem 3.2, we have 1M (T ) > 0 and it contradicts to 1M (T ) ≤ 0. So we consider the following two subcases: Subcase (i): dT (v) = 4. In this subcase either T ⊇ T (1, 1, 1, 3) or T ⊇ T (1, 1, 2, 2). When T ⊇ T (1, 1, 1, 3), by Lemma 3.6(i) we have 1M (T ) = 1M (T (1, 1, 1, 3)) = 0 if and only if T ∼ = T (1, 1, 1, 3). When T ⊇ T (1, 1, 2, 2), by Lemma 3.6(ii) we have 1M (T ) = 1M (T (1, 1, 2, 2)) = 0 if and only if T ∈ Tn (1, 1, 2, 2) ̸= ∅. Subcase (ii): dT (v) = 3. In this subcase T ⊇ T (1, 1, 3) or T ⊇ T (1, 2, 2). First we assume that T ⊇ T (1, 1, 3). Then by Lemma 3.6(i) we have 1M (T ) = 1M (T (1, 1, 3)) = −1 if and only if T ∼ = T (1, 1, 3). Now we have to find trees T for which T ⊃ T (1, 1, 3) and 1M (T ) = 0. Therefore we must have either T ⊇ T (1, 2, 3) or T ⊇ T (1, 1, 4) and hence A = dT (v) µT (v) − dT (v) ≥ 3, a contradiction. Next we assume that T ⊇ T (1, 2, 2). Then by Lemma 3.6(ii) we have 1M (T ) = 1M (T (1, 2, 2)) = −1 if and only if T ∈ Tn (1, 2, 2) ̸= ∅. Now we have to find trees T for which T ⊃ T (1, 2, 2) and 1M (T ) = 0. From the above results, we conclude that n ≥ 7 and T ̸∈ Tn (1, 2, 2) as 1M (T ) = 0. Then there exists at least one vertex, say y (̸= v) of degree 3 in T such that dT (y) µT (y) − dT (y) = 2. If there exists exactly one such vertex y in T , then yr ∈ E (T ) (dT (r ) = 1 or 2) and hence T ∈ T∗n as A = 2. Otherwise, by using Transformation A for the vertices x (̸= v, y) of degree 3 such that dT (x) µT (x) − dT (x) = 2, we arrive at a tree T ′ ∈ T∗n . Then 1M (T ) > 1M (T ′ ) = 0 by Lemma 2.1. Case (4): A = 3. Then we have dT (v) µT (v) = dT (v) + 3. If dT (v) ≥ 4, then
(dT (v) − 2) (µT (v) − 1) = 3 −
6 dT (v)
> 1.
Similarly, by Theorem 3.2, 1M (T ) > 0 and it contradicts to 1M (T ) ≤ 0. Hence dT (v) = 3 as ∆ ≥ 3. Therefore, we must have either T ⊇ T (2, 2, 2) or T ⊇ T (1, 2, 3). When T ⊇ T (2, 2, 2), by Lemma 3.6(ii) we have 1M (T ) = 1M (T (2, 2, 2)) = 0 if and only if T ∈ Tn (2, 2, 2) ̸= ∅. Similarly, when T ⊇ T (1, 2, 3), we get 1M (T ) = 1M (T (1, 2, 3)) = 0 if and only if T ∈ Tn (1, 2, 3) ̸= ∅. This completes the proof. Example 1. By SageMath [17], we characterize all trees of order 9 satisfying
1M (T ) = 0 (see, Fig. 5),
1M (T ) = −1 (see, Fig. 6) and 1M (T ) = −2 (see, Fig. 7).
Let Γ be a class of graphs H = (V , E ) of order n such that H {Cn , Pn , K1, n−1 , T (1, 1, 3), T (1, 1, 1, 3)} and
H ̸∈ Bn , Tn (1, 2, 2), Tn (2, 2, 2), Tn (1, 2, 3), Tn (1, 1, 2, 2), T∗n .
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153
Fig. 5. All the above trees T , 1M (T ) = 0.
Fig. 6. All the above trees T , 1M (T ) = −1.
Fig. 7. All the above trees T , 1M (T ) = −2.
Theorem 3.8. Let G be a connected graph of order n ≥ 3. Then
(i) M1 = M2 if and only if G ∼ = Cn or G ∼ = T (1, 1, 1, 3) or G ∈ Tn (2, 2, 2), Tn (1, 2, 3), Tn (1, 1, 2, 2), T∗n . (ii) M1 (G) > M2 (G) if and only if G ∼ = K1, n−1 or G ∼ = Pn or G ∼ = T (1, 1, 3) or G ∈ Bn or G ∈ Tn (1, 2, 2). (iii) M1 (G) < M2 (G) if and only if G ∈ Γ . Proof. We have 1M (K1, n−1 ) = 1 − n < 0. By Proposition 2.3 and Theorem 3.7, we get the required result.
Acknowledgments The authors are much grateful to two referees for their valuable comments on our paper, which have considerably improved the presentation of this paper. The second author is supported by the National Research Foundation funded by the Korean government with the grant no. 2013R1A1A2009341. The third author is supported by the Korea Foundation for Advanced Studies’ International Scholar Exchange Fellowship for the academic year of 2014–2015. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
A.T. Balaban, I. Motoc, D. Bonchev, O. Mekenyan, Topological indices for structure–activity correlations, Top. Curr. Chem. 114 (1983) 21–55. J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan Press, New York, 1976. G. Caporossi, P. Hansen, Variable neighborhood search for extremal graphs. 1. The auto-graphix system, Discrete Math. 212 (2000) 29–44. K.C. Das, Maximizing the sum of the squares of the degrees of a graph, Discrete Math. 285 (2004) 57–66. K.C. Das, On comparing Zagreb indices of graphs, MATCH Commun. Math. Comput. Chem. 63 (2010) 433–440. K.C. Das, I. Gutman, Some properties of the second Zagreb index, MATCH Commun. Math. Comput. Chem. 52 (2004) 103–112. K.C. Das, I. Gutman, B. Horoldagva, Comparison between Zagreb indices and Zagreb coindices, MATCH Commun. Math. Comput. Chem. 68 (2012) 189–198. K.C. Das, I. Gutman, B. Zhou, New upper bounds on Zagreb indices, J. Math. Chem. 46 (2009) 514–521. K.C. Das, K. Xu, I. Gutman, On Zagreb and harary indices, MATCH Commun. Math. Comput. Chem. 70 (2013) 301–314. B. Furtula, I. Gutman, S. Ediz, On difference of Zagreb indices, Discrete Appl. Math. 178 (2014) 83–88. I. Gutman, B. Ruščić, N. Trinajstić, C.F. Wilcox, Graph theory and molecular orbitals. XII. Acyclic polyenes, J. Chem. Phys. 62 (1975) 3399–3405. P. Hansen, D. Vukičević, Comparing the Zagreb indices, Croat. Chem. Acta 80 (2007) 165–168. B. Horoldagva, K.C. Das, On comparing Zagreb indices of graphs, Hacettepe J. Math. Stat. 41 (2012) 223–230.
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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Complete characterization of graphs for direct comparing Zagreb indices Batmend Horoldagva a,b , Kinkar Ch. Das b,∗ , Tsend-Ayush Selenge b,c a
Department of Mathematics, Mongolian National University of Education, Baga toiruu-14, Ulaanbaatar, Mongolia
b
Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea
c
Department of Mathematics, National University of Mongolia, Baga toiruu-1, Ulaanbaatar, Mongolia
article
abstract
info
The classical2 first and second Zagreb indices of a graph G are defined as M1 (G) = v∈V dG (v) and M2 (G) = uv∈E (G) dG (u) dG (v) , where dG (v) is the degree of the vertex v of graph G. Recently, Furtula et al. (2014) studied the difference between the Zagreb indices and mentioned a problem to characterize the graphs for which M1 (G) > M2 (G) or M1 (G) < M2 (G) or M1 (G) = M2 (G). In this paper we completely solve this problem. © 2016 Elsevier B.V. All rights reserved.
Article history: Received 10 May 2015 Received in revised form 30 June 2016 Accepted 4 July 2016 Available online 27 July 2016 Keywords: Graph First Zagreb index Second Zagreb index
1. Introduction Let G = (V , E ) be a simple graph with |V (G)| = n vertices and |E (G)| = m edges. For vi ∈ V (G), dG (vi ) is the degree of the vertex vi of G, i = 1, 2, . . . , n. The maximum vertex degree is denoted by ∆. The average of the degrees of the vertices adjacent to vertex vi is denoted by µG (vi ). For two vertices vi , vj ∈ V (G), the distance between vi and vj , denoted by dG (vi , vj ), is the length (i.e., the number of edges) of a shortest path connecting them in G. A pendant vertex is a vertex of degree one. In 1972, Gutman and Trinajstić derived a formula for estimating the total π -electron energy of conjugated systems. Within this study, two vertex-degree based invariants were encountered, that eventually were named the first (M1 ) and the second (M2 ) Zagreb indices [1]. Soon after that, M1 and M2 were recognized as measures of the branching of the carbon-atom molecular skeleton [11], and since then these are frequently used for structure–property modeling [18,19]. The first Zagreb index M1 and the second Zagreb M2 of graph G are defined as M1 (G) =
dG (vi )2
and
M2 (G) =
vi ∈V (G)
dG (vi ) dG (vj ).
vi vj ∈E (G)
Some recent mathematical results on the Zagreb indices are reported in [4–10,13–15], where also references to the previous mathematical research in this area can be found. Furtula et al. [10] mentioned that these two Zagreb indices were introduced simultaneously and examined almost always together, but the relations between them were not found until now. Caporossi et al. [3] conjectured the following relation between M1 (G) and M2 (G): M2 (G) M1 (G) n
∗
≤
m
.
Corresponding author. E-mail addresses: [email protected] (B. Horoldagva), [email protected] (K.Ch. Das), [email protected] (T.-A. Selenge).
http://dx.doi.org/10.1016/j.dam.2016.07.008 0166-218X/© 2016 Elsevier B.V. All rights reserved.
B. Horoldagva et al. / Discrete Applied Mathematics 215 (2016) 146–154
147
Fig. 1. Two graphs G and G′ .
In [12], it has been shown that the above conjecture is not true in general. After that several research papers appeared in the literature regarding the above conjecture. Recently, Furtula et al. [10] studied the difference between the Zagreb indices and mentioned the following problem: Problem 1 ([10]). To characterize the graphs for which M1 (G) > M2 (G) or M1 (G) < M2 (G)
or M1 (G) = M2 (G).
For a subset W of V (G), let G − W be the subgraph of G obtained by deleting the vertices of W and the edges incident with them. If W = {vi }, then the subgraph G − W will be written as G − vi for short. For any two nonadjacent vertices vi and vj in graph G, we use G + vi vj to denote the graph obtained from adding a new edge vi vj to graph G. As usual, Pn , Cn and K1, n−1 , denote, respectively, the path, the cycle and the star on n vertices. A cyclic graph is a graph containing at least one graph cycle. The cycle of a graph G is denoted by C (G). For other undefined notations and terminology from graph theory, the readers are referred to [2]. The paper is organized as follows. In Section 2, we obtain the difference of Zagreb indices for cyclic graphs. In Section 3, we find the difference of Zagreb indices for trees and graphs. Moreover, we solve Problem 1. 2. Difference of Zagreb indices for cyclic graphs In this section we give the difference of Zagreb indices for cyclic graphs. Transformation A. Suppose that G is a nontrivial connected graph and x is a given vertex in G which has at least one adjacent vertex of degree one. Also let u and v be two adjacent vertices of degrees greater than one in G. If x1 , x2 , . . . , xk (k ≥ 1) are all pendant vertices adjacent to vertex x, then we transform G into another graph G′ as follows: G′ = G − {uv, xx1 , xx2 , . . . , xxk } + {ux1 , x1 x2 , x2 x3 , . . . , xk v}.
(1)
The above referred graphs are illustrated in Fig. 1. Denote by 1M (G) = M2 (G) − M1 (G). We now give the following result which is useful for our main result in the next section. Lemma 2.1. Let G and G′ be two graphs as shown in Fig. 1. Then
1M (G) − 1M (G′ ) ≥ dG (u) − 2 dG (v) − 2
(2)
with equality holding if and only if there is exactly one non-pendant vertex of degree two adjacent to vertex x in G. Proof. Let G′ be the graph obtained from G by Transformation A. Then from (1), we have dG′ (w) = dG (w) for w ̸= x, xi whereas dG′ (x) = dG (x) − k and dG′ (xi ) = 2 for 1 ≤ i ≤ k. Thus M1 (G) − M1 (G′ ) = 2k dG (x) − k2 − 3k
(3)
and
M2 (G) − M2 (G′ ) = dG (u) − 2
dG (v) − 2 + k dG (x) + k dG (x) µG (x) − k2 − 4k.
(4)
Since G is connected, we have dG (x)µG (x) − dG (x) ≥ 1. From (3) and (4), we get
1M (G) − 1M (G′ ) = dG (u) − 2 dG (v) − 2 + k dG (x) µG (x) − dG (x) − 1 ≥ dG (u) − 2 dG (v) − 2 .
(5)
Moreover, the equality holds in (5) if and only if there is exactly one non-pendant vertex of degree two adjacent to vertex x in G. Thus we complete the proof of the lemma.
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Milošević et al. [16] proved that if a connected graph G is not a star, then 1M (G) ≥ −2. Moreover, it has been proved that if G is a connected graph that is neither a tree nor a cycle, then 1M (G) ≥ 1. The following result is obtained from [14]: Lemma 2.2 ([14]). Let G be a cyclic graph, possessing two adjacent vertices u and v , both of degree greater than or equal to 2. Also let x be a pendant vertex of G, which is adjacent to a vertex y(̸= u, v). Consider the transformation G′ = G − yx − uv + ux + xv . (i) If y ̸∈ V (C (G)), then 1M (G) ≥ 1M (G′ ). (ii) If y ∈ V (C (G)), then 1M (G) ≥ 1M (G′ ) + 1. From the above result, we obtain the following: Proposition 2.3. Let G be a connected cyclic graph. Then 1M (G) ≥ 0 with equality holding if and only if G is isomorphic to a cycle Cn . Proof. If G contains a pendant vertex, then we apply the transformation in Lemma 2.2 and we obtain 1M (G) ≥ 1M (G′ ). Continuing the above procedure several times and we obtain a graph G∗ such that dG∗ (vi ) ≥ 2 for any vi ∈ V (G∗ ). Thus we have
1M (G) ≥ 1M (G′ ) ≥ · · · ≥ 1M (G∗ ). For G = G∗ , we have
1M (G) =
dG (vi ) dG (vj ) − dG (vi ) − dG (vj )
vi vj ∈E (G)
=
dG (vi ) − 1
dG (vj ) − 1 − m
vi vj ∈E (G)
≥ 0. Hence 1M (G) ≥ 0. Moreover, the equality holds if and only if G is isomorphic to a cycle Cn .
3. Difference of Zagreb indices for trees and graphs From Proposition 2.3, it is sufficient to consider our main problem for trees. In other words we characterize all trees T satisfying −2 ≤ 1M (T ) ≤ 0. Let Pn−a be a path of order n − a with end vertices u and v . The broom B(n, a) (n ≥ a + 3) consists of the path Pn−a with a independent vertices adjacent to one pendant vertex u of the path. Then we say that the pendant vertex v is the origin of the broom B(n, a). For a vertex vi ∈ V (T ), the eccentricity of vi , denoted by εT (vi ), is the maximum distance from vi to other vertices in T . Let T1 = T (n1 , n2 , . . . , nd ) be a tree of order n with a vertex v of degree d (εT1 (v) = 2) and adjacent vertices of v have degrees dT1 (v1 ), dT1 (v2 ), . . . , dT1 (vd )(ni = dT1 (vi ), 1 ≤ n1 ≤ n2 ≤ · · · ≤ nd ) such that
d
i=1
ni + 1 = n. Thus we have
T1 − v = K1, n1 −1 ∪ K1, n2 −1 ∪ · · · ∪ K1, nd −1 . Let T2 = T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ) (see Fig. 2) be a tree with a vertex v (εT2 (v) = 3) of degree p + q such that T2 − v = K1, n1 −1 ∪ · · · ∪ K1, np −1 ∪ K1, np+1 ∪ · · · ∪ K1, np+q . If T ∼ = T2 = T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ), then the number of vertices in T is equal to 1 + q + p
M2 (T ) = 2(p + q) q + (p + q)
p
ni +
i =1
q
(ni − 1) ni + 2
i=1
q
np+i +
i =1
(np+i − 1) np+i
i=1
and M1 (T ) = (p + q)2 +
p
n2i + ni − 1 +
i =1
q
n2p+i + np+i − 1 + 4 q,
i=1
we have
1M (T ) = M2 (T ) − M1 (T )
= q2 − p2 + p − 3q + (p + q)
p i=1
ni − 2
p
ni
i=1
= q2 − p2 + p − 3q + (p + q) dT (v) µT (v) − 2q − 2 dT (v) µT (v) − 2q = dT (v)2 µT (v) − 2 dT (v) µT (v) − dT (v)2 + dT (v) as dT (v) = p + q.
p+q i=1
ni . By
B. Horoldagva et al. / Discrete Applied Mathematics 215 (2016) 146–154
149
Fig. 2. Tree T2 .
Hence we get the following result:
1M (T ) = dT (v) dT (v) − 2 µT (v) − 1 − 1 .
(6)
Lemma 3.1. Let T be a tree with n vertices and v be its vertex of degree greater than one. If each connected component of T − v is star or broom with origin w , vw ∈ E (T ), then
1M (T ) = dT (v) dT (v) − 2 µT (v) − 1 − 1 . Proof. For εT (v) = 1, we have T T (n1 , n2 , . . . , nd ). Then
(7)
∼ = K1, n−1 and hence the equality holds in (7). For εT (v) = 2, we have T ∼ =
M2 (T ) = dT (v)2 µT (v) − dT (v) µT (v) +
dT (u)2
u: v u∈E (T )
and M1 (T ) = dT (v)2 + dT (v) µT (v) − dT (v) +
dT (u)2 .
u: v u∈E (T )
Thus, from the above two results, we get (7). For εT (v) = 3, we have T ∼ = T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ). By (6), we get the result in (7). Otherwise, εT (v) ≥ 4. Then there exists a pendant neighbor (say, x) in T such that the distance between v and x is strictly greater than two, that is, dT (v, x) ≥ 3. Let y be the unique vertex of degree two which is adjacent to x. Let T ′ be a tree obtained from T by contracting an edge xy. Then still each connected component of T ′ − v is star or broom with origin w , vw ∈ E (T ′ ) and |V (T ′ )| = |V (T )| − 1, dT ′ (v) = dT (v) and µT ′ (v) = µT (v). Hence
1M (T ′ ) = M2 (T ′ ) − M1 (T ′ ) = M2 (T ) − M1 (T ) = 1M (T ).
(8)
By the above described construction, the value of 1M (T ) remains constant. If T is the tree T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ), then from (6), the result in (7) remains valid. Otherwise, we continue the construction as follows. We choose a pendant neighbor x such that dT ′ (v, x) ≥ 3 and a vertex y of degree 2 is adjacent to x in T ′ . Repeating the above procedure sufficient number of times, we arrive at T (n1 , n2 , . . . , np , np+1 , np+2 , . . . , np+q ) = T2 . Therefore dT2 (v) = dT (v) and µT2 (v) = µT (v). Thus we have ′
1M (T ) = 1M (T ′ ) = · · · = 1M (T2 ) = dT (v) dT (v) − 2 µT (v) − 1 − 1 . This completes the proof of the lemma.
We now obtain the difference of Zagreb indices for any tree T . Theorem 3.2. Let T be a tree with n vertices and v be its vertex of degree greater than one. Then
1M (T ) ≥ dT (v) (dT (v) − 2)(µT (v) − 1) − 1 with equality holding if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ).
(9)
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Proof. If each connected component of T − v is star or broom with origin w , vw ∈ E (T ), then the equality holds in (9) by Lemma 3.1. Otherwise, there is at least one connected component of T − v , say T3 , is different from star and broom. Then there exists a vertex x in T3 which has at least one adjacent vertex of degree one and dT (x)µT (x) − dT (x) > 1. Let now u (a vertex in T3 ) be adjacent to v in T . Then clearly dT (u) ≥ 2. We now apply Transformation A on T and obtain a new tree T 1 with dT 1 (v) = dT (v) and µT 1 (v) = µT (v). Then by Lemma 2.1, we get
1M (T ) − 1M (T 1 ) ≥ dT (u) − 2 dT (v) − 2 ≥ 0,
that is, 1M (T ) ≥ 1M (T 1 ).
(10)
Now we repeat the above procedure sufficient number of times and obtain a tree T r such that its each connected component T r −v is star or broom with origin w , wv ∈ E (T r ). Therefore dT r (v) = dT (v) and µT r (v) = µT (v). By Lemmas 2.1 and 3.1, we have
1M (T ) ≥ 1M (T 1 ) ≥ 1M (T 2 ) ≥ · · · ≥ 1M (T r −1 ) > 1M (T r ) = dT (v) (dT (v) − 2)(µT (v) − 1) − 1 . The inequality in the last step is strict, because there is exactly one non-pendant vertex x of degree greater than two such that dT r −1 (x) µT r −1 (x) ≥ dT r −1 (x) + 2. Consequently, the inequality in (10) is strict. This completes the proof. Corollary 3.3. Let T be a tree of order n with maximum degree ∆. Then
1M (T ) ≥ ∆ (∆ − 2)(µ∆ − 1) − 1 ,
(11)
where µ∆ is the average degree of the adjacent vertices of the maximum degree vertex v in T . Moreover, the equality holds in (11) if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ). Corollary 3.4. Let T be a tree of order n. Also let v be a vertex in T such that dT (v) ≥ 3. If dT (v) µT (v) − dT (v) > 3, then 1M (T ) > 0. Proof. Since dT (v)µT (v) − dT (v) > 3 and dT (v) ≥ 3, we have
(dT (v) − 2) (µT (v) − 1) > (dT (v) − 2) ·
3 dT (v)
Thus by Theorem 3.2, we get the required result.
=3−
6 dT (v)
≥ 1.
Given n and n1 , n2 , . . . , nd (d ≥ 2) are positive integers such that 1 + i=1 ni ≤ n and 1 ≤ n1 ≤ n2 ≤ · · · ≤ nd . Denote by Tn (n1 , n2 , . . . , nd ), the set of trees T of order n with a vertex v of degree d and adjacent vertices of v have degrees n1 , n2 , . . . , nd , and each connected component of T − v is star or broom with origin w , vw ∈ E (T ). Then by Lemma 3.1, we have
d
1M (T ) = d (d − 2)
d 1
d i=1
ni − 1
− 1 = const
for all T ∈ Tn (n1 , n2 , . . . , nd ). Lemma 3.5. Let n and n1 , n2 , . . . , nd (d ≥ 2) be given positive integers such that 1 + i=1 ni < n. Then Tn (n1 , n2 , . . . , nd ) = ∅ if and only if there is a tree T of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni ̸= 2 for all i, 1 ≤ i ≤ d.
d
Proof. Let Tn (n1 , n2 , . . . , nd ) = ∅. Suppose that there is a tree T of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni = 2 for at least one i (1 ≤ i ≤ d). Then there is a vertex u of degree two in T (n1 , n2 , . . . , nd ) which is adjacent to v . We now construct tree T with n vertices from T (n1 , n2 , . . . , nd ) in the following: The r pendant edges are attached to the pendant vertex, neighbor of vertex u in T (n1 , n2 , . . . , nd ) such that r + 1 + dT (v) dT (vi ) = n. Then the degrees of adjacent vertices of v in the constructed tree T are same as the degrees of adjacent i=1 vertices of v in the tree T (n1 , n2 , . . . , nd ) and dT (v) = d. Thus the constructed tree T belongs to Tn (n1 , n2 , . . . , nd ) and it contradicts that Tn (n1 , n2 , . . . , nd ) = ∅. Hence there is a tree T of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni ̸= 2 for all i, 1 ≤ i ≤ d. Conversely, let T be a tree of order n with T ⊇ T (n1 , n2 , . . . , nd ) such that dT (v) = d and ni ̸= 2 for all i, 1 ≤ i ≤ d. Suppose that Tn (n1 , n2 , . . . , nd ) ̸= ∅. Then there is a tree T of order n in Tn (n1 , n2 , . . . , nd ). Thus we have d T ⊃ T (n1 , n2 , . . . , nd ) because 1 + i=1 ni < n. From this result with ni ̸= 2 for all 1 ≤ i ≤ d, there exists a connected component of T − v which is neither star nor broom with origin w , vw ∈ E (T ). Therefore T ̸∈ Tn (n1 , n2 , . . . , nd ) and it contradicts that T ∈ Tn (n1 , n2 , . . . , nd ). Hence Tn (n1 , n2 , . . . , nd ) = ∅. Let n1 , n2 , . . . , nd (d ≥ 2) be given positive integers. In view of the proof of Lemma 3.5, for a given integer n such that d 1 + i=1 ni < n, we can construct all graphs in Tn (n1 , n2 , . . . , nd ) from T (n1 , n2 , . . . , nd ). For example, if n = 17, then all the trees in T17 (1, 2, 2, 3, 5) are shown in Fig. 3. Moreover, Tn (3, 3, 4, 5) = ∅, n > 16.
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Fig. 3. Tree T (1, 2, 2, 3, 5) and all the trees in T17 (1, 2, 2, 3, 5).
Lemma 3.6. Let T be a tree of order n with T ⊇ T (n1 , n2 , . . . , nd ), where dT (v) = d (≥3). Then
1M (T ) ≥ 1M (T (n1 , n2 , . . . , nd )) with equality holding if and only if T ∈ Tn (n1 , n2 , . . . , nd ). Moreover, (i) for all i (1 ≤ i ≤ d), ni ̸= 2, then 1M (T ) = 1M (T (n1 , n2 , . . . , nd )) if and only if T ∼ = T (n1 , n2 , . . . , nd ), (ii) for any i (1 ≤ i ≤ d), ni = 2, then 1M (T ) = 1M (T (n1 , n2 , . . . , nd )) if and only if T ∈ Tn (n1 , n2 , . . . , nd ) ̸= ∅. Proof. Let v be a vertex of T (n1 , n2 , . . . , nd ) in which adjacent vertices of it have degrees n1 , n2 , . . . , nd . Then by Lemma 3.1, we have d 1 ni − 1 − 1 . 1M (T (n1 , n2 , . . . , nd )) = d (d − 2)
(12)
d i=1
Since T ⊇ T (n1 , n2 , . . . , nd ), we have v ∈ V (T ). Then clearly dT (v) = d and µT (v) ≥ Theorem 3.2, we have
1 d
d
i=1
ni . For dT (v) ≥ 3, by
1M (T ) ≥ dT (v) (dT (v) − 2)(µT (v) − 1) − 1
(13)
with equality holding if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ). Using d the result dT (v) = d and µT (v) ≥ 1d i=1 ni , from (12) and (13), we get
1M (T ) ≥ 1M (T (n1 , n2 , . . . , nd )).
(14)
Moreover, the equality holds in (14) if and only if each connected component of T − v is star or broom with origin w , vw ∈ E (T ) and µT (v) = 1d di=1 ni , that is, T ∈ Tn (n1 , n2 , . . . , nd ). (i) Let ni ̸= 2 for all i (1 ≤ i ≤ d). If 1 +
d
i=1
ni < n, then by Lemma 3.5, Tn (n1 , n2 , . . . , nd ) = ∅ and hence the inequality
in (14) is strict. Otherwise, 1 + i=1 ni = n. Then T ∼ = T (n1 , n2 , . . . , nd ) and the equality holds in (14). (ii) Let ni = 2 for any i (1 ≤ i ≤ d). Then by Lemma 3.5, we have Tn (n1 , n2 , . . . , nd ) ̸= ∅. Then it follows that the equality holds in (14) if and only if T ∈ Tn (n1 , n2 , . . . , nd ) ̸= ∅.
d
The double broom DB(n, a, b) consists of the path Pn−a−b together with a independent vertices adjacent to one pendant vertex of the path and b independent vertices adjacent to the other pendant vertex. The graph DB(n, a, b) is said to be a balanced double broom if |a − b| ≤ 1. Let Bn be the class of trees T of order n such that T ∼ = B(n, ∆ − 1) or T ∼ = DB(n, a, b). Then by Lemma 3.1, it is easy to see that if T ∈ Bn , then 1M (T ) = −2. Also we have 1M (Pn ) = −2 and 1M (K1, n−1 ) = 1 − n. We denote by T∗n (see Fig. 4), a class of trees T of order n obtained from double broom DB(n − 2, a, b) (double broom is of order n − 2) by attaching two pendant edges to the two non-adjacent vertices of average degree two. If T ∈ T∗n , then there exist two vertices v and y such that dT (u) = 3, µT (u) = 5/3 (u = v, y) and v y ̸∈ E (T ). By Lemma 3.5, one can see easily that Tn (2, 2, 2) ̸= ∅, Tn (1, 2, 3) ̸= ∅, Tn (1, 1, 2, 2) ̸= ∅ (n ≥ 7) and Tn (1, 2, 2) ̸= ∅ (n ≥ 6). We now obtain the complete characterization of 1M (T ) ≤ 0 for trees. Theorem 3.7. Let T be a tree with n ≥ 3 vertices and different from a star. Then
(i) 1M (T ) = 0 if and only if T ∼ = T (1, 1, 1, 3) or T ∈ Tn (2, 2, 2), Tn (1, 2, 3), Tn (1, 1, 2, 2), T∗n . (ii) 1M (T ) = −1 if and only if T ∼ = T (1, 1, 3) or T ∈ Tn (1, 2, 2). (iii) 1M (T ) = −2 if and only if T ∼ = Pn or T ∈ Bn .
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Fig. 4. Tree T , T ∈ T∗n .
Proof. If T ∼ = Pn , then 1M (T ) = −2. Otherwise, there is a vertex of degree greater than or equal to 3 in T . Let S be a set of vertices v (dT (v) ≥ 3) in T such that dT (v) µT (v) − dT (v) is maximum, that is, S = {v ∈ V (T ) : dT (v) ≥ 3 and dT (v) µT (v) − dT (v) is maximum}. Let v be the maximum degree vertex of degree dT (v) (≥3) in S. Then we have dT (v) ≤ ∆. Also let A = dT (v) µT (v) − dT (v). Then one can see easily that A ≥ 0. If A ≥ 4, then 1M (T ) > 0 by Corollary 3.4. We now consider the following cases: Case (1): A = 0. In this case dT (v) µT (v) = dT (v). Therefore vertex v is adjacent to the pendant vertices in T , that is, T ∼ = K1, n−1 , a contradiction. Case (2): A = 1. In this case dT (v) µT (v) = dT (v) + 1. Therefore vertex v is adjacent to exactly one non-pendant vertex in T . Since dT (v) ≥ 3 and dT (v) µT (v) − dT (v) is maximum with maximum value 1, then all the vertices of degree greater than or equal to 3 in T must be adjacent to exactly one non-pendant vertex. Then T is isomorphic to broom or double broom. Hence T ∈ Bn . Moreover, 1M (T ) = −2 if T ∈ Bn . Case (3): A = 2. Then we have dT (v) µT (v) = dT (v) + 2. If dT (v) ≥ 5, then
(dT (v) − 2) (µT (v) − 1) = 2 −
4 dT (v)
> 1.
By Theorem 3.2, we have 1M (T ) > 0 and it contradicts to 1M (T ) ≤ 0. So we consider the following two subcases: Subcase (i): dT (v) = 4. In this subcase either T ⊇ T (1, 1, 1, 3) or T ⊇ T (1, 1, 2, 2). When T ⊇ T (1, 1, 1, 3), by Lemma 3.6(i) we have 1M (T ) = 1M (T (1, 1, 1, 3)) = 0 if and only if T ∼ = T (1, 1, 1, 3). When T ⊇ T (1, 1, 2, 2), by Lemma 3.6(ii) we have 1M (T ) = 1M (T (1, 1, 2, 2)) = 0 if and only if T ∈ Tn (1, 1, 2, 2) ̸= ∅. Subcase (ii): dT (v) = 3. In this subcase T ⊇ T (1, 1, 3) or T ⊇ T (1, 2, 2). First we assume that T ⊇ T (1, 1, 3). Then by Lemma 3.6(i) we have 1M (T ) = 1M (T (1, 1, 3)) = −1 if and only if T ∼ = T (1, 1, 3). Now we have to find trees T for which T ⊃ T (1, 1, 3) and 1M (T ) = 0. Therefore we must have either T ⊇ T (1, 2, 3) or T ⊇ T (1, 1, 4) and hence A = dT (v) µT (v) − dT (v) ≥ 3, a contradiction. Next we assume that T ⊇ T (1, 2, 2). Then by Lemma 3.6(ii) we have 1M (T ) = 1M (T (1, 2, 2)) = −1 if and only if T ∈ Tn (1, 2, 2) ̸= ∅. Now we have to find trees T for which T ⊃ T (1, 2, 2) and 1M (T ) = 0. From the above results, we conclude that n ≥ 7 and T ̸∈ Tn (1, 2, 2) as 1M (T ) = 0. Then there exists at least one vertex, say y (̸= v) of degree 3 in T such that dT (y) µT (y) − dT (y) = 2. If there exists exactly one such vertex y in T , then yr ∈ E (T ) (dT (r ) = 1 or 2) and hence T ∈ T∗n as A = 2. Otherwise, by using Transformation A for the vertices x (̸= v, y) of degree 3 such that dT (x) µT (x) − dT (x) = 2, we arrive at a tree T ′ ∈ T∗n . Then 1M (T ) > 1M (T ′ ) = 0 by Lemma 2.1. Case (4): A = 3. Then we have dT (v) µT (v) = dT (v) + 3. If dT (v) ≥ 4, then
(dT (v) − 2) (µT (v) − 1) = 3 −
6 dT (v)
> 1.
Similarly, by Theorem 3.2, 1M (T ) > 0 and it contradicts to 1M (T ) ≤ 0. Hence dT (v) = 3 as ∆ ≥ 3. Therefore, we must have either T ⊇ T (2, 2, 2) or T ⊇ T (1, 2, 3). When T ⊇ T (2, 2, 2), by Lemma 3.6(ii) we have 1M (T ) = 1M (T (2, 2, 2)) = 0 if and only if T ∈ Tn (2, 2, 2) ̸= ∅. Similarly, when T ⊇ T (1, 2, 3), we get 1M (T ) = 1M (T (1, 2, 3)) = 0 if and only if T ∈ Tn (1, 2, 3) ̸= ∅. This completes the proof. Example 1. By SageMath [17], we characterize all trees of order 9 satisfying
1M (T ) = 0 (see, Fig. 5),
1M (T ) = −1 (see, Fig. 6) and 1M (T ) = −2 (see, Fig. 7).
Let Γ be a class of graphs H = (V , E ) of order n such that H {Cn , Pn , K1, n−1 , T (1, 1, 3), T (1, 1, 1, 3)} and
H ̸∈ Bn , Tn (1, 2, 2), Tn (2, 2, 2), Tn (1, 2, 3), Tn (1, 1, 2, 2), T∗n .
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Fig. 5. All the above trees T , 1M (T ) = 0.
Fig. 6. All the above trees T , 1M (T ) = −1.
Fig. 7. All the above trees T , 1M (T ) = −2.
Theorem 3.8. Let G be a connected graph of order n ≥ 3. Then
(i) M1 = M2 if and only if G ∼ = Cn or G ∼ = T (1, 1, 1, 3) or G ∈ Tn (2, 2, 2), Tn (1, 2, 3), Tn (1, 1, 2, 2), T∗n . (ii) M1 (G) > M2 (G) if and only if G ∼ = K1, n−1 or G ∼ = Pn or G ∼ = T (1, 1, 3) or G ∈ Bn or G ∈ Tn (1, 2, 2). (iii) M1 (G) < M2 (G) if and only if G ∈ Γ . Proof. We have 1M (K1, n−1 ) = 1 − n < 0. By Proposition 2.3 and Theorem 3.7, we get the required result.
Acknowledgments The authors are much grateful to two referees for their valuable comments on our paper, which have considerably improved the presentation of this paper. The second author is supported by the National Research Foundation funded by the Korean government with the grant no. 2013R1A1A2009341. The third author is supported by the Korea Foundation for Advanced Studies’ International Scholar Exchange Fellowship for the academic year of 2014–2015. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
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