Zagreb eccentricity indices of unicyclic graphs

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Zagreb eccentricity indices of unicyclic graphs Xuli Qi a, *, Bo Zhou b , Jiyong Li a a

College of Mathematics and Information Science, Hebei Normal University, Hebei Key Laboratory of Computational Mathematics and Applications, Shijiazhuang 050024, PR China b Department of Mathematics, South China Normal University, Guangzhou 510631, PR China

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Article history: Received 4 February 2017 Received in revised form 27 July 2017 Accepted 1 August 2017 Available online xxxx Keywords: Zagreb indices Zagreb eccentricity indices Vertex eccentricities Unicyclic graphs

a b s t r a c t For a connected graph, the first Zagreb eccentricity index (ξ1 ) is defined as the sum of the squares of the eccentricities of the vertices, and the second Zagreb eccentricity index (ξ2 ) is defined as the sum of the products of the eccentricities of pairs of adjacent vertices. We determine the n-vertex unicyclic graphs with minimum, second-minimum and third-minimum ξ1 and ξ2 , the n-vertex unicyclic graphs with maximum and secondmaximum ξ1 , and the n-vertex unicyclic graphs with maximum, second-maximum and third-maximum ξ2 . © 2017 Published by Elsevier B.V.

1. Introduction Let G be a simple connected graph with vertex set V (G) and edge set E(G). For u ∈ V (G), degG (u) or deg(u) denotes the degree of u in G. For u ∈ V (G), eG (u) or e(u) denotes the eccentricity of u in G, which is equal to the largest distance from u to other vertices. Two types of Zagreb indices were proposed in [3,4]. The first Zagreb index of G is defined as M1 (G) =

∑

deg 2 (u),

u∈V (G)

while the second Zagreb index of G is defined as M2 (G) =

∑

deg(u)deg(v ).

uv∈E(G)

As summarized by Todeschini and Consonni [7,8], the Zagreb indices and their variants are useful molecular descriptors which found considerable use in QSPR and QSAR studies. Some graph invariants based on vertex eccentricities received much attention recently, see, e.g. [9,11]. Vukičeivc and Graovac [9] introduced two types of Zagreb eccentricity indices. The first Zagreb eccentricity index of G is defined as

ξ1 (G) =

∑

e2 (u),

u∈V (G)

while the second Zagreb eccentricity index of G is defined as

ξ2 (G) =

∑

e(u)e(v ).

uv∈E(G)

*

Corresponding author. E-mail address: [email protected] (X. Qi).

http://dx.doi.org/10.1016/j.dam.2017.08.001 0166-218X/© 2017 Published by Elsevier B.V.

Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

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Fig. 1. The graphs G1 and G2 in Lemma 2.1.

Some mathematical and computational properties of the Zagreb eccentricity indices ξ1 and ξ2 have been established in [2,10]. Among them, various lower and upper bounds for the Zagreb eccentricity indices were given, the trees with the first a few smallest and largest Zagreb eccentricity indices were determined, the trees with minimum and maximum Zagreb eccentricity indices were determined when diameter, number of pendent vertices, and matching number are respectively given, and the trees with maximum Zagreb eccentricity indices were determined when maximum degree is given. Qi and Du [6] determined the trees with minimum Zagreb eccentricity indices when domination number, maximum degree, and bipartition size are respectively given. Das and Lee [1] presented some properties, upper and lower bounds for the Zagreb eccentricity indices and characterized the extremal graphs. Luo and Wu [5] computed the Zagreb eccentricity indices of the generalized hierarchical product. In this paper, we determine the n-vertex unicyclic graphs with minimum, second-minimum and third-minimum ξ1 and ξ2 , the n-vertex unicyclic graphs with maximum and second-maximum ξ1 , and the n-vertex unicyclic graphs with maximum, second-maximum and third-maximum ξ2 . 2. Preliminaries Let Sn , Pn and Cn be the n-vertex star, path and cycle, respectively. Let Un be the set of n-vertex unicyclic graphs. For integers r and n with 3 ≤ r ≤ n, let Un,r be the set of n-vertex unicyclic graphs with cycle length r, and Un,≥r the set of n-vertex unicyclic graphs with cycle length at least r. Note that U3 = U3,3 = U3,≥3 = {C3 }. Therefore, we may suppose that n ≥ 4. Let Cr (T1 , T2 , . . . , Tr ) be the graph constructed as follows. Let the vertices of the cycle Cr be labeled consecutively by v1 , v2 , . . . , vr . Let T1 , T2 , . . . , Tr be vertex–disjoint trees such that Ti and the cycle Cr have exactly one vertex vi in common for i = 1, 2, . . . , r. Denote |G| = ∑|rV (G)| for a graph G. Then any n-vertex unicyclic graph G with a cycle on r vertices is of the form Cr (T1 , T2 , . . . , Tr ), where i=1 |Ti | = n. For a graph G with u ∈ V (G) and a graph H that is vertex–disjoint with G, a graph obtained by attaching H at its vertex v to u of G is such a graph obtained from G and H by adding an edge uv . Lemma 2.1. Let u be a vertex of a connected graph G0 with at least two vertices. For integer t ≥ 1, let G1 be the graph obtained from G0 by attaching a star St +1 at its center v to u, and G2 the graph obtained from G0 by attaching t + 1 pendent vertices to u, see Fig. 1. Then ξ1 (G2 ) < ξ1 (G1 ) and ξ2 (G2 ) < ξ2 (G1 ). Proof. Let w be a pendent neighbor of v in G1 and a pendent neighbor of u in G2 outside G0 . Note that eG2 (x) ≤ eG1 (x) for any x ∈ V (G0 ), eG1 (u) ≤ eG1 (v ) < eG1 (w ) and eG2 (w ) = eG1 (v ). Then

ξ1 (G2 ) − ξ1 (G1 ) =

∑ [

e2G2 (x) − e2G1 (x)

]

x∈V (G0 )

+ (t + 1)e2G2 (w ) − t · e2G1 (w) − e2G1 (v ) ≤ (t + 1)e2G1 (v ) − t · e2G1 (w) − e2G1 (v ) [ ] = t e2G1 (v ) − e2G1 (w ) < 0, ∑ [ ] eG2 (x)eG2 (y) − eG1 (x)eG1 (y) ξ2 (G2 ) − ξ2 (G1 ) = xy∈E(G0 )

+ (t + 1)eG2 (w )eG2 (u) − t · eG1 (w)eG1 (v ) − eG1 (v )eG1 (u) ≤ (t + 1)eG1 (v )eG1 (u) − t · eG1 (w)eG1 (v ) − eG1 (v )eG1 (u) [ ] = t · eG1 (v ) eG1 (u) − eG1 (w) < 0, and thus ξ1 (G2 ) < ξ1 (G1 ) and ξ2 (G2 ) < ξ2 (G1 ).

□

Lemma 2.2. Let u be a vertex of a connected graph G0 with at least two vertices. For integer t ≥ 1, let G1 be the graph obtained from G0 by attaching a star St +2 at its center v to u, and G2 the graph obtained from G0 by attaching t pendent vertices and a path P2 at one terminal vertex v to u, see Fig. 2. If eG0 (u) = 1, then ξ1 (G2 ) = ξ1 (G1 ) and ξ2 (G2 ) = ξ2 (G1 ); If eG0 (u) ≥ 2, then ξ1 (G2 ) < ξ1 (G1 ) and ξ2 (G2 ) < ξ2 (G1 ). Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

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Fig. 2. The graphs G1 and G2 in Lemma 2.2.

Proof. Let w1 be a pendent neighbor of v in G1 and the pendent neighbor of v in G2 , w2 a pendent neighbor of u in G2 outside G0 . Suppose that eG0 (u) = 1. Then eG2 (x) = eG1 (x) for any x ∈ V (G0 ), eG2 (u) = eG1 (u) = eG2 (v ) = eG1 (v ) = 2, eG2 (w1 ) = eG1 (w1 ) = eG2 (w2 ) = 3. Then ξ1 (G2 ) − ξ1 (G1 ) = t [e2G (w2 ) − e2G (w1 )] = 0 and ξ2 (G2 ) − ξ2 (G1 ) = 2 1 t [eG2 (w2 )eG2 (u) − eG1 (w1 )eG1 (v )] = 0, implying that ξ1 (G2 ) = ξ1 (G1 ) and ξ2 (G2 ) = ξ2 (G1 ). Now suppose that eG0 (u) ≥ 2. Then eG2 (x) = eG1 (x) for any x ∈ V (G0 ), eG1 (u) < eG1 (v ) < eG1 (w1 ) and eG2 (w2 ) = eG1 (v ). Then

ξ1 (G2 ) − ξ1 (G1 ) = t · e2G2 (w2 ) − t · e2G1 (w1 ) [ ] = t e2G1 (v ) − e2G1 (w1 ) < 0, ξ2 (G2 ) − ξ2 (G1 ) = t · eG2 (w2 )eG2 (u) − t · eG1 (w1 )eG1 (v ) ] [ = t · eG1 (v ) eG1 (u) − eG1 (w1 ) < 0, and thus ξ1 (G2 ) < ξ1 (G1 ) and ξ2 (G2 ) < ξ2 (G1 ). □ For n ≥ 4, let Pn′ be the n-vertex tree formed by attaching a pendent vertex to the neighbor of one terminal vertex of the path Pn−1 . For n ≥ 6, let Pn′′ be the n-vertex tree formed by attaching a pendent vertex to the vertex that is of distance 2 from one terminal vertex of the path Pn−1 . Lemma 2.3 ([10]). Among the n-vertex trees, Pn for n ≥ 3, Pn′ for n ≥ 4, and Pn′′ for n ≥ 6 are, respectively, the unique trees with maximum, second-maximum and third-maximum ξ1 and ξ2 , where

ξ1 (Pn ) =

ξ2 (Pn ) =

ξ1 (Pn′′ ) =

⎧ 7n3 − 9n2 + 2n ⎪ ⎪ ⎨

12 ⎪ 7n − 9n2 − n + 3 ⎪ ⎩ 12 3

if n is odd,

⎧ 3 2 ⎪ ⎪ 7n − 21n + 20n ⎨

if n is even

⎩

if n is odd,

12 3 2 ⎪ ⎪ 7n − 21n + 17n − 3 12

⎧ 7n3 − 18n2 − 34n + 96 ⎪ ⎪ ⎨ 12

3

2

⎪ 7n − 18n − 31n + 90 ⎪ ⎩ 12

ξ2 (Pn′′ ) =

if n is even

if n is even if n is odd,

⎧ 7n3 − 30n2 − 4n + 96 ⎪ ⎪ ⎨

if n is even

⎩

if n is odd.

12 3 2 ⎪ ⎪ 7n − 30n − n + 96 12

Lemma 2.4. For n-vertex cycle Cn with n ≥ 3, we have

ξ1 (Cn ) = ξ2 (Cn ) =

⎧ n3 ⎪ ⎪ ⎨

4 2 ⎪ ⎪ n(n − 1)

⎩

4

if n is even if n is odd.

Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

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3. Unicyclic graphs with minimum Zagreb eccentricity indices For 3 ≤ r ≤ n − 1, let Sn,r = Cr (T1 , T2 , . . . , Tr ), where |T1 | = n − r + 1, |T2 | = |T3 | = · · · = |Tr | = 1 and T1 is a star with center v1 . Theorem 3.1. Let G ∈ Un,r , where 3 ≤ r ≤ n − 1. Then ξ1 (G) ≥ ξ1 (Sn,r ) and ξ2 (G) ≥ ξ2 (Sn,r ), where the equalities hold if and only if G ∼ = Sn,r ,

ξ1 (Sn,r ) =

ξ2 (Sn,r ) =

⎧ (n − 4)r 2 + 4nr + 4n + 4 ⎪ ⎪ ⎨

if r is even

⎩

if r is odd,

4 2 ⎪ ⎪ (n − 4)r + (2n + 8)r + n 4

⎧ (n − 2)r 2 + (2n + 4)r ⎪ ⎪ ⎨

if r is even

4

2

⎪ (n − 2)r + 10r − n − 4 ⎪ ⎩ 4

if r is odd.

Proof. The case r = n − 1 is obvious. Suppose that 3 ≤ r ≤ n − 2. Let G1 = Cr (T1 , T2 , . . . , Tr ) be a graph in Un,r with minimum ξ1 . By Lemma 2.1, Ti is a star with center vi for each i = 1, 2, . . . , r. Then

ξ1 (G1 ) =

∑ v∈V (Cr )

=

r ∑

v∈V (G)\V (Cr )

(|Ti | − 1)[e(vi ) + 1]2

i=1

=

=

r ∑ [

e2 (v )

r ∑

e2 (vi ) +

r ∑ [

∑

e 2 (v ) +

i=1

|Ti |e2 (vi ) + 2(|Ti | − 1)e(vi ) + |Ti | − 1

]

i=1

] |Ti |e2 (vi ) + 2(|Ti | − 1)e(vi ) + n − r .

i=1

Note that the eccentricity of the vertex vi ∈ V (Cr ) is exactly ⌊ 2r ⌋ or ⌊ 2r ⌋ + 1. If r is even, there exists a vertex in V (Cr ), say vp with eccentricity 2r + 1; if r is odd, there are two adjacent vertices in V (Cr ), say vq and vq+1 with eccentricity r +2 1 . Thus r ∑ [

] |Ti |e2 (vi )

i=1

= |T1 |e2 (v1 ) + |T2 |e2 (v2 ) + · · · + |Tr |e2 (vr ) ⎧ r r ⎪ ⎨( + 1)2 |Tp | + ( )2 (n − |Tp |) 2 2 ≥ r r ⎪ ⎩( + 1)2 (|Tq | + |Tq+1 |) + ( )2 (n − |Tq | − |Tq+1 |) 2 2 ⎧ r r ⎪ if r is even ⎨( + 1)2 + ( )2 (n − 1) 2 2 ≥ ⎪ ⎩2( r + 1 )2 + ( r − 1 )2 (n − 2) if r is odd, 2

if r is even if r is odd

2

where the equalities hold if and only if G1 ∼ = Sn,r . Similarly, we have r ∑

2(|Ti | − 1)e(vi ) ≥

i=1

{

r(n − r)

if r is even

(r − 1)(n − r)

if r is odd,

where the equality holds if and only if G1 ∼ = Sn,r . Therefore, G1 ∼ = Sn,r and

ξ1 (G1 ) ≥

⎧ (n − 4)r 2 + 4nr + 4n + 4 ⎪ ⎪ ⎨

if r is even

⎩

if r is odd.

4 2 ⎪ ⎪ (n − 4)r + (2n + 8)r + n 4

Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

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Let G2 = Cr (T1 , T2 , . . . , Tr ) be a graph in Un,r with minimum ξ2 . By the similar argument as above, we have G2 ∼ = Sn,r and

∑

ξ2 (G2 ) =

e(v )e(w ) +

vw∈E(Cr )

∑

=

=

(|Ti | − 1)e(vi )[e(vi ) + 1]

i=1

e(v )e(w ) +

vw∈E(Cr )

=

r ∑

r ∑

(|Ti | − 1)[e2 (vi ) + e(vi )]

i=1

⎧ r r r ⎪ 2( )( + 1) + (r − 2)( )2 ⎪ ⎪ 2 2 2 ⎪ ⎪ ⎪ r r ⎪ ⎪+(n − r)[( )2 + ] ⎨ 2

if r is even

2 r −1

r +1 2 r +1 r −1 2 ⎪ ⎪ ( ) + 2( )( ) + (r − 3)( ) ⎪ ⎪ 2 2 2 2 ⎪ ⎪ ⎪ ⎪ ⎩+(n − r)[( r − 1 )2 + r − 1 ] 2 2 ⎧ 2 (n − 2)r + (2n + 4)r ⎪ ⎪ ⎨ if r is even

if r is odd

4

⎪ (n − 2)r 2 + 10r − n − 4 ⎪ ⎩ 4

if r is odd.

This completes the proof. □ For 4 ≤ r ≤ n − 2, let Sn′ ,r be the n-vertex unicyclic graph formed by attaching a path P2 and n − r − 2 pendent vertices to a vertex of the cycle Cr . For 4 ≤ r ≤ n − 2, let Snt ,r (i, j) be the n-vertex unicyclic graph formed by attaching t and n − r − t r pendent vertices, respectively, to two vertices vi and vj of the cycle Cr , where 1 ≤ t ≤ ⌊ n− ⌋ and 1 ≤ i < j ≤ r. Let 2

S∗n,r =

⎧ n−r r ⎪ ⎨{Snt ,r (i, j) : 1 ≤ t ≤ ⌊ ⌋, 1 ≤ d(vi , vj ) ≤ − 1}

if r is even

⎪ ⎩{Snt ,r (i, j) : 1 ≤ t ≤ ⌊ n − r ⌋, d(vi , vj ) = 1}

if r is odd.

2

2

2

Theorem 3.2. Let G ∈ Un,r and G ̸ ∼ = Sn,r , where 4 ≤ r ≤ n − 2. Then

ξ1 (G) ≥

⎧ (n − 4)r 2 + (4n + 4)r + 4n + 8 ⎪ ⎪ ⎨

if r is even

⎩

if r is odd ,

4 2 ⎪ ⎪ (n − 4)r + (2n + 12)r + n 4

where the equality holds if and only if G is isomorphic to a graph in S∗n,r . Proof. Let G be a graph in Un,r not isomorphic to Sn,r with minimum ξ1 . By Lemmas 2.1 and 2.2 and Theorem 3.1, G must be (isomorphic to) Sn′ ,r or Cr (T1 , T2 , . . . , Tr ), where Ti is a star with center vi for each i = 1, 2, . . . , r and at least two trees of T1 , T2 , . . . , Tr are nontrivial. First suppose that G ∼ = Cr (T1 , T2 , . . . , Tr ), where Ti is a star with center vi for each i = 1, 2, . . . , r, and G ̸∼ = Sn,r . Note that the eccentricities of the vertices on Cr exactly ⌊ 2r ⌋ or ⌊ 2r ⌋ + 1. If r is even, there exist at least two vertices on Cr with eccentricity 2r + 1; if r is odd, there are at least three adjacent vertices on Cr with eccentricity r +2 1 . By the similar argument as in the proof of Theorem 3.1, we have

ξ1 (G) =

≥

=

r ∑ [ ] |Ti |e2 (vi ) + 2(|Ti | − 1)e(vi ) + n − r i=1 ⎧ r r r ⎪ ⎨2( + 1)2 + (n − 2)( )2 + 2(n − r)( ) + n − r

2 r +1

2 r −1

2

r −1 ⎪ ⎩3( )2 + (n − 3)( )2 + 2(n − r)( )+n−r 2 2 2 ⎧ 2 (n − 4)r + (4n + 4)r + 4n + 8 ⎪ ⎪ ⎨ if r is even

if r is even if r is odd

4

⎪ (n − 4)r 2 + (2n + 12)r + n ⎪ ⎩ 4

if r is odd,

where the equality holds if and only if G is isomorphic to some graph in S∗n,r . To prove our result, we only need to show that ξ1 (Sn′ ,r ) > ξ1 (G) for any G ∈ S∗n,r . Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

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By direct calculations, we have

ξ1 (Sn′ ,r ) =

⎧ (n − 4)r 2 + (4n + 16)r + 4n + 36 ⎪ ⎪ ⎨

if r is even

⎪ (n − 4)r 2 + (2n + 28)r + n + 24 ⎪ ⎩

if r is odd.

4

4

Therefore, for any G ∈ S∗n,r , we have

ξ1 (Sn,r ) − ξ1 (G) = ′

{

3r + 7 4r + 6

if r is even if r is odd

> 0. This completes the proof. □ For n ≥ 6, let Tn,3 (a, b, c) be the n-vertex unicyclic graph obtained from C3 = v1 v2 v3 v1 by attaching a path P2 and a pendent vertices to the vertex v1 , b pendent vertices to the vertex v2 , and c pendent vertices to the vertex v3 , where a ≥ 0, b ≥ 1, c ≥ 0 and a + b + c = n − 5. Lemma 3.1. For integers a ≥ 0, b ≥ 1 and c ≥ 0, let G0 = Tn,3 (a, b, c), G1 = Tn+1,3 (a, b + 1, c) and G2 = Tn+1,3 (a + 1, b, c) . Then ξ1 (G2 ) < ξ1 (G1 ) and ξ2 (G2 ) < ξ2 (G1 ). Proof. Let u be a pendent neighbor of v2 in G1 and a pendent neighbor of v1 in G2 . Note that eG2 (x) = eG1 (x) for any x ∈ V (G0 ), eG2 (u) = 3 and eG1 (u) = 4. Then

ξ1 (G2 ) − ξ1 (G1 ) = e2G2 (u) − e2G1 (u) = 32 − 42 = −7 < 0, ξ2 (G2 ) − ξ2 (G1 ) = eG2 (u)eG2 (v1 ) − eG1 (u)eG1 (v2 ) = 3 · 2 − 4 · 3 = −6 < 0, and thus ξ1 (G2 ) < ξ1 (G1 ) and ξ2 (G2 ) < ξ2 (G1 ). □ For n ≥ 5, let Sn,3 (a, b, c) be the n-vertex unicyclic graph obtained from C3 = v1 v2 v3 v1 by attaching a pendent vertices to the vertex v1 , b pendent vertices to the vertex v2 , and c pendent vertices to the vertex v3 , where a ≥ b ≥ max{c , 1} and a + b + c = n − 3. Let Sn,3 = {Sn,3 (a, b, c) : a ≥ b ≥ max{c , 1}, a + b + c = n − 3}. For integer t with 0 ≤ t ≤ n − 5, (t) let Sn,3 be the graph obtained from C3 = v1 v2 v3 v1 by attaching a star Sn−t −3 at its center and t pendent vertices to v1 . Let (t)

S′n,3 = {Sn,3 : 0 ≤ t ≤ n − 5}.

Theorem 3.3. Among the graphs in Un,3 , Sn,3 for n ≥ 4, the graphs in Sn,3 for n ≥ 5 and the graphs in S′n,3 for n ≥ 5 are, respectively, the unique graphs with minimum, second-minimum and third-minimum ξ1 and ξ2 , the first Zagreb eccentricity indices of which are equal to 4n − 3, 9n − 15 and 9n − 10, respectively, while the second Zagreb eccentricity indices of which are equal to 2n + 2, 6n − 6 and 6n + 1, respectively. Proof. By Theorem 3.1, Sn,3 for n ≥ 4 is the unique graph in Un,3 with minimum ξ1 and ξ2 . By Lemmas 2.1 and 2.2, the graphs in Un,3 with second-minimum ξ1 and ξ2 must be the graphs in Sn,3 or S′n,3 . For any G1 ∈ Sn,3 and G2 ∈ S′n,3 , by direct calculations, we have ξ1 (G1 ) = 9n − 15 < 9n − 10 = ξ1 (G2 ) and ξ2 (G1 ) = 6n − 6 < 6n + 1 = ξ2 (G2 ). Thus the graphs in Sn,3 are the unique graphs with second-minimum ξ1 and ξ2 in Un,3 . Then by Lemma 3.1, the graphs in Un,3 with third-minimum ξ1 and ξ2 must be Tn,3 (n − 6, 1, 0), Tn,3 (n − 7, 1, 1) or the graphs in S′n,3 . Note that ξ1 (Tn,3 (n − 7, 1, 1)) = 9n + 16 > ξ1 (Tn,3 (n − 6, 1, 0)) = 9n + 9 > 9n − 10 and ξ2 (Tn,3 (n − 7, 1, 1)) = 6n + 21 > ξ2 (Tn,3 (n − 6, 1, 0)) = 6n + 15 > 6n + 1. Thus the graphs in S′n,3 are the unique graphs in Un,3 with third-minimum ξ1 and ξ2 . □ Theorem 3.4. Among the graphs in Un , (i) Sn,3 for n ≥ 4 is the unique graph with minimum ξ1 and ξ2 , equal to 4n − 3 and 2n + 2, respectively; (ii) Cn for n = 4, 5, the graphs in Sn,3 , Sn,4 and Sn,5 for n ≥ 6 are the unique graphs with second-minimum ξ1 , equal to 16 for n = 4, 20 for n = 5 and 9n − 15 for n ≥ 6; while Cn for n = 4, 5, the graphs in Sn,3 for n ≥ 6 are the unique graphs with second-minimum ξ2 , equal to 16 for n = 4, 20 for n = 5 and 6n − 6 for n ≥ 6; (iii) S5,3 (1, 1, 0) and S5,4 for n = 5, the graphs in S′6,3 and S61,4 (1, 2) for n = 6, and the graphs in S′n,3 , S∗n,4 and S∗n,5 for n ≥ 7 are the unique graphs with third-minimum ξ1 , equal to 30 for n = 5, 44 for n = 6 and 9n − 10 for n ≥ 7; S5,3 (1, 1, 0) for n = 5 and Sn,4 for n ≥ 6 are the unique graphs with third-minimum ξ2 , equal to 24 for n = 5 and 6n − 4 for n ≥ 6. Proof. The cases n = 4, 5 can be checked easily. Suppose that n ≥ 6. If r is even and 4 ≤ r ≤ n − 3, then by Theorem 3.1, we have ξ1 (Sn,r +2 ) − ξ1 (Sn,r ) = (n − 4)(r + 1) + 2n > 0 and ξ2 (Sn,r +2 ) − ξ2 (Sn,r ) = (n − 2)(r + 1) + n + 2 > 0, implying that ξ1 (Sn,r ) and ξ2 (Sn,r ) are increasing for even r with 4 ≤ r ≤ n − 1. If r is odd and 3 ≤ r ≤ n − 3, then by Theorem 3.1, we have ξ1 (Sn,r +2 ) − ξ1 (Sn,r ) = (n − 4)(r + 1) + n + 4 > 0 and ξ2 (Sn,r +2 ) − ξ2 (Sn,r ) = (n − 2)(r + 1) + 5 > 0, implying Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

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that ξ1 (Sn,r ) and ξ2 (Sn,r ) are increasing for odd r with 3 ≤ r ≤ n − 1. Then by Theorem 3.1 and Lemma 2.4, we find that the graphs in Un with minimum ξ1 (ξ2 , respectively) are just the graphs in {Sn,3 , Sn,4 , Cn } with minimum ξ1 (ξ2 , respectively), which is equal to min{4n − 3, 9n − 15,

2

n3 4

} = 4n − 3 for even n and min{4n − 3, 9n − 15, n(n−4 1) } = 4n − 3 for odd n 3 n(n−1)2 (min{2n + 2, 6n − 4, n4 } = 2n + 2 for even n and min{2n + 2, 6n − 4, } = 2n + 2 for odd n, respectively), proving (i). 4 Then we prove (ii). By Lemma 2.4 and Theorem 3.1, we have ξ1 (Sn,4 ) = ξ1 (Sn,5 ) = 9n − 15 < ξ1 (Cn ) and ξ2 (Sn,4 ) = 6n − 4 < 6n − 1 = ξ2 (Sn,5 ) < ξ2 (Cn ). Then Sn,4 and Sn,5 are the unique graphs in Un,≥4 with minimum ξ1 , and Sn,4 is the unique graph in Un,≥4 with minimum ξ2 . By Theorems 3.1 and 3.3, we find that the graphs in Un with second-minimum ξ1 (respectively, ξ2 ) are just the graphs in Un,3 \ {Sn,3 } ∪ Un,≥4 with minimum ξ1 (ξ2 , respectively), which is equal to min{ξ1 (G)(G ∈ Sn,3 ), ξ1 (Sn,4 ), ξ1 (Sn,5 )} = min{9n − 15, 9n − 15, 9n − 15} = 9n − 15(min{ξ2 (G)(G ∈ Sn,3 ), ξ2 (Sn,4 )} = min{6n − 6, 6n − 4} = 6n − 6, respectively), proving (ii). Now we prove (iii). By Lemma 2.4 and Theorem 3.1, we have ξ1 (S7,6 ) = 77 > 63 = ξ1 (C7 ) for n = 7 and ξ1 (Sn,6 ) = ξ1 (Sn,7 ) = 16n − 35 < ξ1 (Cn ) for n ≥ 8. Then Cn for n = 6, 7, and Sn,6 and Sn,7 for n ≥ 8 are the unique graphs in Un,≥6 with minimum ξ1 . By Theorems 3.1–3.3, we find that the graphs in Un with third-minimum ξ1 are just the graphs in Un,3 \ ({Sn,3 } ∪ Sn,3 ) ∪ (Un,4 \ {Sn,4 }) ∪ (Un,5 \ {Sn,5 }) ∪ Un,≥6 with minimum ξ1 , which is equal to min{ξ1 (G)(G ∈ S′6,3 ), ξ1 (S61,4 (1, 2)), ξ1 (C6 )} = min{44, 44, 54} = 44 for n = 6, min{ξ1 (G)(G ∈ S′7,3 ), ξ1 (G)(G ∈ S∗7,4 ), ξ1 (S71,5 (1, 2)), ξ1 (C7 )} = min{53, 53, 53, 63} = 53 for n = 7 and min{ξ1 (G)(G ∈ S′n,3 ), ξ1 (G)(G ∈ S∗n,4 ), ξ1 (G)(G ∈ S∗n,5 ), ξ1 (Sn,6 ), ξ1 (Sn,7 )} = min{9n − 10, 9n − 10, 9n − 10, 16n − 35, 16n − 35} = 9n − 10 for n ≥ 8. Similarly, the graphs in Un with third-minimum ξ2 are just the graphs in Un,3 \ ({Sn,3 } ∪ Sn,3 ) ∪ Un,≥4 with minimum ξ2 , which is equal to min{ξ2 (G)(G ∈ S′n,3 ), ξ2 (Sn,4 )} = min{6n + 1, 6n − 4} = 6n − 4. Now (iii) follows. □ 4. Unicyclic graphs with maximum Zagreb eccentricity indices For n ≥ 4, let Pn,3 = C3 (T1 , T2 , T3 ), where |T1 | = n − 2, |T2 | = |T3 | = 1, and T1 = Pn−2 is a path with a terminal vertex v1 . For n ≥ 5, let Pn,4 = C4 (T1 , T2 , T3 , T4 ), where |T1 | = n − 3, |T2 | = |T3 | = |T4 | = 1, and T1 = Pn−3 is a path with a terminal vertex v1 . For n ≥ 5, let Pn′ ,3 = C3 (T1 , T2 , T3 ), where |T1 | = n − 3, |T2 | = 2, |T3 | = 1, and T1 = Pn−3 is a path with a terminal vertex v1 . For an edge e of the graph G, G − e denotes the graph obtained from G by deleting the edge e. Theorem 4.1. Among the graphs in Un , Cn for n = 4 and Pn,3 for n ≥ 5 are the unique graphs with maximum ξ1 and ξ2 , where

ξ1 (C4 ) = ξ2 (C4 ) = 16, ⎧ 3 2 ⎪ ⎪ 7n − 18n − 10n + 36 ⎨ 12 ξ1 (Pn,3 ) = ⎪ 7n3 − 18n2 − 7n + 30 ⎪ ⎩

if n is even

⎧ 7n3 − 18n2 − 28n + 72 ⎪ ⎪ ⎨

if n is even

12

ξ2 (Pn,3 ) =

12

3

2

⎪ 7n − 18n − 25n + 72 ⎪ ⎩ 12

if n is odd,

if n is odd.

Proof. The cases n = 4, 5 can be checked directly. Suppose that n ≥ 6. Then

ξ1 (Pn,3 ) = ξ1 (Pn−1 ) + (n − 2)2 =

⎧ 7n3 − 18n2 − 10n + 36 ⎪ ⎪ ⎨

if n is even

⎪ 7n3 − 18n2 − 7n + 30 ⎪ ⎩

if n is odd,

12

12

ξ2 (Pn,3 ) = ξ2 (Pn−1 ) + (n − 2)2 + (n − 2)(n − 3) =

⎧ 7n3 − 18n2 − 28n + 72 ⎪ ⎪ ⎨ 12

3

2

⎪ 7n − 18n − 25n + 72 ⎪ ⎩ 12

if n is even if n is odd.

Combined with the expressions of ξ1 (Pn′′ ) and ξ2 (Pn′′ ) in Lemma 2.3, we have ξ1 (Pn,3 ) − ξ1 (Pn′′ ) = 2n − 5 > 0 and ξ2 (Pn,3 ) − ξ2 (Pn′′ ) = (n − 1)2 − 3 > (n − 2)(n − 3). For G ∈ Un , if G ̸ ∼ = Cn and G ̸∼ = Pn,3 , then we can always find an edge e of G such that G − e is a spanning tree and ′ ∼ G − e ̸ = Pn , Pn . Then the maximum eccentricities of the two vertices on the edge e are n − 2 and n − 3, respectively. Combined with Lemma 2.3 and the fact that deleting an edge in G cannot decrease the eccentricities of vertices in G, we have

ξ1 (G) ≤ ξ1 (G − e) ≤ ξ1 (Pn′′ ) < ξ1 (Pn,3 ), ξ2 (G) ≤ ξ2 (G − e) + (n − 2)(n − 3) ≤ ξ2 (Pn′′ ) + (n − 2)(n − 3) < ξ2 (Pn,3 ). Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

8

X. Qi et al. / Discrete Applied Mathematics (

)

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Now suppose that G ∼ = Cn . It is easily checked that

ξ1 (Pn,3 ) − ξ1 (Cn ) =

⎧ 4n3 − 18n2 − 10n + 36 ⎪ ⎪ ⎨ >0

if n is even

12

3

2

⎪ 4n − 12n − 10n + 30 ⎪ ⎩ >0

if n is odd,

⎧ 4n3 − 18n2 − 28n + 72 ⎪ ⎪ ⎨ >0

if n is even

12

ξ2 (Pn,3 ) − ξ2 (Cn ) =

12

3

2

⎪ 4n − 12n − 28n + 72 ⎪ ⎩ >0

if n is odd.

12

Then the result follows. □ Theorem 4.2. Among the graphs in Un , (i) P4,3 for n = 4, Pn,4 and Pn′ ,3 for n ≥ 5, are the unique graphs with second-maximum ξ1 , where ξ1 (P4,3 ) = 13 and

ξ1 (Pn,4 ) = ξ1 (Pn′ ,3 ) =

⎧ 7n3 − 18n2 − 34n + 96 ⎪ ⎪ ⎨

if n is even and n ≥ 6

12

3

2

⎪ 7n − 18n − 31n + 90 ⎪ ⎩

if n is odd and n ≥ 5; 12 (ii) P4,3 for n = 4, P5,4 for n = 5, P6,4 and C6 for n = 6, and Pn,4 for n ≥ 7 are the unique graphs with second-maximum ξ2 , and Pn′ ,3 for n ≥ 5 is the unique graph with third-maximum ξ2 , where ξ2 (P4,3 ) = 10, ξ2 (C6 ) = 54,

ξ2 (Pn,4 ) =

⎧ 26 ⎪ ⎪ ⎪ ⎪ 7n3 − 18n2 − 64n + 168 ⎨ 12 ⎪ ⎪ 3 2 ⎪ 7n − 18n − 61n + 168 ⎪ ⎩ 12

ξ2 (Pn′ ,3 ) =

⎧ 24 ⎪ ⎪ ⎪ ⎪ ⎨ 7n3 − 18n2 − 76n + 204 12 ⎪ ⎪ 3 2 ⎪ 7n − 18n − 73n + 204 ⎪ ⎩ 12

if n = 5 if n is even and n ≥ 6 if n is odd and n ≥ 7, if n = 5 if n is even and n ≥ 6 if n is odd and n ≥ 7.

Proof. The cases n = 4, 5, 6, 7 can be checked directly. Suppose that n ≥ 8. Then

ξ1 (Pn,4 ) = ξ1 (Pn′ ,3 ) = ξ1 (Pn−1 ) + (n − 3)2 =

⎧ 7n3 − 18n2 − 34n + 96 ⎪ ⎪ ⎨

if n is even

12

3

2

⎪ 7n − 18n − 31n + 90 ⎪ ⎩

if n is odd,

12

ξ2 (Pn,4 ) = ξ2 (Pn−1 ) + (n − 2)(n − 3) + (n − 3)(n − 4) =

⎧ 7n3 − 18n2 − 64n + 168 ⎪ ⎪ ⎨ 12

3

2

⎪ 7n − 18n − 61n + 168 ⎪ ⎩ 12

ξ2 (Pn′ ,3 ) = ξ2 (Pn−1 ) + (n − 3)2 + (n − 3)(n − 4) =

⎧ 7n3 − 18n2 − 76n + 204 ⎪ ⎪ ⎨ 12

3

2

⎪ 7n − 18n − 73n + 204 ⎪ ⎩ 12

if n is even if n is odd,

if n is even if n is odd.

Combined with the expressions of ξ1 (Pn′′ ) and ξ2 (Pn′′ ) in Lemma 2.3, we have ξ1 (Pn,4 ) = ξ1 (Pn′ ,3 ) = ξ1 (Pn′′ ), ξ2 (Pn,4 ) − ξ2 (Pn′ ,3 ) = n − 3 > 0 and ξ2 (Pn′ ,3 ) − ξ2 (Pn′′ ) = (n − 3)2 . For G ∈ Un , if G has an edge e such that G − e is a spanning tree and G − e ̸ ∼ = Pn , Pn′ , Pn′′ , then the maximum eccentricities of the two vertices on the edge e are both n − 3. Combined with Lemma 2.3 and the fact that deleting an edge in G cannot decrease the eccentricities of vertices in G, we have

ξ1 (G) ≤ ξ1 (G − e) < ξ1 (Pn′′ ) = ξ1 (Pn,4 ) = ξ1 (Pn′ ,3 ), ξ2 (G) ≤ ξ2 (G − e) + (n − 3)2 < ξ2 (Pn′′ ) + (n − 3)2 = ξ2 (Pn′ ,3 ) < ξ2 (Pn,4 ). Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.

X. Qi et al. / Discrete Applied Mathematics (

)

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Now suppose that any spanning tree of G is one of Pn , Pn′ , Pn′′ . Then G must be Cn , Pn′ ,3 or Pn,4 . By Lemma 2.4, we have

ξ1 (Pn′ ,3 ) − ξ1 (Cn ) =

⎧ 4n3 − 18n2 − 34n + 96 ⎪ ⎪ ⎨ >0

if n is even

⎪ 4n3 − 12n2 − 34n + 90 ⎪ ⎩ >0

if n is odd,

12 12

ξ2 (Pn′ ,3 ) − ξ2 (Cn ) =

⎧ 4n3 − 18n2 − 76n + 204 ⎪ ⎪ ⎨ >0 12

3

2

⎪ 4n − 12n − 76n + 204 ⎪ ⎩ >0 12

if n is even if n is odd.

Then the results follow easily. □ Acknowledgments This work was supported by the Hebei Provincial Natural Science Foundation of China (Grant Number A2014205149) and the National Natural Science Foundation of China (Grant Number 11401164). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]

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Please cite this article in press as: X. Qi, et al., Zagreb eccentricity indices of unicyclic graphs, Discrete Applied Mathematics (2017), http://dx.doi.org/10.1016/j.dam.2017.08.001.