The difference between remoteness and radius of a graph

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The difference between remoteness and radius of a graph Hongbo Hua a,b,∗ , Yaojun Chen b , Kinkar Ch. Das c a

Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huai’an, Jiangsu 223003, PR China

b

Department of Mathematics, Nanjing University, Nanjing 210093, PR China

c

Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea

article

abstract

info

Article history: Received 17 April 2014 Received in revised form 2 November 2014 Accepted 4 February 2015 Available online xxxx Keywords: Remoteness Radius Diameter Low bound Conjecture

Let G be a connected graph of order n ≥ 3. The remoteness ρ = ρ(G) is the maximum, over all vertices, of the average distance from a vertex to all others. The radius r = r (G) is the minimum, over all vertices, of the eccentricity of a vertex. Aouchiche and Hansen (2011) 2

n if n is odd and ρ − r ≥ 42n(n−−n1) if n is even. In this paper, we conjectured that ρ − r ≥ 3− 4 confirm this conjecture. In addition, we completely characterize extremal graphs attaining the lower bound. © 2015 Elsevier B.V. All rights reserved.

1. Introduction Let G = (V (G), E (G)) be a simple connected graph. For a vertex v in V (G), we let NG (v) be the set of vertices adjacent to v in G, and dG (v) = |NG (v)| be the degree of v . Let G′ be a subgraph of G and v ∈ V (G′ ). Then dG′ (v) is equal to the number of edges incident to v in G′ . Let Pn and Cn be the path and cycle on n vertices, respectively. For vertices u, v ∈ V (G), the distance dG (u, v) between u and v is defined as the length of the shortest path connecting u and v in G and DG (v) is equal to the sum of distances between v and all other vertices in G, that is, DG (v) = u∈V (G) dG (v, u). The eccentricity εG (v) of a vertex v is the maximum distance from v to any other vertex in G, i.e., εG (v) = maxu∈V (G) dG (v, u). The radius of a connected graph G is defined by r = r (G) = minv∈V (G) εG (v), and the diameter of a connected graph G is defined by D = D(G) = maxv∈V (G) εG (v). The proximity π = π (G) of a connected graph G is the minimum, over all vertices, of the average distance from a vertex to all others. The remoteness ρ = ρ(G) of a connected graph G is the maximum of the average distance from a vertex to all others. That is,

π = π (G) = min

v∈V (G)

1 n−1

DG (v)

and

ρ = ρ(G) = max

v∈V (G)

1 n−1

DG (v).

The sum of distances from a vertex to all others is also known as its transmission. So π and ρ can also be seen as the minimum and maximum normalized (divided by the number of vertices implied in the sum) transmission in a graph. It

∗

Corresponding author at: Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huai’an, Jiangsu 223003, PR China. E-mail addresses: [email protected] (H. Hua), [email protected] (Y. Chen), [email protected] (K.Ch. Das).

http://dx.doi.org/10.1016/j.dam.2015.02.007 0166-218X/© 2015 Elsevier B.V. All rights reserved.

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Fig. 1. Graphs Cn1 , Cn2 and Cn3 (n is odd).

is interesting to normalize the transmission in order to get invariants that fit in the same range as the radius, diameter, average distance and average eccentricity, and that preserve the properties of the transmission. The above two invariants were proposed in [2,3]. Nordhaus–Gaddum relations for both of them, i.e., lower and upper bounds on π (G) + π (G), π(G) · π (G), ρ(G) + ρ(G), ρ(G) · ρ(G), are given in [3]. In [4], Aouchiche and Hansen compared these two graph invariants π and ρ to the diameter, radius, average eccentricity, average distance, independence number and matching number. Most bounds so obtained were proved, but a few of them remained as conjectures. These proved results and conjectures stated in [4] were obtained with the use of AutoGraphiX, a conjecture-making system in graph theory [1,5,6]. Recently, some authors including two of the present authors solved several conjectures from AutoGraphiX (see [8,7,9–13]). Among those results and conjectures mentioned in [4], there are two dealing with the difference between the remoteness ρ and radius r. Aouchiche and Hansen have found a sharp upper bound on the difference ρ − r. However, the lower bound on the difference ρ − r was unsolved by them and remained as a conjecture (see [4]), which reads as follows. n Conjecture 1. Let G be a connected graph on n ≥ 3 vertices with remoteness ρ and radius r. Then ρ − r ≥ 3− if n is odd and 4

ρ −r ≥

2n−n2 4(n−1)

if n is even. The inequality is best possible as shown by the cycle Cn if n is even and by the graph composed by the cycle Cn together with two crossed edges on four successive vertices of the cycle. Recently, this conjecture has been proved to be correct for the case of trees by Sedlar [15]: Theorem 1 (Sedlar [15]). Let T be a tree on n ≥ 3 vertices with remoteness ρ and radius r. If n is odd, then ρ − r ≥ 21 with equality if and only if T ∼ = Pn , and if n is even, then ρ − r ≥ 2(nn−1) with equality if and only if T is a tree obtained by attaching a pendent edge to a central vertex of the path Pn−1 . This conjecture is still open so far. In this paper, we confirm that Conjecture 1 holds for all connected graphs of order n ≥ 3. Moreover, we characterize all extremal graphs attaining the lower bound. Our main result is the following. n Theorem 2. Let G be a connected graph on n ≥ 3 vertices with remoteness ρ and radius r. If n is odd, then ρ − r ≥ 3− with 4

2 equality if and only if G ∼ = Cn or Cni , i = 1, 2, 3 (see Fig. 1), and if n is even, then ρ − r ≥ 42n(n−−n1) with equality if and only if G∼ = Cn .

Remark 1. Denote by Cn3 , a graph of odd order n, obtained from the cycle Cn together with two crossed edges on four successive vertices of the cycle. One can see that Cn3 is just an extremal graph mentioned in Conjecture 1 for the case of odd n. n Evidently, ρ(Cn3 ) − r (Cn3 ) = 3− . 4 k Denote by Cn (k = 1, 2) (see Fig. 1.) graphs of odd order n, obtained from the cycle Cn−1 by adding two additional edges between an isolated vertex u and each of two vertices v and w on the cycle Cn−1 such that dCn−1 (v, w) = k. It is easy to check that

ρ(Cnk ) − r (Cnk ) =

3−n 4

,

k = 1, 2.

Thus, each Cnk is also a graph attaining the lower bound on ρ − r of Conjecture 1 for the case of odd n, k = 1, 2. In order to prove Theorem 2, we need additional terminologies and results. Let u, v be two distinct nonadjacent vertices of a graph G and S ⊆ V (G)−{u, v}. If u and v belong to different components of G − S, then we say that S separates u and v . The following is a well known result on connectivity of a graph due to Menger in 1927 [14].

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Theorem 3 (Menger [14]). Let G be a graph and u, v be two distinct nonadjacent vertices of G. Then the maximum number of pairwise internally vertex disjoint paths connecting u and v is equal to the minimum number of vertices in a vertex cut set that separates u and v . 2. The proof of Theorem 2 An edge is said to be a chord if this edge joins two non-adjacent vertices in a cycle. Now, we give the proof of Theorem 2. Proof. It is not difficult to check that Theorem 2 holds for n = 3. So we may assume that n ≥ 4 hereinafter. Let D be the diameter of G and PD+1 = v0 v1 · · · vD a diametrical path in G. Set Xi = {u | dG (v0 , u) = i},

where i = 1, 2, . . . , D.

If D = 1, then G = Kn and the conclusion holds. Thus we may assume that D ≥ 2. For any u ∈ V (G), since D(G) = D, we have dG (u, v0 ) + dG (u, vD ) ≥ D. This implies that 2ρ ≥

1



n−1

nD



DG (v0 ) + DG (vD ) ≥

n−1

,

that is,

ρ≥

nD 2(n − 1)

.

(∗)

1 Obviously, r ≤ D ≤ 2r. If D ≥ 2r − 1, that is, r ≤ D+ , then by (∗) and D ≥ 2, we have 2

ρ−r ≥

nD 2(n − 1)

−

D+1 2

=

D+1−n 2(n − 1)

≥

3−n 2(n − 1)

≥

3−n 6

>

3−n 4

>

2n − n2 4(n − 1)

,

as claimed. Now, we may suppose that D ≤ 2r − 2. We distinguish between two cases. Case 1. Any vertex of PD+1 cannot separate v0 and vD . By Theorem 3, G has two internally vertex disjoint paths connecting v0 and vD . In this case, we have |Xi | ≥ 2 for each i = 1, . . . , D − 1, and so n ≥ 2D. Claim 1. If n = 2D + 1 and r = D, then G ∼ = Cn or Cni for some i with 1 ≤ i ≤ 3. Proof. If G has a C2D , we let C = u1 u2 · · · u2D u1 and V (G) − V (C ) = {u}. When dG (u) = 1, we let v be the unique neighbor of u in G. For any w in V (G) \ {u, v}, we have dG (v, w) ≤ D − 1, since dG (u, w) = dG (v, w) + 1 ≤ D. Thus, r ≤ εG (v) < D, contradicting our assumption that r = D. So dG (u) ≥ 2. When D = 2, we have n = 2D + 1 = 5. By our assumption that G has a cycle C = u1 u2 u3 u4 u1 and V (G) \ V (C ) = {u}. It is obvious that C has at most one chord, for otherwise, G has a vertex of degree 4, and thus, r = 1 < 2 = D, a contradiction. Suppose that C has a chord u2 u4 . Then uu2 ̸∈ E (G) and uu4 ̸∈ E (G), for otherwise r = 1 < 2 = D, a contradiction. Since dG (u) ≥ 2, we have uu1 ∈ E (G) and uu3 ∈ E (G). Hence, G ∼ = Cn3 . Suppose that C has no chords. Since 2 ≤ dG (u) ≤ 3, 3 1 2 ∼ G = Cn or Cn or Cn . Thus, the claim holds. So, we may suppose that D ≥ 3. We first show that C has no chords. Assume to the contrary that u1 ul is a chord of C . We will prove that r < D. Set C ′ = u1 u2 . . . ul u1 and C ′′ = u1 ul ul+1 . . . u2D u1 . Assume that |V (C ′ )| ≥ |V (C ′′ )|. Then D + 1 ≤ l ≤ 2D − 1. For any ver-

≤ |V (2C )| . For any vertex v of V (C ′ ), we have max{dG (u1 , v), dG (ul , v)} ≤ 2 . So we may suppose that NG (u) V (C ′ ) ̸= ∅ and prove that εG (u1 ) < D or εG (ul ) < D. |V (C ′ )| When l ≤ 2D−3, for any vertex v in V (C ′ ), we have dG (u1 , v) ≤ 2 ≤ 2D2−3 < D−1. Thus, dG (u1 , u) < (D−1)+1 = D, and hence r ≤ εG (u1 ) < D. |V (C ′ )| When l = 2D − 2, for any vertex v in V (C ′ ), we have dG (u1 , v) ≤ 2 ≤ 2D2−2 ≤ D − 1 with equality only if v = uD , and ′ |V (C )| dG (ul , v) ≤ 2 ≤ 2D2−2 ≤ D − 1 with equality only if v = uD−1 . If u is not adjacent to uD , then dG (u1 , u) ≤ (D − 2)+ 1 < D. Hence, r ≤ εG (u1 ) < D. If u is adjacent to uD , then dG (ul , u) ≤ (D − 2) + 1 < D. Hence, r ≤ εG (ul ) < D. |V (C ′ )| When l = 2D − 1, for any vertex v in V (C ′ ), we have dG (u1 , v) ≤ ⌊ 2 ⌋ ≤ ⌊ 2D2−1 ⌋ = D − 1 with equality only if  ′ V (C ′ ) \ {uD−1 , v = uD or uD+1 , and dG (ul , v) ≤ ⌊ |V (2C )| ⌋ ≤ ⌊ 2D2−1 ⌋ = D − 1 with equality only if v = uD or uD−1 . If NG (u)   uD , uD+1 } ̸= ∅, then dG (u1 , u) ≤ (D − 2) + 1 < D, and hence r ≤ εG (u1 ) < D. So, we may assume that NG (u) V (C ′ ) \     ′ {uD−1 , uD , uD+1 } = ∅, that is, NG (u) {uD−1 , uD , uD+1 } ̸= ∅ (as NG (u) V (C ) ̸= ∅). If uD−1 ∈ NG (u) {uD−1 , uD , uD+1 }, tex v of V (C ′′ ), we have max{dG (u1 , v), dG (ul , v)} ≤ |V (C ′ )|



|V (C ′′ )| 2

′

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then dG (u1 , u) = dG (u1 , uD−1 ) + 1 ≤ (D − 2) + 1 < D, and hence r ≤ εG (u1 ) < D. If uD+1 ∈ NG (u) {uD−1 , uD , uD+1 }, then d 2) + 1 < D, and hence r ≤ εG (u1 ) < D. Now, we may assume that uD−1 , uD+1 ̸∈ G (ul , u) = dG (ul , uD+1 ) + 1 ≤ (D − NG (u) {uD−1 , uD , uD+1 }. So, uD ∈ NG (u) {uD−1 , uD , uD+1 }. Note that dG (u) ≥ 2 and that l = 2D − 1. We have NG (u) = {uD , u2D }. So dG (u1 , u) = dG (u1 , u2D ) + dG (u2D , u) = 2 < 3 ≤ D. Therefore, r ≤ εG (u1 ) < D. Summarizing above, if C has a chord, then r < D, a contradiction to our assumption that r = D. 1 So C has no chord. Since r = D = n− , u is adjacent to at least two vertices, say v and w , on the cycle C , and 1 ≤ 2 dC (v, w) ≤ 2. Thus, 2 ≤ dC (u) ≤ 3. When dC (u) = 3, u is adjacent to three consecutive vertices on the cycle C , that is, G∼ = Cn3 . When dC (u) = 2, we have G ∼ = Cn1 if dC (v, w) = 1 and G ∼ = Cn2 if dC (v, w) = 2 as 1 ≤ dC (v, w) ≤ 2. If G has no C2D , we let P ′ , P ′′ be two internally vertex disjoint paths connecting v0 and vD with |P ′ | ≥ |P ′′ |. For dG (v0 , vD ) = D and n = 2D + 1, we have |P ′′ | = D + 1 and |P ′ | = D + 2. If |XD | = 2, then since G has no C2D and r = D, we see that G ∼ = Cn . If |XD | = 1, we let P ′ = v0 x1 · · · xi−1 x′i x′′i xi+1 · · · xD−1 vD and P ′′ = v0 y1 y2 · · · yD−1 vD , where xk , yk ∈ Xk for each 1 ≤ k ≤ D − 1 and k ̸= i, and yi , x′i , x′′i ∈ Xi . Noting that G has no C2D , we have i ̸= 1 and x′′i xi−1 ̸∈ E (G). Thus we have x′′i yi−1 ∈ E (G), and hence r ≤ εG (x′′i ) < D, which contradicts r = D. 



We consider the following three possible subcases. Subcase 1.1. |XD | = 1 and n is odd. Since n is odd, we have n ≥ 2D + 1. If n ≥ 2D + 3, then by (∗) we have

ρ≥

nD

≥

2(n − 1)

(2D + 3)D (2D + 2)D 1 > = (D2 + D). 2(n − 1) 2(n − 1) n−1

Noting that r ≤ D, we have

ρ−r >

D2 + D n−1

D2 − (n − 2)D

−D=

n−1

.

3 For n ≥ 2D + 3, we have D ≤ n− . Let 2

f1 (x) =

x2 − (n − 2)x n−1

.

2 It can be seen that f1 (x) is decreasing on the interval [1, n− ]. Hence, 2

ρ − r > f1 (D) ≥ f1



n−3

 =

2

3−n 4

,

as claimed. 1 If n = 2D + 1, then D = n− . By (∗), we have 2

ρ≥

n 4

.

3 If r ≤ D − 1 = n− , then 2

ρ−r ≥

6−n 4

>

3−n 4

,

as expected. If r = D, then by Claim 1, we have G ∼ = Cni for some i with 1 ≤ i ≤ 3. It can be seen from Remark 1 that each of these 3 −n graphs satisfy the equality ρ − r = 4 . Subcase 1.2. |XD | = 1 and n is even. Since r ≤ D ≤

ρ−r ≥

n , 2

by (∗), we have

D2 n−1

−D=

D2 − (n − 1)D n−1

.

Let f2 (x) =

x2 − (n − 1)x n−1

.

1 ] and increasing on the interval [ n−2 1 , Then f2 (x) is decreasing on the interval [1, n− 2

ρ − r ≥ f2 (D) ≥ f2



n−2 2

 = f2

n 2

=

2n − n2 4(n − 1)

.

n 2

].

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If the equality holds, that is,

ρ−r =

2n − n2 4(n − 1)

, n , 2

then by the argument above, we must have r = D =

which implies that G ∼ = Cn .

Subcase 1.3. |XD | ≥ 2. In this case, we have

ρ≥

DG (v0 ) n−1

2(1 + 2 + · · · + D)

≥

n−1

≥

D2 + D n−1

.

Since r ≤ D, we have

ρ−r ≥

D2 + D n−1

−D=

D2 − (n − 2)D n−1

= f1 (D).

Because there exist two internally vertex disjoint paths connecting v0 and vD in G, we have D ≤ ⌊ 2n ⌋. If n is odd, then



ρ − r ≥ f1 (D) ≥ f1

n−1



 = f1

2

n−3 2

 =

3−n 4

,

and if

ρ−r =

3−n 4

,

1 then we must have r = D = n− . By Claim 1, G ∼ = Cn or Cni for some 1 ≤ i ≤ 3. If n is even, then 2



ρ − r ≥ f1 (D) ≥ f1

n−2 2

 =−

(n − 2)2 2n − n2 > . 4(n − 1) 4(n − 1)

Case 2. PD+1 has at least one vertex that separates v0 and vD . As 3−n 4

>

2n − n2 4(n − 1)

,

we need only to prove that

ρ−r >

3−n 4

in this case. Among all vertices separating v0 and vD , we choose vi to be the nearest one to the vertex v⌊ D ⌋ . Suppose without loss of 2

generality that 1 ≤ i ≤ ⌊ D2 ⌋. Further, we let G1 be the component of G − vi containing v0 , G2 the component of G − vi containing vD and G3 = G − V (G1 ) − V (G2 ) − {vi }. We first show that v⌊ D ⌋ cannot separate v0 and vD . To see this, we assume to the contrary that v⌊ D ⌋ separates v0 and vD , 2

2

and let u be a vertex such that dG (v⌊ D ⌋ , u) = εG (v⌊ D ⌋ ). By our previous assumption that D ≤ 2r − 2, we have r ≥ 2

2

D +2 . 2

Since v⌊ D ⌋ is a vertex that separates v0 and vD , we know that v0 and vD cannot belong to the same component of G − v⌊ D ⌋ . 2

2

If u does not belong to the component containing v0 of G − v⌊ D ⌋ , then D(G) ≥ dG (v0 , u) = dG (v0 , v⌊ D ⌋ ) + dG (v⌊ D ⌋ , u) =

⌊ D2 ⌋ + εG (v⌊ D ⌋ ) ≥ ⌊ D2 ⌋ + r ≥ ⌊ D2 ⌋ + 2

D +2 2

2

2

2

> D, a contradiction. If u belongs to the component containing v0 of G − v⌊ D ⌋ , 2

2 then D(G) ≥ dG (vD , u) = dG (vD , v⌊ D ⌋ ) + dG (v⌊ D ⌋ , u) = ⌈ D2 ⌉ + εG (v⌊ D ⌋ ) ≥ ⌈ D2 ⌉ + r ≥ ⌈ D2 ⌉ + D+ > D, a contradiction. 2 2 2 2 D Thus, by the choice of vi , we have 1 ≤ i ≤ ⌊ 2 ⌋ − 1. Before proceeding any further, we need the following claim.

Claim 2. D ≥ r + i. Proof. Let u be a vertex such that dG (u, vi ) = εG (vi ). Since r ≥ D2 + 1, we must have u ̸∈ V (G1 ), for otherwise dG (vD , u) = dG (u, vi ) + dG (vi , vD ) > D, a contradiction. So, u ∈ V (G2 ) ∪ V (G3 ). Since vi separates v0 and u, we have D ≥ dG (v0 , u) = dG (v0 , vi ) + dG (vi , u) = i + εG (vi ) ≥ r + i.  Now, we further consider the following subcases.

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Subcase 2.1. None of the vertices v⌊ D ⌋+1 , . . . , vD−1 can separate v0 and vD . 2

In this case, none of the vertices vi+1 , . . . , vD−1 can separate vi and vD . By Theorem 3, there are two internally vertex disjoint paths connecting vi and vD , which implies that |Xj | ≥ 2 for i + 1 ≤ j ≤ D − 1. Hence,

ρ≥

1 n−1

(1 + 2 + · · · + D) + [(i + 1) + · · · + (D − 1)] n−1 2(1 + 2 + · · · + D) − D − (1 + 2 + · · · + i) = n−1

DG (v0 ) ≥

=

D2 −

i2 2

−

i 2

n−1 2

(i + r )2 − i2 − 2i (D ≥ i + r byClaim 2) ≥ n−1 i(i−1) r 2 + 2ir + 2 r 2 + 2r = ≥ (as i ≥ 1). n−1 n−1 Therefore, r 2 + 2r

ρ−r ≥

−r =

n−1

r 2 − (n − 3)r n−1

.

Let f3 (x) =

x2 − (n − 3)x n−1

.

3 Then f3 (x) is decreasing on the interval [1, n− ] and increasing on the interval [ n−2 3 , 2 n ≥ 4, we have

ρ − r ≥ f3 (r ) ≥ f3

n − 3 2

=−

n 2

]. Since r ≤ ⌊ 2n ⌋,

n −3 2

≤ ⌊ 2n ⌋ and

3−n (n − 3)2 > , 4(n − 1) 4

as desired. Subcase 2.2. At least one of the vertices v⌊ D ⌋+1 , . . . , vD−1 that separates v0 and vD . 2

Among all vertices vl with ⌊ ⌋ + 1 ≤ l ≤ D − 1 that separate v0 and vD , we choose vj to be the one closest to v⌊ D ⌋ in D 2

2

PD+1 . Then dG (vi , vj ) ≥

 D  2

  D   +1 − − 1 = 2. 2

Let H1 be the component of G − vj containing vD , H2 the component of G − vj containing v0 and H3 = G − V (H1 ) − V (H2 ) − {vj }. Let u be a vertex such that dG (u, vj ) = εG (vj ). Since r > D2 , we must have u ̸∈ V (H1 ), otherwise, we have dG (v0 , u) = dG (v0 , vj ) + dG (vj , u) > D. So, u ∈ V (H2 ) ∪ V (H3 ). Since vj separates v0 and vD , we have D ≥ dG (vD , u) = dG (vD , vj ) + dG (vj , u) = (D − j) + εG (vj ) ≥ (D − j) + r , that is, j ≥ r. By the choice of vi and vj , none of the vertices vi+1 , . . . , vj−1 can separate vi and vj . Note that dG (vi , vj ) ≥ 2, by Theorem 3, there are two internally vertex disjoint paths connecting vi and vj , which implies that |Xs | ≥ 2 for i + 1 ≤ s ≤ j − 1. In this case, we have 1 DG (v0 ) n−1  1  ≥ (1 + 2 + · · · + D) + (i + 1) + · · · + (j − 1) n−1

ρ ≥

=

1 n−1

 D(D + 1) 2

+

j(j − 1) 2

−

i2 2

−

i 2

.

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If j ≥ r + 1, then noting that D ≥ i + r by Claim 2 and i ≥ 1, we have

ρ ≥ =

 (i + r )(i + r + 1)

1 n−1

2

r 2 + (i + 1)r n−1

r 2 + 2r

≥

n−1

+

r (r + 1)

−

2

i2 2

−

i 2

.

If i ≥ 2, then noting that D ≥ i + r by Claim 2 and j ≥ r, we have

ρ ≥ =

 (i + r )(i + r + 1)

1 n−1

2

r 2 + ir

≥

n−1

r 2 + 2r n−1

+

r (r − 1)

−

2

i2 2

−

i 2

.

Therefore, if j ≥ r + 1 or i ≥ 2, we have

ρ−r ≥

r 2 + 2r n−1

−r =

r 2 − (n − 3)r

= f3 (r ),

n−1

where f3 (x) is defined as in Subcase 2.1. As proved in Subcase 2.1, we have

ρ−r >

3−n 4

.

If D ≥ i + r + 1, then noting that j ≥ r and i ≥ 1, we have

ρ ≥

 (i + r + 1)(i + r + 2)

1 n−1

2

r + (i + 1)r + i + 1 2

=

n−1

+

r (r − 1)

2

≥

2

r + 2r + 2 n−1

−

i2 2

−

i 2

.

Therefore,

ρ−r ≥

r 2 + 2r + 2 n−1

 ≥ f3

n−3

−r =

 +

2

r 2 − (n − 3)r + 2 n−1

=

= f3 ( r ) +

2 n−1

−(n − 3)2 + 8 4(n − 1)

2 n−1 3−n −n2 + 6n − 1 = > , 4(n − 1) 4

as expected. Noting that D ≥ i + r by Claim 2, i ≥ 1 and j ≥ r, by the arguments above, we are left to consider the case when D = i + r , i = 1 and j = r. In such a case, D = j + 1 and |Xs | ≥ 2 for 2 ≤ s ≤ D − 2, which implies that n ≥ 2(D − 3) + 4 = 2D − 2. 2 If n = 2D − 2, i.e., D = n+ , then 2 DG (v0 ) = 1 + 2(2 + · · · + D − 2) + D − 1 + D = D2 − D. Note that r = D − i = D − 1, we have

ρ−r = =

D2 − D n−1

−D+1

D2 − nD + (n − 1)

n−1 (n + 2)2 − 2n(n + 2) + 4(n − 1) = 4(n − 1) 2 −n + 4n 3−n = > . 4(n − 1) 4 If n ≥ 2D − 1, then there exists some s with 2 ≤ s ≤ D − 2 such that |Xs | ≥ 3, and so we have DG (v0 ) ≥ 1 + [2(2 + · · · + D − 2) + 2] + D − 1 + D = D2 − D + 2.

8

H. Hua et al. / Discrete Applied Mathematics (

)

–

Thus,

ρ−r ≥ =

D2 − D + 2

−D+1

n−1

D2 − nD + (n + 1) n−1

.

Let f4 (x) =

x2 − nx + (n + 1) n−1

.

Then f4 (x) is decreasing on the interval [1,

ρ − r ≥ f4 (D) ≥ f4

n

This completes the proof.

2

=

n 2

] and increasing on the interval [ 2n ,

n +1 2

]. Since D ≤

n +1 , 2

we have

−n2 + 4(n + 1) 3−n > . 4(n − 1) 4



Acknowledgments Hua was supported by Natural Science Foundation of the Higher Education Institutions of Jiangsu Province under Grant No. 12KJB110001, National Natural Science Foundation of China under Grant Nos. 11201227, 11171273 and China Postdoctoral Science Foundation Funded Project under Grant No. 2014M551542. Chen was supported by National Natural Science Foundation of China under Grant Nos. 11071115, 11371193. Das was supported by the National Research Foundation funded by the Korean government under Grant No. 2013R1A1A2009341. References [1] M. Aouchiche, J.-M. Bonnefoy, A. Fidahoussen, G. Caporossi, P. Hansen, L. Hiesse, J. Lacheré, A. Monhait, Variable neighborhood search for extremal graphs. 14. The AutoGraphiX 2 system, in: L. Liberti, N. Maculan (Eds.), Global Optimization: from Theory to Implementation, Springer, New York, 2006, pp. 281–310. [2] M. Aouchiche, G. Caporossi, P. Hansen, Variable neighborhood search for extremal graphs. 20. Automated comparison of graph invariants, MATCH Commun. Math. Comput. Chem. 58 (2007) 365–384. [3] M. Aouchiche, P. Hansen, Nordhaus–Gaddum relations for proximity and remoteness in graphs, Comput. Math. Appl. 59 (2010) 2827–2835. [4] M. Aouchiche, P. Hansen, Proximity and remoteness in graphs: results and conjectures, Networks 58 (2011) 95–102. [5] G. Caporossi, P. Hansen, Variable neighborhood search for extremal graphs. I. The AutoGraphiX system, Discrete Math. 212 (2000) 29–44. [6] G. Caporossi, P. Hansen, Variable neighborhood search for extremal graphs: V. Three ways to automate finding conjectures, Discrete Math. 276 (2004) 81–94. [7] K.C. Das, Conjectures on index and algebraic connectivity of graphs, Linear Algebra Appl. 433 (2010) 1666–1673. [8] K.C. Das, On conjectures involving second largest signless Laplacian eigenvalue of graphs, Linear Algebra Appl. 432 (2010) 3018–3029. [9] K.C. Das, Proofs of conjecture involving the second largest signless Laplacian eigenvalue and the index of graphs, Linear Algebra Appl. 435 (2011) 2420–2424. [10] K.C. Das, Proof of conjectures involving the largest and the smallest signless Laplacian eigenvalues of graphs, Discrete Math. 312 (2012) 992–998. [11] K.C. Das, Proof of conjectures on adjacency eigenvalues of graphs, Discrete Math. 313 (2013) 19–25. [12] H. Hua, K.C. Das, Proof of conjectures on remoteness and proximity in graphs, Discrete Appl. Math. 171 (2014) 72–80. [13] B. Ma, B. Wu, W. Zhang, Proximity and average eccentricity of a graph, Inform. Process. Lett. 112 (2012) 392–395. [14] K. Menger, Zur allgemeinen Kurventheorie, Fund. Math. 10 (1927) 96–115. [15] J. Sedlar, Remoteness, proximity and few other distance invariants in graphs, Filomat 28 (2013) 1425–1435.