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Complete solution to a conjecture on Randić index
European Journal of Operational Research 200 (2010) 9–13
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Discrete Optimization
Complete solution to a conjecture on Randic´ index q Xueliang Li a,*, Bolian Liu b, Jianxi Liu a a b
Center for Combinatorics and LPMC-TJKLC, Nankai University, Tianjin 300071, PR China School of Mathematical Science, South China Normal University, Guangzhou 510631, PR China
a r t i c l e
i n f o
Article history: Received 20 September 2007 Accepted 6 December 2008 Available online 24 December 2008 Keywords: Simple graph Minimum degree Randic´ index Minimum value
a b s t r a c t For a graph G, the Randic´ index RðGÞ of G is defined by RðGÞ ¼
P
u;v
1 pffiffiffiffiffiffiffiffiffiffiffiffi ffi, where dðuÞ is the degree of a dðuÞdðv Þ
vertex u and the summation runs over all edges uv of G. Let Gðk; nÞ be the set of connected simple graphs } s once asked for finding the minimum value of the of order n with minimum degree k. Bollobás and Erdo Randic´ index among the graphs in Gðk; nÞ. There have been many partial solutions for this question. In this paper we give a complete solution to the question. Ó 2008 Elsevier B.V. All rights reserved.
1. Introduction The Randic´ index R ¼ RðGÞ of a graph G is defined as follows:
R ¼ RðGÞ ¼
X u;v
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; dðuÞdðv Þ
ð1:1Þ
where dðuÞ denotes the degree of a vertex u and the summation runs over all edges uv of G. This topological index was first proposed by Randic´ [19] in 1975, suitable for measuring the extent of branching of the carbon-atom skeleton of saturated hydrocarbons. Randic´ himself demonstrated [19] that his index is well correlated with a variety of physico-chemical properties of alkanes. The R became one of the most popular molecular descriptors to which three books are devoted [10,12,13]. Initially, the Randic´ index was studied only by chemists [10,11], but recently it attracted much attention also of mathematicians [13]. One of the mathematical questions asked in connection with R is which graphs in a given class of graphs have maximum and minimum R values [2,18]. Let Gðk; nÞ be the set of connected simple graphs of order n } s asked for finding the minimum value of the Randic´ index with minimum degree k. In [6] Fajtlowitcz mentioned that Bollobás and Erdo among the graphs in Gðk; nÞ. The solution of such problem turned out to be difficult, and only a few partial results have been achieved so } s found that for a connected graph G far. In [2] Bollobás and Erdo
RðGÞ P
pffiffiffiffiffiffiffiffiffiffiffiffi n 1;
ð1:2Þ
and the bound is tight if and only if G is a star. The problem for k ¼ 2 was solved in [5], which gave a stronger result, say, if the minimum degree is greater or equal to 2, then
2n 4 1 ; RðGÞ P pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2n 2 n 1
ð1:3Þ
and the bound is tight if and only if G ¼ K I 2;n2 which arises from the complete bipartite graph K 2;n2 by joining the vertices in the partite set with two vertices by a new edge. In these papers a graph theoretical approach has been used. In other papers [3,4,7–9], a linear programming and a quadratic programming technique [14] for finding extremal graphs has been used. In [15] the problem was solved for k ¼ 1 and k ¼ 2, respectively, by using linear programming. Delorme et al. [10] gave a conjecture about this problem. The conjecture in [5] is that the Randic´ index for graphs in Gðk; nÞ, where 1 6 k 6 n 2, attains its minimum value q
Supported by NSFC No. 10831001, 10771080 PCSIRT and the ‘‘973” program. * Corresponding author. Tel.: +86 22 23502180; fax: +86 22 23509272. E-mail address: [email protected] (X. Li).
0377-2217/$ - see front matter Ó 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2008.12.010
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X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
for the graph K I k;nk which arises from the complete bipartite graph K k;nk by joining every pair of vertices in the partite set with k vertices by a new edge. Conjecture 1 [5]. Let G ¼ ðV; EÞ be a graph of order n with minimum degree k. Then
kðn kÞ RðGÞ P pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ kðn 1Þ
k 1 ; 2 n1
ð1:4Þ
where equality holds if and only if G ¼ K I k;nk . Using again linear programming, Pavlovic´ [16] proved that Conjecture 1 holds when k ¼ ðn 1Þ=2 or k ¼ n=2. See also [14] for further results proved by using quadratic programming. Divnic and Pavlovic´ [17] proved that Conjecture 1 holds when k 6 n=2 and nk P n k, where nk denotes the number of vertices of degree k. Recently in [1], however, Aouchiche and Hansen showed that Conjecture 1 does not hold in general and proposed a modified conjecture as follows. Let the graph Gn;p;k be the complement of a graph Gn;p;k composed of a ðn k 1Þ-regular graph on p vertices together with n p isolated vertices. The minimal counterexample of Conjecture 1 is the graph G7;4;5 , which was given in [1], see Fig. 1. Let
8 nþ2 if > 2 > > nþ3 < if 2 kn ¼ nþ4 > > 2 if > : nþ3 if 2
n 0ðmod 4Þ; n 1ðmod 4Þ; n 2ðmod 4Þ; n 3ðmod 4Þ;
8 n2 > < 2 if n 2ðmod 4Þ and k is even; n p¼ if n 3ðmod 4Þ; 2 > : n otherwise: 2
ð1:5Þ
For such a graph G ¼ Gn;p;k ,
RðGÞ ¼
ðn pÞðn p 1Þ pðp þ k nÞ pðn pÞ þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2ðn 1Þ 2k kðn 1Þ
Using these results, the authors of [1] gave the following Conjecture 2 as a modification of Conjecture 1. Conjecture 2 [1]. Let G ¼ ðV; EÞ be a graph of order n with minimum degree k, and kn and p be given in (1.5). Then
RðGÞ P
8 kðk1Þ kðnkÞ > ffi < 2ðn1Þ þ pffiffiffiffiffiffiffiffiffiffi kðn1Þ
if k < kn ;
> ffiffiffiffiffiffiffiffiffiffiffi if kn 6 k 6 n 2; þ pðpþknÞ þ ppðnpÞ : ðnpÞðnp1Þ 2ðn1Þ 2k kðn1Þ
where equality holds if and only if G is
KI k;nk
for k < kn , and Gn;p;k for k P kn .
}s’ question of finding the minimum value of the Randic´ index for the In this paper, we want to completely solve the Bollobás and Erdo graphs in Gðk; nÞ. As usual, we formulate the question into a mathematical programming problem. Denote by ni the number of vertices of degree i in G, and by xi;j ðxi;j P 0Þ the number of edges joining the vertices of degrees i and j in G. The mathematical description of our problem is as follows:
min RðGÞ ¼
X k6i6n1 i6j6n1
x pi;jffiffiffi ij
subject to: n1 X
xi;j þ 2xi;i ¼ ini
for k 6 i 6 n 1;
ð1:6Þ
j¼k j–i
nk þ nkþ1 þ þ nn1 ¼ n;
ð1:7Þ
xi;j 6 ni nj for k 6 i 6 n 1; i < j 6 n 1; ni xi;i 6 for k 6 i 6 n 1; 2
ð1:8Þ ð1:9Þ
xi;j ; ni are nonnegative integers; for k 6 i 6 j 6 n 1:
ð1:10Þ
Obviously, (1.6)–(1.10) define a nonlinearly constrained optimization problem.
G 7,4,5
G 7,4,5
Fig. 1. The minimal counterexample of Conjecture 1.
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X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
2. Main result Denote
8n > > > 2 > n > n > > > 2 or 2 > > > > > 2n > > < or nþ2 p ¼ n2 2 2 > > >n > > > 2 > > n > > n > > > 2 or 2 > > : n 2
if n 0ðmod 4Þ; if n 1ðmod4Þ and k is even; if n 1ðmod 4Þ and k is odd; if n 2ðmod 4Þ and k is even;
ð2:11Þ
if n 2ðmod 4Þ and k is odd if n 3ðmod 4Þ and k is even if n 3ðmod 4Þ and k is odd:
Theorem 2.1. Let G ¼ ðV; EÞ be a graph of order n with minimum degree k, and p be given in (2.11). Then we have
RðGÞ P
8 kðk1Þ kðnkÞ > ffi < 2ðn1Þ þ pffiffiffiffiffiffiffiffiffiffi kðn1Þ > : ðnpÞðnp1Þ 2ðn1Þ
þ
pðpþknÞ 2k
if k 6 n=2; pðnpÞ
þ pffiffiffiffiffiffiffiffiffiffiffi if k > n=2;
where equality holds if and only if G is
kðn1Þ
KI k;nk
for k 6 n=2, and Gn;p;k for k > n=2.
Proof. It is easy to see that nn1 6 k, or the minimum degree of a graph in Gðk; nÞ would be larger than k. Therefore we only need to consider the case when nn1 6 k. Let nn1 ¼ k t for some integer t such that 0 6 t 6 k, and let Rkt denote the Randic´ index for any graph in Gðk; nÞ with nn1 ¼ k tð0 6 t 6 kÞ. Since xi;n1 ¼ ni nn1 for k 6 i 6 n 2 and xn1;n1 ¼ nn1 ðnn1 1Þ=2, we have
Rkt ¼
X k6i6n1 i6j6n1
n2 X x ni nn1 n ðn 1Þ pi;jffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ n1 n1 2ðn 1Þ iðn 1Þ ij i¼k
n2 x 1X xj1;j x xjþ1;j xj;n2 pk;jffiffiffiffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ 2 pj;jffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ 2 j¼k ðj 1Þj jðj þ 1Þ jðn 2Þ jj kj
P
n2 X i¼k
þ
¼
! ni nn1 nn1 ðnn1 1Þ 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2ðn 1Þ 2 iðn 1Þ kðk þ 1Þ kðn 2Þ kk
n2 xk;j þ þ xj1;j þ 2xj;j þ xj;jþ1 þ þ xj;n2 1 X pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 j¼kþ1 jðn 1Þ
n2 X i¼k
þ
!
ð2:12Þ
ni nn1 n ðn 1Þ 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ n1 n1 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2ðn 1Þ 2 iðn 1Þ kðk þ 1Þ kðn 2Þ kk
!
n2 jnj nj nn1 1 X pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 j¼kþ1 jðn 1Þ
ð2:13Þ
n2 pffi n2 X X 1 nn1 n nn1 ðnn1 1Þ pjffi þ jnj þ pffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffi 2ðn 1Þ 2 n 1 j¼kþ1 j 2 n 1 j¼kþ1 ! 1 2xk;k xk;kþ1 xk;n2 nk nn1 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 2 kðk þ 1Þ kðn 2Þ kðn 1Þ kk 1 ffi where inequality (2.12) holds because p1ffii P pffiffiffiffiffiffi for k þ 1 6 i 6 n 2, and equality (2.13) holds because of (1.6). After substitution of n1 nn1 ¼ k t and nk ¼ n k þ t nkþ1 nkþ2 nn2 into the last equality, we have
Rkt P
Since ptffiffi k
! ! n2 pffi pffiffiffi ðk tÞðk t 1Þ ðk tÞðn k þ tÞ X 1 1 t nj pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi j k ðk tÞ pffiffiffi pffi þ pffiffiffi þ þ 2ðn 1Þ kðn 1Þ 2 n 1 j k k j¼kþ1 ! 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : þ 2 kðk þ 1Þ kðn 2Þ kk
pffi pffiffi pffiffiffiffi pffi pffiffiffi pffi pffiffiffi j k kj > k t for k þ 1 6 j 6 n 2, we have j k > ðk tÞ pffiffiffi , i.e., j k > ðk tÞ
> 0. Since if nj ¼ 0 then xi;j ¼ 0 for k 6 i 6 j 6 n 2, we then have
kj
p1ffiffi k
pffi pffiffiffi 1ffi . Thus j k ðk tÞ p1ffiffik p1 ffi þ p j
j
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X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
! ! n2 pffi pffiffiffi ðk tÞðk t 1Þ ðk tÞðn k þ tÞ X 1 1 t nj pffiffiffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ j k ðk tÞ pffiffiffi pffi þ pffiffiffi 2ðn 1Þ kðn 1Þ j k k 2 n1 j¼kþ1 ! 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 kðk þ 1Þ kðn 2Þ kk P
ðk tÞðk t 1Þ ðk tÞðn k þ tÞ xk;k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ : 2ðn 1Þ k kðn 1Þ
Notice that equalities hold in all the above inequalities if and only if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2. Thus,
Rkt P
ðk tÞðk t 1Þ ðk tÞðn k þ tÞ xk;k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ; 2ðn 1Þ k kðn 1Þ
where equality holds if and only if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2. We know that if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2, then nk ¼ n k þ t; nn1 ¼ k t; xk;k ¼ ðn k þ tÞt=2; xn1;n1 ¼ ðk tÞðk t 1Þ=2 and all other xi;j and xi;i are equal to zero. Therefore,
Rkt P
ðk tÞðk t 1Þ ðk tÞðn k þ tÞ ðn k þ tÞt pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ; 2ðn 1Þ 2k kðn 1Þ
where equality holds if and only if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2. Let
f ðk; tÞ ¼
ðk tÞðn k þ tÞ ðk tÞðk t 1Þ ðn k þ tÞt pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ : 2ðn 1Þ 2k kðn 1Þ
We only need to get the minimum value of f ðk; tÞ, by distinguishing two cases. Case 1. k 6 n=2. Since of ðk; tÞ=ot ¼ nþ2t2k 2
p1ffiffi k
2 1 ffi pffiffiffiffiffiffi P 0, and of ðk; tÞ=ot > 0 strictly holds except for k t ¼ n=2, i.e., k ¼ n=2 þ t, we know that f ðk; tÞ n1
attains its minimum if and only if t ¼ 0 since k 6 n=2 in this case. So, in this case we can conclude that the Randic´ index attains its minimum in Gðk; nÞ if and only if all the above equalities hold, which k and all other xi;j ; xi;i are equal to zero. Therefore, a graph G in means that nk ¼ n k; nn1 ¼ k; nj ¼ 0; j ¼ k þ 1; . . . ; n 2; xn1;n1 ¼ 2 Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ K k;nk for k 6 n=2. Case 2. n 2 P k > n=2.
2
2 1 ffi 1 ffi p1ffiffi pffiffiffiffiffiffi Let of ðk; tÞ=ot ¼ nþ2t2k ¼ 0. Then t ¼ k n=2. Since o2 f ðk; tÞ=ot2 ¼ p1ffiffik pffiffiffiffiffiffi > 0, f ðk; tÞ attain its minimum if and only if 2 n1 n1 k t ¼ k n=2. Then we have nk ¼ n=2; xk;k ¼ nð2k nÞ=2 and xn1;n1 ¼ nðn 2Þ=8. Next, we need to check whether they are integers or not, since the obtained solutions have no graph theoretical meaning when one of the three values, namely, xk;k ¼ ðn k þ tÞt=2, xn1;n1 ¼ ðk tÞðk t 1Þ=2 and t, is not an integer.
Subcase 2.1. n 0ðmod 4Þ. We can easily check that t ¼ k n=2; xk;k ¼ n=2 and xn1;n1 ¼ nðn 2Þ=8 are integers in this case. Therefore, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ Gn;n=2;k in the case n 0ðmod 4Þ. Subcase 2.2. n 1ðmod 4Þ. We see first that t ¼ k n=2 is not an integer in this case. Therefore, the obtained solutions have no graph theoretical meaning. Then t can not attain k n=2 if we want to get the minimum value of the Randic´ index in Gðk; nÞ. We then let t 6 k nþ1 or t P k n1 . 2 2 , we have n þ 2t 2k 6 1 < 0. Thus of ðk; tÞ=ot < 0. Therefore, f ðk; tÞ attains its minimum if and only if t ¼ k nþ1 in this For t 6 k nþ1 2 2 case. And then nk ¼ ðn 1Þ=2; nn1 ¼ ðn þ 1Þ=2; xk;k ¼ ðn 1Þð2k n 1Þ=8; xn1;n1 ¼ ðn þ 1Þðn 1Þ=2 all are integers. So, pffiffiffiffiffiffiffiffiffiffiffi þ ðnþ1Þðn1Þ þ ðnþ1Þðn1Þ . For t P k n1 , we have n þ 2t 2k P 1 > 0. Thus of ðk; tÞ=ot > 0. Therefore, f ðk; tÞ attains its minf ðk; tÞ ¼ ðn1Þð2kn1Þ 8k 8ðn1Þ 2 4
kðn1Þ
in this case. And then nk ¼ ðn þ 1Þ=2; nn1 ¼ ðn 1Þ=2; xk;k ¼ ðn þ 1Þð2k n þ 1Þ=8; xn1;n1 ¼ minimum if and only if t ¼ k n1 2 pffiffiffiffiffiffiffiffiffiffiffi þ ðn1Þðn3Þ pffiffiffiffiffiffiffiffiffiffiffi þ ðnþ1Þðn1Þ þ ðnþ1Þðn1Þ ¼ ðn1Þð2kn1Þ þ ðnþ1Þðn1Þ , which is ðn 1Þðn 3Þ=2 all are integers if k is even. So, minf ðk; tÞ ¼ ðnþ1Þð2knþ1Þ 8k 8ðn1Þ 8k 8ðn1Þ 4
kðn1Þ
4
kðn1Þ
. the same as the minimum value for the above case when t 6 k nþ1 2 Therefore, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G is Gn;bnc;k for k both even and odd, or Gn;dne;k 2 2 for k even in the case n 1ðmod4Þ. Subcase 2.3. n 2ðmod 4Þ. We can easily check that t ¼ k n=2; xk;k ¼ nð2k nÞ=8 and xn1;n1 ¼ nðn 2Þ=8 are integers if k is odd in this case. Therefore, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G is Gn;n=2;k for k odd in the case n 1ðmod 4Þ. If k is even, then xk;k ¼ nð2k nÞ=8 is not an integer in this case, which means that t can not attain k n=2 if we want to get the minor t P k n2 . imum value of the Randic´ index in Gðk; nÞ. We then let t 6 k nþ2 2 2
X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
13
For t 6 k nþ2 , we have n þ 2t 2k 6 2 < 0. Thus of ðk; tÞ=ot < 0. Therefore, f ðk; tÞ attains its minimum if and only if t ¼ k nþ2 in this 2 2 case. And then nk ¼ ðn 2Þ=2; nn1 ¼ ðn þ 2Þ=2; xk;k ¼ ðn 2Þð2k n 2Þ=8; xn1;n1 ¼ nðn þ 2Þ=2 all are integers. So, minf ðk; tÞ ¼ ðn2Þð2kn2Þ nðnþ2Þ pffiffiffiffiffiffiffiffiffiffiffi þ 8ðn1Þ þ ðn2Þðnþ2Þ . 8k 4
kðn1Þ
For t P k n2 , we have n þ 2t 2k P 2 > 0. Thus of ðk; tÞ=ot > 0. Therefore, f ðk; tÞ attains its minimum if and only if t ¼ k n2 in this 2 2 case. And then nk ¼ ðn þ 2Þ=2; nn1 ¼ ðn 2Þ=2; xk;k ¼ ðn þ 2Þð2k n þ 2Þ=8; xn1;n1 ¼ ðn 2Þðn 4Þ=2 all are integers. So, minf ðk; tÞ ¼ ðnþ2Þð2knþ2Þ nðnþ2Þ pffiffiffiffiffiffiffiffiffiffiffi þ ðn2Þðn4Þ pffiffiffiffiffiffiffiffiffiffiffi þ 8ðn1Þ þ ðn2Þðnþ2Þ ¼ ðn2Þð2kn2Þ þ ðn2Þðnþ2Þ , which is the same as the minimum value for the above case when 8k 8ðn1Þ 8k 4 kðn1Þ 4 kðn1Þ . t 6 k nþ2 2 Therefore, in the case n 2ðmod 4Þ and k is odd, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ Gn;2n;k . In the case n 2ðmod 4Þ and k is even, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ Gn;n2;k or 2 Gn;nþ2;k . 2
Subcase 2.4. n 3ðmod 4Þ. Similar to the proof of Subcase 2.2, we can get that a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G is Gn;dne;k for k both even and odd, or Gn;bnc;k for k even. 2
2
The proof is now complete. h
References [1] M. Aouchiche, P. Hansen, On a conjecture about the Randic´ index, Discrete Mathematics 307 (2) (2007) 262–265. } s, Graphs of extremal weights, Ars Combinatoria 50 (1998) 225–233. [2] B. Bollobás, P. Erdo [3] G. Caporossi, I. Gutman, P. Hansen, Variable neighborhood search for extremal graphs IV: Chemical trees with extremal connectivity index, Computers and Chemistry 23 (1999) 469–477. [4] G. Caporossi, I. Gutman, P. Hansen, Lj. Pavlovic´, Graph with maximum connectively index, Computational Biology and Chemistry 27 (2003) 85–90. [5] C. Delorme, O. Favaron, D. Rautenbach, On the Randic´ index, Discrete Mathematics 257 (1) (2002) 29–38. [6] S. Fajtlowicz, Written on the Wall: Conjectures Derived on the Basis of the Program Galatea Gabriella Graffiti, University of Houston, 1998. [7] M. Fischermann, A. Hoffmann, D. Rautenbach, L. Volkmann, A linear-programming approach to the generalized Randic´ index, Discrete Applied Mathematics 128 (2003) 375–385. [8] I. Gutman, O. Miljkovic, G. Caporossi, P. Hansen, Alkanes with small and large Randic´ connectivity indices, Chemical Physics Letters 306 (1999) 366–372. [9] I. Gutman, Lj. Pavlovic´, O. Miljkovic, On graphs with extremal connectivity indices, Bulletin de l’Academie Serbe des Sciences er des Arts 25 (2000) 1–14. [10] L.B. Kier, L.H. Hall, Molecular Connectivity in Chemistry and Drug Research, Academic Press, New York, 1976. [11] L.B. Kier, L.H. Hall, The nature of structure–activity relationships and their relation to molecular connectivity, European Journal of Medical Chemistry 12 (1977) 307–312. [12] L.B. Kier, L.H. Hall, Molecular Connectivity in Structure–Activity Analysis, Research Studies Press-Wiley, Chichester(UK), 1986. [13] X. Li, I. Gutman, Mathematical aspects of Randic-type molecular structure descriptors, Mathematical Chemistry Monographs, No. 1, Kragujevac, 2006, pp. VI + 330. [14] N. Limic, H. Psasgic, C. Rnjak, Linearno i nelinearno programiranje, Informator, Zagreb(HR), 1978. [15] Lj. Pavlovic´, Graphs with extremal Randic´ index when the minimum degree of vertices is two, Kragujevac Journal of Mathematics 25 (2003) 55–63. [16] Lj. Pavlovic´, On the conjecture of Delorme, Favaron and Rautenbach about the Randic´ index, European Journal of Operational Research 180 (2007) 369–377. [17] Lj. Pavlovic´, T. Divnic, A quadratic programming approach to the Randic´ index, European Journal of Operational Research 176 (2007) 435–444. [18] Lj. Pavlovic´, I. Gutman, Graph with extremal connectivity index, Novi Sad Journal of Mathematics 31 (2) (2001) 53–58. [19] M. Randic´, On characterization of molecular branching, Journal of the American Chemical Society 97 (1975) 6609–6615.
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European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Discrete Optimization
Complete solution to a conjecture on Randic´ index q Xueliang Li a,*, Bolian Liu b, Jianxi Liu a a b
Center for Combinatorics and LPMC-TJKLC, Nankai University, Tianjin 300071, PR China School of Mathematical Science, South China Normal University, Guangzhou 510631, PR China
a r t i c l e
i n f o
Article history: Received 20 September 2007 Accepted 6 December 2008 Available online 24 December 2008 Keywords: Simple graph Minimum degree Randic´ index Minimum value
a b s t r a c t For a graph G, the Randic´ index RðGÞ of G is defined by RðGÞ ¼
P
u;v
1 pffiffiffiffiffiffiffiffiffiffiffiffi ffi, where dðuÞ is the degree of a dðuÞdðv Þ
vertex u and the summation runs over all edges uv of G. Let Gðk; nÞ be the set of connected simple graphs } s once asked for finding the minimum value of the of order n with minimum degree k. Bollobás and Erdo Randic´ index among the graphs in Gðk; nÞ. There have been many partial solutions for this question. In this paper we give a complete solution to the question. Ó 2008 Elsevier B.V. All rights reserved.
1. Introduction The Randic´ index R ¼ RðGÞ of a graph G is defined as follows:
R ¼ RðGÞ ¼
X u;v
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; dðuÞdðv Þ
ð1:1Þ
where dðuÞ denotes the degree of a vertex u and the summation runs over all edges uv of G. This topological index was first proposed by Randic´ [19] in 1975, suitable for measuring the extent of branching of the carbon-atom skeleton of saturated hydrocarbons. Randic´ himself demonstrated [19] that his index is well correlated with a variety of physico-chemical properties of alkanes. The R became one of the most popular molecular descriptors to which three books are devoted [10,12,13]. Initially, the Randic´ index was studied only by chemists [10,11], but recently it attracted much attention also of mathematicians [13]. One of the mathematical questions asked in connection with R is which graphs in a given class of graphs have maximum and minimum R values [2,18]. Let Gðk; nÞ be the set of connected simple graphs of order n } s asked for finding the minimum value of the Randic´ index with minimum degree k. In [6] Fajtlowitcz mentioned that Bollobás and Erdo among the graphs in Gðk; nÞ. The solution of such problem turned out to be difficult, and only a few partial results have been achieved so } s found that for a connected graph G far. In [2] Bollobás and Erdo
RðGÞ P
pffiffiffiffiffiffiffiffiffiffiffiffi n 1;
ð1:2Þ
and the bound is tight if and only if G is a star. The problem for k ¼ 2 was solved in [5], which gave a stronger result, say, if the minimum degree is greater or equal to 2, then
2n 4 1 ; RðGÞ P pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2n 2 n 1
ð1:3Þ
and the bound is tight if and only if G ¼ K I 2;n2 which arises from the complete bipartite graph K 2;n2 by joining the vertices in the partite set with two vertices by a new edge. In these papers a graph theoretical approach has been used. In other papers [3,4,7–9], a linear programming and a quadratic programming technique [14] for finding extremal graphs has been used. In [15] the problem was solved for k ¼ 1 and k ¼ 2, respectively, by using linear programming. Delorme et al. [10] gave a conjecture about this problem. The conjecture in [5] is that the Randic´ index for graphs in Gðk; nÞ, where 1 6 k 6 n 2, attains its minimum value q
Supported by NSFC No. 10831001, 10771080 PCSIRT and the ‘‘973” program. * Corresponding author. Tel.: +86 22 23502180; fax: +86 22 23509272. E-mail address: [email protected] (X. Li).
0377-2217/$ - see front matter Ó 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2008.12.010
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X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
for the graph K I k;nk which arises from the complete bipartite graph K k;nk by joining every pair of vertices in the partite set with k vertices by a new edge. Conjecture 1 [5]. Let G ¼ ðV; EÞ be a graph of order n with minimum degree k. Then
kðn kÞ RðGÞ P pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ kðn 1Þ
k 1 ; 2 n1
ð1:4Þ
where equality holds if and only if G ¼ K I k;nk . Using again linear programming, Pavlovic´ [16] proved that Conjecture 1 holds when k ¼ ðn 1Þ=2 or k ¼ n=2. See also [14] for further results proved by using quadratic programming. Divnic and Pavlovic´ [17] proved that Conjecture 1 holds when k 6 n=2 and nk P n k, where nk denotes the number of vertices of degree k. Recently in [1], however, Aouchiche and Hansen showed that Conjecture 1 does not hold in general and proposed a modified conjecture as follows. Let the graph Gn;p;k be the complement of a graph Gn;p;k composed of a ðn k 1Þ-regular graph on p vertices together with n p isolated vertices. The minimal counterexample of Conjecture 1 is the graph G7;4;5 , which was given in [1], see Fig. 1. Let
8 nþ2 if > 2 > > nþ3 < if 2 kn ¼ nþ4 > > 2 if > : nþ3 if 2
n 0ðmod 4Þ; n 1ðmod 4Þ; n 2ðmod 4Þ; n 3ðmod 4Þ;
8 n2 > < 2 if n 2ðmod 4Þ and k is even; n p¼ if n 3ðmod 4Þ; 2 > : n otherwise: 2
ð1:5Þ
For such a graph G ¼ Gn;p;k ,
RðGÞ ¼
ðn pÞðn p 1Þ pðp þ k nÞ pðn pÞ þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2ðn 1Þ 2k kðn 1Þ
Using these results, the authors of [1] gave the following Conjecture 2 as a modification of Conjecture 1. Conjecture 2 [1]. Let G ¼ ðV; EÞ be a graph of order n with minimum degree k, and kn and p be given in (1.5). Then
RðGÞ P
8 kðk1Þ kðnkÞ > ffi < 2ðn1Þ þ pffiffiffiffiffiffiffiffiffiffi kðn1Þ
if k < kn ;
> ffiffiffiffiffiffiffiffiffiffiffi if kn 6 k 6 n 2; þ pðpþknÞ þ ppðnpÞ : ðnpÞðnp1Þ 2ðn1Þ 2k kðn1Þ
where equality holds if and only if G is
KI k;nk
for k < kn , and Gn;p;k for k P kn .
}s’ question of finding the minimum value of the Randic´ index for the In this paper, we want to completely solve the Bollobás and Erdo graphs in Gðk; nÞ. As usual, we formulate the question into a mathematical programming problem. Denote by ni the number of vertices of degree i in G, and by xi;j ðxi;j P 0Þ the number of edges joining the vertices of degrees i and j in G. The mathematical description of our problem is as follows:
min RðGÞ ¼
X k6i6n1 i6j6n1
x pi;jffiffiffi ij
subject to: n1 X
xi;j þ 2xi;i ¼ ini
for k 6 i 6 n 1;
ð1:6Þ
j¼k j–i
nk þ nkþ1 þ þ nn1 ¼ n;
ð1:7Þ
xi;j 6 ni nj for k 6 i 6 n 1; i < j 6 n 1; ni xi;i 6 for k 6 i 6 n 1; 2
ð1:8Þ ð1:9Þ
xi;j ; ni are nonnegative integers; for k 6 i 6 j 6 n 1:
ð1:10Þ
Obviously, (1.6)–(1.10) define a nonlinearly constrained optimization problem.
G 7,4,5
G 7,4,5
Fig. 1. The minimal counterexample of Conjecture 1.
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X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
2. Main result Denote
8n > > > 2 > n > n > > > 2 or 2 > > > > > 2n > > < or nþ2 p ¼ n2 2 2 > > >n > > > 2 > > n > > n > > > 2 or 2 > > : n 2
if n 0ðmod 4Þ; if n 1ðmod4Þ and k is even; if n 1ðmod 4Þ and k is odd; if n 2ðmod 4Þ and k is even;
ð2:11Þ
if n 2ðmod 4Þ and k is odd if n 3ðmod 4Þ and k is even if n 3ðmod 4Þ and k is odd:
Theorem 2.1. Let G ¼ ðV; EÞ be a graph of order n with minimum degree k, and p be given in (2.11). Then we have
RðGÞ P
8 kðk1Þ kðnkÞ > ffi < 2ðn1Þ þ pffiffiffiffiffiffiffiffiffiffi kðn1Þ > : ðnpÞðnp1Þ 2ðn1Þ
þ
pðpþknÞ 2k
if k 6 n=2; pðnpÞ
þ pffiffiffiffiffiffiffiffiffiffiffi if k > n=2;
where equality holds if and only if G is
kðn1Þ
KI k;nk
for k 6 n=2, and Gn;p;k for k > n=2.
Proof. It is easy to see that nn1 6 k, or the minimum degree of a graph in Gðk; nÞ would be larger than k. Therefore we only need to consider the case when nn1 6 k. Let nn1 ¼ k t for some integer t such that 0 6 t 6 k, and let Rkt denote the Randic´ index for any graph in Gðk; nÞ with nn1 ¼ k tð0 6 t 6 kÞ. Since xi;n1 ¼ ni nn1 for k 6 i 6 n 2 and xn1;n1 ¼ nn1 ðnn1 1Þ=2, we have
Rkt ¼
X k6i6n1 i6j6n1
n2 X x ni nn1 n ðn 1Þ pi;jffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ n1 n1 2ðn 1Þ iðn 1Þ ij i¼k
n2 x 1X xj1;j x xjþ1;j xj;n2 pk;jffiffiffiffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ 2 pj;jffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ 2 j¼k ðj 1Þj jðj þ 1Þ jðn 2Þ jj kj
P
n2 X i¼k
þ
¼
! ni nn1 nn1 ðnn1 1Þ 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2ðn 1Þ 2 iðn 1Þ kðk þ 1Þ kðn 2Þ kk
n2 xk;j þ þ xj1;j þ 2xj;j þ xj;jþ1 þ þ xj;n2 1 X pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 j¼kþ1 jðn 1Þ
n2 X i¼k
þ
!
ð2:12Þ
ni nn1 n ðn 1Þ 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ n1 n1 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2ðn 1Þ 2 iðn 1Þ kðk þ 1Þ kðn 2Þ kk
!
n2 jnj nj nn1 1 X pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 j¼kþ1 jðn 1Þ
ð2:13Þ
n2 pffi n2 X X 1 nn1 n nn1 ðnn1 1Þ pjffi þ jnj þ pffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffi 2ðn 1Þ 2 n 1 j¼kþ1 j 2 n 1 j¼kþ1 ! 1 2xk;k xk;kþ1 xk;n2 nk nn1 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 2 kðk þ 1Þ kðn 2Þ kðn 1Þ kk 1 ffi where inequality (2.12) holds because p1ffii P pffiffiffiffiffiffi for k þ 1 6 i 6 n 2, and equality (2.13) holds because of (1.6). After substitution of n1 nn1 ¼ k t and nk ¼ n k þ t nkþ1 nkþ2 nn2 into the last equality, we have
Rkt P
Since ptffiffi k
! ! n2 pffi pffiffiffi ðk tÞðk t 1Þ ðk tÞðn k þ tÞ X 1 1 t nj pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi j k ðk tÞ pffiffiffi pffi þ pffiffiffi þ þ 2ðn 1Þ kðn 1Þ 2 n 1 j k k j¼kþ1 ! 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : þ 2 kðk þ 1Þ kðn 2Þ kk
pffi pffiffi pffiffiffiffi pffi pffiffiffi pffi pffiffiffi j k kj > k t for k þ 1 6 j 6 n 2, we have j k > ðk tÞ pffiffiffi , i.e., j k > ðk tÞ
> 0. Since if nj ¼ 0 then xi;j ¼ 0 for k 6 i 6 j 6 n 2, we then have
kj
p1ffiffi k
pffi pffiffiffi 1ffi . Thus j k ðk tÞ p1ffiffik p1 ffi þ p j
j
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X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
! ! n2 pffi pffiffiffi ðk tÞðk t 1Þ ðk tÞðn k þ tÞ X 1 1 t nj pffiffiffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ j k ðk tÞ pffiffiffi pffi þ pffiffiffi 2ðn 1Þ kðn 1Þ j k k 2 n1 j¼kþ1 ! 1 2xk;k xk;kþ1 xk;n2 pffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 kðk þ 1Þ kðn 2Þ kk P
ðk tÞðk t 1Þ ðk tÞðn k þ tÞ xk;k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ : 2ðn 1Þ k kðn 1Þ
Notice that equalities hold in all the above inequalities if and only if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2. Thus,
Rkt P
ðk tÞðk t 1Þ ðk tÞðn k þ tÞ xk;k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ; 2ðn 1Þ k kðn 1Þ
where equality holds if and only if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2. We know that if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2, then nk ¼ n k þ t; nn1 ¼ k t; xk;k ¼ ðn k þ tÞt=2; xn1;n1 ¼ ðk tÞðk t 1Þ=2 and all other xi;j and xi;i are equal to zero. Therefore,
Rkt P
ðk tÞðk t 1Þ ðk tÞðn k þ tÞ ðn k þ tÞt pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ; 2ðn 1Þ 2k kðn 1Þ
where equality holds if and only if nj ¼ 0; j ¼ k þ 1; k þ 2; . . . ; n 2. Let
f ðk; tÞ ¼
ðk tÞðn k þ tÞ ðk tÞðk t 1Þ ðn k þ tÞt pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ : 2ðn 1Þ 2k kðn 1Þ
We only need to get the minimum value of f ðk; tÞ, by distinguishing two cases. Case 1. k 6 n=2. Since of ðk; tÞ=ot ¼ nþ2t2k 2
p1ffiffi k
2 1 ffi pffiffiffiffiffiffi P 0, and of ðk; tÞ=ot > 0 strictly holds except for k t ¼ n=2, i.e., k ¼ n=2 þ t, we know that f ðk; tÞ n1
attains its minimum if and only if t ¼ 0 since k 6 n=2 in this case. So, in this case we can conclude that the Randic´ index attains its minimum in Gðk; nÞ if and only if all the above equalities hold, which k and all other xi;j ; xi;i are equal to zero. Therefore, a graph G in means that nk ¼ n k; nn1 ¼ k; nj ¼ 0; j ¼ k þ 1; . . . ; n 2; xn1;n1 ¼ 2 Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ K k;nk for k 6 n=2. Case 2. n 2 P k > n=2.
2
2 1 ffi 1 ffi p1ffiffi pffiffiffiffiffiffi Let of ðk; tÞ=ot ¼ nþ2t2k ¼ 0. Then t ¼ k n=2. Since o2 f ðk; tÞ=ot2 ¼ p1ffiffik pffiffiffiffiffiffi > 0, f ðk; tÞ attain its minimum if and only if 2 n1 n1 k t ¼ k n=2. Then we have nk ¼ n=2; xk;k ¼ nð2k nÞ=2 and xn1;n1 ¼ nðn 2Þ=8. Next, we need to check whether they are integers or not, since the obtained solutions have no graph theoretical meaning when one of the three values, namely, xk;k ¼ ðn k þ tÞt=2, xn1;n1 ¼ ðk tÞðk t 1Þ=2 and t, is not an integer.
Subcase 2.1. n 0ðmod 4Þ. We can easily check that t ¼ k n=2; xk;k ¼ n=2 and xn1;n1 ¼ nðn 2Þ=8 are integers in this case. Therefore, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ Gn;n=2;k in the case n 0ðmod 4Þ. Subcase 2.2. n 1ðmod 4Þ. We see first that t ¼ k n=2 is not an integer in this case. Therefore, the obtained solutions have no graph theoretical meaning. Then t can not attain k n=2 if we want to get the minimum value of the Randic´ index in Gðk; nÞ. We then let t 6 k nþ1 or t P k n1 . 2 2 , we have n þ 2t 2k 6 1 < 0. Thus of ðk; tÞ=ot < 0. Therefore, f ðk; tÞ attains its minimum if and only if t ¼ k nþ1 in this For t 6 k nþ1 2 2 case. And then nk ¼ ðn 1Þ=2; nn1 ¼ ðn þ 1Þ=2; xk;k ¼ ðn 1Þð2k n 1Þ=8; xn1;n1 ¼ ðn þ 1Þðn 1Þ=2 all are integers. So, pffiffiffiffiffiffiffiffiffiffiffi þ ðnþ1Þðn1Þ þ ðnþ1Þðn1Þ . For t P k n1 , we have n þ 2t 2k P 1 > 0. Thus of ðk; tÞ=ot > 0. Therefore, f ðk; tÞ attains its minf ðk; tÞ ¼ ðn1Þð2kn1Þ 8k 8ðn1Þ 2 4
kðn1Þ
in this case. And then nk ¼ ðn þ 1Þ=2; nn1 ¼ ðn 1Þ=2; xk;k ¼ ðn þ 1Þð2k n þ 1Þ=8; xn1;n1 ¼ minimum if and only if t ¼ k n1 2 pffiffiffiffiffiffiffiffiffiffiffi þ ðn1Þðn3Þ pffiffiffiffiffiffiffiffiffiffiffi þ ðnþ1Þðn1Þ þ ðnþ1Þðn1Þ ¼ ðn1Þð2kn1Þ þ ðnþ1Þðn1Þ , which is ðn 1Þðn 3Þ=2 all are integers if k is even. So, minf ðk; tÞ ¼ ðnþ1Þð2knþ1Þ 8k 8ðn1Þ 8k 8ðn1Þ 4
kðn1Þ
4
kðn1Þ
. the same as the minimum value for the above case when t 6 k nþ1 2 Therefore, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G is Gn;bnc;k for k both even and odd, or Gn;dne;k 2 2 for k even in the case n 1ðmod4Þ. Subcase 2.3. n 2ðmod 4Þ. We can easily check that t ¼ k n=2; xk;k ¼ nð2k nÞ=8 and xn1;n1 ¼ nðn 2Þ=8 are integers if k is odd in this case. Therefore, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G is Gn;n=2;k for k odd in the case n 1ðmod 4Þ. If k is even, then xk;k ¼ nð2k nÞ=8 is not an integer in this case, which means that t can not attain k n=2 if we want to get the minor t P k n2 . imum value of the Randic´ index in Gðk; nÞ. We then let t 6 k nþ2 2 2
X. Li et al. / European Journal of Operational Research 200 (2010) 9–13
13
For t 6 k nþ2 , we have n þ 2t 2k 6 2 < 0. Thus of ðk; tÞ=ot < 0. Therefore, f ðk; tÞ attains its minimum if and only if t ¼ k nþ2 in this 2 2 case. And then nk ¼ ðn 2Þ=2; nn1 ¼ ðn þ 2Þ=2; xk;k ¼ ðn 2Þð2k n 2Þ=8; xn1;n1 ¼ nðn þ 2Þ=2 all are integers. So, minf ðk; tÞ ¼ ðn2Þð2kn2Þ nðnþ2Þ pffiffiffiffiffiffiffiffiffiffiffi þ 8ðn1Þ þ ðn2Þðnþ2Þ . 8k 4
kðn1Þ
For t P k n2 , we have n þ 2t 2k P 2 > 0. Thus of ðk; tÞ=ot > 0. Therefore, f ðk; tÞ attains its minimum if and only if t ¼ k n2 in this 2 2 case. And then nk ¼ ðn þ 2Þ=2; nn1 ¼ ðn 2Þ=2; xk;k ¼ ðn þ 2Þð2k n þ 2Þ=8; xn1;n1 ¼ ðn 2Þðn 4Þ=2 all are integers. So, minf ðk; tÞ ¼ ðnþ2Þð2knþ2Þ nðnþ2Þ pffiffiffiffiffiffiffiffiffiffiffi þ ðn2Þðn4Þ pffiffiffiffiffiffiffiffiffiffiffi þ 8ðn1Þ þ ðn2Þðnþ2Þ ¼ ðn2Þð2kn2Þ þ ðn2Þðnþ2Þ , which is the same as the minimum value for the above case when 8k 8ðn1Þ 8k 4 kðn1Þ 4 kðn1Þ . t 6 k nþ2 2 Therefore, in the case n 2ðmod 4Þ and k is odd, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ Gn;2n;k . In the case n 2ðmod 4Þ and k is even, a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G ¼ Gn;n2;k or 2 Gn;nþ2;k . 2
Subcase 2.4. n 3ðmod 4Þ. Similar to the proof of Subcase 2.2, we can get that a graph G in Gðk; nÞ attains the minimum value of the Randic´ index if and only if G is Gn;dne;k for k both even and odd, or Gn;bnc;k for k even. 2
2
The proof is now complete. h
References [1] M. Aouchiche, P. Hansen, On a conjecture about the Randic´ index, Discrete Mathematics 307 (2) (2007) 262–265. } s, Graphs of extremal weights, Ars Combinatoria 50 (1998) 225–233. [2] B. Bollobás, P. Erdo [3] G. Caporossi, I. Gutman, P. Hansen, Variable neighborhood search for extremal graphs IV: Chemical trees with extremal connectivity index, Computers and Chemistry 23 (1999) 469–477. [4] G. Caporossi, I. Gutman, P. Hansen, Lj. Pavlovic´, Graph with maximum connectively index, Computational Biology and Chemistry 27 (2003) 85–90. [5] C. Delorme, O. Favaron, D. Rautenbach, On the Randic´ index, Discrete Mathematics 257 (1) (2002) 29–38. [6] S. Fajtlowicz, Written on the Wall: Conjectures Derived on the Basis of the Program Galatea Gabriella Graffiti, University of Houston, 1998. [7] M. Fischermann, A. Hoffmann, D. Rautenbach, L. Volkmann, A linear-programming approach to the generalized Randic´ index, Discrete Applied Mathematics 128 (2003) 375–385. [8] I. Gutman, O. Miljkovic, G. Caporossi, P. Hansen, Alkanes with small and large Randic´ connectivity indices, Chemical Physics Letters 306 (1999) 366–372. [9] I. Gutman, Lj. Pavlovic´, O. Miljkovic, On graphs with extremal connectivity indices, Bulletin de l’Academie Serbe des Sciences er des Arts 25 (2000) 1–14. [10] L.B. Kier, L.H. Hall, Molecular Connectivity in Chemistry and Drug Research, Academic Press, New York, 1976. [11] L.B. Kier, L.H. Hall, The nature of structure–activity relationships and their relation to molecular connectivity, European Journal of Medical Chemistry 12 (1977) 307–312. [12] L.B. Kier, L.H. Hall, Molecular Connectivity in Structure–Activity Analysis, Research Studies Press-Wiley, Chichester(UK), 1986. [13] X. Li, I. Gutman, Mathematical aspects of Randic-type molecular structure descriptors, Mathematical Chemistry Monographs, No. 1, Kragujevac, 2006, pp. VI + 330. [14] N. Limic, H. Psasgic, C. Rnjak, Linearno i nelinearno programiranje, Informator, Zagreb(HR), 1978. [15] Lj. Pavlovic´, Graphs with extremal Randic´ index when the minimum degree of vertices is two, Kragujevac Journal of Mathematics 25 (2003) 55–63. [16] Lj. Pavlovic´, On the conjecture of Delorme, Favaron and Rautenbach about the Randic´ index, European Journal of Operational Research 180 (2007) 369–377. [17] Lj. Pavlovic´, T. Divnic, A quadratic programming approach to the Randic´ index, European Journal of Operational Research 176 (2007) 435–444. [18] Lj. Pavlovic´, I. Gutman, Graph with extremal connectivity index, Novi Sad Journal of Mathematics 31 (2) (2001) 53–58. [19] M. Randic´, On characterization of molecular branching, Journal of the American Chemical Society 97 (1975) 6609–6615.