Two Sufficient Conditions
for Dominating Cycles
Mei Lu,1 Huiqing Liu,2 and Feng Tian2
1
DEPARTMENT OF MATHEMATICAL SCIENCES
TSINGHUA UNIVERSITY
BEIJING 100084, CHINA
E-mail: mlu@math.tsinghua.edu.cn
2
INSTITUTE OF SYSTEMS SCIENCE
ACADEMY OF MATHEMATICS AND SYSTEMS SCIENCE
CHINESE ACADEMY OF SCIENCES
BEIJING 100080, CHINA
E-mail: liuhuiqing@eyou.com and ftian@ftian.iss.ac.cn
Received April 8, 2002; Revised August 31, 2004
Published online in Wiley InterScience(www.interscience.wiley.com).
DOI 10.1002/jgt.20070
Abstract: A cycle C of a graph G is dominating if each component of GnC
is edgeless. In the paper, we will give two sufficient conditions for each
longest cycle of a 3-connected graph to be a dominating cycle.
ß 2005 Wiley Periodicals, Inc. J Graph Theory 49: 135–150, 2005
Keywords: dominating cycle; connectivity; insertible vertex
1. INTRODUCTION
We use Bondy and Murty [4] for terminology and notation not defined here and
consider simple undirected graphs only. Let G be a graph. If S VðGÞ, then
NðSÞ denotes the neighbors of S. For a subgraph H of G and S VðGÞ, let
——————————————————
Contract grant sponsor: NNSFC (to M.L.); Contract grant number: 60172005;
Contract grant sponsor: NNSFC.
ß 2005 Wiley Periodicals, Inc.
135
136 JOURNAL OF GRAPH THEORY
NH ðSÞ ¼ NðSÞ \ VðHÞ. When H ¼ G, we write NðSÞ instead of NG ðSÞ. If
S ¼ fs1 ; s2 ; . . . ; sl g, then NH ðSÞ is written as NH ðs1 ; s2 ; . . . ; sl Þ. For a graph G,
we denote by ðGÞ and ðGÞ the independence number and the connectivity of G,
respectively. We define k ðGÞ by the minimum value of the degree sum of any k
independent vertices of G if k ðGÞ; if k > ðGÞ, we set k ðGÞ ¼ þ1.
A cycle C of G is said to be a D -cycle, if jHj < for any component H of
GnC. Obviously, a D1 -cycle is a hamiltonian cycle and a D2 -cycle is a dominating
cycle. For a graph G, we denote by pðGÞ and cðGÞ the order of a longest path and
the order of a longest cycle in G, respectively. Let diffðGÞ ¼ pðGÞ cðGÞ. Then a
connected graph G is hamiltonian if and only if diffðGÞ ¼ 0 and if diffðGÞ 1,
then each longest cycle of G is a dominating cycle.
Bondy [3] gives the following sufficient condition for each longest cycle of a
2-connected graph G to be a dominating cycle.
Theorem 1 [3]. Let G be a 2-connected graph of order n 3 with 3 ðGÞ n þ 2. Then each longest cycle of G is a dominating cycle.
A much stronger result is given in [6].
Theorem 2 [6].
diffðGÞ 1.
A 2-connected graph G of order n with 3 ðGÞ n þ 2 satisfies
When involved in connectivity, we have the following theorem.
Theorem 3 [2]. Let G be a 2-connected graph on n vertices with 3 ðGÞ n þ . Then G is hamiltonian.
A short proof of Theorem 3 was given in [10]. In [8], a problem proposed by
Tian in [9] is solved and a related conjecture is proposed.
Theorem 4 [8]. Let G be a 3-connected graph of order n 3 with
4 ðGÞ n þ 2. Then G contains a longest cycle which is a dominating cycle.
Conjecture A. Let G be a k-connected graph of order n with kþ1 ðGÞ n þ ðk 1Þ. Then each longest cycle of G is a Dk1 -cycle.
In the paper, we show that under the condition of Theorem 4, each longest
cycle is a dominating cycle.
Theorem 5. Let G be a 3-connected graph of order n 3 with 4 ðGÞ n þ 2.
Then each longest cycle of G is a dominating cycle.
Theorems 3 and 5 show that Conjecture A holds when k ¼ 2 and 3.
Li et al. proposed a conjecture about the difference diffðGÞ as follows.
Conjecture B [7]. Let G be a 3-connected graph of order n. If 4 ðGÞ 43 n þ 53 ,
then diffðGÞ 1.
And we will propose a problem also about the difference diffðGÞ and is
supported by Theorem 5.
CONDITIONS FOR DOMINATING CYCLES
137
Problem C. Let G be a 3-connected graph of order n 3 with
4 ðGÞ n þ 2. Is it true that diffðGÞ 1?
The following class of graphs shows the sharpness of Conjecture B. Let
G ¼ kK1 þ ðk þ 1ÞK2 , (k 3). Then jGj ¼ n ¼ 3k þ 2 and 4 ðGÞ ¼ 4ðk þ 1Þ ¼
4
4
3 n þ 3 , but diffðGÞ ¼ 2.
In the paper, we obtain the following result which supports Conjecture B.
Theorem 6. Let G be a 3-connected graph of order n 13. If 4 ðGÞ 43 n þ 53 ,
then each longest cycle of G is a dominating cycle.
The class of graphs described above also shows that Theorem 6 is best
possible.
Next, we will use two examples to show that the conditions of Theorems 5 and
6 are independent.
Example 1. Let G ¼ kK1 þ ðkK2 [ K1 Þ with k 3. Then n ¼ 3k þ 1, ðGÞ ¼ k
and 4 ðGÞ ¼ 43 n þ 53. Hence by Theorem 6 each longest cycle of G is a
dominating cycle. But n þ 2 ¼ 5n2
3 , that is, the conditions of Theorem 5 are not
satisfied.
c
Example 2. Let F ¼ ðG1 þ G2 Þ þ K2r
with G1 ¼ G2 ¼ Kr ; where r 3. The
graph G is defined as: VðGÞ ¼ VðFÞ [ VðK2r Þ and EðGÞ ¼ EðFÞ [ EðK2r Þ [
fuv : u 2 VðK2r Þ; v 2 VðG1 Þg. It is easy to check that n ¼ 6r, ðGÞ ¼ r and
4 ðGÞ ¼ 8r. Obviously, 4 ðGÞ ¼ n þ 2. Hence by Theorem 5 each longest cycle
4nþ5
of G is a dominating cycle. But 4 ðGÞ ¼ 4n
3 < 3 . That is, the conditions of
Theorem 6 are not satisfied.
Let C ¼ v 1 v 2 v t v 1 be a cycle of G with a given orientation. For v i 2 VðCÞ,
we use v þ
i ; v i to denote the successor and predecessor of v i on C, respectively.
The i-th successor and i-th predecessor of a vertex v on C are denoted by v þi and
v i , respectively. If A VðCÞ, then A and Aþ are the sets fv : v 2 Ag and
fv þ : v 2 Ag, respectively. If u; v 2 VðCÞ, we denote by uCv the subpath
uuþ v v of C. The same subpath, in reverse order, is denoted by vCu. We will
consider uCv and vCu both as paths and as vertex sets. We also use analogous
notations for a path P in G.
2. LEMMAS
In this section, we give some lemmas which will be used in Sections 3 and 4.
Lemma 1 [11]. Let G be a 3-connected graph of order n 3 with 4 n þ 6
and C be a longest cycle of G. Then C is a D3 -cycle.
Next, we assume that G is a 3-connected nonhamiltonian graph. Let C be a
longest cycle of G with a given orientation and H a component of GnC. Assume
that VðHÞ ¼ fu; vg. Suppose that NC ðHÞ ¼ fx1 ; x2 ; . . . ; xt g, and the occurrence
138 JOURNAL OF GRAPH THEORY
of the vertices on C agrees with the given orientation of C. A vertex u 2 xþ
i Cxiþ1
þ
þ
is insertible if there exist vertices v; v 2 xiþ1 Cxi such that uv; uv 2 EðGÞ;
then vv þ 2 EðGÞ is an insertion edge of u. We denote the set of insertion
edges of u by IðuÞ, and denote the first noninsertible vertex occurring on xþ
i Cxiþ1
þ
by ui. Let Ui ¼ xi Cui , T ¼ fu1 ; u2 ; . . . ; ut g, Tu ¼ fui : xi 2 NC ðuÞg and Tv ¼
fuj : xj 2 NC ðvÞg. Then we have the following lemmas.
Lemma 2 [1][5]. (1) ui exists.
(2) ui 62 NðuÞ [ NðvÞ for any ui 2 T.
(3) There is no path whose internal vertices (if any) in GnC joining a vertex
of Ui and a vertex of Uj , and IðxÞ \ IðyÞ ¼ ; for all x 2 xþ
i Cui and
þ
y 2 xj Cuj . In particular, GnT is connected.
(4) For any ui ; uj 2 T; uj w 62 EðGÞ whenever ui w 2 EðGÞ and w 2 uþ
j Cxi .
By using the proof of Lemma 2 ([1]), we can show the following lemma
similarly.
þ
Lemma 3. Let ui 2 Tu and uj 2 Tv . Then ð1Þ ui uþ
j 62 EðGÞ and ui uj 62 EðGÞ;
þ
ð2Þ uj w 62 EðGÞ for w 2 uj Cxi and ui w2 2 EðGÞ; ð3Þ ui w 62 EðGÞ for w 2 uþ
i Cxj
and uj w2 2 EðGÞ.
Proof. We only give the proof of (2), and parts (1) and (3) can be proved in a
similar way.
2
2 EðGÞ. By Lemma 2,
(2) Assume uj w 2 EðGÞ for some w 2 uþ
j Cxi and ui w
þ
þ
2
ww ; w w 62 IðuÞ for each u 2 ðxi Cui Þ [ ðxj Cuj Þ. Then we can insert the
þ
2
vertices of ðxþ
i Cui Þ [ ðxj Cuj Þ in the cycle xi Cwuj Cw ui Cxj vuxi to derive a
contradiction with the choice of C.
&
Lemma 4. If ui 2 Tu ðui 2 Tv , resp.Þ and uj ; uk 2 Tv ðuj ; uk 2 Tu , resp.Þ, then
dðui Þ þ dðuj Þ þ dðuk Þ n jWj þ 1;
where W ðNðuÞ \ NðvÞÞnNðui ; uj ; uk Þ.
Proof. Set C1 ¼ ui Cu
j , C2 ¼ uj Cuk , C3 ¼ uk Cui , and Wh ¼ W \ Ch ,
1 h 3. Since W ðNðuÞ\ NðvÞÞnNðui ; uj ; uk Þ, fw; wþ; wþ2 g\ Nðui ; uj ; uk Þ ¼
; for each w 2 W by Lemmas 2 and 3.
2
(1) NC21 ðui Þ [ NC1 ðuj Þ [ NC1 ðuk Þ ðC1nW1 Þ [ fu
i g, and NC1 ðui Þ, NC1 ðuj Þ and
NC1 ðuk Þ are pairwise disjoint by Lemmas 2 and 3. Hence
dC1 ðui Þ þ dC1 ðuj Þ þ dC1 ðuk Þ jC1 j jW1 j þ 1:
(2) NC3 ðui Þ [ NC3 ðuj Þ [ NC23 ðuk Þ ðC3nW3 Þ [ fu
k g, and
2
NC3 ðuk Þ are pairwise disjoint by Lemmas 2 and 3. Hence
ð2:1Þ
NC3 ðui Þ, NC3 ðuj Þ and
dC3 ðui Þ þ dC3 ðuj Þ þ dC3 ðuk Þ jC3 j jW3 j þ 1:
ð2:2Þ
CONDITIONS FOR DOMINATING CYCLES
139
(3) Now we want to show that
dC2 ðui Þ þ dC2 ðuj Þ þ dC2 ðuk Þ jC2 j jW2 j þ 1:
ð2:3Þ
First we note that if uk 2 Tv \ Tu , then NC2 ðui Þ [ NC22 ðuj Þ [ NC2 ðuk Þ 2
ðC2nW2 Þ [ fu
j g, and NC2 ðuj Þ, NC2 ðui Þ, and NC2 ðuk Þ are pairwise disjoint by
Lemmas 2 and 3, and hence ð2:3Þ holds. So in the following proof, we assume
that uk 2 TvnTu .
Claim 2.1. xk 2
= W2 :
Proof. Since uk 2 TvnTu , we have xk 2 NðvÞnNðuÞ. Thus xk 2
= W2 :
&
uþ
j Cxk
Set P ¼
¼ bl bl1 b1 . For each r, 1 r l, define AðrÞ ¼
fb1 ; b2 ; . . . ; br g and W2 ðrÞ ¼ W2 \ AðrÞ. Let Uk0 ¼ ðUknfuk gÞ [ fuj g. Then
W2 ¼ W2 ðlÞ and C2 ¼ AðlÞ [ Uk0 . By Lemma 2, Nðui Þ \ Uk0 ¼ ; and
Nðuj Þ \ Uk0 ¼ ;. For each r, 1 r l, if
dAðrÞ ðui Þ þ dAðrÞ ðuj Þ þ dAðrÞ ðuk Þ jAðrÞj jW2 ðrÞj þ 2;
ð2:4Þ
then
dC2 ðui Þ þ dC2 ðuj Þ þ dC2 ðuk Þ ¼ dAðlÞ ðui Þ þ dAðlÞ ðuj Þ þ dAðlÞ ðuk Þ þ dUk0 ðuk Þ
jAðlÞj jW2 ðlÞj þ 2 þ jUk0 j 1 ¼ jC2 j jW2 j þ 1:
Hence, we just need to show that ð2:4Þ holds for each r, 1 r l.
Before giving the proof of (2.4), we point out the following facts.
= NðuÞ, and hence we get
Since b1 ¼ xk 2 NðvÞ, we get b2 ; b3 2
= W2 , and hence if bt 2 W2 , then t 4.
Fact 1. b2 ; b3 2
By the definition of W and C being the longest cycle, we have
Fact 2. If bt 2 W2 , then bt 62 Nðui ; uj ; uk Þ; btþ1 ; btþ2 2
= W2 ; and bt2 ; bt1 2
=
W2 [ Nðui ; uj ; uk Þ:
It is easy to check that (2.4) holds for r 3; (in this case, jW2 ðrÞj ¼ 0 by
Fact 1). Next we assume that ð2:4Þ is true for AðsÞ, where 3 s r < l. We will
prove that ð2:4Þ is true for Aðr þ 1Þ. Let J ¼ fui ; uj ; uk g.
Claim 2.2. fbrþ1 ; br g \ W2 ¼ ;:
Proof. (i) If brþ1 2 W2 , (in this case, r 3 by Fact 1), then jW2 ðrÞj ¼
jW2 ðr þ 1Þj 1 and dJ ðbrþ1 Þ ¼ 0. By induction,
dAðrþ1Þ ðui Þ þ dAðrþ1Þ ðuj Þ þ dAðrþ1Þ ðuk Þ
¼ dAðrÞ ðui Þ þ dAðrÞ ðuj Þ þ dAðrÞ ðuk Þ þ dJ ðbrþ1 Þ
jAðrÞj jW2 ðrÞj þ 2 ¼ jAðr þ 1Þj jW2 ðr þ 1Þj þ 2;
and hence (2.4) holds for Aðr þ 1Þ.
140 JOURNAL OF GRAPH THEORY
(ii) If br 2 W2 , (in this case, r 4), then brþ1 2
= W2 and br2 ; br1 2
=
[ Nðui ; uj ; uk Þ by Fact 2. Thus jW2 ðr 3Þj ¼ jW2 ðr þ 1Þj 1 and
W
P2rþ1
t¼r2 dJ ðbt Þ ¼ dJ ðbrþ1 Þ 3: By induction,
dAðrþ1Þ ðui Þ þ dAðrþ1Þ ðuj Þ þ dAðrþ1Þ ðuk Þ
¼ dAðr3Þ ðui Þ þ dAðr3Þ ðuj Þ þ dAðr3Þ ðuk Þ þ dJ ðbrþ1 Þ
jAðr 3Þj jW2 ðr 3Þj þ 5 ¼ jAðr þ 1Þj jW2 ðr þ 1Þj þ 2;
and hence (2.4) holds for Aðr þ 1Þ.
&
Claim 2.3. dJ ðbrþ1 Þ 2.
Proof. Suppose dJ ðbrþ1 Þ 1. Then by Claim 2.2, jW2 ðrÞj ¼ jW2 ðr þ 1Þj. By
induction,
dAðrþ1Þ ðui Þ þ dAðrþ1Þ ðuj Þ þ dAðrþ1Þ ðuk Þ
¼ dAðrÞ ðui Þ þ dAðrÞ ðuj Þ þ dAðrÞ ðuk Þ þ dJ ðbrþ1 Þ
jAðrÞj jW2 ðrÞj þ 3 ¼ jAðr þ 1Þj jW2 ðr þ 1Þj þ 2;
and hence (2.4) holds for Aðr þ 1Þ.
&
Claim 2.4. br1 2
= W2 .
Proof. Suppose that br1 2 W2 , (in this case,
r 5). Noting that
Prþ1
jW2 ðr 3Þj ¼ jW2 ðr þ 1Þj 1 by Claim 2.2 and
t¼r2 dJ ðbt Þ ¼ dJ ðbr Þ þ
dJ ðbrþ1 Þ 3 by the definition of uk ; ui and C being the longest cycle, we get
&
that (2.4) holds for Aðr þ 1Þ by a similar argument as Claim 2.2.
Now by Claim 2.3, we finish the proof by considering the following two cases.
Case 1. dJ ðbrþ1 Þ ¼ 2.
In this case, we first have that brþ1 uk 2
= EðGÞ. Otherwise, by Lemma 2 and the
definition of uk , br 62 Nðui ; uj ; uk Þ. Thus jW2 ðr 1Þj ¼ jW2 ðr þ 1Þj by Claim 2.2,
and hence ð2:4Þ holds by induction. Hence, we assume that brþ1 2 Nðui Þ \ Nðuj Þ
= Nðui ; uj ; uk Þ. Thus
and then br 62 Nðui ; uj Þ. Note that if br 2 Nðuk Þ, then br1 2
= Nðuk Þ, and
jW2 ðr tÞj ¼ jW2 ðr þ 1Þj by Claims 2.2 and 2.4, where t ¼ 1 if br 2
t ¼ 2 if br 2 Nðuk Þ, and ð2:4Þ holds for Aðr þ 1Þ by induction.
Case 2. dJ ðbrþ1 Þ ¼ 3.
In this case, br 2
= Nðui ; uj ; uk Þ and br1 2
= Nðui ; uj Þ by Lemma 3. If
= Nðuk Þ, then jW2 ðr 2Þj ¼ jW2 ðr þ 1Þj by Claims 2.2 and 2.4, and hence
br1 2
= Nðui ; uj ; uk Þ, and
ð2:4Þ holds.
On the other hand, if br1 2 Nðuk Þ, then br2 2
P
d
ðb
Þ
¼
4.
We
may
assume
that
r
4.
Note
that jW2 ðr tÞj ¼
hence rþ1
t¼r2 J t
= W2 and t ¼ 4 if br2 2 W2 , (in this
jW2 ðr þ 1Þj ðt 3Þ, where t ¼ 3 if br2 2
case, r 6). Thus ð2:4Þ holds for Aðr þ 1Þ by induction.
CONDITIONS FOR DOMINATING CYCLES
141
By (2.1), (2.2), (2.3) and noting dR ðui Þ þ dR ðuj Þ þ dR ðuk Þ n jCj 2, we
&
have dðui Þ þ dðuj Þ þ dðuk Þ n jWj þ 1:
The proof of the following lemma is very trivial, so we will omit it here.
Lemma 5. Let P be a path and v i ; v j 62 VðPÞ. If NP ðv i Þ \ NP ðv j Þ ¼ ;, then
dP ðv i Þ þ dP ðv j Þ jPj þ 1.
3. PROOF OF THEOREM 6
Suppose Theorem 6 fails to hold, then there exists a longest cycle C in G which is
not a dominating cycle. Since n 13, 4 ðGÞ 43 n þ 53 n þ 6. By Lemma 1,
each component of GnC has at most two vertices. Let R ¼ GnC. Then we can
choose a component H of R with H ¼ fu; vg. Give the cycle C an orientation.
Suppose that NC ðu; vÞ ¼ fx1 ; x2 ; . . . ; xt g, and the occurrence of the vertices on C
agrees with the given orientation of C. Denote the first noninsertible vertex
occurring on xþ
i Cxiþ1 by ui. The other notations such as Ui ; T; Tu , and Tv are the
same as those in Section 2. By Lemma 2, we know that T [ fug and T [ fvg are
both independent sets of G. Since G is 3-connected, jTu j 2, jTv j 2 and
jTu [ Tv j 3. Choose ui ; uj ; uk 2 T with ui 2 Tu and uj ; uk 2 Tv . By Lemma 4
and letting W ¼ ;, we have the following claim.
Claim 3.1. dðui Þ þ dðuj Þ þ dðuk Þ n þ 1.
Now we have another claim as follows.
Claim 3.2. dðuj Þ þ dðuk Þ þ dðuÞ n þ 1.
þ
Proof. Let P1 ¼ uþ
j Cxk and P2 ¼ uk Cxj . By
= NP1 ðuj Þ [ NP1 ðuk Þ
NP1 ðuj Þ \ NP1 ðuk Þ ¼ ; and uh 2
= NP2 ðuk Þ [ NP2 ðuj Þ
NP2 ðuk Þ \ NP2 ðuj Þ ¼ ; and uh 2
Lemmas 2 and 3, we have
for uh 2 Tu \ P1. Similarly,
for uh 2 Tu \ P2. Hence, by
Lemmas 2 and 5,
dC ðuj Þ þ dC ðuk Þ ¼
2
X
ðdPl ðuj Þ þ dPl ðuk ÞÞ þ dUj ðuj Þ þ dUk ðuk Þ dC ðuÞ þ 2
l¼1
ðjP1 j þ 1Þ þ ðjP2 j þ 1Þ þ ðjUj j 1Þ þ ðjUk j 1Þ dC ðuÞ þ 2
¼ jCj dC ðuÞ þ 2:
Since dR ðuj Þ þ dR ðuk Þ þ dR ðuÞ n jCj 1, Claim 3.2 holds.
&
Claim 3.3. dðui Þ þ dðuj Þ þ dðuÞ n þ 1.
þ
Proof. Let P1 ¼ uþ
i Cxj and P2 ¼ uj Cxi . By Lemma 2, we have
þ
\ NP1 ðuj Þ ¼ ;. If uh 2 Nðui Þ for some uh 2 Tu \ P1 , then uþ2
= Nðui Þ
h 2
þ
= Nðuj Þ. Hence, we have that
by the definition of ui . By Lemma 3, uh 2
NP1 ðui Þ
142 JOURNAL OF GRAPH THEORY
uh 62 NP1 ðui Þ [ NP1 ðuj Þ when uþ
= Nðui Þ and uþ
h 2
h 62 NP1 ðui Þ [ NP1 ðuj Þ when
þ
uh 2 Nðui Þ. On the other hand, by Lemmas 2 and 3, NP2 ðuj Þ \ NP2 ðui Þ ¼ ; and
uh 62 NP2 ðuj Þ [ NP2 ðui Þ for uh 2 Tu \ P2. Hence, by Lemmas 2 and 5, we have that
dC ðui Þ þ dC ðuj Þ ¼
2
X
ðdPl ðui Þ þ dPl ðuj ÞÞ þ dUi ðui Þ þ dUj ðuj Þ dC ðuÞ þ 2
l¼1
ðjP1 j þ 1Þ þ ðjP2 j þ 1Þ þ ðjUi j 1Þ þ ðjUj j 1Þ dC ðuÞ þ 2
¼ jCj dC ðuÞ þ 2:
Since dR ðui Þ þ dR ðuj Þ þ dR ðuÞ n jCj 1, Claim 3.3 holds.
&
By a similar argument as Claim 3.3, we obtain
Claim 3.4. dðui Þ þ dðuk Þ þ dðuÞ n þ 1.
Since fu; ui ; uj ; uk g is an independent set and by Claims 3.1–3.4, we have
4
4
dðui Þ þ dðuj Þ þ dðuk Þ þ dðuÞ n þ ;
3
3
a contradiction.
&
4. PROOF OF THEOREM 5
The main ideas of the proof of Theorem 5 come from [10] and [8].
Let G be a graph of order n with connectivity 3 satisfying 4 ðGÞ n þ 2, and G has a longest cycle C which is not a dominating cycle.
We denote GnC by R. Since 3 and by Lemma 1, each component of GnC
has at most two vertices. Let H be chosen as in Section 2. Suppose that
NC ðu; vÞ ¼ fx1 ; x2 ; . . . ; xt g, and the occurrence of the vertices on C agrees with
the given orientation of C. Denote the first noninsertible vertex occurring on
xþ
i Cxiþ1 by ui. The other notations such as T, Tu , and Tv are the same as those in
Section 2. By Lemma 2, T [ fug and T [ fvg are both independent sets of G. Let
S be any vertex cut set with jSj ¼ , and B1 ; B2 ; . . . ; Bp ; p 2 the components of
GnS, where p is the number of components of GnS. By Lemma 2(3), GnT is
connected.
Obviously, we have the following claim, which will be used in the next often.
= Bi for some i; 1 i p. If there exists a path
Claim 4.0. Let v s 2 Bi and v t 2
Pðv s ; v t Þ in G joining v s and v t , then S \ Pðv s ; v t Þ 6¼ ;.
Claim 4.1. minfdðuÞ; dðvÞg þ 2.
Proof. Suppose that dðuÞ þ 1. Since G is 3-connected, jTj 3 and
minfdðuÞ; dðvÞg 3. So such three vertices as those of Lemma 4 exist. Assume,
without loss of generality, that ui 2 Tu and uj ; uk 2 Tv , then fu; ui ; uj ; uk g is an
CONDITIONS FOR DOMINATING CYCLES
143
independent set. By Lemma 4 and letting W ¼ ;, we have dðuÞ þ dðui Þ þ dðuj Þ þ
dðuk Þ n þ 1 þ dðuÞ n þ þ 2; a contradiction. Therefore, dðuÞ þ 2.
&
Similarly, dðvÞ þ 2.
The proofs of Claims 6 and 5 in [8] can be used to prove the following
Claims 4.2 and 4.3, respectively. For the sake of completeness, we still provide
the proofs of them below.
Claim 4.2. If Bi \ T ¼ ;, then Bi \ fu; vg ¼ ;, where 1 i p.
Proof. Otherwise, without loss of generality, let u 2 Bi . By Claim 4.1, we
may assume that dðuÞ ¼ d þ 1, where d þ 1. Suppose that NC ðuÞ ¼
fx1 ; x2 ; . . . ; xd g. Thus we have d paths: uxj Cuj ; 1 j d, which are from u to
vertices of T, and pairwise disjoint except for u. Since Bi \ T ¼ ;, each of these d
paths contains at least one vertex of S by Claim 4.0. So jSj d þ 1, a
&
contradiction.
Claim 4.3. jfBi : Bi \ T 6¼ ;; 1 i pgj 2:
Proof. Otherwise, without loss of generality, suppose that Bi \ T 6¼ ;; i ¼ 1;
2; 3: Clearly,
dðyi Þ jBi j þ jSj jðBi [ SÞ \ ðT [ fu; vgÞj;
where yi 2 Bi \ T; i ¼ 1; 2; 3. So we have
dðy1 Þ þ dðy2 Þ þ dðy3 Þ 3
X
jBi j þ 3jSj i¼1
n þ 2jSj 3
X
jðBi [ SÞ \ ðT [ fu; vgÞj
i¼1
p
X
i¼4
jBi j j
3
[
ðBi [ SÞ \ ðT [ fu; vgÞj
i¼1
n þ 2jSj ðjTj þ 2Þ n þ 2 dðuÞ 1:
That is, dðy1 Þ þ dðy2 Þ þ dðy3 Þ þ dðuÞ n þ 2 1; a contradiction.
&
By Claim 4.3 we may assume, without loss of generality, that jBi \ Tj ¼ 0 for
i 3. Then by Claim 4.2, we get T [ fu; vg B1 [ B2 [ S. Since GnT is
connected, we have S 6 T, and hence jS \ Tj 1.
Claim 4.4. If jS \ Tj ¼ 1 and u; v 2 Br , then jBs \ Tj ¼ 0, where
fr; sg ¼ f1; 2g.
Proof. Assume, without loss of generality, that u; v 2 B1 . Suppose that
jB2 \ Tj 1 and let ui 2 B2 \ T. Since u; v 2 B1 and ui 2 B2 , S \ ðxi Cu
i Þ 6¼ ;
Cx
Þ
¼
6
;.
Thus
jSnTj
2.
Since
jS
\
Tj
¼
1,
we
have
that
and S \ ðuþ
iþ1
i
&
jSj þ 1, a contradiction.
144 JOURNAL OF GRAPH THEORY
Claim 4.5. If jBr \ Tj 2 for some r 2 f1; 2g, then there exists some s,
1 s 2, such that jBs \ Tu j 2 or jBs \ Tv j 2.
Proof. Assume, without loss of generality, that jB1 \ Tj 2. Suppose, to the
contrary, that jBi \ Tu j 1 and jBi \ Tv j 1 for i ¼ 1; 2. Then jB1 \ Tj ¼ 2, and
we can assume that jB1 \ ðTunTv Þj ¼ jB1 \ ðTvnTu Þj ¼ 1. Obviously, jB2 \ Tj jB2 \ Tu j þ jB2 \ Tv j 2. Moreover, when jB2 \ Tj ¼ 2, we can assume that
jB2 \ ðTunTv Þj ¼ jB2 \ ðTvnTu Þj ¼ 1.
(a) jS \ Tj ¼ 1, and hence jSnTj ¼ 1.
Note that T S [ B1 [ B2 . Since jTj dC ðuÞ þ jB1 \ ðTvnTu Þj þ jB2 \
ðTvnTu Þj þ 2 þ jB2 \ ðTvnTu Þj; we get that jS \ Tj ¼ jTj jB1 \ Tj jB2 \ Tj þ jB2 \ ðTvnTu Þj jB2 \ Tj 1. Therefore, jS \ Tj ¼ 1.
(b) u; v 2 B1 .
By (a), we may assume, without loss of generality, that u 2
= S. If u 2 B2 , then
v2
= B1 . Let ui 2 B1 \ ðTunTv Þ and uj 2 B1 \ ðTvnTu Þ. Thus S \ ðuxi Cu
i Þ 6¼ ;
Þ
¼
6
;,
which
contradict
jSnTj
¼
1.
Thus
u
2
B
and
hence
and S \ ðvxj Cu
1
j
v 2 B1 [ S. If v 2 S, then SnT ¼ fvg. Thus we get jB2 \ Tu j ¼ 0, and hence
jTu j ¼ jS \ Tu j þ jB1 \ Tu j þ jB2 \ Tu j ð 1Þ þ 1 ¼ ; which contradicts
Claim 4.1. Thus v 2 B1 .
By (a), (b), and Claim 4.4, we have jB2 \ Tj ¼ 0: But in this case,
jTj dC ðuÞ þ jfuj gj þ 2 and then jS \ Tj ð þ 2Þ 2 ¼ . This contra&
dicts that jS \ Tj 1:
Claim 4.6. If ui ; uj 2 B1 \ Tu , (ui ; uj 2 B1 \ Tv , resp.), then
dC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ jB2 \ Tj dðvÞ þ þ 1;
ðdC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ jB2 \ Tj dðuÞ þ þ 1; resp:Þ:
Symmetrical statements of Claim 4.6 hold when exchanging B1 and B2 .
h
h
Proof. Set P1 ¼ ui Cu
j , P2 ¼ uj Cui , Tv ¼ Tv \ Ph and B2 ¼ B2 \ Ph ,
ðh ¼ 1; 2Þ.
By Lemma 2, we have NP1 ðui Þ \ NP1 ðuj Þ ¼ ;. By Lemma 3, we have
NP1 ðui Þ [ NP1 ðuj Þ ðP1nðB12 [ ðTv1 nfui gÞÞÞ [ P1 ðui Þ;
ð4:1Þ
where P1 ðui Þ ¼ fw 2 B12 : wþ 2 NP1 ðui Þg.
Similarly, we have
NP2 ðuj Þ [ NP2 ðui Þ ðP2nðB22 [ ðTv2 nfuj gÞÞÞ [ P2 ðuj Þ;
ð4:2Þ
CONDITIONS FOR DOMINATING CYCLES
145
where P2 ðuj Þ ¼ fw 2 B22 : wþ 2 NP2 ðuj Þg.
Set y1 ¼ ui and y2 ¼ uj . Thus by (4.1) and (4.2), we have
dC ðui Þ þ dC ðuj Þ
2
X
ðjPh j jBh2 j jTvh nfyh gj þ jBh2 \ ðTvh nfyh gÞj þ jPh ðyh ÞjÞ
h¼1
jCj jB2 \ Cj þ jB2 \ Tv j jTvnfui ; uj gj þ jP1 ðui Þj þ jP2 ðuj Þj:
ð4:3Þ
For convenience, we set P1 ¼ P1 ðui Þ and P2 ¼ P2 ðuj Þ. Now, we will prove the
following facts.
Fact 1. jP1 j þ jP2 j jS \ Cj 1. Moreover, jP1 j þ jP2 j jS \ Cj 2 if Ph 6¼
; for h ¼ 1; 2.
Proof of Fact 1. (a) By the definition of P1 and ui 2 B1 , we get that ðP1 Þþ S \ P1 , and hence jP1 j ¼ jðP1 Þþ j jS \ P1 j. Similarly, we have jP2 j jS \ P2 j.
(b) Suppose that P1 6¼ ;. Thus B2 \ P1 6¼ ;. We denote by w the first vertex of
P1 occurring in B2 . From Claim 4.0 we obtain that w 2 SnðP1 Þþ , and hence
jðS \ P1 ÞnðP1 Þþ j 1: Thus jP1 j jS \ P1 j 1. Similarly, if P2 6¼ ; then jP2 j jS \ P2 j 1.
Note that since ðGÞ 3, we have jS \ Cj 1 by Lemma 1. Thus, Fact 1
follows from (a) and (b) immediately.
&
Note that jTv j ¼ dðvÞ 1, jB2 \ Tv j jB2 \ Tj. Thus by Fact 1 and (4.3) we
may assume, without loss of generality, that
(a) jS \ Cj ¼ ; and hence S C:
(b) jP1 j ¼ jS \ Cj 1 ¼ 1 and jP2 j ¼ 0:
= Tv ,
(c) ui ; uj 2 Tv ; (otherwise, assume, without loss of generality, that ui 2
then Tvnfui ; uj g ¼ Tvnfuj g. Thus the claim holds by (4.3) and Fact 1.)
Since jTv j ¼ dðvÞ 1, by (4.3) and (a)–(c), we have
dC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ jB2 \ Tv j dðvÞ þ þ 2:
ð4:4Þ
We set P1 ¼ fw1 ; w2 ; . . . ; w1 g and their occurrence on C agrees with the
given orientation of C. Since ðP1 Þþ S, we get that jSnðP1 Þþ j ¼ 1 by (b). Since
ui 2 B1 and w1 2 B2 , there exists one and only one vertex, denoted by w0, in
þ
uþ
i Cw1 such that w0 2 SnðP1 Þ . Thus we get that
þ
þ
þ
(d) S ¼ fw0 ; wþ
1 ; w2 ; . . . ; w1 g ui Cxj :
Noting that ui ; uj 2 B1 and w1 ; w2 ; . . . ; w1 2 B2 , by (d) and Claim 4.0, we
have the following fact.
146 JOURNAL OF GRAPH THEORY
þ2
2
þ2
Fact 2. ðwþ
0 Cw1 Þ [ ð[t¼1 ðwt Cwtþ1 ÞÞ B2 and w1 Cw0 B1 , and hence
C ¼ ðB1 \ CÞ [ ðB2 \ CÞ [ S.
Fact 3. B1 \ Tu B1 \ Tv .
Proof of Fact 3. Otherwise, suppose that there exists a vertex uk 2 B1 \
ðTu nTv Þ. Then uk 6¼ ui ; uj by (c). By Lemmas 2 and 3, uk 62 NP1 ðui Þ [ NP1 ðuj Þ (or
uk 62 NP2 ðuj Þ [ NP2 ðui Þ, resp.). Hence, we can replace (4.1) (or (4.2), resp.) by
NP1 ðui Þ [ NP1 ðuj Þ ðP1nðB12 [ ðTv1 nfui gÞ [ fuk gÞÞ [ P1 ðui Þ;
ð4:10 Þ
NP2 ðuj Þ [ NP2 ðui Þ ðP2nðB22 [ ðTv2 nfuj gÞ [ fuk gÞÞ [ P2 ðuj Þ;
ð4:20 Þ
(or
resp.).
Thus applying a similar argument to (4.10 ) and (4.2), (or (4.1) and (4.20 ), resp.),
we have
dC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ jB2 \ Tv j dðvÞ þ þ 2 jfuk gj;
and hence the claim holds.
&
Fact 4. jB2 \ Tv j 1.
Proof of Fact 4. Otherwise, suppose that jB2 \ Tv j 2. Let ue ; uf 2 B2 \ Tv .
Now as the beginning of the proof, we define Ph ; Tuh ; Bh1 ; ðh ¼ 1; 2Þ and
P1 ðue Þ; P2 ðuf Þ for ue ; uf . By a similar argument, we have the following inequality
similar to (4.3):
dC ðue Þ þ dC ðuf Þ jCj jB1 \ Cj þ jB1 \ Tu j jTunfue ; uf gj
þ jP1 ðue Þj þ jP2 ðuf Þj:
Similar to Fact 1, we also have jP1 ðue Þj þ jP2 ðuf Þj jS \ Cj 1. Thus, we get
the following inequality similar to (4.4):
dC ðue Þ þ dC ðuf Þ jCj jB1 \ Cj þ jB1 \ Tu j dðuÞ þ þ 2:
Therefore,
dC ðui Þ þ dC ðuj Þ þ dC ðue Þ þ dC ðuf Þ 2jCj jB1 \ Cj jB2 \ Cj þ jB1 \ Tu j
þ jB2 \ Tv j dðvÞ dðuÞ þ 2 þ 4:
CONDITIONS FOR DOMINATING CYCLES
147
On the other hand, we have
dR ðui Þ þ dR ðuj Þ þ dR ðue Þ þ dR ðuf Þ n jCj 2:
By Fact 2 and Fact 3, we have
dðui Þ þ dðuj Þ þ dðue Þ þ dðuf Þ n þ jCj jB1 \ Cj jB2 \ Cj þ jB1 \ Tu j
þ jB2 \ Tv j dðvÞ dðuÞ þ 2 þ 2
n þ ðjCj jB1 \ Cj jB2 \ CjÞ þ ðjB1 \ Tv j
þ jB2 \ Tv jÞ dðvÞ dðuÞ þ 2 þ 2
n þ jS \ Cj þ jTv j dðvÞ dðuÞ þ 2 þ 2
¼ n þ 3 dðuÞ þ 1:
Thus, by Claim 4.1, we have that dðui Þ þ dðuj Þ þ dðue Þ þ dðuf Þ n þ 2 1; a
&
contradiction.
Fact 5. uj wþ
= EðGÞ.
1 2
Proof of Fact 5. Otherwise, suppose uj wþ
1 2 EðGÞ: In this case, we first show
þ
= EðGÞ. By Fact 2, we have w1 ; w2 2 B2 and ðxþ
that w1 w2 2
i Cui Þ [ ðxj Cuj Þ þ
þ
þ
B1 . Thus IðaÞ 6¼ fwi ; wi g for all a 2 ðxi Cui Þ [ ðxj Cuj Þ, where i ¼ 1; 2.
þ
If w1 w2 2 EðGÞ, then, by inserting the vertices ðxþ
i Cui Þ [ ðxj Cuj Þ into the
þ
þ
cycle xi Cuj w1 Cw2 w1 Cui w2 Cxj uvxi , we can have a cycle longer than C, a con= EðGÞ, and hence we have
tradiction. Thus w1 w2 2
dC ðw1 Þ jB2 \ Cj þ jSj 2:
ð4:5Þ
By (d) and Fact 2, xi 2 B1 and hence v 2 B1 . Since w1 2 B2 \ C, the set fui ; uj ;
v; w1 g is independent. By Fact 4, (4.4) and (4.5), we have dC ðui Þ þ dC ðuj Þ þ
dC ðw1 Þ þ dðvÞ jCj þ 2 þ 1: Noting that S C, we get dR ðui Þ þ dR ðuj Þ þ
dR ðw1 Þ n jCj 2: Therefore, we obtain that dðui Þ þ dðuj Þ þ dðw1 Þ þ dðvÞ &
n þ 2 1; a contradiction.
Now, we complete the proof of Claim 4.6. By Fact 2, we have that wþ2
1 2 B2 ,
þ
2
=
EðGÞ;
and
by
Fact
5,
we
get
w
2
=
N
ðu
Þ.
Thus
wþ
=
and hence ui wþ2
P1 j
1
1
1 2
þ
þ
NP1 ðui Þ [ NP1 ðuj Þ. On the other hand, since w1 2 S and ui w1 2 EðGÞ,
= B12 [ Tv . Thus, we can replace (4.1) by
wþ
1 2
NP1 ðui Þ [ NP1 ðuj Þ ðP1nðB12 [ ðTv1nfui gÞ [ fwþ
1 gÞÞ [ P1 ðui Þ:
ð4:100 Þ
Applying a similar argument to (4.100 ) and (4.2), we have
dC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ jB2 \ Tv j þ ðjP1 j þ jP2 jÞ jTv j þ 2 jfwþ
1 gj
jCj jB2 \ Cj þ jB2 \ Tj þ ð 1Þ dðvÞ þ 2
¼ jCj jB2 \ Cj þ jB2 \ Tj dðvÞ þ þ 1;
which is just that we want to prove.
&
148 JOURNAL OF GRAPH THEORY
Claim 4.7. jT nSj 2. Moreover, if jT nSj ¼ 2, then jS \ Tj ¼ 1 and
jBr \ Tj ¼ 0 for some r 2 f1; 2g.
Proof. If NC ðuÞ 6¼ NC ðvÞ, then jTj jTv j þ jTu nTv j þ 2 by Claim 4.1.
Note that jS \ Tj 1 and hence jT nSj 3.
If NC ðuÞ ¼ NC ðvÞ, then jTjð¼ jTu j ¼ jTv jÞ þ 1, and hence jT nSj 2.
When jT nSj ¼ 2, we have jTj ¼ þ 1 and jS \ Tj ¼ 1. Assume, without
loss of generality, that jB1 \ Tj 6¼ 0. Let ui 2 B1 \ T. We will show that u; v 2 B1 .
Since NC ðuÞ ¼ NC ðvÞ and jS \ Tj ¼ 1, u; v 62 S. If u 2 B2 , then
þ
S \ ðxi Cu
i Þ 6¼ ; and S \ ðui Cxiþ1 Þ 6¼ ; which contradicts to jSj ¼ . Hence
&
u 2 B1 and similarly, v 2 B1 . By Claim 4.4, we have jB2 \ Tj ¼ 0.
Claim 4.8. jBr \ Tj ¼ 0 for some r 2 f1; 2g.
Proof. If jT nSj ¼ 2, then jBr \ Tj ¼ 2 for some r 2 f1; 2g by Claim 4.7 and
thus Claim 4.8 holds. So we suppose that jT nSj 3. Thus for some r 2 f1; 2g;
jBr \ Tj 2. By Claim 4.5, we have that, for some s 2 f1; 2g; jBs \ Tu j 2
or jBs \ Tv j 2. Assume, without loss of generality, that jB1 \ Tu j 2. Let
ui ; uj 2 B1 \ Tu . We will show that jB2 \ Tj ¼ 0: Otherwise, let uk 2 B2 \ T.
Clearly dC ðuk Þ jB2 \ Cj þ jS \ Cj jB2 \ Tj; and by Claim 4.6 and jS \ Cj , dC ðui Þ þ dC ðuj Þ þ dC ðuk Þ jCj þ 2 dðvÞ þ 1: Note that dR ðui Þ þ dR ðuj Þ þ
dR ðuk Þ n 2 jCj: Therefore, dðui Þþ dðuj Þ þ dðuk Þ n 2 þ 2 dðvÞ þ 1.
Hence, dðui Þ þ dðuj Þ þ dðuk Þ þ dðvÞ n þ 2 1; a contradiction. Thus jB2 \
&
Tj ¼ 0.
Noting that for each k 3, jBk \ Tj ¼ 0, by Claims 4.8 and 4.2, we have
fu; vg [ T B1 [ S or fu; vg [ T B2 [ S.
Claim 4.9. If jBr \ Tj ¼ 0, then BrnðNðuÞ \ NðvÞÞ 6¼ ;, where r 2 f1; 2g.
Proof. Assume, without loss of generality, that jB2 \ Tj ¼ 0. Thus, we have
fu; vg [ T B1 [ S. We want to show that B2 nðNðuÞ \ NðvÞÞ 6¼ ;. Otherwise,
assume that B2 NðuÞ \ NðvÞ. Then u; v 2 S and thus jS \ Tj 2. Since
jTu j þ 1, jTv j þ 1 and T S [ B1 , jB1 \ Tu j 3 and jB1 \ Tv j 3.
Hence, we can assume that ui 2 B1 \ Tu and uj ; uk 2 B1 \ Tv . Noting that
B2 ðNðuÞ \ NðvÞÞnNðui ; uj ; uk Þ, we get dðui Þ þ dðuj Þ þ dðuk Þ n þ 1 jB2 j
by letting W ¼ B2 in Lemma 4. Let x 2 B2 , then fx; ui ; uj ; uk g is an independent
set and dðxÞ jB2 j 1 þ ; hence dðui Þ þ dðuj Þ þ dðuk Þ þ dðxÞ n þ ; a
&
contradiction.
We assume, without loss of generality, that B2 \ T ¼ ; by Claim 4.8, and
hence u; v 2 S [ B1 by Claim 4.2, and B2nðNðuÞ \ NðvÞÞ 6¼ ; by Claim 4.9.
Now we complete the proof of Theorem 5 by considering two cases.
Case 1. B2nC 6¼ ;.
Let x 2 B2nC. Since B2 \ fu; vg ¼ ;, x 2
= fu; vg. Since dR ðxÞ 1 by Lemma 1,
dðxÞ jB2 \ Cj þ jS \ Cj þ 1. By Claim 4.7, jT \ B1 j 2 and then we have that
CONDITIONS FOR DOMINATING CYCLES
149
jB1 \ Tu j 2 or jB1 \ Tu j 2 by Claim 4.5. Note that for any ui ; uj 2 B1 \ Tu or
ui ; uj 2 B1 \ Tv , dR ðui Þ þ dR ðuj Þ n 3 jCj; since x 2 B2nC.
If jB1 \ Tu j 2, then we have dC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ dðvÞ þ
1 by Claim 4.6. Thus dðui Þ þ dðuj Þ n jB2 \ Cj þ dðvÞ 2: Therefore,
we have dðui Þ þ dðuj Þ þ dðxÞ n þ 2 dðvÞ 1: That is, dðui Þ þ dðuj Þ þ
dðxÞ þ dðvÞ n þ 2 1, a contradiction to the fact that fui ; uj ; x; vg is an
independent set.
If jB1 \ Tv j 2, then we have dC ðui Þ þ dC ðuj Þ jCj jB2 \ Cj þ dðuÞ þ
1 by Claim 4.6. Thus dðui Þ þ dðuj Þ n jB2 \ Cj þ dðuÞ 2: Therefore,
we have dðui Þ þ dðuj Þ þ dðxÞ n þ 2 dðuÞ 1: That is, dðui Þ þ dðuj Þ þ
dðxÞ þ dðuÞ n þ 2 1, a contradiction to the fact that fui ; uj ; x; ug is an
independent set.
Case 2. B2nC ¼ ;.
Let x 2 B2nðNðuÞ \ NðvÞÞ. Assume, without loss of generality, that x 2
= NðuÞ.
Clearly dðxÞ jB2 j þ jSj 1 ¼ jB2 \ Cj þ jSj 1:
Since jS \ Tj 1, jTv j þ 1 and Tv S [ B1 , we have jB1 \ Tv j 2.
Assume ui ; uj 2 B1 \ Tv , then fui ; uj ; x; ug is an independent set. Note that for
ui ; uj 2 B1 \ Tv , dR ðui Þ þ dR ðuj Þ n 2 jCj: By Claim 4.6, we have dC ðui Þ þ
dC ðuj Þ jCj jB2 \ Cj þ dðuÞ þ 1: Thus dðui Þ þ dðuj Þ n jB2 \ Cj þ
dðuÞ 1: Hence dðui Þ þ dðuj Þ þ dðxÞ n þ 2 dðuÞ 2. That is,
&
dðui Þ þ dðuj Þ þ dðxÞ þ dðuÞ n þ 2 2, our final contradiction.
ACKNOWLEDGMENTS
Many thanks to the anonymous referees for their many helpful comments and
suggestions, which have considerably improved the presentation of the paper.
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