Volume 5, Issue 9, September 2015
ISSN: 2277 128X
International Journal of Advanced Research in
Computer Science and Software Engineering
Research Paper
Available online at: www.ijarcsse.com
A New Lower Bound for the Circumference of Tough Graphs
Nguyen Huu Xuan Truong*
Department of Economic Information System,
Academy of Finance, Vietnam
Vu Dinh Hoa
Department of Information Technology
HaNoi National University of Education, Vietnam
Abstract – A set of the vertices 𝑨 ⊆ 𝑽(𝑮) and not empty is called an independent set iff every vertex of it is not
adjacent to any other. We denote 𝑵𝑪𝟐 𝑮 = 𝒎𝒊𝒏⁡{ 𝑵 𝒖 ∪ 𝑵 𝒗 : 𝒅 𝒖, 𝒗 = 𝟐}, where 𝒅(𝒖, 𝒗) is length of the
shortest path joining 𝒖 and 𝒗 if 𝑮 is not complete. Moreover, we define 𝑵𝑪𝒌 𝑮 = 𝒎𝒊𝒏⁡
{| 𝒌𝒊=𝟏 𝑵 𝒗𝒊 |} and
𝝈𝒌 𝑮 = 𝒎𝒊𝒏⁡{𝒅 𝒗𝟏 + 𝒅 𝒗𝟐 + ⋯ + 𝒅 𝒗𝒌 } where {𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒌 } is an independent set of vertices in case there
exists the independent set {𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒌 }. A graph 𝑮 is called to be 1-tough if for every nonempty set 𝑺 ⊆ 𝑽(𝑮), the
number component of 𝑮 − 𝑺, denoted by 𝝎(𝑮 − 𝑺), less than or equal to |𝑺|. In this paper, we consider 1-tough graph
𝑮 with 𝒏 vertices and 𝝈𝟑 (𝑮) ≥ 𝒏. Denote by 𝒄(𝑮) the circumference (the length of a longest cycle) of 𝑮, Bauer, Fan
and Veldman proved that 𝒄(𝑮) ≥ 𝒎𝒊𝒏⁡{𝒏, 𝟐𝑵𝑪𝟐 𝑮 } . Vu Dinh Hoa strengthened this result by showing that
𝒄(𝑮) ≥ 𝒎𝒊𝒏⁡
{𝒏, 𝟐𝑵𝑪𝝈𝟑−𝒏+𝟓 + 𝟐. Our goal in this paper is to prove the conjecture posed by Bauer, Veldman and Fan
that 𝒄(𝑮) ≥ 𝒎𝒊𝒏⁡
{𝒏, 𝟐𝑵𝑪𝟐 𝑮 + 𝟒}.
Keywords – 1-tough graphs, dominating cycles, longest cycle, circumference.
I. INTRODUCTION
In this paper, we only consider finite undirected graph 𝐺 with 𝑛 ≥ 3 vertices. We use the notation and terms in [3]
and denote vertex and edge set of it by 𝑉(𝐺)and 𝐸(𝐺), respectively.
For any vertex 𝑣 ∈ 𝑉(𝐺) , let 𝑁 𝑣 and 𝑑 𝑣 = |𝑁 𝑣 | denote the set of neighbors and the degree of 𝑣 ,
respectively. Moreover, for any subset 𝑋 ⊆ 𝑉(𝐺) we define 𝑁 𝑋 = 𝑣∈𝑋 𝑁(𝑣) − 𝑋. A set of vertices 𝐴 ⊆ 𝑉(𝐺) is
called an independent set iff every vertex of it is not adjacent to any other. We use 𝛼to denote the cardinality of a
maximum independent set of vertices of 𝐺. A cycle 𝐶 of 𝐺 is called a dominating cycle, or briefly 𝐷-cycle, if
𝑉 𝐺 − 𝑉(𝐶) is an independent set of vertices in 𝐺 . We denote 𝑁𝐶 𝐺 = min⁡
{ 𝑁 𝑢 ∪ 𝑁 𝑣 : 𝑢𝑣 ∉ 𝐸 𝐺 }and
𝑁𝐶2 𝐺 = min⁡
{ 𝑁 𝑢 ∪ 𝑁 𝑣 : 𝑑 𝑢, 𝑣 = 2}, where 𝑑(𝑢, 𝑣) is length of the shortest path joining 𝑢 and 𝑣 if 𝐺 is not
complete. We set 𝑁𝐶 𝐺 = 𝑁𝐶2 𝐺 = 𝑛 − 1 if 𝐺 is complete. Moreover, we define 𝑁𝐶𝑘 𝐺 = min⁡
{| 𝑘𝑖=1 𝑁 𝑣𝑖 |}
and 𝜎𝑘 𝐺 = min⁡{𝑑 𝑣1 + ⋯ + 𝑑 𝑣𝑘 } where {𝑣1 , … , 𝑣𝑘 } is an independent set of vertices in case there exists the
independent set {𝑣1 , … , 𝑣𝑘 }, otherwise 𝑁𝐶𝑘 𝐺 = 𝑛 − 𝛼 and 𝜎𝑘 𝐺 = 𝑘(𝑛 − 𝛼). Sometimes we write 𝑁(𝑎, 𝑏) instead
of 𝑁(𝑎) ∪ 𝑁(𝑏).
A graph 𝐺 is called to be 1-tough if for every nonempty set𝑆 ⊆ 𝑉(𝐺), the number component of 𝐺 − 𝑆, denoted
by𝜔(𝐺 − 𝑆), less than or equal to|𝑆|. Two disjoint paths of𝐺 is said to be to adjacent if there is at least one vertex of a path
adjacent to an vertex of the other.Denote by 𝑐(𝐺) the circumference (the length of a longest cycle) of 𝐺, the following
bound on𝑐(𝐺) due to Bauer, Fan and Veldman
Theorem 1 (Theorem 26 in [1]).If 𝐺is 1-tough and𝜎3 ≥ 𝑛, then𝑐(𝐺) ≥ min⁡
{𝑛, 2𝑁𝐶2 𝐺 }.
Vu Dinh Hoa strengthened this result in [4].
Theorem 2 (Theorem 2 in [4]).If𝐺 is a 1-tough graph with𝜎3 ≥ 𝑛 ≥ 3, then𝑐(𝐺) ≥ min⁡
{𝑛, 2𝑁𝐶𝜎3 −𝑛+5 + 2}.
Bauer, Veldman and Fan also conjectured :
Conjecture 1 (Conjecture 27 in [1]).If 𝐺is 1-tough and𝜎3 ≥ 𝑛, then𝑐(𝐺) ≥ min⁡
{𝑛, 2𝑁𝐶2 𝐺 + 4}.
The conjecture remains unsolved even some people have tried to prove it (see [9] ...). The proof of Tri Lai [9] is not
correct, for example the proof for the Proposition 26, and the definition of bad path is not strong enough. In the mean
time, a lot of new bounds for the length of longest cycle was found. For examples:
3 𝑁𝐶+1
Theorem 3 (Theorem 3 in [5]).Let𝐺 be a graph of order𝑛. If 𝐺is 3-connected then𝑐(𝐺) ≥ min⁡
{𝑛,
}; if𝐺 is
2
4-connected, then𝑐(𝐺) ≥ min⁡{𝑛, 2𝑁𝐶}. Furthermore, these bounds are both sharp.
Theorem 4 (Theorem 2 in [8]).If 𝐺 is 3-connected𝐾1,3 -free graph of order𝑛 such that𝑁𝐶2 ≥ (2𝑛 − 6)/3 then𝐺 is
hamiltonian.
Theorem 5 (Theorem 4 in [6]) Let𝐺 be a 2-connected graph such thatmax⁡
{𝑑 𝑢 , 𝑑 𝑣 } ≥ 𝑐/2 for each pair of
nonadjacent vertices𝑢 and𝑣 in an induded claw, and|𝑁 𝑥 ∩ 𝑁 𝑦 | ≥ 2 for each pair of nonadjacent vertices𝑥 and 𝑦in
an induced modified claw. Then𝐺 contains either Hamiltonian cycle or a cycle of length at least𝑐.
Our goal in this paper is to prove the conjecture posed by Bauer, Veldman and Fan in [1].
© 2015, IJARCSSE All Rights Reserved
Page | 643
Truong et al., International Journal of Advanced Research in Computer Science and Software Engineering 5(9),
September- 2015, pp. 643-649
II. PRELIMINARIES
For what follows, we assume that𝐺 is1-tough with𝜎3 ≥ 𝑛 and nonhamiltonian. For a longest cycle𝐶, we denote
by𝜇 𝐶 = max⁡
{𝑑 𝑣 : 𝑣 ∉ 𝐶} and 𝜇 𝐺 = max⁡{𝜇 𝐶 : 𝑙 𝐶 = 𝑐 𝐺 } where 𝑙 𝐶 is the length of 𝐶 . We consider a
longest cycle𝐶 and a vertex𝑣 ∉ 𝐶 such that𝜇 𝐶 = 𝜇(𝐺) and 𝑑 𝑣 = 𝜇(𝐶). On𝐶 (𝐶 with a given orientation), we
denote the predecessor and successor of𝑥 ∈ 𝐶 (along the direction of𝐶) by𝑥 − , 𝑥 + and𝑥 ++ = (𝑥 + )+ , similarly, 𝑥 −− =
(𝑥 −)− . In general, 𝑥 +𝑖 = (𝑥 + 𝑖−1 )+ and 𝑥 −𝑖 = (𝑥 − 𝑖−1 )− . Let𝑆 = 𝑁(𝑣)+ ∩ 𝑁(𝑣)− . The arc joining two vertices𝑥 and 𝑦
of𝐶, along the direction of𝐶 , is denoted by𝑥𝐶 𝑦. Similarly, the arc joining two vertices𝑥 and 𝑦 of𝐶, along the direction
of𝐶 , is denoted by𝑥𝐶 𝑦. Moreover, for any𝐴 ⊆ 𝑉(𝐺), we write𝐴+ = {𝑥 + : 𝑥 ∈ 𝐴} and𝐴− = {𝑥 − : 𝑥 ∈ 𝐴}.We begin with
following lemmas:
Lemma 1 (Theorem 5 in [2]).Every longest cycle𝐶 is a 𝐷-cycle.
𝜎
Lemma 2 (Proof of Theorem 9 in [2]). 𝑑(𝑣) ≥ 3 .
3
Lemma 3 (Lemma 9 in [4]). 𝑆 ≥ 𝜎3 − 𝑛 + 4.
For what follows we use a lemma of Woodall, sometimes called Hopping Lemma, in [7]:
Lemma 4.Let𝐶 be a cycle of length𝑚 in a graph𝐺. Assume that𝐺 contains no cycle of length𝑚 + 1 and no cycle𝐶 ∗ of
length𝑚 with𝜔 𝐺 − 𝐶 ∗ < 𝜔(𝐺 − 𝐶) and𝑣 is an isolated vertices of𝐺 − 𝐶. Set𝑌0 = ∅ and for𝑖 > 1:
𝑋𝑖+1 = 𝑁(𝑌𝑖 ∪ 𝑣 ),
𝑌𝑖+1 = 𝑋𝑖 ∩ 𝑉(𝐶) + ∩ 𝑋𝑖 ∩ 𝑉(𝐶) −.
Set𝑋 = ∞𝑖=1 𝑋𝑖 and𝑌 = ∞𝑖=1 𝑌𝑖 . Then
(a) 𝑋 ⊆ 𝑉(𝐶),
(b) If𝑥1 , 𝑥2 ∈ 𝑋, then𝑥1+ ≠ 𝑥2 .
(c)𝑋 ∩ 𝑌 = ∅.
By (a), (b) and (c) of Lemma 4, respectively, we easily conclude that:
Corollary 1.Assume that𝐺, 𝐶, 𝑋, 𝑌 satisfy the hypothesis of Lemma 4, then:
(a) There are no edges joining a vertex of𝑌 with a vertex of𝐺 − 𝐶.
(b)𝑋 ∩ 𝑋 + = ∅, and therefore𝑙(𝐶) ≥ 2|𝑋|.
(c) 𝑌is an independent set of vertices.
Assume that𝑋, 𝑌 are the vertex sets mentioned in the assumption of Lemma 4. First, we fixed a vertex𝑠0 ∈ 𝑆 and denote
the vertices of𝑋 by 𝑥1 , 𝑥2 , … , 𝑥𝑚 (𝑚 = 𝑋 ), occurring consecutively on𝐶 such that𝑥1 = 𝑠0+ and𝑥𝑚 = 𝑠0− . For1 ≤ 𝑖 ≤
−
𝑚 − 1 we write𝑢𝑖 = 𝑥𝑖+ and𝑤𝑖 = 𝑥𝑖+1
. By Lemma 4,𝑥𝑖+ ≠ 𝑥𝑖+1 and therefore𝐶 − 𝑋 is in fact the union of |𝑋| disjoint
paths𝑢𝑖 𝐶 𝑤𝑖 (𝑖 = 1, 𝑚) on𝐶, calling arcs. An arc with𝑝 vertices is called a 𝑝-arc.𝑢𝑖 𝐶 𝑤𝑖 is an 1-arc if and only if
𝑢𝑖 ≡ 𝑤𝑖 ∈ 𝑌. The next lemma can be proved using Corollary 1 and Lemma 1.
Fig.1 Illustrating the vertices on cycle 𝐶.
Lemma 5.
(a)𝑌 ∪ 𝑉(𝐺 − 𝐶) is an independent set of vertices.
(b)𝑙 𝐶 ≥ 2 𝑋 + 2.
Proof.
(a) By (a) of Corollary 1, there is no edge joining a vertex of𝑌 with a vertex of𝑉(𝐺 − 𝐶). By (c) of Corollary 1,𝑌 is an
independent set of vertices. By Lemma 1, 𝑉(𝐺 − 𝐶) is an independent set of vertices. Thus, 𝑌 ∪ 𝑉(𝐺 − 𝐶) is an
independent set of vertices.
(b) By (b) of Corollary 1,𝑙(𝐶) ≥ 2|𝑋|. If𝑙 𝐶 ≤ 2 𝑋 + 1 then𝐶 − 𝑋 is an union of|𝑋| 1-arcs or an union of one 2-arc and
( 𝑋 − 1) 1-arcs. The 1-arcs of𝐶 − 𝑋 are in fact the vertices of 𝑌. By (a), the graph𝐺 − 𝑋 contains at least 𝑋 + 1
components, a contradiction to the toughness of 𝐺.
By the toughness of𝐺, by 𝑋 = 𝑁(𝑌) and by (a) of Lemma 5, we easily conclude that:
Corollary 2.𝐶 − 𝑋 contains at least two arcs with length≥ 2.
By𝑑 𝑣, 𝑠 = 2 for any𝑠 ∈ 𝑆 and by𝑁(𝑠) ∪ 𝑁(𝑣) ⊆ 𝑋, 𝑁𝐶2 ≤ |𝑁 𝑠 ∪ 𝑁 𝑣 | ≤ |𝑋|. If𝑁𝐶2 ≠ |𝑋|then by Lemma
5,𝑐 𝐺 ≥ 2 𝑋 + 2 ≥ 2𝑁𝐶2 + 4 and Conjecture 1 is proved. For what follows we assume that 𝑁𝐶2 = |𝑋| and𝑐 𝐺 ≤
2 𝑋 + 3 and will show that it leads to a contradiction. By𝑁𝐶2 = |𝑋|, 𝑋 = 𝑁(𝑠) ∪ 𝑁(𝑣) for any 𝑠 ∈ 𝑆 and we conclude
from Lemma 4 in [4] that:
Lemma 6.𝑉(𝐺 − 𝐶) ∪ 𝑋 +is an independent set. Similarly, 𝑉(𝐺 − 𝐶) ∪ 𝑋 −is an independent set.
© 2015, IJARCSSE All Rights Reserved
Page | 644
Truong et al., International Journal of Advanced Research in Computer Science and Software Engineering 5(9),
September- 2015, pp. 643-649
The following definition will help us to get shorter proofs.
Definition 1.A path𝑃 in𝐺 is called a bad-path if it has one of the following two forms:
1. 𝑃consists of all vertices in𝑥1 𝐶 𝑤𝑗 −1 , and the ends of𝑃 are𝑥𝑘 ∈ 𝑁(𝑣) ∩ 𝑁(𝑠0 ) and𝑥𝑖 ∈ 𝑋 for1 ≤ 𝑘, 𝑖 < 𝑗 ≤ 𝑚 and
𝑖 ≠ 𝑘.
2.𝑃 consists of all vertices in 𝑥𝑚 𝐶 𝑢𝑖+1 , and the ends of 𝑃 are 𝑥𝑘 ∈ 𝑁(𝑣) ∩ 𝑁(𝑠0 ) and 𝑥𝑗 ∈ 𝑋 for 1 ≤ 𝑖 < 𝑗, 𝑘 ≤ 𝑚
and 𝑗 ≠ 𝑘.
Lemma 7.There are no bad-paths in 𝐺.
Proof. Assume the contrary that𝐺 has a bad-path𝑃. Without loss of generality, we assume that𝑃 is a bad-path of form 1.
We will show a cycle𝐶′ is longer than 𝐶, then get a contradiction. As remark before that𝑋 = 𝑁(𝑠) ∪ 𝑁(𝑣) for any𝑠 ∈ 𝑆
and specially𝑋 = 𝑁(𝑠0 ) ∪ 𝑁(𝑣). Therefore, there are only four possible cases as follows. In each case we will get a
contradiction by constructing a cycle𝐶′ which is longer than 𝐶.
Case 1:𝑥𝑖 , 𝑥𝑗 ∈ 𝑁(𝑠0 ), then𝐶 ′ = (𝑥𝑘 𝑃𝑥𝑖 𝑠0 𝑥𝑗 𝐶 𝑥𝑚 𝑣𝑥𝑘 ).
Case 2: 𝑥𝑖 , 𝑥𝑗 ∈ 𝑁(𝑣), then𝐶 ′ = (𝑥𝑘 𝑃𝑥𝑖 𝑣𝑥𝑗 𝐶 𝑠0 𝑥𝑘 ).
Case 3:𝑥𝑖 ∈ 𝑁 𝑠0 , 𝑥𝑗 ∈ 𝑁(𝑣), then𝐶 ′ = (𝑥𝑘 𝑃𝑥𝑖 𝑠0 𝐶 𝑥𝑗 𝑣𝑥𝑘 ).
Case 4: 𝑥𝑖 ∈ 𝑁 𝑣 , 𝑥𝑗 ∈ 𝑁(𝑠0 ), then𝐶 ′ = (𝑥𝑘 𝑃𝑥𝑖 𝑣𝑥𝑚 𝐶 𝑥𝑗 𝑠0 𝑥𝑘 ).
Lemma 8.If𝑤𝑖 𝑢𝑗 ∈ 𝐸(𝐺) for1 ≤ 𝑖 < 𝑗 < 𝑚 then𝑥𝑖+1 , 𝑥𝑗 ∈ 𝑁 𝑣 − 𝑁(𝑠0 ) or 𝑥𝑖+1 , 𝑥𝑗 ∈ 𝑁 𝑠0 − 𝑁(𝑣).
Proof. Assume to the contrary that 𝑥𝑖+1 ∈ 𝑁 𝑠0 , 𝑥𝑗 ∈ 𝑁(𝑣) or 𝑥𝑖+1 ∈ 𝑁 𝑣 , 𝑥𝑗 ∈ 𝑁(𝑠0 ) then we have a cycle 𝐶 ′ =
(𝑣𝑥𝑗 𝐶 𝑥𝑖+1 𝑠0 𝐶 𝑤𝑖 𝑢𝑗 𝐶 𝑥𝑚 𝑣) or𝐶 ′ = (𝑣𝑥1 𝐶 𝑤𝑖 𝑢𝑗 𝐶 𝑠0 𝑥𝑗 𝐶 𝑥𝑖+1 𝑣) respectively, is longer than 𝐶, a contradiction.
Lemma 9.If𝑤𝑖 𝑢𝑗 ∈ 𝐸(𝐺) for1 ≤ 𝑖 < 𝑗 < 𝑚 then𝑥𝑖+1 , 𝑥𝑗 ∉ 𝑁(𝑢1 ) ∪ 𝑁(𝑤𝑚 −1 ).
Proof.
If
𝑥𝑖+1 ∈ 𝑁(𝑢1 ) then 𝑃 = (𝑥𝑚 𝐶 𝑢𝑗 𝑤𝑖 𝐶 𝑢1 𝑥𝑖+1 𝐶 𝑥𝑗 ) is
a
bad-path
of
form
2,
and
if𝑥𝑗 ∈ 𝑁(𝑢1 )then𝑃 = (𝑥𝑚 𝐶 𝑢𝑗 𝑤𝑖 𝐶 𝑢1 𝑥𝑗 𝐶 𝑥𝑖+1 ) is also a bad-path of form 2, which contradicts to Lemma 7. Thus, we have
𝑥𝑖+1 , 𝑥𝑗 ∉ 𝑁(𝑢1 ). Similarly, by reversing the direction of𝐶, we have𝑥𝑖+1 , 𝑥𝑗 ∉ 𝑁(𝑤𝑚 −1 ).
Lemma 10.Assume that 𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺) for some 1 ≤ 𝑖 < 𝑗 < 𝑚 . Then 𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑢𝑘 ) for every 𝑖 < 𝑘 ≤ 𝑗 .
Analogously,𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑤𝑘 ) for every𝑖 ≤ 𝑘 < 𝑗.
Proof. If𝑥𝑖 𝑢𝑘 ∈ 𝐸(𝐺) for some𝑖 < 𝑘 ≤ 𝑗 then 𝑃 = (𝑥1 𝐶 𝑥𝑖 𝑢𝑘 𝐶 𝑤𝑗 𝑢𝑖 𝐶 𝑥𝑘 ) is a bad-path, which contradicts to Lemma 7.
If 𝑥𝑗 +1 𝑢𝑘 ∈ 𝐸(𝐺) for some 𝑖 < 𝑘 ≤ 𝑗 then 𝑃 = (𝑥𝑚 𝐶 𝑥𝑗 +1 𝑢𝑘 𝐶 𝑤𝑗 𝑢𝑖 𝐶 𝑥𝑘 ) is a bad-path, a contradiction. Thus, we
have𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑢𝑘 )for every 𝑖 < 𝑘 ≤ 𝑗. Similarly, by reversing the direction of𝐶, we have𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑤𝑘 ) for every
𝑖 ≤ 𝑘 < 𝑗.
Lemma 11.Assume that 𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺) for some1 ≤ 𝑖 < 𝑗 < 𝑚. If there exists a vertex𝑧 ∈ 𝑢𝑖+ 𝐶 𝑤𝑗− such that𝑧𝑢1 ∈ 𝐸(𝐺)
and𝑖 > 1 then𝑧 + , 𝑧 − ∉ 𝑁(𝑥1 ). Analogously, if there exists a vertex𝑧 ∈ 𝑢𝑖+ 𝐶 𝑤𝑗− such that 𝑧𝑤𝑚 −1 ∈ 𝐸(𝐺) and𝑗 < 𝑚 − 1
then𝑧 + , 𝑧 − ∉ 𝑁(𝑥𝑚 ).
Proof.Assume that 𝑧 is vertex in 𝑢𝑖+ 𝐶 𝑤𝑗− such that 𝑧𝑢1 ∈ 𝐸 𝐺 and 𝑖 > 1 . If 𝑧 − 𝑥1 ∈ 𝐸 𝐺 then we have
𝑃 = 𝑥1 𝑧 − 𝐶 𝑢𝑖 𝑤𝑗 𝐶 𝑧𝑢1 𝐶 𝑥𝑖
is
a
bad-path,
which
contradicts
to
Lemma
7.
If𝑧 + 𝑥1 ∈ 𝐸 𝐺 then𝑃 = (𝑥1 𝑧 + 𝐶 𝑤𝑗 𝑢𝑖 𝐶 𝑧𝑢1 𝐶 𝑥𝑖 ) is a bad-path, a contradiction. Thus, we have𝑧 + , 𝑧 − ∉ 𝑁(𝑥1 ). Similarly,
if𝑧𝑤𝑚 −1 ∈ 𝐸(𝐺)and𝑗 < 𝑚 − 1then𝑧 + , 𝑧 − ∉ 𝑁(𝑥𝑚 ).
Lemma 12.Assume that 𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺) for some1 ≤ 𝑖 < 𝑗 < 𝑚. If there exists a vertex𝑥𝑘 ∈ 𝑢𝑖 𝐶 𝑤𝑗 such that𝑥𝑘 ∈ 𝑁(𝑣) ∩
𝑁(𝑠0 ) then𝑥1 , 𝑥𝑚 ∉ 𝑁(𝑤𝑘−1 ) ∪ 𝑁(𝑢𝑘 ).
Proof. Let 𝑥𝑘 be a vertex in𝑁(𝑣) ∩ 𝑁(𝑠0 ) ∩ 𝑢𝑖 𝐶 𝑤𝑗 . If𝑥1 𝑤𝑘 −1 ∈ 𝐸(𝐺) then𝑃 = (𝑥𝑘 𝐶 𝑤𝑗 𝑢𝑖 𝐶 𝑤𝑘 −1 𝑥1 𝐶 𝑥𝑖 ) is a bad-path,
which contradicts to Lemma 7. If𝑥𝑚 𝑤𝑘 −1 ∈ 𝐸(𝐺) then𝑃 = (𝑥𝑘 𝐶 𝑤𝑗 𝑢𝑖 𝐶 𝑤𝑘 −1 𝑥𝑚 𝐶 𝑥𝑗 +1 )is a bad-path, a contradiction.
Thus, 𝑥1 , 𝑥𝑚 ∉ 𝑁(𝑤𝑘−1 ). Similarly, we have𝑥1 , 𝑥𝑚 ∉ 𝑁(𝑢𝑘 ).
Lemma 13.Assume that𝐶 has a 𝑝-arc𝑢𝑖 𝐶 𝑤𝑖 and a 𝑞-arc𝑢𝑗 𝐶 𝑤𝑗 with𝑚𝑖𝑛 𝑝, 𝑞 = 2, 𝑖 ≠ 𝑗. Then{𝑢𝑖 𝑤𝑗 , 𝑢𝑗 𝑤𝑖 } ⊈ 𝐸(𝐺).
Proof. Assume to the contrary that𝑢𝑖 𝑤𝑗 , 𝑢𝑗 𝑤𝑖 ∈ 𝐸(𝐺). Without loss of generality, we assume that𝑝 = 2, then we have𝑃 =
(𝑥1 𝐶 𝑢𝑖 𝑤𝑗 𝐶 𝑢𝑗 𝑤𝑖 𝐶 𝑥𝑗 ) is a bad-path, which contradicts to Lemma 7.
Lemma 14. Assume that 𝐶 has a 𝑝-arc𝑢𝑖 𝐶 𝑤𝑖 and a 𝑞-arc𝑢𝑗 𝐶 𝑤𝑗 with𝑝, 𝑞 ≥ 2, 𝑖 < 𝑗, such that𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺) and there
are no 𝑘-arcs (𝑘 ≥ 2) on𝑥𝑖+1 𝐶 𝑥𝑗 . If𝑁 𝑤𝑖 ⊆ 𝑋 ∪ {𝑤𝑖− } or 𝑁(𝑢𝑗 ) ⊆ 𝑋 ∪ {𝑢𝑗+ }, then𝑥𝑖+1 ≡ 𝑥𝑗 .
Proof. Assume to the contrary that𝑥𝑖+1 ≠ 𝑥𝑗 . Without loss of generality, we assume that𝑁 𝑤𝑖 ⊆ 𝑋 ∪ {𝑤𝑖− }. Since there
are no 𝑘-arcs (𝑘 ≥ 2) on𝑥𝑖+1 𝐶 𝑥𝑗 ,𝑢𝑖+1 ∈ 𝑌. By Lemma 6, 𝑑 𝑤𝑖 , 𝑢𝑖+1 = 2. By Lemma 10 for𝑘 = 𝑖 + 1, we have
𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁 𝑤𝑖 , 𝑢𝑖+1 .
Thus,
by
Lemma
6
we
have𝑁 𝑤𝑖 , 𝑢𝑖+1 ⊆ 𝑋 ∪ 𝑤𝑖− − {𝑥𝑖 , 𝑥𝑗 +1 }.Therefore, 𝑋 − 1 ≥ 𝑁 𝑤𝑖 , 𝑢𝑖+1 ≥ 𝑁𝐶2 𝐺 = |𝑋|, a contradiction.
Lemma 15.Assume that𝐶 has 𝑝-arc𝑢𝑖 𝐶 𝑤𝑖 and a 𝑞-arc𝑢𝑗 𝐶 𝑤𝑗 with𝑝 ≥ 2, 𝑞 ≥ 2, 𝑖 < 𝑗 such that there are no𝑘-arcs
(𝑘 ≥ 2) on𝑥1 𝐶 𝑥𝑖 ∪ 𝑥𝑖+1 𝐶 𝑥𝑚 . If𝑁(𝑢𝑖 ) ⊆ 𝑋 ∪ {𝑢𝑖+} or𝑁 𝑤𝑗 ⊆ 𝑋 ∪ {𝑤𝑗− } then𝑤𝑖 𝑢𝑗 ∉ 𝐸(𝐺).
Proof. Assume to the contrary that 𝑤𝑖 𝑢𝑗 ∈ 𝐸(𝐺). Without loss of generality, we assume that𝑁 𝑤𝑗 ⊆ 𝑋 ∪ {𝑤𝑗− }. By
Lemma 8, we have 𝑥𝑖+1 , 𝑥𝑗 ∈ 𝑁 𝑣 − 𝑁(𝑠0 ) or 𝑥𝑖+1 , 𝑥𝑗 ∈ 𝑁 𝑠0 − 𝑁(𝑣). Let𝑦 be a vertex in{𝑣, 𝑠0 } such that 𝑥𝑖+1 , 𝑥𝑗 ∉
© 2015, IJARCSSE All Rights Reserved
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𝑁(𝑦). Clearly,𝑑 𝑦, 𝑢1 = 𝑑 𝑦, 𝑤𝑚 −1 = 2, so 𝑁 𝑦, 𝑢1 , 𝑁 𝑦, 𝑤𝑚 −1 ≥ 𝑁𝐶2 𝐺 = |𝑋|. If 𝑥𝑖 ≠ 𝑥1 , then𝑢1 ∈ 𝑌. By
Lemma 9, we have𝑥𝑖+1 , 𝑥𝑗 ∉ 𝑁(𝑢1 ), and therefore𝑁 𝑦, 𝑢1 ⊆ 𝑋 − {𝑥𝑖+1 , 𝑥𝑗 }, a contradiction. Thus,𝑥𝑖 ≡ 𝑥1 . Similarly, we
have 𝑥𝑗 +1 ≡ 𝑥𝑚 and𝑤𝑗 ≡ 𝑤𝑚 −1 .By Lemma 9, we have𝑥𝑖+1 , 𝑥𝑗 ∉ 𝑁 𝑤𝑚 −1 ,so 𝑁 𝑦, 𝑤𝑚 −1 ⊆ 𝑋 ∪ 𝑤𝑚− −1 − {𝑥𝑖+1 , 𝑥𝑗 }. If
𝑥𝑖+1 ≠ 𝑥𝑗 , then 𝑁 𝑦, 𝑤𝑚 −1 ≤ 𝑋 ∪ 𝑤𝑚− −1 − 𝑥𝑖+1 , 𝑥𝑗 ≤ 𝑋 − 1 , a contradiction. Thus 𝑥𝑖+1 ≡ 𝑥𝑗 , and therefore
𝑆 = 1, which contradicts to Lemma 3 that 𝑆 ≥ 𝜎3 − 𝑛 + 4 ≥ 4.
By 𝑐 𝐺 ≤ 2 𝑋 + 3 and by (b) of Lemma 5, either𝑐 𝐺 = 2 𝑋 + 3 or 𝑐 𝐺 = 2 𝑋 + 2. We consider two cases:
Case 1.𝑐 𝐺 = 2 𝑋 + 2. In this case, by Corollary 2,𝐶 − 𝑋 contains two 2-arc and some 1-arcs.
Case 2.𝑐 𝐺 = 2 𝑋 + 3. In this case, by Corollary 2, there are only two possible cases:
Case 2.1.𝐶 − 𝑋 contains one 2-arc, one 3-arc and some 1-arcs.
Case 2.2.𝐶 − 𝑋 contains three 2-arc and some 1-arcs.
III. PROOF FOR THE CASES
A. Case 1.𝒄 𝑮 = 𝟐 𝑿 + 𝟐, 𝑪 − 𝑿contains two 2-arc and some 1-arcs.
Assume that the two 2-arcs are𝑢𝑖 𝑤𝑖 and 𝑢𝑗 𝑤𝑗 with 1 ≤ 𝑖 < 𝑗 < 𝑚. We have the following Claims (from Claim 1 to
Claim 4):
Claim 1.𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺)and𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺).
Proof. Assume that𝑢𝑖 𝑤𝑗 ∉ 𝐸(𝐺), then𝑢𝑗 𝑤𝑖 ∈ 𝐸(𝐺), otherwise, by Lemma 6,𝜔 𝐺 − 𝑋 ≥ 𝑋 + 1, which contradicts
the toughness of𝐺.By Lemma 6,𝑁 𝑢𝑖 ⊆ 𝑋 ∪ 𝑤𝑖 = 𝑋 ∪ 𝑢𝑖+ , and by Lemma 15,𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺), a contradiction.
Thus,𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺), and by Lemma 13,𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺).
By Lemma 6, Lemma 14 and Claim 1, we get:
Claim 2.𝑥𝑖+1 ≡ 𝑥𝑗 .
Note: by Claim 1 and Claim 2,𝑑 𝑤𝑖 , 𝑢𝑗 = 2. By Lemma 10, we have𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑤𝑖 , 𝑢𝑗 ).
Claim 3.𝑥𝑗 ∉ 𝑁(𝑣) ∩ 𝑁(𝑠0 ).
Proof. Assume the contrary that𝑥𝑗 ∈ 𝑁(𝑣) ∩ 𝑁(𝑠0 ). By Lemma 12,𝑥1 , 𝑥𝑚 ∉ 𝑁(𝑤𝑖 , 𝑢𝑗 ). By Lemma 6, 𝑁 𝑤𝑖 , 𝑢𝑗 ⊆ 𝑋 ∪
𝑢𝑖 , 𝑤𝑗 − {𝑥1 , 𝑥𝑚 , 𝑥𝑖 , 𝑥𝑗 +1 } . By 𝑁 𝑤𝑖 , 𝑢𝑗 = 𝑁 𝑤𝑖 ∪ 𝑁 𝑢𝑗 ≥ 𝑁𝐶2 𝐺 = |𝑋| , we have 𝑥𝑖 ≡ 𝑥1 and 𝑥𝑗 +1 ≡ 𝑥𝑚 .
Thus, 𝑆 = 1, which contradicts to Lemma 3 that 𝑆 ≥ 𝜎3 − 𝑛 + 4 ≥ 4.
Claim 4.𝑥𝑖 ≡ 𝑥1 and𝑥𝑗 +1 ≡ 𝑥𝑚 .
Proof. Assume that𝑥𝑖 ≠ 𝑥1 , then𝑢1 ∈ 𝑌. By Claim 3, there exists 𝑦 ∈ {𝑣, 𝑠0 } such that𝑥𝑗 𝑦 ∉ 𝐸(𝐺). Clearly,𝑑 𝑢1 , 𝑦 =
2 . If 𝑢1 𝑥𝑗 ∉ 𝐸(𝐺) , then 𝑁 𝑢1 , 𝑦 ⊆ 𝑋 − {𝑥𝑗 } , and therefore 𝑁 𝑢1 , 𝑦 ≤ 𝑋 − 1 , which contradicts 𝑁 𝑢1 , 𝑦 ≥
𝑁𝐶2 𝐺 = |𝑋|. Thus, 𝑢1 𝑥𝑗 ∈ 𝐸(𝐺). By Lemma 11,𝑥1 ∉ 𝑁 𝑤𝑖 , 𝑢𝑗 , and by Lemma 6, 𝑁 𝑤𝑖 , 𝑢𝑗 ⊆ 𝑋 ∪ {𝑢𝑖 , 𝑤𝑗 } −
{𝑥1 , 𝑥𝑖 , 𝑥𝑗 +1 }. Therefore, 𝑁 𝑤𝑖 , 𝑢𝑗 = 𝑋 − 1, which contradicts 𝑑 𝑤𝑖 , 𝑢𝑗 = 2 as noted above.
Thus 𝑥𝑖 ≡ 𝑥1 . Similarly, we have 𝑥𝑗 +1 ≡ 𝑥𝑚 .
By Claim 2 and Claim 4, 𝑆 = 1, which contradicts Lemma 3. So, the Case 1 does not happen.
B. Case 2.𝒄 𝑮 = 𝟐 𝑿 + 𝟑. In this case, there are only two possible cases:
Case 2.1. 𝐶 − 𝑋 contains one 2-arc, one 3-arc and some 1-arcs.
Assume that the 3-arc is𝑢𝑖 𝑢𝑖+𝑤𝑖 and the 2-arc is 𝑢𝑗 𝑤𝑗 . Without loss of generality, we assume that𝑖 < 𝑗. We have two
following Propositions:
Proposition 1. 𝑢𝑖 𝑤𝑖 ∉ 𝐸(𝐺).
Proof. Assume to the contrary that𝑢𝑖 𝑤𝑖 ∈ 𝐸(𝐺). We have the following Claims (from Claim 5 to Claim 10):
Claim 5.𝑢𝑖+ 𝑢𝑗 , 𝑢𝑖+ 𝑤𝑗 ∉ 𝐸(𝐺).
Proof. If𝑢𝑖+𝑢𝑗 ∈ 𝐸(𝐺) then𝑃 = (𝑥𝑚 𝐶 𝑢𝑗 𝑢𝑖+ 𝑢𝑖 𝑤𝑖 𝐶 𝑥𝑗 ) is a bad-path, which contradicts Lemma 7. If 𝑢𝑖+ 𝑤𝑗 ∈ 𝐸(𝐺)
then𝑃 = (𝑥1 𝐶 𝑢𝑖 𝑤𝑖 𝑢𝑖+ 𝑤𝑗 𝐶 𝑥𝑖+1 ) is a bad-path, which contradicts Lemma 7.
Claim 6.𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺)and𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺).Moreover,𝑥𝑖+1 ≡ 𝑥𝑗 .
Proof. If 𝑢𝑖 𝑤𝑗 , 𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺) , then by Lemma 6 and by Claim 5, 𝜔 𝐺 − 𝑋 ≥ 𝑋 + 1 , so 𝐺 is not 1-tough, a
contradiction.
If𝑢𝑗 𝑤𝑖 ∈ 𝐸(𝐺), then by Lemma 13,𝑢𝑖 𝑤𝑗 ∉ 𝐸(𝐺). Therefore, by Lemma 6 and by Claim 5,𝑁 𝑤𝑗 ⊆ 𝑋 ∪ {𝑢𝑗 }. By
Lemma 15 we have𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺), a contradiction.
Thus,𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺). By the toughness of 𝐺, by Claim 5 and by Lemma 6,𝑢𝑖 𝑤𝑗 ∈ 𝐸(𝐺).By𝑢𝑗 𝑤𝑖 ∉ 𝐸(𝐺) and by Lemma
6, 𝑁 𝑢𝑗 ⊆ 𝑋 ∪ {𝑤𝑗 }. By Lemma 14,𝑥𝑖+1 ≡ 𝑥𝑗 .
Claim 7.𝑑 𝑤𝑖 , 𝑢𝑗 = 2 and𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑤𝑖 , 𝑢𝑗 ).
Proof. By Claim 6,𝑑 𝑤𝑖 , 𝑢𝑗 = 2. By Lemma 10,𝑥𝑖 , 𝑥𝑗 +1 ∉ 𝑁(𝑤𝑖 , 𝑢𝑗 ).
Claim 8.𝑥𝑗 ∉ 𝑁(𝑣) ∩ 𝑁(𝑠0 ).
Proof. Assume to the contrary that𝑥𝑗 ∈ 𝑁(𝑣) ∩ 𝑁(𝑠0 ). Since Lemma 12 we have𝑥1 , 𝑥𝑚 ∉ 𝑁(𝑤𝑖 , 𝑢𝑗 ).By Lemma 6 and
by Claim 7, 𝑁 𝑤𝑖 , 𝑢𝑗 ⊆ 𝑋 ∪ {𝑢𝑖 , 𝑢𝑖+ , 𝑤𝑗 } − {𝑥1 , 𝑥𝑚 , 𝑥𝑖 , 𝑥𝑗 +1 } . Because 𝑁 𝑤𝑖 , 𝑢𝑗 ≥ 𝑁𝐶2 𝐺 = |𝑋| , so 𝑥𝑖 ≡
𝑥1 or𝑥 𝑗 +1 ≡ 𝑥 𝑚.
If 𝑥 𝑖 ≡ 𝑥 1 , then𝑥 𝑗 +1 ≠ 𝑥 𝑚 because of|𝑆 | ≥ 4, and there are at least three 1-arcs on𝑥 𝑗 +1 𝐶 𝑥 𝑚. Since𝑁 𝑤 𝑖 , 𝑢 𝑗 ⊆
© 2015, IJARCSSE All Rights Reserved
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𝑋 ∪ {𝑢 𝑖 , 𝑢 +𝑖 , 𝑤 𝑗 } − {𝑥 1 , 𝑥 𝑚, 𝑥 𝑗 +1 }, so every vertex𝑥 𝑘 (𝑗 + 1 < 𝑘 < 𝑚) ∈ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ), and we observe that𝑢 𝑘 ∉
𝑁 (𝑥 𝑗 +1 ). Indeed, suppose otherwise that𝑢 𝑘 𝑥 𝑗 +1 ∈ 𝐸 (𝐺 ), by Claim 6 we have:
(1) If 𝑥 𝑘 𝑤 𝑖 ∈ 𝐸 (𝐺 ) then𝑃 = (𝑥 𝑚𝐶 𝑢 𝑘 𝑥 𝑗 +1 𝐶 𝑥 𝑘 𝑤 𝑖 𝑢𝑖+ 𝑢 𝑖 𝑤 𝑗 𝑢 𝑗 𝑥 𝑗 ) is a bad-path, a contradiction.
(2) If 𝑥 𝑘 𝑢 𝑗 ∈ 𝐸 (𝐺 ) then 𝑃 = (𝑥 𝑚𝐶 𝑢 𝑘 𝑥 𝑗 +1 𝐶 𝑥 𝑘 𝑢 𝑗 𝑤 𝑗 𝑢 𝑖 𝐶 𝑥 𝑗 ) is a bad-path, a contradiction.
Therefore, 𝑢 𝑘 ∉ 𝑁 (𝑥 𝑗 +1 ). Clearly that 𝑢 𝑘 ∈ 𝑌 for𝑗 + 1 < 𝑘 < 𝑚 , so𝑑 𝑢 𝑗 +2 , 𝑢 𝑗 +3 = 2 and 𝑁 𝑢 𝑗 +2 , 𝑢 𝑗 +3 ⊆
𝑋 − {𝑥 𝑗 +1 }, this implies that 𝑁𝐶 2 𝐺 ≤ 𝑋 − 1, a contradiction.
Similarly, if 𝑥 𝑗 +1 ≡ 𝑥 𝑚, then we get a contradiction.
Note:By Claim 8, let𝑦 ∈ {𝑣 , 𝑠 0 } such that 𝑥 𝑗 ∉ 𝑁 (𝑦 ).
Claim 9.𝑥 1 , 𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Proof. If𝑥 𝑖 ≡ 𝑥 1 then by Claim 7, it is clear that𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ). We consider the case𝑥 𝑖 ≠ 𝑥 1 .
Assume that𝑢 1 𝑥 𝑗 ∉ 𝐸 (𝐺 ). Clearly that 𝑑 𝑦 , 𝑢 1 = 2 and𝑢 1 ∈ 𝑌 , so 𝑁 𝑦 , 𝑢 1 ⊆ 𝑋 − {𝑥 𝑗 }, this implies that 𝑋 −
1 ≥ 𝑁 𝑦 , 𝑢 1 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, a contradiction.Therefore,𝑢 1 𝑥 𝑗 ∈ 𝐸 (𝐺 ). By Lemma 11, we have𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Similarly, we have𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Claim 10.𝑥 𝑖 ≡ 𝑥 1 or𝑥 𝑗 +1 ≡ 𝑥 𝑚.
Proof. Assume to the contrary that 𝑥 𝑖 ≠ 𝑥 1 and 𝑥 𝑗 +1 ≠ 𝑥 𝑚 . By Claim 7 and Claim 9, 𝑥 𝑖 , 𝑥 𝑗 +1 , 𝑥 1 , 𝑥 𝑚 ∉
𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) . By Lemma 6, Claim 5 and Claim 6, 𝑁 𝑤 𝑖 , 𝑢 𝑗 ⊆ 𝑋 ∪ {𝑢 𝑖 , 𝑢 +𝑖 , 𝑤 𝑗 } − {𝑥 1 , 𝑥 𝑚, 𝑥 𝑖 , 𝑥 𝑗 +1 } . Thus,
𝑋 − 1 ≥ 𝑁 𝑤 𝑖 , 𝑢 𝑗 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, a contradiction.
If 𝑥 𝑖 ≡ 𝑥 1 , then by 𝑆 ≥ 𝜎 3 − 𝑛 + 4 ≥ 4 and by 𝑥 𝑖 +1 ≡ 𝑥 𝑗 (see Claim 6), we have 𝑥 𝑗 +1 ≠ 𝑥 𝑚 and there are at
least three 1-arcs on 𝑥 𝑗 +1 𝐶 𝑥 𝑚. Applying similar arguments of the proof for 𝑢 𝑘 ∈ 𝑥 𝑗 +1 𝐶 𝑥 𝑚 of Claim 8, we have𝑢 𝑘 ∈
𝑌 and 𝑢 𝑘 ∉ 𝑁 (𝑥 𝑗 +1 ) for every 𝑗 + 1 < 𝑘 < 𝑚 ,so 𝑑 𝑢 𝑗 +2 , 𝑢 𝑗 +3 = 2 and 𝑁 𝑢 𝑗 +2 , 𝑢 𝑗 +3 ⊆ 𝑋 − {𝑥 𝑗 +1 } , this
implies that 𝑁𝐶 2 𝐺 ≤ 𝑋 − 1, a contradiction.Thus 𝑥 𝑖 ≠ 𝑥 1 . Similarly, 𝑥 𝑗 +1 ≠ 𝑥 𝑚, which contradicts Claim 10.
This finishes the proof of the Proposition 1.
Proposition 2.𝑢 𝑗 𝑤 𝑖 ∉ 𝐸 (𝐺 ) and𝑢 𝑖 𝑤 𝑗 ∈ 𝐸 (𝐺 ).
Proof. If 𝑢 𝑗 𝑤 𝑖 ∈ 𝐸 (𝐺 ), by Lemma 13, 𝑢 𝑖 𝑤 𝑗 ∉ 𝐸 (𝐺 ). Therefore, by Lemma 6 and by Proposition 1, 𝑁 𝑢 𝑖 ⊆ 𝑋 ∪
{𝑢 +
𝑖 }. By Lemma 15, 𝑢 𝑗 𝑤 𝑖 ∉ 𝐸 (𝐺 ), a contradiction. Thus, 𝑢 𝑗 𝑤 𝑖 ∉ 𝐸 (𝐺 ).
If 𝑢 𝑖 𝑤 𝑗 ∉ 𝐸 (𝐺 ) , then by Lemma 6 and by Proposition 1, 𝜔 𝐺 − 𝑋 ∪ 𝑢 +
≥ 𝑋 + 2 , which contradicts the
𝑖
toughness of𝐺 . Thus, 𝑢 𝑖 𝑤 𝑗 ∈ 𝐸 (𝐺 ).
−
By Lemma 6 and by Proposition 1,𝑁 𝑤 𝑖 ⊆ 𝑋 ∪ 𝑢 +
𝑖 = 𝑋 ∪ {𝑤 𝑖 }. . By Lemma 14,𝑥 𝑖 +1 ≡ 𝑥 𝑗 , and by Proposition
2, 𝑑 (𝑤 𝑖 , 𝑢 𝑗 ) = 2. By Lemma 10,𝑥 𝑖 , 𝑥 𝑗 +1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
By using similar arguments in the proofs of Claim 8 and of Claim 9, 𝑥 𝑗 ∉ 𝑁 (𝑣 ) ∩ 𝑁 (𝑠 0 ) , and 𝑥 1 , 𝑥 𝑚 ∉
𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).Therefore, 𝑁 𝑤 𝑖 , 𝑢 𝑗 ⊆ 𝑋 ∪ {𝑢 +𝑖 , 𝑤 𝑗 } − {𝑥 1 , 𝑥 𝑚, 𝑥 𝑖 , 𝑥 𝑗 +1 }. By 𝑁𝐶 2 𝐺 = |𝑋 |, we have𝑥 𝑖 ≡ 𝑥 1 and
𝑥 𝑗 +1 ≡ 𝑥 𝑚. This implies that 𝑆 = 1, which contradicts Lemma 3 that 𝑆 ≥ 𝜎 3 − 𝑛 + 4 ≥ 4.
Thus, the Case 2.1. does not happen.
Case 2.2. 𝐶 − 𝑋 contains three 2-arcs and some 1-arcs on 𝐶 .
Assume that the three 2-arcs are𝑢 𝑖 𝑤 𝑖 , 𝑢 𝑗 𝑤 𝑗 , 𝑢 𝑘 𝑤 𝑘 with𝑖 < 𝑗 < 𝑘 . We will prove that the three 2-arcs are not
adjacent. We have two following Propositions:
Proposition 3.𝑢 𝑖 𝑤 𝑖 is not adjacent to 𝑢 𝑘 𝑤 𝑘 .
Proof. Assume the contrary that𝑢 𝑖 𝑤 𝑖 is adjacent to𝑢 𝑘 𝑤 𝑘 .We have the following Claims (from Claim 11 to Claim 15):
Claim 11.𝑢 𝑖 𝑤 𝑘 ∈ 𝐸 (𝐺 ) and𝑤 𝑖 𝑢 𝑘 ∉ 𝐸 (𝐺 ).
Proof. Assume otherwise that𝑢 𝑖 𝑤 𝑘 ∉ 𝐸 (𝐺 ), then𝑤 𝑖 𝑢 𝑘 ∈ 𝐸 (𝐺 ) since 𝑢 𝑖 𝑤 𝑖 is adjacent to𝑢 𝑘 𝑤 𝑘 .
If 𝑤 𝑘 𝑢 𝑗 ∉ 𝐸 (𝐺 ), then by Lemma 6, 𝑁 𝑤 𝑘 ⊆ 𝑋 ∪ {𝑢 𝑘 }. By Lemma 15, applying for 𝑢 𝑖 𝐶 𝑤 𝑖 and𝑢 𝑘 𝐶 𝑤 𝑘 we
have 𝑤 𝑖 𝑢 𝑘 ∉ 𝐸 (𝐺 ) , which contradicts to the fact that 𝑤 𝑖 𝑢 𝑘 ∈ 𝐸 (𝐺 ) . Thus, 𝑤 𝑘 𝑢 𝑗 ∈ 𝐸 (𝐺 ) . Similarly, we
have𝑢 𝑖 𝑤𝑗 ∈ 𝐸 (𝐺 ).
By Lemma 9 applying for 𝑤 𝑖 𝑢 𝑘 ∈ 𝐸 (𝐺 ), 𝑥 𝑖 +1 , 𝑥 𝑘 ∉ 𝑁 (𝑢 1 , 𝑤 𝑚−1 ). By Lemma 8 we have 𝑥 𝑖 +1 , 𝑥 𝑘 ∈ 𝑁 𝑣 −
𝑁 (𝑠 0 ) or 𝑥 𝑖 +1 , 𝑥 𝑘 ∈ 𝑁 𝑠 0 − 𝑁 (𝑣 ). Let𝑦 ∈ {𝑣 , 𝑠 0 } such that 𝑥 𝑖 +1 , 𝑥 𝑘 ∉ 𝑁 (𝑦 ).
If 𝑥 𝑖 ≠ 𝑥 1 then 𝑢 1 ∈ 𝑌 and𝑁 𝑦 , 𝑢 1 ⊆ 𝑋 − {𝑥 𝑖 +1 , 𝑥 𝑘 }, so 𝑁 𝑦 , 𝑢 1 ≤ 𝑋 − 2. However, since 𝑑 𝑦 , 𝑢 1 = 2,
we get 𝑁 𝑦 , 𝑢 1 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, a contradiction. Thus, 𝑥 𝑖 ≡ 𝑥 1 . Similarly, we have 𝑥 𝑘 +1 ≡ 𝑥 𝑚.
Because 𝑆 ≥ 4, 𝑥 𝑖 +1 𝐶 𝑥 𝑗 or 𝑥 𝑗 +1 𝐶 𝑥 𝑘 has at least two 1-arcs. Without loss of generality, we assume that
𝑥 𝑖 +1 𝐶 𝑥 𝑗 has at least two 1-arcs, so 𝑢 𝑖 +1 , 𝑢 𝑖 +2 ∈ 𝑌 . By 𝑢 𝑖 𝑤 𝑗 ∈ 𝐸 (𝐺 ) and by Lemma
10, 𝑥 1 , 𝑥 𝑗 +1 ∉ 𝑁 (𝑢 𝑖 +1 , 𝑢 𝑖 +2 ) . Therefore, 𝑁 𝑢 𝑖 +1 , 𝑢 𝑖 +2 ⊆ 𝑋 − {𝑥 1 , 𝑥 𝑗 +1 } , so 𝑁 𝑢 𝑖 +1 , 𝑢 𝑖 +2 ≤ 𝑋 − 2 .
Because 𝑑 𝑢 𝑖 +1 , 𝑢 𝑖 +2 = 2, we have 𝑁 𝑢 𝑖 +1 , 𝑢 𝑖 +2 ≥ 𝑁𝐶 2 𝐺 = 𝑋 , a contradiction.
Thus,𝑢 𝑖 𝑤 𝑘 ∈ 𝐸 (𝐺 ). By Lemma 13, we have𝑤 𝑖 𝑢 𝑘 ∉ 𝐸 (𝐺 ).
Claim 12.𝑢 𝑘 𝑤 𝑗 ∉ 𝐸 (𝐺 ) and 𝑤 𝑖 𝑢 𝑗 ∉ 𝐸 (𝐺 ).
Proof. If 𝑢 𝑘 𝑤 𝑗 ∈ 𝐸 (𝐺 ) then𝑃 = (𝑥 𝑚𝐶 𝑤 𝑘 𝑢 𝑖 𝐶 𝑤 𝑗 𝑢 𝑘 𝐶 𝑥 𝑗 +1 ) is a bad-path, a contradiction.
If𝑤 𝑖 𝑢 𝑗 ∈ 𝐸 (𝐺 ) then𝑃 = (𝑥 1 𝐶 𝑢 𝑖 𝑤 𝑘 𝐶 𝑢 𝑗 𝑤 𝑖 𝐶 𝑥 𝑗 )is a bad-path, a contradiction.
Claim 13.𝑥 𝑖 +1 ≡ 𝑥 𝑗 and𝑥 𝑗 +1 ≡ 𝑥 𝑘 .
Proof. Assume that 𝑥 𝑖 +1 ≠ 𝑥 𝑗 . Then𝑢 𝑖 +1 ∈ 𝑌 and 𝑑 𝑤 𝑖 , 𝑢 𝑖 +1 = 2. By Lemma 10,𝑥 𝑖 , 𝑥 𝑘 +1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑖 +1 ).
By Lemma 6 and by Claims 11, 12, we have 𝑁 𝑤 𝑖 , 𝑢 𝑖 +1 ⊆ 𝑋 ∪ 𝑢 𝑖 − {𝑥 𝑖 , 𝑥 𝑘 +1 } , therefore 𝑋 − 1 ≥
𝑁 𝑤 𝑖 , 𝑢 𝑖 +1 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, a contradiction.
Thus𝑥 𝑖 +1 ≡ 𝑥 𝑗 . Similarly, we have 𝑥 𝑗 +1 ≡ 𝑥 𝑘 .
Note:Now we consider the pair {𝑤 𝑖 , 𝑢 𝑗 } with distance 2. By Lemma 10, we have 𝑥 𝑖 , 𝑥 𝑘 +1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
© 2015, IJARCSSE All Rights Reserved
Page | 647
Truong et al., International Journal of Advanced Research in Computer Science and Software Engineering 5(9),
September- 2015, pp. 643-649
Claim 14.𝑥 1 , 𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Proof. First, we prove that 𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ). The proof for the case 𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) is similar.If 𝑥 𝑖 ≡ 𝑥 1 then by
Lemma 10, 𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ). For what follows, we assume that 𝑥 𝑖 ≠ 𝑥 1 . Clearly, 𝑢 1 ∈ 𝑌 .
If 𝑥 𝑗 ∈ 𝑁 (𝑣 ) ∩ 𝑁 (𝑠 0 ), then by Lemma 12, 𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
If 𝑥 𝑗 ∉ 𝑁 (𝑣 ) ∩ 𝑁 (𝑠 0 ) . Let 𝑦 ∈ {𝑠 0 , 𝑣 } such that 𝑥 𝑗 ∉ 𝑁 (𝑦 ) . Clearly, 𝑑 𝑦 , 𝑢 1 = 2 . If 𝑢 1 𝑥 𝑗 ∉ 𝐸 (𝐺 ) ,
then 𝑁 𝑦 , 𝑢 1 ⊆ 𝑋 − {𝑥 𝑗 } and therefore 𝑁 𝑦 , 𝑢 1 ≤ 𝑋 − 1 , which contradicts to the fact that 𝑁 𝑦 , 𝑢 1 ≥
𝑁𝐶 2 𝐺 = 𝑋 . Thus,𝑢 1 𝑥 𝑗 ∈ 𝐸 (𝐺 ). By Lemma 11, we have𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Claim 15.𝑥 𝑖 ≡ 𝑥 1 or𝑥 𝑘 +1 ≡ 𝑥 𝑚.
Proof. Assume to the contrary that 𝑥 𝑖 ≠ 𝑥 1 and 𝑥 𝑘 +1 ≠ 𝑥 𝑚 . By Lemmas 6, 10 and by Claims 12, 14, we
have 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) ⊆ 𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 − {𝑥 1 , 𝑥 𝑚, 𝑥 𝑖 , 𝑥 𝑘 +1 } . By Claim 13, 𝑑 𝑤 𝑖 , 𝑢 𝑗 = 2 , we get 𝑋 − 1 ≥
𝑁 𝑤 𝑖 , 𝑢 𝑗 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, a contradiction.
Assume without loss of generality that 𝑥 𝑖 ≡ 𝑥 1 . By Claim 13 and by 𝑆 ≥ 4, 𝑥 𝑘 +1 ≠ 𝑥 𝑚 and there are at least three
1-arcs on 𝑥 𝑘 +1 𝐶 𝑥 𝑚 .Since 𝑋 = 𝑁𝐶 2(𝐺 ) ≤ 𝑁 𝑤 𝑖 , 𝑢 𝑗 ≤ |𝑋 ∪ {𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 } − 𝑥 1 , 𝑥 𝑚, 𝑥 𝑘 +1 | = |𝑋 | , so 𝑥 𝑡 ∈
𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) for every 𝑘 + 1 < 𝑡 < 𝑚 , and we will show that 𝑢 𝑡 𝑥 𝑘 +1 ∉ 𝐸 (𝐺 ) . Indeed, assume otherwise
that𝑢 𝑡 𝑥 𝑘 +1 ∈ 𝐸 (𝐺 ). We have two cases:
(1) If 𝑥 𝑡 𝑤 𝑖 ∈ 𝐸 (𝐺 ) then𝑃 = (𝑥 𝑚𝐶 𝑢 𝑡 𝑥 𝑘 +1 𝐶 𝑥 𝑡 𝑤 𝑖 𝑢 𝑖 𝑤 𝑘 𝐶 𝑥 𝑗 ) is a bad-path, a contradiction.
(2) If 𝑥 𝑡 𝑢 𝑗 ∈ 𝐸 (𝐺 ) then 𝑃 = (𝑥 𝑚𝐶 𝑢 𝑡 𝑥 𝑘 +1 𝐶 𝑥 𝑡 𝑢 𝑗 𝐶 𝑤 𝑘 𝑢 𝑖 𝑤 𝑖 𝑥 𝑗 )is a bad-path, a contradiction.
Clearly, 𝑢 𝑘 +2 , 𝑢 𝑘 +3 ∈ 𝑌 and 𝑁 𝑢 𝑘 +2 , 𝑢 𝑘 +3 ⊆ 𝑋 − {𝑥 𝑘 +1 } and 𝑑 𝑢 𝑘 +2 , 𝑢 𝑘 +3 = 2 , this implies that 𝑋 =
𝑁𝐶 2 𝐺 ≤ 𝑁 𝑢 𝑘 +2 , 𝑢 𝑘 +3 ≤ 𝑋 − 𝑥 𝑘 +1 = 𝑋 − 1, a contradiction.
This finishes the proof of the Proposition 3.
Proposition 4.𝑢 𝑖 𝑤 𝑖 , 𝑢 𝑘 𝑤 𝑘 are not adjacent to𝑢 𝑗 𝑤 𝑗 .
Proof. We need only to prove that 𝑢 𝑖 𝑤 𝑖 is not adjacent to𝑢 𝑗 𝑤 𝑗 . The proof that𝑢 𝑘 𝑤 𝑘 is not adjacent to𝑢 𝑗 𝑤 𝑗 is similar.
Assume to the contrary that 𝑢 𝑖 𝑤 𝑖 is adjacent to𝑢 𝑗 𝑤 𝑗 .We have the following Claims (from Claim 16 to Claim 20):
Claim 16.𝑢 𝑖 𝑤 𝑗 ∈ 𝐸 (𝐺 ) and𝑤 𝑖 𝑢 𝑗 ∉ 𝐸 (𝐺 ). Moreover,𝑥 𝑖 +1 ≡ 𝑥 𝑗 .
Proof. Assume to the contrary that𝑢 𝑖 𝑤 𝑗 ∉ 𝐸 (𝐺 ), then by the assumption that 𝑢 𝑖 𝑤 𝑖 is adjacent to𝑢 𝑗 𝑤 𝑗 and by
Lemma 6,𝑤 𝑖 𝑢 𝑗 ∈ 𝐸 (𝐺 ). By Lemma 6 and by Proposition 3, 𝑁 𝑢 𝑖 ⊆ 𝑋 ∪ {𝑤 𝑖 }. Arguing similarly to the proofs of
Lemma 15, we have𝑥 𝑖 ≡ 𝑥 1 and𝑥 𝑘 +1 ≡ 𝑥 𝑚 and𝑥 𝑖 +1 ≡ 𝑥 𝑗 . Because of|𝑆 | ≥ 4,𝑥 𝑗 +1 ≠ 𝑥 𝑘 and there are at least
three 1-arcs on 𝑥 𝑗 +1 𝐶 𝑥 𝑘 . We conclude that 𝑢 𝑡 ∈ 𝑌 and 𝑢 𝑡 ∉ 𝑁 (𝑥 𝑗 ) for any 𝑗 < 𝑡 < 𝑘 , otherwise 𝑃 =
(𝑥 1 𝑢 𝑖 𝑤 𝑖 𝑢 𝑗 𝐶 𝑢 𝑡 𝑥 𝑗 ) is a bad-path, a contradiction. Therefore𝑁 𝑢 𝑗 +1 , 𝑢 𝑗 +2 ⊆ 𝑋 − {𝑥 𝑗 }and 𝑁 𝑢 𝑗 +1 , 𝑢 𝑗 +2 ≤
𝑋 − 1. By𝑑 (𝑢 𝑗 +1 , 𝑢 𝑗 +2 ) = 2, 𝑁 𝑢 𝑗 +1 , 𝑢 𝑗 +2 ≥ 𝑁𝐶 2 𝐺 = 𝑋 , a contradiction.
Thus,𝑢 𝑖 𝑤 𝑗 ∈ 𝐸 (𝐺 ). By Lemma 13,𝑤 𝑖 𝑢 𝑗 ∉ 𝐸 (𝐺 ). By Lemma 6 and by Proposition 3,𝑁 𝑤 𝑖 ⊆ 𝑋 ∪ {𝑢 𝑖 }. By
Lemma 14, we have 𝑥 𝑖 +1 ≡ 𝑥 𝑗 .
Claim 17.𝑑 𝑤 𝑖 , 𝑢 𝑗 = 2 and𝑥 𝑖 , 𝑥 𝑗 +1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Proof. By Claim 16, 𝑑 𝑤 𝑖 , 𝑢 𝑗 = 2. By Lemma 10, 𝑥 𝑖 , 𝑥 𝑗 +1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Claim 18.𝑥 𝑗 ∉ 𝑁 (𝑣 ) ∩ 𝑁 (𝑠 0 ).
Proof. Assume to the contrary that 𝑥 𝑗 ∈ 𝑁 (𝑣 ) ∩ 𝑁 (𝑠 0 ). Since Lemma 12, we have𝑥 1 , 𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ). By Lemma
6 and by Claim 17, 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) ⊆ 𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 − {𝑥 1 , 𝑥 𝑚, 𝑥 𝑖 , 𝑥 𝑗 +1 }. Because of|𝑁 (𝑤 𝑖 , 𝑢 𝑗 )| ≥ 𝑁𝐶 2(𝐺 ) =
|𝑋 |, we get𝑥 𝑖 ≡ 𝑥 1 and𝑢 𝑗 𝑤 𝑘 ∈ 𝐸 (𝐺 ).
Since Lemma 13, we have 𝑢 𝑘 𝑤 𝑗 ∉ 𝐸 (𝐺 ), by Lemma 6 and by Proposition 3,𝑁 𝑢 𝑘 ⊆ 𝑋 ∪ {𝑤 𝑘 }. By Lemma
14,𝑥 𝑗 +1 ≡ 𝑥 𝑘 .
By|𝑆 | ≥ 4, 𝑥 𝑘 +1 ≠ 𝑥 𝑚 and there are at least three 1-arcs on 𝑥 𝑘 +1 𝐶 𝑥 𝑚 . Since 𝑋 = 𝑁𝐶 2 𝐺 ≤ 𝑁 𝑤 𝑖 , 𝑢 𝑗 ≤
𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 − 𝑥 1 , 𝑥 𝑚, 𝑥 𝑗 +1 = |𝑋 | , so 𝑥 𝑡 ∈ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) for every 𝑘 + 1 < 𝑡 < 𝑚 .Moreover, 𝑢 𝑡 ∈ 𝑌
and𝑢 𝑡 𝑥 𝑘 +1 ∉ 𝐸 (𝐺 ). Indeed, assume otherwise that 𝑢 𝑡 𝑥 𝑘 +1 ∈ 𝐸 (𝐺 ), we distinguish two cases:
(1) If 𝑥 𝑡 𝑤 𝑖 ∈ 𝐸 (𝐺 ) then 𝑃 = (𝑥 𝑚𝐶 𝑢 𝑡 𝑥 𝑘 +1 𝐶 𝑥 𝑡 𝑤 𝑖 𝑢 𝑖 𝑤 𝑗 𝐶 𝑤 𝑘 𝑢 𝑗 𝑥 𝑗 ) is a bad-path, a contradiction.
(2) If 𝑥 𝑡 𝑢 𝑗 ∈ 𝐸 (𝐺 ) then 𝑃 = (𝑥 𝑚𝐶 𝑢 𝑡 𝑥 𝑘 +1 𝐶 𝑥 𝑡 𝑢 𝑗 𝑤 𝑘 𝐶 𝑤 𝑗 𝑢 𝑖 𝑤 𝑖 𝑥 𝑗 ) is a bad-path, a contradiction.
Clearly, 𝑢 𝑘 +2 , 𝑢 𝑘 +3 ∈ 𝑌 and 𝑁 𝑢 𝑘 +2 , 𝑢 𝑘 +3 ⊆ 𝑋 − {𝑥 𝑘 +1 } and 𝑑 𝑢 𝑘 +2 , 𝑢 𝑘 +3 = 2, we conclude that 𝑋 =
𝑁𝐶 2 𝐺 ≤ 𝑁 𝑢 𝑘 +2 , 𝑢 𝑘 +3 ≤ 𝑋 − 𝑥 𝑘 +1 = 𝑋 − 1, a contradiction.
Note:Let 𝑦 ∈ {𝑣 , 𝑠 0 }such that𝑦 ∉ 𝑁 (𝑥 𝑗 ). Clearly that𝑑 𝑦 , 𝑢 1 = 𝑑 𝑦 , 𝑤 𝑚−1 = 2.
Claim 19.If𝑥 𝑖 ≠ 𝑥 1 then𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ). Similarly, if𝑥 𝑘 +1 ≠ 𝑥 𝑚 then𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Proof. Assume that 𝑥 𝑖 ≠ 𝑥 1 , then 𝑢 1 ∈ 𝑌 . If 𝑢 1 𝑥 𝑗 ∉ 𝐸 (𝐺 ) then 𝑁 𝑦 , 𝑢 1 ⊆ 𝑋 − {𝑥 𝑗 } , so 𝑋 = 𝑁𝐶 2(𝐺 ) ≤
𝑁 𝑦 , 𝑢 1 ≤ 𝑋 − 1, a contradiction. Therefore,𝑢 1 𝑥 𝑗 ∈ 𝐸 (𝐺 ). By Lemma 11, we have𝑥 1 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ). Similarly,
if 𝑥 𝑘 +1 ≠ 𝑥 𝑚then𝑥 𝑚 ∉ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ).
Claim 20.𝑥 𝑖 ≡ 𝑥 1 and𝑥 𝑘 +1 ≡ 𝑥 𝑚.
Proof. Assume that 𝑥 𝑖 ≠ 𝑥 1 and 𝑥 𝑘 +1 ≠ 𝑥 𝑚. By Lemma 6 and by Proposition 3 and by Claims 16, 17, 19, we
have 𝑁 𝑤 𝑖 , 𝑢 𝑗 ⊆ 𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 − 𝑥 1 , 𝑥 𝑚, 𝑥 𝑖 , 𝑥 𝑗 +1 , so 𝑋 = 𝑁𝐶 2 𝐺 ≤ 𝑁 𝑤 𝑖 , 𝑢 𝑗 ≤ 𝑋 − 1 , a
contradiction. Thus,𝑥 𝑖 ≡ 𝑥 1 or 𝑥 𝑘 +1 ≡ 𝑥 𝑚.
If𝑥 𝑖 ≠ 𝑥 1 then𝑥 𝑘 +1 ≡ 𝑥 𝑚 and𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) ⊆ 𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 − {𝑥 1 , 𝑥 𝑖 , 𝑥 𝑗 +1 }.
Because 𝑁 𝑤 𝑖 , 𝑢 𝑗 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, so 𝑁 𝑤 𝑖 , 𝑢 𝑗 = 𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 , 𝑤 𝑘 − {𝑥 1 , 𝑥 𝑖 , 𝑥 𝑗 +1 } and by Proposition
3,𝑢 𝑗 𝑤 𝑘 ∈ 𝐸 (𝐺 ). Since Lemma 13,𝑤 𝑗 𝑢 𝑘 ∉ 𝐸 (𝐺 ), so by Lemma 6 and by Proposition 3,𝑁 (𝑢 𝑘 ) ⊆ 𝑋 ∪ 𝑤 𝑘 . By
Lemma 14, we have𝑥 𝑗 +1 ≡ 𝑥 𝑘 .
Because|𝑆 | ≥ 4, there are at least three 1-arcs on 𝑥 1 𝐶 𝑥 𝑖 . Clearly that 𝑥 𝑡 ∈ 𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) for every 1 < 𝑡 < 𝑖 .
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Truong et al., International Journal of Advanced Research in Computer Science and Software Engineering 5(9),
September- 2015, pp. 643-649
Moreover, 𝑤 𝑡 −1 ∈ 𝑌 and𝑤 𝑡 −1 𝑥 𝑖 ∉ 𝐸 (𝐺 ). Indeed, assume that 𝑤 𝑡 −1 𝑥 𝑖 ∈ 𝐸 (𝐺 ). Then:
(1) If 𝑥 𝑡 ∈ 𝑁 (𝑤 𝑖 )then 𝑃 = (𝑥 1 𝐶 𝑤 𝑡 −1 𝑥 𝑖 𝐶 𝑥 𝑡 𝑤 𝑖 𝑢 𝑖 𝑤 𝑗 𝑢 𝑗 𝑥 𝑗 ) is a bad-path, a contradiction.
(2) If 𝑥 𝑡 ∈ 𝑁 (𝑢 𝑗 ) then 𝑃 = (𝑥 1 𝐶 𝑤 𝑡 −1 𝑥 𝑖 𝐶 𝑥 𝑡 𝑢 𝑗 𝑤 𝑗 𝑢 𝑖 𝑤 𝑖 𝑥 𝑗 ) is a bad-path, a contradiction.
Therefore, 𝑁 𝑤 𝑖 −2 , 𝑤 𝑖 −3 ⊆ 𝑋 − 𝑥 𝑖 and 𝑁 𝑤 𝑖 −2 , 𝑤 𝑖 −3 ≤ 𝑋 − 1. Because 𝑑 𝑤 𝑖 −2 , 𝑤 𝑖 −3 = 2, this implies
that 𝑁 𝑤 𝑖 −2 , 𝑤 𝑖 −3 ≥ 𝑁𝐶 2 𝐺 = |𝑋 |, a contradiction.
Thus,𝑥 𝑖 ≡ 𝑥 1 . Similarly, we have 𝑥 𝑘 +1 ≡ 𝑥 𝑚.
By Claim 16 and by Claim 20, we conclude that𝑥 𝑖 ≡ 𝑥 1 , 𝑥 𝑘 +1 ≡ 𝑥 𝑚 and 𝑥 𝑖 +1 ≡ 𝑥 𝑗 . By |𝑆 | ≥ 4,𝑥 𝑗 +1 ≠ 𝑥 𝑘 and
there are at least three 1-arcs on 𝑥 𝑗 +1 𝐶 𝑥 𝑘 .
If𝑢 𝑗 𝑤 𝑘 ∈ 𝐸 (𝐺 ) then by Lemma 13,𝑤 𝑗 𝑢 𝑘 ∉ 𝐸 (𝐺 ). By Lemma 6 and by Proposition 3,𝑁 (𝑢 𝑘 ) ⊆ 𝑋 ∪ 𝑤 𝑘 . By
Lemma 14, 𝑥 𝑗 +1 ≡ 𝑥 𝑘 , a contradiction. Therefore𝑢 𝑗 𝑤 𝑘 ∉ 𝐸 (𝐺 ). By Lemma 6 and by Proposition 3 and by Claim 17,
we have𝑁 (𝑤 𝑖 , 𝑢 𝑗 ) ⊆ 𝑋 ∪ 𝑢 𝑖 , 𝑤 𝑗 − 𝑥 1 , 𝑥 𝑗 +1 .
Arguing similarly to the proofs of Claim 18, we have 𝑢 𝑡 ∈ 𝑌 and𝑢 𝑡 𝑥 𝑗 +1 ∉ 𝐸 (𝐺 )for every𝑗 + 1 < 𝑡 < 𝑘 . Therefore
𝑁 𝑢 𝑗 +2 , 𝑢 𝑗 +3 ⊆ 𝑋 − {𝑥 𝑗 +1 } and 𝑁 𝑢 𝑗 +2 , 𝑢 𝑗 +3 ≤ 𝑋 − 1, which contradicts to the fact that 𝑁 𝑢 𝑗 +2 , 𝑢 𝑗 +3 ≥
𝑁𝐶 2 𝐺 = |𝑋 |.This finishes the proof of the Proposition 4.
Finally, we conclude that 𝑢 𝑖 𝑤 𝑖 , 𝑢 𝑗 𝑤 𝑗 , 𝑢 𝑘 𝑤 𝑘 are pairwise non-adjacent arcs. By Lemma 6,𝜔 𝐺 − 𝑋 ≥ 𝑋 + 1,
which contradicts to the assumption that 𝐺 is 1-tough.
Thus, the Case 2.2. does not happen.
IV. CONCLUSIONS
In this paper, we present some lower bounds for the circumference of tough graphs and prove the conjecture posed by
Bauer, Fan and Veldman [1], namely, that: “If 𝐺 is 1-tough graph with 𝜎 3 ≥ 𝑛 , then 𝑐 (𝐺 ) ≥ 𝑚𝑖𝑛 ⁡
{𝑛 , 2𝑁𝐶 2 𝐺 +
4}”.
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http://arxiv.org/pdf/1309.5379.pdf
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