Information Processing Letters 109 (2009) 585–588
Contents lists available atScienceDirect
Information Processing Letters
www.elsevier.com/locate/ipl
Conditional fault hamiltonian connectivity of the complete graph
Tung-Yang Ho
a,
∗
, Yuan-Kang Shih
b, Jimmy J.M. Tan
b, Lih-Hsing Hsu
caDepartment of Information Management, Ta Hwa Institute of Technology, Hsinchu, Taiwan 30740, ROC bDepartment of Computer Science, National Chiao Tung University, Hsinchu, Taiwan 30010, ROC
cDepartment of Computer Science and Information Engineering, Providence University, Taichung, Taiwan 43301, ROC
a r t i c l e i n f o a b s t r a c t
Article history:
Received 22 August 2008
Received in revised form 15 January 2009 Available online 20 February 2009 Communicated by A.A. Bertossi
Keywords:
Complete graph Hamiltonian Hamiltonian connected Fault tolerance
A path in G is a hamiltonian path if it contains all vertices of G. A graph G is hamiltonian connected if there exists a hamiltonian path between any two distinct vertices of G. The degree of a vertex u in G is the number of vertices of G adjacent to u. We denote by δ(G)the minimum degree of vertices of G. A graph G is conditional k edge-fault tolerant hamiltonian connected if G−F is hamiltonian connected for every F⊂E(G)with|F| k
and δ(G−F)3. The conditional edge-fault tolerant hamiltonian connectivity HC3e(G)
is defined as the maximum integer k such that G is k edge-fault tolerant conditional hamiltonian connected if G is hamiltonian connected and is undefined otherwise. Let n4. We use Kn to denote the complete graph with n vertices. In this paper, we show that
HC3
e(Kn)=2n−10 for n∈ {/ 4,5,8,10},HC3e(K4)=0,HC3e(K5)=2,HC3e(K8)=5, and HC3
e(K10)=9.
©2009 Elsevier B.V. All rights reserved.
1. Introduction
For the graph definitions and notations, we follow [1]. Let G
= (
V,
E)be a graph if V is a finite set and E is a sub-set of{(
u,v)| (
u,v) is an unordered pair of V}
. We say that V is the vertex set and E is the edge set. Two vertices u and v are adjacent if(u,
v)
∈
E. The complete graph Knis the graph with n vertices such that any two distinct vertices are adjacent. The degree of a vertex u in G, de-noted by degG
(u)
, is the number of vertices adjacent to u.We use
δ(G)
to denote min{
degG(u)
|
u∈
V(G)
}
. A path of length m−
1,v0,
v1, . . . ,
vm−1, is an ordered list of dis-tinct vertices such that viand vi+1are adjacent for 0i m−
2. We also write the path v0, . . . ,
vk,
P,
vl, . . . ,
vmfor P
=
vk, . . . ,
vl. A cycle is a path with at least threevertices such that the first vertex is the same as the last one. A hamiltonian cycle of G is a cycle that traverses every vertex of G exactly once. A graph is hamiltonian if it has a hamiltonian cycle. A hamiltonian path is a path of length V
(G)
−
1.*
Corresponding author.E-mail address:[email protected](T.-Y. Ho).
A hamiltonian graph G is k edge-fault tolerant hamilto-nian if G
−
F remains hamiltonian for every F⊂
E(G)with|
F|
k. The edge-fault tolerant hamiltonicity,H
e(G)
, isde-fined as the maximum integer k such that G is k edge-fault tolerant hamiltonian if G is hamiltonian and is undefined otherwise. Assume that G is a hamiltonian graph, and x is a vertex such that degG
(x)
= δ(
G). We arbitrary choose degG(x)
−
1 edges from those edges incident to x to form an edge faulty set F . Obviously, degG−F(x)
=
1; hence, G−
F is not hamiltonian. Therefore,H
e(G)
δ(
G)−
2 ifH
e(G)
is defined. Assume that n is an integer with n3.It is proved by Ore [9] that any n-vertex graph with at least C(n,2
)
− (
n−
3)
edges is hamiltonian. Moreover, there ex-ists a non-hamiltonian n-vertex graph with C(n,
2)
−(
n−
2)
edges. In other words,H
e(K
n)
=
n−
3 for n3. In [5],it is proved that
H
e(Q
n)
=
n−
2 for n2 where Qnis the n-dimensional hypercube. In [6], it is proved that
H
e(S
n)
=
n−
3 for n3 where Sn is the n-dimensionalstar graph.
Chan and Lee [2] began the study of the existence of hamiltonian cycle in a graph such that each vertex is inci-dent to at least two fault-free edges. A graph G is condi-tional k edge-fault tolerant hamiltonian if G
−
F is hamilto-0020-0190/$ – see front matter ©2009 Elsevier B.V. All rights reserved.586 T.-Y. Ho et al. / Information Processing Letters 109 (2009) 585–588
nian for every F
⊂
E(G)
with|
F|
k andδ(G
−
F)
2. The conditional edge-fault tolerant hamiltonicity,H
2e
(G)
, isdefined as the maximum integer k such that G is condi-tional k edge-fault tolerant hamiltonian if G is hamiltonian and is undefined otherwise. Chan and Lee [2] proved that
H
2e
(Q
n)
=
2n−
5 for n3. Recently, Fu [3] studies theconditional edge-fault tolerant hamiltonicity of the com-plete graph.
Fault tolerant hamiltonian connectivity is another im-portant parameter for graphs [4]. A graph G is hamilto-nian connected if there exists a hamiltohamilto-nian path between any two distinct vertices of G. It is easy to see that a hamiltonian connected graph with at least three vertices is hamiltonian. It is proved by Moon [7] that the degree of any vertex in a hamiltonian connected graph with at least four vertices is at least 3. A graph G is k edge-fault tol-erant hamiltonian connected if G
−
F remains hamiltonian connected for any F⊂
E(G)
with|
F|
k. The edge-fault tolerant hamiltonian connectivity of a graph G,HC
e(G)
, isdefined as the maximum integer k such that G is k edge-fault tolerant hamiltonian connected if G is hamiltonian connected and is undefined otherwise. Assume that G is a hamiltonian connected graph with at least four vertices and x is a vertex such that degG
(x)
= δ(
G). We arbitrary choose degG(x)
−
2 edges from those edges incident tox to form an edge faulty set F . Obviously, degG−F
(x)
=
2; hence, G
−
F is not hamiltonian connected. Therefore,HC
e(G)
δ(
G)−
3 ifHC
e(G)
is defined. Again, Ore [8]proved that
HC
e(K
n)
=
n−
4 for n4.In this paper, we study the concept of conditional edge-fault tolerant hamiltonian connectivity. Since the degree of any vertex in a hamiltonian connected graph with at least four vertices is at least 3, it is natural to assume that each vertex is incident to at least three fault-free edges. A graph G is conditional k edge-fault tolerant hamiltonian connected if G
−
F is hamiltonian connected for every F⊂
E(G)
with|
F|
k andδ(G
−
F)
3. The conditional edge-fault toler-ant hamiltonian connectivity,HC
3e(G)
, is defined to be themaximum integer k such that G is conditional k edge-fault tolerant hamiltonian connected if G is hamiltonian con-nected and is undefined otherwise.
Assume that n is an integer with n
4. In this pa-per, we prove thatHC
3e(K
n)
=
2n−
10 for n∈ {
/
4,
5,
8
,
10}
,HC
3e(K
4)
=
0,HC
3e(K
5)
=
2,HC
3e(K
8)
=
5, andHC
3e
(K
10)
=
9. To reach this goal, we present some pre-liminary in the following section. In Section 3, we prove our main result.2. Preliminary
Let F be a faulty edge set. We define Kn
(F)
be a graphwith E
(K
n(F))
=
F and V(K
n(F))
=
V(K
n)
. The followingstatement is proved in [3]:
Suppose F
⊂
E(Kn)
andδ(K
n−
F)
2, where n4. If n∈
/
{
7,
9}
(respectively, n∈ {
7,
9}
) then Kn−
F is hamiltonian,where
|
F|
2n−
8 (respectively,|
F|
2n−
9).In the conclusion of [3], it is claimed that the above statement is optimal. Using our terminology, we obtain the following statement.
H
2e
(K
n)
=
2n−
8 for n∈ {
/
7,
9}
and n4,H
2e(K
7)
=
5, andH
2e
(K
9)
=
9.Yet, it is easy to check that
H
2e(K
3)
is 0 andH
2e(K
4)
is 2 (not 0.) Thus, we have the following theorem.Theorem 1.
H
2e(K
n)
=
2n−
8 for n∈ {
/
7,
9}
and n 5,H
2e
(K
3)
=
0,H
2e(K
4)
=
2,H
2e(K
7)
=
5, andH
e2(K
9)
=
9.Lemma 1. Assume that n is an integer with n
6 and F is any subset of E(Kn)
with|
F| =
2n−
10 if n∈ {
/
8,
10}
and|
F| =
2n
−
11 if n∈ {
8,
10}
. There exists a vertex w in Kn(F
)
suchthat 1
degKn(F)(w)
n−21
−
1.Proof. Suppose that the lemma is false. Then degKn(F)
(w)
n−12
for every vertices with degKn(F)(w)
=
0. Obviously,there are at least
n−21+
1 vertices with degKn(F)(w)
=
0.Hence,
|
F| (
n−21(
n−21+
1))/
2. However,(
n−21×
(
n−21+
1))/
2>
2n−
10 for n∈ {
/
8,
10}
and(
n−21(
n−21+
1))/
2>
2n−
11 for n∈ {
8,
10}
. It is a contradiction. The lemma is proved.2
The following theorem can be found in [1].
Theorem 2. (See [1].) Let D
= (
d1,
d2, . . . ,d
n)
be anonin-creasing sequence with d1
1 and di0 for 2in. Weset D
= (
d1,
d2, . . . ,
dn−1)
= (
d2−
1,
d3−
1, . . . ,d
d1+1−
1,
dd1+2, . . . ,d
n). Then there exists a graph G with vertex set
{
x1,
x2, . . . ,
xn}
such that degG(x
i)
=
di for 1in if andonly if there exists a graph Gwith vertex set
{
y1,
y2, . . . ,
yn−1} such that degG(y
j)
=
djfor 1jn−
1.By the above theorem, we know that there is a graph G with degree sequence D if and only if there is a graph G with degree sequence D. If di
<
0 for some i, then D is not the degree sequence of any graph, neither is D.Lemma 2. Let F be a subset of E
(K
9)
with|
F| =
8 andδ(
K9−
F)
3. Let u and v be any two distinct vertices in K9such that degK9(F)(u)
=
0 and degK9(F)(v)
=
0. Then there exists a ver-tex w with degK9(F)(w)
∈ {
2,
3}
.Proof. Let
{
x1,
x2, . . . ,
x8=
u,x9=
v}
be the vertex set of K9 such that degK9(F)(x
i)
=
di and d1d2· · ·
d9. Ob-viously,9
i=1di=
16. Assume that the lemma is false.Then degK9(F)
(x
i)
∈ {
0,
1,
4,
5}
for 1i9. By brute force, all such sequences are listed below:(
5,
5,
5,
1,
0,
0,
0,
0,
0)
,(
5,
5,
4,
1,
1,
0,
0,
0,
0)
,(
5,
4,
4,
1,
1,
1,
0,
0,
0)
,(
4,
4,
4,
4,
0,
0,
0,
0,
0)
, and(
4,
4,
4,
1,
1,
1,
1,
0,
0)
. By Theorem 2, we can check that such a graph does not exist. Hence, the lemma is proved.2
Lemma 3. Let F be a subset of E
(K
11)
with|
F| =
12 andδ(K
11−
F)
3. Let u and v be any two distinct vertices in K11 such that degK11(F)(u)
=
0 and degK11(F)(v)
=
0. Then there exists a vertex w with degK11(F)(w)
∈ {
2,
3,
4}
.Proof. Let
{
x1,
x2, . . . ,
x10=
u,x11=
v}
be the vertex set of K11 such that degK11(F)(x
i)
=
di and d1d2· · ·
T.-Y. Ho et al. / Information Processing Letters 109 (2009) 585–588 587
d11. Obviously,
11
i=1di=
24. Assume that the lemmais false. Then degK11(F)
(x
i)
∈ {
0,
1,
5,
6,
7}
for 1 i11. By brute force, all such sequences are listed below:
(
7,
7,
7,
1,
1,
1,
0,
0,
0,
0,
0)
,(
7,
7,
6,
1,
1,
1,
1,
0,
0,
0,
0)
,(
7,
7,
5,
5,
0,
0,
0,
0,
0,
0,
0)
,(
7,
7,
5,
1,
1,
1,
1,
1,
0,
0,
0)
,(
7,
6,
6,
5,
0,
0,
0,
0,
0,
0,
0)
,(
7,
6,
6,
1,
1,
1,
1,
1,
0,
0,
0)
,(
7,
6,
5,
5,
1,
0,
0,
0,
0,
0,
0)
,(
7,
6,
5,
1,
1,
1,
1,
1,
1,
0,
0)
,(
7,
5,
5,
5,
1,
1,
0,
0,
0,
0,
0)
,(
6,
6,
6,
6,
0,
0,
0,
0,
0,
0,
0)
,(
6,
6,
6,
5,
1,
0,
0,
0,
0,
0,
0)
,(
6,
6,
6,
1,
1,
1,
1,
1,
1,
0,
0)
,(
6,
6,
5,
5,
1,
1,
0,
0,
0,
0,
0)
,(
6,
5,
5,
5,
1,
1,
1,
0,
0,
0,
0)
, and(
5,
5,
5,
5,
1,
1,
1,
1,
0,
0,
0)
. By Theorem 2, we can check that such a graph does not exist. The lemma is proved.2
We can easily obtain the following lemma.
Lemma 4. Let k
2. Let G be a hamiltonian connected graph. Then deleting any set S of k vertices from G, the resulting graph G−
S contains at most k−
1 connected components.By the above lemma, we have a simple observation.
Lemma 5. Let k
2. Let G be a graph. If there is a set S of k ver-tices such that G−
S contains k or more connected components, then G is not hamiltonian connected.3. Main result
Lemma 6. Let n
4 and F⊂
E(K
n)
withδ(K
n−
F)3.Then Kn
−
F is hamiltonian connected if|
F|
2n−
10 forn
∈ {
/
4,
5,
8,
10}
,|
F| =
0 for n=
4,|
F|
2 for n=
5, and|
F|
2n−
11 for n∈ {
8,
10}
.Proof. We prove this lemma by induction on n. Yet, we
should be very careful because the size of
|
F|
is depending on n. Without loss of generality, we assume that|
F| =
2n−
10 for n∈ {
/
4,
5,
8,
10}
,|
F| =
0 for n=
4,|
F| =
2 for n=
5, and|
F| =
2n−
11 for n∈ {
8,
10}
. The induction bases are n=
4, n=
5, and n=
6. Suppose n=
4 and|
F| =
0. It is easy to see that the complete graph K4 is hamiltonian connected. Suppose n=
5 and|
F| =
2. To keepδ(K
5−
F)
3, F forms two independent edges. By brute force, it is easy to check whether K5−
F is hamiltonian connected. Suppose that n=
6 and|
F| =
2. Obviously, F is either two adjacent edges or two independent edges. Again, by brute force, we can check that K6−
F is hamiltonian connected. Now, we assume that n7. Let u and v be any two vertices of Kn. The lemma follows if we can find ahamil-tonian path of Kn
−
F between u and v.Case 1. degKn(F)
(u)
=
0 or degKn(F)(v)
=
0. Without lossof generality, we assume that degKn(F)
(u)
=
k=
0. Let i1, . . . ,
ikbe the vertices such that(u,
ij)
∈
F for 1jk.Let F
= (
F−{(
u,i1), . . . , (u,
ik)
})∪{(
v,i1), . . . , (v,
ik)
}
.Ob-viously,
|
F| |
F|
. Now, we consider Kn−{
u}
as a completegraph of
(n
−
1)
vertices with faulty edge set F. Obviously,|
F|
2(n
−
1)
−
8 for n∈ {
/
8,
10}
and|
F|
2(n
−
1)
−
9 for n∈ {
8,
10}
. Moreover,δ(K
n− {
u} −
F)
2. Thus, wecan apply Theorem 1 to obtain a hamiltonian cycle C in Kn
− {
u} −
F. Without loss of generality, we write C as v,x, . . . ,y,v. Then, u,x, . . . ,y,v forms a hamiltonian path of Kn−
F joining u to v.Case 2. degKn(F)
(u)
=
0 and degKn(F)(v)
=
0. By Lem-mas 1, 2, and 3, there exists a vertex w such that 2degKn(F)(w)
n−1
2
−
1 for n∈ {
9,
11}
and 1 degKn(F)(w)
n−21
−
1 for n∈ {
/
9,
11}
.Obviously,
δ(K
n−
F− {
w})
2. Suppose thatδ(K
n−
F
− {
w}) =
2. Let x be any vertex in Kn− {
w}
such thatdegKn−{w}−F
(x)
=
2. Obviously,(x,
w) /∈
F , degKn−F(x)
=
3,and degKn(F)
(x)
=
n−
4. We claim that x is the only vertex in Kn− {
w}
with degKn−{w}−F(x)
=
2. If otherwise, let zbe another vertex in Kn
− {
w}
with degKn−{w}−F(z)
=
2.Then
|
F|
degKn(F)(x)
+
degKn(F)(z)
−
1=
2n−
9. This is impossible because|
F|
2n−
10. Thus, x is the only vertex in Kn−{
w}
such that degKn−{w}−F(x)
=
2. Thus,δ(K
n−
F−
{
u,x})
3.Let F
=
F− {(
x,i)|
i∈
V(K
n)
}
. We consider Kn− {
u,x}
as a complete graph of
(n
−
2)
vertices with faulty edge set F. Obviously,|
F| =
12 for n=
7,|
F| =
n−
72(n
−
2)
−
10 for n∈ {
/
10,
12}
, and|
F| =
n−
72(n
−
2)
−
11 for n∈ {
10,
12}
. By induction, we have a hamiltonian path P of Kn− {
u,x} −
F joining w to v. Sou,x,w,P,
vformsa hamiltonian path of Kn
−
F joining u to v.Now, we consider
δ(K
n− {
w} −
F)
3. Since 2degKn(F)
(w)
n−1
2
−
1 for n∈ {
9,
11}
and 1 degKn(F)(w)
n−21
−
1 for n∈ {
/
9,
11}
, there exists(x,
y)∈
F such that{(
w,x), (w,y)} ∩
F= ∅
. We set F as F− {(
w,z)| (
w,z)∈
F} − {(
x,y)}
and consider Kn− {
w}
with faulty set F. We have
|
F| =
2n−
10−
degKn(F)(w)
−
1
2n−
13 for n∈ {
9,
11}
and|
F| =
2n−
10−
degKn(F)(w)
−
1
2n−
12 for n∈ {
/
9,
11}
. By induction, there exists a hamiltonian path P=
u=
x1,
x2, . . . ,
xn−1=
v of Kn−
{
w} −
F joining u to v. Suppose that(x,
y)∈
P . There exists an integer i such that{
xi,
xi+1} = {x,y}
for some i.Suppose that
(x,
y) /∈
P . Since degKn(F)(w)
n−21
−
1 and degKn(F)(w)
+
degKn−F(w)
=
n−
1, degKn−F(w)
n
2
+
1. Hence, there exists an integer i such that(x
i,
xi+1)
∈
P and{(
w,xi), (w,
xi+1)
}∩
F= ∅
. Therefore,u=
x1,
x2, . . . ,
xi,
w,xi+1
,
xi+2, . . . ,
vforms a hamiltonian path of Kn−
Fjoin-ing u to v.
2
Theorem 3. Let n
4. ThenHC
3e(K
n)
=
2n−
10 for n∈
/
{
4,
5,
8,
10}
,HC
3e(K
4)
=
0,HC
3e(K
5)
=
2,HC
3e(K
8)
=
5, andHC
3e
(K
10)
=
9.Proof. Let F be any subset of E
(K
n)
withδ(K
n−
F)
3.Since
δ(K
n−
F)
3,|
F| =
0 for n=
4 and|
F|
2 for n=
5.Thus,
HC
3e(K
4)
=
0 andHC
3e(K
5)
=
2.Suppose n
=
8. Let V(K
8)
= {
x1,
x2, . . . ,
x8}. We set R=
{
x1, . . . ,
x4}, S= {
x5, . . . ,
x8}, and F= {(
u,v)|
u,v∈
R}
. We can check thatδ(
K8−
F)
3,|
F| =
6 and(K
8−
F)−
S has four connected components. By Lemma 5, K8−
F is not hamiltonian connected. See Fig. 1(a) for illustration. Thus,HC
3e
(K
8) <
6. By Lemma 6,HC
3e(K
8)
=
5.Suppose n
=
10. Let V(K
10)
= {
x1,
x2, . . . ,
x10}. We set R= {
x1, . . . ,
x5}, S= {
x6, . . . ,
x10}, and F= {(
u,v)|
u,v∈
588 T.-Y. Ho et al. / Information Processing Letters 109 (2009) 585–588
Fig. 1. All white vertices are in R, all black vertices are in S, and all gray
vertices are in T . All dashed lines are in F .
R
}
. Then,δ(K
10−
F)
3,|
F| =
10, and(K
10−
F)−
S has five connected components. By Lemma 5, K10−
F is not hamiltonian connected. See Fig. 1(b) for illustration. Thus,HC
3e
(K
10) <
10. By Lemma 6,HC
3e(K
10)
=
9.Suppose that n
∈ {
6,
7,
9} ∪ {
i|
i 11}
. Let V(K
n)
=
{
x1,
x2, . . . ,
xn}
. We set R= {
x1,
x2}, S= {
x3,
x4,
x5}, T=
{
x6, . . . ,
xn}
, and F= {(
u,v)|
u∈
R,v∈
R∪
T}
. Obviously,δ(
Kn−
F)3,|
F| =
2(n
−
5)
+
1=
2n−
9, and(K
n−
F)
−
Shas three connected components. See Fig. 1(c) for illustra-tion for case n
=
9. By Lemma 5, Kn−
F is nothamil-tonian connected. Thus,
HC
3e(K
n) <
2n−
9. By Lemma 6,HC
3e
(K
n)
=
2n−
10.The theorem is proved.
2
References
[1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North-Holland, New York, 1980.
[2] M.Y. Chan, S.J. Lee, On the existence of Hamiltonian circuits in faulty hypercubes, SIAM Journal on Discrete Mathematics 4 (1991) 511–527. [3] J.S. Fu, Conditional fault hamiltonicity of the complete graph,
Informa-tion Processing Letters 107 (2008) 110–113.
[4] W.T. Huang, J.J.M. Tan, C.N. Hung, L.H. Hsu, Fault-tolerant hamiltonic-ity of twisted cubes, Journal of Parallel and Distributed Computing 62 (2002) 591–604.
[5] S. Latifi, S.Q. Zheng, N. Bagherzadeh, Optimal ring embedding in hy-percubes with faulty links, in: Proceedings of the IEEE Symposium on Fault-Tolerant Computing, 1992, pp. 178–184.
[6] T.K. Li, J.J.M. Tan, L.H. Hsu, Hyper Hamiltonian laceability on the edge fault star graph, Information Sciences 165 (2004) 59–71.
[7] J.W. Moon, On a problem of Ore, Math. Gaz. 49 (1965) 40–41. [8] O. Ore, Hamilton connected graphs, Journal of Mathematic Pures
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