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International Journal of Computer
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The construction of mutually
independent Hamiltonian cycles in
bubble-sort graphs
Yuan-Kang Shih a , Cheng-Kuan Lin a , D. Frank Hsu b , Jimmy J. M. Tan a & Lih-Hsing Hsu c
a
Department of Computer Science , National Chiao Tung University , Hsinchu, Taiwan, R.O.C.
b
Department of Computer and Information Science , Fordham University , New York, NY, USA
c
Department of Computer Science and Information Engineering , Providence University , Taichung, Taiwan, R.O.C.
Published online: 20 Apr 2010.
To cite this article: Yuan-Kang Shih , Cheng-Kuan Lin , D. Frank Hsu , Jimmy J. M. Tan & Lih-Hsing Hsu (2010) The construction of mutually independent Hamiltonian cycles in bubble-sort graphs, International Journal of Computer Mathematics, 87:10, 2212-2225, DOI:
10.1080/00207160802512700
To link to this article: http://dx.doi.org/10.1080/00207160802512700
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International Journal of Computer Mathematics
Vol. 87, No. 10, August 2010, 2212–2225
The construction of mutually independent Hamiltonian cycles
in bubble-sort graphs
Yuan-Kang Shiha, Cheng-Kuan Lina, D. Frank Hsub, Jimmy J. M. Tanaand Lih-Hsing Hsuc*
aDepartment of Computer Science, National Chiao Tung University, Hsinchu, Taiwan, R.O.C.; bDepartment of Computer and Information Science, Fordham University, New York, NY, USA;cDepartment
of Computer Science and Information Engineering, Providence University, Taichung, Taiwan, R.O.C.
(Received 20 March 2008; final version received 4 September 2008)
A Hamiltonian cycle C= u1, u2, . . . , un(G), u1 with n(G) = number of vertices of G, is a cycle C(u1; G), where u1is the beginning and ending vertex and ui is the ith vertex in C and ui= uj for
any i= j, 1 ≤ i, j ≤ n(G). A set of Hamiltonian cycles {C1, C2, . . . , Ck} of G is mutually independent if any two different Hamiltonian cycles are independent. For a hamiltonian graph G, the mutually
indepen-dent Hamiltonianicity number of G, denoted by h(G), is the maximum integer k such that for any vertex
uof G there exist k-mutually independent Hamiltonian cycles of G starting at u. In this paper, we prove that h(Bn)= n − 1 if n ≥ 4, where Bnis the n-dimensional bubble-sort graph.
Keywords: Hamiltonian cycle; bubble-sort networks; interconnection networks; mutually independent
Hamiltonian cycles; Cayley graph
2000 AMS Subject Classifications: 05C38; 05C45; 05C75; 05C90; 68M10
1. Introduction
Let H be a group, and let S be a generating set of H with S−1= S. The Cayley graph on a group
H with generating set S, denoted by Cay(H; S), is the graph with vertex set H , and for two vertices u and v in H , u is adjacent to v if and only if v= us for some s ∈ S. Hamiltonian cycles in Cayley graphs exist naturally in computing and communication [10], in the study of word-hyperbolic groups and automatic groups [6], in changing–ringing [13], in creating Escher-like repeating patterns in hyperbolic plane 1 [4], and in combinatorial designs [4]. It is conjectured that every connected Cayley graph with more than three vertices is Hamiltonian [3]. Up to now, this conjecture is unsolved. Yet, some Cayley graphs have many more Hamiltonian cycles than we expected. In this paper, we introduce and study the concept of mutually independent Hamiltonian (MIH) cycles in Cayley graphs.
For graph definitions and notations we follow [2]. G= (V, E) is a graph if V is a finite set and
Eis a subset 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. We use n(G) to denote|V |. Let S be a nonempty subset of V (G). The subgraph *Corresponding author. Email: lhhsu@pu.edu.tw
ISSN 0020-7160 print/ISSN 1029-0265 online © 2010 Taylor & Francis
DOI: 10.1080/00207160802512700 http://www.informaworld.com
induced by S is the subgraph of G with its vertex set S and with its edge set consisting of all
edges of G joining any two vertices in S. We use G− S to denote the subgraph of G induced by
V − S. Two vertices u and v are adjacent if (u, v) is an edge of G. The set of neighbours of u,
denoted by NG(u), is{v | (u, v) ∈ E}. The degree of a vertex u of G, degG(u), is the number of
edges incident with u. The minimum degree of G, δ(G), is min{degG(x)| x ∈ V }. A graph G is k-regular if degG(u)= k for every vertex u in G. A path between vertices v0and vkis a sequence
of vertices represented byv0, v1, . . . , vk with no repeated vertex and (vi, vi+1)is an edge of G
for every i, 0≤ i ≤ k − 1. We use Q(i) to denote the ith vertex viof Q= v1, v2, . . . , vk. We also write the pathv0, v1, . . . , vk as v0, . . . , vi, Q, vj, . . . , vk, where Q is a path form vi to vj. A cycle is a path with at least three vertices such that the first vertex is the same as the last. A Hamiltonian cycle of G is a cycle that traverses every vertex of G. A graph is Hamiltonian if it has a Hamiltonian cycle. A graph G= (B ∪ W, E) is bipartite with bipartition B and W if
V (G)= B ∪ W, B ∩ W = ∅, and E(G) is a subset of {(u, v) | u ∈ B and v ∈ W}. Let G be a
bipartite graph with bipartition B and W . We say that a Hamiltonian bipartite graph is Hamiltonian
laceable if there is a Hamiltonian path between any pair of vertices{x, y}, where x in B and y in W. Let a, b, m∈ Z with m > 0. Then a is said to be congruent to b modulo m, denoted a ≡ b mod m, if m|(a − b).
A Hamiltonian cycle C(u1; G) of a Hamiltonian graph G is described as C(u1; G) =
u1, u2, . . . , un(G), u1 to emphasize the order of vertices in C. Thus, u1 is the beginning ver-tex and ui is the ith vertex in C. Two Hamiltonian cycles of G beginning at the vertex x, C1= C(u1; G) = u1, u2, . . . , un(G), u1 and C2 = C(v1; G) = v1, v2, . . . , vn(G), v1, are
inde-pendent if x= u1 = v1and ui = vi for every i, 2≤ i ≤ n(G). Let G be a Hamiltonian graph.
A set of Hamiltonian cycles{C1, C2, . . . , Ck} of G is mutually independent if any two different Hamiltonian cycles are independent. The mutually independent Hamiltonianicity number of a Hamiltonian graph G, called the MIH number of G and denoted by h(G), is the maximum inte-ger k such that for any vertex u of G there exist k-mutually independent Hamiltonian cycles of
Gstarting at u. Obviously, h(G)≤ δ(G) for a Hamiltonian graph G. The concept of MIH cycles can be applied in many different areas. Interested readers can refer to [7, 9, 11, 12] for a more detailed introduction.
In this paper, we study MIH cycles of n-dimensional bubble-sort graph Bn. In the following
section, we give some basic properties for the n-dimensional bubble-sort graph. In Section 3, we construct MIH cycles in Bnand compute h(Bn), the MIH number of Bn.
2. The bubble-sort graphs
We setn = {1, 2, . . . , n} if n is a positive integer and we set 0 being the empty set. The
n-dimensional bubble-sort graph, Bn, is the graph with vertex set V (Bn)= {u1, . . . , un| ui ∈ n
and ui= uj for i= j}. The adjacency is defined as follows: u1, . . . , ui−1, ui, . . . , unis adjacent to v1, . . . , vi−1, vi, . . . , vnthrough an edge of dimension i with 2≤ i ≤ n if vj = uj for every j ∈ n − {i − 1, i}, vi−1= ui, and vi = ui−1, i.e., swap ui−1and ui. The bubble-sort graphs B2, B3, and B4are illustrated in Figure 1. It is known that the connectivity of Bnis (n− 1). We use
boldface to denote vertices in Bn. Hence, u1,u2, . . . ,undenote a sequence of vertices in Bn.
By definition, Bnis an (n− 1)-regular graph with n! vertices. We use e to denote the vertex
1, 2, . . . , n. It is known that Bnis a bipartite graph with one partite set containing those vertices
corresponding to odd permutations and the other containing those vertices corresponding to even permutations. We use white vertices to represent those even permutation vertices and use black vertices to represent those odd permutation vertices. Let u= u1, u2, . . . , unbe an arbitrary vertex
of the bubble-sort graph Bn. We say that ui is the ith coordinate of u, (u)i, for 1≤ i ≤ n. For
Figure 1. The graphs B2, B3, and B4.
1≤ i ≤ n, let Bn{i}−1be the subgraph of Bninduced by those vertices u with (u)n= i. Then Bncan
be decomposed into n subgraphs Bn{i}−1, 1≤ i ≤ n, and each Bn{i}−1 is isomorphic to Bn−1. Thus,
the bubble-sort graph can also be constructed recursively. Let I be any subset ofn. We use BI
n−1
to denote the subgraph of Bninduced by
i∈IV (B
{i}
n−1). For any two distinct elements i and j
inn, we use Eni,j−1to denote the set of edges between Bn{i}−1and Bn{j}−1. By the definition of Bn,
there is exactly one neighbour v of u such that u and v are adjacent through an i-dimensional edge with 2≤ i ≤ n. For this reason, we use (u)ito denote the unique i-neighbour of u. We have
((u)i)i = u and (u)n∈ B{(u)n−1}
n−1 .
Lemma1 Let i and j be any two distinct elements inn with n ≥ 3. Then |Ei,j
n−1| = (n − 2)!. Moreover, there are (n− 2)!/2 edges joining black vertices of Bn{i}−1to white vertices of Bn{j}−1.
Theorem1 (See [8]) The bubble-sort graph Bnis Hamiltonian laceable if and only if n= 3. Theorem2 (See [1]) Let x be a black vertex in Bnwith n≥ 4. Suppose that u and v are two
distinct white vertices in Bn. There is a Hamiltonian path of Bn− {x} joining u to v.
Lemma2 Let I = {a1, a2, . . . , ar} be a subset of n for some r ∈ n with n ≥ 5. Assume that
u is a white vertex in B{a1}
n−1 and v is a black vertex in B{a
r}
n−1. Then there is a Hamiltonian path
u = x1, H1,y1,x2, H2,y2, . . . ,xr, Hr,yr= v of BnI−1 joining u to v such that x1 = u, yr=v, and Hiis a Hamiltonian path of Bn{a−1i} joining xi to yi for every i, 1≤ i ≤ r.
Proof We set x1= u and yr= v. By Theorem 1, this lemma holds for r = 1. Suppose that r≥ 2. By Lemma 1, there are (n − 2)!/2 ≥ 3 edges joining black vertices of B{ai}
n−1 to white
vertices of B{ai+1}
n−1 for every i∈ r − 1. We can choose an edge (yi,xi+1)∈ Enai−1,ai+1 with yi
being a black vertex and xi+1 being a white vertex for every i∈ r − 1. By Theorem 1,
there is a Hamiltonian path Hi of Bn{a−1i} joining xi to yi for every i∈ r. Then the path
u = x1, H1,y1,x2, H2,y2, . . . ,xr, Hr,yr= v is the desired path.
Lemma3 Let Bn{a}−1and Bn{b}−1be two distinct subgraphs of Bnwith n≥ 5. Let s be a black vertex
in Bn{a}−1, let t be a white vertex in Bn{b}−1, let u be a white vertex in Bn{a}−1, and let v be a black vertex in Bn{b}−1. Then there is a Hamiltonian path of Bn{a,b}−1 − {s, t} joining u to v.
Proof Let x be a white vertex in Bn{a}−1− {u} with (x)nbeing a black vertex in Bn{b}−1− {v}. By Theorem 2, there are Hamiltonian paths: Q1 of Bn{a}−1− {s} joining u to x; Q2 of Bn{b}−1− {t}
joining (x)nto v. Thenu, Q
1,x, (x)n, Q2,v is the Hamiltonian path of Bn{a,b}−1 − {s, t} joining u
to v.
Lemma4 For n≥ 5, let u be a black vertex in B{n}
n−1and let v be a white vertex in B {1} n−1. Then there is a Hamiltonian path of Bn− {e, (e)n} joining u to v.
Proof Let y be a white vertex in Bn{n−1}−1 with (y)n−1= n − 2. By Lemma 3, there is a Hamiltonian
path Q1of Bn{n−1,n}−1 − {e, (e)n} joining u to y. By Lemma 2, there is a Hamiltonian path Q
2 of
n−2
i=1 Bn{i}−1joining the black vertex (y)nto v. Thenu, Q1,y, (y)n, Q2,v is a Hamiltonian path
of Bn− {e, (e)n} joining u to v.
3. The MIH property of Bn
For every i inn − 1 with n ≥ 5, we set zi0= (e)i+1and we set zji = (zij−1)i+j+1 for any j in
n − i − 1. Let A5
5be the empty set, let A
i
5= {z
i
j | j ∈ 4 − i ∪ {0}} for any i in 4, and let Ain= {zij | j ∈ n − i − 1 ∪ {0}} for any i in n − 1 for n ≥ 6. We set Xn1= A1n∪ A2n∪ {e}, X2
n= A2n∪ A4n∪ {e}, Xn3= A3n∪ A4n∪ {e}, and X4n= A3n∪ A5n∪ {e} for any n ≥ 5. We set Yni = Ai
n∪ Ain+1∪ {e} for n ≥ 6 and for 3 ≤ i ≤ n − 2.
Lemma5 There is a Hamiltonian path of B5− X15joining a vertex u with (u)5= 5 to a vertex
v with (v)5= 1 such that the colour of u and the colour of v are distinct.
Proof We set Q1= 12435, 21435, 24135, 24315, 42315, 42135, 41235, 14235, 14325, 41325, 43125, 34125, 31425, 31245, 32145, 32415, 34215, 43215. Note that Q1is a Hamiltonian path of B4{5}− (X1
5− {z 1
3 = 23451, z22= 13452}) joining the black vertex 12435 to the white vertex 43215.
Case 1 Suppose that u is a black vertex and v is a white vertex. We set u= 12435 and x = 43215. Let w be a black vertex in B4{1}with (w)4= 2. By Theorem 2, there is a Hamiltonian path Q2of
B4{1}− {z13} joining the black vertex (x)5to w. Let y be any black vertex in B4{1}with (y)4= 4. Without loss of generality, we write Q2= (x)5, R1,y, m, R2,w. By Theorem 2, there is a Hamiltonian path Q3of B4{2}− {z22} joining a white vertex s with (s)4= 3 to (w)5. By Lemma 2, there is a Hamiltonian path Q4of B4{3,4}joining the white vertex (y)5to the black vertex (s)5. We let v= m. Then u, Q1,x, (x)5, R1,y, (y)5, Q4, (s)5,s, Q3, (w)5,w, R2−1,m= v is the desired path.
Case 2 Suppose that u is a white vertex and v is a black vertex. We set u= 43215 and x = 12435. Let v be a black vertex in B4{1}, and let s be a white vertex in B4{2} with (s)4= 4. By Lemma 3, there is a Hamiltonian path Q2of B4{1,2}− {z
1 3,z
2
2} joining s to v. By Lemma 2, there is a Hamiltonian path Q3 of B4{3,4}joining the white vertex (x)5to the black vertex (s)5. Then
u, Q−1
1 ,x, (x) 5, Q
3, (s)5,s, Q2,v is the desired path.
Lemma6 For n≥ 5, there is a Hamiltonian path of Bn− X1njoining a vertex u with (u)n= n
to a vertex v with (v)n= 1 such that the colour of u and the colour of v are distinct.
Proof We prove this statement by induction on n. By Lemma 5, this statement holds for n= 5. We suppose that this statement holds for n− 1 with n ≥ 6. We have the following cases.
Case 1 Suppose that u is a black vertex and v is a white vertex.
Case 1.1 Suppose that n is even. Thus, z1
n−2is a black vertex in Bn{1}−1and z2n−3is a white vertex
in Bn{2}−1. By induction, there is a Hamiltonian path Q1 of Bn{n}−1− (X1
n− {z1n−2,z2n−3}) joining
u to a white vertex q with (q)n−1= 1. Obviously, (q)nis the black vertex in Bn{1}−1. Let s and
w be two white vertices in Bn{1}−1 with (s)n−1= n − 1 and (w)n−1= 2. By Theorem 2, there
is a Hamiltonian path Q2 of Bn{1}−1− {z1
n−2} joining w to s. Without loss of generality, we write Q2= w, R1,m, (q)n, R2,s. Let t be a black vertex in Bn{2}−1with (t)n−1= 3. By Theorem 2, there
is a Hamiltonian path Q3of Bn{2}−1− {z2
n−3} joining t to the black vertex (w)n. By Lemma 2, there is
a Hamiltonian path Q4ofni=3−1Bn{i}−1joining the black vertex (s)nin B{n−1}
n−1 to the white vertex (t)n
in Bn{3}−1. We set v= m. Then u, Q1,q, (q)n, R2,s, (s)n, Q4, (t)n,t, Q3, (w)n,w, R1,m= v is the desired path.
Case 1.2 Suppose that n is odd. Thus, z1
n−2is a white vertex in B {1} n−1and z
2
n−3is a black vertex
in Bn{2}−1. The proof of this case is similar to Case 1.1.
Case 2 Suppose that u is a white vertex and v is a black vertex.
Case 2.1 Suppose that n is even. Thus, z1
n−2is a black vertex in B {1} n−1and z
2
n−3is a white vertex
in Bn{2}−1. By induction, there is a hamiltonian path Q1 of Bn{n}−1− (X
1
n− {z1n−2,z
2
n−3}) joining
u to a black vertex p with (p)n−1 = n − 1. Let s and t be any two white vertices in Bn{1}−1 with (s)n−1= 2 and (t)n−1= 2. By Theorem 2, there is a Hamiltonian path Q2of Bn{1}−1− {z
1
n−2} joining
s to t. Let y be any white vertex in Bn{1}−1− {s, t} with (y)n−1= 3. Without loss of generality,
we write Q2= s, R1,y, m, R2,t. By Theorem 2, there is a Hamiltonian path Q3 of Bn{2}−1−
{z2
n−3} joining the black vertex (s)
nto the black vertex (t)n. By Lemma 2, there is a Hamiltonian
path Q4of
n−1
i=3 B
{i}
n−1 joining the white vertex (p)
nto the black vertex (y)n. Let v= m. Then
u, Q1,p, (p)n, Q4, (y)n,y, R−11 ,s, (s)n, Q3, (t)n,t, R2−1,m= v is the desired path.
Case 2.2 Suppose that n is odd. Thus, z1n−2is a white vertex in Bn{1}−1and z2n−3is a black vertex in Bn{2}−1. Note that n is odd. The proof of this case is similar to Case 2.1.
Lemma7 There is a Hamiltonian path of B5− X25 joining a white vertex u with (u)5= 5 to a
white vertex v with (v)5= 1.
Proof We set Q1= 24135, 42135, 41235, 41325, 14325, 14235, 12435, 21435, 21345, 23145, 32145, 31245, 31425, 34125, 43125, 43215, 42315, 24315, 23415, 32415, 34215 being a hamil-tonian path of B4{5}− (X2
5− {z 2
2= 13452, z40= 12354}) joining the black vertex p = 24135 to the black vertex q= 34215. Let r be any black vertex in Q1with (r)4= 4. Obviously, (q)5is a white vertex in B4{1}. Without loss of generality, we rewrite Q1= p, R1,m, r, R2,q. Let s be a white vertex in B4{4}with (s)4= 1, and let w be a black vertex in B4{3}with (w)4= 2. By Theorem 1, there is a Hamiltonian path Q2of B4{3}joining the white vertex (p)5to w. By Lemma 3, there is a
Hamiltonian path Q3of B4{1,2}− {z2
2, (q)5} joining the white vertex (w)5to the black vertex (s)5. By Theorem 2, there is Hamiltonian path Q4of B4{4}− {z4
0} joining s to (r)
5. We set v= (q)5and
u= m. Then u = m, R1−1,p, (p)5, Q
2,w, (w)5, Q3, (s)5,s, Q4, (r)5,r, R2,q, (q)5 = v is the
desired path.
Lemma8 For n≥ 5, there is a Hamiltonian path of Bn− X2njoining a vertex u with (u)n= n
to a vertex v with (v)n= 1, where both u and v are black vertices if n is even, and both u and v are white vertices if n is odd.
Proof We prove this statement by induction on n. By Lemma 7, this statement holds on n= 5. We suppose that this statement holds on n− 1 with n ≥ 6.
Case 1 Suppose that n is even. It is easy to know that z2
n−3and z
4
n−5are two white vertices. By
induction, there is a Hamiltonian path Q1of Bn{n}−1− (X2n− {z2n−3,z4n−5}) joining a white vertex p with (p)n−1 = n − 1 to a white vertex q with (q)n−1= 1. Obviously, (q)nis a black vertex
in Bn{1}−1. Let t be a white vertex in Q1 with (t)n−1= 2. We rewrite Q1= p, R1,m, t, R2,q. Let s be a black vertex in Bn{2}−1 with (s)n−1= 1, and let w be a black vertex in Bn{4}−1 with (w)n−1= 3. By Lemma 2, there is a Hamiltonian path Q2 of (∪ni=5−1B
{i}
n−1
Bn{3}−1)joining the black vertex (p)n to the white vertex (w)n. By Lemma 3, there is a Hamiltonian path Q3 of Bn{1,4}−1 − {z4
n−5, (q)
n} joining w to white vertex (s)n. By Theorem 2, there is a
Hamilto-nian path Q4 of Bn{2}−1− {z2n−3} joining s to the black (t)n. We set v= (q)nand u= m. Then
u = m, R−11 ,p, (p)
n, Q
2, (w)n,w, Q3, (s)n,s, Q4, (t)n,t, R2,q, (q)n= v is the desired path.
Case 2 Suppose that n is odd. It is easy to know that z2n−3and z4n−5are two black vertices. The proof of this case is similar to Case 1.
This completes the proof.
Lemma9 There is a Hamiltonian path of B5− X3
5joining a vertex u with (u)5= 5 to a vertex
v with (v)5= 1 such that the colour of u and the colour of v are distinct.
Proof We set Q1 = 21435, 21345, 23145, 23415, 32415, 32145, 31245, 13245, 13425, 31425, 34125, 43125, 41325, 14325, 14235, 41235, 42135, 24135, 24315, 42314, 43215, 34215. Obviously, Q1is a Hamiltonian path of B4{5}− (X53− {z
3
1 = 12453, z40= 12354)} joining the white vertex 21435 to the black vertex 34215.
Case 1 Suppose u is a white vertex and v is a black vertex. We set u= 21435 and x = 34215. Obviously, (x)5is the white vertex in B{1}
4 . Let w be a black vertex in B
{1}
4 with (w)4= 2. Note that (w)5is the white vertex in B{2}
4 . By Theorem 1, there is a Hamiltonian path Q2of B4{1}joining
(x)5to w. Let y be any white vertex in B4{1}with (y)4= 3. Without loss of generality, we write
Q2= (x)5, R1,y, m, R2,w. Let t be a white vertex in B4{4}with (t)4= 2. By Lemma 3, there is a Hamiltonian path Q3of B4{3,4}− {z13,z40} joining the black vertex (y)5 to t. By Theorem 1, there is a Hamiltonian path Q4of B4{2}joining the black vertex (t)5to (w)5. Let v= m. Then
u, Q1,x, (x)5, R1,y, (y)5, Q3,t, (t)5, Q4, (w)5,w, R−12 ,m= v is the desired path.
Case 2 Suppose u is a black vertex and v is a white vertex. We set u= 34215 and x = 21435. Let y be a white vertex in B4{4}with (y)4= 2. By Lemma 3, there is a Hamiltonian path Q2 of
B4{3,4}− {z3
1,z40} joining the black vertex (x)5to y. Let v be any white vertex in B4{1}. By Lemma 2,
there is a Hamiltonian path Q3of B4{1,2}joining the black vertex (y)5to the white vertex v. Then
u, Q−1
1 ,x, (x)5, Q2,y, (y)5, Q3,v is the desired path.
Lemma10 For n≥ 5, there is a hamiltonian path of Bn− X3
njoining a vertex u with (u)n= n to a vertex v with (v)n= 1, where the colour of u and the colour of v are distinct.
Proof We prove this statement by induction on n. By Lemma 9, this statement holds for n= 5. We suppose that this statement holds for n− 1 with n − 1 ≥ 5.
Case 1 Suppose that u is a white vertex and v is a black vertex.
Case 1.1 Suppose that n is even. Thus, z3n−4is a black vertex in Bn{3}−1and z4n−5is a white vertex in Bn{4}−1. By induction, there is a Hamiltonian path Q1of Bn{n}−1− (X3
n− {z
3
n−4,z
4
n−5}) joining a
white vertex u to a black vertex q with (q)n−1 = 1. Obviously, (q)nis a white vertex in Bn{1}−1.
Let w be a black vertex in Bn{1}−1 with (w)n−1 = 2. By Theorem 1, there is a Hamiltonian path Q2of Bn{1}−1 joining (q)nto w. Let y be any white vertex in Bn{1}−1 with (y)n−1= 4. Without loss
of generality, we rewrite Q2= (q)n, R
1,y, m, R2,w. Let t be a white vertex in Bn{3}−1 with (t)n−1= 5. By Lemma 3, there is a Hamiltonian path Q3 of Bn{3,4}−1 − {z3n−4,z4n−5} joining the
black vertex (y)nto t. By Lemma 2, there is a Hamiltonian path Q
4 of ( n−1 i=5B {i} n−1 Bn{2}−1)
joining the black vertex (t)n in Bn{5}−1 to the white vertex (w)n in Bn{2}−1. We let v= m. Then
u, Q1,q, (q)n, R1,y, (y)n, Q3,t, (t)n, Q4, (w)n,w, R−12 ,m= v is the desired path.
Case 1.2 Suppose that n is odd. Thus, z3
n−4is a white vertex in Bn{3}−1and z4n−5is a black vertex
in Bn{4}−1. The proof of this case is similar to Case 1.1.
Case 2 Suppose that u is a black vertex and v is a white vertex.
Case 2.1 Suppose that n is even. Thus, z3
n−4is a black vertex in Bn{3}−1and z4n−5is a white vertex in Bn{4}−1. By induction, there is a Hamiltonian path Q1of Bn{n}−1− (X
3
n− {z3n−4,z
4
n−5}) joining a black
vertex u to a white vertex q with (q)n−1= n − 1. Obviously, (q)nis a black vertex in Bn{n−1}−1 . Let y
be a white vertex in Bn{2}−1with (y)n−1= 4, and let s be a white vertex in Bn{3}−1with (s)n−1= 1. By
Lemma 2, there is a Hamiltonian path Q2of (ni=5−1Bn{i}−1∪ Bn{2}−1)joining (q)nto y. By Lemma 3, there is a Hamiltonian path Q3 of Bn{3,4}−1 − {z3n−4,z4n−5} joining the black vertex (y)nto s. Let
v be any white vertex in Bn{1}−1. By Theorem 1, there is a Hamiltonian path Q4 of Bn{1}−1joining the black vertex (s)nto v. Thenu, Q1,q, (q)n, Q2,y, (y)n, Q3,s, (s)n, Q4,v is the desired path.
Case 2.2 Suppose that n is odd. Thus, z3
n−4is a white vertex in Bn{3}−1and z4n−5is a black vertex
in Bn{4}−1. The proof of this case is similar to Case 2.1.
Lemma11 There is a Hamiltonian path of B5− X54joining a black vertex u with (u)5= 5 to a
black vertex v with (v)5 = 1.
Proof We set Q1= 34215, 43215, 42315, 24315, 24135, 42135, 41235, 14235, 14325, 41325, 43125, 34125, 31425, 13425, 13245, 31245, 32145, 32415, 23415, 23145, 21345, 21435. Obvi-ously, Q1is a Hamiltonian path of B4{5}− (X45− {z
3
1 = 12453}) joining the black vertex 34215
to the white vertex 21435. Let u= 34215 and x = 21435. Let y be a black vertex in B4{3} with
(y)4= 4. By Theorem 2, there is a Hamiltonian path Q2of B4{3}− {z31} joining the black vertex
(x)5to y. Let v be a black vertex in B{1}
4 . By Lemma 2, there is a hamiltonian path Q3of B4{1,2,4} joining the white vertex (y)5to v. Thenu, Q
1,x, (x)5, Q2,y, (y)5, Q3,v is the desired path.
Lemma12 For n≥ 5, there is a Hamiltonian path of Bn− X4njoining a vertex u with (u)n= n
to a vertex v with (v)n= 1, where both u and v are white vertices if n is even and both u and v are black vertices if n is odd.
Proof We prove this statement by induction for n. By Lemma 11, this statement holds for n= 5. We suppose that this statement holds on n− 1 with n ≥ 6.
Case 1 Suppose that n is even. It is easy to know that z5
n−6and z
3
n−4are two black vertices. By
induction, there is a Hamiltonian path Q1of Bn{n}−1− (X
4
n− {z
3
n−4,z
5
n−6}) joining a black vertex
p with (p)n−1 = n − 1 to a black vertex q with (q)n−1 = 1. Let t be a black vertex in Q1 with
(t)n−1= 3. We rewrite Q1= p, R1,m, t, R2,q. Let s be a white vertex in Bn{3}−1with (s)n−1= 1,
and let w be a white vertex in Bn{5}−1with (w)n−1= 4. By Lemma 2, there is a Hamiltonian path Q2 of (ni=6−1Bn{i}−1∪ Bn{2,4}−1)joining the white vertex (p)nto the black vertex (w)n. By Lemma 3, there is a Hamiltonian path Q3of Bn{1,5}−1 − {z5n−6, (q)n} joining w to the black vertex (s)n. By Theorem 2, there is a Hamiltonian path Q4of Bn{2}−1− {z3n−4} joining s to the white vertex (t)n. We set v= (q)n and u= m. Then u = m, R−11 ,p, (p)n, Q2, (w)n,w, Q3, (s)n,s, Q4, (t)n,t, R2,q, (q)n= v is the desired path.
Case 2 Suppose that n is odd. It is easy to know that z3
n−4and z
5
n−6are two white vertices. The
proof of this case is similar to Case 1.
Lemma13 For n≥ 6, there is a Hamiltonian path of Bn− Yn−2
n joining a vertex u with (u)n= n to a vertex v with (v)n= 1, where the colour of u and the colour of v are distinct.
Proof We know that Yn−2
n = Ann−2∪ Ann−1∪ {e}, where Ann−2 = {(e)n−1, ((e)n−1)n} and Ann−1 =
{(e)n}. By Lemma 4, there is a Hamiltonian path Q
1of Bn{n}−1− {(e)
n−1,e} joining a black vertex
p with (p)n−1= n − 1 to a white vertex q with (q)n−1= 1.
Case 1 Suppose that u is a black vertex and v is a white vertex. Let y be a black ver-tex in Bn{1}−1 with (y)n−1= n − 1, and let s be a white vertex in Bn{1}−1 with (s)n−1= 2. By
Theorem 1, there is a Hamiltonian path Q2 of Bn{1}−1 joining the black vertex (q)nto s. Without
loss of generality, we write Q2= (q)n, R
1,y, m, R2,s. Let w be a black vertex in Bn{n−2}−1 with (w)n−1= n − 3. By Lemma 3, there is a Hamiltonian path Q3of Bn{n−2,n−1}−1 − {((e)n−1)n, (e)n}
joining the white vertex (y)nto w. By Lemma 2, there is a Hamiltonian path Q4ofn−3
i=2B
{i}
n−1
joining the white vertex (w)n to the black vertex (s)n. We set u= p and v = m. Then u =
p, Q1,q, (q)n, R1,y, (y)n, Q3,w, (w)n, Q4, (s)n,s, R2−1,m= v is the desired path.
Case 2 Suppose that u is a white vertex and v is a black vertex. Let w be a black vertex in Bn{n−2}−1 with (w)n−1= n − 3. By Lemma 3, there is a Hamiltonian path Q2 of Bn{n−2,n−1}−1 −
{((e)n−1)n, (e)n} joining the white vertex (p)nto w. Let v be a black vertex in B{1}
n−1. By Lemma 2,
there is a Hamiltonian path Q3ofni=1−3Bn{i}−1joining the white vertex (w)nto v. We set u= q. Thenu = q, Q−11 ,p, (p)n, Q2,w, (w)n, Q3,v is the desired path.
Lemma14 Let n≥ 6. For every 3 ≤ i ≤ n − 2, there is a Hamiltonian path of Bn− Ynijoining
a vertex u with (u)n= n to a vertex v with (v)n= 1, where the colour of u and the colour of v are distinct.
Proof We prove this statement by induction on n. We have Y3
6 = X36. By Lemmas 10 and 13, the statement holds on n= 6. We suppose that this statement holds on n − 1 with n ≥ 7.
By induction, there is a Hamiltonian path Q1of Bn{n}−1− (Yi
n− {z
i
n−i−1,z
i+1
n−i−2}) joining a white
vertex p with (p)n−1= n − 1 to a black vertex q with (q)n−1= 1. Case 1 Suppose that u is a white vertex and v is a black vertex.
Case 1.1 Suppose that zi
n−i−1 is a white vertex in Bn{i}−1and z
i+1
n−i−2 is a black vertex in Bn{i+1}−1 .
Let y be a white vertex in Bn{1}−1 with (y)n−1= i + 2 and x be a black vertex in Bn{1}−1 with (x)n−1 = i + 1. By Theorem 1, there is a Hamiltonian path Q2of Bn{1}−1joining the white vertex (q)nto x. Without loss of generality, we rewrite Q2= (q)n, R
1,y, m, R2,x. Let w be a white vertex in Bn{i−1}−1 with (w)n−1= i. By Lemma 2, there is a Hamiltonian path Q3 of BnI−1 with I = n − 1 − {1, i, i + 1} joining the black vertex (y)nto w. By Lemma 3, there is a Hamiltonian
path Q4of Bn{i,i+1}−1 − {zin−i−1,zin+1−i−2} joining the black vertex (w)nto the white vertex (x)n. We set
u= p and v = m. Then u = p, Q1,q, (q)n, R1,y, (y)n, Q3,w, (w)n, Q4, (x)n,x, R−12 ,m= v is the desired path.
Case 1.2 Suppose that zni−i−1 is a black vertex in Bn{i}−1and zni+1−i−2is a white vertex in Bn{i+1}−1 . The proof of this case is similarly to Case 1.1.
Case 2 Suppose that u is a black vertex and v is a white vertex.
Case 2.1 Suppose that zi
n−i−1is a white vertex in B {i} n−1and z
i+1
n−i−2is a black vertex in B {i+1} n−1 . Let
w be a white vertex in Bn{n−1}−1 with (w)n−1= i. By Theorem 1, there is a Hamiltonian path Q2of
Bn{n−1}−1 joining the black vertex (p)nto w. Let y be a white vertex in B{i+1}
n−1 with (y)n−1 = i + 2. By
Lemma 3, there is a Hamiltonian path Q3of Bn{i,i+1}−1 − {z i
n−i−1,z
i+1
n−i−2} joining the black vertex (w)nto the white vertex y. Let v be any white vertex in Bn{1}−1. By Lemma 2, there is a hamiltonian path Q4of BnI−1with I = n − 2 − {i, i + 1} joining the black vertex (y)
nto v. We set u= q.
Thenu = q, Q−11 ,p, (p)n, Q
2,w, (w)n, Q3,y, (y)n, Q4,v is the desired path.
Case 2.2 Suppose that zni−i−1 is a white vertex in Bn{i}−1and zni+1−i−2 is a black vertex in Bn{i+1}−1 . The proof of this case is similarly to Case 2.1. Thus, this lemma is proved.
Theorem3 For the bubble-sort graph B5with e the vertex denoting identity permutation, there
exist four MIH cycles starting at vertex e.
We give the proof of Theorem 3 in Appendix 1. Now, we can find the MIH of the bubble-sort graph Bn.
Theorem4 Let n≥ 6. We have h(Bn)≥ n − 1.
Proof Since Bnis vertex transitive, we show that there are (n− 1)-mutually independent
Hamil-tonian cycles of Bn form e. Suppose that n≥ 6. Let v11, v12, . . ., v
n 1 be the vertices of Bn{1}−1, Bn{2}−1, . . . , Bn{n}−1 with (v2 1)n−1= 4, (v31)n−1 = 5, (v41)n−1= 3, (vn1)n−1= 1 and (v j 1)n−1= j + 1
for 5≤ j ≤ n − 1, respectively. By Theorem 1, there are Hamiltonian paths: H2
1 of Bn{2}−1joining
z2
n−3to v21; H14of Bn{4}−1joining (v21)nto v41, H13of Bn{3}−1joining (v41)nto v13and H15of Bn{5}−1joining (v3
1)n to v51. By Theorem 1, there is a Hamiltonian path H1i of Bn{i}−1 joining (v
i−1
1 )n to v1i for 6≤ i ≤ n − 1. By Lemma 6, there is a Hamiltonian path Hn
1 of Bn{n}−1− Xn1−1joining (v
n−1
1 )nto
vn
1. By Theorem 1, there is a hamiltonian path H11 of Bn{1}−1joining (v1n)nto z1n−2. We set C1 =
e, z2 0, . . . ,z2n−3, H12,v21, (v21)n, H14,v41, (v41)n, H13,v31, (v13)n, H15,v15, (v51)n, H16,v61, . . . , (v n−2 1 )n, H1n−1,vn1−1, (vn1−1)n, Hn 1,v n 1, (v n 1)n, H 1 1,z 1 n−2,z 1 n−3, . . . ,z 1
0,e being a Hamiltonian cycle of Bn
form e.
Let v12, v22, . . ., v2n be the vertices of Bn{1}−1, Bn{2}−1, . . . , Bn{n}−1 with (v22)n−1= 4, (v23)n−1= 5, (v24)n−1= 3, (v2n)n−1= 1, and (vj2)n−1 = j + 1 for 5 ≤ j ≤ n − 1, respectively. By Theorem 1,
there are Hamiltonian paths H4
2 of Bn{4}−1 joining z4n−5 to v42; H23 of Bn{3}−1 joining (v24)n to v3 2; and H 5 2 of B {5} n−1 joining (v 3 2) n to v5
2. By Theorem 1, there is a Hamiltonian path H
i
2 of Bn{i}−1 joining (vi2−1)n to vi
2 for 6≤ i ≤ n − 1. By Lemma 8, there is a Hamiltonian path
Hn 2 of B {n} n−1− X 2 n−1 joining (v n−1 2 ) n to vn
2. By Theorem 1, there are Hamiltonian paths:
H1 2 of B {1} n−1 joining (v n 2) n to v1 2 and H 2 2 of B {2} n−1 joining (v 1 2) n to z2 n−3. We set C2= e, z4 0, . . . ,z 4 n−5, H 4 2, v 4 2, (v 4 2) n, H3 2, v 3 2, (v 3 2) n, H5 2, v 5 2, (v 5 2) n, H6 2, v 6 2, . . . , (v n−2 2 ) n, Hn−1 2 , v n−1 2 , (v2n−1)n, Hn 2,v n 2, (v n 2) n, H1 2,v 1 2, (v 1 2) n, H2 2,z 2 n−3,z 2 n−4, . . . ,z 2
0,e being a Hamiltonian cycle of Bn
form e. Let l = (n − 1)(n − 1)! − (n − 2) + 1. The lth vertex of C1 is (vn1)
n, which is in B{1}
n−1,
and the lth vertex of C2 is v21, also in B
{1} n−1. Obviously, ((v n 1) n) n−1= n and (v12)n−1= 2, then (vn 1) n= v1 2. Let v1 3, v 2 3, . . . ,v n 3 be the vertices of B {1} n−1, B {2} n−1, . . . , B {n} n−1 with (v 2 3)n−1 = 4, (v33)n−1= 5, (v34)n−1= 3, (v3n)n−1= 1, and (vj3)n−1 = j + 1 for 5 ≤ j ≤ n − 1, respectively. By Theorem 1,
there is a Hamiltonian path H33 of Bn{3}−1 joining z3n−4 to v33 and a Hamiltonian path H35of Bn{5}−1 joining (v33)n to v5
3. By Theorem 1, there is a Hamiltonian path H
i
3 of Bn{i}−1 joining (v
i−1
3 )n to vi3 for 6≤ i ≤ n − 1. By Lemma 10, there is a Hamiltonian path H3n of Bn{n}−1− Xn3−1 joining (vn3−1)n to vn
3. Let v 1
3 be a vertex in Bn{1}−1 with (v
1
3)n−1= 2 such that the vertex
v13∈ N((v/ n2)n) and there exists a vertex s∈ N(v13) with s= (vn2)n. By Theorem 2, there is a Hamiltonian path H1
3 of Bn{1}−1− v31 joining the vertex (v
n
3)n to s. Let v23 be a vertex in
Bn{2}−1 with (v2
3)n−1 = 4 such that the vertex v23∈ N((v/ 21)n)and there exists a vertex t∈ N(v32) with t= (v1
2)n. By Theorem 2, there is a Hamiltonian path H32 of Bn{2}−1 joining the vertex (v1
3)n to t. By Theorem 1, there are Hamiltonian paths: H31 of Bn{1}−1 joining (v n
3)n to v13;
H2
3 of Bn{2}−1 joining (v31)n to v23; and H34 of Bn{4}−1 joining (v32)n to z4n−5. We set C3= e,
z3 0, . . . ,z3n−4, H33, v33, (v33)n, H35, v53, (v53)n, H36, v63, . . . , (v n−2 3 )n, H n−1 3 , v n−1 3 , (v n−1 3 )n, H n 3, vn3, (vn3)n, H1
3,s, v13, (v31)n, H32,t, v23, (v23)n, H34,z4n−5,zn4−6, . . . ,z40,e being a Hamiltonian cycle of Bnform e. Let l = m(n − 1)! + (n − 4) + 1 for 1 ≤ m ≤ n − 4. The lth vertices are coincided
in the same subgraph between C2and C3. Obviously, the (n− 1)-th position of the l-th vertices in C2and C3are different.
Let v14, v24, . . ., v4n be the vertices of Bn{1}−1, Bn{2}−1, . . . , Bn{n}−1 with (v24)n−1= 4, (v43)n−1= 5, (v44)n−1= 3, (v4n)n−1= 1, and (vj4)n−1 = j + 1 for 5 ≤ j ≤ n − 1, respectively. By Theorem 1,
there is a Hamiltonian path H5
4 of Bn{5}−1 joining z5n−6 to v45. By Theorem 1, there is a Hamiltonian path Hi
4 of Bn{i}−1 joining (v
i−1
4 )n to v4i for 6≤ i ≤ n − 1. By Lemma 12, there is a Hamiltonian path Hn
4 of Bn{n}−1− X4n−1 joining (v
n−1
4 )n to v4n. By Theorem 1, there are Hamiltonian paths: H1
4 of Bn{1}−1 joining (vn4)n to v41; H42 of Bn{2}−1 joining (v14)n to v2
4; H44 of Bn{4}−1 joining (v24)n to v44; H43 of Bn{3}−1 joining (v44)n to z3n−4. We set C4= e, z50, . . . ,z5n−6, H45,v45, (v54)n, H46,v46, . . . , (v n−2 4 )n, H n−1 4 ,v n−1 4 , (v n−1 4 )n, H n 4,v n 4, (v n 4)n,
H1
4,v41, (v14)n, H42, (v24)n, H44,v44, (v44)n, H43,zn3−4,z3n−5, . . . ,z30,e being a Hamiltonian cycle of
Bnform e.
Case 3.1 Suppose that n= 6. Let v1 5, v 2 5, v 3 5, v 4 5and v 6 5be the vertices of B {1} 5 , B {2} 5 , B {3} 5 , B {4} 5 , and B5{6}with (v1
5)5= 2, (v25)5= 4, (v45)5= 3, (v35)5= 5, and (v56)5= 1, respectively. By Theorem 1, there is a Hamiltonian path H56of B5{6}joining e to v65. Let v15, v25, v54, and v35be the vertices in Bn{1}−1,
Bn{2}−1, Bn{4}−1, and Bn{3}−1, with (v15)n−1 = 2, (v52)n−1= 4, (v45)n−1= 3, and (v35)n−1= 5, such that
v51∈ N((v/ 64)n), v25∈ N((v/ 14)n), v54∈ N((v/ 24)n), and v35∈ N((v/ 44)n), respectively. And there exist
s15)∈ N(v15), s25∈ N(v52), s45∈ N(v45), and s35∈ N(v35). By Theorem 2, there are Hamiltonian paths: H1 5 of B5{1}− v51joining (v n 5)nto s15; H52of B5{2}− v25joining (v15)nto s25; H54of B5{4}− v45 joining (v2 5) nto s4 5; and H 3 5 of B {3} 5 − v 3 5joining (v 4 5) nto s3
5. By Theorem 1, there is a Hamiltonian path H55of Bn{5}−1joining (v35)nto z5 0. We set C5= e, H 6 5, v 6 5, (v 6 5)6, H 1 5, s 1 5, v 1 5, (v 1 5)6, H 2 5, s 2 5, v 2 5,
(v25)6, H54, s45, v45, (v45)6, H53, s35, v35, (v35)6, H55, z50, e being a Hamiltonian cycle of Bnform e.
Then{C1, C2, . . . , C5} forms a set of five-mutually independent Hamiltonian cycles.
Case 3.2 Suppose that n > 6. Let v1 5, v 2 5, . . . ,v n 5 be the vertices of B {1} n−1, B {2} n−1, . . . , B {n} n−1with (v2 5)n−1 = 4, (v35)n−1= 5, (v45)n−1= 3, (vn5)n−1= 1, and (v j 5)n−1= j + 1 for 5 ≤ j ≤ n − 1,
respectively. By Theorem 1, there is a Hamiltonian path H56 of Bn{6}−1 joining z6n−7 to v65. By Theorem 1, there is a Hamiltonian path H5i of Bn{i}−1joining (v5i−1)nto vi5for 6≤ i ≤ n − 1. By Lemma 14, there is a Hamiltonian path H5nof Bn{n}−1− Yn5−1joining (vn5−1)nto vn5. Let v51, v25, v45, and
v53be the vertices in Bn{1}−1, Bn{2}−1, Bn{4}−1, and Bn{3}−1, with (v15)n−1= 2, (v25)n−1= 4, (v45)n−1 = 3,
and (v35)n−1= 5, such that v15∈ N((v/ 64)n), v52∈ N((v/ 14)n), v45∈ N((v/ 42)n), and v35∈ N((v/ 44)n), respectively. And there exist s15 ∈ N(v15), s25∈ N(v52), s45 ∈ N(v45), and s35∈ N(v35). By Theorem 2, there are Hamiltonian paths: H1
5 of Bn{1}−1− v15 joining (v
n
5)n to s15; H52 of Bn{2}−1− v25 joining
(v1
5)nto s25; H54of Bn{4}−1− v54joining (v25)n to s54; and H53of Bn{3}−1− v53joining (v45)n to s35. By Theorem 1, there is a Hamiltonian path H5
5 of B {5} n−1 joining (v35) n to z5 n−6. We set C5= e, z6 0, z61, . . . ,zn−76, H56, v65, (v56)n, H57, v75, . . . , (vn5−1)n, H5n, vn5, (v5n)n, H51, s51, v15, (v51)n, H52, s25, v2 5, (v25)n, H54,s45,v45, (v54)n, H53,s53,v35, (v53)n, H55,z5n−6,z5n−7, . . . ,z50,e is a Hamiltonian cycle of Bnform e.
Assume that 6≤ i ≤ n − 2. Let v1
i, v2i, . . ., vinbe the vertices of Bn{1}−1, Bn{2}−1, . . . , Bn{n}−1with (v2
1)n−1 = 4, (v3i)n−1= 5, (v4i)n−1= 3, (vni)n−1= 1, and (v j
i)n−1= j + 1 for 5 ≤ j ≤ n − 1,
respectively. By Theorem 1, there is a Hamiltonian path Hii+1 of Bn{i+1}−1 joining zin+1−i−2 to
vii+1. By Theorem 1, there is a Hamiltonian path Hii of Bn{i}−1 joining (vii−1)n to zin−i−1. By Theorem 1, there is a Hamiltonian path Hij of Bn{j}−1joining (vji−1)nto vj
i for 6≤ j ≤ n − 1 and j /∈ {i, i + 1}. By Lemma 14, there is a Hamiltonian path Hinof Bn{n}−1− Yni−1 joining (vni−1)n
to vni. By Theorem 1, there are Hamiltonian paths: H1
i of Bn{1}−1 joining (v n
i)n to v15; Hi2 of
Bn{2}−1 joining (v1
i)n to v52; Hi4 of Bn{4}−1 joining (v2i)n to v45; and Hi3 of Bn{3}−1 joining (v4i)n to
v3 5. We set Ci= e, zi0+1,z i+1 1 , . . . ,z i+1 n−i−2, H i+1 i ,v i+1 i , (v i+1 i )n, H i+2 i ,v i+2 i , . . . , (v n−1 i )n, H n i,v n i, (vni)n, H1 i,v1i,(v1i)n, Hi2,v2i,(v2i)n, Hi4,v4i,(v4i)n, Hi3,v3i, (vi3)n, Hi5,vi5, . . . , (v i−1 i )n, H i i,z i n−i−1,
zin−i−2, . . . ,zi0,e being a Hamiltonian cycle of Bnform e.
Let v1
n−1, v2n−1, . . . ,v n
n−1 be the vertices of Bn{1}−1, Bn{2}−1, . . . , Bn{n}−1 with (vn2−1)n−1 = 4, (v3
n−1)n−1= 5, (v4n−1)n−1 = 3, (vnn−1)n−1 = 1, and (v j
n−1)n−1= j + 1 for 5 ≤ j ≤ n − 1,
respec-tively. By Theorem 1, there is a Hamiltonian path Hn
n−1of Bn{n}−1joining e to vnn−1. Again, there
is a Hamiltonian path Hi
n−1of Bn{i}−1joining v
i−1
n−1 to vni−1 for 6≥ i ≥ n − 2. Moreover, there is
a Hamiltonian path Hnn−1−1 of Bn{n−1}−1 joining (vnn−2−1)nto zn−1
0 . By Theorem 1, there are Hamilto-nian paths: H1
n−1 of Bn{1}−1joining (v n
n−1)nto v1n−1; Hn2−1 of Bn{2}−1joining (v1n−1)nto v2n−1; Hn4−1
of Bn{4}−1 joining (v2
n−1)n to v4n−1; Hn3−1 of Bn{3}−1 joining (v4n−1)n to v3n−1. We set Cn−1= e, Hnn−1, vnn−1, (vnn−1)n, H1
n−1, v1n−1, (vn1−1)n, Hn2−1, v2n−1, (vn2−1)n, Hn4−1, v4n−1, (v4n−1)n, Hn3−1,
v3n−1, (v3n−1)n, Hn5−1,v5n−1, . . . , (vnn−1−2)n, Hnn−1−1,zn0−1,e being a Hamiltonian cycle of Bnform e.
Then{C1, C2, . . . , Cn−1} is a set of (n − 1)-mutually independent Hamiltonian cycles for Bn
from e.
Corollary1 For n≥ 4, we have h(Bn)= n − 1. Moreover, h(B3)= 1.
Proof Since δ(Bn)= n − 1, h(Bn)≤ n − 1. Since B3is a cycle with six vertices, it is easy to check that h(B3)= 1. To show h(Bn)= n − 1 for n ≥ 4, we need to construct (n − 1)-mutually
independent Hamiltonian cycles of Bnfrom every vertex u. Since Bnis vertex transitive, we show
that there are (n− 1)-mutually independent Hamiltonian cycles of Bnfrom e. Case 1 Suppose that n= 4. We set
C1= 1234, 2134, 2143, 2413, 2431, 2341, 2314, 3214, 3241, 3421, 4321, 4231, 4213, 4123, 4132, 4312, 3412, 3142, 3124, 1324, 1342, 1432, 1423, 1243, 1234, C2= 1234, 1243, 1423, 1432, 4132, 4123, 4213, 4231, 4321, 4312, 3412, 3421, 3241, 2341, 2431, 2413, 2143, 2134, 2314, 3214, 3124, 3142, 1342, 1323, 1234, and C3= 1234, 1324, 3124, 3142, 1342, 1432, 1423, 1243, 2143, 2413, 4213, 4123, 4132, 4312, 3412, 3421, 4321, 4231, 2431, 2341, 3241, 3214, 2314, 2134, 1234. Then{C1, C2, C3} is a set of three-mutually independent Hamiltonian cycles for B4from e.
Case 2 Suppose that n≥ 5. By Theorems 3 and 4, there is a set of (n − 1)-mutually independent Hamiltonian cycles on Bnfrom e.
Summarily, Case 1 and Case 2, we have h(Bn)= n − 1 for n ≥ 4.
Acknowledgements
The authors are grateful to the referees for their thorough reviews of the paper and many helpful suggestions. Jimmy J. M. Tan was partially supported by the National Science Council of the Republic of China under contract NSC 96-2221-E-009-137-MY3 and also by the Aiming for the Top University and Elite Research Center Development Plan.
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