We need the following lemmas.
We extend the above lemma by considering the occurrence of edge faults.
Proof.
We prove this lemma by induction on n. By Lemma 2.3, this lemma is true for n=3. Thus, we assume n≥4 and F≥1. For 1≤i≤n, let Fi denote the set of idimensional edges in F. Thus, ∑i=1nFi=F. Without loss of generality, we assume that Fn≥1. For i=0, 1, we use Fi to denote the set E(Qni)∩F. Obviously, Fi≤n4. Let Si={u,v,x,y}∩V(Qni) for i=0, 1. Without loss of generality, we can assume that S0≥S1. We have the following cases.
Case 1 (S0=4). By induction, there exist two spanning disjoint paths R1 and R2 of Qn0F0 such that R1 joins u to v, and R2 joins x to y. Since 2n12>2(n3), there exists an edge (s,t) of R1 or R2 such that {(s,(s)n),(t,(t)n)}∩F=∅. Without loss of generality, we can assume that (s,t) is in R1. Thus, R1 can be written as 〈u,H1,s,t,H2,v〉. By Lemma 2.2, there exists a Hamiltonian path Q joining (s)n to (t)n of Qn1F1. We set P1 as 〈u,H1,s,(s)n,Q,(t)n,t,H2,v〉 and set P2 as R2.
Case 2 (S0=3). Without loss of generality, we can assume that y∈Qn1. Since there are 2n2 black vertices in Qn0 and 2n21>n3 for n≥4, there exists a black vertex z of Qn0 such that z≠v and (z,(z)n)∉F. By induction, there exist two spanning disjoint paths P1 and R1 of Qn0F0 such that P1 joins u to v, and R1 joins x to z. By Lemma 2.2, there exists a Hamiltonian path R2 joining (z)n to y of Qn1F1. We set P2 as 〈x,R1,z,(z)n,R2,y〉. Thus, P1 and P2 form the required paths.
Case 3 (S0=2 and Si={u,v} for some i=0,1). Without loss of generality, we can assume that S0={u,v}. By Lemma 2.2, there exists a Hamiltonian path P1 of Qn0F0 joining u to v, and there exists a Hamiltonian path P2 of Qn1F1 joining x to y. Obviously, P1 and P2 form the required paths.
Case 4 (S0=2 and Si≠{u,v} for each i=0,1). Without loss of generality, we can assume that S0={u,x}. Since there are 2n2 black vertices in Qn0 and 2n2>n2 for n≥4, there exist two black vertices s and t of Qn0 such that {(s,(s)n),(t,(t)n)}∩F=∅. By induction, there exist two spanning disjoint paths Q1 and Q2 of Qn0F0 such that Q1 joins u to s and Q2 joins x to t. Similarly, there exist two spanning disjoint paths R1 and R2 of Qn1F1 such that R1 joins (s)n to v, and R2 joins (t)n to y. We set P1 as 〈u,Q1,s,(s)n,R1,v〉 and set P2 as 〈x,Q2,t,(t)n,R2,y〉.
Proof.
Let w be any faultfree white vertex and b be any faultfree black vertex. We need to construct a Hamiltonian path of QnF joining w to b by induction on n. Let H=Fe∪{(bi,wi)∣{bi,wi}⊂Fav}. For 1≤i≤n, let Hi denote the set of idimensional edges in H. Thus, ∑i=1nHi=fe+fav. Without loss of generality, we assume that Hn=0.
By brute force, we can check that the required paths exist for n=3,4. By Lemmas 2.2 and 2.6, the required paths exist when fav=0 or fe+fav≤n3. Therefore, we only need to consider the case fav≥1 and fav+fe=n2 for n≥5. Thus, Fe≠∅. Let Fei=Fe∩E(Qni), Favi=Fav∩V(Qni), and Fi=Favi∪Fei for i=0, 1. Let b1 and w1 be a pair of Fav0 where b1 is a black vertex.
Case 1
(
F
⊄
Q
n
i
and

F
a
v
i

≤
n

4
,
for
i
=
0,1
)
.
We first consider the case that w and b are in the same subcube. Without loss of generality, we can assume that both w and b are in Qn0. By induction, there exists a Hamiltonian path P1 of Qn0F0 joining w to b. Note that l(P1)4fav1=2n12fav014fav1≥2n14n+15>0. We can write P1 as 〈w,R1,u,v,R2,b〉 for some u and v such that {(u)n,(v)n}∩Fav1=∅. By induction, there exists a Hamiltonian path R3 of Qn1F1 joining (u)n to (v)n. Thus, 〈w,R1,u,(u)n,R3,(v)n,v,R2,b〉 is a desired path.
Thus we consider the case that w and b are in different subcubes. Without loss of generality, we can assume that w∈Qn0 and b∈Qn1. Since there are 2n2 black vertices in Qn0 and 2n2≥n3 for n≥5, there exists a black vertex x in Qn0 such that {x,(x)n}∩Fav=∅. By induction, there exists a Hamiltonian path P1 of Qn0F0 joining w to x and there exists a Hamiltonian path P2 of Qn1F1 joining (x)n to b. Thus, 〈w,P1,x,(x)n,P2,b〉 is a desired path.
Case 2
(
F
⊂
Q
n
i
or Favi=n3 for some i=0,1). Without loss of generality, we can assume that F⊂Qn0 or Fav0=n3. Thus, Fav1=0.
Assume that both w and b are in Qn0. By induction, there exists a Hamiltonian path P1 of Qn0(F{b1,w1}) joining w to b. Suppose that the edge (b1,w1) is in P1. Without loss of generality, we can write P1 as 〈w,Z1,x,b1,w1,y,Z2,b〉. By Lemma 2.2, there exists a Hamiltonian path Z3 of Qn1Fe1 joining (x)n to (y)n. Obviously, 〈w,Z1,x,(x)n,Z3,(y)n,y,Z2,b〉 is a desired Hamiltonian path. Thus, we consider the case that (b1,w1) is not in P1. Without loss of generality, P1 can be written as 〈w,R1,u,b1,v,R2,x,w1,y,R3,b〉. By Lemma 2.4, there are two disjoint spanning paths R4 and R5 of Qn1Fe1 such that R4 joins (u)n to (x)n and R5 joins (v)n to (y)n of Qn1Fe1. Obviously, 〈w,R1,u,(u)n,R4,(x)n,x,R21,v,(v)n,R5,(y)n,y,R3,b〉 is a desired path.
Now, we consider the case that w and b are in different subcubes. Without loss of generality, we can assume that w∈Qn0 and b∈Qn1. By induction, there exists a Hamiltonian path P1 of Qn0(F{b1,w1}) joining w and b1. Suppose that the edge (b1,w1) is in P1. We can write P1 as 〈w,Z1,y,w1,b1〉. By Lemma 2.2, there exists a Hamiltonian path Z2 of Qn1Fe1 joining (y)n to b. Obviously, 〈w,Z1,y,(y)n,Z2,b〉 is a desired path. Thus, we consider the case that (b1,w1) is not in P1. We can write P1 as 〈w,R1,u,w1,v,R2,x,b1〉. Assume that (x)n≠b. By Lemma 2.4, there are two disjoint spanning paths R3 and R4 of Qn1Fe1 such that R3 joins (u)n to (x)n and R4 joins (v)n to b. Obviously, 〈w,R1,u,(u)n,R3,(x)n,x,R21,v,(v)n,R4,b〉 is a desired path. Assume that (x)n=b. By Lemma 2.2, there exists a Hamiltonian path R5 of Qn1Fe1{b} joining (u)n to (v)n. Obviously, 〈w,R1,u,(u)n,R5,(v)n,v,R2,x,(x)n=b〉 is a desired path.
Finally, we consider the case that w and b are in Qn1. Let e=(x,y) be any faulty edge of F, where x is a white vertex. By Lemma 2.5, there exists a Hamiltonian cycle C1 of Qn0(F{e}).
Suppose that e is not an edge of C1. Since the length of C is at least 2n12(n3), we can write C1 as 〈u,Z1,v,u〉 such that {(u)n,(v)n}∩{w,b}=∅ and u is a white vertex. By Lemma 2.4, there exist two disjoint spanning paths Z2 and Z3 of Qn1Fe1 such that Z2 joins w to (u)n and Z3 joins (v)n to b. Obviously, 〈w,Z2,(u)n,u,Z1,v,(v)n,Z3,b〉 is a desired path.
Thus, we consider the case that e is in C1. We can write C1 as 〈x,P1,y,x〉. Suppose that {(x)n,(y)n}∩{w,b}=∅. By Lemma 2.4, there exist two disjoint spanning paths P2 and P3 of Qn1Fe1 such that P2 joins w to (x)n and P3 joins (y)n to b. Obviously, 〈w,P2,(x)n,x,P1,y,(y)n,P3,b〉 is a desired path. Suppose that {(x)n,(y)n}∩{w,b}=1. Without loss of generality, we assume that (x)n=b. By Lemma 2.2, there exists a Hamiltonian path P4 of Qn1Fe1{b} joining w to (y)n. Thus, 〈w,P4,(y)n,y,P11,x,(x)n=b〉 is a Hamiltonian path of QnF. Suppose that {(x)n,(y)n}={w,b}. Obviously, C1 can be written as 〈y,P5,u,v,P6,x,y〉 for some black vertex u. By induction, there exists a Hamiltonian path P7 of Qn1Fe1{w,b} joining (u)n to (v)n. Obviously, 〈w=(y)n,y,P5,u,(u)n,P7,(v)n,v,P6,x,(x)n=b〉 is a desired path.