Proof.
We proceed by induction on n⩾3.
Since G3≅Q3 or VQ3, which is vertex-transitive, it is easy to check the conclusion is true for n=3. Suppose now that n⩾4 and the result holds for any integer less than n. Let F⊂E(Gn)∪V(Gn) with F⩽n-3, and let x and y be two distinct vertices in Gn-F. We need to prove that Gn-F contains an xy-Hamilton path. Without loss of generality, we can assume F⊂V(Gn). Let Gn=L⊕MR, where (1)L=Gn-200⊕M1Gn-201, R=Gn-210⊕M2Gn-211,and let(2)FL=F∩L, FR=F∩R.By symmetry of structure of Gn, we may assume FL⩾FR.
Case 1 (
|
F
L
|
⩽
n
-
4
). In this case, by the hypothesis, we have |FR|⩽|FL|⩽n-4.
Subcase 1.1 (
x
,
y
∈
L
or
x
,
y
∈
R
). Without loss of generality, assume x,y∈R.
Since R=Gn-1 and FR⩽n-4=(n-1)-3, by the induction hypothesis R-FR contains an xy-Hamilton path, say PR. Since VPR=2n-1-FR⩾2n-1-(n-4)>2(n-3)⩾2F, there is an edge uRvR in PR such that the neighbors uL and vL of uR and vR in L are not in F. Since L=Gn-1 and FL⩽n-4=(n-1)-3, by the induction hypothesis L-FL contains a uLvL-Hamilton path, say PL. Thus, PR-uRvR+uRuL+vRvL+PL is an xy-Hamilton path in Gn-F (see Figure 3(a)).
Subcase 1.2 (
x
∈
L
and
y
∈
R
). Since M=2n-1 and 2n-1-2>2(n-3)⩾2F, there is an edge uLuR∈M such that uL and uR are not in F∪{x,y}. By the induction hypothesis, let PL be an xuL-Hamilton path in L-FL, and let PR be a yuR-Hamilton path in R-FR. Then PL+uLuR+PR is an xy-Hamilton path in Gn-F (see Figure 3(b)).
Case 2 (
F
L
=
n
-
3
). In this case, |FR|=0.
Subcase 2.1 (
x
,
y
∈
L
). Arbitrarily take a vertex u∈FL. Since FL-u=n-4=(n-1)-3, by the induction hypothesis L-(FL-u) contains an xy-Hamilton path, say PL. Without loss of generality, assume u∈V(PL). Let uL and vL be two neighbors of u in PL, and let uLuR,vLvR∈M. By the induction hypothesis, R contains a uRvR-Hamilton path, say PR. Then PL-u+uLuR+vLvR+PR is an xy-Hamilton path in Gn-F.
Subcase 2.2 (
x
∈
L
and
y
∈
R
). If n=4, then L≅R≅Q3 or VQ3. Since |FL|=1 and L is vertex-transitive, we can assume FL={u}={000} unless x=000. It is easy to check that L-u contains a Hamilton cycle, say CL. Choose a neighbor uL of x in CL such that its neighbor uR in R is not y. By the induction basis, R contains a yuR-Hamilton path, say PR. Then, CL-xuL+uLuR+PR is an xy-Hamilton path in G4-F.
Assume now n⩾5; that is, n-2⩾3. Let F00=FL∩V(Gn-200), F01=FL∩V(Gn-201). Without loss of generality, we can assume F00≠∅.
(a)
y
∈
G
n
-
2
11
(See Figure 4(a)). Arbitrarily take z11∈Gn-211 with z11≠y, and let z01z11∈M. Since n-2⩾3, by the induction hypothesis Gn-211 contains a z11y-Hamilton path, say P11. Arbitrarily take a vertex u∈F00. Since n⩾5, by the induction hypothesis L-(FL-u) contains an xz01-Hamilton path, say PL. If u is in PL, then let u00 and w00 be two neighbors of u in PL; if u is not in PL, then let u00v00 be an edge in PL. Let u00u10,v00v10∈M. By the induction hypothesis, Gn-210 contains a u10v10-Hamilton path, say P10. Let PL′=PL-u if u is in PL and PL′=PL-u00v00 if u is not in PL. Then P10+u00u10+v00v10+PL′+z01z11+P11 is an xy-Hamilton path in Gn-F (see Figure 4(a)).
(b)
y
∈
G
n
-
2
10
(See Figure 4(b)). Arbitrarily take a vertex z01 in Gn-201-FL with z01≠x. Let z11 be the neighbor of z01 in Gn-211. Arbitrarily take a vertex u∈F00. Since n⩾5, by the induction hypothesis L-(FL-u) contains an xz01-Hamilton path, say PL. If u is in PL, then let u00 and w00 be two neighbors of u in PL; if u is not in PL, then let u00v00 be an edge in PL. Let u00u10,v00v10∈M. By the induction hypothesis, Gn-210 contains a u10v10-Hamilton path, say P10. Since y∈P10, we can write P10=P10(v10,y)+yw10+P10(w10,u10). Let w11 be the neighbor of w10 in Gn-211. By the induction hypothesis, Gn-211 contains a z11w11-Hamilton path, say P11. Let PL′=PL-u if u is in PL and PL′=PL-u00v00 if u is not in PL. Then PL′+u00u10+v00v10+P10-yw10+w10w11+P11+z01z11 is an xy-Hamilton path in Gn-F (see Figure 4(b)).
Subcase 2.3 (
x
,
y
∈
R
). If n=4, then L≅R≅G3. By the induction basis, R contains an xy-Hamilton path, say PR. Since G3 is vertex-transitive and |FL|=1, it is easy to check that L-FL contains a Hamilton cycle, say CL. Since L and R are 3-regular and isomorphic, there is an edge uRvR in PR which is not incident with x and y such that the corresponding edge eL in L is contained in CL. By Definition 2 eL=uLvL, where uL and vL are neighbors of uR and vR in L, respectively. Thus, PR-uRvR+uLuR+vLvR+CL-eL is an xy-Hamilton path in G4-F (as a reference, see Figure 3(a)).
Assume n⩾5 below; that is, n-2⩾3.
(a)
x
,
y
∈
G
n
-
1
11
(See Figure 5(a)). By the induction hypothesis, Gn-211 contains an xy-Hamilton path, say P11. Take u11v11∈E(P11), and let u01 and v01 be neighbors of u11 and v11 in Gn-201, respectively. Take a vertex u in F00. By the induction hypothesis, L-(FL-u) contains a u01v01-Hamilton path, say PL. If u is in PL, then let w00 and z00 be two neighbors of u in PL; if u is not in PL, then let w00z00 be an edge in PL. Let w10 and z10 be neighbors of w00 and z00 in Gn-210, respectively. By the induction hypothesis, Gn-210 contains a w10z10-Hamilton path, say P10. Let PL′=PL-u if u is in PL and PL′=PL-w00z00 if u is not in PL. Thus, P10+w00w10+z00z10+PL′+P11-u11v11+u01u11+v01v11 is an xy-Hamilton path in Gn-F (see Figure 5(a)).
(b)
x
∈
G
n
-
1
11
and
y
∈
G
n
-
2
10
(See Figure 5(b)). Arbitrarily take a vertex u in F00 and an edge u00v00 in Gn-200. By the induction hypothesis, L-(FL-u) contains a u00v00-Hamilton path, say PL. If u is in PL, then let P′=PL-u+u00v00; if u is not in PL, then let P′=PL. Without loss of generality, assume that u is in PL and let u00 and v00 be two neighbors of u in PL.
Let u10 and v10 be neighbors of u00 and v00 in Gn-210, respectively. By the induction hypothesis, Gn-210 contains a u10v10-Hamilton path, say P10. Since y is in P10, we can write P10=P10(v10,y)+yw10+P10(w10,u10) (see Figure 5(b)). Let w11 be the neighbor of w10 in Gn-211. By the induction hypothesis, Gn-211 contains an xw11-Hamilton path, say P11. Then PL′+P10-yw10+w10w11+P11 is an xy-Hamilton path in Gn-F (see Figure 5(b)).
(c)
x
,
y
∈
G
n
-
2
10
(See Figure 6)
(c1)
F
01
≠
0
. By the induction hypothesis, Gn-210 contains an xy-Hamilton path, say P10. Take w10z10∈E(P10), and let w00 and z00 be neighbors of w10 and z10 in Gn-200, respectively. Take a vertex u in F01. By the induction hypothesis, L-(FL-u) contains a w00z00-Hamilton path, say PL. If u is in PL, let u00 and v00 be two neighbors of u in PL; if u is not in PL, let u00v00 be an edge in PL. Let PL′=PL-u if u is in PL and PL′=PL-u00v00 if u is not in PL.
Let u11 and v11 be neighbors of u01 and v01 in Gn-211, respectively. By the induction hypothesis, Gn-211 contains a u11v11-Hamilton path, say P11. Thus, P10-w10z10+w00w10+z00z10+PL′+u01u11+v01v11+P11 is an xy-Hamilton path in Gn-F (see Figure 6(a)).
(c2)
|
F
01
|
=
0
. In this case, F00=F=n-3⩾2 since n⩾5. Consider the subgraph H of Gn induced by V(Gn-200)∪V(Gn-210). By Definition 2, it is easy to check that H=Gn-200⊕MGn-210. Let u∈F. By the induction hypothesis, H-(F-u) contains an xy-Hamilton path, say PH. Without loss of generality, assume that u is in PH. Let u00 and v00 be two neighbors of u in PH, and let u01 and v01 be two neighbors of u00 and v00 in Gn-201. Then there is a u01v01-Hamilton path in Gn-201, say P01. Take an edge w01z01 in P01, and let w11 and z11 be neighbors of w01 and z01 in Gn-211. Then there is a w11z11-Hamilton path in Gn-211, say P11. Thus, PH-u+P01-w01z01+P11 is an xy-Hamilton path in Gn-F (see Figure 6(b)).
The theorem follows.