Lemma 14.
Assume that F=HPK(n1,n2,n3,n4)[Kt], n1,n2,n3,n4≥4. x,y,z,w is pairwise nonadjacent in F, let x=(x1,x2,x3,x4,x5), y=(y1,y2,y3,y4,y5), z=(z1,z2,z3,z4,z5), w=(w1,w2,w3,w4,w5), then
(1) N(x)∩N(y)∩N(z)∩N(w)∩N*(X)={(w1,x2,y3,z4,r∣r∈V5)}, and X contains x2,y3,z4 54 set intersection, and
(7)N*(X) =N(x1,x2,x3,y4,r)∩N(x1,x2,z3,x4,r) ∩N(x1,x2,z3,y4,r) ∩N(y1,x2,x3,x4,r)∩N(y1,x2,x3,y4,r) ∩N(y1,x2,z3,x4,r)∩N(y1,x2,z3,y4,r) ∩N(z1,x2,x3,x4,r)∩N(z1,x2,x3,y4,r) ∩N(z1,x2,z3,x4,r)∩N(z1,x2,z3,y4,r) ∩N(x1,y2,y3,x4,r)∩N(x1,y2,y3,y4,r) ∩N(x1,z2,y3,x4,r)∩N(x1,z2,y3,y4,r) ∩N(y1,y2,y3,x4,r)∩N(y1,z2,y3,x4,r) ∩N(y1,z2,y3,y4,r)∩N(z1,y2,y3,x4,r) ∩N(z1,y2,y3,y4,r)∩N(z1,z2,y3,x4,r) ∩N(z1,z2,y3,y4,r)∩N(x1,y2,x3,z4,r) ∩N(x1,y2,z3,z4,r)∩N(x1,z2,x3,z4,r) ∩N(x1,z2,z3,z4,r)∩N(y1,y2,x3,z4,r) ∩N(y1,y2,z3,z4,r)∩N(y1,z2,x3,z4,r) ∩N(y1,y2,z3,z4,r)∩N(z1,y2,x3,z4,r) ∩N(z1,y2,z3,z4,r)∩N(z1,z2,x3,z4,r) ∩N(x1,x2,y3,x4,r)∩N(x1,x2,y3,y4,r) ∩N(y1,x2,y3,x4,r)∩N(y1,x2,y3,y4,r) ∩N(z1,x2,y3,x4,r)∩N(z1,x2,y3,y4,r) ∩N(x1,x2,x3,z4,r)∩N(x1,x2,z3,z4,r) ∩N(y1,x2,x3,z4,r)∩N(y1,x2,z3,z4,r) ∩N(z1,x2,x3,z4,r)∩N(z1,x2,z3,z4,r) ∩N(x1,y2,y3,z4,r)∩N(x1,z2,y3,z4,r) ∩N(y1,y2,y3,z4,r)∩N(y1,z2,y3,z4,r) ∩N(z1,y2,y3,z4,r)∩N(z1,z2,y3,z4,r) ∩N(x1,x2,y3,z4,r)∩N(y1,x2,y3,z4,r) ∩N(z1,x2,y3,z4,r),(2)(8)N(w1,w2,w3,w4,r)∩N(x1,x2,x3,x4,r) ∩N(w1,y2,y3,y4,r)∩N(y1,w2,y3,y4,r) ∩N(w1,z2,z3,z4,r)∩N(z1,w2,z3,z4,r) -N(y1,y2,y3,y4,r)-N(z1,z2,z3,z4,r) ={(w1,w2,x3,x4,r)∣ r∈V5},(3)(9)M=N(w1,w2,w3,w4,r)∩N(x1,x2,x3,x4,r) ∩N(w1,y2,y3,y4,r)∩N(y1,w2,y3,y4,r) ∩N(y1,y2,w3,y4,r)∩N(w1,z2,z3,z4,r) ∩N(z1,w2,z3,z4,r)∩N(z1,z2,w3,z4,r) -N(y1,y2,y3,y4,r)-N(z1,z2,z3,z4,r) ={(w1,w2,w3,x4,r)∣ r∈V5}.
Proof of Theorem 13.
We are now ready to prove the following quantitative version of our theorem, we have four cases as follows.
Case 1. For any u∈V(G), let F1=G-N*(u)=HPK(n1,n2,n3,n4)[Kt], by Definition 6, V(F1)={(x1,x2,x3,x4,x5)∣1≤xi≤ni, i=1,2,3,4;1≤x5≤t}, x=(x1,x2,x3,x4,x5) is adjacent to y=(y1,y2,y3,y4,y5) in F1, if and only if x≠y and for any xi=yi, i=1,2,3,4.
Case 2. Let F2=G-N*(n1,n2,n3,n4,1)=HPK(n1,n2,n3,n4)[Kt], by Lemma 14, using V(F1)∩V(F2) in the name of the vertex of V(F2)-V(F1) in the naming of vertices. First, prove that u=(0,0,0,0,1), for any x,y∈V(F1)∩V(F2), x=(x1,x2,x3,x4,1), y=(y1,y2,y3,y4,1), xi≠yi, i=1,2,3,4. For convenience, N(x1,x2,x3,x4,x5) is the closed neighborhood of a vertex (x1,x2,x3,x4,x5), N*(x1, x2, x3, x4, x5)=N(x1, x2, x3, x4, x5)∪(x1,x2,x3,x4,x5), by Lemma 14, let
(11)N(u)∩N(x)∩N(y)∩N(z)∩N*(X) ={(0,x2,y3,z4,r)∣1≤r≤t},N(u)∩N(x)∩N(y)∩N(z)∩N*(Y) ={(x1,0,y3,z4,r)∣1≤r≤t},N(u)∩N(x)∩N(y)∩N(z)∩N*(Z) ={(x1,y2,0,z4,r)∣1≤r≤t},N(u)∩N(x)∩N(y)∩N(z)∩N*(W) ={(x1,y2,z3,0,r)∣1≤r≤t},(12)N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(0,y2,y3,y4,1)∩N(y1,0,y3,y4,1) ∩N(0,z2,z3,z4,1)∩N(z1,0,z3,z4,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(0,0,x3,x4,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(y1,0,y3,y4,1)∩N(y1,y2,0,y4,1) ∩N(z1,0,z3,z4,1)∩N(z1,z2,0,z4,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(x1,0,0,x4,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(y1,y2,0,y4,1)∩N(y1,y2,y3,0,1) ∩N(z1,z2,0,z4,1)∩N(z1,z2,z3,0,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(x1,x2,0,0,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(0,y2,y3,y4,1)∩N(y1,y2,0,y4,1) ∩N(0,z2,z3,z4,1)∩N(z1,z2,0,z4,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(0,x2,0,x4,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(0,y2,y3,y4,1)∩N(y1,y2,y3,0,1) ∩N(0,z2,z3,z4,1)∩N(z1,z2,z3,0,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(0,x2,x3,0,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(y1,0,y3,y4,1)∩N(y1,y2,y3,0,1) ∩N(z1,0,z3,z4,1)∩N(z1,z2,z4,0,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(x1,0,x3,0,r)1≤r≤t},(13)N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(0,y2,y3,y4,1)∩N(y1,0,y3,y4,1) ∩N(y1,y2,0,y4,1)∩N(0,z2,z3,z4,1) ∩N(z1,0,z3,z4,1)∩N(z1,z2,0,z4,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(0,0,0,x4,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(y1,0,y3,y4,1)∩N(y1,y2,0,y4,1) ∩N(y1,y2,y3,0,1)∩N(z1,0,z3,z4,1) ∩N(z1,z2,0,z4,1)∩N(z1,z2,z3,0,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(x1,0,0,0,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(0,y2,y3,y4,1)∩N(y1,0,y3,y4,1) ∩N(y1,y2,y3,0,1)∩N(0,z2,z3,z4,1) ∩N(z1,0,z3,z4,1)∩N(z1,z2,z3,0,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(0,0,x3,0,r)∣1≤r≤t},N(0,0,0,0,1)∩N(x1,x2,x3,x4,1) ∩N(0,y2,y3,y4,1)∩N(y1,y2,0,y4,1) ∩N(y1,y2,y3,0,1)∩N(0,z2,z3,z4,1) ∩N(z1,z2,0,z4,1)∩N(z1,z2,z3,0,1) -N(y1,y2,y3,y4,1)-N(z1,z2,z3,z4,1) ={(0,x2,0,0,r)∣1≤r≤t}.
Lastly let {v∈V(F2)∣NF2*(v)=NF2*(u)}={(0,0,0,0,r)∣1≤r≤t}, so far V(F2)-V(F1) vertex already all named, the uniqueness of the name. By Lemma 14 and Definition 6, then F2≅HPK(n1,n2,n3,n4)[Kt], V(F2)={(x1,x2,x3,x4,x5)∣0≤xi≤xni,i=1,2,3,4;1≤x5≤t}. Two vertices x=(x1,x2,x3,x4,x5) and y=(y1,y2,y3,x4,x5) are adjacent to each other, if and only if exist x≠y and xi=yi for i=1,2,3,4.
Case 3. Let F3=G-N*(1,1,1,1,1)≅HPK(n1,n2,n3,n4)[Kt], to F3 not named point naming, these point sets are composed of
(14)V(F3)-V(F1)-V(F2) =V(F3)∩(V(F1)∪V(F2))¯ =V(F3)∩(N*(0,0,0,0,1)∩N*(n1,n2,n3,n4,1) ∩ N*(m1,m2,m3,m4,1)) =N*(0,0,0,0,1)∩N*(n1,n2,n3,n4,1) ∩N*(m1,m2,m3,m4,1)-N*(1,1,1,1,1),
for any x=(x1,x2,x3,n4,1)∈V(F1)∩V(F2)∩V(F3), 2≤xi≤ni-1, i=1,2,3,4. By Lemma 14 and (11), we can name
(15){(0,x2,n3,m4,r),(x1,0,n3,m4,r),(x1,n2,0,m4,r), (x1,n3,m4,0,r)∣1≤r≤t}.
By Lemma 14 and (12), we can name
(16){(0,0,n3,m4,r),(0,n2,m3,0,r),(0,n2,0,m4,r), (n1,0,0,m4,r),(n2,m2,0,0,r), (n1,0,m3,0,r)∣1≤r≤t}.
By Lemma 14 and (13), we can name
(17){(0,0,0,m4,r),(0,0,m3,0,r), (0,m2,0,0,r),(m1,0,0,0,r)∣1≤r≤t}.
Similarly, we can name
(18){(0,n2,n3,n4,r),(n1,0,n3,n4,r), (n1,n2,0,n4,r),(n1,n2,n3,0,r)∣1≤r≤t}.
For the V(F2)∩V(F3) point rename, by Lemma 14 and the uniqueness of the name get the name and in F2 name is the same, then V(F3)={(x1,x2,x3,x4,x5)∣0≤xi≤ni, xi≠1, i=1,2,3,4;1≤x4≤t}, and two vertices x=(x1,x2,x3,x4,x5), y=(y1,y2,y3,y4,y5) are adjacent to each other in the F3, if and only if x≠y and, for any xi=yi, for i=1,2,3,4.
Case 4. Named G all unnamed point, this kind of composition is set N*(0,0,0,0,1)∩N*(n1,n2,n3,n4,1)∩N*(1,1,1,1,1). Let F4=G-N*(2,2,2,2,1)=HPK(n1,n2,n3,n4)[Kt]. Obviously, (0,0,0,0,1),(n1, n2, n3, n4, 1), (1,1,1,1,1)∈V(F4). By Lemmas 9 and 14, then
(19)|NF4*(0,0,0,0,0,1)∩NF4*(n1,n2,n3,n4,1) ∩ NF4*(1,1,1,1,1)|=24t,NF4*(0,0,0,0,1)∩NF4*(n1,n2,n3,n4,1) ∩NF4*(1,1,1,1,1)⊂V(F4).
As described in Case 3, in accordance with the Lemma 14, we name V(F2)∩V(F4), V(F3)∩V(F4) the same as in F2,F3. Similarly, by Lemma 14 and (11), we can name
(20)(0,1,n3,n4,r),(0,n2,1,n4,r),(0,n2,n3,1,r),(1,0,n3,n4,r),(1,n2,n3,0,r),(1,n2,0,n4,r), 1≤r≤t.
Hence,
(21)V(F4)={(x1,x2,x3,x4,x5)∣0≤xi≤ni,xi≠2, i=1,2,3,4,1≤x4≤t},
two vertices x=(x1,x2,x3,x4,x5), y=(y1,y2,y3,y4,y5) are adjacent to each other in F4, if and only if, exist x≠y and xi=yi for i=1,2,3,4.
So far, we have the G of each vertex named and V(G)={(x1,x2,x3,x4,x5)∣0≤xi≤ni,i=1,2,3,4,1≤x4≤t}). According to the naming rules, for any x=(x1,x2,x3,x4,x5), y=(y1,y2,y3,y4,y5), if x and y in the same Fi,i=1,2,3,4,5, x is adjacent to y, if and only if x≠y and for any xi=yi, for i=1,2,3,4. Otherwise, for any z∈V(G), z is not adjacent x, y. Let G1=G-N*(z)≅HPK(n1,n2,n3,n4)[Kt], then x,y∈V(G1), the uniqueness of the name also get x and y is adjacent to in G. So x is adjacent to y in G, if and only if x≠y and for any xi=yi, for i=1,2,3,4.
With comprehensive discussed above, then G≅HPK(n1+1,n2+1,n3+1,n4+1)[Kt]. The proof is complete.