Profile minimization on product of graphs

3.2.1 Profile of K

^m

× K

_n

This subsection establishes the profile of Km× Kn.

Theorem 3.2.1 If m = 1 or n ≥ max{m, 4}, then P (Km× Kn) = ¹₂(m − 1)(mn²+ n²− n − 4).

Proof. As the case of m = 1 is obvious, we may assume that m ≥ 2 and n ≥ max{m, 4}.

First, consider a proper numbering g of Km× Kn satisfying

g(vi,j) =

while the other vertices are assigned numbers arbitrarily, see Figure 3.1 for g of K5× K₉ in which the edges are not drawn for simplicity.

K5

The profile width of vertex vi,j is

wg(vi,j) =

Therefore,

P (Km× Kn) ≤ Pg(Km× Kn)

= (mn − n − m + 1) +

mn−1X

k=n

(k − 1) − (m − 1)

= 1

2(m − 1)(mn²+ n²− n − 4).

Next, we shall prove that P (Km × Kn) ≥ ¹₂(m − 1)(mn² + n² − n − 4). Choose a profile numbering f of Km× Kn. Notice that P (Km× Kn) = |E((Km× Kn)f)|. Without loss of generality, we may assume that f (v1,1) = 1. For positive integers a and b, let ea,b = 2 ^a_2
b

2

+ (a − 1) ^b₂

+ (b − 2) ^a₂

+ 2 ^a−1₂

. We consider the following three cases.

Case 1. f⁻¹(2) ∈ R₁, say f (v_1,j) = j for 1 ≤ j ≤ r but f (vs,t) = r + 1 with s 6= 1 for some r ≥ 2.

We shall count the number of edges in (Km× Kn)f. Notice that besides the edges in Km×Kn, extra edges are due to the following cliques in (Km×Kn)f which are independent sets in Km× Kn.

Each row Ri with 2 ≤ i ≤ m is a clique in (Km× Kn)f, since for vi,p, vi,q ∈ Ri with f (vi,p) < f (vi,q), we can choose k ∈ {1, 2} − {q}, such that f (v1,k) = k < f (vi,p) < f (vi,q) and v1,kvi,q ∈ E(Km × Kn) ⊆ E((Km × Kn)f), which imply vi,pvi,q ∈ E((Km × Kn)f).

Notice that we use the interval property (3.1) in this implication. As the property will be used frequently, we shall not mention it every time.

Each column Cj with 2 ≤ j ≤ r is a clique in (Km × Kn)f, since for vp,j, vq,j ∈ Cj

with f (vp,j) < f (vq,j), we have q ≥ 2, and so f (v1,1) = 1 < f (vp,j) < f (vq,j) and v1,1vq,j ∈ E(Km× Kn) ⊆ E((Km× Kn)f), which imply vp,jvq,j ∈ E((Km× Kn)f).

For the case r + 1 ≤ n, any column Cj with j ≥ r + 1 but j 6= t is a clique in (Km× Kn)f, since for vp,j, vq,j ∈ Cj with f (vp,j) < f (vq,j), we can choose x = v1,1 (when q 6= 1) or vs,t (when q = 1), such that f (x) < f (vp,j) < f (vq,j) and xvq,j ∈ E(Km× Kn) ⊆ E((Km× Kn)f), which imply vp,jvq,j ∈ E((Km× Kn)f).

Similarly, Cj − {v1,j} is cliques in (Km × Kn)f for 1 ≤ j ≤ n. In particular, this is true for j = 1, t.

Therefore, totally the graph (Km× Kn)f has at least em,n = 2 ^m_2
n

3.2.2 Profile of (K

^s

∨ G ) × K

ⁿ

This subsection determines the profile of (Ks∨ G) × Kn with |V (G)| = t ≤ s.

The notations we use in this subsection are the same as above except now we let m = s + t and V ((Ks∨ G)) = S ∪ T , where S = {x1, x2, . . . , xs} = V (Ks) and T =

+ (n²− 2)st, consider the proper numbering g of (Ks∨ G) × Kn defined by

Notice that two vertices vi,j, vi⁰,j⁰ are adjacent in (Ks∨ G) × Kn if and only if one is in Sj and the other in Tj⁰ for some j 6= j⁰ or one is in Tj and the other is in Tj⁰ with xixi⁰ ∈ E(G). As no vertex in Si is adjacent to a vertex with smaller numbering in (Ks∨ G) × Kn, S × V (Kn) is an independent set in ((Ks∨ G) × Kn)g.

For any two vertices vi,jand vi⁰,j⁰in T ×Knwith g(vi,j) < g(vi⁰,j⁰), we may choose k from {1, 2} such that k 6= j⁰. So, g(v1,k) < g(vi,j) < g(vi⁰,j⁰) and v1,kvi⁰,j⁰ ∈ E((Ks∨ G) × Kn) ⊆ E(((Ks∨G)×Kn)g) imply that vi,jvi⁰,j⁰ ∈ E(((Ks∨G)×Kn)g). This proves that T ×V (Kn) is a clique in ((Ks∨ G) × Kn)g, which gives ^nt₂

edges.

For any vi,j ∈ Sj and vi⁰,j ∈ Tj with 2 ≤ j ≤ n − 1, we have g(v1,1) < g(vi,j) < g(vi⁰,j) and v1,1vi⁰,j ∈ E((Ks∨ G) × Kn) ⊆ E(((Ks∨ G) × Kn)g) implying that vi,jvi⁰,j ∈ E(((Ks∨ G) × Kn)g). It is also the case that no vertex in Sj is adjacent to a vertex in Tj in ((Ks ∨ G) × Kn)g for j = 1 or n. So, vertices in Sj are adjacent to vertices in Tj⁰ in ((Ks∨ G) × Kn)g for all j and j⁰ except j = j⁰ ∈ {1, n}. These give (n² − 2)st edges.

Therefore, P ((Ks∨ G) × Kn) ≤ |E(((Ks∨ G) × Kn)g)| = ^nt₂

+ (n²− 2)st.

Next, we shall prove that P ((Ks ∨ G) × Kn) ≥ ^nt₂

+ (n² − 2)st. Choose a profile numbering f of (Ks∨ G) × Kn. For the first case, assume that f (v1,1) = 1. Let f (va,b) = min{f (vi,j) : vi,j ∈ T₂∪ . . . ∪ Tn}.

For any vertices vi,j ∈ Sj and vi⁰,j⁰ ∈ Tj⁰, by the definition, vi,jvi⁰,j⁰ ∈ E((Ks∨ G) × Kn) ⊆ E(((Ks ∨ G) × Kn)f) if j 6= j⁰. Suppose j = j⁰ 6∈ {1, b}. If f (vi,j) < f (vi⁰,j⁰), then f (v_1,1) < f (vi,j) < f (vi⁰,j⁰) and v_1,1vi⁰j⁰ ∈ E(((Ks∨ G) × Kn)f) imply that vi,jvi⁰j⁰ ∈ E(((Ks∨ G) × Kn)f). If f (vi,j) > f (vi⁰,j⁰), then f (va,b) < f (vi⁰,j⁰) < f (vi,j) and va,bvi,j ∈ E(((Ks∨ G) × Kn)f) imply that vi,jvi⁰j⁰ ∈ E(((Ks∨ G) × Kn)f). So, vertices in Sj are adjacent to vertices in Tj⁰ for all j and j⁰ except j = j⁰ ∈ {1, b}. These give (n² − 2)st edges.

Consider any two vertices vi,j and vi⁰,j⁰ in T1∪ T₂∪ . . . ∪ Tn such that f (vi,j) < f (vi⁰,j⁰).

For j⁰ ≥ 2, we have f (v_1,1) < f (vi,j) < f (vi⁰,j⁰) and v1,1vi⁰,j⁰ ∈ E(((Ks∨G)×Kn)f) implying vi,jvi⁰,j⁰ ∈ E(((Ks∨ G) × Kn)f). So, T₂∪ T₃∪ . . . ∪ Tn is a clique in ((Ks∨ G) × Kn)f. This gives ^(n−1)t₂

edges. If T₁∪T₂∪. . . ∪Tn is a clique in ((Ks∨G)×Kn)f, then these give ^nt₂
edges. Therefore, P ((Ks∨ G) × Kn) ≥ ^nt₂

+ (n²− 2)st. Now, we may assume that there

are two non-adjacent vertices vp,q and vp⁰,q⁰ in T1 ∪ T2 ∪ . . . ∪ Tn with f (vp,q) < f (vp⁰,q⁰) and q⁰ = 1.

For any two vertices vi,j and vi⁰,j⁰ in S2 ∪ S₃ ∪ . . . ∪ Sn such that f (vi,j) < f (vi⁰,j⁰).

If f (vp,q) > f (vi,j), then f (vi,j) < f (vp,q) < f (vp⁰,q⁰) and vi,jvp⁰,q⁰ ∈ E(((Ks∨ G) × Kn)f) imply vp,qvp⁰,q⁰ ∈ E(((Ks ∨ G) × Kn)f), a contradiction. Therefore, it is always the case that f (vp,q) < f (vi,j) < f (vi⁰,j⁰). Except for the case when q = j⁰ = b, we have vp,qvi⁰,j⁰ ∈ E(((Ks∨ G) × Kn)f), which together with the above inequalities gives that vi,jvi⁰,j⁰ ∈ E(((Ks∨ G) × Kn)f).

Now, if q 6= b, we have that S2 ∪ S₃ ∪ . . . ∪ Sn is a clique in ((Ks ∨ G) × Kn)f. This gives ^(n−1)s₂

edges. And so P ((Ks∨ G) × Kn) ≥ ^(n−1)s₂

+ ^(n−1)t₂

+ (n² − 2)st ≥ 2 ^(n−1)t₂

+ (n²−2)st as n ≥ 4. Hence we may assume that if vp,q and vp⁰,q⁰ are nonadjacent in T1 ∪ T₂ ∪ . . . ∪ Tn with f (vp,q) < f (vp⁰,q⁰), then q = b and q⁰ = 1. In this case, S₂∪ S₃ ∪ . . . ∪ S_b−1∪ S_b+1 ∪ S_b+2∪ . . . ∪ Sn is a clique and T₁ ∪ T₂ ∪ . . . ∪ Tn is a clique in ((Ks∨ G) × Kn)f except that vertices in T1 are not necessarily adjacent to vertices in Tb. This gives P ((Ks∨ G) × Kn) ≥ ^(n−2)s₂

+ ^nt₂

− t²+ (n²− 2)st ≥ ^nt₂

+ (n²− 2)st as n ≥ 4.

For the second case, assume that f (vs+1,1) = 1. By symmetric argument of the first case, we also obtain P ((Ks∨ G) × Kn) ≥ ^nt₂

+ (n²− 2)st as n ≥ 4.

3.2.3 Profile of P

^m

× K

_n

Finally in this section, we study the profile of Pm× Kn.

The results in the previous subsections cover the case for P1×Kn= K1×Kn, P2×Kn = K2 × Kn = K1,1 × Kn and P3× Kn = K1,2× Kn. In the following we consider only for m ≥ 4.

Theorem 3.2.3 If m, n ≥ 4, then P (Pm× Kn) = (m − 2) ⁿ₂

+ (m − 1)(n²− 1).

Proof. For P (Pm× Kn) ≤ (m − 2) ⁿ₂

+ (m − 1)(n²− 1), consider the proper numbering

g of Pm× Kn defined by

see Figure 3.4 for g of P5× K9 in which the edges are not drawn for simplicity.

P5

The profile width of vertex vi,j is

wg(vi,j) =

Let

ai = min

vi,j∈Ri

f (vi,j) and f (vi,bi) = ai for 1 ≤ i ≤ m.

A = {i : 2 ≤ i ≤ m − 1 and Ri is not a clique in (Pm× Kn)f} and p = |A|.

B = {i : 2 ≤ i ≤ m − 1 and ai < min{a_i−1, a_i+1}} and q = |B|.

Λi,i⁰ = {vi,jvi⁰,j⁰ ∈ E((Pm× Kn)f) : 1 ≤ j, j⁰ ≤ n}.

λi,i⁰ = |Λi,i⁰| for 1 ≤ i, i⁰ ≤ m.

Λ⁼_i,i0 = {vi,jvi⁰,j⁰ ∈ E((Pm× Kn)f) : 1 ≤ j = j⁰ ≤ n}.

λ⁼_i,i0 = |Λ⁼_i,i0| for 1 ≤ i, i⁰ ≤ m.

Λ^≤_i,i0 = {vi,jvi⁰,j⁰ ∈ E((Pm× Kn)f) : 1 ≤ j ≤ j⁰ ≤ n}.

λ^≤_i,i0 = |Λ^≤_i,i0| for 1 ≤ i, i⁰ ≤ m.

Claim 1. Suppose |i − i⁰| = 1. Then λ⁼_i,i0 ≥ n − 2 and so λi,i⁰ ≥ n² − 2. Furthermore, if bi = bi⁰, or f (vi,b_i0) < f (vi⁰,b_i0), or Ri is a clique in (Pm × Kn)f with ai < ai⁰, then λ⁼_i,i0 ≥ n − 1 and so λi,i⁰ ≥ n²− 1.

Proof of Claim 1. Consider any j /∈ {bi, bi⁰}. If f (vi,j) < f (vi⁰,j), then f (vi,bi) < f (vi,j) <

f (vi⁰,j) and vi,bivi⁰,j ∈ E(Pm × Kn) ⊆ E((Pm× Kn)f) imply vi,jvi⁰,j ∈ E((Pm × Kn)f).

If f (vi,j) > f (vi⁰,j), then f (vi⁰,b_i0) < f (vi⁰,j) < f (vi,j) and vi⁰,b_i0vi,j ∈ E(Pm × Kn) ⊆ E((Pm× Kn)f) imply vi⁰,jvi,j ∈ E((Pm× Kn)f). In any case, vi,jvi⁰,j ∈ E((Pm× Kn)f) for j /∈ {bi, bi⁰}, which give λ⁼_i,i0 ≥ n − 2. There are already other n(n − 1) edges between Ri

and Ri⁰ in E(Pm× Kn), so we have λi,i⁰ ≥ n²− 2. For the case bi = bi⁰, there are at least n − 1 edges vi,jvi⁰,j ∈ E((Pm× Kn)f) for j /∈ {bi, bi⁰}. So, λ⁼_i,i0 ≥ n − 1 and λi,i⁰ ≥ n²− 1.

Now suppose bi 6= bi⁰. For the case f (vi,b_i0) < f (vi⁰,b_i0), besides the n − 2 edges vi,jvi⁰,j

for j /∈ {bi, bi⁰}, we also have the edge vi,b_i0vi⁰,b_i0, since f (vi,bi) < f (vi,b_i0) < f (vi⁰,b_i0) and vi,bivi⁰,b_i0 ∈ E(Pm× Kn) ⊆ E((Pm× Kn)f) implying vi,b_i0vi⁰,b_i0 ∈ E((Pm× Kn)f). For the case when f (vi,b_i0) > f (vi⁰,b_i0) and Ri is a clique with ai < ai⁰, again f (vi,bi) = ai < ai⁰ = f (vi⁰,b_i0) < f (vi,b_i0) and vi,bivi,b_i0 ∈ E((Pm× Kn)f) imply vi⁰,b_i0vi,b_i0 ∈ E((Pm× Kn)f). In any case, vi,jvi⁰,j ∈ E((Pm× Kn)f) for j 6= bi, which gives λ⁼_i,i0 ≥ n − 1 and λi,i⁰ ≥ n²− 1.



Claim 2. If i ∈ A, then λ^≤_i−1,i+1 ≥ ⁿ⁻¹₂

≥ 3.

Proof of Claim 2. As Ri is not a clique in (Pm × Kn)f, we may choose c 6= d such that vi,cvi,d ∈ E((P/ m× Kn)f). Consider any j, j⁰ ∈ {c, d} with 1 ≤ j ≤ j/ ⁰ ≤ n. In the 4-cycle (vi,c, vi−1,j, vi,d, vi+1,j⁰, vi,c), we have vi,cvi,d ∈ E((P/ m× Kn)f) implying vi−1,jvi+1,j⁰ ∈ E((Pm× Kn)f) by the chordality property (3.3). This gives that λ^≤_i−1,i+1 ≥ (1 + 2 + . . . + (n − 2)) = ⁿ⁻¹₂

≥ 3. 

Claim 3. If i ∈ B, then λ^≤_i−1,i+1 ≥ ⁿ₂

≥ 6.

Proof of Claim 3. For any j, j⁰ ∈ {b/ i} with 1 ≤ j ≤ j⁰ ≤ n, since f (vi,bi) = ai < ai−1 ≤ f (vi−1,j) with vi,bivi−1,j ∈ E(Pm × Kn) ⊆ E((Pm × Kn)f) and f (vi,bi) = ai < ai+1 ≤ f (vi+1,j⁰) with vi,b_ivi+1,j⁰ ∈ E(Pm×Kn) ⊆ E((Pm×Kn)f), by perfect elimination property (3.2), vi−1,jvi+1,j⁰ ∈ E((Pm× Kn)f). These give λ^≤_i−1,i+1 ≥ 1 + 2 + . . . + (n − 1) = ⁿ₂

≥ 6.



Having these three claims in mind, we are ready to prove the theorem. As n ≥ 4, there is a bijection from {{j, k} : 1 ≤ j < k ≤ n} to itself such that {j, k} is disjoint from its image {j⁰, k⁰}. This can be done by setting {j⁰, k⁰} = {(j + δ) mod n, (k + δ) mod n}, where δ = 2 when j and k are consecutive under modulo n, and δ = 1 otherwise. We may assume that j⁰ > k⁰ for our convenience. Consider the following (m − 2) ⁿ₂

disjoint sets:

Si,j,k = {vi,jvi,k, v_i−1,j⁰v_i+1,k⁰},

where 2 ≤ i ≤ m − 2 and 1 ≤ j < k ≤ n. In the 4-cycle (vi,j, vi−1,j⁰, vi,k, vi+1,k⁰, vi,j) (see Figure 3.5), at least one of the edge in Si,j,k must exist. These give totally at least (m − 2) ⁿ₂

edges.

r r r r r r

r r r r r r

r r r r r r











@@

@

R_i−1 Ri

R_i+1 v_i−1,j⁰

vi,j

vi,k

vi+1,k⁰

Figure 3.5: The 4-cycle (vi,j, vi−1,j⁰, vi,k, vi+1,k⁰, vi,j).

Among the m − 2 rows R2, R3, . . . , Rm−1, there are p rows that are not cliques in (Pm× Kn)f and the other m − 2 − p rows are cliques. Among the m − 2 − p clique rows, let there be p⁰ consecutive pairs, that is, cliques Ri and Ri⁰ with |i − i⁰| = 1. By Claim 1, λi,i⁰ ≥ n²− 1 for these p⁰ pairs and λi,i⁰ ≥ n²− 2 for the remaining m − 1 − p⁰ pairs of

i and i⁰ with |i − i⁰| = 1. These give totally at lease p⁰(n²− 1) + (m − 1 − p⁰)(n² − 2) = (m − 1)(n²− 1) + (p⁰+ 1 − m) edges.

By Claim 3, there are at least 6q extra edges from the sets Λ^≤_i−1,i+1 for i ∈ B. By Claim 2, there are at least 3(p − q) extra edges from the sets Λ^≤_i−1,i+1 for i ∈ A \ B. These give at least 3p + 3q extra edges. So, we have

P (Pm× Kn) ≥ (m − 2)

n 2



+ (m − 1)(n²− 1) + (p⁰+ 1 − m + 3p + 3q).

In particular, P (Pm× Kn) ≥ (m − 2) ⁿ₂

+ (m − 1)(n²− 1) when p⁰+ 1 − m + 3p + 3q ≥ 0.

So, now assume that p⁰+ 1 − m + 3p + 3q ≤ −1 or p⁰ ≤ m − 3p − 3q − 2.

Notice that there are p non-clique rows Ri with 2 ≤ i ≤ m − 1. These rows separate the other rows into p + 1 runs. Each run with α clique rows in R2, R3, . . . , Rm−1 has max{0, α − 1} ≥ α − 1 consecutive pairs of cliques. Therefore, p⁰ ≥ m − 2 − p − (p + 1) = m − 2p − 3 with equality holds if and only if α ≥ 1 for each run of clique rows.

Or equivalently, any two rows in A ∪ {R1, Rm} are not consecutive, which implies that 3 ≤ i ≤ m − 2 for i ∈ A.

Now, m − 2p − 3 ≤ p⁰ ≤ m − 3p − 3q − 2 imply that p + 3q ≤ 1. This is possible only when p ≤ 1 and q = 0. Suppose p = 1, say A = {Ri}. Then, the above inequalities are in fact equalities, i.e., m − 2p − 3 = p⁰ and so 3 ≤ i ≤ m − 2. Therefore, Ri−1 and Ri+1 are clique rows. As q = 0, we have i /∈ B and so either ai−1 < ai or ai+1 < ai. By Claim 1, either λ⁼_i−1,i ≥ n − 1 or λ⁼_i,i+1 ≥ n − 1. So in the above calculation, we in fact have p⁰+ 1, rather than p⁰, consecutive pairs of i and i⁰ with λi,i⁰ ≥ n²− 1. Thus,

P (Pm× Kn) ≥ (m − 2)

n 2



+ (m − 1)(n²− 1) + (p⁰+ 2 − m + 3p + 3q),

where p⁰ + 2 − m + 3p + 3q ≥ (m − 2p − 3) + 2 − m + 3p + 3q = p + 3q − 1 = 0 and so again P (Pm× Kn) ≥ (m − 2) ⁿ₂

+ (m − 1)(n²− 1).

Now we may suppose that p = q = 0. In other words, R₂, R₃, . . . , R_m−1 are cliques and

a1 < a2 < . . . < ar−1 < ar and ar > ar+1 > ar+2 > . . . > am (3.4) for some r. By Claim 1, we have

λ1,2 ≥ n²− 2, λi,i+1 ≥ n²− 1 for 2 ≤ i ≤ m − 2, λm−1,m≥ n²− 2.

These together with the m − 2 clique rows gives at least (m − 2) ⁿ₂

+ (m − 1)(n²− 1) − 2 edges. In the following, two extra edges, one with an end vertex in R1 and the other with an end vertex in Rm, are to be found to make P (Pm× Kn) ≥ (m − 2) ⁿ₂

+ (m − 1)(n²− 1).

Assume, by symmetric, there is no such extra edge with a vertex in R1 which we call an R₁-edge, we shall either get a contradiction or find two other extra edges.

First, we may assume that b1 6= b2 and a1 < a2 and f (v1,b2) > f (v2,b2), for otherwise Claim 1 gives that λ1,2 ≥ n² − 1 rather than only λ1,2 ≥ n² − 2 which give an extra R1-edge, a contradiction. Notice that the two non-edges between R1 and R2 are v1,b1v2,b1

and v1,b2v2,b2.

We claim that in fact a1 = 1. Suppose to the contrary that a1 > 1. By (3.4), we have am = 1. This together with am < a1 < a2 ≤ ar implies that there is some i such that ar ≥ ai−1 > a1 > ai ≥ am = 1. Then, for each j 6= bi, we have f (vi,bi) < f (v1,b1) < f (vi−1,j) and vi,biv_i−1,j ∈ E((Pm × Kn)f) implying v_1,b1v_i−1,j ∈ E((Pm× Kn)f), which gives n − 1 extra R1-edges, a contradiction. Thus, a1 = 1.

As a1 = 1 and f (v1,b2) > a2, without loss of generality, we may assume that f (v1,j) = j for 1 ≤ j ≤ ` − 1 but f<sup>−1</sup>(`) = vi^∗,j^∗ 6∈ R1, where ` ≤ n. Notice that we assume b1 = 1 now. By the inequalities in (3.4), we have ` = am or ` = a2. For the case

` = am, for any j 6= 1, we have f (v1,1) = 1 < ` = am = f (vm,bm) < f (v2,j) and v_1,1v_2,j ∈ E ((Pm× Kn)f), implying vm,bmv_2,j⁰ ∈ E ((Pm× Kn)f), which are n − 1 ≥ 2 extra edges as desired. For the case ` = a<sub>2</sub>, we may assume that b<sub>2</sub> = n. If ` < n, then for any j < n, we have f (v2,n) < f (v1,`) with v2,nv1,` ∈ E((Pm× Kn)f) and f (v2,n) < f (v3,j) with v2,nv3,j ∈ E((Pm×Kn)f), implying v1,`v3,j ∈ E((Pm×Kn)f) by the perfect elimination property (3.2). This gives n − 1 ≥ 2 extra edges as desired. So, we may assume that

` = n.

Next, f (v1,n) > f (v3,1), for otherwise, f (v1,n) < f (v3,1) gives that f (v2,n) < f (v1,n) <

f (v3,1), this together with v2,nv3,1 ∈ E((Pm× Kn)f) implying v1,nv3,1 ∈ E((Pm× Kn)f), which is an extra R₁-edge, a contradiction. Similarly, for each j with 2 ≤ j ≤ n−1 we have f (v_2,j) > f (v_3,1), for otherwise, f (v_2,j) < f (v_3,1) gives that f (v_2,j) < f (v_3,1) < f (v_1,n), this together with v2,jv1,n ∈ E((Pm× Kn)f) implying v3,1v1,n∈ E((Pm× Kn)f), which is

an extra R1-edge, a contradiction. Also, f (v4,2) > f (v3,1), for otherwise, f (v4,2) < f (v3,1) gives that for each j with 2 ≤ j ≤ n − 1, we have f (v1,1) < f (v4,2) < f (v3,1) < f (v2,j), this together with v1,1v2,j ∈ E((Pm× Kn)f) implying v4,2v2,j ∈ E((Pm× Kn)f), which are n − 2 ≥ 2 extra edges as desired. Now, for each j with 2 ≤ j ≤ n − 1, we have f (v3,1) <

f (v_2,j) with v_3,1v_2,j ∈ E((Pm×Kn)f), and f (v_3,1) < f (v_4,2) with v_3,1v_4,2 ∈ E((Pm×Kn)f), implying v2,jv4,2 ∈ E((Pm× Kn)f), which are n − 2 ≥ 2 extra edges as desired.

在文檔中 圖形的帶寬與輪廓 (頁 49-60)