Then there exist two vertices such that.
Suppose that there exist two vertices such that.
Otherwise, if there exist two vertices a 1∈A and b 1∈B such that a 1 ≁ b 1 Open image in new window, a contradiction with diam(G =2 (because d(a 1,b 1)≥3, d(a 1,b ≥3 or d a,b 1)≥3).
Observation 2(2) implies that G has no pendant edge, and there exist two vertices u and v such that G[V(G)−{u,v}] is the union of some connected graph of order at least 2 and p+1≤deg G (u)≤n−2, q+1≤deg G (v ≤n−2, otherwise, a contradiction with γ t r (G =2.
If C r and P ( G σ ) have no common arc, then there exist three vertices, say u, v and w, such that each of them is not contained in P ( G σ ) and v adjacent to both u and w.
Then, there exist three vertices v r,v s,v t ∈V(G) such that v r
r is even and r > 4. Then there exist three vertices, say u, v and w, such that each of them is not contained in P ( G σ ) and v adjacent to both u and w.
Then there must exist two vertices u ∈ U, v ∈ V not adjacent in G. Let W 1 = u u 1 ⋯ u 2 k 1 v be a walk of G. Now, we join the vertices u and v and suppose the resulting graph is G 1. Obviously, we can give the edge uv a direction (for instance, from u to v) such that the closed 2 ( k 1 + 1 ) -walk W = W 1 + u v satisfies sgn ( W ς ) = ( − 1 ) k 1 + 1 in G 1 ς = G σ + ( u, v ).
Then k = d, and either v 0 or v d is the pendent vertex of G σ. Recall that p ≥ 3, then there exist two adjacent vertices, say u and v, are not contained in P ( G σ ).
In one version of this problem, we make use of the graph theory to reformulated it as follows: Given a graph G with n vertices, do there exist two θ-powers of paths G S =(V S,E S ) and G T = V T,E T ) such that G S ∩G T contains G as an induced subgraph?
1. G F is a minimal triangulation of G. 2. Let (T, B) be a clique tree of G F. There exists a minimal separator F ∈ F if and only if there exist two adjacent vertices x and y in T where B x)∩ B y)= F. 3. △ H is a maximal set of pairwise parallel minimal separators of G and G △ H = H.
