Báo cáo toán học: "Laplacian Integral Graphs with Maximum Degree 3" docx

... graph G, we let ∆(G) denote its maximum degree. One of the natural lines of investigation for Laplacian integral graphs is to focus on graphs for which the maximum degree is not very large. It is ... (3) Figure 5: connected, non-regular, Laplacian integral graphs with maximum degree 3 the electronic journal of combinatorics 15 (2008), #R120 6 Figure 41: graphs arising...

Ngày tải lên: 07/08/2014, 21:20

... 1996 Abstract Connected graphs with minimum degree δ and at least 2δ + 1 vertices have paths with at least 2δ + 1 vertices. We provide a characterization of all such graphs which have no longer ... the sense that there exist graphs of minimum degree δ with no longer paths. Our purpose in this article is to characterize all such extremal graphs. For a graph G with n vertice...

Ngày tải lên: 07/08/2014, 06:20

... (9) and (10) with V replaced by V ′ . Then ∆(V ′ m ) = C ′ σ with σ being the identity permutation and E ′ m the union of a ll C ′ σ with σ ∈ Sym(m) where C ′ σ = Cyl m (T ′ σ ) with T ′ σ = ... may exist even fo r non-isomorphic graphs. In any o f these cases the graphs are called equivalent (Definition 3.3). For example any two strongly regular graphs with the same parameters...

Ngày tải lên: 08/08/2014, 01:20

... s i vertices of degree i then it has at least  k−1 i=0 (k − i) s i distinct k-colorings and (b) with extremely high probability, for every fixed i, G(n, cn)hasΩ(n) vertices of degree i.IfX is ... the argument to uniformly random pseudographs on a given degree sequence (for a definition see also [15]). In particular, they proved that such a pseudograph with ρn edges is w.h.p. non-k-co...

Ngày tải lên: 07/08/2014, 06:20

... infinite class of MNT graphs. We call the graphs in these classes Zelinka graphs, and we call MNT graphs that cannot be constructed by one of Zelinka’s constructions non-Zelinka MNT graphs. By consulting ... of non-Zelinka MNT graphs that have cut-vertices and hence have toughness at most 1/2. We shall call these graphs DKW graphs. In this paper we investigate the structure of MNT...

Ngày tải lên: 07/08/2014, 15:23

... the Plane problem. In this paper we construct a new class of graphs with ambiguous chromatic numb er. They are unit distance graphs with vertex set R n , and hence may be seen as further evidence ... present a new family of graphs with ambiguous chro- matic number. Unlike Sz´ekely’s example, the new examples have all of R n as their vertex set, and unlike Shelah and Soifer’s grap...

Ngày tải lên: 07/08/2014, 21:21

... of R with copies of the complete bi partite graph K N,N and a sparser graph G is constructed by repl acing the edges o f R with som e (ε, δ)-super-regular pairs. If a graph H with maximum degree ... Hladk´y and Schacht [10] determined the minimum degree threshold for K s,t -factors with s < t. Let G r (N) denote the family of r-partite graphs with N vertices in each of it...

Ngày tải lên: 08/08/2014, 01:20

... this conjecture holds. Keywords: Graphs with cyclic defect, Moore bound, defect, repeat. 1 Nonexistence of graphs with cyclic defect Let G be a graph of maximum degree ∆, diameter D and order M(∆, ... consider graphs of maximum degree ∆, diameter D and order M(∆, D) − 2, where M(∆, D) is the Moore bound, that is, graphs of defect 2. In [1] Delorme and Pineda-Villavicencio...

Ngày tải lên: 08/08/2014, 14:22

... adjacency matrices of integral graphs with n vertices is less than or equal to 2 n(n−1) 2 − n 400 for a sufficiently large n. For the background and some known results about integral graphs, we ref er ... characterizing integral graphs seems to be very difficult and so it is wise to restrict ourselves to certain families of graphs. Here we are interested to study Cayley graphs. Let...

Ngày tải lên: 08/08/2014, 01:20

... inequalities with Luexmburg norms in L φ (m)-averaging domains. J Inequal Appl (2010). Article ID 241759 doi:10.1186/1029-242X-2011-90 Cite this article as: Wen: The integral estimate with Orlicz ... u ϕ 1 ( Q ) ≤ C  u ϕ 2 ( σ Q ) , (2:18) where Q is any cube with sQ ⊂ Ω, s is constant with 1<s < ∞. Proof. Using Hölder inequality with 1 = 1/p +(p - 1)/p, we have  u ϕ 1...

Ngày tải lên: 20/06/2014, 22:20