Báo cáo toán học: "List-Distinguishing Colorings of Graphs" ppt

... Oχ(K 5,5 )=5andOχ(K 5,4 )=5. ProofofTheorem1.4: Consider a d-degenerate ordering {v 1 , ,v n } of the vertices of G. We need to create a set {c 1 , ,c k } of k-orthogonal colorings. We prove the theorem by ... spanning subgraph of G r,r has a strong orthogonal scheme. Proof: We begin with the proof of the first part. It suffices to show that G r,r−1 has a SPOC. We define two colorings...

Ngày tải lên: 07/08/2014, 06:20

... member of the second coordinate of the label of y, delete the edge x j y.If the digit of z is 3 and y is not a member of the second coordinate of the label of z, delete the electronic journal of ... Subclaims, the proof of our lemma is complete. 5 Proof of the Main Theorem In this section, we present the proof of the main theorem. Let C be a topologically closed class o...

Ngày tải lên: 07/08/2014, 06:23

... version of Graffiti made number of conjectures about operations R and B (derived from the chromatic number) and we included a sample of them in the first draft of this paper. Since then, most of these ... proved it in some cases and established all lower bounds. Examples of critical P (r, b) graphs are disjoint unions of b − 1 copies of of complete k-partite graphs in which every...

Ngày tải lên: 07/08/2014, 06:20

... parts are all of size one. Alternately, the complement of such a graph either has no edges or consists of the the disjoint union of three or fewer complete graphs. Proof. The complement of K − 5 is ... similar proofs for complexes of dimension 2 and higher. We conjecture that the answer is “yes,” but the complexity of the proof would be astronomical, even for dimension 2. Hidden...

Ngày tải lên: 07/08/2014, 06:20

... class of maximal K 3,3 -minor-free graphs and we can use the number of planar graphs and the number of triangulations as lower bounds. Determining the number (of graphs of sub-classes) of planar graphs ... according to y. We omit the proof as it follows closely the lines of proof of Lemma 5 in [7]. Lemma 3.5. Let W (x, z) = z(1+U(x, z)). The generating function B(x, y) of 2-c...

Ngày tải lên: 07/08/2014, 21:20

... path of three edges to triangulate each hexagonal tile, contributing to the degrees of the vertices in the two central rows of vertices in the strip of hexagons. When the central edges of the ... may equivalently study the maximum multiplicity of the weights of edges of G. An edge is light in G if its weight is bounded by a specified constant. If the sum of the weights of t...

Ngày tải lên: 07/08/2014, 21:21

... examples of fixing sets of non-trivial finite groups are of the form {1, . . . , k } for some k. If the fixing set of every non-trivial finite group is of this form, then the computation of a fixing ... element of the stabilizer of S. This clearly requires at least ⌈ n−1 k+1 ⌉ vertices, so fix(Inf k (K n )) = ⌈ n−1 k+1 ⌉. It seems likely that the proof of Theorem 1 6 could extend...

Ngày tải lên: 07/08/2014, 21:21

... colouring of U can be extended to a U-proper (L, 2) ∗ -colouring of G 1 , and that the resulting colouring of X can be extended to an X-pro per (L, 2) ∗ -colouring of G 2 , and the union of these ... G such that one endvertex of P is an internal vertex of P ab , the other endvertex is a vertex of H ′ that is not in P ab , and no other vertex of P i s in H ′ . Proof. If there...

Ngày tải lên: 08/08/2014, 01:20

... 2-coloring the edges of K n we can break every vertex automorphism of K n , since every automorphism of the vertices of K n is an automorphism of the edges as well. Let G consist of a path of n vertices ... subgroup of Aut(G), each element of which preserves the labeling c i of the vertices of G i .Clearly Aut(G i+1 ) is a subgroup of Aut(G i ). We claim Aut(G i+1 ) is...

Ngày tải lên: 07/08/2014, 06:20

... hypercube Q of dimension is defined as the Cartesian product of copies of K 2 . Its vertex set is the set of all functions of a set S of cardinality into {0, 1}, where the preimage of 1 is finite, ... all neighbors of v  in T are labeled; the electronic journal of combinatorics 14 (2007), #R36 5 Assume next that the degree of G is not bounded. The vertex set of G is the u...

Ngày tải lên: 07/08/2014, 15:22