Báo cáo toán học: "Graphs with chromatic roots in the interval (1, 2)" doc

... continuity that the chromatic polynomial has a root in the interval (1, 2). We can perform a similar analysis on the graphs obtained by adding the single edge v 1 v 4 to X(s, t) with the following ... an in nite family of 3-connected non-bipartite graphs with chromatic roots in the interval (1, 2) thus resolving a conjecture of Jackson’s in the negat...

Ngày tải lên: 07/08/2014, 13:21

... position in the BFD-ordering of G to the vertex w i in the i-th position in the BFD-ordering of G  . By the properties (B1) and (B2) φ is an isomorphism, as v i and w i have the same degree and the ... prove these theorems by means of a technique of rearranging graphs which has been developed in [2] for the problem of minimizing the first Dirichlet eigenvalue within...

Ngày tải lên: 07/08/2014, 21:20

... : (there exists g ∈ L extending h i ) & (g(v r ) = v)  . That is to say, each vertex in W i together with the vertices of h i (H  ) induces a copy of H. Write d for the degree of v r in ... Hence, L covers G  , completing the induction step and the proof of Theorem 3. ✷ References [1] B. Bollob´as, Modern Graph Theory, Graduate Texts in Mathematics, 184, Springer- Verl...

Ngày tải lên: 07/08/2014, 15:23

... well. Note that the A 1 B-path and the BA 2 -path can only have the vertex B in common, since their lengths sum up to the distance between A 1 and A 2 . Case 2: The vertex B is not contained in any shortest ... = 2 then X 2×c is the subgraph of N 1×c obtained by removing the edge between the vertices (0, 0) and ( 1 , 0), and the edge between the vertices (0, c) and (...

Ngày tải lên: 08/08/2014, 12:23

... Colorings without rainbow k-matchings. In the coloring (a) of Figure 1, a complete subgraph K 2k− 3 of G is rainbow and the other edges are colored with exactly one color, namely, monochromatic. In ... monochromatic. In the coloring (b), a complete subgraph K n−(k−2) of G is monochromatic and the other edges are rainbow. In each coloring, it is clear that there is no rainbow...

Ngày tải lên: 07/08/2014, 21:21

... uced. Proof. In order to use Lemma 2.2, take any intervening neighbours word w, start anywhere in this word, and follow the correspo nding path in the root automaton. To begin with, the 1 in the initial ... paths in the poset: The path starts at the unit root corresponding to the first letter of the word. For each subsequent letter, follow the the corresponding e...

Ngày tải lên: 08/08/2014, 01:20

... to picking a subset of vertices, and coloring them with the non-primary colors. The rest of the set composition and the term  p (π)−1  comes from choosing color classes, and assigning them colors. ... any other hyperedge, we end up counting all colorings by taking the the product, over all e ∈ E J of the ways to color the non-primary edges. The term t J comes from the...

Ngày tải lên: 08/08/2014, 14:23

... has to find them). So, in the guided composition mode, the system has to compute/&/'as they have been defined in §2. The objects are moving during the game so, as opposed to the ira'st ... correct sub-string (from the beginning of the string) and proposes all the possible words that can continue this sub-string. If the sentence is empty, then the s...

Ngày tải lên: 31/03/2014, 04:20

... (7.3) with (6.8) we then see that A n (q, X)= inv A n (q, X). In other words, the generating polynomial for S n by (maj, lowerp) is the same as the generating polynomial by (inv, lowerp). A combinatorial ... using the “fmaj” interpretation, on the one hand, and the “finv” geometry, on the other. In each case we will go beyond the three-variable case, as we consider the...

Ngày tải lên: 07/08/2014, 08:22

... B(x, y)wedenotethe generating functions of the sets M, A and B, respectively, according to the parameters l and r. Using the partition {A, B} of M, we obtain at once the following equation ([17], Proposition ... − r r − γ − 1  k n−2r . In the following result we consider the remaining three cases. Though the produced formulae are more complicated, their proofs are similar to...

Ngày tải lên: 07/08/2014, 08:22