Báo cáo toán học: "Dense H-free graphs are almost (χ(H) − 1)-partite" docx

... and including the edge (α, β) if α −1 β ∈ S ∪ S −1 , where S −1 = {s −1 | s ∈ S}. As the set S ∪ S −1 is closed under inverse, α −1 β ∈ S ∪ S −1 ⇔ β −1 α ∈ S ∪ S −1 so that we may naturally treat ... 1, Pr  1 k  i p i (ρ) ∈  1 − ε 2 , 1+ε 2   ≤ 2d ρ exp  − kε 2 4ln2  =2d ρ exp  − ln(2)[log D + b +1]  = d ρ D 2 −b . Finally, Pr[λ 2 (X(...

Ngày tải lên: 07/08/2014, 08:20

... e 4πi 3 e 2πi 3 e 2πi 3 e 4πi 3 χ 4 3 3 −1 0 0 0 0 χ 5 2 −2 0 −1 −1 1 1 χ 6 2 −2 0 −e 2πi 3 −e 4πi 3 e 4πi 3 e 2πi 3 χ 7 2 −2 0 −e 4πi 3 −e 2πi 3 e 2πi 3 e 4πi 3 g 2 =  −1 0 0 −1  , g 3 =  0 1 −1 0  , g −1 5 = g 4 =  1 ... g −1 7 = g 6 =  −1 1 0 −1  Character Table of V 24 1 b 2 a a 3 a 5 a 2 a 2 b 2 b ab χ 1 1 1 1 1 1 1 1 1 1 χ 2 1 1 1 1 1 1...

Ngày tải lên: 08/08/2014, 01:20

... two odd numbered vertices of C are adja- cent. Suppose 1 ≤ i<j≤ 2δ − 1withi and j odd and v i adjacent to v j .Then the vertices x, v i−1 ,v i ,v j ,v j−1 ,v j−2 , ···,v i+1 ,y,v j+1 ,v j+2 , ... non-hamiltonian graphs which barely miss satisfying Dirac’s condition. We are sure this must be known, but we have not found it in the literature. We are grateful to Arthur Hobbs [6],...

Ngày tải lên: 07/08/2014, 06:20

... n        1 2 − 1 2 1 − 1 2 1 2        −a        − 1 2 1 2 0 − 1 2 1 2        + c        1 3 − 4 3 2 − 4 3 1 3        +        −5 10 −1 0 5 −1        + (δ −7 2)        1 24 − 1 6 1 4 − 1 6 1 24        +        7 2 −1 3 18 −1 1 5 2        . (3.5) Note ... n  ...

Ngày tải lên: 07/08/2014, 21:20

... − 1. Since R ∪ R s and R ϕ ∪ R s are the edge sets of the graphs G and G ′ respectively, and ϕ is an m-similarity of the scheme [G] = [G ′ ] to itself, we conclude that the graphs G and G ′ are ... observe that by Theorem 4.2 the graphs G {m} and G m are isomorphic. By equality (8) this implies that they are equivalent. Similarly, the graphs (G ′ ) {m} and G ′ m are equi...

Ngày tải lên: 08/08/2014, 01:20

... ∆(G ′ [X 1 ])  εn, v(N) > 3r−4 3r−1 n + 2εn. Now consider u ∈ N. We have d N (u) > v(N) −  3 3r − 1 − 4ε  n > v(N) −  3 3r − 1 − 4ε  3r − 1 3r − 4 v(N) >  3r − 7 3r − 4 + 4ε  v(N) . By ... |V 1 | − n 3r−1 choices for v 1 . At the second step we have more than |V 2 | − n 3r − 1 −  3 3r − 1 − 4ε  n + (|V 1 |...

Ngày tải lên: 08/08/2014, 11:20

... of combinatorics 1 (1994),#R12 5 Similarly, (2) implies that (2 k−1 − 1 −2 x)(2 k−1 − 2y) − (1 + 2(x −y)) 2 4 −9 · 2 k −3 ·2 k/2 − 1/4. This completes the proof of part 2 of the theorem. Part ... S n = U 0 + U 1 ,whereU i are defined as above. Let A 0 be the 3k by 2 k−1 −1 binary matrix whose columns are all vectors of U 0 ,andletA 1 be the 3k by 2 k−1 matrix who...

Ngày tải lên: 07/08/2014, 06:20

... 2∆(k) − 1)giveninSection5, we have L(4∆(k)) = 2k − 2+ √ 3andL(2∆(k + 1) + 2∆(k) − 1) = 2k − 1+ √ 3, from which it follows that both lim k→∞ δ(4∆(k)) and lim k→∞ δ(2∆(k + 1) + 2∆(k) −1 ) are at ... do exist), then all three limits are equal. In this sense the classes ∆(2k) + 1, ∆(2k + 1) + 1, and ∆(3k +1)+ 2are “loose”. Similarly, the classes n =∆(k +2 )− 2and∆(2k +3...

Ngày tải lên: 07/08/2014, 06:20

... occurs only once is  n + k − 2 j  j!(k − j)!S(n + k − j − 2,k− j); so the required number of trees is k−1  j=0 (−1 ) j  k j  n + k − 2 j  j!(k − j)!S(n + k − j − 2,k− j). Dividing by k!andsummingoverk ... number y n of series-reduced trees with n labelled leaves is y n = n−2  k=1 k−1  j=0 (−1 ) j  n + k − 2 j  S(n + k − j − 2,k− j)....

Ngày tải lên: 07/08/2014, 06:20

... N C (u)andN C (v) (and hence also N C (T − u)andN C (T − v)) are nonempty by 2-connectivity. Assume that N C (T − v)containsnon- consecutive vertices a and b along C. We contract T − v to a vertex w, and let ... a, b; (b): c = a. Property 11 If u and v arethetwoendsofthepathT and u  = v, then N C (u) and N C (v) are nonempty, and each of N C (T − u) and N C (T − v...

Ngày tải lên: 07/08/2014, 06:20