Báo cáo toán học: "Zero Capacity Region of Multidimensional Run Length Constraints" ppt

... Capacity Region of Multidimensional Run Length Constraints Hisashi Ito Department of Information Science Toho University, Chiba 274-8510, Japan his@kuro.is.sci.toho-u.ac.jp Akiko Kato Dept. of ... e n hasrowsoflengthd + 2, each containing a single one. For any collection of bi-subcubes, on each of which e n is a permutation bi-subcube, any row of S n+1 can intersect at m...

Ngày tải lên: 07/08/2014, 06:20

... 2011:60 http://www.advancesindifferenceequations.com/content/2011/1/60 Page 8 of 14 comes down to zero. The greatest values of the considered functions are infinite, when k 0 = ∞. 2.7. The first derivatives of the ch1 function of HC in relation to the degree of freedom ... 2)ψ 2 (υ +1)+ψ 4 (k + n − 2)  . In general, the HC of the mth derivative of n can be defined on the basis...

Ngày tải lên: 20/06/2014, 21:20

Ngày tải lên: 05/08/2014, 10:20

... adjacency matrix of Γ D .IfA D belongs to the algebra generated by A, then A D = J −R(A), where R := r 2 P − r 2 +1 and P is the alternating polynomial of Γ. P roof. The first part of the proof goes as ... vectors by the vertices of Γ. Moreover, for any vertex u ∈ V , we denote by e u the u-th unitary vector of the canonical basis of R n . Thus, the characteristic vector of a ver...

Ngày tải lên: 07/08/2014, 06:22

... A proof of Julian West’s conjecture that the number of two-stack-sortable permutations of length n is 2(3n)!/((n + 1)!(2n + 1)!), Discrete Math. 102 (1992) 85–93. the electronic journal of combinatorics ... introduced in the proof of Proposition 2.1. We obtain the number of two-stack-sortable permutations of length n having m right-to- left maxima, as given in [8]. the elect...

Ngày tải lên: 07/08/2014, 06:22

... polarizes to a non-singular B,thenQ must be of one of two types χ(Q)=±1, where Q has exactly 2 2t−1 + χ(Q)2 t−1 zeroes. Next, let {B 1 ,B 2 , , B r } be a set of alternating bilinear forms on V ;ifB i + ... eigenmatrix Q on page 326 of [2], by setting n =2 2t , r =2 2t−1 , a =2 t+1 , θ =2 t −1andτ = −2 t −1inthat paper; the adjacency eigenvalues of G t are then given by the fourth c...

Ngày tải lên: 07/08/2014, 06:22

... polynomial time, and the proof follows. Proof of Corollary 6: Compute D = D (P ) by taking the least common multiple of the denominators of all of the coordinates of the vertices of P . Assume that we ... [GC50]), where N is defined in terms of the lengths of λ and µ. Then #(P ∩ Z N ) is the dimension of the weight µ subspace of the irreducible repre- sentation of GL n C...

Ngày tải lên: 07/08/2014, 13:21

... either a union of copies of C 4 , or a union of copies of C 4 and a copy of C i for i = 5, 6, 7, respectively. Equalities in (2.16-2.18) hold if and only if G is either a union of copies of C 3 , or ... the numerator of the right-hand side of (4.10). Divide this number of choices by the number of permutations of {1, . . . , rn} to deduce the lemma. ✷ Using the methods...

Ngày tải lên: 07/08/2014, 21:20

... hypergraphs. Therefore it is ∗ This research forms part of the Ph.D thesis written by the author under the supervision of Prof. Benny Sudakov. † Department of Mathematics, UCLA, Los Angeles, CA, 90095. ... {x 1 , . . . , x n } and clauses C as collection of disjunction of literals V ∪ V . A formula is called satisfiable if there exists an assignment of values to variables so that t...

Ngày tải lên: 07/08/2014, 21:21

... n. In 1999, Bollob´as and Leader investigated the problem of determining the minimal cardinality of |  n (S)| in terms of the length of |S| assuming that 0 ∈  n (S). For every positive integer ... consists of all elements in G which can be expressed as the sum over a subsequence of S of length n, i.e.,  n (S) = {a i 1 + . . . + a i n : 1 ≤ i 1 < . . . < i n ≤ k}. I...

Ngày tải lên: 07/08/2014, 15:22