Báo cáo toán học: "On unimodality problems in Pascal’s triangle" potx

... decreasing (resp. increasing), and so the log-concavity implies the unimodality. The unimodality problems, including concavity (convexity) and log- concavity (log-convexity), arise naturally in many ... combinatorial interpretation. Lattice path techniques have been shown to be useful in solving the unimodality problem. As an example, we present a combinatorial proof of Theorem 1 (i...

Ngày tải lên: 07/08/2014, 21:20

... edges joining u to its ≤ A -smallest non-neighbors in A. This is possible, since by our assumptions, deg M (v) + deg M (u) + ε √ n ≤ ∆(G M ) + 2ε √ n ≤ q. Finally, Breaker uses his remaining moves ... following ho l ds. 1. If q ≤ c 0 n 1/m(G) , then Maker has a winning strategy i n the game G(G; n, q). 2. If q ≥ C 0 n 1/m(G) , then Breaker ha s a wi nning strategy in the game G(G; n, q)....

Ngày tải lên: 08/08/2014, 14:22

Ngày tải lên: 05/08/2014, 15:20

... arithmetical analysis which appeared in Chinese [6] in 1988 and in English [7] in 1989. In the following result, we prove that any set of p non-negative integers s 1 ,s 2 , ,s p with s 1 <s 2 < ... the inequalities in the above mentioned formula for those values of k,forwhichs k <s k+1 hold, need to be checked [5, p.608]. The following results can be seen in [5]. Theore...

Ngày tải lên: 06/08/2014, 05:20

... A in I whichisjoinedtofewerthanq/2 trees of odd order. As in the preceding paragraph we obtain an independent set of G of size greater than (n − 1)/2, contradicting the assumption that G is in ... β]isthesubmatrixofA determined by the rows indexed by α and columns indexed by β. Problem 1 is equivalent to the following. Problem 3 Let t and n be positive integers with n ≥ t ≥ 2. Determine...

Ngày tải lên: 07/08/2014, 06:20

... elementary construction using binary codes gives new record kissing numbers in dimensions from 32 to 128. 1. Introduction Let τ n denote the maximal kissing number in dimension n, that is, the ... 30,828. In this way we obtain a kissing number of 331,737,984. n=80. By taking 4 orbits under L 2 (79) we obtain A(80, 16, 16) ≥ 143,780. We take n 0 = 64, n 1 = 16, n 2 =4,n 3 = 1 and obtain...

Ngày tải lên: 07/08/2014, 06:22

... us in (1) k(|D|−|S|)+d|S|−q +2≤ e G (D, S) ≤ d|D|−q, leading to (d − k)(|D|−|S|) ≥ 2. ✷ Proof of Theorem 1.1. The first case follows from the well-known Theorem of Petersen [6]. In the remaining ... disjoint subsets D, S of V (G), whether G has a k-factor or not. In 1985, Enomoto, Jackson, Katerinis and Saito [3] proved the following result. Lemma 2.2 Let G be a graph and k a positive in...

Ngày tải lên: 07/08/2014, 08:20

... so that the resulting digraph D  is acyclic. If true, this conjecture would imply the following strengthening of Lemma 5: Each digraph D obtained from a tournament by deleting k edges, that ... outdegree less than √ k. This in turn would imply some improvements in the bounds of Theorems 2 and 3. 3 A sketch of Shen’s proof In this section, we outline the proof in [5]. Assume that ther...

Ngày tải lên: 07/08/2014, 13:21

... two non-codewords in F(i, j) are in the lower-left and upper-right corners (in the other case, we just go up instead of going down in what follows: as explained above, the voting rule is symmetric in this ... Lobstein, The minimum density of an identifying code in the king lattice, Discrete Math. 276 (2004), pp. 95–109. [5] I. Charon, O. Hudry and A. Lobstein, Extremal cardinalitie...

Ngày tải lên: 07/08/2014, 15:22

... upper bound given in [2] by 1, the proof is much more involved. Another problem we address in this paper, is that of determining the smallest n for which Maker has a winning strategy in H M n . Though ... organized as follows. In Section 3 we prove Theorem 1.1 and in Section 4 we prove Theorem 1.2. Finally, in Section 5 we present some concluding remarks and related open problems...

Ngày tải lên: 07/08/2014, 21:21