Báo cáo toán học: " LONGEST INCREASING SUBSEQUENCES OF RANDOM COLORED PERMUTATIONS" pps

... order on the set of Young diagrams as follows: λ  µ if and only if there exists a sequence ν 1 ,ν 2 , ,ν k of Young diagrams such that LONGEST INCREASING SUBSEQUENCES OF RANDOM COLORED PERMUTATIONS Alexei ... the (centered and scaled) length of the longest increasing subsequence of random colored permutations. The limit distribution function is a power of that f...

Ngày tải lên: 07/08/2014, 06:20

... the length of the longest increasing subsequence in the class S n (321) of (321)-avoiding permutations of n letters. Let f(n, k) denote the number of such permutations whose longest increasing ... distribution. Theorem 2. In the class of (132)-avoiding permutations of n letters, and in the class of (213)-avoiding permutations of n letters, the length of the longest...

Ngày tải lên: 07/08/2014, 07:21

... Borodin, Longest increasing subsequences of random colored permutations, Electron. J. Combin. 6 (1999) #R13. [4] J. Baik, P. Deift and K. Johansson, On the distribution of the length of the longest increasing ... pattern avoidance and longest increasing subsequences by deriving the limiting distribution of the longest increasing subsequences in τ-avoiding per...

Ngày tải lên: 07/08/2014, 15:22

... denote the number of times that π ∈ S m is contained in σ ∈ S n by ν(π, σ). If we divide this number by the total number of subsequences of σ of length m (for m ≤ n) we get the density of π in σ: d(π, ... electronic journal of combinatorics 9(2) (2002), #R1 9 Remark 2.2. This result follows from [4, Theorem 3.1], but since that paper is not easily available, a simple proof is gi...

Ngày tải lên: 07/08/2014, 07:21

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Ngày tải lên: 05/08/2014, 15:20

... numbers: 05A05, 05A15 A board B is a subset of [d] × [d](where[d]isdefinedtobe{1, 2, ,d})andtherook numbers r B k of a board are the number of subsets of B of size k such that no two elements have ... more elegant formula, as a corollary of more general reciprocity theorems. Here, fol- lowing a suggestion of Goldman, we provide a direct combinatorial proof of this new formula....

Ngày tải lên: 07/08/2014, 06:20

... (Note that the number of the first equals the number of all diagram squares, and the number of the latter counts all diagram squares of rank 1). We start by considering increasing subsequences. In ... the number of subsequences of type 12 ···k in π ∈S n (1243, 2143) is defined for k ≥ 2by  s∈{E,D}  h(s) k − 1  where h(s) denotes the height of the starting point of step s...

Ngày tải lên: 07/08/2014, 07:21

... purposes of proving that the game coloring number of Cartesian products is not bounded by a function of game coloring numbers of factor graphs. However the class of Cartesian products of stars ... the proof of this result more transparent, we introduce the so-called game of combinations which could also be of independent interest. At the end of this section we present some...

Ngày tải lên: 07/08/2014, 21:20

... Proofs of Auxiliary Lemmas 7.1 Proof of Lemma 2.2 The proof of Lemma 2.2 relies on the following general tail bound, which is a consequence of Azuma’s inequality (cf. [17, p. 38] for a proof). Lemma ... 3. The proof of Proposition 1.1 shows that the basic reason why the spectral gap of a sparse random graph G(n, d) is small actually is the existence of vertices of degree ≪ ¯...

Ngày tải lên: 08/08/2014, 01:20

... representatives of g–orbits of edges. In other words, every pair of cycles of g of length k 1 , k 2 , with (k 1 , k 2 ) ∈ J, gives gcd(k 1 , k 2 ) distinct representatives. Since the number of cycles of ... of k–cycles in the representation of g as a product of disjoint cycles. Given a permutation g ∈ G, when we will speak of a “k–cycle of g” we will always mean a k–...

Ngày tải lên: 08/08/2014, 12:22