Báo cáo toán học: "Enumeration of derangements with descents in prescribed positions" ppt

... interpretation. This interpretation is in line with the rest of this art icle, counting permutations having an initial descending segment and λ-coloured fixed points to the right of the initial segment. In addition, ... n, generalising the results with λ = 0 from [9]. In our proofs, we will use the following convent io ns. If λ is a positive integer, and (π,C) is a permutation w...

Ngày tải lên: 07/08/2014, 21:21

... lozenge in exactly one-third of the tilings. Proof. We compare the result of setting m = n in Corollary 12 with the result of setting m = n in 15 and evaluating the sum by Lemma 16. 8. Computing ... Aztec diamond of order 3. We are interested in tilings of hexagons with lozenges, which are rhombi with unit sides and angles of 120 and 60 degrees, and tilings of Azte...

Ngày tải lên: 07/08/2014, 06:20

Ngày tải lên: 05/08/2014, 15:21

... monotonic uniformly continuous function of x and y on the unit square. In fact, it is increasing in x if and only if α begins with d and is increasing in y if and only if α ends with u. (b) f(x, y, ... in (8.1), but it is needed now because we normalize φ to have  1 0 φ = 1 and we incorporate a factor of λ in C.) One can interpret finding the eigenfunction φ in terms of...

Ngày tải lên: 07/08/2014, 08:20

... A perfect match- ing of a graph G is a set o f independent edges of G covering all vertices of G. Problems involving enumeration of perfect mat chings of a graph were first examined by chemists and ... Enumeration of perfect matchings of Q m,n,r 3.1 The sign of pfB(x 1 , x 2 ) In order to decide the sign of pfaffians of B(x 1 , x 2 )(x 1 , x 2 = ±1), we distinguish the perf...

Ngày tải lên: 08/08/2014, 12:22

... q-enumeration of ASM’s, which consists in counting ASM’s (or classes of symmetry of ASM’s) with respect to their number of −1 entries (to the number of orbits of −1 entries in the case of symmetric ... symmetric in all its variables. Stroganov proved this surprising symmetry property by a study of Izergin-Korepin determinant. A proof only involving Yang-Baxter equation...

Ngày tải lên: 08/08/2014, 12:22

... theorem. Furthermore, letting x = 0 in (10), we deduce that the number of permutation triples of S n with “RRR” forbidden satisfies the following binomial relation n  k=0 (−1) k  n k  3 C k = 0 with n > ... 6 Enumeration of Restricted Permutation Triples Xiaojing Chen School of Mathematics and Computational Science China University of Petroleum Dongying 257061, P. R. Chin...

Ngày tải lên: 08/08/2014, 12:22

... choice of the origin p 0 , a pin representation is not associated with a unique pin word, but with at most 8 pin words (see Figure 4). The set of pin words is the set of all encodings of pin-permutations. ... by keeping in the same order points p i that form an occurrence of π in σ. Instead of random patterns of a pin-permutat io n σ, we will often be interested in pat...

Ngày tải lên: 08/08/2014, 14:22

... defect. Interest in such graphs is part of the general study of the ∗ mirka.miller @newcastle.edu.au † This research was supported by a Marie Curie International Incoming Fellowship within the ... graphs of defect 2. In such a graph G any vertex v can reach within D steps either two vertices (called repeats of v) in two different ways each, or one vertex (called double repeat of...

Ngày tải lên: 08/08/2014, 14:22

... element, as in a classical structure, but a subset of H. One of the interests of the researchers in the field of hyperstructur es is to construct new hyperoperations using g raphs [18], binary relations [2, ... examples of join spaces have been constructed in connection with binary relations, graphs and lattices. In order to de- fine a join space, we need the following notat...

Ngày tải lên: 08/08/2014, 14:23