Báo cáo toán học: "Longest Induced Cycles in Circulant Graphs" pps

... maximum lengh of a cycle in unit circulant graphs on a Dedekind domain quotiented by an ideal with r distinct maximal factors. Dedekind domains are exactly those integral domains in which every ideal ... have a new induced k-cycle in X n , since we have not changed the coincidences between any of the rows in C. We now use the Observation 2.5 to define isomorphisms between induced...

Ngày tải lên: 07/08/2014, 13:21

... endpoint in S, the other a distinct endpoint not in S and of a colour different to that of any edge incident with a vertex of SMALL. This set of edges contains at least |M|/k colours. If M contains ... combinatorics 2 (1995), #R19 6 5 Binomial tails We make use of the following estimates of tails of the Binomial distribution several times in subse- quent proofs. Let X be a random variabl...

Ngày tải lên: 07/08/2014, 06:20

... vertex v of in nite degree we can only demand all or nothing. Setting u(v) to a finite value in Theorem 2 is the same as requiring in nitely many outgoing edges at v, whereas putting u(v) = ∞ ... the in nite star. We immediately encounter the problem that the set L of all leaves is incident with in nitely many edges but has in nite demand for ingoing edges, i.e. l(L) = ∞. This results...

Ngày tải lên: 07/08/2014, 21:20

... Erd˝os-Szekeres theorem Finally we ask for the limiting distribution function for the length of the longest increasing subsequence in the class S n (321) of (321)-avoiding permutations of n letters. ... enough n, the question of the limiting distribution function of the LIS remains. Theorem 4. The number a(n, k) of (321)-avoiding permutations having longest increasing subsequence of length...

Ngày tải lên: 07/08/2014, 07:21

... endpoint to the corresponding endpoint in level i − 1. Similarly, if an endpoint in level i has no surrounding circle then there is an edge going from such an endpoint to a corresponding endpoint ... we join the endpoints with red arrows pointing away from each other to create the new wrapped edges in G + P . We thus obtain a bent Hamilton cycle in Q  (5, 3, 3), and by repeatedly joi...

Ngày tải lên: 07/08/2014, 07:21

... avoidance and longest increasing subsequences by deriving the limiting distribution of the longest increasing subsequences in τ-avoiding permutations, for each of the possible τ ∈ S 3 . Inspired by the ... S n (132). Essentially, the inverse of f fixes the position of each left-right minimum, and fills the remaining positions in decreasing order. Recall that π i is called a left-right minim...

Ngày tải lên: 07/08/2014, 15:22

... endpoints x, y. As C is closed, every point in P \ C is contained in a C-arc in P. Suppose no C-arc in P is a shortcut of C. We can find a family (W i ) i∈N of countably many internally disjoint ... disjoint arcs in P , such that fo r every i, W i is either a C-arc or an arc cont ained in C, and every edge in P lies in some W i (there may, however, exist ends in P that are no...

Ngày tải lên: 08/08/2014, 01:20

... (1) Theorem 4 is proved in Janson [4] (Theorem 11) and in Molloy, Robalewska-Szalat, Robin- son and Wormald [8]. 2 Proof of Theorem 1 We begin by letting Y be a set of size 2 rm consisting of r = O(1) ... Y j , p 1 ) contains a perfect matching M j . (We need the split into r copies of H n,p 2 ;3 to allow a “colour” to appear several times in a matching). Now each perfect matching M j g...

Ngày tải lên: 08/08/2014, 11:20

... partitioned into 2m cells X x , x ∈ X of d points. Analogously Y is represented as dκm points partitioned into 2κm cells Y y , y ∈ Y of d/2 points. To construct Λ d we take a random pairing of X into ... ways of defining Hamilton cycles in hypergraphs, depending on the size of the intersection of successive edges. As far as we know, when these intersections have more than one vertex, nothi...

Ngày tải lên: 08/08/2014, 12:23

Ngày tải lên: 05/08/2014, 15:20