Báo cáo toán học: "Parking functions, empirical processes, and the width of rooted labeled trees" pps

... motivations of Odlyzko and Wilf, when they study the width of labeled trees, is to give a tight estimate for the average bandwidth of this class of tree. 2 The breadth first search random walk From ... initially (0). Then, at each of the n following stages of the BFS, the vertex x at the head of the queue is removed from the queue, and all “new” neighbor...

Ngày tải lên: 07/08/2014, 06:22

... Λ(π i−1 ,π i )andminB  . It is easy to see that Γ m is a tree. Root Γ m at the vertex 1 and erase the vertex labels. If v i is the vertex of e i farthest from the root, then move the label i of the edge ... ,Λ(π n−1 ,π n )). 3.1 Theorem. The labels Λ(m) of the maximal chains of NC n+1 consist of the parking functions of length n,eachoccuringonce. Proof. If Λ(π...

Ngày tải lên: 07/08/2014, 06:22

... have the same number of elements. First we remark that there is a natural action of S n on the set of parking functions, which permutes the a j . There is also an action of S n on the set of factorizations, ... w]. If W = S n is the group of permutations of [1,n], then the reflections are the transposi- tions, and w is the cycle (1 2 n). To any permutation σ ∈...

Ngày tải lên: 07/08/2014, 06:23

... arrive at the first element of p, p k . At this point, the top of the stack is the index of the last element of q encountered. We move the top of the stack out, as the first entry of the right ... arrive at the element p s of p, the top of the stack is the index of the smallest element of q which is not smaller than p s , other than the elements...

Ngày tải lên: 07/08/2014, 07:21

... pictures of weight k.Acard of weight k is a pair consisting of a picture of weight k and a k-element subset of N that we will call the label set of the card. A hand of weight n and size l is a set of ... bijection between the set of compositions of n −2by{1, 2} and the set of compositions of n − 1 by odd numbers. The bijection works by prepending a 1 to...

Ngày tải lên: 07/08/2014, 06:20

... self-avoiding then we append q0andq1 to the end of the queue. the electronic journal of combinatorics 8 (2001), #N2 2 Weakly Self-Avoiding Words and a Construction of Friedman Jeffrey Shallit ∗ and Ming-wei ... subword of “introduction”. 1 We use the notation x[k]todenotethek’th letter chosen from the string x. We write x[a b] to denote the subword of x of length b...

Ngày tải lên: 07/08/2014, 06:22

... Given a matching m of the Aztec diamond graph A, define the weight of the matching (m) to be the product of the weights of all the edges in the matching. Then the weight of the Aztec graph A is (A)=  m (m), ... graph of order n is the dual graph to the Aztec diamond of order n in which the vertices are the squares and an edge joins two vertices if...

Ngày tải lên: 07/08/2014, 08:20

... faces of ∆ for 0 ≤ i ≤ d − 1andf −1 =1. We say that ∆ is a triangulation of a polytopal complex F if the union of the faces of ∆ is equal to the union of the faces of F and every face of ∆ is ... from the description of the facets of P that if I 0 ⊂ I 1 ⊂···⊂I e is a chain of nonempty ideals of Ω 0 as in the definition of the ideal height for Ω then t...

Ngày tải lên: 07/08/2014, 08:22

... |E|. Theorem 1 can also be extended to hypergraphs, but the appropriate condition on |f(v)| then depends on the upper rank of the hypergraph – that is, the size of the largest hyperedge. The proofs ... 1}, then F is the edge set of a non-trivial partial f-factor of G. 3 Concluding Remarks We mention a possible direction for extending the results of Theorem 1 and Th...

Ngày tải lên: 07/08/2014, 15:22

... some c ∈ R and realize the above as a quadratic polynomial in c. Since the coefficient of c 2 is negative, the function is concave down. By finding the zeros of this polynomial in terms of n and then ... B m+ for m ≤ . Then by making superficial changes to the remainder of the proof we establish the result for the nonsymmetric case. Thus we have the following: the...

Ngày tải lên: 07/08/2014, 15:23