Báo cáo toán học: "On the Number of Matchings in Regular Graphs" pptx

... +1) 6 49n 6 . 7. The number of ascendants of a given node in a LBST As in the case of the number of ascendants in a random BST, computing the probability that the j th node in a random LBST has m ascendants ... other random variables: the number of descendants D n and the number of ascendants A n of a randomly chosen internal node...

Ngày tải lên: 07/08/2014, 06:22

... n)k −1 φ(k). In either case, the inequality in part 1 of the lemma holds. On the number of perfect matchings and Hamilton cycles in -regular non-bipartite graphs Alan Frieze ∗ Department of Mathematical ... of combinatorics 7 (2000), #R57 4 3 Hamilton Cycles A Hamilton cycle is the union of two perfect matchings and so h(G) ≤ 1 2 m(G) 2 a...

Ngày tải lên: 07/08/2014, 06:22

... in the case when m is prime. Whether a permutation of n admits an m-th root can be read on the partition of n determined by the lengths of the permutation’s cycles, because the class of such the ... is the exponential generating function (EGF, formal series) of the m-thpowersinthegroups n . This means that the number of m-thpowersin n is p n (m)ìn!foreachn....

Ngày tải lên: 07/08/2014, 06:23

... we are not in one of these cases, then there is exactly one edge e attached to v which is below l.Theedgee is on a loop of the configuration. Let us start with e and follow the part of the loop ... which is based on a variation of the above reasoning, see Section 5. Finally, for the proof of the more specific assertions in Conjectures 1.1 and 1.2 on the i...

Ngày tải lên: 07/08/2014, 08:22

... x and y to be row and column sums of a matrix is that they have the same weight. Clearly, the row sums of a member of M are at most n.Conversely,ifx =(x 1 , ,x m ) and 0 x i n,letR = R(x)bethem ... (5) the electronic journal of combinatorics 13 (2006), #N8 5 3 Start of the proof Let N = {0, 1, }. Define the weight of a matrix N to be the sum of its e...

Ngày tải lên: 07/08/2014, 13:21

... listed in π s,m n−1 .Sincen is the largest number, this forces the insertion point to be right at the beginning of the list. Since π  is precisely π s,m n−1 with m = n−1−s 2 , this yields the permutation ... from the proof of Theorem 1.3, as if the structure of a permutation is ‘close’ to that of π s,m n then the configuration of the first type in Figure 2...

Ngày tải lên: 07/08/2014, 13:21

... (2.6) and (2.8), the number of copies of K k in E is (1 + o(1)) |E| k k! q − ( k 2 ) . . This implies that the number of the number of k-tuples of k mutually orthogonal vectors in E is also (1 ... o(1)) |E| k k! q − ( k 2 ) , completing the proof of Theorem 1.2. Acknowledgments The research is performed during the authors visit at the Erwin Schrăodinger I...

Ngày tải lên: 07/08/2014, 15:22

... such that the tree together with the circle it is drawn on can be embedded in a surface of genus one, but not of genus zero. Hough [3] observed that the number of genus one labeled circle trees ... offspring of C by definition. Proposition 4. The genus of the reduced form T 3 of a u-c-tree T is one if and only if the genus of T is one. Proof. L...

Ngày tải lên: 07/08/2014, 15:23

... N i denote the number of Eulerian orientations of G 0 which have i valid cross over boxes, that is N  is the number of Eulerian orientations of G. The number of Eulerian orientations of G k is EO(G k ) ... bound for the number of almost alternating Eulerian orientations yields a lower bound for the number of 2 -orientations of G  k, . For the...

Ngày tải lên: 07/08/2014, 21:20

... get the numerator of the right-hand side of (4.10). Divide this number of choices by the number of permutations of {1, . . . , rn} to deduce the lemma. ✷ Using the methods in the proof of Theorem ... := r i=1 A i . Let m ∈ min(p, q). Then perm m A is the number of m -matchings of G := ∨ r i=1 G i , which is equal to the number of m -matchings o...

Ngày tải lên: 07/08/2014, 21:20

... y 2l = true) will make the remaining clauses true. Next we prove the linear upper bound of number of clauses in minimal unsatisfiable 2-SAT formulas. Proposition 1. Minimal unsatisfiable 2-SAT formulas ... deletion of any edge makes it 2 colorable. In this analogy, minimal unsatisfiable formulas correspond to critically 3-chromatic hypergraphs. Therefore it is ∗ This resear...

Ngày tải lên: 07/08/2014, 21:21

... . 2 The number of matchings in a tree In t his section, we turn to the number of matchings in a graph. This is a lso known as the Hosoya index, or the Z-index in mathematical chemistry. For a rooted ... An interesting paper of Wagner [5] looks at the number of independent sets modulo m. Wagner showed that the pr oportion of trees on n vertices w...

Ngày tải lên: 08/08/2014, 11:20

... 05C05 60C05 Abstract By a theorem of Dobrow and Smythe, the depth of the kth no de in very simple families of increasing trees (which includes, among others, binary increasing trees, recursive trees ... Dobrow/Smythe, Theorem 6) In a random increasing tree with n nodes from a simple family, the distribution of the distance between nodes i and k (i < k) is the...

Ngày tải lên: 08/08/2014, 12:22

... realization having parame ters ˜r 1 , ˜r 2 , ˜s 1 and ˜s 2 such that ˜r 1 < ˜s 1 or ˜r 1 = ˜s 1 and ˜r 2 ≤ ˜s 2 . Then Q 1 is mutation equivalent to Q 2 if and only if r 1 + 2r 2 = ˜r 1 + 2˜r 2 and s 1 + ... Combinatorial Grammar In this section we describe the elements of the mutation classes of type ˜ A n−1 by a combinatorial grammar. This can be viewed as an exer...

Ngày tải lên: 08/08/2014, 14:23

... subsequences with a given sum in a nite abelian group Gerard Jennhwa Chang, 123 Sheng-Hua Chen, 13 Yongke Qu, 4 Guoqing Wang, 5Đ and Haiyan Zha ng 6ả 1 Department of Mathematics, National Taiwan University, ... Wang, A quantitative aspect of non-unique factorizations: the Narkiewicz constants, International Journal of Number Theory, to appear. [13] W.D. Gao and J.T...

Ngày tải lên: 08/08/2014, 14:23