Báo cáo toán học: "Positively Curved Combinatorial 3-Manifolds" potx

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Ngày tải lên: 05/08/2014, 15:20

... independent computations of the number of Latin squares of order 10. We also give counts of Latin rectangles with up to 10 columns, and estimates of the number of Latin squares of orders up to 15. Mathematics ... subgraph of degree one. A one-factorization of G is a partition of E(G) into one-factors. Clearly, the rows of a Latin rectangle R correspond to the one-fac...

Ngày tải lên: 07/08/2014, 06:20

... S of subsets of a finite set S whose union is S is connected if and only if there is an evolutionary path in S between any two members of S. Proof:IfS is a finite connected family of subsets of ... the members of i S [S] account for at least five of the seven nonempty subsets of S. Without losing generality we may assume that the complement in P(S) − {∅} of i S [S] is a subset...

Ngày tải lên: 07/08/2014, 06:20

... August 3, 2000 Abstract Let d(n) count the lattice paths from (0, 0) to (n, n) using the steps (0,1), (1,0), and (1,1). Let e(n) count the lattice paths from (0, 0) to (n, n)with permitted steps ... of lattice paths defined on various sets of permitted steps which yield path counts related to the Narayana polynomials. Mathematical Reviews Subject Classification: 05A15 Key words: La...

Ngày tải lên: 07/08/2014, 06:20

... ,q |Q| }. 4.Ateverystageofthisconstructionthenumberofpolytopesgrowsbythenumberof clonings;ifatthatstageonehadRpolytopesthepointchosenwillbeinatleasthR ofthemandtherewillbeatmost(1−h)Rclonings.Thenumberofpolytopesinthat stagewillgrowtoatmost(2−h)R. 5.ThusthefinalnumberofpolytopeswillbeatmostN(2−h) |Q| . 6.Wewillusethefactthatthenumberofclonedescendantsofanyoneofouroriginal polytopescannotexceed...

Ngày tải lên: 07/08/2014, 06:22

... played the game with his father.) Two pieces of cut potato can serve as the medium on which to carve the motif and its reflected image, and then these potato stamps are inked and used to produce a ... translations of the translation block. These symmetries add a geometric layer of complexity to the combinatorial scheme. Escher's Case (1) We first consider Escher's case (1), in whi...

Ngày tải lên: 07/08/2014, 06:22

... graph as a planar embedding of a maximal hypergraph such that the faces of the graph correspond to the edges of the hypergraph. 3 Colouring planar mixed hypergraphs Let H =(V,C, D) be a mixed hypergraph. ... colouring properties of general planar mixed hypergraphs was first raised in [13] (problem 8, p.43). It is evident that every planar D-hypergraph is colourable, just as every...

Ngày tải lên: 07/08/2014, 06:22

... chromatically unique if a k <a 1 + a 2 , and find examples showing that θ(a 1 ,a 2 , ···,a k )maynotbe chromatically unique if a k = a 1 + a 2 . Keywords: Chromatic polynomials, χ -unique, χ-closed, ... θ(a 1 ,a 2 , ···,a k ) is χ -unique if a i ≥ k − 1 for i = 1, 2, ···,k. ✷ Theorem 2.3 ([11]) Let h ≥ s +1≥ 2 or s = h +1. Then for k ≥ 5, θ(a 1 ,a 2 , ···,a k ) is χ -unique if a 2 − 1...

Ngày tải lên: 07/08/2014, 08:20

... Orthogonal Polynomials and Applications (VII SPOA) (Granada, 1991), J. Comput. Appl. Math. 49:1–3 (1993), 243–249. [6] A. L. Schmidt, “Legendre transforms and Ap´ery’s sequences”, J. Austral. Math. ... On a combinatorial problem of Asmus Schmidt W. Zudilin ∗ Department of Mechanics and Mathematics Moscow Lomonosov State University Vorobiovy Gory, GSP-2, Moscow 119992 RUSSIA URL:...

Ngày tải lên: 07/08/2014, 08:20

... vertex. 2. The initial vertex of the next edge is v(j, ∓k). 3. Determine the direction of the next edge by the sign of the entry ∓k in list j and the direction of the preceding edge according to the ... combining the information about the direction of the first edge with the corresponding crossing yields a rule for the direction of the second edge. If...

Ngày tải lên: 07/08/2014, 08:20

... d, and p i > p j . In particular, 1 -descents correspond to descents in the traditional sense, and (n − 1) -descents correspond to inversions. This concept was introduced in [2] by De Mari and ... Y n,k and the vertices of G, and 2. If V 1 and V 2 are two disjoint sets of vertices of G so that no edge of G has one end- point in V 1 and another one in V 2 , then the corres...

Ngày tải lên: 07/08/2014, 15:22

... general problem that we solve, the “coupon collector’s problem with quotas,” is for a payoff word with letters in the set L which appear with probabilities p = p  :  ∈ L and quotas q = q  : ... journal of combinatorics 15 (2008), #N31 5 with a uniform distribution of the coupons. Wilf and Myers in [3] re-derived the result of Newman and Shepp, but with a generating function...

Ngày tải lên: 07/08/2014, 15:22

... image partition regular if given any pair (S, A), M is image partition regular over S with respect to A. In the following discussions we shall observe that for matrices of finite order image partition ... In this paper we introduce a new notion which we call universally image partition regular matrices, which are in fact image partition regular over all semigroups and every...

Ngày tải lên: 07/08/2014, 21:21

... 2 Proofs The multiplicative magic squares can be described in terms of the additive ones in the following way: We write n A for the multiplicative magic square given by n A = n A(i,j) 1i,jr . If ... additive -multiplicative magic squares of order r = 5 with distinct entries? We have seen that the answer is negative for r = 4. On the contrary, Horner [4] found an additive and mul...

Ngày tải lên: 08/08/2014, 01:20

... construct a positively curved M 3 with vertices v and w at edge-distance five and Lk(v) = L. proof. Suppose L is a positively curved combinatorial 2-sphere. We wish to find a positively curved 3-manifold ... 15 Theorem 1.2 (Part 2) Let M be a positively curved combinatorial 3 -manifold with vertices v, w at edge-distance five. If M ′ is another positively curved 3-manifold with ve...

Ngày tải lên: 08/08/2014, 12:22