Báo cáo toán học: "A Small Trivalent Graph of Girth 14" pot

... Constructions for cubic graphs of large girth. Electron. J. Combin., 5 (1998) A1. [2] M. J. Hoare, On the girth of trivalent Cayley graphs. Graphs and Other Combinatorial Topics (Proceedings of the Third ... semiregularly. According to [1], this graph is the smallest known trivalent graph of girth 14, improving the previous bound of 406 [2]. An electronic copy of the...

Ngày tải lên: 07/08/2014, 06:23

... lists. Proof of Theorem 2. For each x ∈ V ,letl (x) be an assigned list of colors. The general idea of the proof is to list color all copies of K ∆+1 \ e in G which share a vertex of degree ∆ ... disconnects the graph, i.e., a cutvertex of a graph. An end-block is a block with exactly one separating vertex. A Gallai tree is a graph all of whose blocks are either complete gra...

Ngày tải lên: 07/08/2014, 07:21

... = dr dτ (≥ 0) = marginal response of cold-drink consumption to a change in τ(temperature) C 1 = opportunity cost of money tied up in inventory. C 2 = rate of change of inventory cost with respect ... − τ ∗ τ  2 p(τ) Vietnam Journal of Mathematics 33:4 (2005) 437–442 A Stochastic EOQ Policy of Cold-Drink-For a Retailer Shib Sankar Sana 1 and Kripasindhu Chaudhuri 2 1 Departme nt...

Ngày tải lên: 06/08/2014, 05:20

... prove the second statement of Theorem 1. ProofofStatement(ii) of Theorem 1. By the same argument in the proof of Stament (i) of Theorem 1, we can assume without loss of generality that A is complete. Assume ... ideal of M  and let (x 1 , ,x d ) be a system of parameters of M  such that q  =(x 1 , ,x d )A. Then the lemma is proved if we can show the existence of a system...

Ngày tải lên: 06/08/2014, 05:20

... concrete example, let S bethesetofallwordsoflength7overΣ={a, b}. Each word of S is 2-avoidable, but any binary word of length 7 or more encounters an element of S. An image of xxx or xyxyx under a non-erasing ... palindrome, we need only show that no prefix of m n is an overlap. First note that no prefix of m n is a square of length 12 or greater; otherwise the prefix of m n of...

Ngày tải lên: 07/08/2014, 06:20

... [4] to each of the 96 × 72 × 240 = 1658880 cases above, we find the smallest values of a and b to be: (a, b) = (407389224418, 76343678551) with the triples below: the electronic journal of combinatorics ... coverings of the integers given the m k ) and minimize the a’s and b’s con- structed by Knuth’s method. Thus the minimum values of a and b are determined for the pairs (p k ,m k )...

Ngày tải lên: 07/08/2014, 06:20

... A n (π B ,π G ,) be the set of ASMs of order n in which the blue subgraph induces pairing π B , the green subgraph induces pairing π G , and the sum of the number of cycles in the two subgraphs is .Ifπ  B is ... of the colorings of the edges of S and reverses the colors in the other 14. Figure 6 shows the effect of H S on S for a complete (up to rotations) set of colorings...

Ngày tải lên: 07/08/2014, 06:20

... graphs, and some distance-regular graphs with diameter d =3. The main result in this paper is a characterization of these graphs (among regular graphs with d distinct eigenvalues), in terms of ... by-product of such a proof, we can also give bounds for the sum of “even multiplicities” σ e in any regular graph, in terms of its eigenvalues and harmonic mean H. Corollary 2.3 Let Γ be...

Ngày tải lên: 07/08/2014, 06:20

... will meet each of the original sets. Proof of Lemma 1. If we can find a set of f (d, g) points that meets the convex hull of every subset of g|Q| of the elements of Q,itwillmeeteveryoneofourpolytopes. In ... hull of every set of 9g 2 |Q| 2 /100 of these intersection points, by an appropriate induction hypothesis. 4. Every set of g|Q| points of Q that has at least g|Q|/1...

Ngày tải lên: 07/08/2014, 06:22

... by Lunnon. No proof of Theorem 2.2 will be given because the proof of Theorem 2.2 is ex- tremely similar to the proof of Theorem 2.1 given below. 3 Proof of Theorem 2.1 For ease of notation let ... electronic journal of combinatorics 5(1998), #R3 12 4 Calculating the new upper bound Our first step is to construct the sequence of greatest elements of the given sequence of sets...

Ngày tải lên: 07/08/2014, 06:22