Báo cáo toán học: "The minimum size of complete caps in (Z/nZ)2" ppt

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Ngày tải lên: 05/08/2014, 09:46

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Ngày tải lên: 05/08/2014, 10:20

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Ngày tải lên: 05/08/2014, 10:20

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Ngày tải lên: 05/08/2014, 10:20

... a strongly rpp semigroup. Again by Lemma 3.7 and the definition of a strongly right (left) adequate semigroup, we can formulate our main theorem. Theorem 3.8. (i) The translational hull of a strongly ... the translational hull of a strongly right (left) adequate semigroup is still the same type. Thus, the result obtained by Fountain and Lawson in...

Ngày tải lên: 06/08/2014, 05:20

... Hopf algebras in the graded case. In order to study tame basic Hopf algebras or generally tame Hopf algebras, we need more examples of tame Hopf algebras. The Andruskiewitsch-Schneider algebra is ... modules for the quantum double of a special Taft algebra in [3]. From this, we can deduce immediately that the quantum double of this special Taft algebra is ta...

Ngày tải lên: 06/08/2014, 05:20

... for the independence number of general graphs of a given odd girth. We shall prove some similar bounds to those we prove for regular graphs in terms of a measure of the concentration of edges. Monien ... prove our results for graphs of odd girth at least 7. Dealing with graphs with larger odd girth will simply require a generalisation of this argument. Theo...

Ngày tải lên: 07/08/2014, 06:20

... Denley’s Theorem 4. References 1. Denley, T., The Independence number of graphs with large odd girth, The Electronic Journal of Combinatorics 1 (1994) #R9. 2 We can now prove the following theorem. Theorem ... is sharp for the regular complete bipartite graphs K aa . The above results are readily extended to graphs of larger odd girth. Lemma 2: Let G have odd girth...

Ngày tải lên: 07/08/2014, 06:20

... the number of partitions of (the number) n,andQis its ordinary generating function. Also, define p(n, k) to be the number of partitions of n with largest part at most k. We will have use of the ... values of n, we shall use elementary estimates and some well-known bounds for ordinary partitions (of a number) to express most values of b n in terms of the number...

Ngày tải lên: 07/08/2014, 06:20

... monotone subsequences of length 3 in L, R are also such subsequences of S.There are no monotone subsequences of S containing both 0 and n − 1. There are no monotone subsequences of S of the form a, ... up of k monotone descending subsequences, each of length n/k or n/k; clearly it has no monotone ascending subsequences of length k +1, and any monotone descendin...

Ngày tải lên: 07/08/2014, 07:21

... complete cap in PG(2,q) containing at most √ q(ln q) 10 points. A similar bound holds for the minimum size of a complete cap in AG(2,q) since we can obtain AG(2,q)fromPG(2,q) by removing a line ... any line in (Z/p 2 Z) 2 which contains p ∗x + p ∗(1,t) the electronic journal of combinatorics 13 (2006), #R58 12 The minimum size of complete caps in (Z/nZ) 2 Jack...

Ngày tải lên: 07/08/2014, 13:21

... to the root, and then assigning the polarity of additional vertices in the order given by a breadth-first search of the minimum spanning tree.) If this function is the one that comes up, then the ... variable Z n be the cost of the minimum spanning tree when the edges of the complete graph K n have costs selected independently and uniformly at random in t...

Ngày tải lên: 07/08/2014, 13:21

... square all-1 matrix. Proof of Theorem 1.1: The square of the number of perfect matchings of G counts ordered pairs of such matchings. We claim that this is the number of spanning 2-regular subgraphs ... Cuckler and J. Kahn, Entropy bounds for perfect matchings and Hamiltonian cycles, to appear. [3] S. Friedland, An upper bound for the number of perfect matching...

Ngày tải lên: 07/08/2014, 15:22

... several properties of a maximal IC-coloring f of K m,n . Proposition 3.3. If f is a maximal IC-coloring of G = (A, B) = K m,n , then all the colorings of vertices of G are distinct. Proof. Suppose, ... proof is complete. the electronic journal of combinatorics 15 (2008), #R43 2 n ≥ 3, we have f 4 = 10 ≤ 3 · 2 4−2 − 2 4−(n+2) − 1. This is a contradiction to that f is a maximal I...

Ngày tải lên: 07/08/2014, 15:23

... f points in the polar space P is (q d+ 1+ + 1) (q d+1 − 1)/ (q − 1). The size of a spread in P is equal to q d+ 1+ + 1, which is of course, also an upper bound on the size of a partial spread. The ... showing that this construction does in fact yield a maximal partia l spread of size q 2n+1 + 1 in all spaces H(4n + 1,...

Ngày tải lên: 07/08/2014, 21:21