Báo cáo toán học: "Factoring (16, 6, 2) Hadamard difference sets" docx

... the Hadamard difference sets we seek in groups of order 16. All ( 16, 6, 2) difference sets are constructed in this manner; in section 4 we provide the fac- toring for each of the 27 ( 16, 6, 2) difference ... Each of the 27 difference sets discovered by Kibler can be rediscovered in this manner. The results are listed below. 4 All ( 16,6 ,2) Difference sets We list al...

Ngày tải lên: 07/08/2014, 21:20

... n(n +1)/2. So if S is a subset of the differences, we have:  X ijk ∈S X ijk ≥ n(n +1)/2 where |S| = n (2) (since X ijk =  j+k−1 h=k X i1h the inequalities (2) can be expressed in terms of our ... (I,J) difference triangle set, ∆, as a set of integers {a ij | 1 ≤ i ≤ I,0≤j≤J}such that all the differences a ij − a ik ,1≤i≤I,0≤k<j≤J are positive and distinct. Let m = m(∆) be the maximum d...

Ngày tải lên: 07/08/2014, 06:20

... |A| 15 14 . Proof. We start with |A + B| ≤ K|A| and |AB| ≤ K|A|. By using Pl¨unnecke’s inequality (see Ch 6, [14]), we have |A+A| ≤ K 2 |A| and |B +B + B +B| ≤ K 4 |A|. First, by Cauchy-Schwarz inequality, ... completes the proof. We note that from the result in [11] and Pl¨unnecke sumset inequality ( see Ch 6, [14]), we have that if |B| ∼ |A| < p 1/2 then max(|A + B|, |AB|)  |A| 25/2...

Ngày tải lên: 07/08/2014, 15:22

... Ψ(f 1 ) β = 1 2 Ψ(f 0 ) − Ψ(f n ), Ψ(f 1 ) − Ψ(f n ), Ψ(f 2 ) − Ψ(f n ), . . . , Ψ(f n−1 ) − Ψ(f n ) β . By Lemma 4 .6, we obtain that Ψ  f 0 +f 1 2  , Ψ(f 0 ), and Ψ(f 1 ) are 2- collinear. By Lemma 4.5, we get ... Mordeson, JN: Fuzzy linear operators and fuzzy normed linear spaces. Bull. Calcutta Math. Soc. 86, 429–436 (1994) [8] Bag, T, Samanta, SK: Finite dimensional fuzzy normed l...

Ngày tải lên: 20/06/2014, 21:20

Ngày tải lên: 05/08/2014, 15:21

... c 2,1 = b, c 2,3 = r and we are done. Now we may assume C(1, 2) ∩ C(2, 2) = ∅ by (iii). Suppose there exists a ∈ C(1, 2) and r ∈ C(2, 2) such that a is missing from at least one of C(1, 1) or C(1, ... c 2,3 ,c 2,2 ,c 2,1 ,c 1,1 ,c 1,2 .This coloring will complete a proper list coloring since C(1, 2) ∩ C(2, 2) = ∅. ✷ the electronic journal of combinatorics 9 (20 02), #N8 5 [5] F. G...

Ngày tải lên: 07/08/2014, 06:23

... 2-connected, by Condition (3). As G(t)−f t (tt  ) is 2-connected unless G(t) is a polygon, Condition (2) implies at least one of H and K is 2-connected. We must show that one of H and K is an (H ∩ K)-bridge ... an (H ∩ K)-bridge in G. We claim that G(t) is a bond. The same argument applies to K, so Condition (2) implies at least one of H and K is an (H ∩ K)-bridge in G. Let [x, y]=H ∩ K.If...

Ngày tải lên: 07/08/2014, 08:20

... also see that k-COLORABILITY can be transformed into STAR k-COLORABILITY of bipartite graphs (t = 2). To obtain the case for 3 ≤ t ≤ k simply add in a disjoint K t . 8 Questions We conclude with ... Kostochka, D. R. Woodall, Acyclic colourings of planar graphs with large girth. J. London Math. Soc. (2) 60 (1999), 344–352. [7] R. Diestel, Graph Theory, Springer graduate texts (1997). the ele...

Ngày tải lên: 07/08/2014, 08:20

... we need only verify that for each of the 15 pairs (n  , j  ) (0, 1), (0, 2), (1, 0), (1, 3), (1, 4), (2, 1), (2, 2), (2, 4), (3, 0), (3, 3), (4, all), mod 5 it is true that the right side of ... put x = 1 here, and read off the coefficients of like powers of q, we have (2). ✷ 3 Bijective proof A bijective proof of (2) follows from the theory of 2-cores. The 2-core of a partition π is obtai...

Ngày tải lên: 07/08/2014, 13:21

... W and by V (G) − W are connected.) Note that ∂ W =  w∈W ∂{w}, where the addition is symmetric difference. In our situation, let B i = ∂ {u, i}, i ∈ {1, 2, 3}, and B 0 = ∂ {u, 1, 2, 3}. Then B 0 , ... not dual realizable by Lemma 10. By Lemma 13, G is also not dual realizable. Concerning statement (2), if G is a graph of type G ∗ 0 , G ∗ 1 or G ∗ 2 , then it follows from Lemma 10 that G is du...

Ngày tải lên: 07/08/2014, 13:21