Báo cáo toán học: "Strings with maximally many distinct subsequences and substrings" ppsx

... absence of each character, is distinct. Note that (2/3 − d ) d ≤ C (d) 1 ≤ (2 +  d ) d , for some  d → 0. 3 Strings with maximally many distinct substrings We close with a solution to a simpler ... string, but for d = 2 no such infinite string exists. 2 Strings with maximally many distinct subsequences Let Σ be a finite alphabet of size d; without loss of generality we t...

Ngày tải lên: 07/08/2014, 08:20

... graph with parts A and B. Let 2 ≤ r ≤ n, (ln n) −1/r 2 ≤ c ≤ 1/2, |A| = m, |B| = n, and s =  c r 4 −r 2 +r ln n  . If s ≤ (c/2 r )m + 1 and e(F ) ≥ (c/2 r−1 )mn, then F contains a K 2 (s, t) with ... for the complete bipartite graph with parts of size s and t; - f| X for the restriction of a map f to a set X. Specific notation Suppose that G and H are graphs, and let X be a...

Ngày tải lên: 07/08/2014, 15:23

Ngày tải lên: 05/08/2014, 15:20

Ngày tải lên: 05/08/2014, 15:21

... Cabrerizo, A. Carriazo, L. M. Fern´andez and M. Fern´andez studied slant and semi-slant submanifolds of a contact manifold [3, 4], and presented existence and uniqueness theorems for slant submanifolds ... L. M. Fern´andez and M. Fern´andez, Semi-slant sub- manifolds of a Sasakian manifold, Geom. Dedicata 78 (1999) 183–199. 4. J. L. Cabrerizo, A. Carriazo, L. M. Fern´andez and M....

Ngày tải lên: 06/08/2014, 05:20

... each of the pops could have been handled as a push and the whole sequence of elements could have been handled with a single pop. Otherwise, we discard notes and continue fresh. The only situation ... pop impossible, and also if the preceding operation was a push, and l>1orl = 1 and after the pop the top plate of the stack is not red. The accepting states are any state with an...

Ngày tải lên: 07/08/2014, 07:21

... surface of genus g. Alon, Mohar, and Sanders [3] have shown that a(G)=O(g 4/7 ) and this is nearly best possible since there are graphs with a(G)=Ω(g 4/7 / log 1/7 g). Together with Heawood’s bound χ(G)=O(g 1/2 ) ... that ω(G t )=t and χ s (G t )=  t+1 2  . Moreover, G 3 is outerplanar and G 4 is planar. Proof. We give a recursive construction with base cases G 1 = K 1 and G...

Ngày tải lên: 07/08/2014, 08:20

... Math. 40 (1982), 141–152. [2] E. Boros and Z. F¨uredi, The number of triangles covering the center of an n-set, Geom. Dedicata 17 (1984), 69–77. [3] R. C. Buck and E. F. Buck, Equipartition of convex ... 2006 Mathematics Subject Classification: 52C10, 52C30 Abstract We give a simpler proof of the result of Boros and F¨uredi that for any finite set of points in the plane in general positio...

Ngày tải lên: 07/08/2014, 13:21

... Γ = Cay(H, S) over H with connection set S is the graph with vertex set H so that the vertices x and y are adjacent if and only if x −1 y ∈ S. Then Γ is undirected graph without loops. Accordingly, ... Golemac, J. Mandi´c and T. Vuˇciˇci´c, One (96,20,4) Symmetric Design and Related Nonabelian Difference Sets, Designs, Codes and Cryptography, 37 (2005), 5-13. [5] L.K. Jorgensen...

Ngày tải lên: 07/08/2014, 13:21