the number of authors presented

... informatics and soft skills They said that students of the modern age prove to be dynamic and eager to learn, but they still have a lot of shortcomings Especially, they lack professional skills ... under the Ministry of Education and Training The development and application of ICT was an essential part of reforming the national education system, he said.The Government wiould foot the bill ... stationaries for the new coming school year The new targets aim to eliminate manipulation of exam results, and to keep students in school, reducing the number of drop-outs Trang 8Minister of Education

Ngày tải lên: 19/07/2013, 16:34

... four types, one of which is that of slide homeomorphisms In [9], the author considered the Nielsen numbers and fixed points of The fixed point index of the complement of the union of the sliding ... reducing the number of fixed points is equivalent to reducing the number of intersection points between the sliding paths and sliding spheres The minimal numberMI( { α1, , α m },{ S1, , S m }) of ... of the in-tersection of sliding paths and sliding spheres gives a possible number of fixed points for N( f ) is a lower bound of the number of fixed points for maps in the homotopy class of f

Ngày tải lên: 22/06/2014, 22:20

... performed on the behavior of the four criteria, with respect to the dynamic range of the amplitudes of the two sinusoids (Figure 6a) and to the whiteness of the noise (Figure 6b) S/N ratios of 10 dB ... measures the relative slope variation of the eigenvalues {λk }k=1, ,M The difference between the current eigenvalue and the mean of the next ones has been preferred to the simple subtraction of the ... ratios the number of significant eigenvalues equals the number of harmonic components, the others taking values close to zero, as it can be seen on Figure 1a When the noise level increases, the N

Ngày tải lên: 23/06/2014, 01:20

... IDENTITIES AND A GENERALIZATION OF THE NUMBER OF TOTALLY SYMMETRIC SELF-COMPLEMENTARY PLANE PARTITIONS C Krattenthaler Institut făr Mathematik der Universităt Wien, u a Strudlhofgasse 4, A-1090 Wien, ... Math 158 (1993), 1–14) to compute this number explicitly The evaluation of the generalized determinant is independent of Andrews and Burge’s computations, and therefore in particular constitutes ... originally hoping to find a proof of the following conjecture of Robbins and Zeilberger [16, Conjecture C’=B’] (caution: in the quotient defining B it should read (m + + 2j) instead of (m + + j)), which

Ngày tải lên: 07/08/2014, 06:20

... An Asymptotic Expansion for the Number of Permutations with a Certain Number of Inversions Lane Clark Department of Mathematics Southern Illinois University Carbondale Carbondale, ... replace the o(n −7/2 ) error term in the asymptotic expansion of B(n)/n!withO(n −9/2 ln 19 n). The following extension of Theorem 1 (the case m = 3) giving a complete asymptotic expansion of I(n, ... complete asymptotic expansion of the integral. Among the consequences, we have a complete asymptotic expansion for b(n, k)/n!fora range of k including the maximum of the b(n, k)/n!. AMS Subject

Ngày tải lên: 07/08/2014, 06:20

... subpath of the other The root of T (R, v) is the vertex corresponding to the empty path Let ηv,r be the number of closed walks of length r in T (R, v) starting at v and define wr (R) = ηv,2r v The ... maximising the number of perfect matchings in an (n − k)-regular subgraph of Kn,n The other incarnation of the problem is in (0, 1)-matrices Let Λk denote the set of n (0, 1)-matrices of order ... Now since the rth moment of the roots of ρ(R, x) dominates the rth moment of the roots of ρ(S, x) we conclude that λR ≥ λS , otherwise taking r sufficiently large yields a contradiction The original

Ngày tải lên: 07/08/2014, 06:22

... define the number of descendants D n,j as the size of the subtree rooted at the j th node, so we count the j th node as a descendant of itself. The number of ascendants A n,j is the number of internal ... Abstract. The number of descendants of a node in a binary search tree (BST) is the size of the subtree having this node as a root; the number of ascendants is the number of nodes on the path connecting ... further explanation. THE ELECTRONIC JOURNAL OF COMBINATORICS 5 (1998), #R20 8 3. The number of descendants in random BSTs The number of the descendants D n,j of the j th node of a BST of size n is recursively

Ngày tải lên: 07/08/2014, 06:22

... completing the proof of part (a) of Theorem 1 Trang 43 Hamilton CyclesA Hamilton cycle is the union of two perfect matchings and so h(G) ≤ 1 2m(G)2 and the upper bound in part (b) of Theorem 1 ... n!. In both cases the bounds are “close” to the expected number of in the random graph G n,d The results here are strongly related to the result of Alon, R¨odl and Ruci´nski [2] They considered ... observe that the edge (x, x 0 ) may be chosen in n i ways (minimality of x fixes the orientation of the edge), and that the choice of (x, x 0) combined with the information provided by the sequence

Ngày tải lên: 07/08/2014, 06:22

... 4.3 Statement and proof of the main theorem The situation is the following: we look for the asymptotics of the coefficients p n(m) of the formal series Pm = Cm R m where the coefficients cn(m) ... natural boundary: the singularities of R m form a dense subset of the unit circle The argument given to reach the desired asymptotics uses the convergence of the series of coefficients of Rm, and a ... where q is the quadratfrei radical of m The order of the singularity 1 is clearly P µ(k)/k, where the sum extends to all divisors of q Let ϕ be the Euler function, i.e ϕ(q) is the number of all positive

Ngày tải lên: 07/08/2014, 06:23

... lattices, one rather considers the vertical sum in that case, where the only difference to the former is that now the top element  1 of the lower summand and the bottom element ⊥ 2 of the upper summand ... inequalities are the numbers v k of vertically indecomposable distributive lattices of size k. We present the explicit values of the numbers d k and v k for k<50 and prove the following exponential ... components of a lattice are intervals of that lattice. For graph theorists it may be of interest that the ordinal decomposition of a poset into indecomposable summands corresponds to the partition of the

Ngày tải lên: 07/08/2014, 06:23

... problems on the square lattice L n There the authors gave the following bounds for the asymp-totics off(n), the number of forests of L n, andα(n), the number of acyclic orienta-tions ofL n: 3.209912 ... orientation of G is an assignment of a direction to every edge in E such that there is no directed cycle We denote by α(G) the number of acyclic orientations of G and by f (G) the number of spanning ... of length c m defined above, and ~1 is the vector of length c m with all entries equal to 1. Proof By definition, the vector X m encodes the contribution to the rank polynomial of the edges of

Ngày tải lên: 07/08/2014, 07:21

... “large” m, see the end of Section 6 for the precise statement. There we also give an explanation of the difficulty of closing the gap We conclude the introduction by outlining the proofs of our results, ... of the Lindstr¨om–Gessel–Viennot theorem (recalled in Lemma 2.3), to obtaina determinant for the number of rhombus tilings, see the proof of Lemma 6.4 In bothcases, the polynomial nature of the ... rhombus tilings in the statement of the theorem and these semistandard tableaux Proof of Theorem 2.2 It should be observed that, due to the nature of the region R(λ, d, h), there are several “forced”

Ngày tải lên: 07/08/2014, 08:22

... a formula for the total number of monic permutation binomials of degree less than 4p over F 4p+1, where p and 4p + 1 are primes, in terms of the numbers of three special types of permutation ... characterization to study the form and the number of monic permutation binomials over particular finite fields We ∗The second and the third authors are partially funded by NSERC of Canada. Trang 2describe ... described as one of the above types. Trang 10Proof We first show in detail the cases (i) (1, 2, p) and (i) (1, 4, 3p), since they arerepresentatives of the technique used to prove the remaining

Ngày tải lên: 07/08/2014, 13:21

... uppermost cell of the rightmost column of the polyomino has the maximal ordinate among all the cells of the polyomino; U2 : the lowest cell of the rightmost column of the polyomino has the minimal ... {fn}n≥1 of positive integers, where fn is the number of nodes at level n of the generating tree, assuming that the root is at level 1 2 Generation of convex permutominoes Let Cn be the set of convex ... polyomino the difference between the number of salient and reentrant points is equal to 4 Let us turn to consider the class of convex permutominoes In a convex permutomino of size n the length of the

Ngày tải lên: 07/08/2014, 15:22

... outermost edges of it Also, call the set of crossededges of the e-graph the set of parallel edges of the e-graph and call the set of edges of thee-graph that are uncrossed the set of uncrossed edges ... intersection of {E1, E2, .,Ek+1} and the points of one of the e-graphs, and point Ek+1 is the intersection of {E1, E2, , Ek+1} and the points of the other of the e-graphs Theorem 1 There can ... if the u-c-graph is an offspring of C.The final offspring of a u-c-tree C is the offspring which has no edges which could be deleted by the execution of operations 1) and 2) (Some offsprings of

Ngày tải lên: 07/08/2014, 15:23

... both, the Schnyder woods and the spanning trees, as α-orientations also gives an immediate proof of this bijection We now turn to the proof of the upper bound stated in Theorem 7 The proof uses the ... edges,i.e., of edges of T which are not in M If n = 3 the conclusion of the lemma is true and we may thus assume n > 3 for the rest of the proof Hence, there are no vertices of degree 2 in ... isbecause the three spanning trees have to cover all 3n−6 edges of the triangulation and theedges of the outer triangle must be bidirected because of the rule of vertices Theorem 3says, that the edge

Ngày tải lên: 07/08/2014, 21:20

... 3)C4∪ P5.Furthermore, if G6= F then ΦF ≺ ΦG and if G6= H then ΦG≺ ΦH Proof The proof of this theorem is similar to the proof of Theorem 3.7, and webriefly point out the different arguments one should ... implies the second part of the theorem of copies of C4 and a copy of Ci for i = 5, 6, 7, respectively Equalities in (2.16-2.18) hold if and only if G is either a union of copies of C3, or a union of ... even then pi−1qj+1 ≺ piqj This shows the first inequality in the second line of (3.9).If i is odd then piqj ≺ pi−1qj+1 This shows the inequality between the last term of thefirst line and the

Ngày tải lên: 07/08/2014, 21:20

... asymptotic estimates for the number of planar graphs Already in 1962, the asymptotic number of triangulations was given by Tutte [15] Investigating how much the number of planar graphs (triangulations) ... asymptotic estimates for the number (and properties) of the graphs in these classes The last section contains the enumeration of graphs not containing K3,3+ as a minor, where K3,3+ is the graph obtained ... Kuratowski’s theorem [10] planar graphs are K3,3- and K5 -minor-free Hence, the class of (maximal) planar graphs is contained in the class of maximal K3,3-minor-free graphs and we can use the number of

Ngày tải lên: 07/08/2014, 21:20

... crossing number e(Mǫa,b, Nǫ c,d) is the minimal number of intersection of represen-tations of Mǫ a,b and Nǫ ′ c,d in the interior of the punctured polygon When a = b and c = d, we let the crossing number ... triangulations ofthe punctured n-gon However, their method of counting triangulations is very differentfrom ours They use the classification of quivers of mutation type Dn, recently given in[V] The authors ... in [T] The Catalan number C(i) can be defined as the number of triangulations of an i+2-gonwith i − 1 diagonals It is given by C(i) = 1 i + 1 2ii  The number of equivalence classes in the mutation

Ngày tải lên: 07/08/2014, 21:21

... in terms of ∆ L . There are only finitely many such polynomials, and counting them gives the theorem of [18]. The main idea of Theorem 1.1 is to count r-tuples of integers in L instead of single ... bounds on the number of Galois extensions of Q with bounded discriminant. In combination with the lower bound in Theorem 1.1 for the total number of extensions, this yields the fact that “most number ... |Q|≥2. Let b M (Y ) be the number of H-extensions of M such that N M Q D L/M = Y . Let S be the set of primes of Q dividing Y , let G S (M) be the Galois group of the maximal extension of M unramified...

Ngày tải lên: 06/03/2014, 08:21