... 1.c indicate the hook ofthe cell (2, 1) The hook of a cell ρ ofthe shifted Ferrers diagram of λ again includes all cells that are either inthe same row as ρ and to the right of ρ, or inthe ... cell ρ ofthe ordinary Ferrers diagram of λ is the set of cells that are either inthe same row as ρ and to the right of ρ, or inthe same column as ρ and below ρ, ρ included The dots in Figure ... r The components of λ are called the parts of λ The integer n, the sum of all the parts of λ, is called the norm of λ and is denoted by n(λ) The (ordinary) Ferrers diagram of λ is an array of...
... averse, and deposits began to increase once again. In Japan, the ongoing trend toward a declining birthrate and an aging population is making it increasingly difficult to maintain the current ... Consequently, Japan has the largest amount of cash and deposits inthe world Household Assets in Major Countries % 100 Cash and deposits Insurance and pension Securities excluding stocks Stocks and other equities Others ... Massive amount of Deposits the Largest Balance of Deposits inthe World The Bank has the largest amount of deposits inthe world, even in comparison with overseas banks, and is one ofthe world’s largest financial institutions...
... for the future, pioneering the development and provision of relevant training in our Institutes and Centres and working in partnership where possible - Maintain our current investment in training ... Since then temozolomide plus radiotherapy has become the international standard of care for the brain cancer glioblastoma, leading to a pronounced increase in survival Informing and influencing ... has on the lives of hundreds of thousands of people every year We have made huge steps forward in improving survival andin preventing thousands of new cases ofthe disease, both inthe UK and across...
... tearing inthe internal structure ofthe paper and blistering inthe printed areas (see figures) binding agents and coating pigments used, the amount of binding agents contained inthe paper and ... The situation in paper production and printing industry ˿ The situation in printing and finishing industry ˿ Circumstances inthe workshop ˿ Paper handling 11 ˿ ˿ ˿ ˿ Blistering Breaking inthe ... Climate and Paper The interaction between climate andthe processing of coated papersin printing and finishing Table of contents V Circumstances within the printer’s power to control I Introduction...
... dishes in each ofwhichthe individual flavors are blended and disguised The former — and this book — reveals the individual ingredients and explains how they are prepared and combined Another purpose ... elliptic integral ofthe second kind Carlson’s elliptic integral ofthe third kind Carlson’s degenerate elliptic integral Legendre elliptic integral ofthe first kind Legendre elliptic integral ofthe ... think about your programming tasks, insofar as possible, exclusively in terms of these standard control structures In writing programs, you should get into the habit of representing these standard...
... construction, and management of training institutions Outside ofthe MoH, the Cabinet Office andthe Ministry of Finance control the annual budget for all ministries The Ministry of Education andthe Ministry ... library, dining, and office space The three largest public multicadre training institutions faced the unique challenge of expanding their infrastructure in space-constrained, urban Lusaka Each of these ... across the five years of scale-up including expanding thenumberof nursing bachelors and masters students graduating from the University of Zambia from 35 to 50 students annually and introducing...
... [24] The probability of selecting an institution was inversely proportional to thenumberof institutions in its stratum and proportional to its numberof beds Eight subjects were picked at random ... (number of subjects inthe institution /number of available beds), andthe answer refusal rate (higher in psychiatric centres) In 1998, 2,075 institutions were selected and 155 of them (7.5%) refused ... size of household, type of household and geographical area size based on the latter survey For the institution survey, weights included size of strata, the institution occupation rate (number of...
... image processing at the Naval Academy in Brest In 1992 he joined theEngineering School ENSIETA of Brest, where he held the positions of Senior Researcher and Head ofthe Electronics and Informatics ... from the Military Technical Academy of Bucharest in 1992 In 1997, he received the M.S degree in electronic engineering, andin 1999 he received the Ph.D degree in signal processing, both from the ... behavior ofthe four criteria, with respect to the dynamic range ofthe amplitudes ofthe two sinusoids (Figure 6a) and to the whiteness ofthe noise (Figure 6b) S/N ratios of 10 dB and 15 dB,...
... Using the linearity of CTw , manipulating the series, using the definition (Carl) ofthe q−binomial coefficients, and simplifying, brings the left side of (Sasha ) to be CTw ... complete the proof ofthe theorem, we use lemma to evaluate the inner sum of (SumAnna), then to get A2n (q), we replace n by 2n, and then replace l by l + n, and finally use the binomial theorem ... is odd and r = (n − 1)/2, inwhich case it is a monomial in q times (1−w)(1+w) , wr+1 and applying CTw kills it all the same, thanks to the symmetry ofthe Chu-Pascal triangle Summing the expression...
... 7(b + 2) 49n6 Thenumberof ascendants of a given node in a LBST As inthe case ofthenumberof ascendants in a random BST, computing the probability that the j th node in a random LBST has ... corresponding internal node has > b descendants, andthe claim inthe proposition follows On the other hand, let j be thenumberof comparisons made betweenthe j th element and other elements, during ... moments, including the expected value and variance (Theorem 3.2) Then we analyze thenumberof descendants of a random node, obtaining the probability that Dn = m, as well as the moments of Dn (Theorems...
... (β ln n)k −1 φ(k) In either case, the inequality in part ofthe lemma holds the electronic journal of combinatorics (2000), #R57 Part ofthe lemma follows from the chain of inequalities (β ln ... contain three edges — (x, x ), (y, y ) and (z, z ) — from a single cycle Ci of F , the remaining edges coming from F − The labeling of vertices in C can be made canonical inthe following way: ... segment containing x, the segment containing x , andthe remaining segment Going round the cycle Ci , starting at x and ending at x, the vertices x, x , y, y , z, z may appear in one of eight possible...
... vertices, and (G) denote thenumberof edges In this paper, we construct a graph G having no two cycles with the same length which leads to the following result Theorem Let t = 27720r + 169 (r ≥ 1), then ... 32t − 17t+1 Then f (n) ≥ n + 32t − 1, for n ≥ nt This completes the proof ofthe theorem From the above theorem, we have f (n) − n 2562 √ ≥ 2+ , n 6911 √ 487 which is better than the previous ... College(Natural Science Edition) 4(1)(1990) 29,30-34 [5] Chunhui Lai, On the size of graphs with all cycle having distinct length, Discrete Math 122(1993) 363-364 [6] Chunhui Lai, The edges in a graph in which...
... + 1, the electronic journal of combinatorics (2001), #N6 giving inequality (a) To prove inequality (b), let M be the submatrix of M consisting of rows βm+1, , m and all columns of M Since ... Disjoint cycles in Eulerian digraphs andthe diameter of interchange graphs, J Combin Theory Ser B, to appear [4] D.W Walkup, Minimal interchanges of (0, 1)-matrices and disjoint circuits in a graph, ... loss of generality, assume v ∗ ∈ A (since otherwise we can interchange the roles of m and n, as we did not impose any cardinality constraints upon them) It clearly suffices to prove the following:...
... given in Fig Then the following lemma is clear, since from the knowledge of σ(B) we can update thenumberof components inthe union B ∪ C Inthe example in Fig we have r(B) = 26, r(C) = and δ(B, ... corresponds to the edge set of a spanning forest of Lkp+1 i=0 The reason is the following Suppose there is a cycle C in B, then it would intersect some ofthe subgraphs Bi The cycle C cannot be inside ... α(n)1/n n→∞ n→∞ The proof ofthe second equality is very similar; it relies again on the formula T (G∗ ; x, y) = T (G; y, x), andthe fact that thenumberof forests andthenumberof connected...
... above argument with the portion of R beginning at um−1 andthe vertex um Theorem Let hp (n) be the minimum numberof distinct hamiltonian paths in a strong tournament of order n Then · β n−3 ... assume the result for all paths Q of order at most m−1 Let Q = u1 u2 · · · um−1 and apply the induction hypothesis using the paths P and Q to obtain a path R satisfying the lemma Next, we repeat the ... uj on R Proof Note that we allow the special case where m = 0; in this case the path Q is a path on vertices, and R = P satisfies the lemma trivially The remainder ofthe proof is by induction...
... numberof edges in Γπ,0 equals thenumberof permutations σ that are covered by π inthe strong Bruhat1 order on Sn , andthenumberof edges in Gπ,0 equals the degree ofthe vertex π inthe Hasse2 ... case analysis andthe proof of Theorem 1.2 Edge bounds in Gπ,s In this section we prove Theorem 1.3, which gives bounds on the maximum numberof edges in Gπ,s with π ∈ Sn Proof of Theorem 1.3 ... neighbors ofand t must lie inthe union ofthe marked rectangular regions A, B, C, D, and E (but we not need E if we are inthe case ofthe second diagram) Since l and t are adjacent, thenumber of...
... This implies that thenumberofthenumberof k-tuples of k mutually orthogonal vectors in E is also |E|k −(k) (1 + o(1)) q , k! completing the proof of Theorem 1.2 Acknowledgments The research is ... Jnq,d is the all one matrix and Inq,d is the identity matrix, both of size nq,d × nq,d ) Therefore, the largest eigenvalue of V is Dq,d andthe absolute values of all other eigenvalues are either ... satisfies n ∆ m λ d Then, for every subset U ⊂ V of cardinality m, thenumberof (not necessarily induced) copies of H in U is (1 + o(1)) ms | Aut(H)| the electronic journal of combinatorics 15 (2008),...
... even then pi−1 qj+1 pi qj This shows the first inequality inthe second line of (3.9) If i is odd then pi qj pi−1 qj+1 This shows the inequality betweenthe last term ofthe first line andthe ... all these choices to get the numerator ofthe right-hand side of (4.10) Divide this numberof choices by thenumberof permutations of {1, , rn} to deduce the lemma Using the methods inthe ... appearing inthe first line of (3.1-3.2) Assume now that i ≥ is even and j ≥ i So (−1)i−2 = Hence pi−2 pj+2 − pi pj This explains the ordering ofthe polynomials appearing inthe second line of (3.1-3.2)...