Báo cáo toán học: "On the Number of Matchings in Regular Graphs" pptx
Báo cáo toán học: " ON THE NUMBER OF DESCENDANTS AND ASCENDANTS IN RANDOM SEARCH TREES" ppsx
... +1)649n6.7. The number of ascendants of a given node in a LBSTAs in the case of the number of ascendants in a random BST, computing the probability that the jthnode in a random LBST has m ascendants ... other random variables: the number of descendants Dn and the number of ascendants An of a randomly chosen internal node in a random search tree of size n. This corresponds to averagingDn,j and ... mentioned in the introduction, the standard model for random LBSTs statesthat a random LBST of size n is the result of n random insertions into an initially empty tree.Equivalently, a random...
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Báo cáo toán học: "On the number of perfect matchings and Hamilton cycles in -regular non-bipartite graphs" ppsx
... n)k−1φ(k). In either case, the inequality in part 1 of the lemma holds. On the number of perfect matchings and Hamilton cycles in -regular non-bipartite graphsAlan Frieze∗Department of Mathematical ... of combinatorics 7 (2000), #R57 43 Hamilton Cycles A Hamilton cycle is the union of two perfect matchings and so h(G) ≤12m(G)2 and the upper bound in part (b) of Theorem 1 follows from the ... segment containing x, the segment containing x, and the remaining segment.Going round the cycle Ci, starting at x and ending at x, the vertices x, x,y,y,z,zmay appear in one of eight...
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Báo cáo toán học: "On the number of permutations admitting an m-th root" pptx
... in the case when m is prime.Whether a permutation of nadmits an m-th root can be read on the partition of n determined by the lengths of the permutation’s cycles, because the class of such the ... is the exponential generating function (EGF, formal series) of the m-thpowersinthegroupsn. This means that the number of m-thpowersinnispn(m)ìn!foreachn.Inthesameway,Cmis the EGF of the ... EGF of the permutations having onlycycles of length prime to m in their canonical decomposition (they admit a m-th root)and Rm the EGF of the rectangular m-th powers, that is the m-th powers...
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Báo cáo toán học: " ON THE NUMBER OF FULLY PACKED LOOP CONFIGURATIONS WITH A FIXED" pot
... we are not in one of these cases, then there is exactly one edge e attached to v which isbelow l.Theedgee is on a loop of the configuration. Let us start with e and follow the part of the loop ... which is based on a variation of the above reasoning, seeSection 5. Finally, for the proof of the more specific assertions in Conjectures 1.1 and 1.2 on the integrality of the coefficients of the polynomials ... of the region, and as well in eachmid-point along the downward edges which form part of a notch on the top of the region,and along the downward edges along the top-right side of the region....
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Báo cáo toán học: "On the number of possible row and column sums of 0,1-matrices" pps
... x and y to be row and column sums of a matrixis that they have the same weight.Clearly, the row sums of a member of M are at most n.Conversely,ifx =(x1, ,xm) and 0 xi n,letR = R(x)bethem ... (5) the electronic journal of combinatorics 13 (2006), #N8 5 3 Start of the proofLet N = {0, 1, }. Define the weight of a matrix N to be the sum of its entries, and write |N| for the weight of ... Rij=1if1≤ j ≤ xi and Rij= 0 otherwise. Then R lies in M and has row sums equal to x.Thisproves:Lemma 5 Let x =(x1, ,xm) ∈ Nm. Then x is the vector of row sums of an m ì nmatrix with...
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Báo cáo toán học: "Bounding the number of edges in permutation graphs" pdf
... listed in πs,mn−1.Sincen is the largest number, this forces the insertion point to be right at the beginning of the list. Since πisprecisely πs,mn−1with m =n−1−s2, this yields the permutation ... from the proof of Theorem 1.3, as if the structure of a permutation is ‘close’ to that of πs,mnthen the configuration of the first type in Figure 2 will not occur for most pairs of vertices. Then ... combinatorial problem was motivatedby the observation that the number of edges in Γπ,0equals the number of permutationsσ that are covered by π in the strong Bruhat1order on Sn, and the number...
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Báo cáo toán học: "On the number of orthogonal systems in vector spaces over finite fields" ppsx
... (2.6) and (2.8), the number of copies of Kk in E is(1 + o(1))|E|kk!q−(k2)..This implies that the number of the number of k-tuples of k mutually orthogonal vectors in E is also(1 ... o(1))|E|kk!q−(k2),completing the proof of Theorem 1.2.Acknowledgments The research is performed during the authors visit at the Erwin Schrăodinger InternationalInstitute for Mathematical Physics. The author ... (2008), arXiv:0807.0592. the electronic journal of combinatorics 15 (2008), #N32 4 On the number of orthogonal systems in vector spaces over finite fieldsLe Anh VinhMathematics DepartmentHarvard...
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Báo cáo toán học: "On the number of genus one labeled circle trees" doc
... such that the tree together with the circle it is drawn on can be embedded in a surface of genus one, but not of genus zero.Hough [3] observed that the number of genus one labeled circle trees ... offspring of C by definition.Proposition 4. The genus of the reduced form T3 of a u-c-tree T is one if and only if the genus of T is one. Proof. Let T2be the pre-reduced form of u-c-tree T . Then ... describe all of the possible reduced forms of genus one u-c-trees. The interrelations of a u-c-tree andits reduced form enables us to get an insight into the behavior of the number of genus one l-c-trees...
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Báo cáo toán học: "On the Number of Planar Orientations with Prescribed Degrees" pptx
... Nidenote the number of Eulerian orientations of G0which have i valid cross overboxes, that is Nis the number of Eulerian orientations of G. The number of Eulerian orientations of GkisEO(Gk) ... bound for the number of almost alternating Eulerian orientations yields a lower bound for the number of 2 -orientations of Gk,.For the sake of simplicity we will work with alternating orientations ... either none or both of the edges.Let cGbe the number of 2-factors of G. Among the 2-factors of G(k) let S denote the set of those, which induce a 2-factor of G, and Sc the other 2-factors....
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Báo cáo toán học: "On the Number of Matchings in Regular Graphs" pptx
... get the numerator of the right-hand side of (4.10). Dividethis number of choices by the number of permutations of {1, . . . , rn} to deduce the lemma.✷Using the methods in the proof of Theorem ... :=ri=1Ai. Let m ∈ min(p, q). Then permmA is the number of m -matchings of G := ∨ri=1Gi, which is equal to the number of m -matchings obtained in the following way. Consider m1, . ... k ≥ 3. All matchings of Pk, where the vertex k is not in the matching, generate the polynomial pk−1(x). All matchings of Pk, where the vertex k is in the matching, generate the polynomial...
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Báo cáo toán học: "On the size of minimal unsatisfiable formulas" pptx
... y2l= true) will make the remainingclauses true.Next we prove the linear upper bound of number of clauses in minimal unsatisfiable 2-SAT formulas.Proposition 1. Minimal unsatisfiable 2-SAT formulas ... deletion of any edge makes it 2 colorable. In this analogy, minimal unsatisfiable formulas correspond to critically 3-chromatic hypergraphs. Therefore it is∗This research forms part of the Ph.D thesis ... Samsung Scholarship. the electronic journal of combinatorics 16 (2009), #N3 1 By the same process as above one can verify that F(k)0is minimal unsatisfiable. Thenreplace the m-clauses by k-clauses...
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Báo cáo toán học: "On the number of independent sets in a tree" pdf
... .2 The number of matchings in a tree In t his section, we turn to the number of matchings in a graph. This is a lso known as the Hosoya index, or the Z-index in mathematical chemistry. For a rooted ... An interesting paper of Wagner [5] looks at the number of independent sets modulo m.Wagner showed that the pr oportion of trees on n vertices with the number of independent sets divisible ... 47–55.[3] V. Linek, Bipartite graphs can have any number of independent sets, Discrete Math. 76(1989), 131–136.[4] H. Prodinger and R.F. Tichy, Fibonacci numbers of graphs, Fibonacci Quart. 20 (1982),16–21.[5]...
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Báo cáo toán học: "On the distribution of depths in increasing trees" ppsx
... 05C05 60C05AbstractBy a theorem of Dobrow and Smythe, the depth of the kth no de in very simplefamilies of increasing trees (which includes, among others, binary increasing trees,recursive trees ... Dobrow/Smythe, Theorem 6) In a random increasing tree with n nodes from a simple family, the distribution of the distance between nodes i and k (i < k) is the same as the distribution of the sum of ... random increasing tree with n nodes from a simple family, the distribution of the depth of node k is the same as the distribution of the number of 1-edges with labels k.Furthermore, the probability...
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Báo cáo toán học: "Counting the number of elements in the ˜ mutation classes of An −quiver" ppt
... realization having parame ters˜r1, ˜r2, ˜s1and ˜s2such that ˜r1< ˜s1or ˜r1= ˜s1and ˜r2≤ ˜s2. Then Q1is mutation equivalent to Q2if and only if r1+ 2r2= ˜r1+ 2˜r2and s1+ ... Combinatorial Grammar In this section we describe the elements of the mutation classes of type ˜ An−1by acombinatorial grammar. This can be viewed as an exercise in the theory of species(introduced ... we obtaina symmetric quiver of type ˜ A2r−1by taking L and rev(L), and gluing together the end point of L with the start point of rev(L) and vice versa. E.g., the symmetricquiver in Figure...
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Báo cáo toán học: "On the number of subsequences with a given sum in a finite abelian grou" pptx
... subsequences with a given sum in a nite abelian groupGerard Jennhwa Chang,123Sheng-Hua Chen,13Yongke Qu,4Guoqing Wang,5Đand Haiyan Zha ng6ả1Department of Mathematics, National Taiwan University, ... Wang, A quantitative aspect of non-uniquefactorizations: the Narkiewicz constants, International Journal of Number Theory,to appear.[13] W.D. Gao and J.T. Peng, On the number of zero -sum subsequences ... 278, Chapman & Hall/CRC,2006.[15] A. Geroldinger, Additive group theory and non-unique factorizations, Combinatorial Number Theory and Additive Group Theory, Advanced Courses in Mathematics...
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