Báo cáo toán học: " Long heterochromatic paths in edge-colored graphs" potx

... of combinatorics 12 (2005), #R33 28 Long heterochromatic paths in edge-colored graphs He Chen and Xueliang Li Center for Combinatorics and LPMC Nankai University Tianjin 300071, P.R. China lxl@nankai.edu.cn Submitted: ... through fixed vertices in edge-colored graphs, Math. Inf. Sci. Hun. 32 (1994), 49–58. the electronic journal of combinatorics 12 (2005), #R33 31 3. Long hete...

Ngày tải lên: 07/08/2014, 13:21

... endpoint in S, the other a distinct endpoint not in S and of a colour different to that of any edge incident with a vertex of SMALL. This set of edges contains at least |M|/k colours. If M contains ... separating C  1 and C  2 along ˆ P = P (ˆa, ˆ b). We now consider the two half paths P 1 ,P 2 obtained by splitting ˆ P at x.We consider rotations of P i ,i=1, 2withx as a fixed endpoint....

Ngày tải lên: 07/08/2014, 06:20

... us in (1) k(|D|−|S|)+d|S|−q +2≤ e G (D, S) ≤ d|D|−q, leading to (d − k)(|D|−|S|) ≥ 2. ✷ Proof of Theorem 1.1. The first case follows from the well-known Theorem of Petersen [6]. In the remaining ... disjoint subsets D, S of V (G), whether G has a k-factor or not. In 1985, Enomoto, Jackson, Katerinis and Saito [3] proved the following result. Lemma 2.2 Let G be a graph and k a positive in...

Ngày tải lên: 07/08/2014, 08:20

... 1-factor containing e,so does G ∗ . By Theorem 1, G ∗ has a k-factor containing e and another k-factor avoiding e for all k,1≤ k ≤ 2r − 1. Deleting the matching M from G ∗ ,weobtaina[k, k + 1]-factor the ... 1)-factor containing e and G − F 1 is an (r − 1)-factor avoiding e. By Theorem 2, the odd-factors F 1 and G − F 2 containing e, respectively, imply the existence of k-factors containing e...

Ngày tải lên: 07/08/2014, 08:22

... is not completely contained inside R ∗ , the center of S is inside R ∗ , S does not contain any point in Q as its interior point, and there is no line segment L i ∈ U that intersects two nonadjacent ... discussed in the subsequent subsections. 5.1 Case-2 In this case, S is not completely contained inside R ∗ , the center of S is inside R ∗ , S does not contain any point in Q as its in...

Ngày tải lên: 07/08/2014, 13:21

... Denote the length of the longest induced cycle in G ∈G S by M(S), and define µ(r)=max S M(S)to be the length of the longest induced cycle in all graphs in G S ,whereS contains r integers. Theorem 4.2. ... maximum lengh of a cycle in unit circulant graphs on a Dedekind domain quotiented by an ideal with r distinct maximal factors. Dedekind domains are exactly those integral domains in...

Ngày tải lên: 07/08/2014, 13:21

... able to show that in the domains θ = o  √ λ min /a(r) 2  and 1/θ = O  √ λ min  the inequality (V) is certainly true 1 . But since √ λ min /a(r) 2 = o  1/ √ λ min  there is a gap ... B(r) ε , we obtain det B(r) = O  r σd(dL+1) min  . On setting η = r − e 3 min this implies max θ∈∆(r) |ε(r, θ)| = O  r σd(Ld+1)ε min · r − e 2 min r − e 3 min  → 0 for r min → ∞ because of ......

Ngày tải lên: 07/08/2014, 13:21

... vertex v of in nite degree we can only demand all or nothing. Setting u(v) to a finite value in Theorem 2 is the same as requiring in nitely many outgoing edges at v, whereas putting u(v) = ∞ ... the in nite star. We immediately encounter the problem that the set L of all leaves is incident with in nitely many edges but has in nite demand for ingoing edges, i.e. l(L) = ∞. This results...

Ngày tải lên: 07/08/2014, 21:20

... efficiently removes transient noise while preserving the slowly varying component in the input signal. In other words, the slowly varying component of desired speech remains in the output of the median filter. ... ms. Since τ p (l) in Eq. (5) is the integer multiple of the sampling period of the input signal, the estimation error of the pitch period depends on the sampling f re- quenc...

Ngày tải lên: 20/06/2014, 21:20

... b β min +1 +1andb  β min +1 = b β min − 1. The remaining b’s are unchanged, thus b  β = b β , β = β min ,β min + 1. Hence the value of b is unchanged. Since T  β min terminates at v f β min +1 , T  β min +1 terminates ... Φ ω . Moving the bridge on T β min to the right decreases b β min by one and moving the bridge on T β min +1 to the left increases b β min +1 by one and hence b  β min...

Ngày tải lên: 07/08/2014, 08:20