Báo cáo toán học: "Packing Graphs: The packing problem solved " potx

... neighbours in G ′′ . Thus by Lemma 3 there is a packing σ ′′ of G ′′ such that σ ′′ (W ) = W . Then (yz)σ ′′ is a packing of G. Therefore, we get a contradiction again, so the proof is completed.  References [1] ... packable. Thus, by Lemma 3, there is a packing σ ′ of G ′ such that σ ′ (U) = U. This packing can be easily modified in order to obtain a packing of G. Namely, note th...

Ngày tải lên: 07/08/2014, 21:21

... 05C15 Abstract The concept of a packing colouring is related to a frequency assignment problem. The packing chromatic number χ p (G) of a graph G is the smallest integer k such that the vertex set ... s shown t hat the packing chromatic number of the infinite square lat tice is at most 22. Using a computer we found the f ollowing result improving the upper bound of χ p (...

Ngày tải lên: 08/08/2014, 11:20

... Indeed, there are many problems in this field remaining unsolved. Here we solve an open problem which was posed by Xu in [4], which extends the area of integer sum graphs. Let N ∗ denote the set ... Xu posed an open problem as follows. Problem 1. Is it true that any odd cycle is an  Σ-graph ? In this section, we give an affirmative answer to the above open problem, namely Theorem...

Ngày tải lên: 06/08/2014, 05:20

... abnormally off their mean then it affects the change in deg t (v). (If v  ,v have a common edge e then deg(v  ) affects the number of (e, e  ) which affects the expected change of deg(v).) The N different ... (almost) environment so that the errors are basically independent of each other. With that the square of the total error will be close to the sum of the squares of the...

Ngày tải lên: 07/08/2014, 06:22

... on H  2,4 , then this only extends to a 2-coloring of the larger graph, whereas from the 4-coloring we obtain all other values in the spectrum. It remains to prove the necessity of the condition. ... C = ∅ and the D = E(G). The spectrum of G as a mixed hypergraph is the interval {χ(G),χ(G)+1, ,t},whereχ(G) ≤ 4 denotes the usual chromatic number of G. This shows the sufficie...

Ngày tải lên: 07/08/2014, 06:22

... B m and  < m ≤ 2, then (v j , v i ) ∈ B m− . Thus we have B m ˘ = B m+ for m ≤ . Then by making superficial changes to the remainder of the proof we establish the result for the nonsymmetric ... maximizing (over n) the greatest of them, we can find the c which will guarantee the inequality in (6). For n ≥ 2, it is sufficient to take c ≥ 5.2. For such k, we have that P < 1...

Ngày tải lên: 07/08/2014, 15:23

... either 8 divides m or 4 divides n. If we color the squares of the lattice in a checkerboard pattern, then any placement of the tile covers 3 squares of one color and 5 of the other color. Therefore this ... even under more complex box -packing, which is the content of the next theorem. Theorem 3.6. (Barnes) Suppose that every box in the set S satisfies the restriction the ele...

Ngày tải lên: 07/08/2014, 21:20

... totically, i.e. in the limit when the side length o f the cube tends to infinity. While the proof of Lemma 2 in [9] (o r the equivalent statement in [7]) is not hard, it is nevertheless interesting ... integers. By consequence the spectrum of Y(w) is just the union of the spectra of tho se T α (w) for which y α is nonzero. 2 We define the set of all symmetric elements in CG an...

Ngày tải lên: 07/08/2014, 21:21

... nm (MNP-PEI); the size further increased upon pegylation to 361 nm. The zeta potential of the ‘naked’ MNPs was negative (−38.2 mV), due to the -COOH groups on the surface of the particles ... prevent the polymer-coated nanoparticles from attaching to the plates [31]. The results showed that using the 0.2 % w/v silica solution did decrease the adhesive effect of the...

Ngày tải lên: 20/06/2014, 20:20

... in the 5 wt.% ATN than those of other samples and the weight of Ru is 0.42 at.%. So, the ATN on the TiO 2 film gave

Ngày tải lên: 20/06/2014, 20:20