Do you want to publish a course? Click here

Tiling with Cuisenaire Rods

129   0   0.0 ( 0 )
 Added by Malcolm Connolly
 Publication date 2016
  fields
and research's language is English
 Authors M. Connolly




Ask ChatGPT about the research

In this paper a closed form expression for the number of tilings of an $ntimes n$ square border with $1times 1$ and $2times1$ cuisenaire rods is proved using a transition matrix approach. This problem is then generalised to $mtimes n$ rectangular borders. The number of distinct tilings up to rotational symmetry is considered, and closed form expressions are given, in the case of a square border and in the case of a rectangular border. Finally, the number of distinct tilings up to dihedral symmetry is considered, and a closed form expression is given in the case of a square border.



rate research

Read More

Let K_4^3-2e denote the hypergraph consisting of two triples on four points. For an integer n, let t(n, K_4^3-2e) denote the smallest integer d so that every 3-uniform hypergraph G of order n with minimum pair-degree delta_2(G) geq d contains floor{n/4} vertex-disjoint copies of K_4^3-2e. Kuhn and Osthus proved that t(n, K_4^3-2e) = (1 + o(1))n/4 holds for large integers n. Here, we prove the exact counterpart, that for all sufficiently large integers n divisible by 4, t(n, K_4^3-2e) = n/4 when n/4 is odd, and t(n, K_4^3-2e) = n/4+1 when n/4 is even. A main ingredient in our proof is the recent `absorption technique of Rodl, Rucinski and Szemeredi.
Komlos [Komlos: Tiling Turan Theorems, Combinatorica, 2000] determined the asymptotically optimal minimum-degree condition for covering a given proportion of vertices of a host graph by vertex-disjoint copies of a fixed graph H, thus essentially extending the Hajnal-Szemeredi theorem which deals with the case when H is a clique. We give a proof of a graphon version of Komloss theorem. To prove this graphon version, and also to deduce from it the original statement about finite graphs, we use the machinery introduced in [Hladky, Hu, Piguet: Tilings in graphons, arXiv:1606.03113]. We further prove a stability version of Komloss theorem.
Partitioning a set into similar, if not, identical, parts is a fundamental research topic in combinatorics. The question of partitioning the integers in various ways has been considered throughout history. Given a set ${x_1, ldots, x_n}$ of integers where $x_1<cdots<x_n$, let the {it gap sequence} of this set be the nondecreasing sequence $d_1, ldots, d_{n-1}$ where ${d_1, ldots, d_{n-1}}$ equals ${x_{i+1}-x_i:iin{1,ldots, n-1}}$ as a multiset. This paper addresses the following question, which was explicitly asked by Nakamigawa: can the set of integers be partitioned into sets with the same gap sequence? The question is known to be true for any set where the gap sequence has length at most two. This paper provides evidence that the question is true when the gap sequence has length three. Namely, we prove that given positive integers $p$ and $q$, there is a positive integer $r_0$ such that for all $rgeq r_0$, the set of integers can be partitioned into $4$-sets with gap sequence $p, q$, $r$.
Komlos [Tiling Turan theorems, Combinatorica, 20,2 (2000), 203{218] determined the asymptotically optimal minimum degree condition for covering a given proportion of vertices of a host graph by vertex-disjoint copies of a fixed graph. We show that the minimum degree condition can be relaxed in the sense that we require only a given fraction of vertices to have the prescribed degree.
A fundamental result of Kuhn and Osthus [The minimum degree threshold for perfect graph packings, Combinatorica, 2009] determines up to an additive constant the minimum degree threshold that forces a graph to contain a perfect H-tiling. We prove a degree sequence version of this result which allows for a significant number of vertices to have lower degree.
comments
Fetching comments Fetching comments
Sign in to be able to follow your search criteria
mircosoft-partner

هل ترغب بارسال اشعارات عن اخر التحديثات في شمرا-اكاديميا