No Arabic abstract
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 $p, q$, $r$.
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.
The status of a vertex $x$ in a graph is the sum of the distances between $x$ and all other vertices. Let $G$ be a connected graph. The status sequence of $G$ is the list of the statuses of all vertices arranged in nondecreasing order. $G$ is called status injective if all the statuses of its vertices are distinct. Let $G$ be a member of a family of graphs $mathscr{F}$ and let the status sequence of $G$ be $s.$ $G$ is said to be status unique in $mathscr{F}$ if $G$ is the unique graph in $mathscr{F}$ whose status sequence is $s.$ In 2011, J.L. Shang and C. Lin posed the following two conjectures. Conjecture 1: A tree and a nontree graph cannot have the same status sequence. Conjecture 2: Any status injective tree is status unique in all connected graphs. We settle these two conjectures negatively. For every integer $nge 10,$ we construct a tree $T_n$ and a unicyclic graph $U_n,$ both of order $n,$ with the following two properties: (1) $T_n$ and $U_n$ have the same status sequence; (2) for $nge 15,$ if $n$ is congruent to $3$ modulo $4$ then $T_n$ is status injective and among any four consecutive even orders, there is at least one order $n$ such that $T_n$ is status injective.
Given a dense subset $A$ of the first $n$ positive integers, we provide a short proof showing that for $p=omega(n^{-2/3})$ the so-called {sl randomly perturbed} set $A cup [n]_p$ a.a.s. has the property that any $2$-colouring of it has a monochromatic Schur triple, i.e. a triple of the form $(a,b,a+b)$. This result is optimal since there are dense sets $A$, for which $Acup [n]_p$ does not possess this property for $p=o(n^{-2/3})$.
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.
In this article, we construct explicit examples of pairs of non-isomorphic trees with the same restricted $U$-polynomial for every $k$; by this we mean that the polynomials agree on terms with degree at most $k+1$. The main tool for this construction is a generalization of the $U$-polynomial to rooted graphs, which we introduce and study in this article. Most notably we show that rooted trees can be reconstructed from its rooted $U$-polynomial.