2-factors with k cycles in Hamiltonian graphs

A well known generalisation of Dirac's theorem states that if a graph $G$ on $n\ge 4k$ vertices has minimum degree at least $n/2$ then $G$ contains a $2$-factor consisting of exactly $k$ cycles. This is easily seen to be tight in terms of the bound on the minimum degree. However, if one assumes in addition that $G$ is Hamiltonian it has been conjectured that the bound on the minimum degree may be relaxed. This was indeed shown to be true by S\'ark\"ozy. In subsequent papers, the minimum degree bound has been improved, most recently to $(2/5+\varepsilon)n$ by DeBiasio, Ferrara, and Morris. On the other hand no lower bounds close to this are known, and all papers on this topic ask whether the minimum degree needs to be linear. We answer this question, by showing that the required minimum degree for large Hamiltonian graphs to have a $2$-factor consisting of a fixed number of cycles is sublinear in $n.$Comment: 13 pages, 6 picture

2-factors with k cycles in Hamiltonian graphs

A well known generalisation of Dirac’s theorem states that if a graph G on n ≥ 4k vertices has minimum degree at least n/2 then G contains a 2-factor consisting of exactly k cycles. This is easily seen to be tight in terms of the bound on the minimum degree. However, if one assumes in addition that G is Hamiltonian it has been conjectured that the bound on the minimum degree may be relaxed. This was indeed shown to be true by S´ark¨ozy. In subsequent papers, the minimum degree bound has been improved, most recently to (2/5 + ε)n by DeBiasio, Ferrara, and Morris. On the other hand no lower bounds close to this are known, and all papers on this topic ask whether the minimum degree needs to be linear. We answer this question, by showing that the required minimum degree for large Hamiltonian graphs to have a 2-factor consisting of a fixed number of cycles is sublinear in n

A Tight Lower Bound for Counting Hamiltonian Cycles via Matrix Rank

For even $k$, the matchings connectivity matrix $\mathbf{M}_k$ encodes which pairs of perfect matchings on $k$ vertices form a single cycle. Cygan et al. (STOC 2013) showed that the rank of $\mathbf{M}_k$ over $\mathbb{Z}_2$ is $\Theta(\sqrt 2^k)$ and used this to give an $O^*((2+\sqrt{2})^{\mathsf{pw}})$ time algorithm for counting Hamiltonian cycles modulo $2$ on graphs of pathwidth $\mathsf{pw}$. The same authors complemented their algorithm by an essentially tight lower bound under the Strong Exponential Time Hypothesis (SETH). This bound crucially relied on a large permutation submatrix within $\mathbf{M}_k$, which enabled a "pattern propagation" commonly used in previous related lower bounds, as initiated by Lokshtanov et al. (SODA 2011). We present a new technique for a similar pattern propagation when only a black-box lower bound on the asymptotic rank of $\mathbf{M}_k$ is given; no stronger structural insights such as the existence of large permutation submatrices in $\mathbf{M}_k$ are needed. Given appropriate rank bounds, our technique yields lower bounds for counting Hamiltonian cycles (also modulo fixed primes $p$) parameterized by pathwidth. To apply this technique, we prove that the rank of $\mathbf{M}_k$ over the rationals is $4^k / \mathrm{poly}(k)$. We also show that the rank of $\mathbf{M}_k$ over $\mathbb{Z}_p$ is $\Omega(1.97^k)$ for any prime $p\neq 2$ and even $\Omega(2.15^k)$ for some primes. As a consequence, we obtain that Hamiltonian cycles cannot be counted in time $O^*((6-\epsilon)^{\mathsf{pw}})$ for any $\epsilon>0$ unless SETH fails. This bound is tight due to a $O^*(6^{\mathsf{pw}})$ time algorithm by Bodlaender et al. (ICALP 2013). Under SETH, we also obtain that Hamiltonian cycles cannot be counted modulo primes $p\neq 2$ in time $O^*(3.97^\mathsf{pw})$, indicating that the modulus can affect the complexity in intricate ways.Comment: improved lower bounds modulo primes, improved figures, to appear in SODA 201

Single-Strip Triangulation of Manifolds with Arbitrary Topology

Triangle strips have been widely used for efficient rendering. It is NP-complete to test whether a given triangulated model can be represented as a single triangle strip, so many heuristics have been proposed to partition models into few long strips. In this paper, we present a new algorithm for creating a single triangle loop or strip from a triangulated model. Our method applies a dual graph matching algorithm to partition the mesh into cycles, and then merges pairs of cycles by splitting adjacent triangles when necessary. New vertices are introduced at midpoints of edges and the new triangles thus formed are coplanar with their parent triangles, hence the visual fidelity of the geometry is not changed. We prove that the increase in the number of triangles due to this splitting is 50% in the worst case, however for all models we tested the increase was less than 2%. We also prove tight bounds on the number of triangles needed for a single-strip representation of a model with holes on its boundary. Our strips can be used not only for efficient rendering, but also for other applications including the generation of space filling curves on a manifold of any arbitrary topology.Comment: 12 pages, 10 figures. To appear at Eurographics 200

Parity balance of the ii-th dimension edges in Hamiltonian cycles of the hypercube

Let $n\geq 2$ be an integer, and let $i\in\{0,...,n-1\}$. An $i$-th dimension edge in the $n$-dimensional hypercube $Q_n$ is an edge ${v_1}{v_2}$ such that $v_1,v_2$ differ just at their $i$-th entries. The parity of an $i$-th dimension edge \edg{v_1}{v_2} is the number of 1's modulus 2 of any of its vertex ignoring the $i$-th entry. We prove that the number of $i$-th dimension edges appearing in a given Hamiltonian cycle of $Q_n$ with parity zero coincides with the number of edges with parity one. As an application of this result it is introduced and explored the conjecture of the inscribed squares in Hamiltonian cycles of the hypercube: Any Hamiltonian cycle in $Q_n$ contains two opposite edges in a 4-cycle. We prove this conjecture for $n \le 7$, and for any Hamiltonian cycle containing more than $2^{n-2}$ edges in the same dimension. This bound is finally improved considering the equi-independence number of $Q_{n-1}$, which is a concept introduced in this paper for bipartite graphs

Families of graph-different Hamilton paths

Let D be an arbitrary subset of the natural numbers. For every n, let M(n;D) be the maximum of the cardinality of a set of Hamiltonian paths in the complete graph K_n such that the union of any two paths from the family contains a not necessarily induced cycle of some length from D. We determine or bound the asymptotics of M(n;D) in various special cases. This problem is closely related to that of the permutation capacity of graphs and constitutes a further extension of the problem area around Shannon capacity. We also discuss how to generalize our cycle-difference problems and present an example where cycles are replaced by 4-cliques. These problems are in a natural duality to those of graph intersection, initiated by Erd\"os, Simonovits and S\'os. The lack of kernel structure as a natural candidate for optimum makes our problems quite challenging

The number of Hamiltonian decompositions of regular graphs

A Hamilton cycle in a graph $\Gamma$ is a cycle passing through every vertex of $\Gamma$. A Hamiltonian decomposition of $\Gamma$ is a partition of its edge set into disjoint Hamilton cycles. One of the oldest results in graph theory is Walecki's theorem from the 19th century, showing that a complete graph $K_n$ on an odd number of vertices $n$ has a Hamiltonian decomposition. This result was recently greatly extended by K\"{u}hn and Osthus. They proved that every $r$-regular $n$-vertex graph $\Gamma$ with even degree $r=cn$ for some fixed $c>1/2$ has a Hamiltonian decomposition, provided $n=n(c)$ is sufficiently large. In this paper we address the natural question of estimating $H(\Gamma)$, the number of such decompositions of $\Gamma$. Our main result is that $H(\Gamma)=r^{(1+o(1))nr/2}$. In particular, the number of Hamiltonian decompositions of $K_n$ is $n^{(1-o(1))n^2/2}$