Contents Online

# Journal of Combinatorics

## Volume 10 (2019)

### Number 1

### Packing without some pieces

Pages: 1 – 25

DOI: https://dx.doi.org/10.4310/JOC.2019.v10.n1.a1

#### Author

#### Abstract

Erdös and Hanani proved that for every fixed integer $k \geq 2$, the complete graph $K_n$ can be almost completely packed with copies of $K_k$; that is, $K_n$ contains pairwise edge-disjoint copies of $K_k$ that cover all but an $o_n(1)$ fraction of its edges. Equivalently, elements of the set $\mathcal{C}(k)$ of all red-blue edge colorings of $K_k$ can be used to almost completely pack every red-blue edge coloring of $K_n$.

The following strengthening of the result of Erdös and Hanani is considered. Suppose $\mathcal{C}^{\prime} \subset \mathcal{C}(k)$. Is it true that we can use elements only from $\mathcal{C}^{\prime}$ and almost completely pack every red-blue edge coloring of $K_n$? An element $\mathcal{C} \in \mathcal{C}(k)$ is *avoidable* if $\mathcal{C}^{\prime} = \mathcal{C}(k) \setminus \mathcal{C}$ has this property and a subset $\mathcal{F} \subset \mathcal{C}(k)$ is avoidable if $\mathcal{C}^{\prime} = \mathcal{C}(k) \setminus \mathcal{F}$ has this property.

It seems difficult to determine all avoidable graphs as well as all avoidable families. We prove some nontrivial sufficient conditions for avoidability. Our proofs imply, in particular, that (i) almost all elements of $\mathcal{C}(k)$ are avoidable (ii) all Eulerian elements of $\mathcal{C}(k)$ are avoidable and, in fact, the set of all Eulerian elements of $\mathcal{C}(k)$ is avoidable.

#### Keywords

packing, avoidable graph, edge-coloring

#### 2010 Mathematics Subject Classification

05C35, 05C70

This research was supported by the Israel Science Foundation (grant No. 1082/16).

Received 22 February 2017

Published 7 December 2018