---
tag: SVP75YY386
kind: conjecture
name: Hadwiger covering conjecture
msc: [52C17, 52A20]
uses: [437FSG3419, 0ZG1PNEP35]
about: [437FSG3419]
sources:
  - url: https://en.wikipedia.org/wiki/Hadwiger_conjecture_(combinatorial_geometry)
same_as:
  wikidata: Q5638115
  wikipedia: Hadwiger conjecture (combinatorial geometry)
history:
  posed_by: "Hugo Hadwiger"
  first_stated: "1957"
---

## Statement

Let $n \ge 1$, and let $K \subseteq \R^n$ be a [[convex body|437FSG3419]].
Then there are $\lambda \in (0, 1)$ and points $x_1, \ldots, x_{2^n} \in \R^n$ with
$$K \subseteq \bigcup_{i=1}^{2^n} (x_i + \lambda K),$$
that is, $K$ can be covered by $2^n$ smaller [[homothetic|0ZG1PNEP35]] copies of itself.
Moreover, $2^n$ copies are needed only when $K$ is a parallelotope.

## Remarks

The conjecture is open for $n \ge 3$; Levi proved the case $n = 2$.
The cube $[0, 1]^n$ shows that $2^n$ cannot be lowered.
A translate of $\lambda [0, 1]^n$ with $\lambda < 1$ has all side lengths below $1$, so it contains at most one of the $2^n$ vertices.
General upper bounds of exponential order, about $4^n$ times a polynomial factor in $n$, are known.
Boltyanski showed that the conjecture is equivalent to a statement about illuminating the boundary of $K$ by $2^n$ directions.

## Explanations

### L1 human

The conjecture says that any convex shape in $n$-dimensional space can be covered by $2^n$ slightly shrunk copies of itself.
In the plane four copies are enough, and a square really needs four, one for each corner.
In space it predicts eight copies, but this is still unproved.