Math-3356 · mathematics · Graph theory

Math-3356

¿Are critical k-forests tight?

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Fix $k\geq2$ and a finite set $V$ with $|V|\geq k$. A $k$-uniform hypergraph is $H=(V,E)$ with $E\subseteq\binom Vk$. A $k$-colouring means a surjection $c:V\to[k]=\{1,\ldots,k\}$. An edge is rainbow when $c(e)=[k]$. The hypergraph is a $k$-forest if each $e\in E$ is the unique rainbow edge for some $k$-colouring. It is critical (saturated) if adding any missing $k$-element edge on the same vertex set destroys this property. It is tight if every surjective $k$-colouring has a rainbow edge; a tight $k$-forest is a $k$-tree.

Statement recorded in the supplied source.
Must every critical $k$-forest be tight, and hence be a $k$-tree? This is the historical conjecture; the cited primary paper gives a saturated $3$-forest on six vertices which is not tight. [S2] [S3]

Short English statement

Does being inclusion-maximal among these hypergraph forests force the corresponding connectivity property? The historical assertion has a six-vertex counterexample.

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.