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Hypergraphs are generalizations of graphs, where edges can be made of more than two vertices.
20
votes
3
answers
987
views
Does the hypergraph of subgroups determine a group?
A hypergraph is a pair $H=(V,E)$ where $V\neq \emptyset$ is a set and $E\subseteq{\cal P}(V)$ is a collection of subsets of $V$. …
8
votes
1
answer
694
views
Edge chromatic number of hypergraphs
This is question Selection problem in a collection of non-empty sets with a simplification in criterion 3.
Is there a set $X\neq\emptyset$ and a collection ${\cal F}\subseteq {\cal P}(X)\setminus\{\e …
8
votes
1
answer
478
views
Scheduling "parent talks" at school
Real life motivation. In my younger son's class, there are $18$ students. His teacher provided $18$ time slots for the parents of each child to have a 30-minute conversation of their kid's progress in …
7
votes
2
answers
382
views
Infinite projective plane with small edges
Let $\kappa$ be an infinite cardinal. We say $E\subseteq {\cal P}(\kappa)$ is an infinite projective plane on $\kappa$ if
$e_1\neq e_2\in E$ implies $|e_1\cap e_2| = 1$, and
whenever $n\neq m\in \ka …
6
votes
1
answer
285
views
Singular cardinal $\kappa$ with projective plane such that all edges have cardinality $<\kappa$
Is there an infinite singular cardinal $\kappa$ such that there is a set $E\subseteq{\cal P}(\kappa)$ with the following properties?
$|e| < \kappa$ for all $e\in E$,
whenever $\alpha\neq\beta\in \kap …
6
votes
1
answer
186
views
Chromatic number of a connected Hausdorff space
(Note that this coloring notion comes from hypergraph coloring.) …
6
votes
1
answer
110
views
Tameable hypergraphs
Let $H=(V,E)$ be a hypergraph. We say that $I\subseteq V$ is an independent set if $e\not\subseteq I$ for all $e\in E$. … Every graph is tameable and, more generally, so is every hypergraph with finite edges. …
6
votes
0
answers
128
views
Edge-coloring number of a linear hypergraph
A linear hypergraph is a hypergraph $H=(V,E)$ such that
$|e|\geq 2$ for all $e\in E$,
$|e_1\cap e_2|\leq 1$ for all $e_1, e_2\in E$ with $e_1\neq e_2$. … If $H=(V,E)$ is a linear hypergraph such that $|V| = 2n$ for some positive $n\in\mathbb{N}$ and edge coloring number $|V|$, does this imply that $H$ is a near-pencil? …
6
votes
3
answers
228
views
Refinement-minimal intersecting covers
Motivation. Yesterday I was sitting idly in the train, contemplating the train network. I noticed that a lot of lines (not all) intersected, and some pairs of lines intersected in quite a few stations …
5
votes
0
answers
166
views
Cardinals realizable by the chromatic number of a regular hypergraph
If $H=(V,E)$ is a hypergraph, and $\kappa$ is a cardinal, we say that a map $c:V\to \kappa$ is a coloring if for every $e\in E$ with $|e|\geq 2$ the restriction $c|_e$ is non-constant. …
5
votes
1
answer
210
views
Is the category of hypergraphs cartesian-closed?
If $H_i = (V_i, E_i)$ for $i=1,2$ are hypergraphs then a map $f:V_1\to V_2$ is said to be a hypergraph homomorphism if $f(e_1)\in E_2$ for all $e_1\in E_1$. … Hypergraphs together with hypergraph homomorphisms form a category. Is this category cartesian closed? …
5
votes
1
answer
80
views
Sizes of matchings and transversals in hypergraphs
Let $H=(V,E)$ be a hypergraph. We call $H$ proper if $E\neq\emptyset, \emptyset \notin E$ and for no $e_1\neq e_2\in E$ we have $e_1\subseteq e_2$. … Given infinite cardinals $\alpha < \beta$ is there a proper hypergraph $H=(V,E)$ with the following properties? …
5
votes
1
answer
397
views
4-color theorem for hypergraphs
Does every hypergraph that does not admit a complete minor with $5$ elements have a coloring with $4$ colors?
Below are the definitions to make this precise. … If $H = (V, E)$ is a hypergraph and $W \subseteq V$, then we let the induced sub-hypergraph of $W$ be $H|_W := (W, E|_W)$, where $E|_W := \{e \cap W: e \in E \text{ and }e\cap W \neq \emptyset\}$. …
4
votes
1
answer
159
views
Are two "perfectly dense" hypergraphs on $\mathbb{N}$ necessarily isomorphic?
We say that a hypergraph $(\mathbb{N}, E)$ where $E\subseteq {\cal P}(\mathbb N)$ is perfectly dense if
$\mathbb{N}\notin E$,
all $e\in E$ are infinite,
$e_1, e_2 \in E$ implies $|e_1\cap e_2| = 1$ …
4
votes
0
answers
156
views
Does $\mathbb{R}$ have a partite subbase?
Let $H = (V,E)$ be a hypergraph with $V \neq \varnothing$ and $\bigcup E = V$. … It is easy to see that every edge in a partite hypergraph $H$ has the same cardinality $\kappa$, and that $\kappa$ is also the cardinality of every splitting partition of $H$. …