I knitted a scarf! It is the “Oak Park” pattern on Ravelry.

Well, it’s senior year, which means I have a thesis! It’s about the Baranyai Theorem. Here is a little bit of introduction… look at this post on my actual blog (instead of the dash) to see the full glory of LaTex in action.

Let \(r \in \mathbb{N}\). If \(X\) is a set, then we will denote the set of all \(r\)-subsets (subsets of size \(r\)) by \(X^{(r)}\).

A hypergraph (on \(V\))} is a pair \((V,E)\) where \(V\) is a set and \(E \subset \mathcal{P}(V)\) (where \(\mathcal{P}(V)\) denotes the power set of \(V\)).

A 1-factor of a hypergraph \(H=(V,E)\) is a set of edges \(E’ \subset E\) such that every vertex \(v \in V\) belongs to exactly one of the \(e \in E’\). A factorization of \(H\) is a partition of \(E\) into 1-factors.

Denote the complete \(r\)-graph of order \(n\) with \(K_n^{(r)}\). Then \(K_n^{(r)} = (V, V^{(r)})\), where \(|V| = n\).

Note that if \(K_n^{(r)}\) is factorizable, then \(r \mid n\). It turns out that this necessary condition is also sufficient. This assertion had been around in a vague form for more than 100 years before it was proved by Baranyai in 1975. The theorem states that if \(K_n^{(r)}\) is factorizable if and only if \(r \mid n\).

Here is a factorization of \(K_6^{(2)}\), where the factors are represented by different colors.

"It occurred to me that we could completely solve the Oxford Comma controversy just by always using set notation."

Modern Geometry professor (via mathprofessorquotes)