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Here is a short proof. Thanks to David Speyer for simplifying an earlier proof of mine (see the comments below). For each colour $i \in [n]$, let $a_i$ be the number of vertices incident to an edge of …

Pilipczuk and Siebertz proved that every planar graph has such a partition with an even stronger property. Namely, each part $V_i$ is a geodesic path, and the graph obtained by contracting each part …

This becomes true at $a = \frac{1}{3}$. Claim. If $G$ is an $n$-vertex bipartite graph such that $\delta(G) \geq \frac{1}{3}n$, then $\delta(G)=\kappa(G)$. Proof. Let $(A,B)$ be the bipartition of $G$ …

The answer to Question I is yes. It is true that if a bipartite graph has a unique perfect matching, then the biadjacency matrix can be permuted to be lower triangular. This was proved by Chris Godsi …

Let $G=(U,V,E)$ be a bipartite graph where $U=[n], V=\binom{[n]}{3}$, and there is an edge between $u \in U$ and $v \in V$ if and only if $u \in v$. Then $\deg(u)=\binom{n-1}{2}$ for all $u \in V$ an …

This is the weighted MAX CUT problem, and it is NP-hard to compute exactly. Note that the case of $\{0,1\}$-weights corresponds to computing a MAX CUT in an arbitrary graph. This later problem has a …

Edit. My previous upper bound was not correct. Thanks to Gilad for pointing that out. If $m<k$, then of course it is not possible. Otherwise, for an upper bound start with a matching $M$ saturating …

In the case that $2d$ divides $n$, one can take $G$ to be $\frac{n}{2d}$ disjoint copies of $K_{d,d}$. This graph has $(d!)^{n/2d}$ perfect matchings, and as Gjergji's answer shows, this is the worst …

If one instead starts with the definition of bipartite as not containing any odd cycles, then there are results for hypergraphs in this direction. Indeed we can define a cycle in a hypergraph as a se …