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Consider a graph on vertex set $V_1\cup V_2\cup V_3\cup V_4\cup \{a,b,c,d\}$ where $|V_1|=|V_2|=|V_3|=|V_4|=n$. The edge set $E$ is covered by $C=\{V_1\cup\{a,c\},V_2\cup\{a,d\},V_3\cup\{b,c\},V_4\cup\{b,d\},\{a,b\},\{c,d\}\}$. When $n$ is large enough, any minimalist cover must contain the four maximum cliques $V_1\cup\{a,c\}$ and so on, so it is not hard ...


3

...aha, found it! This is the bipartite dimension problem, and yes it is NP-hard without further assumptions. Previously asked here: https://cs.stackexchange.com/questions/49266/finding-a-minimal-cover-of-a-subset-of-a-finite-cartesian-product-by-cartesian-p


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An $r$-regular expander should do it. The following is a simple observation that I first saw in Li (arXiv:2106.05513): if an $r$-regular graph has conductance $\phi$, then the smaller side $S$ of a minimum cut contains at most $|S| \leq 1/\phi$ vertices. Indeed, by definition of conductance we have that $|E(S,S^c)| \geq \phi r |S|$. Since this defines a ...


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How about the following counter-example? $m=2$, $n=4$. $A=\{a_1,a_2\}$, $B=\{b_1,b_2\}$, $C=\{c_1,c_2\}$. $T=\{(a_1,b_1,c_1), (a_1,b_2,c_2), (a_2,b_1,c_2), (a_2,b_2,c_1)\}$. With the partition $A\cup B\cup C = \{a_1,b_1\}\cup\{a_2,b_2\}\cup\{c_1\}\cup\{c_2\}$.


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My two cents: The worst case of building $G_\top$ is in $\Omega(n^2)$ time and space: assume $\bot$ contains a single node linked to all nodes in $\top$. Maybe you are not looking for a worst case complexity? Then, $O(\sum_{u\in\bot}(d_u)^2)$ time to build $G_\top$ by listing all edges $u,v$ such that $u$ and $v$ are neighbors of the same node in $\bot$. ...


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