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Nov 7, 2017 at 8:43 history edited Dominic van der Zypen CC BY-SA 3.0
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Jun 15, 2015 at 6:40 history edited Dominic van der Zypen CC BY-SA 3.0
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Jun 15, 2015 at 6:38 comment added Dominic van der Zypen Consider the graph $G$ with $V(G) = \{0,1\}$ and $E(G) = \emptyset$. Then the cover $M:=\{\{0\},\{1\}\}$ is not strongly minimal: Let $K:=\{\{0,1\}\}$. Then we have $|M\setminus K| = 2>1= |K\setminus M|$.
Jun 13, 2015 at 15:55 comment added Alex Ravsky I am sorry, but why a trivial cover of $V(G)$ by one-vertex sets is not strongly minimal? It violates some restrictions? For instance, the family $\text{Ind(G)}$ consists of maximal independent subsets?
Jun 11, 2015 at 6:59 history asked Dominic van der Zypen CC BY-SA 3.0