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Jun 1, 2018 at 22:20 history bounty ended geogeek
Jun 1, 2018 at 16:15 comment added Nat Let us continue this discussion in chat.
Jun 1, 2018 at 16:08 comment added Ami Tavory @Nat Clearly, $x, y, z$ - this is stated in the question.
Jun 1, 2018 at 15:51 comment added Nat The set cover problem is to identify the smallest sub-collection of $S$. What is the collection of sets $S$ in this example?
Jun 1, 2018 at 15:39 comment added Ami Tavory @Nat E.g., $x = \{a, b, c\}$ and $U = \{a, b, c, d\}$.
Jun 1, 2018 at 15:36 comment added Nat The union of $\left\{a, b, c\right\}$ and $\left\{d\right\}$ is equal to the $\left\{a, b, c, d\right\}$, but the union of $x$ and $z$ is not $\left\{x, y, z\right\}$. Therefore, I was asking that you clarify what is $U$ and what are your sets.
Jun 1, 2018 at 15:34 vote accept geogeek
Jun 1, 2018 at 15:17 comment added Ami Tavory I think you've misunderstood the question. Aside from some curious choices about active and passive tenses in the phrasing, the question is exactly set cover, "So the minimum number of groups that could cover the whole data set are 2" shows this, and the union of $x$ and $z$ is indeed $U$, as you can see from the start of the question ("a belongs to the classes...").
Jun 1, 2018 at 14:44 comment added Nat Can you clarify how the example in the question is a set-cover problem? It appears the universe is $U=\left\{x, y, z\right\}$ and the collection of sets is $S=\left\{\left\{x,y,z\right\}, \left\{x,y\right\}, \left\{x\right\}, \left\{z\right\}\right\}$. The solution to the example is $x$ and $z$. The union of $x$ and $z$ does not equal the universe $U$.
May 25, 2018 at 17:35 history answered Ami Tavory CC BY-SA 4.0