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Does the following approximation algorithm for vertex cover also have an approximation ratio of 2? Why? Why not?

Input: $G = (V,E)$

  1. Set $C \gets \emptyset$ and $E' \gets E$.
  2. while $E' \neq \emptyset$ do:
    • Pick any edge $(u,v) \in E'$.
    • $C \gets C \cup \{u\}$.
    • Remove from $E'$ all edges incident to $u$.
  3. return $C$.
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    – John L.
    Apr 29, 2020 at 23:24
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    – John L.
    May 15, 2020 at 23:16

1 Answer 1

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No, this is a terrible algorithm. Consider a star graph: the vertices are $x,y_1,\ldots,y_n$, and the edges are $(x,y_1),\ldots,(x,y_n)$. Your algorithm goes over the edges in an arbitrary order, and always picks the $y$ vertices. It ends up with $n$ vertices rather than the optimal $1$ vertex solution.

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