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Here is an approximation algorithm that finds vertex cover of a graph.

 C = {}
 E' = {Edge set}
 while E' =/ 0
   Let (u,v) be an arbitrarily edge of E'
    C = C U {u,v}
    remove E' incident on u and v.
return C

A variant: what if instead of removing edges incident on both $u$ and $v$, we removed only $u$. Would this affect the optimal vertex cover? If so, how?

I somehow feel the optimal vertex cover remains the same whereas only the number of steps to remove the edges would increase increasing time and space complexity. Am I right?

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  • $\begingroup$ What do you think? Have you tried running your modification on some examples? $\endgroup$ – Yuval Filmus Nov 15 '18 at 9:09
  • $\begingroup$ Nothing that the algorithm does can affect the optimal vertex cover. Are you interested in the approximation ratio of the algorithm, perchance? $\endgroup$ – Yuval Filmus Nov 15 '18 at 9:11
  • $\begingroup$ i wanted to check the quality of the algorithm? $\endgroup$ – khushboo shah Nov 15 '18 at 19:35
  • $\begingroup$ This “quality” is known as the approximation ratio. $\endgroup$ – Yuval Filmus Nov 15 '18 at 20:28
  • $\begingroup$ What is an approximation ratio?And how to calculate? $\endgroup$ – khushboo shah Nov 15 '18 at 20:33
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Consider a star consisting of a center $x$ connected to the vertices $y_1,\ldots,y_n$. Your modified algorithm could act as follows:

  1. Choose the edge $(x,y_1)$; add $x,y_1$ to the vertex cover; remove all edges incident to $y_1$.
  2. Choose the edge $(x,y_2)$; add $x,y_2$ to the vertex cover; remove all edges incident to $y_2$.
  3. ...

I'll let you take it from here.

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