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May
8
awarded  Self-Learner
May
8
awarded  Teacher
May
8
revised Hardness of approximation of Dominating Set
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May
8
awarded  Scholar
May
8
accepted Hardness of approximation of Dominating Set
May
8
revised Hardness of approximation of Dominating Set
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May
8
revised Hardness of approximation of Dominating Set
added 613 characters in body
May
8
answered Hardness of approximation of Dominating Set
May
7
comment Hardness of approximation of Dominating Set
I agree this is sort of confusing. I guess this is why people tend to say nonchalantly that Set Cover and Dominating Set are equivalent as approximation problems (because they L-reduce to each other), and THEREFORE Dominating Set is not approximable within $\Omega(\log n)$, either. Well, these are two different $n$'s, so we should pay attention...
May
7
revised Hardness of approximation of Dominating Set
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May
7
revised Hardness of approximation of Dominating Set
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May
7
awarded  Autobiographer
May
7
awarded  Editor
May
7
revised Hardness of approximation of Dominating Set
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May
7
answered Hardness of approximation of Dominating Set
May
7
comment Hardness of approximation of Dominating Set
Using my notation, the problem size of Set Cover is $m\cdot n$. This is correct, but all the approximation bounds are always given just in terms of $n$. That is, there is a greedy algorithm that achieves a $\ln n$ approximation ratio (no $m$ involved), and it is NP-hard to achieve a $c\cdot\log n$ approximation ratio (no $m$ involved). Hence, when reducing to Dominating Set, $n$ cannot be the number of vertices, but should be its logarithm.
May
6
awarded  Student
May
6
asked Hardness of approximation of Dominating Set