71 reputation
6
bio website giovanniviglietta.com
location Ottawa, Canada
age 32
visits member for 1 year, 6 months
seen Sep 30 at 7:55
Research Associate at Carleton University, Ottawa, working on distributed algorithms and computational geometry.

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awarded  Self-Learner
May
8
awarded  Teacher
May
8
revised Hardness of approximation of Dominating Set
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May
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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
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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
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revised Hardness of approximation of Dominating Set
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May
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revised Hardness of approximation of Dominating Set
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awarded  Autobiographer
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awarded  Editor
May
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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