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From: tchow on 15 Jun 2010 15:03 In article <d8bd6348-2009-4651-8d2f-d5e8a6d38634(a)c33g2000yqm.googlegroups.com>, Sadeq <msdousti(a)gmail.com> wrote: >> Ah, I understand your confusion now. �When people talk about "relativizing >> NP^NP to the oracle C" they mean "NP^(NP^C)" and not "(NP^NP)^C" since >> that's the definition that makes sense. � > >I'm afraid that is not the case. Here's a snippet of the >aforementioned paper: >http://i50.tinypic.com/110kqrd.png > >As you can see, the base is \Sigma_2 \cap \Pi_2. The notation for oracles is terribly confusing. When people write \Sigma_2 ^ C, they mean \Sigma_2 relativized to the oracle C, which means NP^(NP^C). If you think about it more closely, this *has* to be the right way to define relativization. What are we doing when we relativize a complexity class anyway? We're assuming that computational entities have access to a particular oracle. In a relativized world where access to an oracle C is freely available, what is the meaning of something like P^NP? It must mean things that you can compute in polynomial time given access to C and to an NP^C-complete language. NP^C already contains C so the relativized version of P^NP must be P^(NP^C). This is the only definition that will let you relativize arguments smoothly. >>"(NP^NP)^C" is not a well-behaved entity. >Again, I beg to differ. In an earlier post, you said that NP^NP is a >semantic class. I object, since the class has complete problems. I should have stated this more carefully. NP^NP is a syntactic class, but the way you were proposing to define (NP^NP)^C was treating it as a semantic class. You were talking about the class of machines that recognizes the language NP^NP. This is not a good way to define a class. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
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