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PCP theorem states that $$PCP(O(\log n),O(1))=NP.$$ Could we not run through $O(\log n)$ bits deterministically?

Does PCP theorem statement mean any set of $O(\log n)$ random bits out of $2^{O(\log n)}$ choices of such bits suffice? If so, why cannot I choose all $0$ bits? If not why not?

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    $\begingroup$ Make sure you understand the definition of $PCP(\_,\_)$! $\endgroup$ – Raphael Apr 24 '15 at 5:47
  • $\begingroup$ You're not making any sense. Please read a good survey on PCPs before asking any questions on them. $\endgroup$ – Yuval Filmus Apr 24 '15 at 13:33
  • $\begingroup$ if the number of random bits is small you can indeed usually run through all possibilities while staying in polynomial time. However in this case, the random bits are used to access the certificate, which we don't want to access n times. $\endgroup$ – Albert Hendriks Jan 4 '18 at 8:34
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The statement of the PCP theorem is essentially equivalent to the following statement:

For every problem in NP there is a reduction $f$ to $O(1)$-FUNCTION-SAT such that if $x$ is a YES instance then $f(x)$ is satisfiable, whereas if $x$ is a NO instance then at most half of the clauses of $f(x)$ can be satisfied.

Here an instance of $k$-FUNCTION-SAT consists of a collection of "clauses", each an arbitrary function on $k$ inputs.

Where did the parameters $O(\log n),O(1)$ in the PCP theorem disappear? The first parameter is the logarithm of the number of clauses, so there are $2^{O(\log n)}$ clauses; the second parameter is the arity of the clauses. Together, these parameters guarantee that the FUNCTION-SAT instance has polynomial size, and so the reduction runs in polynomial time.

PCPs are not always described this way, but the descriptions are equivalent. Proof of the equivalence left to the reader.

You are enumerating over all choices of $O(\log n)$ when you are constructing the FUNCTION-SAT instance. But it only allows you to construct the instance, not to tell whether it is satisfiable or not; that is an NP task.

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  • $\begingroup$ So log n bits index clause location. correct? $\endgroup$ – T.... Apr 24 '15 at 12:47
  • $\begingroup$ If we fix T=10^10..^10^10 locations, the fraction of SAT problems where soundness cannot be guaranteed is less than 0.5^T. Soundness per SAT problem cannot be guaranteed however soundness in average sense seems to holds true? $\endgroup$ – T.... Apr 24 '15 at 12:54
  • $\begingroup$ I don't follow what you're saying, but rest assured that either it doesn't quite make sense, of it has been thought of. As a start, I suggest you read some nice survey on PCPs to make sure that you understand the terminology. If you still have questions after you do that, you can ask them again here, in a more informed way. $\endgroup$ – Yuval Filmus Apr 24 '15 at 13:32
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The first parameter is the number of random bits, and the second the number of queries to the oracle string $\Pi$ the verifier may use.

The length of $\Pi$ is not bounded; therefore, having only access to $O(1)$ symbols of $\Pi$ (and being allowed only relatively few random bits for choosing which) is a non-trivial restriction.

See here for an example illustrating this.

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  • $\begingroup$ I understand that. My question is what difference would it make if instead of using $O(\log n)$ bits, I just pick $0000..0000$ as $O(\log n)$ bits? $\endgroup$ – T.... Apr 24 '15 at 10:21
  • $\begingroup$ @Turbo That would derandomize the whole thing, which will cause everything to not work. The example I link illustrates this (in "ad 3"); if we don't pick the position to check randomly, we don't get the correct acceptance/reject probabilities and the algorithm is not of the required type. $\endgroup$ – Raphael Apr 24 '15 at 10:33
  • $\begingroup$ So even if we pick 10^10^10^10...^10 positions apriori, it will be useless? $\endgroup$ – T.... Apr 24 '15 at 10:41
  • $\begingroup$ If I understand everything correctly (I'm by no means an expert in this), yes. That's true for any probabilistic algorithm. $\endgroup$ – Raphael Apr 24 '15 at 10:46
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there are multiple comments stating the question is "uninformed" etc and makes no logical sense and suggesting "read a survey on PCP" but not actually (helpfully) volunteering/ citing any. here is a nice one run across by Ryan O'Donnell an expert on boolean functions. its more of an overview but states the basics, possibly enough to remove some beginner misconceptions, and points to many other papers with more details. consider this something like a "hint" answer.

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