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As there are no strings in $\emptyset$, any machine that computes it always rejects, so we can't map Yes-instance of other problems to anything. Similarly for $\Sigma^{\ast}$ there's nothing to map No …

There are at least a few reasons that NPC is interesting: The class NP contains many problems that are interesting (both practically and theoretically), moreover many of these problems turn out to b …

The "parent" of the problem you're looking at is sometimes called Weighted Satisfiability (WSAT, particularly in parameterized complexity) or Min-Ones (though this is normally an optimization version, …

One reason that we see different approximation complexities for NP-complete problems is that the necessary conditions for NP-complete constitute a very coarse grained measure of a problem's complexity …

Your intuition about "relative hardness" is correct, the underlying mathematics is why III. is true. However your justification about I. is a little off (not wrong, but the reasoning is possibly not a …

Computational complexity is just a more general term, as time is not the only resource we might want to consider. The next most obvious is the space that an algorithm uses, and hence we can talk about …

There is actually a stronger result; A problem is in the class $\mathrm{FPTAS}$ if it has an fptas1: an $\varepsilon$-approximation running in time bounded by $(n+\frac{1}{\varepsilon})^{\mathcal{O}(1 …

Unless I'm missing something, it's trivially in P as the length of the formula is exponential in the number of variables. Hence all $2^{n}$ truth assignments can be generated and checked in polynomial …

There's a similar question over on cstheory, with lots of examples ranging from the "realistically impractically slow" algorithms with exponents of 6 or 7 upwards. That question also discusses large c …

In the most formal sense, the size of the input is measured in reference to a Turing Machine implementation of the algorithm, and it is the number of alphabet symbols needed to encode the input. This …

Only in one direction. As $\mathsf{P}=\text{co-}\mathsf{P}$, if $\mathsf{NP}\neq\text{co-}\mathsf{NP}$ then we would know that $\mathsf{P}\neq\mathsf{NP}$. However the reverse implication doesn't hold …

I would be very dubious that it has (in the sense of the proof of existence of a polynomial time algorithm). While it is not impossible that the paper is correct, there are a number of warning signs: …

The short way to say it is that these problems are $\mathcal{NP}$-complete. Of course this only has meaning to those who understand what $\mathcal{NP}$-complete means. Not only does this say these …

If you're not absolutely set on the normal SAT, your idea is already a reduction to MIN-ONES (over positive CNF formulae), which is basically SAT, but where you can set at most $k$ variables to true ( …

The (very) simplified version is that they convert any verification algorithm $A$ for a language $L \in \text{NP}$ into a circuit. What they end up with is a circuit $C$ that, given a (binary) string …