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In a trivial sense, no, because circuits have a fixed input size. To compute the same function as a Turing machine, you need an infinite family of circuits, one for each input size. So a single circui …

Unless you want to be more rigid in what you allow as a "description", every decidable language has a finite description - precisely the Turing machine that decides it. This Turing machine must exist …

No, a running time like $f(k)n^{g(n,k)}$ is not suitable for any non-constant function $g$. $\mathrm{FPT}$ is the class of problems computable in time $O(f(k)\cdot n^{c})$ for some fixed $c \in \math …

Understanding how Bubble Sort works is of course really the key here. The main part for unraveling this is that Bubble Sort goes back to the start if it ever performs a swap. If it never performs a s …

Most immediately, if they publish it, they would win the Clay Mathematics Institute prize for solving this problem, so there's at least $1 million in it. They would also almost certainly be able to ob …

The basic definitions of computational complexity are normally phrased in terms of plain Turing Machines, so one has to be careful about transferring results to different models. Naturally, if you can …

Yes, and with many different algorithms. See https://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations One important thing to notice is that the size of the input is the num …

Firstly, there is an important part of the quoted statement you are missing - the parameterization. It can only be classified with respect to some parameterization. For example, Dominating Set is $\ma …

One of the things to remember when dealing with natural numbers (and others, but naturals are the central things here) is the encoding, and that the definitions of $P$ and $NP$ reference the length of …

A $\mathrm{No}$ answer means that there is no solution, not that there is a non-solution. So for Subset-Sum this means there is no subset of the input integers that sum to zero, or to put it another …

There's nothing wrong with your solution, the exercise is just easier than you expected. Your analysis correct, apart from missing out the $-1$, so it should be $k^{\mathcal{O}(1)}\cdot\binom{r}{k-1} …

Reading Toda's original paper, it's not clear that a general application of $\cdot$ has meaning, however it may have been extended later. Toda introduces three operators: $\oplus\cdot$, $\mathsf{BP}\ …

Question (1) is a tricky one that I have never seen a good reason for. One suggestion may be that we're much more likely to comprehend and find the answers for simpler problems, the harder ones take s …

Once you get the hang of these, they are quite easy, so the first thing is not to over think things. To prove containment in NP (in this manner) you need (as you mention): A witness. This is just a …

There's two trivial answers: No. For a problem to be NP-complete it must be in NP. To be in NP it must be a decision problem and promise problems aren't decision problems (they don't have to answer …