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Well, all graphs are colorable with $\geq k$ colors: Assign each vertex a different color (and leave the other color classes empty if $k$ is larger than the number of vertices of the given graph). As …

Yes. Let $L$ be any undecidable language (e.g. the halting problem), w.l.o.g. encoded over the alphabet $\{0, 1\}$ and such that all $x \in L$ start with $1$. Further, let $\leq_\text{lex}$ denote the …

Suppose $f(n) = 2^n$. Then $f(n / 2) = 2^{n / 2} = \sqrt{2^n}$ and setting e.g. $g(n) = 2^n$ as well shows $f(n) \in \Theta(g(n))$ but $f(n / 2) \notin \Theta(g(n))$, so this does not hold in general. …

For your first question, I'll give an elementary proof for the $\mathsf{NP}$-hardness of $\mathrm{MSAT}$ by reduction from $\mathrm{SAT}$. Let $\varphi$ be a propositional formula in CNF (i.e., a $\ma …

The output size does not scale with (time) complexity of problems in general. In fact, complexity theorists are largely interested in decision problems for which the output will either be 0 or 1.

You can map any $\text{HamiltonianCycle}$ instance $G$ to an $L$-instance $(G, 1)$, yielding a polynomial time reduction from an $\mathsf{NP}$-complete problem to $L$ which implies the $\mathsf{NP}$-h …

Your $M_1$ is not a polynomial time NTM unless $\mathsf{NP} = \mathsf{coNP}$, I think. In the last step (checking that $G_{ij}$ contains no HC), you essentially want to decide the complement of HC, wh …

The reason for why you are having problems is because your solution does not work. In particular, the problem is that your formula does not capture the VC problem statement. More precisely, the formul …

Consider your binary tree as a binary search tree on the integers $1$ through $n$ to find that the answer for some given list $L \subseteq [1, n]$ will be $\{\min L, \max L\}$. In case you are not giv …

Usually, when we talk about the complement of a set we have some reference set to compare to. In the setting of languages over some alphabet $\Sigma$, this means that the complement of some language $ …

Welcome to the site! A Karp reduction from $\mathrm{Hamiltonian Cycle}$ could work as follows: Given a graph $G = (V, E)$, we construct $\mathrm{Long Path}$-instance $(G', v, v', |V|)$ where $G'$ is o …

The statement is correct for exactly the reason you started with: $\mathsf P$ is a subset of $\mathsf{NP}$ which means that every problem in $\mathsf P$ is also in $\mathsf{NP}$. You can also go throu …

We give a Turing reduction from the $\mathrm{SubsetSum}$ problem. Suppose we are given a $\mathrm{SubsetSum}$ instance $(A, k)$ where w.l.o.g. $A$ only contains positive integers, i.e. we want to find …

There are many ways to encode Turing machines. Essentially, this is possible because a TM $M$ is specified by a finite amount of information just like a computer program is completely specified by its …

There is a simple reduction from the Clique problem, i.e. given some (undirected, simple) graph $G = (V, E)$ and some number $k$ we are asked whether or not $G$ contains a clique of size at least $k$. …