Search Results

$NP$ however is a class of decision problems, so the only valid outputs are Yes and No. … It should be easier to see that this version is in $NP$. …

So we can't have an $\mathrm{NP}$-complete algorithm, only an $\mathrm{NP}$-complete problem. … On the flip side, we can have an algorithm for an $\mathrm{NP}$ problem that is not the one that shows that the problem is in $\mathrm{NP}$. …

To prove containment in NP (in this manner) you need (as you mention): A witness. This is just a "string" that proves an instance is a Yes-instance. So it can just be the solution. … For NP, we have the additional restriction that the verifier has to run in deterministic polynomial time. So for this problem we can represent a solution as a collection of sets of boxes. …

Given a graph $G$, we want to do two things: Find the smallest $k$ such that $G$ is $k$-colourable, Given such a $k$, find a colouring of $G$ with $k$ colours. If we have an oracle for the decisio …

What is normally meant in cases like this is there is a simple, obvious algorithm which demonstrates that the problem is in $\mathcal{NP}$, with an implicit but we don't have enough space, … Before returning to the Steiner Tree problem, recall the two common definitions of $\mathcal{NP}$ membership: A problem is in $\mathcal{NP}$ if, given the input and a solution, we can check that the …

Of course you have to show that the problem is in NP, but from other posts I gather that's no issue. The sticking point lies with the reduction from an NP-hard problem. … Hamiltonian Cycle is NP-hard, we map that to a subproblem of the Wheel Subgraph Problem (where $n=k$), so this is also NP-hard, and hence the Wheel Subgraph Problem is in general NP-hard (as one of its …

Every $\mathsf{NP}$-complete problem is polynomial-time many-one reducible to every other $\mathsf{NP}$-complete problem, so the two reducibility conditions hold trivially. … If $\mathsf{P} \neq \mathsf{NP}$, then $B$ can't be in $\mathsf{P}$, and if $\mathsf{NP} \neq co\mathsf{NP}$, then $C$ can't be in $co\mathsf{NP}$. …

Let $(G,k)$ be an instance of Near-Clique, construct an instance $(G',k')$ of Clique as follows. For each $uv \notin E(G)$, add to $G'$ a subgraph corresponding to $(N(u)\cap N(v))$, i.e., for eac …

I know of no compendium that is completely satisfactory in any regard. It seems to be too much of a thankless task maintaining these things, thus I suspect that broad answer to your question is, unfor …

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. … You may be familiar with the basic definition of a problem $\Pi$ being NP-complete: $\Pi$ is in NP, and For every other problem $\Xi$ in NP, we can turn an instance $x$ of $\Xi$ into an instance $y$ …

Recall that for a problem $\Pi$ to be $\mathsf{NP}$-complete we need the following: $\Pi \in \mathsf{NP}$. … However we know unconditionally that $\mathsf{NP}\neq \mathsf{NEXP}$, so SUCCINCT 3-SAT is not in $\mathsf{NP}$. …

The easiest way (for me) to understand co-NP is as the class of problems where certificates for "No" answers can be quickly verified (i.e. certificates of non-membership). … So if we look at NP as a class of nondeterministic Turing Machines which quickly (in polynomial time) determine that their input satisfies some property $\Pi$, co-NP is the class of nondeterministic Turing …

NP-completeness is NP-hardness, plus the additional property that the problem is in NP. So saying problem A is NP-complete means problem A is NP-hard and A is in NP. … us whether B is in NP or not. …

Working from the "machine" definition (that a problem is in $\mathcal{NP}$ if and only if it can be decided by a non-deterministic Turing Machine in polynomial time), we can take a similar approach as … Given a problem $P \in \mathcal{NP}$, there is, by definition, a non-deterministic Turing Machine $M_{P}$ that correctly decides whether $x \in P$ or not in time bounded by $|x|^{c}$ for every $x \in \ …

Although it's split into lots of classes, the $W$-hierachy bears a practical relationship to $FPT$ in a similar manner to that of $NP$ to $P$, in that if a problem is $W[t]$-hard for some $t$, then it … has no $FPT$ algorithm unless $W[t] = FPT$, and we conjecture that $FPT \subset W[1]$ for similar reasons to $P \subset NP$. …