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Questions about decision problems that can be solved on nondeterministic Turing machines in time polynomial in the length of the input.
1
vote
Accepted
A NP-complete problem with difficult NP membership proof
Integer Linear Programming is such a problem.
Input: two matrices $A\in \mathcal{M}_n(\mathbb{Z})$ and $B\in \mathcal{M}_{n,1}(\mathbb{Z})$.
Question: is there a matrix $X\in \mathcal{M}_{n,1}(\mathb …
1
vote
Reduction from Clique to IS degree at most 4
As you stated, this is not a correct reduction, because you would need to solve Clique for any graph (or prove that Clique with graphs of degree at most 4 is $NP$-complete). … I think the intention of the author for solving this exercice is to use the remark stated at the beginning of the exercice (which you cropped): "$\textsf{3SAT}$ remains $\textsf{NP}$-complete even when …
1
vote
Accepted
K-Path-Problem is in $P$ or $NPC$
This problem is $\textsf{NP}$-complete. This is more or less the longest path problem. …
1
vote
Accepted
NP-Complete Reduction
I think a reduction from $\texttt{Directed Hamiltonian Path}$ (not cycle) would work quite well.
Given a digraph $G = (V, E)$ where $V = \{v_1, …, v_n\}$, consider the digraph $G' = (V', E')$ where:
…
1
vote
Accepted
Practicability of enumerating all Turing machines with fewer than $s$ states to find a poten...
There are a lot of Turing machines, so no, this idea is completely unpractical (I don't think there is really a loophole, since you are considering many restrictions). For example, there are roughly $ …
2
votes
Accepted
Is the traveling salesman on a map NP-hard?
It was proved to be NP-complete by Garey, Graham and Johnson in 1976 and Papadimitriou in 1977.
Source: Computers and Intractability (Garey and Johnson) …
2
votes
Accepted
Show that $PLANAR \in co-NP \cap NP$
Such an embedding is a $\mathsf{NP}$ certificate and is of polynomial size. It can be verified in polynomial time that no two edges cross. …
1
vote
$L_1= (1$ { $0, 1$ }$^∗) \cup ${ $0x | x \in L$} is NP- complete
I am sure that you can prove that $L_1\in \mathsf{NP}$. …
2
votes
Accepted
If A reduces to B and B reduces to C, does that mean A reduces to C?
$A\in \mathsf{NP}$ because $\mathsf{P}\subset \mathsf{NP}$. … {NP}$-complete and $B\leqslant_m^p C$. …
3
votes
Why does nondeterministically estimating a number have exponential time complexity?
That means that it is solving $\overline{\texttt{PRIME}}$ (the complement problem), and shows that $\texttt{PRIME}\in \text{co}\mathsf{NP}$, not that $\texttt{PRIME}\in \mathsf{NP}$. …
1
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Clarification for binary search in solving optimal TSP when a polynomial algorithm with a bu...
A $NP$-complete problem is a $NP$-hard problem which also needs to be in $NP$. So when speaking about a $NP$-complete problem, it is not false to state that it is $NP$-hard. … However, $NP$ is a set of decision problems, therefore, TSP-OPT being an optimization problem cannot be in $NP$. …
1
vote
How to reduce the hamiltonian path problem to 1/2 hamiltonian path problem
You just need to add $n$ (where $n$ is the order of the graph) vertices with no additional edges.
2
votes
How can I prove the following problem is NP complete?
Since knapsack is $NP$-complete, this proves the $NP$-hardness of your problem. …
0
votes
Accepted
Why is SET PACKING in NP?
Sure, $\mathsf{NP}$ is defined considering the executing time of a Turing machine, however we, human beings, are not made to think in TM. …
1
vote
Karp's reduction strategy
$(3)\Rightarrow (1)$ should just be a consequence of the definition of $\textsf{NP}$-completeness. …