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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.
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how to show that if $L'\in \text{P}\iff \text{P}=\text{NP}?$
If $L'\in \text{NP}$ and for all $L\in \text{NP}$ such that $L\leq_p L'.$
My question is, how to show that if $L'\in \text{P}\iff \text{P}=\text{NP}? … $
It's obvious by intuition, since $L$ is in $\text{NP}$, and $L$ polynomially reduces to $L'$, $L'$ also belongs to $L$, altogether implies $\text{P}=\text{NP}$.
But I came looking for formal proof. …
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$\textbf{NP}\neq \textbf{DTIME}(2^{\sqrt{n}})$
I want to prove that $\textbf{NP}\neq \textbf{DTIME}(2^{\sqrt{n}}).$
My thoughts is:
if I try to prove $\textbf{NP}\not\subseteq \textbf{DTIME}(2^{\sqrt{n}})$ would imply $\textbf{NP $\neq$ P}$. … if I try to prove $\textbf{DTIME}(2^{\sqrt{n}})\not\subseteq \textbf{NP}$ would imply $\textbf{NP $\neq$ EXP}$.
I think both are open questions, is there any other way to prove it? …
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Graph contains two disjoint cliques
I have tried to prove it is $\texttt{NP-complete}$.
NP Membership: A candidate solution consists of two sets of vertices: $C_1 \subseteq V$ , with $|C_1| = k_1$ . … $\texttt{NP-hardness}$ Reduction:
To prove $\text{2disjCLIQUEs}$ is $\text{NP-complete}$ , we reduce a known $\text{NP-complete}$ problem to it. A natural candidate is the Clique problem. …