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Add LAAI3 NP proofs

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Tian Cheng (Riccardo) Xia
2024-03-28 09:56:22 +01:00
parent 692acb2888
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src/languages-and-algorithms-for-ai/module3/sections/_complexity.tex
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@ -132,11 +132,11 @@
\end{description}
\begin{theorem}
\begin{theorem} \label{th:P_NP_EXP_relationship}
$\P \subseteq \NP \subseteq \EXP$.
\begin{proof}
\phantom{}
We have to prove that $\P \subseteq \NP$ and $\NP \subseteq \EXP$:
\begin{description}
\item[$\P \subseteq \NP$)]
Given a language $\mathcal{L} \in \P$, we want to prove that $\mathcal{L} \in \NP$.
@ -192,6 +192,28 @@
\begin{theorem}
The relation $\leq_p$ is a pre-order (i.e. reflexive and transitive).
\begin{proof}
We want to prove that $\leq_p$ is reflexive and transitive:
\begin{descriptionlist}
\item[Reflexive)]
Given a language $\mathcal{L}$, we want to prove that $\mathcal{L} \leq_p \mathcal{L}$.
We have to find a poly-time function $f: \{0, 1\}^* \rightarrow \{0, 1\}^*$ such that:
\[ x \in \mathcal{L} \iff f(x) \in \mathcal{L} \]
We can choose $f$ as the identity function.
\item[Transitive)]
Given the languages $\mathcal{L}, \mathcal{H}, \mathcal{J}$, we want to prove that:
\[ (\mathcal{L} \leq_p \mathcal{H}) \land (\mathcal{H} \leq_p \mathcal{J}) \Rightarrow (\mathcal{L} \leq_p \mathcal{J}) \]
By hypothesis, it holds that $\mathcal{L} \leq_p \mathcal{H}$ and $\mathcal{H} \leq_p \mathcal{J}$.
Therefore, there are two poly-time functions $f, g: \{0, 1\}^* \rightarrow \{0, 1\}^*$ such that:
\[ x \in \mathcal{L} \iff f(x) \in \mathcal{H} \text{ and } y \in \mathcal{H} \iff f(y) \in \mathcal{J} \]
We want to find a poly-time mapping from $\mathcal{L}$ to $\mathcal{J}$. This function can be the composition $(g \circ f)(z) = g(f(z))$.
$(g \circ f)$ is poly-time as $f$ and $g$ are poly-time.
\end{descriptionlist}
\end{proof}
\end{theorem}
\item[\NP-hard] \marginnote{\NP-hard}
@ -206,9 +228,36 @@
\begin{theorem}
\phantom{}
\begin{enumerate}
\item If $\mathcal{L}$ is \NP-hard and $\mathcal{L} \in \P$, then $\P = \NP$.
\item\label{th:np_hard_p} If $\mathcal{L}$ is \NP-hard and $\mathcal{L} \in \P$, then $\P = \NP$.
\item If $\mathcal{L}$ is \NP-complete, then $\mathcal{L} \in \P \iff \P = \NP$.
\end{enumerate}
\begin{proof}
\phantom{}
\begin{enumerate}
\item Let $\mathcal{L}$ be \NP-hard and $\mathcal{L} \in \P$.
We want to prove that $\P = \NP$:
\begin{descriptionlist}
\item[$\P \subseteq \NP$)] Proved in \Cref{th:P_NP_EXP_relationship}.
\item[$\NP \subseteq \P$)]
Let $\mathcal{H}$ be a language in $\NP$.
As $\mathcal{L}$ is \NP-hard, by definition it holds that $\mathcal{H} \leq_p \mathcal{L}$.
Moreover, by hypothesis, it holds that $\mathcal{L} \in \P$.
Therefore, we can conclude that $\mathcal{H} \in \P$ as it can be reduced to a language in $\P$.
\end{descriptionlist}
\item Let $\mathcal{L}$ be \NP-complete.
We want to prove that $\mathcal{L} \in \P \iff \P = \NP$:
\begin{descriptionlist}
\item[$(\mathcal{L} \in \P) \Rightarrow (\P = \NP)$)]
Trivial for \hyperref[th:np_hard_p]{Point 1} as $\mathcal{L}$ is also \NP-hard.
\item[$(\mathcal{L} \in \P) \Leftarrow (\P = \NP)$)]
Let $\P = \NP$.
As $\mathcal{L}$ is \NP-complete, it holds that
$\mathcal{L} \in \NP=\P$.
\end{descriptionlist}
\end{enumerate}
\end{proof}
\end{theorem}
\begin{theorem}