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【离散数学】MIT 6.042J - Fall 2010 - Note for Lecture 7
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\title{\huge \bfseries Mathematics for Computer Science, MIT 6.042J - Fall 2010 - Note for Lecture 7}
\author{\Large Teddy van Jerry}
\date{\today}
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\chead{MIT 6.042J Lecture 6 - Matching Problems}
\begin{document}
\maketitle
{\Large Lecture 7: Matching Problems}
\section{Matching}
\begin{definition}
Given Graph $G:(V,E)$, a \textbf{matching} is a subgraph of G where every node has edge 1.
\end{definition}
\begin{definition}
A matching is perfect if it has size $\dfrac{|V|}{2}$.
\end{definition}
\begin{definition}
The \textbf{weight} of a matching $\mathfrak{M}$ is the sum of the weights on the edges in $\mathfrak{M}$.
\end{definition}
\begin{definition}
A \textbf{minimum-weight matching} for graph G is a perfect matching for G with the minimum weight.
\end{definition}
\begin{definition}
Given a matching $\mathfrak{M}, x$ and $y$ form a \textbf{rogue couple} if they prefer each other to their mates in $\mathfrak{M}$.
\end{definition}
\begin{definition}
A matching is \textbf{stable} if there are no rogue couples.
\end{definition}
Goal: Find a perfect matching that is stable.
\section{Stable Marriage Problem}
\subsection{Problem}
\begin{itemize}
\item $N$ boys and $N$ girls.
\item Each boy has his own ranked list of all the girls.
\item Each girl has his own ranked list of all the boys.
\end{itemize}
Goal: Find perfect matching with no rogue couples.
\subsection{Mating Algorithm}
Need to show:
\begin{itemize}
\item TMA terminates (quickly)
\item Everyone gets married
\item No rogue couples
\item Fairness
\end{itemize}
\begin{theorem}
TMA terminates in $\leqslant N^2+1$ days.
\end{theorem}
\begin{proof}
(By contradiction)
Suppose TMA does not terminate in $N^2+1$ days.
Claim: If we don't terminate on a day, then one boy crosses a girl off his least that night.
There are at most $N^2$ cross-outs, so there is the contradiction.
\end{proof}
\begin{theorem}
Everyone gets married in the end.
\label{Thm}
\end{theorem}
Let $P = $ ``If a girl ever rejected a boy B, then G has a suitor who she prefers to be."
\begin{lemma}
$P$ is an invariant for TMA.
\label{Lem}
\end{lemma}
\begin{proof}[Proof of Lemma \ref{Lem}]
(By induction on days)
Base Case: Day 0: No one rejected yet, so it is vacuously true.
Inductive Step: Assume $P$ holds at the end of Day $d$.
Case 1: G rejected B on day $d+1$. Then there was someone better. So $P$ is true.
Case 2: G rejects B before $d+1$. $P$ implies G holds better one than B.
G has same or better suitor on day $d+1$. So $P$ is true.
\end{proof}
\begin{proof}[Proof of Theorem \ref{Thm}]
(By contradiction)
Assume that somebody B that is not married at end.
$\therefore$ B was rejected by every girl.
$\therefore$ Every girl has suitor better than B. (Lemma \ref{Lem})
$\therefore$ Every girl has married.
However, it is impossible.
\end{proof}
\begin{theorem}
TMA produces a stable matching.
\end{theorem}
\begin{proof}
Let Bob and Gail be any pair that are not married.
Case 1: Gail rejected Bob. Gail marries someone she thinks than Bob. (Lemma \ref{Lem}) So Gail and Bob is not rogue.
Case 2: Gail did not reject Bob. So Gail never serenades Gail. So Gail is lower on Bob's list than Bob's wife. So Gail and Bob is not rogue.
$\therefore$ TMA is stable.
\end{proof}
\begin{theorem}
TMA favors boys.
\label{Thm_}
\end{theorem}
Let $S = $ ``set of all stable matchings". $S \neq \varnothing$.
For each person $P$, we define the realm of possibility for $P$ to be $\left\{Q|\exists \mathfrak{M}\in S, \{P,Q\}\in\mathfrak{M}\right\}$
\begin{definition}
A person's optimal mate is his/her favorite from the realm of possibility.
\end{definition}
\begin{definition}
A person's pessimal mate is his/her least favorite from the realm of possibility.
\end{definition}
\begin{theorem}
TMA marries every boy to his optimal mate.
\label{THM_1}
\end{theorem}
\begin{theorem}
TMA marries every girl to her pessimal mate.
\label{THM_2}
\end{theorem}
Assume Theorem \ref{THM_1} to be true, now prove Theorem \ref{THM_2}.
\begin{proof}
(By contradiction)
Suppose that $\exists$ stable matching $\mathfrak{M}$ where $\exists$ girl G who fares worse than in TMA.
Let $B'$ be the mate of $G$ in $\mathfrak{M}$.
Let $B$ be the mate of $G$ in TMA.
Let $G$ be the mate of $G'$ in TMA.
However, according to Theorem \ref{THM_1}, $B$ loves $G$ better and $G$ loves $B$ better, so $B$ and $G$ is a rogue couple.
$\therefore \mathfrak{M}$ is not stable.
\end{proof}
\section*{Appendix}
~
This note is written in \LaTeX.
\LaTeX\ code in my blog: \url{https://blog.csdn.net/weixin_50012998/article/details/113818419}\\
The webpage of the video: \url{https://www.bilibili.com/video/BV1zt411M7D2?p=7}
\end{document}
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