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\documentclass{article}
\input{structure.tex}

\title{CENG567: Homework \#1}
\author{Yiğit Sever}
\date{\today}

%----------------------------------------------------------------------------------------

\begin{document}

\maketitle

\section{Stable Matching}%
\label{sec:stable_matching}

% question a ✅ {{{1 %

\begin{question}%
    \label{q:1_a}
    Use \emph{Gale-Shapley} algorithm to find a stable matching for the following set of four colleges, four students and their preference lists.
\end{question}

\begin{commandline}[Gale-Shapley algorithm, from lecture slides, edited for the context]
    \begin{verbatim}
    Initialize each person to be free.
    while (some student is free and hasn't applied to every college) {
        Choose such a student m
        c = 1st college on m's list to whom m has not yet applied
        if (c is free)
            assign c to m for potential application (a)
        else if (c prefers m to their current applicant m')
            assign m and c for potential application, and m' to be free (b)
        else
            c rejects m (c)
    }
    \end{verbatim}
 \end{commandline}

A quick trace of the algorithm;

\begin{enumerate}
    \item $S_1$ is free;
        \begin{enumerate}
            \item applies to first college on their preference list $C_4$;
            \item $C_4$ is free so it accepts and is matched with $S_1$ (a).
        \end{enumerate}
    \item $S_2$ is free
        \begin{enumerate}
            \item applies to first college on their preference list; $C_1$
            \item $C_1$ is free so it accepts and is matched with $S_2$ (a).
        \end{enumerate}
    \item $S_3$ is free;
        \begin{enumerate}
            \item applies to first college on their preference list; $C_1$
            \item $C_1$ rejects $S_3$ because it prefers $S_2$ to $S_3$ (c).
            \item applies to second college on their preference list; $C_2$
            \item $C_2$ is free so it accepts and is matched with $S_3$ (a).
        \end{enumerate}
    \item $S_4$ is free;
        \begin{enumerate}
            \item applies to first college on their preference list; $C_4$
            \item $C_4$ rejects $S_4$ because it prefers $S_1$ to $S_4$ (c).
            \item applies to second college on their preference list; $C_3$
            \item $C_3$ is free so it accepts and is matched with $S_4$ (a).
        \end{enumerate}
        \item There are no more free students to match, algorithm terminates.
\end{enumerate}

The final matching and the answer to Question~\ref{sec:stable_matching}(a) is;

\begin{align*}
    S_1 &\rightarrow C_4 \\
    S_2 &\rightarrow C_1 \\
    S_3 &\rightarrow C_2 \\
    S_4 &\rightarrow C_3
\end{align*}

% 1}}} %

% question b ✅ {{{1 %

\begin{question}
    Find another stable matching with the same algorithm.
\end{question}

All executions of \emph{Gale-Shapley} yield the same stable matching (that is proposer-optimal) and cannot produce \emph{another} stable matching like the question text asks for.

% 1}}} %

% question c ✅ {{{1 %

\begin{question}
    Consider a pair of man $m$ and woman $w$ where $m$ has $w$ at the top of his preference list and $w$
    has $m$ at the top of her preference list. Does it always have to be the case that the pairing $(m, w)$ exist
    in every possible stable matching? If it is true, give a short explanation. Otherwise, give a counterexample.
\end{question}

\emph{Proof by contradiction}.
Assume that in the resulting matching of \emph{Gale-Shapley}, we have $S'$, where $m$ is matched with $w'$ and $w$ is matched with $m'$.

The definition of stable matching dictates that there is \emph{no incentive to exchange}, yet in $S'$ $m$ can trade up to $w$ since $m$ prefers $w$ to $w'$ and $w$ can trade up since $w$ prefers $m$ to $m'$.

$S'$ could not have occurred since men propose in accordance to their preference list, which $w$ is on top of for $m$ and no other men that may propose to $w$ can make $w$ switch since they are not more preffered than $m$.

% 1}}} %

% question d ✅ {{{1 %

\begin{question}
    Give an instance of $n$ colleges, $n$ students, and their preference lists so that the Gale-Shapley algorithm requires only $O(n)$ iterations, and prove this fact.
\end{question}

For a proposer agnostic formation, arrange the preference lists of colleges and students as follows;

\begin{gather*}%
    \label{eq:student_prefs}
    C_{1} \rightarrow \left\{ S_{1}, S_{n}, S_{n-1}, \dots, S_{2} \right\} \\
    C_{2} \rightarrow \left\{ S_{2}, S_{n}, S_{n-1}, \dots, S_{3} \right\} \\
    \dots  \\
    C_{k} \rightarrow \left\{ S_{k}, S_{n}, S_{n-1}, \dots, S_{k+1} \right\} \\
    \dots \\
    C_{n} \rightarrow \left\{ S_{n}, S_{n-1}, S_{n-2}, \dots, S_{1} \right\} \\
\end{gather*}

\begin{gather*}%
    \label{eq:student_prefs}
    S_{1} \rightarrow \left\{ C_{1}, C_{n}, C_{n-1}, \dots, C_{2} \right\} \\
    S_{2} \rightarrow \left\{ C_{2}, C_{n}, C_{n-1}, \dots, C_{3} \right\} \\
    \dots  \\
    S_{k} \rightarrow \left\{ C_{k}, C_{n}, C_{n-1}, \dots, C_{k+1} \right\} \\
    \dots \\
    S_{n} \rightarrow \left\{ C_{n}, C_{n-1}, C_{n-2}, \dots, C_{1} \right\} \\
\end{gather*}

Where the student list $\{S_1, S_2, \dots, S_{n}\}$ and college list $\{C_1, C_2, \dots, C_{n}\}$ are shifted.

In this setup, ever proposer will propose to the first suitor in their preference list which is guaranteed to be free since they are not the first on any other suitor's preference list.

The algorithm in this instance runs in $\mathcal{O}(n)$ iterations, every proposer will follow the (a) branch in the algorithm given under Question~\ref{q:1_a}.

% 1}}} %

% question e TODO {{{1 %

\begin{question}
    Give another instance for which the algorithm requires $\Omega(n^{2})$ iterations (that is, it requires at least $cn^{2}$ iterations for some constant $0 < c \le 1)$, and prove this fact.
\end{question}

Arrange the preference lists such as every student arranges their preference \emph{exactly the same};

\begin{equation}
    \label{eq:student_prefs_2}
    C_1, C_2, \dots, C_{n}
\end{equation}

Whereas every college should arrange their preference list as;

\begin{equation}
    \label{eq:college_prefs_2}
    S_n, S_{n-1}, S_{n-2}, \dots, S_{1}
\end{equation}

% 1}}} %

\section{Stable Matching Variation}%
\label{sec:stable_matching_variation}

% stable matching question ALMOST {{{1 %

\begin{question}
    Consider a Stable Matching problem with men and women.
    Consider a woman $w$ where she prefers man $m$ to $m'$, but both $m$ and $m'$ are low on her list of preferences.
    Can it be the case that by switching the order of $m$ and $m'$ on her list of preferences (i.e., by falsely claiming that she prefers $m'$ to $m$) and running the algorithm with this modified preference list, $w$ will end up with a man $m''$ that she prefers to both $m$ and $m'$?
    Either give a proof that shows such an improvement is impossible, or give an example preference list for which an improvement for $w$ is possible.
\end{question}

Example preference list for men; $M = {x,y,z}$ and women: $W = {a,b,c}$.

\begin{table}[!htb]
    \caption{Preference lists of women.}
    \begin{minipage}{.5\linewidth}
        \caption{Everyone is being truthful}
        \centering
        \begin{tabular}{l|lll}
            \textbf{a} & y & x & z \\
            \textbf{b} & x & y & z \\
            \textbf{c} & x & y & z
        \end{tabular}
    \end{minipage}%
    \begin{minipage}{.5\linewidth}
        \centering
        \caption{a is lying about their preference}
        \begin{tabular}{l|lll}
            \textbf{a} & y & z & x \\
            \textbf{b} & x & y & z \\
            \textbf{c} & x & y & z
        \end{tabular}
        \begin{tikzpicture}[overlay]
         \draw[red, line width=1pt] (-0.72,0.52) ellipse (0.70cm and 0.2cm);
        \end{tikzpicture}
    \end{minipage}
\end{table}

% TODO: final polish, highlight the lies <06-11-20, yigit> %

\begin{table}[htb]
    \centering
    \begin{tabular}{l|lll}
        \textbf{x} & a & b & c \\
        \textbf{y} & b & a & c \\
        \textbf{z} & a & b & c
    \end{tabular}
    \caption{The preference lists of the men.}%
    \label{tab:men_pref}
\end{table}

The trace of \emph{Gale-Shapley} algorithm for the truth telling case;

\begin{enumerate}
    \item Matching x with a (a)
    \item Matching y with b (a)
    \item z proposes to a but gets rejected because a prefers x more (c)
    \item z proposes to b but gets rejected because b prefers y more (c)
    \item Matching z with c (a)
\end{enumerate}

\begin{align*}
    x &\rightarrow a \\
    y &\rightarrow b \\
    z &\rightarrow c
\end{align*}

The trace of \emph{Gale-Shapley} algorithm when \emph{a} lies about their preferences;

\begin{enumerate}
    \item Matching y with b (a)
    \item Matching z with a (a)
    \item x proposes to a but gets rejected because a prefers z more (c)
    \item Matching x with b, y is now free (b)
    \item Matching y with a, z is now free (b)
    \item z proposes to b but gets rejected because b prefers x more (c)
    \item Matching z with c (a)
\end{enumerate}

\begin{align*}
    x &\rightarrow b \\
    y &\rightarrow a \\
    z &\rightarrow c
\end{align*}

% 1}}} %

\section{Asymptotics}%
\label{sec:asymptotics}

% asymptotics ✅ {{{1 %

\begin{question}
    What is the running time of this algorithm as a function of $n$? Specify a function $f$ such that the running time of the algorithm is $\Theta(f(n))$.
\end{question}

{\centering
    \begin{minipage}{.7\linewidth}
        \begin{algorithm}[H]
            \For(\tcc*[f]{outer loop}){$i = 2$; $i < n$; $i += 1$ }{
                \For(\tcc*[f]{inner loop}){$j=1$; $j < n$; $j * = i$}{
                    Some $\Theta(1)$ operation\;
                }
            }
            \caption{Equivalent algorithm to one given in Question 3, edited for brevity}%
            \label{alg:question_3}
\end{algorithm}
\end{minipage}
\par
}

The outer loop runs in linear time $\mathcal{O}(n)$ whereas the inner loop requires some deconstruction;

Take the first iteration of the inner loop, $i = 2$ and $j$ is initialized at $1$.

\begin{align*}%
    \label{eq:3_iterations}
    1^{\text{st}} ~ \text{iteration} &\rightarrow  j = j * i \implies  j = 2 \\
    2^{\text{nd}} ~ \text{iteration} &\rightarrow  j = 4 \\
    3^{\text{rd}} ~ \text{iteration} &\rightarrow  j = 8 \\
    \dots \\
    m^{\text{th}} ~ \text{iteration} &\rightarrow  j = i^{m} < n \\
\end{align*}

$m$ is the last iteration of the inner loop because it hit the stopping condition $i^{m} < n$.
Using the Equations above we can find the running time for the inner loop to hit stopping condition;

\begin{gather*}
    i^{m} < n \\
    m < \log_{i} n
\end{gather*}

In other words, the inner loop has a running time in the order of $\mathcal{O}(\log{n})$. By the product property of $\mathcal{O}$-notation we have the running time of $\mathcal{O}(n\log_{a}{n}), a > 1$ for the entire algorithm. The base $a$ is useful to answer the rest of the question;

We can specify a function $f$ such as $f = n \log_{b}(n)$ where $b > 1$. From the course slides;

\begin{equation*}
    \lim_{n \to \infty}\frac{n\log_{a}{n}}{n \log_{b}{n}} = c
\end{equation*}

And when the limit of two functions $f, g$ converge to some constant $c$, then $f(n) = \Theta(g(n))$.

% 1}}} %

\section{Big $\mathcal{O}$ and $\Omega$}%
\label{sec:big_o_and_omega_}

% question a ALMOST DONE {{{1 %

\begin{question}
    Let $f(n)$ and $g(n)$ be asymptotically positive functions. Prove or disprove the following conjectures.
\end{question}

\begin{info}[]
    $f(n) = \mathcal{O}(g(n))$ implies $g(n) = \mathcal{O}(f(n))$
\end{info}

From the definition of Big $\mathcal{O}$ notation given in the lecture slides;

\begin{equation}%
    \label{eq:a_bigo}
    \exists ~ c > 0 \quad \text{and} \quad n_{0} \ge 0 \quad \mid \quad 0 \le f(n) \le c \cdot g(n) \quad \forall n \ge n_{0}
\end{equation}

If we rearrange the right hand side of the Proposition~\ref{eq:a_bigo};

\begin{equation*}%
    0 \le \frac{1}{c} f(n) \le g(n) \\
\end{equation*}

or simply,

\begin{equation}
    \label{eq:a_bio_rearranged}
    0 \le c' f(n) \le g(n)
\end{equation}

By the definition of Big $\mathcal{O}$ notation, in order for $g(n) = \mathcal{O}(f(n))$ to be true,

\begin{equation*}
    \exists ~ k > 0 \quad \text{and} \quad n_{0}' \ge 0 \quad \mid \quad 0 \le g(n) \le k \cdot f(n) \quad \forall n' \ge n_{0}'
\end{equation*}

Has to be true, yet from Equation~\ref{eq:a_bio_rearranged} we know that $f(n)$ multiplied by some constant $c'$ is strictly smaller than $g(n)$. The conjecture is \emph{false}.

\begin{info}[]
    $f(n) = \mathcal{O}((f(n))^{2})$
\end{info}

% TODO: this is dumb ask this <05-11-20, yigit> %

% https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-046j-introduction-to-algorithms-sma-5503-fall-2005/assignments/ps1sol.pdf

The question text states that $f(n)$ is a positive function.

By the definition of Big $\mathcal{O}$ notation, we have;

\begin{equation*}%
    \exists ~ c > 0 \quad \text{and} \quad n_{0} \ge 0 \mid ~ 0 \le f(n) \le c \cdot g(n) \quad \forall n \ge n_{0}
\end{equation*}

For $c > 1$ and $n_{0} > 1$ we have;

% TODO: this makes zero sense? <06-11-20, yigit> %

\begin{align*}%
    f(n) &\le f(n)^{2} \quad \forall n \ge n_{0} \\
    \text{take} ~ g(n) &= f(n)^{2} \\
    f(n) &\le c\cdot g(n) \quad \forall n \ge n_{0}
\end{align*}

\begin{info}[]
    $f(n) + o(f(n)) = \Theta(f(n))$
\end{info}

From the definition of Big Theta notation, we are trying to prove a relation such that;

\begin{equation}%
    \label{eq:a_big_theta}
    0 \le c_{1} \cdot f(n) \le f(n) + o(f(n)) \le c_{2} \cdot f(n)
\end{equation}

First, let's give the little-o notation to remove $o(f(n))$ from Equation~\ref{eq:a_big_theta};

\begin{equation}%
    \label{eq:a_little_o}
    \forall ~ c > 0 \quad \exists ~ n_{0} > 0 \mid ~ 0 \le g(n) < c \cdot f(n) ~ \forall n \ge n_{0}
\end{equation}

So we can rewrite Equation~\ref{eq:a_big_theta} using Equation~\ref{eq:a_little_o};

\begin{equation}%
    \label{eq:a_rewritten}
    c_1 \cdot f(n) \le f(n) + g(n) \le c_2 \cdot f(n)
\end{equation}

It's trivial to pick $c_1 < 1$ to deal with the left hand side of the inequality;

\begin{equation*}
    f(n) \le f(n) + g(n)
\end{equation*}

For the right hand side, we can pick $n_{0} = 1$ and $c = 1$ in the Equation~\ref{eq:a_little_o}; keeping $g(n) < f(n)$.

Finally, picking $c2 > 2$ proves the conjecture.


% 1}}} %

% question b {{{1 %
% https://people.cs.umass.edu/~sheldon/teaching/cs311/hw/hw01.pdf

\begin{question}
    For each function $f(n)$ below, find (and prove that)
    \begin{enumerate}
        \item the smallest integer constant H such that $f(n) = \mathcal{O}(n^H)$
        \item the largest positive real constant L such that $f(n) = \Omega(n^L)$
    \end{enumerate}
    Otherwise, indicate that H or L do not exist.
\end{question}

\begin{info}[]
    $f(n) = \frac{n(n+1)}{2}$
\end{info}

(1) For $n_{0} \ge 1$; $\frac{n(n+1)}{2}$ is strictly smaller than $n^{2}$, yet $H=1$ is not possible since through the definition of Big-$\mathcal{O}$ notation it implies that for a \emph{constant} $c$;

\begin{align*}
    0 \le \frac{n(n+1)}{2} &\le c\cdot n \\
    \frac{\cancel{n}(n+1)}{2} &\le c\cdot \cancel{n} \\
    \frac{n+1}{2} &\cancel{\le} c
\end{align*}

Hence $f(n) = \mathcal{O}(n^{2})$ and $H=2$.

% TODO: reread proof, strengthen? <06-11-20, yigit> %
(2)$L=2$. We can pick the constant $c = \frac{1}{2}$ and write the Big-$\Omega$ notation;

\begin{gather*}
    c > 0 ~ \text{and} ~ n_{0} \ge 0 \\
    \frac{n(n+1)}{2} \ge \frac{1}{2} n^{2} \\
    \frac{\cancel{n}(n+1)}{\cancel{2}} \ge \frac{1}{\cancel{2}} n^{\cancel{2}} \\
    n+1 \ge n
\end{gather*}

Does a $L>2$ work? Pick any $\ell > 2$;

\begin{gather*}
    c > 0 ~ \text{and} ~ n_{0} \ge 0 \\
    \frac{n(n+1)}{2} \ge \frac{1}{2} n^{\ell} \\
    \frac{n(n+1)}{\cancel{2}} \ge \frac{1}{\cancel{2}} n^{\ell} \\
    n(n+1) \ge n^{\ell} \\
    n+1 \ge n^{\ell - 1} \\
    1 \ge n(n^{\ell - 2} - 1)
\end{gather*}

The final equation does not hold $n>1$ so $L>2$ cannot be true.

\begin{info}[]
    $f(n) = \sum^{\lceil{\log{n}}\rceil}_{k=0} \frac{n}{2^{k}}$
\end{info}

\begin{info}[]
    $f(n) = n(\log n)^{2}$
\end{info}


% 1}}} %

\end{document}