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% 10/10/2024 10AM

% 1. Combinatorial methods
% 2. Fourier-analytic techniques
% 3. Probabilistic tools
% 4. Further topics

\newpage
\section{Combinatorial methods}

\begin{fcdefn}[Sumset]
  \glsnoundefn{sumset}{sumset}{sumsets}%
  \glssymboldefn{sumset}%
  \glssymboldefn{diffset}%
  Let $G$ be an abelian group. Given $A, B
  \subseteq G$, define the \emph{sumset} $A + B$
  to be
  \[ A + B \defeq \{a + b : a \in A, b \in B\} \]
  and the \emph{difference set} $A - B$ to be
  \[ A - B \defeq \{a + b : a \in A, b \in B\} .\]
\end{fcdefn}

If $A$ and $B$ are finite, then certainly
\[ \max\{|A|, |B|\} \le |A \plus B| \le |A||B| .\]

\begin{example}
  % Example 1.1
  Let $A = [n] \defeq \{1, 2, \ldots, n\}
  \subseteq \Zbb$. Then
  \[ |A \plus A| = |\{2, \ldots, 2n\}| = 2n - 1 =
  2|A| - 1 .\]
\end{example}

\begin{fclemma}[]
  % Lemma 1.2
  Assuming:
  - $A \subseteq \Zbb$ is finite.
  Then:
  $|A \plus A| \ge 2|A| - 1$, with equality if and
  only if $A$ is an arithmetic progression.
\end{fclemma}

\begin{proof}
  Let $A = \{a_1, a_2, \ldots, a_n\}$ with $a_1 <
  a_2 < \cdots < a_n$. Then
  \[ a_1 + a_1 < a_1 + a_2 < a_1 + a_2 < \cdots <
  a_1 + a_n < a_2 + a_n < \cdots < a_n + a_n ,\]
  so $|A \plus A| \ge 2|A| - 1$. But we could also
  have written
  \[ a_1 + a_1 < a_1 + a_2 < a_2 + a_2 < a_2 + a_3
  < a_2 + a_4 < \cdots < a_2 + a_n < a_3 + a_n <
  \cdots < a_n + a_n .\]
  When $|A \plus A| = 2|A| - 1$, these two
  orderings must be the same. So $a_2 + a_i = a_1
  + a_{i + 1}$ for all $i = 2, \ldots, n - 1$.
\end{proof}

\textbf{Exercise:} If $A, B \subseteq \Zbb$, then
$|A \plus B| \ge |A| + |B| - 1$ with equality if
and only if $A$ and $B$ are arithmetic
progressions with the same common difference.

\begin{example}
  % Example 1.3
  Let $A, B \subseteq \Zp$ with $p$
  prime. Then $|A \plus B| \ge p + 1 \implies A
  \plus B = \Zp$. Indeed, $g \in A \plus
  B \iff A \cap (g \minus B) \neq \emptyset$ (note
  that $g - B$ means $\{g\} - B$). But $\forall g
  \in \Zp$,
  \[ |A \cap (g \minus B)| = |A| + |g \minus B| -
  |A \cup (g \minus B)| \ge |A| + |B| - p \ge 1 .\]
\end{example}

\begin{fcthm}[Cauchy-Davenport]
  % Theorem 1.4
  Assuming:
  - $p$ is a prime
  - $A, B \subseteq \Zp$ nonempty
  Then:
  \[ |A \plus B| \ge \min\{p, |A| + |B| - 1\} .\]
\end{fcthm}

\begin{proof}
  Assume $|A| + |B| \le p + 1$. Without loss of
  generality assume that $1 \le |A| \le |B|$ and
  that $0 \in A$. Apply induction on $|A|$. The
  case $|A| = 1$ is trivial. Suppose $|A| \ge 2$,
  and let $0 \neq a \in A$.

  Since
  $\{a, 2a, 3a, \ldots, (p - 1)a, pa\} = \Zp$ and
  $|A| + |B| \le p + 1$, there must exist $m \ge
  0$ such that $ma \in B$ but $(m + 1)a \notin B$.
  Let $B' = B - ma$, so $0 \in B'$, $a \notin B'$,
  $|B'| = |B|$.

  But $1 \le |A \cap B'| < |A|$, so the inductive
  hypothesis applies to $A \cap B'$ and $A \cup
  B'$. Since
  \[ (A \cap B') \plus (A \cup B') \subseteq A \plus
  B' ,\]
  we have
  \[ |A \plus B| = |A \plus B'| \ge |(A \cap B')
  \plus (A \cup B')| \ge |A \cap B'| + |A \cup B'|
  + 1 = |A| + |B| + 1 . \qedhere \]
\end{proof}

This fails for general abelian groups (or even
general cyclic groups).

\begin{example}
  % Example 1.5
  Let $p$ be (fixed, small) prime, and let $V \le
  \Fp^n$ be a subspace. Then $V + V = V$, so $|V +
  V| = |V|$. In fact, if $A \subseteq \Fp^n$ is
  such that $|A + A| = |A|$, then $A$ must be a
  coset of a subspace.
\end{example}

\begin{example}
  % Example 1.6
  \label{example_1_6}
  Let $A \subseteq \Fp^n$ be such that $|A \plus
  A| < \frac{3}{2}|A|$. Then there exists $V \le
  \Fp^n$ a subspace such that $|V| < \frac{3}{2}
  |A|$ and $A$ is contained in a coset of $V$.
  See Example Sheet 1.
\end{example}

\begin{fcdefn}[Ruzsa distance]
  % Definition 1.7
  \glsnoundefn{rd}{Rusza distance}{N/A}%
  \glssymboldefn{rd}%
  Given finite sets $A, B \subseteq G$, we define
  the \emph{Ruzsa distance} $d(A, B)$ between $A$
  and $B$ by
  \[ d(A, B) = \log \frac{|A \minus
  B|}{\sqrt{|A||B|}} \]
\end{fcdefn}