%! TEX root = Combi.tex
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% 03/12/2024 09AM

% Example class: Friday 24th January,
% 1:30pm, MR3
% Handin Q1, Q5 (by 5pm Thursday)
% And Q10 if you manage it.

\begin{remark*}
  Do need $p$ prime. Grolmusz constructed, for
  each $n$, a value of $r \equiv 0 \pmod 6$ and a
  family $A \subset [n]\rsubs$ such that for all
  distinct $x, y \in A$ we have $|x \cap y|
  \not\equiv 0 \pmod 6$ with $|A| > n^{c\log n /
  \log \log n}$. This is not a polynomial in $n$.
\end{remark*}

\begin{corollary}
  \label{coro_3_5}
  % Corollary 3.5
  Let $A \subset [n]\rsubs$ with $|x \cap y|
  \not\equiv r \pmod p$, for each distinct $x, y
  \in A$, where $p < r$ is prime.
\end{corollary}

\begin{proof}
  We are allowed $p - 1$ values of $|x \cap y|
  \pmod p$, so done by \nameref{thm_3_4}.
\end{proof}

Two $\frac{n}{2}$-sets in $[n]$ typically meet in
about $\frac{n}{4}$ points -- but $|x \cap y|$
exactly equaling $\frac{n}{4}$ is very unlikely.
But remarkably:

\begin{corollary}
  \label{coro_3_6}
  % Corollary 3.6
  Let $p$ be prime, and let $A \subset
  [4p]\rsubs[2p]$ have $|x \cap y| \neq p$ for all
  distinct $x, y \in A$ (``this is not much of a
  constraint'').
  Then $|A| \le 2{4p \choose p - 1}$.
\end{corollary}

\begin{note*}
  ${4p \choose p - 1}$ is a \emph{tiny}
  (exponentially small) proportion
  of ${4p \choose 2p}$. Indeed, ${n \choose n / 2}
  \sim c \cdot \frac{2^n}{\sqrt{n}}$ (for some
  $c$) whereas ${n \choose n / 4} \le 2e^{-n / 32}
  2^n$.
\end{note*}

\begin{proof}
  Halving $|A|$ if necessary, may assume that no
  $x, x^c \in A$ (any $x \in [4p]\rsubs[2p]$).

  Then $x, y \in A$ distinct implies $|x \cap y|
  \neq 0, p$, so $|x \cap y| \not\equiv 0 \pmod
  p$.

  So $|A| \le {4p \choose p - 1}$ by
  \cref{coro_3_5}.
\end{proof}

\subsection{Borsuk's Conjecture}

Let $S$ be a bounded subset of $\Rbb^n$.
\begin{center}
  \includegraphics[width=0.4\linewidth]{images/dc9a0f93396348ec.png}
\end{center}
How few pieces can we break $S$ into such that
each piece has smaller diameter than that of $S$?

The example of a regular simplex in $\Rbb^n$ ($n +
1$ points, all at distance $1$) shows that we may
need $n + 1$ pieces.
\begin{center}
  \includegraphics[width=0.4\linewidth]{images/dd48d455cf9e4c29.png}
\end{center}

\begin{conjecture*}[Borsuk's conjecture (1920s)]
  \label{borsuk}
  $n + 1$ pieces always sufficient.
\end{conjecture*}

Known for $n = 1, 2, 3$. Also known for $S$ a
smooth convex body in $\Rbb^n$ or a symmetric
convex body in $\Rbb^n$ (convex means $x \in S$
implies $-x \in S$).

However, Borsuk is massively false:

\begin{fcthm}[Kahn, Kalai 1995]
  \label{thm_3_7}
  % Theorem 3.7
  Assuming:
  - $n \in \Nbb$
  Then:
  there exists bounded $S \subset \Rbb^n$ such
  that to break $S$ into pieces of smaller
  diameter we need $\ge C^{\sqrt{n}}$, for some
  constant $c > 1$ (not depending on $n$).
\end{fcthm}

\begin{note*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item Our proof will show \nameref{borsuk} is
      false for $n \ge 2000$.
    \item We'll prove it for $n$ of the form ${4p
      \choose 2}$, where $p$ is prime. Then done
      for all $n$ (with a different $c$, e.g.
      because there exists a prime $p$ with
      $\frac{n}{2} \le p \le n$).
  \end{enumerate}
\end{note*}

\begin{proof}
  We'll find $S \subset \Q_n \subset \Rbb^n$ -- in
  fact $S \subset [n]\rsubs$ for some $r$. We have
  already had two genuine ideas from this
  sentence: first that we think about having $S
  \subset \Q_n$, and second that we go for $S
  \subset [n]\rsubs$.

  Have $S \subset [n]\rsubs$, so $\forall x, y \in
  S$:
  \[
    \|x - y\|^2
    = \# \text{coordinates where $x$ and $y$
    differ}
    = 2(r - |x \cap y|)
  .\]
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/78ef6dad14fc4765.png}
  \end{center}
  So seek $S$ with diameter $\min|x \cap y| = k$,
  but every subset of $S$ with $\min|x \cap y| >
  k$ is very small (hence we will need many
  pieces).

  Identify $[n]$ with the edge-set of $K_{4p}$,
  the complete graph on $4p$ points.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/4ceecae9ec2541ad.png}
  \end{center}
  For each $x \in [4p]\rsubs[2p]$ let $G_x$ be the
  complete bipartite graph, with vertex classes
  $x, x^c$. Let $S = \{G_x : x \in
  [4p]\rsubs[2p]\}$. So $S \subset
  [n]\rsubs[4p^2]$, and $|S| = \half {4p \choose
  2p}$.

  Now
  \begin{align*}
    |G_x \cap G_y|
    &= |x \cap y||x^c \cap y^c| + |x^c \cap y||x
    \cap y^c| \\
    &= |x \cap y|^2 + |x^c \cap y|^2 \\
    &= d^2 + (2p - d)^2
  \end{align*}
  where $d = |x \cap y|$.
  \begin{center}
    \includegraphics[width=0.4\linewidth]{images/a15bf97b4b374d0b.png}
  \end{center}
  This is minimised when $d = p$, i.e. when $|x
  \cap y| = p$.

  Now let $S' \subset S$ have smaller diameter
  than that of $S$: say $S' = \{G_x : x \in A\}$.
  So must have $\forall x, y \in A$ distinct: $|x
  \cap y| \neq p$ (else diameter of $S'$ is the
  diameter of $S$).

  Thus
  \[
    |A| \le 2{4p \choose p - 1}
  .\]
  Conclusion: the number of pieces needed is $\ge
  \frac{\half {4p \choose 2p}}{2 {4p \choose p -
  1}} \ge \frac{c \cdot 2^{4p} / \sqrt{p}}{e^{-p /
  8} 2^{4p}}$ (for some $c$). This is $\ge
  (c')^p$, for some $c' > 1$, which is at least
  $(c'')^{\sqrt{n}}$ for some $c'' > 1$.
\end{proof}
