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\subsection{The Giant Component}

\begin{fcprop}[]
  Let $\eps > 0$.
  If $G \sim G(n, p)$, then
  \[
    \lim_{n \to \infty} \Pbb(\text{$G$ is connected})
    = \begin{cases}
      1
      & p \ge (1 + \eps) \frac{\log n}{n} \\
      0
      & p \le (1 - \eps) \frac{\log n}{n}
    \end{cases}
  \]
  and
  \[
    \lim_{n \to \infty} \Pbb(G \supset H)
    = \begin{cases}
      1
      & p^{m(H)} n \gg 1 \\
      0
      & p^{m(H)} n \ll 1
    \end{cases}
  \]
\end{fcprop}

\begin{remark*}
  This ``threshold'' for $p$ looks different from what we saw from the threshold
  of containing a given $H$ -- it has a ``sharp'' jump from $0$ to $1$.
\end{remark*}

\begin{fcdefn}[c1]
  \glssymboldefn{c1}%
  For a graph $G$, let
  \[
    c_1(G)
    = \text{\# vertices in the largest connected component}
  .\]
\end{fcdefn}

\begin{fcthm}[Erdös-Renyi]
  Let $\eps > 0$.
  If $G \sim G(n, p)$ then
  \[
    \bigcomp(G)
    = \begin{cases}
      O_\eps(\log n)
      & \text{if $p \le \frac{1 - \eps}{n}$} \\
      \Omega_\eps(n)
      & \text{if $p \ge \frac{1 + \eps}{n}$}
    \end{cases}
  \]
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/f5e02b2aba404785.png}
\end{center}

\begin{fcdefn}[DFS-sequence]
  \glsnoundefn{dfsseq}{DFS-sequence}{DFS-sequences}%
  Let $G$ be a graph.
  I define the \emph{DFS-sequence} from $G$
  \[
    X_1, X_2, X_3, X_4, \ldots \in \{0, 1\}
  \]
  as follows.
  We process the graph using a Depth first search.
  We also imagine $V(G)$ have an order on them.
  As we saw before, we maintain a partition $V(G) = A \cup B \cup P$ where $P$
  is a path.

  We start with $A = V(G)$, $B = \emptyset$, $P = \emptyset$.
  We then repeat the following:
  \begin{enumerate}[(1)]
    \item If $P$ is empty, move a vertex from $A$ to $P$.
    \item If $P$ is not empty, let $x$ be the last vertex of $P$, and now query
      the edges from $x$ to $A$ in order until I get a ``yes''.

      Now move the neighbour of $v$ to the end of $P$ and append to our DFS
      sequence the outcomes of the queries.

      If $v$ has no neighbour in $A$, then move $v$ to $B$ and append all the
      ``no'' outcomes to the DFS-sequence.
  \end{enumerate}
\end{fcdefn}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/744313451cd74dda.png}
\end{center}

\begin{note*}
  \leavevmode
  \begin{itemize}
    \item We don't query an edge more than once.
    \item We don't query all edges in $G$.
      But if $uv \in G$ is not queried, then $u, v$ are in the same component.
  \end{itemize}
\end{note*}

\begin{fclemma}[]
  Let $G$ be a graph on $n$ vertices, with a component of size $k$.
  Let $X_1, X_2, \ldots$ be the \gls{dfsseq}.
  Then there exists $t$ such that
  \[
    \sum_{i = t}^{t + kn} X_i
    \ge k - 1
  .\]
\end{fclemma}

\begin{proof}
  Let $C = \{x_1, \ldots, x_k\}$ be a component of size $k$, $C \subset V(G)$.
  Say we first encounter $x_1$ at time $t$.
  From time $t$ up until $C \subset B$, we only query edges incident to $v \in
  C$.
  So we make at most
  \[
    {k \choose 2} + k(n - k) \le kn
  \]
  exposures.
  And during this time we must have seen $\ge k - 1$ $1$'s since $V(C)$ is a
  component.
\end{proof}

\begin{fclemma}[]
  For $\eps > 0$, let $p = \frac{1 - \eps}{n}$.
  Then if $G \sim G(n, p)$,
  \[
    \bigcomp(G)
    \le \frac{8\log n}{\eps^2}
  .\]
\end{fclemma}

\begin{proof}
  Let $X_1, X_2, \ldots$ be the corresponding \gls{dfsseq}.
  Since we never query an edge more than once in the definition of the
  \gls{dfsseq}, we have that $X_1, X_2, X_3, \ldots$ is a sequence of iid random
  variables, with each being $X_i \sim \Ber(p)$.

  So let
  \[
    Y_t
    = \sum_{i = t}^{t + kn} X_i
  \]
  for $t \le n^2$.

  We will use \emph{Chernoof's inequality}:

  Let $X \sim \Bin(n, p)$ defined by
  \[
    \Pbb(X = k)
    = {n \choose k} p^k (1 - p)^{n - k}
  .\]
  Then for $0 \le t \le pn$,
  \[
    \Pbb (|X - pn| \ge t)
    \le 2\exp \left( -\frac{t^2}{3pn} \right)
  .\]
  
  Note $Y_t \sim \Bin(kn, p)$ so
  \[
    \Pbb(Y_t \ge k - 1)
    \le \Pbb(|Y_t - pkn| \ge \eps k - 1)
    \le 2 \exp \left( -\frac{(\eps k - 1)^2}{3k} \right)
    \le \exp \left( -\frac{\eps^2 k}{3.5} \right)
  .\]
  Note that if $k \ge \frac{8}{\eps^2} \log n$, then the above is $\le n^{-2 - c}$.
  Then union bound over all $\le n^2$ choices for $t$ to see no such $Y_t$
  satisfies $Y_t \ge k - 1$.
\end{proof}