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\lecture{18}{April 4, 2008}{Ronitt Rubinfeld, lecture given by Krzysztof Onak}{Curtis Fonger}

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%We know the following:

\section{Review of Current Knowledge}

Let USTCONN be the problem of checking if two nodes $s$ and $t$ in an undirected graph $G$ are
connected. STCONN is the same problem for all graphs, including directed graphs.

\begin{enumerate}
\item ${\rm USTCONN} \in {\rm RL}$ (we've already seen this result in class)
\item ${\rm STCONN} \in {\rm NL}$ (non-deterministic log-space)
\item ${\rm NL} \subset {\rm L^2}$ (by Savitch's Theorem)
\item ${\rm RL} \subset {\rm L^{3/2}}$ (Saks \& Zhou 1999)
\item ${\rm USTCONN} \in {\rm L}$ (Reingold in STOC 2005, we will see this result today)
\item There is strong evidence that ${\rm L} = {\rm RL}$ (see Reingold, Trevisan, Vadhan 2006), but this is still an open problem.
\end{enumerate}

\begin{definition}
We say that a graph is an \emph{$(N,D,\lambda)$-graph} if it is a $D$-regular graph on $N$ nodes such that the absolute values of all eigenvalues but one are bounded by $\lambda$.
\end{definition}


\begin{fact}
%\label{thm1}
For all $\lambda < 1$, there exists an $\epsilon > 0$ such that for all $(N, D, \lambda)$-graphs $G$ and $S \subset G$ such that $\abs{S} < N/2$, $\abs{S \cup {\rm Neighborhood}(S)} \geq (1+\epsilon)\abs{S}$.
\end{fact}

\noindent The above fact implies the following.

\begin{fact}\label{fact:small_diam}
%\label{thm2}
Let $\lambda < 1$ be a constant.
Each $(N,D,\lambda)$-graph has $O(\log N)$ diameter.
\end{fact}

\section{Proof Idea}

We apply several graph operations to the input graph $G$ and turn every component of the graph
into a $(N_i,D,\lambda)$-graph, where $N_i$ is the final number of nodes in the $i$-th component
after the transformation. Let $G'$ be the transform	ed graph. It holds that $s$ and $t$ are connected in $G$ if and only if
some nodes $s'$ and $t'$ in $G'$ that correspond to $s$ and $t$, respectively, are connected in $G'$.
$G'$ has a number of nodes which is polynomial in the number of nodes of $G$, and hence, by Fact~\ref{fact:small_diam}, it suffices to explore all paths of logarithmic length that start at $s'$ to check if $s'$ and $t'$ lie in the same connected component. Obviously, we cannot compute and store $G'$ explicitly. Nevertheless, it turns out that in small space we can compute edges of $G'$ whenever we need them to explore $G'$.

\section{Graph Operations}

\subsection{Operation 1: Squaring}

If we have a $D$-regular graph $G = (V,E)$, then we define the $D^2$-regular graph $G^2$ as follows: $G^2 = (V,E')$, where $E' = \{(v,w) : \text{ there exists } t \text{ where } (v,t), (t,w) \in E\}$.  Every edge in $G^2$ is a length-2 path in $G$ (and we drop all length-1 paths in $G$).

\begin{lemma}
%\label{thm3}
If we have an $(N,D,\lambda)$-graph $G$, then $G^2$ is a $(N,D^2,\lambda^2)$-graph.
\end{lemma}
\begin{proof}
If $M$ is a random walk matrix of $G$, then for all eigenvectors $v$ perpendicular to $[1,1,...,1] = u$:
\[
	\|vM^2\| \leq \lambda \|vM\| \leq \lambda^2 \|v\|.
\]
\end{proof}

\subsection{Operation 2: Zig-zag Product}

We define the zig-zag product $G \textcircled{\sffamily z} H$, a $D^2$-regular graph, as the following: if we have $H = (V_2, E_2)$ (a $D$-regular graph) and $G = (V_1, E_1)$ (a $\abs{V_2}$-regular graph), then:
\begin{itemize}
\item Replace each node of $G$ with a copy of $H$ (thin edges).

\item Associate each node of $H$ with one edge of $G$ (thick edges).
That is, for each edge $e=(v,w)$ in $G$, there is a thick edge that connects a node in the copy of $H$ that replaces $v$
to a node in the copy of $H$ that replaces $w$, and moreover, each node in the new graph graph is incident to at most one thick edge.

\item $G \textcircled{\sffamily z} H = (V_1 \times V_2, E_3)$, where we define each edge of $E_3$ as the result of 3 steps:
\begin{enumerate}
\item Start at a node, then make one step in a copy of $H$ along a thin edge.
\item Make one step along a thick edge.
\item Repeat step 1: make one step in a copy of $H$ along a thin edge.
\end{enumerate}

\end{itemize}

\noindent The following lemma states the properties of the zig-zag product (we omit its proof).

\begin{lemma}
%\label{thm4}
If we have a $(N_1, D_1, \lambda_1 = 1-\gamma_1)$-graph $G$ and a $(D_1, D_2, \lambda_2 = 1-\gamma_2)$-graph $H$, then we have that $G {\normalfont\textcircled{\sffamily z}} H$ is a $(N_1 D_1, D_2^2, \lambda_3 = 1-\gamma_1 \gamma_2^2)$-graph.
\end{lemma}

\subsection{Algorithm}

The following fact holds.
\begin{fact}
There exists $D > 1$ and a graph $H$ such that $H$ is a $(D^4, D, 1/4)$-graph.
\end{fact}
From now on, we assume that $H$ is an arbitrary fixed graph of the above properties. We now describe the algorithm.
\begin{enumerate}
\item Reduce the input $(G,s,t)$ to $(G_0, s_0, t_0)$ where $G_0$ is a $D^2$-regular, non-bipartite and $s_0$ and $t_0$ are connected if and only if $s$ and $t$ are connected in $G$.
\item For each $k = 1,...,L$, where $L = O(\log N)$:
\begin{enumerate}
\item $G_k = G_{k-1}^2 \textcircled{\sffamily z} H$
\item Let $s_k$ (and $t_k$) be any vertex in the copy of $H$ that replaces $s_{k-1}$ ($t_{k-1}$, respectively) in $G_k$.
\end{enumerate}
\item Try all paths in $G_L$ of length $O(\log N)$ that start from $s_L$, and accept the input if any visits $t_L$.
\end{enumerate}

\noindent To see how the graph operations change the degree and the number of nodes, note that:
\begin{itemize}
\item $G_0$ is a $D^2$-regular graph on $N'$ nodes,
\item $G_0^2$ is a $D^4$-regular graph on $N'$ nodes,
\item $G_1 = G_0^2 \textcircled{\sffamily z} H$ is a $D^2$-regular graph on $D^4 \cdot N'$ nodes.
\end{itemize}
Hence in general, $G_i$ is a $D^2$-regular graph on $D^{4i} \cdot N'$ nodes. The following fact is relatively easy to prove.

\begin{fact}
%\label{thm5}
For any non-bipartite, undirected graph $G$ with $N$ nodes, we have that
\[
	\lambda(G) \leq 1 - \frac{1}{\poly(N)}.
\]
\end{fact}

It says that the initial spectral gap of each component of the graph is large enough to be amplified in $O(\log N)$ graph operations to at least a constant.
Let $C_0$ be any connected component in $G_0$, and then let $C_{i+1}$ be the component corresponding to $C_i$ in $G_{i+1}$.
We have
\[  \gamma(C_0) \geq \frac{1}{\poly(N)}.\]
By the properties of squaring,
\[  1-\gamma(C_{k-1}^2) \leq (1-\gamma(C_{k-1}))^2, \]
and hence,
\[  \gamma(C_{k-1}^2) \geq 2\gamma(C_{k-1})\left(1-\frac{1}{2}\gamma(C_{k-1})\right).  \]
Finally,
\begin{align*}
\gamma(C_k) = \gamma(C_{k-1}^2 \textcircled{\sffamily z} H) &\geq \gamma(H)^2 \cdot \gamma(C_{k-1}^2) \\
                                                            &\geq \left(\frac{3}{4}\right)^2 \cdot 2 \cdot \gamma(C_{k-1})\cdot\left(1-\frac{1}{2}\gamma(C_{k-1})\right) \\
                                                            &\geq \min\left\{\frac{17}{16}\gamma(C_{k-1}), \frac{1}{18}\right\}.
\end{align*}
We have the last inequality because we have two cases: if $\gamma(C_{k-1}) \leq \frac{1}{18}$, then $\gamma(C_k) \ge \frac{17}{16}\gamma(C_{k-1})$, and if $\gamma(C_{k-1}) \geq \frac{1}{18}$, then 
$\gamma(C_k) \ge \frac{1}{18}$. This implies that after $L=O(\log N)$ iterations we must have $\gamma(C_L) \ge \frac{1}{18}$.

\section{Implementation}

Why can this algorithm be implemented in logarithmic space?  Because:

\begin{itemize}
\item We keep the current path throughout the algorithm. We only need $O(1)$ bits per each step, since $G'$ has bounded degree. Besides, the length of the path is $O(\log N)$.
\item When going back, we can compute the location from scratch.
\item We can compute $G_0$ in logarithmic space and use it as a subroutine.
\item We can compute each $G_k$ recursively with $O(1)$ space per recursion level. We omit the details here.
\end{itemize}

\section{Open Problem}

The following challenging question remains open: ${\rm RL} = {\rm L}$?

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