---
title: "STCONN"
type: concept
related: [Space Class L, Stconn]
source: https://www.jemoka.com/posts/kbhstconn/
confidence: high
status: active
---

\begin{equation} \text{STCONN} = \qty {\langle G, S, t \rangle : \text{$\exists^{?}$ a path from s $\to$ t $\in$ G}} \end{equation}
we solve this with BFS or DFS; but those algorithms&rsquo; working sets require linear space. Turns out, we can solve this algorithm \(\text{STCONN} \in \text{SPACE}\qty(\log^{2}\qty(n))\)
for directed \(G\), we are not sure if its in L. for undirected \(G\), Omer showed that its in L open problem: Savitch&rsquo;s Algorithm is really really slow; we are not sure if there is an algorithm in which we can solve it in better time.
Savitch&rsquo;s Algorithm Savitch&rsquo;s Algorithm is a middle-first search scheme for solving STCONN.
key observation: if there is a path form \(s \to t\) of length \(n\), then there must be a point \(w\) such that \(s\to w\) in \(\frac{n}{2}\) steps and \(w \to t\) in \(\frac{n}{2}\) steps.
Goal: brute force search for this \(w\). There is \(n\) possibilities for \(w\), we shall just try all of them! This engenders a more general problem named PATH: \(\text{STCONN}\qty(s,t) = \text{PATH}\qty(s,t,\ceil{\log n})\)
PATH \begin{equation} \text{PATH}\qty(x,y,k): \text{does there exist path of length $\leq 2^{k}$ from $x$ to $y$} \end{equation}
solution:
if \(k = 0\), then return TRUE IFF \(\qty(x,y) \in E \vee x = y\) (that \(x\) and \(y\) are adjacent or identical) else for each \(w \in V\) if \(\text{PATH}\qty(x,w,k-1) \wedge \text{PATH}\qty(w,y,k-1)\), return TRUE return FALSE space of this:
\begin{equation} S\qty(k) = O\qty(\log n) + S\qty(k-1) \end{equation}
because for every \(w\), we need to track \(\log n\) possible \(w\), and then also need to recurse down to keep track the two \(\text{PATH}\) calls; in particular, we recuse the two recurse PATH calls to prevent exponential growth. This gives \(S(t) = O\qty(k \log n)\), so for \(k = \log n\), we have \(S \qty(\log n) = O \qty(\log^{2} n)\)
time of this:
\begin{equation} T(k) = n 2 T(k-1) = \qty(2n)^{k} = n^{O(k)} \end{equation}
so for \(k = \log n\) (n nodes), then \(T \qty(\log n) = n^{O \qty(\log n)}\)
because we can&rsquo;t recuse time, so the 2 times factor exsits.
STCONN is in NL see STCONN is in NL
