$\require{mathtools}
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Adapted from a talk by Prasanna Ramakrishnan at Stanford’s teach-us-anything seminar.
You probably know the drill about $s$-$t$ connectivity:
- DFS/BFS gets you $\TISP(m,n)$,
- Savitch’s gets you $\TISP(n^{O(\log n)}, \log^2 n)$.
Can you get the best of both worlds? Some believe $\TISP(\poly(n), \polylog(n))$ is possible. But for now, the best we can get in polynomial time is $n / 2^{\Omega(\sqrt{\log n})}$ space.
Idea
Pras claims that if you’re given one week and the hint “find something in the middle of BFS and Savitch’s”, you could come up with it. I doubt I could.
The idea is to split the problem into two parts:
- one BFS-like part that “splits the path from $s$ to $t$ into $n/\lambda$ segments of length $\lambda$”;
- one Savitch-like part that can check whether two nodes are at distance $\leq \lambda$ (as long as $\lambda$ is small).
The BFS-like part makes a lot of concessions in time in order to cut space. The Savitch-like part makes a lot of concessions in space in order to cut time.
BFS-like part: cut it up
BFS requires a lot of space to store all the “equidistance shells”: for each $d$, it stores the set$V_d$ of all vertices at distance $d$ from node $s$. One natural way to save on this is to only store one out of $\lambda$ of those shells: there must be some remainder $r$ such that the shells $V_r, V_{\lambda+r}, V_{2\lambda+r}, \ldots$ contain only $O(n/\lambda)$ nodes.
How do you compute $V_{(i+1)\lambda+r}$ from the previous ones though? The idea is that $v \in V_{(i+1)\lambda+r}$ if and only if:
- it’s at distance exactly $\lambda$ from $V_{i\lambda+r}$;
- it’s not in any of the earlier shells $V_0, \ldots, V_{i\lambda+r-1}$.
The first part is easy to check using two small-distance queries from each $w \in V_{i\lambda+r}$. And you can check the second part simply by checking that $v$ is not a distance $<\lambda$ from the previous shells we memorized $V_{r}, \ldots, V_{(i-1)\lambda + r}$. That’s a lot of queries, but we’re fine with polynomial blow-ups in time!
Overall, we only make a polynomial number of queries, and we need $O(\frac{n \log n}{\lambda})$ bits of memory to store all the shells $V_r, V_{\lambda+r}, \ldots$. So if there was some $\TISP(t(n,\lambda),s(n,\lambda))$ way to check whether two points are at distance at most $\lambda$, we could solve STCON in
\[\TISP(\poly(n)\cdot t(n,\lambda), \frac{n \log n}{\lambda} + s(n,\lambda)).\]
This means that to get poly-time sublinear-space connectivity, we need
- $\lambda \geq \omega(\log n)$;
- $t(n, \lambda) \leq \poly(n)$;
- $s(n, \lambda) \leq o(n)$.
Savitch-like part
Let’s figure what time-space trade-off we can get from Savitch assuming $s$ and $t$ are at distance $\leq \lambda$. Out of the box, it gets you $n^{O(\log \lambda)}$ time and $O(\log n \log \lambda)$ space. Since $\lambda$ is superconstant, this is definitely not going to be polynomial-time, but this is already something nontrivial: for example, setting $\lambda = (\log n)^{100}$ gets you $\TISP(n^{O(\log \log n)}, n/(\log n)^{99})$, which is incomparable with BFS and Savitch’s.
But really, out-of-the-box Savitch’s cares way too much about optimizing space. We only want sublinear space (we have this annoying $\frac{n \log n}{\lambda}$ anyway), so we should be able to do much better!
The trick is to reduce Savitch’s branching factor at the expense of space. Instead of choosing a specific middle vertex among all $n$ possible vertices, we’ll choose a collection of possible middle vertices among $k$ collections of $n/k$ vertices (decided arbitrarily, say by their remainder modulo $k$).
The specification of one recursion of Savitch’s then becomes:
$\mathrm{Savitch}(d, i \in [k],j \in [k],x \in \zo^{n/k})$: returns a vector $y \in \zo^{n/k}$ representing which nodes from collection $j$ are reachable within distance $\leq d$ from at least one of the nodes in collection $i$ that are indicated by a $1$ in $x$.
To solve this recursive problem, first initialize $y \gets 0^{n/k}$, then for each possible “middle collection” $l \in [k]$:
- make a recursive call to find out the vector $z \in \zo^{n/k}$ of each possible node in collection $l$ that is reachable within distance $\leq d/2$ from at least one of the nodes in collection $i$ indicated by $x$;
- make a second recursive call to find out the vector $w \in \zo^{n/k}$ of each possible node in collection $j$ that is reachable within distance $\leq d/2$ from at least one of the nodes in collection $l$ indicated by $z$;
- set $y \gets y \vee w$ (component-wise).
Overall, there are $\log \lambda$ recursion levels, the branching factor is $k$, and we need $\poly(n)$ time and $O(n/k + \log k)$ memory at each recursion level, so this gives
\[\TISP\left(\poly(n) \cdot k^{O(\log \lambda)}, (\log \lambda)(n/k + \log k)\right).\]
Choosing parameters
Let’s ignore the $+ \log k$, it will end up being small anyway. We’re left with
\[\TISP\left(\poly(n) \cdot k^{O(\log \lambda)}, \frac{n \log n}{\lambda} + \frac{n \log \lambda}{k}\right).\]
The log factors in the right-hand side are probably not going to be very different, so a reasonable first step is to equalize the denominators by setting $k \coloneqq \lambda$, and get
\[\TISP\left(\poly(n) \cdot \lambda^{O(\log \lambda)}, \frac{n \log n}{\lambda}\right).\]
Then, it only remains to solve for $\lambda^{O(\log \lambda)} \leq\poly(n)$, which holds as long as $\lambda \le 2^{O(\sqrt{\log n})}$. So the best bound we can get for polynomial time is
\[
\TISP\left(\poly(n), \frac{n}{2^{\Omega\p{\sqrt{\log n}}}}\right).
\]