Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 59 / 2 / a / Solution

With a read-only input tape, deterministic space complexity class $\mathrm{SPACE}(s(n))$ consists of languages decidable by deterministic Turing machines using $O(s(n))$ work-tape cells. The nondeterministic space complexity class $\mathrm{NSPACE}(s(n))$ uses Nondeterministic Turing machines, with the bound holding on every computation path and acceptance meaning that some path accepts. Input storage is not charged. We use the standard finite-state, fixed-alphabet model.

The Savitch theorem isFirst suppose a work-space budget $m\ge \lceil {\log }_2(n+2)\rceil$ is available. A configuration graph for computations restricted to $m$ cells has configurations encoded in $O(m)$ bits: work contents, control state and head positions, including $O(\log n)$ bits for the input head. Its size is at most $2^{cm}$ for a machine-dependent constant $c$. Add accept and exit sinks if needed; the size bound merely changes $c$. A reachable configuration has a simple path of length smaller than the number of configurations.

For encoded configurations $u,v$, define $R(u,v,k)$ to ask whether a path of length at most $2^k$ exists. At $k=0$, test equality or one transition. For $k>0$, enumerate every candidate middle configuration $w$ and testAny path of the given length can be split into two halves of length at most $2^{k-1}$; conversely concatenation gives a path of length at most $2^k$. This proves the recursion. Taking $k=\lceil cm\rceil$ suffices. Depth is $O(m)$, and each recursive frame stores only its endpoints, the current middle configuration and counters in $O(m)$ bits. The two recursive calls are made sequentially and reuse their space. The total is $O(m^2)$, even though the time can be very large.

The printed hypothesis does not say that $s$ is computable. The space-bound discovery by exit reachability removes that issue. Begin with $m=\lceil {\log }_2(n+2)\rceil$ and use the preceding recursion to test both whether an accepting state is reachable within the budget and whether a reachable state has a transition leaving the budget. Accept in the first case. If acceptance is absent and an exit is reachable, double $m$ and repeat. If neither is reachable, reject: all computations remain within this finite configuration graph, and none accepts.

Every path of the original machine uses at most $Cs(n)$ cells. Once $m$ reaches that bound there can be no reachable exit, so this procedure terminates. The final budget is $O(s(n))$ by doubling, and all stages reuse the same storage. Hence its deterministic space is $O(s(n{)}^2)$ without requiring a machine that first computes $s(n)$. This proves the stated general form.