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When Savitch's famous theorem is stated, one often sees the requirement that $S(n)$ be space constructible (interestingly, it is omitted in Wikipedia). My simple question is: Why do we need this? I understand the requirement for $S(n)$ being in $\Omega(\log n)$, which is clear from the proof. But no proof I have seen so far explicitly uses that $S(n)$ is space constructable.

My explanation: in order to call the procedure REACH (or PATH or whatever you like to call it), the last parameter needs to be "spelled out", and in order not to leave our space bounds of S(n) for one call, we must not need more than $S(n)$ space to write it down.

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I believe your explanation is correct. The st-REACH subroutine gets $(s,t,\ell)$ as an input, and finds whether or not $t$ is reachable from $s$ by $\ell$ steps. $s$ and $t$ will be the initial and final configurations, and $\ell=2^{O(s(n))}$, the upper bound on the number of different configurations.

In order to set $\ell$ one needs to be able to compute $s(n)$ (or rather, $2^{O(s(n))}$). If this process takes more than $O(s^2(n))$ space, then the entire machine will have space more than the allowed. It is possible that even $O(s^2(n))$ is too much because of the recursive call to st-REACH (is there any other possible reason?), but I didn't check that.

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This is elaborated nicely in Dexter Kozen's Theory of Computation textbook, in chapter 2.

Savitch's Theorem (Theorem 1 in his paper) says: if $S(n) \ge \log n$, then $\text{NSPACE}(S(n)) \subseteq \text{DSPACE}(S(n)^2)$. Space-constructibility often seems to be assumed in a proof, but this requirement can be removed by restarting the search with a fixed space bound that is allowed to increase with each attempt.

The confusion perhaps arises because the original Savitch paper is largely about investigating whether $\text{NSPACE}(S(n)) \not\subseteq \text{DSPACE}(S(n))$. It therefore spends a lot of effort on space-constructible functions, due to the following observation from the paper:

It is natural to ask if there is any storage function whose deterministic and nondeterministic complexity classes are equal. The answer was given by Manuel Blum and is "yes". Blum showed that there are arbitrarily large storage functions L(n) such that a set is accepted within deterministic storage L(n) if, and only if, it is accepted within nondeterministic storage L(n). These functions L(n) are not, however, "well-behaved" and Theorem 3 does not apply to them.

(Here "well-behaved" refers to the space-constructible functions, called measurable by Savitch.)

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