Consider alternating TMs with polynomial space and at a linear number of alternations. Is the halting problem for this class EXPTIME-complete, or does the linear alternation bound place it somewhere between PSPACE and EXPTIME? Any references or insights would be much appreciated. Thanks, Bart
1 Answer
A polynomial-space ATM with polynomially many alternations can be simulated in deterministic PSPACE. More generally, if $s(n)\ge\log n$ and $a(n)\ge1$, then $$\begin{align*} \mathsf{ALTSPACE}(a(n),s(n))&\subseteq\mathsf{DSPACE}\bigl(s(n)a(n)+s(n)^2\bigr),\\ \mathsf{ALTSPACE}(a(n),s(n))&\subseteq\mathsf{NSPACE}\bigl(s(n)a(n)\bigr). \end{align*}$$ The first of these is stated as Theorem 4.2 in the classic paper
A. K. Chandra, D. C. Kozen, L. J. Stockmayer: Alternation, Journal of the ACM 28 (1981), no. 1, pp. 114–133, doi 10.1145/322234.322243.
where it is attributed to Borodin; the nondeterministic improvement is a consequence of the Immerman–Szelepcsényi theorem.
To show this, the following recursive procedure can check if the ATM accepts from a given configuration $C$ using $\le a$ alternations and $\le s$ work space:
Accept(C,s,a):
if C is in an ∃ state:
for all space-s configurations C':
if C' is reachable from C using only ∃ states (excluding C' itself)
and (C' is accepting or (a>0 and Accept(C',s,a-1)))
then ACCEPT
REJECT
if C is in a ∀ state:
for all space-s configurations C':
if C' is reachable from C using only ∀ states (excluding C' itself)
and (C' is rejecting or (a>0 and ¬Accept(C',s,a-1)))
then REJECT
ACCEPT
The main deterministic part of the algorithm has recursion depth $a$, and each instance uses $O(s)$ space, hence it uses $O(sa)$ space in total.
The reachability test can be done either in nondeterministic space $O(s)$ (NB: since the test effectively appears negated in the $\forall$ case, we need here the Immerman–Szelepcsényi theorem), or in deterministic space $O(s^2)$ using Savitch’s theorem (NB: the work space of the reachability test can be reused; it does not need to be saved on each recursive invocation).
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$\begingroup$ Thanks a lot for providing the full answer! $\endgroup$Bartosz Bednarczyk– Bartosz Bednarczyk2026-03-31 09:05:05 +00:00Commented 18 hours ago
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$\begingroup$ You’re welcome. $\endgroup$Emil Jeřábek– Emil Jeřábek2026-03-31 09:06:33 +00:00Commented 18 hours ago
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1$\begingroup$ Ah, my déja vu feeling was not unjustified: cstheory.stackexchange.com/a/55811 . $\endgroup$Emil Jeřábek– Emil Jeřábek2026-03-31 09:11:42 +00:00Commented 18 hours ago