Theoretical Computer Science Stack Exchange is a question and answer site for theoretical computer scientists and researchers in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading this paper, about the complexity of decision problems for reversal bounded counter machines. I got to Theorem 1 on Page 6. The theorem shows that there's a log-space NTM which can determine if a non deterministic reversal-bounded counter machine is empty or not. (A log-space NTM can be converted into a polynomial time DTM).

The proof shows that, for input machine represented as a string of length $n$, with $m$ counters, that $O(m\log n)$ space is required.

Here's where I get lost. The paper says that, since $m$ is fixed, we can consider the machine to take $O(\log n)$ space.

Does this mean that the algorithm is only uses log-space if $m$ is fixed? Would the corresponding deterministic algorithm then be exponential in terms of $m$?

share|cite|improve this question
It is not necessarily the case that the best algorithm is to determinize the NDTM, but certainly this approach may be exponential in $m$. – András Salamon Jul 6 '13 at 17:36
The problem given in Theorem 1 is NCM(m,r). So, both $ m $ and $ r $ are fixed. When you design an algorithm for this problem, you can embed the values of $ m $ and $ r $ into the finite state set. Moreover, you can squeeze the work tape to $O(\log n)$ space by using more tape symbols (depending the value of $ m $). – Abuzer Yakaryilmaz Jul 8 '13 at 8:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.