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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 $m$?

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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

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