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This language I think is not accepted by a deterministic one counter but accepted by a non-deterministic one counter :

$L = \{a^{i}b^{j}c^{k} \mid (i=j) \vee (j=k) \text{ such that } i\geq0, j\geq0, k\geq0\}$

But how to prove this claim?

Or is it the case that they are equivalent?

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You are right: the language $L$ from your question is accepted by a (nondeterministic) one-counter automaton.

Now for the deterministic case. Counter automata are a special case of push-down automata, where one only uses a single stack symbol (apart from a bottom-of-stack). Every (deterministic) one-counter language is also a (deterministic) push-down language. So in order to demonstrate it is not a deterministic one-counter language, I show a stronger result: it is even not accepted by a deterministic PDA.

According to Ogden, in his paper introducing Ogden's Lemma, that same language $L$ is inherently ambiguous context-free, meaning that every possible context-free grammar for the language is ambiguous, i.e., it has a string with two different succesful derivation trees. On the other hand, it is known that every deterministic PDA can be transformed into a non-ambiguous CFG. In fact I believe the standard construction PDA to CFG does the trick (but that is not obvious). Hence as there is no non-ambiguous CF grammar for $L$ (Ogden), there can be no deterministic PDA for $L$, and in particular no deterministic one-counter automaton.

Conclusion: for one-counter automata determinism is less powerful than the general model, using your example.

(In an edit I moved info from comments into this text.) Reference, see also Wikipedia: Ogden, W. (1968). A helpful result for proving inherent ambiguity. Math. Syst. Theory 2, 191–194.

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  • $\begingroup$ Which is that example? the example I found was the following cstheory.stackexchange.com/questions/9673/… palindrome cannot be recognized by pda with one stack symbol; be it non deterministic. $\endgroup$ – emmy Dec 10 '12 at 13:50
  • $\begingroup$ I think you need to be more elaborate. The second paragraph is misleading in particular: finite automata are a special case of PDA, too, but there nondeterminism is not more powerful. $\endgroup$ – Raphael♦ Dec 11 '12 at 0:20
  • $\begingroup$ thank you HendricJan that was helpful. Can you point me to the paper that you are referring to? $\endgroup$ – emmy Dec 11 '12 at 11:56
  • $\begingroup$ I am glad I could help you a little here. The full reference is added to the main answer, see also the wikipedia lemma. $\endgroup$ – Hendrik Jan Dec 11 '12 at 21:44
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All non-deterministic finite-state automata (NFA) can be represented with an equivalent deterministic finite-state automata (DFA). If you can make a NFSA that recognizes the language, you can convert it to a DFA. Have a look here for more information on how to do that.

In this particular case, I believe that that language cannot be represented even by a NFA, as it would require an infinite number of states to represent the different values of i, j and k (as FSA do not have "memory" so a machine cannot remember that it's counted i a's to count that many b's).

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    $\begingroup$ The question is about one counter automata. NFA + a counter. There are finite no of states and on each transition the counter value can be incremented, decremented or checked for zero $\endgroup$ – emmy Dec 10 '12 at 6:58

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