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It is mentioned in Which languages are recognized by one-counter machines? that one counter machine can accept $\{a^n b^n c^n\mid n\geq 0\}$. Can someone please explain how this is done?

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  • $\begingroup$ single counter can keep any fixed amount of counters by bit-slicing $\endgroup$ – Bulat Apr 6 at 10:10
  • $\begingroup$ According to the linked question, one-counter machines accept a subset of the context-free languages. In particular, they don’t accept non-context-free languages. $\endgroup$ – Yuval Filmus Apr 6 at 10:12
  • $\begingroup$ The answer states that $a^nb^nc^n$ can be accepted by a counter machine; that machine would have to use more than one counter. $\endgroup$ – Yuval Filmus Apr 6 at 12:01
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    $\begingroup$ "One counter machine" is ambiguous. As "one-counter machine", it means a counter machine that has one counter. As "one counter-machine", it means a counter machine that has one or more counters. Please edit your question to clarify. "one counter machine" should be avoided. Use either "a one-counter machine" or "a counter machine". Or use either "a one-counter automaton" or "a counter automaton". $\endgroup$ – Apass.Jack Apr 6 at 13:18
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    $\begingroup$ @Bulat, no, that's actually not possible, because in counter machines, the only operations allowed on the counter are "increment", "decrement", and "compare to zero", so you can't implement bit-slicing. (There is no "extract the least significant bit" or "divide by zero".) $\endgroup$ – D.W. Apr 6 at 16:47
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Be careful where the hyphens are! One counter-machine can recognize $a^n b^n c^n$, but a one-counter machine cannot.

From the answer you linked to:

Languages recognized by one[-]counter automata form a proper subset of the context free languages.

(There's a proof for this, but it's long and boring, so I'm going to leave it out; you can find it online if you're interested.)

Since $a^n b^n c^n$ is not context-free, it cannot be recognized by a one-counter automaton.

It can, however, be recognized by a two-counter machine (a machine with two counters). The basic structure goes something like this:

  • When you see an $a$, increment $X$, increment $Y$
  • When you see a $b$, decrement $X$
  • When you see a $c$, decrement $Y$
  • Accept iff the letters are in the right order (a trivial state machine), AND $X = 0$, AND $Y = 0$

This is "one counter-machine" (in the sense that it's a single machine using counters). But you need at least two counters to do it.

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