Is there a strictly non-deterministic one-counter language whose complement is one-counter?

Question

Let $A= \{L \mid L \;\text{is one-counter and $\bar{L}$ is also one-counter} \}$

Clearly, $\text{Deterministic one-counter} \subseteq A$

Is it the case that $ A = \text{Deterministic one-counter}$?

I know that for context-free languages the analogue is not the case. For example, let $P =\{ ww^r\}$. Then both $P$ and $\bar{P}$ are context-free but $P$ is not deterministic. Hence $A$ defines a (strict) subset of the context-free languages.

The question is: can we construct a similar one-counter example for which the same holds?

Something like $L={x\in{a,b}^*\mid x_{\lfloor|c|/2\rfloor}=a}$? — frafl, Mar 19 '13 at 16:19
A PDA with only one kind of symbols (aside from the bottom symbol) on its stack. — frafl, Mar 19 '13 at 16:27
how about this language : $L = { a^ib^j \mid i=j \ or\ i=2j } $ and $\bar{L} = {a^ib^j \mid i \neq j \ and\ i\neq 2j}$ — e_noether, Mar 20 '13 at 05:38
@emmy: How would a (nondeterministic) 1CM decide $\overline{L}$? — Shaull, Mar 20 '13 at 08:08
@emmy: Ooops, typo: I meant of course $x_{\lfloor x/2\rfloor}$ i.e. the symbol at the $\lfloor x/2\rfloor$-th position. — frafl, Mar 20 '13 at 13:07

score -1 · Answer 1 · answered Mar 20 '13 at 13:11

-1

In response to Shaull's comment above : enter image description here

The first one is an image of 1-counter accepting $a^ib^j$ s.t. $j<i$

The secong one is an image of 1-counter accepting $a^ib^j$ s.t. $j>i,\ j<2i$

The third one is an image of 1-counter accepting $a^ib^j$ s.t. $j>2i$

Here a/-/plus means on seeing a, irrespective of the counter value, increment the counter. b/>1?/sub means on seeing b, if counter value is greater than 1, then decrement the counter.

nop => no operation

$\lambda $ => empty string

answered Mar 20 '13 at 13:11

e_noether

1,289
2
12
19

1

Quite a long answer which is essentially just "Because its the union of three languages, each recognizing one interval of $i$s relative to $j$". – frafl Mar 20 '13 at 13:18
1

yeah :) just wanted to prove it can be done by 1-counter automata – e_noether Mar 20 '13 at 13:20
2

It's OK to elaborate a bit especially if it's a nice exercise for you, but please add a short summary. Additionally you should not use an answer to reply to a comment, but in this case this reply may become an actual answer to your question, so I think it's OK, too. – frafl Mar 20 '13 at 13:26
3

Unfortunately, there is no proof here. – Raphael Mar 20 '13 at 13:40

Is there a strictly non-deterministic one-counter language whose complement is one-counter?

1 Answers1