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I'm interested in the computational complexity of deciding equivalence of 2DFAs.

It is known that converting 2DFA to DFA can incur a blow up in states. However I'm not sure whether this automatically tells us something about the complexity of 2DFA equivalence.

Which leads to my question: Is there a hardness result for 2DFA equivalence like the PSPACE-completeness of language equivalence of NFAs and regular expressions. Conversely, I am interested in literature I might have missed that algorithmically solves the equivalence problem directly on 2DFA, without translating to DFAs first.

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According to the answer to this question: https://cs.stackexchange.com/questions/13456/what-is-the-complexity-of-the-emptiness-problem-for-2-way-dfas

The complexity of emptiness for 2-WAY DFAs is already PSPACE complete, so equivalence is also PSPACE-hard (and membership in PSPACE is easy by the single-exponential translation to DFA).

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  • $\begingroup$ Thanks! This is the reference I needed. $\endgroup$
    – Janmar
    Commented Oct 24, 2022 at 12:07

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