I too was unable to find this Problem in any of the modern books or any related reduction in our modern form either. Seems a lot complicated as compared to our standardized form too. ]]>

I had some more doubts on yesterday’s discussion. As you explained “2-way NFA with N states, there’s an equivalent nfa with n^(n+1) states. So that’s probably where you get the exponential time from.”

I am clear about that, but if I understand correctly the size of the string accepted by the 2Way NFA would still be (in the worst case) polynomial to number of states in the 2Way NFA. Cause if the smallest string accepted was exponential in size, the verification will also take exponential time and thus the problem by definition would not be NP Complete ? Or is there a mistake somewhere in understanding.

Since I don’t have the access to paper I am not sure if an exponential sized accepting sting can be verified in polynomial time (or have the mentioned about input size verification in general)

]]>I await your post on this as its still unclear why it should be NPComplete. On the face of it, doesn’t seem like it.

]]>It’s weird that G&J used “general” to mean “nondeterministic” though…

]]>1. “INSTANCE: A two-way nondeterministic finite state automaton A =(……”

2. “Comment: PSPACE-complete, even if |l|=2 and A is deterministic. If |l| = l the general problem is NP-complete [Galil, 1976]. ”

By general I suppose they mean the nondeterministic variant of the problem. They needed to specify it as in the previous line for a 2 alphabet version they mentioned that ‘even in deterministic case’ its PSPACE complete.

]]>So the question is what are they generalizing in the “general” problem? I’m not sure. My guess is that they’re generalizing from “two-way” to “any-way”. So instead of mapping to Qx{-1,0,1}, you can map into QxZ.

But I’m not sure. Have you read the Galil paper that they reference?

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