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Deterministic Linear Bounded Automaton (LBA) is a single-tape TM that is not allowed to move its head past the right end of the input (but it can read and write on the portion of the tape that originally contained the input).

How can I prove that it is undecidable whether a Deterministic LBA M accepts an infinite number of inputs?

Dan Webster
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1 Answers1

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Given a Turing machine M and a string w, you can show that the following language is accepted by some LBA:

L = { x#y | x is a computation trace of M accepting w and y is any string }

Intuitively, this can be checked by an LBA by having it do the following:

  • Reject if x is not syntactically correct.
  • Reject if x doesn't start off in a correct initial configuration.
  • Reject if any step of the computation trace is incorrect.
  • Reject if x is a trace showing M rejects w.
  • Otherwise accept

It's possible for a TM to construct a description of the LBA that does this.

If M accepts w, then this language is infinite and so the LBA will accept infinitely many inputs. If M does not accept w, then this language is empty. Therefore, if a TM could decide whether the LBA had an infinite language, it could decide whether M accepts w, contradicting that this is impossible.

Hope this helps!

templatetypedef
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  • plz cite source if known to be (originally?) published somewhere – vzn Apr 26 '15 at 16:43
  • @vzn I got the main idea for this proof from the earlier result that testing whether the language of an LBA is undecidable (the idea is that you can build an LBA that accepts all possible accepting computations of a TM M run on a string w, so the halting problem reduces to testing whether the LBA has an empty language). The modification I made above makes it so that the language is either empty or infinite. It's been almost a year since I posted this and to the best of my knowledge I haven't seen this proof anywhere else, but it's not too hard to derive once you know the basic idea. – templatetypedef Apr 26 '15 at 20:30