# week4 - ’ accepts the input which simulates the accepting...

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Week 4 Reduction via the history of Computation: Linear Bounded Turing Machine ( Automata) Definition: A linear bounded automaton is a restricted type of Turing machine where in the tape head isn’t permitted to move off the portion of the tape containing the input. M X Theorem: A LBA ={<M,x>|M is a LBA and M accepts x} is decidable. Proof: First Let us see the following lemma: Configuration(tape contents, state, head position) g: the number of symbols; q: the number of states; n: the length of a input string, so the length of the tape is also n. Then the distinct configuration of M for a tape of length n =(g n .q.n). We now know that the depth of computation tree is at most (g n .q.n) We can decide if M accept x or not by following the computation of all nodes in the computation tree. A NE ={<M>|M is a LBA. L(M) ≠φ } is un-decidable. Proof: Let <M,w> is an instance of L U ( L U ={<M,w>|TM M accepts w} ). We will construct a LBA M such that <M,w> L U iff <M > A NE. (More precisely, M

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Unformatted text preview: ’ accepts the input which simulates the accepting computation history of M with input x.) M M ’ configuration # C # C 1 # … C f # X LBA M ’ works as follows: On input x: (1) Check C is the initial configuration of M with input w; (2) C i C i+1; (3) C f is an accepting configuration for M. If all true then M ’ accepts the input x, otherwise M ’ reject x. M ’ accepts x iff M accepts w ( x is an accepting computation history for M on w) L(M ’ ) ≠φ iff <M, w> ∈ L U <M ’ > ∈ A NE iff <M, w> ∈ L U Now we are ready to state the reduction of A TM to A NE . Suppose that TM R decides A NE. Construct TM S that decides A TM as follows. S=“ On input <M,w>,where M is a TM and w is a string: (1)Construct LBA M ’ from M and w as described above. (2)Run R on input < M ’ >. (3)If R rejects, accept; if R accepts, reject.” Theorem: PCP is undecidable Proof: PCP simulates TM computation...
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week4 - ’ accepts the input which simulates the accepting...

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