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Unformatted text preview: CS324W8 – (1) RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8 L. van Zijl Department of Computer Science University of Stellenbosch 2010 RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8 University of Stellenbosch CS324W8 – (2) Reducibility Sipser p 191 I Now know that some problems are not solvable. I But, how do we prove for a given problem that it is unsolvable? I Method: reducibility – convert one problem into another in such a way that a solution to 2nd problem can be used to solve 1st problem. I Thus, if problem A is undecidable and reducible to problem B , then B is undecidable. I Known undecidable problems??? A TM RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8 University of Stellenbosch CS324W8 – (3) Reducibility Sipser p 192 HALT TM = { < M , w >  M is a TM and M halts on input w } Theorem 5.1: HALT TM is undecidable – proof idea I By contradiction: Assume is decidable, use that assumption to show then A TM must be decidable. I Assume is decidable, and TM R is decider. Now use R to construct TM S which decides A TM . I Given input < M , w > , build S which accepts if M accepts w , and rejects if M either rejects or loops. I But now S is not decider, so use R to decide if M halts on w . I (This is possible, because we assumed R is decider.) I If M doesn’t halt, then let S reject. I Therefore, if R exists, then A TM is decidable – contradiction. RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8 University of Stellenbosch CS324W8 – (4) Reducibility Sipser p 192 HALT TM = { < M , w >  M is a TM and M halts on input w } Theorem 5.1: HALT TM is undecidable – formal proof I Assume TM R decides HALT TM . Construct TM S to decide A TM as follows: I On input < M , w > : 1. Run TM R on input < M , w > 2. If R rejects, reject. 3. If R accepts, simulate M on w until it halts. 4. If M has accepted, accept; if M has rejected, reject. I Therefore, if R decides HALT TM , then S decides A TM . But A TM undecidable, therefore HALT TM undecidable. RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8 University of Stellenbosch CS324W8 – (5) Reducibility Sipser p 193 E TM = { < M >  M is a TM and L ( M ) = ∅} Theorem 5.2: E TM is undecidable – proof idea I Assume E TM is decidable – show that this implies A TM is decidable – contradiction. I Let R be TM that decides E TM and construct S using R . I Let S run R on modification of < M > : let < M 1 > reject all strings except w . I Use R to determine whether < M 1 > recognizes empty language. I Language will be nonempty iff it accepts w ....
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This note was uploaded on 04/13/2010 for the course CS 235 taught by Professor Sabnis during the Spring '10 term at Punjab Engineering College.
 Spring '10
 Sabnis
 Computer Science

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