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**Linear Bounded Automata LBAs**

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Linear Bounded Automata (LBAs) are the same as Turing Machines with one difference: The input string tape space is the only tape space allowed to use Courtesy Costas Busch - RPI

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Linear Bounded Automaton (LBA) Input string Working space in tape Left-end marker Right-end marker All computation is done between end markers Courtesy Costas Busch - RPI

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We define LBA’s as NonDeterministic Open Problem: NonDeterministic LBA’s have same power with Deterministic LBA’s ? Courtesy Costas Busch - RPI

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Example languages accepted by LBAs: LBA’s have more power than NPDA’s LBA’s have also less power than Turing Machines Courtesy Costas Busch - RPI

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The Chomsky Hierarchy Courtesy Costas Busch - RPI

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Unrestricted Grammars: Productions String of variables and terminals String of variables and terminals Courtesy Costas Busch - RPI

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Example unrestricted grammar: Courtesy Costas Busch - RPI

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Theorem: A language is recursively enumerable if and only if is generated by an unrestricted grammar Courtesy Costas Busch - RPI

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Context-Sensitive Grammars: Productions String of variables and terminals String of variables and terminals and: Courtesy Costas Busch - RPI

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The language is context-sensitive: Courtesy Costas Busch - RPI

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Theorem: A language is context sensistive if and only if is accepted by a Linear-Bounded automaton Observation: There is a language which is recursive but not context-sensitive Courtesy Costas Busch - RPI

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The Chomsky Hierarchy Non-recursively enumerable Recursively-enumerable Recursive Context-sensitive Context-free Regular Courtesy Costas Busch - RPI

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Decidability Courtesy Costas Busch - RPI

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Consider problems with answer YES or NO Examples: Does Machine have three states ? Is string a binary number? Does DFA accept any input? Courtesy Costas Busch - RPI

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A problem is decidable if some Turing machine decides (solves) the problem Decidable problems: Does Machine have three states ? Is string a binary number? Does DFA accept any input? Courtesy Costas Busch - RPI

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The Turing machine that decides (solves) a problem answers YES or NO for each instance of the problem Input problem instance YES Turing Machine NO Courtesy Costas Busch - RPI

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The machine that decides (solves) a problem: If the answer is YES then halts in a yes state If the answer is NO then halts in a no state These states may not be final states Courtesy Costas Busch - RPI

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Turing Machine that decides a problem YES states NO states YES and NO states are halting states Courtesy Costas Busch - RPI

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Difference between Recursive Languages and Decidable problems For decidable problems: The YES states may not be final states Courtesy Costas Busch - RPI

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Some problems are undecidable: which means: there is no Turing Machine that solves all instances of the problem A simple undecidable problem: The membership problem Courtesy Costas Busch - RPI

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The Membership Problem Input: Turing Machine String Question: Does accept ? Courtesy Costas Busch - RPI

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Theorem: The membership problem is undecidable (there are and for which we cannot decide whether ) Proof: Assume for contradiction that the membership problem is decidable Courtesy Costas Busch - RPI

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Thus, there exists a Turing Machine that solves the membership problem accepts YES NO rejects Courtesy Costas Busch - RPI

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Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input Courtesy Costas Busch - RPI

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Turing Machine that accepts and halts on any input YES accept accepts ? NO reject Courtesy Costas Busch - RPI

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Therefore, is recursive Since is chosen arbitrarily, every recursively enumerable language is also recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Courtesy Costas Busch - RPI

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Therefore, the membership problem is undecidable END OF PROOF Courtesy Costas Busch - RPI

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Another famous undecidable problem: The halting problem Courtesy Costas Busch - RPI

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The Halting Problem Input: Turing Machine String Question: Does halt on input ? Courtesy Costas Busch - RPI

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Theorem: The halting problem is undecidable (there are and for which we cannot decide whether halts on input ) Proof: Assume for contradiction that the halting problem is decidable Courtesy Costas Busch - RPI

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Thus, there exists Turing Machine that solves the halting problem YES halts on NO doesn’t halt on Courtesy Costas Busch - RPI

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Construction of Input: initial tape contents YES NO Encoding of String Courtesy Costas Busch - RPI

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Construct machine : If returns YES then loop forever If returns NO then halt Courtesy Costas Busch - RPI

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Loop forever YES NO Courtesy Costas Busch - RPI

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Construct machine : Input: (machine ) If halts on input Then loop forever Else halt Courtesy Costas Busch - RPI

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copy Courtesy Costas Busch - RPI

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Run machine with input itself: Input: (machine ) If halts on input Then loop forever Else halt Courtesy Costas Busch - RPI

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on input : If halts then loops forever If doesn’t halt then it halts NONSENSE !!!!! Courtesy Costas Busch - RPI

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Therefore, we have contradiction The halting problem is undecidable END OF PROOF Courtesy Costas Busch - RPI

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Another proof of the same theorem: If the halting problem was decidable then every recursively enumerable language would be recursive Courtesy Costas Busch - RPI

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Theorem: The halting problem is undecidable Proof: Assume for contradiction that the halting problem is decidable Courtesy Costas Busch - RPI

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There exists Turing Machine that solves the halting problem YES halts on NO doesn’t halt on Courtesy Costas Busch - RPI

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Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input Courtesy Costas Busch - RPI

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Turing Machine that accepts and halts on any input NO reject halts on ? YES accept Halts on final state Run with input reject Halts on non-final state Courtesy Costas Busch - RPI

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Therefore is recursive Since is chosen arbitrarily, every recursively enumerable language is also recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Courtesy Costas Busch - RPI

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Therefore, the halting problem is undecidable END OF PROOF Courtesy Costas Busch - RPI

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