# l8 - Lecture 8 Turing Machines and Undecidability David...

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Lecture 8: Turing Machines and Undecidability David Dill Department of Computer Science 1

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Outline Weakening the TM model semi-infinite tape multi-stack PDAs two-counter machines Recursively enumerable vs. RE. A language that is not RE. Properties of recursive languages 2
Equivalence of NTMs and DTMs Thm Every language accepted by a DTM is also accepted by an NTM. Sketch Obviously, NTMs are no less powerful than DTMs. We can simulate an NTM on a multi-tape DTM using breadth-first search on the tree if IDs generated by the possibilities of the NTM. If the NTM accepts an input (either by final state or by halting), there is at least one finite path in the tree. Why not depth-first search ? Because the machine can go down an infinite possible computation forever, and never get around to doing one of the finite (halting) ones. The DTM accepts iff the simulated NTM enters a final state. (Rejection? No path of the NTM has an accepting state. Paths may be finite or infinite.) Simulation overhead: Suppose the maximum number of alternative moves in the NTM is m , and that the NTM takes n steps to accept its input. Then the number of nodes in the tree that must be explored is nm n . The cost is “single exponential” O (2 p ( n ) ) , where p ( n ) is a polynomial function. 3

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P = NP Running time of an NTM on an input: Shortest computation to an accepting state (either a final state, or by halting). Time complexity of an NTM – maximum running time over all inputs of length n . Can we simulate an NTM in polynomial time on a DTM?
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## This note was uploaded on 01/12/2010 for the course CS 154 at Stanford.

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l8 - Lecture 8 Turing Machines and Undecidability David...

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