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CSCI 2670 Introduction to Theory of Computing

CSCI 2670 Introduction to Theory of Computing. October 13, 2004. Agenda. Yesterday Variants of Turing machines Allow “stay put” state Multiple tapes Nondeterministic Today Prove equivalence of deterministic and nondeterministic Turing machines Enumerators. Announcements. Quiz tomorrow

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CSCI 2670 Introduction to Theory of Computing

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  1. CSCI 2670Introduction to Theory of Computing October 13, 2004

  2. Agenda • Yesterday • Variants of Turing machines • Allow “stay put” state • Multiple tapes • Nondeterministic • Today • Prove equivalence of deterministic and nondeterministic Turing machines • Enumerators October 13, 2004

  3. Announcements • Quiz tomorrow • High-level description of TM • Trace through a TM’s operation • Tape notation & configuration notation • High-level description of equivalences • Reminder: tutorial sessions are back • Office hours return to normal • Tuesday 3:00 – 4:00 • Wednesday 3:00 – 4:00 October 13, 2004

  4. Nondeterministic Turing machines • Same as standard Turing machines, but may have one of several choices at any point δ : Q × Г→ P(Q× Г × {L,R}) October 13, 2004

  5. Equivalence of machines Theorem: Every nondeterministic Turing machine has an equivalent deterministic Turing machine Proof method: construction Proof idea: Use a 3-tape Turing machine to deterministically simulate the nondeterministic TM. First tape keeps copy of input, second tape is computation tape, third tape keeps track of choices. October 13, 2004

  6. 0 1 0 1 ~ ~ ~ ~ M a a # 0 1 ~ ~ ~ 1 1 2 1 3 ~ ~ ~ Proof idea Input tape never changes Computation tape Decision path Try new decision paths until the string is accepted October 13, 2004

  7. Intuition • Consider the nondeterministic calculations as a tree • Each node represents a configuration • A node for configuration C1 has one child for each configuration C2 such that C1 yields C2 • Root of tree is configuration q1w • A configuration may appear more than once in the tree October 13, 2004

  8. Example • Nondeterministic TM that accepts {ww | w  {a,b}*} • Use 2 tapes • Copy input to tape 2 • Position heads at beginning of tapes • Move both heads right simultaneously October 13, 2004

  9. Nondeterministic solution • Nondeterministically choose the midpoint • Mark this point on tape 2 and return tape 2’s head to beginning • Compare strings • If tape head points to ~ on tape 1 and midpoint marker on tape 2 then accept • Otherwise, if all possible midpoints have been tried then reject • Otherwise, try a new midpoint October 13, 2004

  10. (a,~) → {a,a},{R,R} {b,~} → {b,b},{R,R} (a,x) → {a,a},{R,R} {b,x} → {b,b},{R,R} qreject (a,a) → (L,L) (b,b) → (L,L) (a,a) → (R,R) (b,b) → (R,R) (a,~)→ (a,a),(R,R) (b,~)→(b,b),(R,R) (~,~) → (L,L) (a,a) → (a,x),(S,Res) (b,b) → (b,x),(S,Res) (a,b), (b,a), (a,x), (b,x), (~,a), (~,b) → (Res,Res) (~,~) → (L,L) qaccept (~,x) → (S,S) Nondeterministic TM October 13, 2004

  11. Move right Compare Right Accept Compare Reset Right Right Accept Reset Compare Compare … Tree representation October 13, 2004

  12. Deterministic equivalent • Assume midpoint is at beginning • If so accept • If not … • Assume midpoint is after first symbol • If so accept • If not … • Assume midpoint is after second symbol • If so accept • If not … • etc. … October 13, 2004

  13. How should it search the tree? • Breadth first search • Search all possibilities involving k steps before searching any possibilities involving (k+1) steps • What’s wrong with depth first search? • If some sequence of choices results in no halting, we will never get to the accept state October 13, 2004

  14. When does it halt? • When it reaches an accept state or • Return accept October 13, 2004

  15. Will it halt on strings in the language? • Yes if the TM accepts the input string • Let b be the largest number of children of any node • Can we be sure b is finite? • Let k be the minimum number of steps it takes to get to the accept state • This method will take at most bk steps to get to the accept state October 13, 2004

  16. What about strings not in the language? • Won’t halt • That’s okay October 13, 2004

  17. Equivalence of approaches Corollary: A language is Turing-recognizable if and only if some nondeterministic Turing machine recognizes it. October 13, 2004

  18. Equivalence of approaches Corollary: A language is Turing-decidable if and only if some nondeterministic Turing machine decides it. Proof: Homework • Modify proof in the book October 13, 2004

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