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Turing Machine accepting { zz | z ∈ {a, b}∗}

Turing Machine accepting { zz | z ∈ {a, b}∗} . Tracing strings aabb & bb by Brian Morales. let’s trace this TM on the string aabb to show rejection. Begin from the initial configuration:. q 0 ∆ aabb. ∆/∆, R  input symbol is ∆ / replace it with ∆ and move the state to the right.

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Turing Machine accepting { zz | z ∈ {a, b}∗}

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  1. Turing Machine accepting{zz | z ∈ {a, b}∗} Tracing strings aabb & bb by Brian Morales

  2. let’s trace this TM on the string aabbto show rejection. Begin from the initial configuration: q0∆aabb

  3. ∆/∆, R  input symbol is ∆ / replace it with ∆ and move the state to the right q0∆aabb Ⱶ ∆q1aabb

  4. a/A, R  input symbol is a / replace it with A and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb

  5. a/a, R  input symbol is a / replace it with aand move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb

  6. b/b, R  input symbol is b / replace it with b and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b

  7. b/b, R  input symbol is b / replace it with b and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆

  8. ∆/∆, L input symbol is ∆ / replace it with ∆ and move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b

  9. b/B, L input symbol is b / replace it with Band move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB

  10. b/b, L input symbol is b / replace it with b and move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB

  11. a/a, L input symbol is a / replace it with aand move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB

  12. A/A, R  input symbol is A / replace it with A and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB

  13. a/A, R  input symbol is a / replace it with A and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB

  14. b/b, R  input symbol is b / replace it with b and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B

  15. B/B, L  input symbol is B / replace it with B and move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB

  16. b/B, L  input symbol is b / replace it with B and move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB

  17. A/A, R  input symbol is A / replace it with A and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB

  18. B/B, L input symbol is B / replace it with Band move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB

  19. A/a, L input symbol is A / replace it with a and move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB Ⱶ ∆q5AaBB

  20. A/a, L input symbol is A / replace it with a and move the state to the left q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB Ⱶ ∆q5AaBB Ⱶ q5∆aaBB

  21. ∆/∆, R  input symbol is ∆ / replace it with ∆ and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB Ⱶ ∆q5AaBB Ⱶ q5∆aaBB Ⱶ ∆q6aaBB

  22. a/A, R  input symbol is a / replace it with A and move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB Ⱶ ∆q5AaBB Ⱶ q5∆aaBB Ⱶ ∆q6aaBB Ⱶ ∆Aq8aBB

  23. a/a, R  input symbol is a / replace it with aand move the state to the right q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB Ⱶ ∆q5AaBB Ⱶ q5∆aaBB Ⱶ ∆q6aaBB Ⱶ ∆Aq8aBB Ⱶ ∆Aaq8BB

  24. There is nowhere to go from q8 with input B, therefore the string aabb is rejected by this TM q0∆aabb Ⱶ ∆q1aabb Ⱶ ∆Aq2abb Ⱶ ∆Aaq2bb Ⱶ ∆Aabq2b Ⱶ ∆Aabbq2∆ Ⱶ ∆Aabq3b Ⱶ ∆Aaq4bB Ⱶ ∆Aq4abB Ⱶ ∆q4AabB Ⱶ ∆Aq1abB Ⱶ ∆AAq2bB Ⱶ ∆AAbq2B Ⱶ ∆AAq3bB Ⱶ ∆Aq4ABB Ⱶ ∆AAq1BB Ⱶ ∆Aq5ABB Ⱶ ∆q5AaBB Ⱶ q5∆aaBB Ⱶ ∆q6aaBB Ⱶ ∆Aq8aBB Ⱶ ∆Aaq8BB Ⱶ ∆AahrBB(reject) Accept only if state ha is reached!

  25. let’s trace this TM on the string bb to show acceptance. Begin from the initial configuration: q0∆bb

  26. ∆/∆, R  input symbol is ∆ / replace it with ∆ and move the state to the right q0∆bb Ⱶ ∆q1bb

  27. b/B, R  input symbol is b / replace it with B and move the state to the right q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b

  28. b/b, R  input symbol is b / replace it with b and move the state to the right q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆

  29. ∆ / ∆, L  input symbol is ∆ / replace it with ∆ and move the state to the left q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b

  30. b / B, L  input symbol is b / replace it with B and move the state to the left q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB

  31. B / B, R  input symbol is B / replace it with B and move the state to the right q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B

  32. B / B, L  input symbol is B / replace it with B and move the state to the left q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB

  33. B / b, L  input symbol is B / replace it with b and move the state to the left q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB Ⱶ q5∆bB

  34. ∆ / ∆, R  input symbol is ∆ / replace it with∆and move the state to the right q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB Ⱶ q5∆bB Ⱶ ∆q6bB

  35. b / B, R  input symbol is b / replace it withBand move the state to the right q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB Ⱶ q5∆bB Ⱶ ∆q6bB Ⱶ ∆Bq7B

  36. B / ∆, L  input symbol is B / replace it with∆and move the state to the left q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB Ⱶ q5∆bB Ⱶ ∆q6bB Ⱶ ∆Bq7B Ⱶ ∆q9B

  37. B / B, R  input symbol is B / replace it withBand move the state to the right q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB Ⱶ q5∆bB Ⱶ ∆q6bB Ⱶ ∆Bq7B Ⱶ ∆q9B Ⱶ ∆Bq6∆

  38. ∆ / ∆, S  input symbol is ∆ / replace it with∆and halt. String bb is accepted by this TM q0∆bb Ⱶ ∆q1bb Ⱶ ∆Bq2b Ⱶ ∆Bbq2∆ Ⱶ ∆Bq3b Ⱶ ∆q4BB Ⱶ ∆Bq1B Ⱶ ∆q5BB Ⱶ q5∆bB Ⱶ ∆q6bB Ⱶ ∆Bq7B Ⱶ ∆q9B Ⱶ ∆Bq6∆ Ⱶ ∆Bha∆ (accept)

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