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Kleene Theorem II

Kleene Theorem II. Equivalence of FA, NDFA,NDFA- . Announcement. Microsoft is coming to RIT Integrating Technology for Performance Xcellence” Mike Adams, Director of Global Performance Xcellence Monday, April 1 – 3pm Webb Auditorium (Rm 1350) Booth Bld Reception will follow

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Kleene Theorem II

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  1. Kleene Theorem II Equivalence of FA, NDFA,NDFA-

  2. Announcement • Microsoft is coming to RIT • Integrating Technology for Performance Xcellence” • Mike Adams, Director of Global Performance Xcellence • Monday, April 1 – 3pm • Webb Auditorium (Rm 1350) Booth Bld • Reception will follow • RSVP 475-2199 or dlsbbbu@rit.edu

  3. Announcement • CS Colloquium • Many to Many Invocation: A New Paradigm for Ad Hoc Collaborative Systems • Hans-Peter Bischof • Tuesday, April 2 • 1-2 pm • 10-1124

  4. Homework • Homework #1 returned today • Homework #2 due today

  5. Homework • Speaking of Homework • Here’s Homework #3 (Due April 4) • From the textbook • 4.8b,c • 4.13b,c • 4.16b,c • 4.31c,d • 4,32a

  6. Before We Start • Any questions?

  7. Plan for today • Last time • We introduced two variants of the FA • Today we will show that each of three FAs are equivalent w.r.t. the class of languages that they accept

  8. Languages • Recall. • What is a language? • What is a class of languages?

  9. Regular Languages • Regular languages • Means of defining: Regular Expressions • Machine for accepting: Finite Automata

  10. Finite Automata • A finite automaton (finite-state machine) is a 5-tuple (Q, , qo, , A) where • Q is a finite set (of states) •  is a finite alphabet of symbols • qo Q is the start state • A Q is the set of accepting states •  is a function from Q x  to Q (transition function)

  11. Kleene Theorem • A language L over  is regular iff there exists an FA that accepts L. • If L is regular there exists an FA M such that L = L(M) • For any FA, M, L(M) is regular L(M), the language accepted by the FA can be expressed as a regular expression.

  12. Proving Kleene Theorem • Approach • Define 2 variants of the Finite Automata • Nondeterministic Finite Automata (NDFA) • Nondeterministic Finite Automata with  transitions (NDFA- ) • Prove that FA, NDFA, and NDFA-  are equivalent w.r.t. the languages they accept • For a regular expression, build a NDFA-  that accepts the same language • For an FA build a regular expression that describes the language accepted by the FA.

  13. Tonight • Last time we defined these two FA variants: • Nondeterministic Finite Automata (NDFA) • Nondeterministic Finite Automata with  transitions (NDFA- ) • Today we show they are equivalent • Questions?

  14. Equivalence • If L is a language over *, then the following 3 statements are equivalent: • L is accepted by a FA • L is accepted by a NDFA • L is accepted by a NDFA-

  15. Equivalence • How we will show this • Given an NDFA that accepts L, create an FA that also accepts L • Given an NDFA-  that accepts L, create an NDFA that also accepts L • Given an FA that accepts L, create a NDFA-  that also accepts L. Are we ready?

  16. Step 1: NDFA->FA • Given NDFA find FA • Let M = (Q, , q0, A, ) be a NDFA then • There exists a FA, M1 = (Q1, , q1, A1, 1) • Such that L(M) = L(M1)

  17. Step 1: NDFA -> FA • Basic idea • Recall that for a NDFA, : Q x   2Q • Use the states of M1 to represent subsets of Q. • If there is one state of M1 for every subset of Q, then the non-determinism of M can be eliminated. • This technique, called subset construction, is a primary means for removing non-determinism from an NDFA.

  18. Step 1: NDFA -> FA • Formal definition • M = (Q, , q0, A, ) be a NDFA • We define FA, M1 = (Q1, , q1, A1, 1) • Q1 = 2Q • q1 = {q0} • For q  Q1 and a  , • A1 = {q  Q1 | q  A   } • Note that we need only include states on M1 (subsets of Q) if the state is reachable.

  19. Step 1: NDFA -> FA • Algorithm for building M1 • Add {q0} to Q1 • While there are states of Q1 whose transitions are yet to be defined • Let q  Q1 • For each a  , determine the set of states, P, in M that are reachable from q on input a • If there is no state in Q1 corresponding to P, add one. • Define 1 (q, a) = state in Q1 corresponding to P • Define A1 as any state in Q1 that corresponds to a subset containing any of the final states of M

  20. Step 1: NDFA -> FA • Example

  21. Step 1: NDFA -> FA q0q3 0 1 0 0 1 q0q2 q0q1q3 0 0 0 1 q0 q0q1 1 1 1 0 q0q2q3 q0q1q2 1 0 1 q0q1q2q3

  22. Step 1: NDFA -> FA • Now we must show that M1 accepts the same language as M • Will show that for all x  * • *1 (q1, x) = *(q0, x) • Show by structural induction • Base case when x =  • *1 (q1, ) = q1 definition of *1 • = {q0} definition of q1 • = *(q0, ) definition of * • = *(q0, x) since x = 

  23. Step 1: NDFA -> FA • Show by structural induction • Induction: • Assume *1 (q1, x) = *(q0, x) • Show *1 (q1, xa) = *(q0, xa) • *1 (q1, xa) = 1 (*1 (q1, x) , a) def of *1 • = 1 (*(q0, x), a) induction hypothesis • = def of 1 • = *(q0, xa) def of *

  24. Step 1: NDFA -> FA • Show that M and M1 recognize the same language • x is accepted by M1 iff *1 (q1, x)  A1 • x is accepted by M1 iff *(q0, x)  A1 • By def of A1, • x is accepted by M1 iff *(q0, x)  A   • Thus • x is accepted by M1 iff x is accepted by M

  25. What have we shown • In Step 1 we’ve shown: • Given a NDFA • There exists an FA that accepts the same language • Non-determinism can be removed from an NDFA by using a subset construction algorithm. • Questions?

  26. What have we shown FA NDFA If L  NDFA then L  FA

  27. Step 2: NDFA- -> NDFA • Given NDFA- find NDFA • Let M = (Q, , q0, A, ) be a NDFA- then • There exists a NDFA, M1 = (Q1, , q1, A1, 1) • Such that L(M) = L(M1)

  28. Step 2: NDFA- -> NDFA • Basic idea • Recall that a NDFA- is still non-deterministic • Replace  transitions with non- transitions • Let 1(q,a) be the set of states reachable from q, reading symbol a, including those reachable via  transitions • Which is just * (q,a) 0 0  0

  29. Step 2: NDFA- -> NDFA • Basic idea • Accepting states • If q0 is not an accepting state of M but it is possible to get to an accepting state by just using  transitions, then q0 must be added to the set of accepting states of M1

  30. Step 2: NDFA- -> NDFA • Formal definition • M = (Q, , q0, A, ) be a NDFA- • We define NDFA, M1 = (Q1, , q1, A1, 1) • Q1 = Q • q1 = q0 • 1 (q,a) = * (q,a) • A1 = A  {qo} if ({qo} )  A   in M • = A otherwise

  31. Step 2: NDFA- -> NDFA • Algorithm for constructing M1 • Set of states is the same as M • Start state is the same as M • Accepting states is the same as M • For each state, q, in M • compute the  closure • For each state in the  closure,p, and for each symbol a, add all elements of  (p,a) to the set 1(q, a) • If you can get to to an accepting state in M from qo by only using  transitions, add q0 to A1

  32. Step 2: NDFA- -> NDFA • Example

  33. Step 2: NDFA- -> NDFA

  34. Step 2: NDFA- -> NDFA r p s w q0 t v u

  35. Step 2: NDFA- -> NDFA • Now we must show that M1 accepts the same language as M • Will show that for all x  * • *1 (q, x) = *(q, x) • Show by structural induction • Base step: show true for x = a • *(q, a) = 1(q,a) def of 1 • *1 (q, a) = def of *1 • = 1(q,a) since *1 (q, ) = {q}

  36. Step 2: NDFA- -> NDFA • Show by structural induction • Assume: *1 (q, x) = *(q, x) • Show: *1 (q, xa) = *(q, xa) def of *1 induction hypothesis def of 1

  37. Where we are • In Step 2 we’ve shown: • Given an NDFA-  • There exists an NDFA that accepts the same language •  transitions can be removed from an NDFA-  by replacing  transitions with non-  transitions in an already non-deterministic NDFA.

  38. Where we are NDFA NDFA-  If L  NDFA-  then L  NDFA

  39. Where we are • We’ve shown that: • If L is accepted by an NDFA-  it is also accepted by an NDFA • If L is accepted by an NDFA it is also accepted by an FA • So if L is accepted by an NDFA- , it is accepted by an FA. • We still need to show: • If L is accepted by an FA, it is also accepted by an NDFA-  • Questions?

  40. Where we are NDFA FA NDFA- 

  41. Step 3: FA-> NDFA- • Given FA find NDFA- • Let M = (Q, , q0, A, ) be a FA then • There exists a NDFA -, M1 = (Q1, , q1, A1, 1) • Such that L(M) = L(M1)

  42. Step 3: FA-> NDFA- • Basic idea • Since FAs are more restrictive than NDFA-s, any FA is essentially an NDFA- that doesn’t take advantage of: • Non-determinism • -transitions • Must consider details of transition functions

  43. Step 3: FA-> NDFA- • Formal definition • M = (Q, , q0, A, ) be a FA • We define NDFA-, M1 = (Q1, , q1, A1, 1) • Q1 = Q • q1 = q0 • A1 = A • 1 (q, ) =  for all q  Q • 1 (q, a) = { (q, a)} for all q  Q, a  

  44. Step 3: FA-> NDFA- • Now we must show that M1 accepts the same language as M • Will show that for all x  * • *1 (q, x) = {*(q, x) } • Show by structural induction • Base step: show true for x =  • *1 (q, ) = (q) def of *1 • = {q} since 1 (q, ) =  • = *(q, ) def of *

  45. Step 3: FA-> NDFA- • Show by structural induction • Assume: *1 (q, x) = {*(q, x) } • Show: *1 (q, xa) = {*(q, xa) } • def of *1 • since 1 (q, ) =  • induction hyp. • = { (*(q, x), a) } • = {*(q, xa) } def of *

  46. Step 3: FA-> NDFA- • Show that M and M1 recognize the same language • x is accepted by M1 iff *1 (q0, x)  A   • x is accepted by M1 iff {*(q0, x)}  A   • That’s the same as saying • *(q0, x)  A or • x is accepted by M • Thus • x is accepted by M1 iff x is accepted by M

  47. Equivalence! • What we have shown • Given an NDFA that accepts L, create an FA that also accepts L • Given an NDFA-  that accepts L, create an NDFA that also accepts L • Given an FA that accepts L, create a NDFA-  that also accepts L. • All 3 are equivalent

  48. Equivalence! NDFA FA NDFA- 

  49. Equivalence! NDFA FA NDFA- 

  50. Questions • break

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