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Lecture 22

Lecture 22. Closure Properties for LFSA using NFA’s union (second proof) concatenation Kleene closure. LFSA closed under set union (again). LFSA closed under set union. Let L 1 and L 2 be arbitrary languages in LFSA Let M 1 and M 2 be NFA ’s s.t. L(M 1 ) = L 1 , L(M 2 ) = L 2

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Lecture 22

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  1. Lecture 22 • Closure Properties for LFSA using NFA’s • union (second proof) • concatenation • Kleene closure

  2. LFSA closed under set union(again)

  3. LFSA closed under set union • Let L1 and L2 be arbitrary languages in LFSA • Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2 • M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA • Construct NFA M3 from NFA’s M1 and M2 • Argue L(M3) = L1 union L2 • There exists NFA M3 s.t. L(M3) = L1 union L2 • L1 union L2 is in LFSA

  4. L1 L1 union L2 L2 M1 M3 M2 NFA’s Visualization • Let L1 and L2 be arbitrary languages in LFSA • Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2 • M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA • Construct NFA M3 from NFA’s M1 and M2 • Argue L(M3) = L1 union L2 • There exists NFA M3 s.t. L(M3) = L1 union L2 • L1 union L2 is in LFSA LFSA

  5. A Algorithm Specification • Input • Two NFA’s M1 and M2 • Output • NFA M3 such that L(M3) = L(M1) union L(M2) NFA M1 NFA M2 NFA M3

  6. a l a,b l a,b a,b A M3 NFA M1 NFA M2 NFA M3 Use l-transition a M1 a,b a,b a,b M2

  7. M2 M1 l l A M3 NFA M1 NFA M2 NFA M3 General Case

  8. Input NFA M1 = (Q1, S1, q1, d1, A1) NFA M2 = (Q2, S2, q2, d2, A2) Output NFA M3 = (Q3, S3, q3, d3, A3) What is Q3? Q3 = Q1 union Q2 union {p} where p is not in Q1 union Q2 Also, we need to rename states in Q2 if any share the same name with states in Q1 What is S3? S3 = S1 = S2 What is q3? q3 = p, the new state not in Q1 or Q2 A NFA M1 NFA M2 NFA M3 Construction

  9. Input NFA M1 = (Q1, S1, q1, d1, A1) NFA M2 = (Q2, S2, q2, d2, A2) Output NFA M3 = (Q3, S3, q3, d3, A3) What is A3? A3 = A1 union A2 What is d3? d3 = d1 union d2 union {(p,l,q1) ,(p,l,q2)} That is, all the old transitions plus d3(p,l) = {q1, q2} A NFA M1 NFA M2 NFA M3 Construction

  10. Comments • You should be able to execute this algorithm • You should understand the idea behind this algorithm • You should understand how this algorithm can be used to simplify design • You should be able to design new algorithms for new closure properties • You should understand how this helps prove result that regular languages and LFSA are identical • In particular, you should understand how this is used to construct an NFA M from a regular expression r s.t. L(M) = L(r) • To be seen later

  11. LFSA closed under set concatenation

  12. LFSA closed under set concatenation • Let L1 and L2 be arbitrary languages in LFSA • Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2 • M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA • Construct NFA M3 from NFA’s M1 and M2 • Argue L(M3) = L1concatenate L2 • There exists NFA M3 s.t. L(M3) = L1 concatenate L2 • L1 concatenate L2 is in LFSA

  13. L1 L1 concatenate L2 L2 M1 M3 M2 NFA’s Visualization • Let L1 and L2 be arbitrary languages in LFSA • Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2 • M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA • Construct NFA M3 from NFA’s M1 and M2 • Argue L(M3) = L1concatenate L2 • There exists NFA M3 s.t. L(M3) = L1 concatenate L2 • L1 concatenate L2 is in LFSA LFSA

  14. A Algorithm Specification • Input • Two NFA’s M1 and M2 • Output • NFA M3 such that L(M3) = L(M1) concatenate L(M2) NFA M1 NFA M2 NFA M3

  15. a l a,b a,b a,b A M3 NFA M1 NFA M2 NFA M3 Use l-transition a M1 a,b a,b a,b M2

  16. M2 M1 l l A M3 NFA M1 NFA M2 NFA M3 General Case

  17. Input NFA M1 = (Q1, S1, q1, d1, A1) NFA M2 = (Q2, S2, q2, d2, A2) Output NFA M3 = (Q3, S3, q3, d3, A3) What is Q3? Q3 = Q1 union Q2 Also, we need to rename states in Q2 if any share the same name with states in Q1 What is S3? S3 = S1 = S2 What is q3? q3 = q1, the initial state of Q1 A NFA M1 NFA M2 NFA M3 Construction

  18. Input NFA M1 = (Q1, S1, q1, d1, A1) NFA M2 = (Q2, S2, q2, d2, A2) Output NFA M3 = (Q3, S3, q3, d3, A3) What is A3? A3 = A2 What is d3? d3 = d1 union d2 union {(p,l,q2) | p in A1} That is, all the old transitions plus d3(p,l) = {q2} for all p in A1 A NFA M1 NFA M2 NFA M3 Construction

  19. Comments • You should be able to execute this algorithm • You should understand the idea behind this algorithm • You should understand how this algorithm can be used to simplify design • You should be able to design new algorithms for new closure properties • You should understand how this helps prove result that regular languages and LFSA are identical • In particular, you should understand how this is used to construct an NFA M from a regular expression r s.t. L(M) = L(r) • To be seen later

  20. LFSA closed under Kleene Closure

  21. LFSA closed under Kleene Closure • Let L be arbitrary language in LFSA • Let M1 be an NFA s.t. L(M1) = L • M1 exists by definition of L1 in LFSA and the fact that every FSA is an NFA • Construct NFA M2 from NFA M1 • Argue L(M2) = L1* • There exists NFA M2 s.t. L(M2) = L1* • L1* is in LFSA

  22. L1 L1* M1 M2 NFA’s Visualization • Let L be arbitrary language in LFSA • Let M1 be an NFA s.t. L(M1) = L • M1 exists by definition of L1 in LFSA and the fact that every FSA is an NFA • Construct NFA M2 from NFA M1 • Argue L(M2) = L1* • There exists NFA M2 s.t. L(M2) = L1* • L1* is in LFSA LFSA

  23. A Algorithm Specification • Input • NFA M1 • Output • NFA M2 such that L(M2) = L(M1)* NFA M1 NFA M2

  24. NFA M1 NFA M2 l a l M2 A Use l-transition a M1

  25. NFA M1 NFA M2 l M1 A l M3 General Case

  26. Input NFA M1 = (Q1, S1, q1, d1, A1) Output NFA M2 = (Q2, S2, q2, d2, A2) What is Q2? Q2 = Q1 union {p} where p is not in Q1 What is S2? S2 = S1 What is q2? q3 = p, the state not in Q1 NFA M1 NFA M2 A Construction

  27. Input NFA M1 = (Q1, S1, q1, d1, A1) Output NFA M2 = (Q2, S2, q2, d2, A2) What is A2? A2 = {p} What is d2? d2 = d1 union {(p,l,q1)} union {(q,l,p) | q in A1} That is, all the old transitions plus d2(p,l) = {q1} and d2(q,l) = d1(q,l) union {p} for all q in A1 NFA M1 NFA M2 A Construction

  28. Comments • You should be able to execute this algorithm • You should understand the idea behind this algorithm • Why do we need to make an extra state p? • You should understand how this algorithm can be used to simplify design • You should be able to design new algorithms for new closure properties • You should understand how this helps prove result that regular languages and LFSA are identical • In particular, you should understand how this is used to construct an NFA M from a regular expression r s.t. L(M) = L(r) • To be seen later

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