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CSCI 2670 Introduction to Theory of Computing

CSCI 2670 Introduction to Theory of Computing. September 6, 2005. Announcements. Homework due next Tuesday (9/13) 1.7(d,h), 1.16b, 1.21b (show the GNFA steps),1.19a, 1.46c, 1.55(c,f,j)

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CSCI 2670 Introduction to Theory of Computing

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  1. CSCI 2670Introduction to Theory of Computing September 6, 2005

  2. Announcements • Homework due next Tuesday (9/13) • 1.7(d,h), 1.16b, 1.21b (show the GNFA steps),1.19a, 1.46c, 1.55(c,f,j) • Old text numbers 1.5(d,g), 1.12b, 1.16b, 1.14a, 1.23d, 1.38(same instructions on (0010*1*, , and Σ*) • If you ask me a question on a homework problem by email, please email me the problem in question • Quiz tomorrow • Section 1.2 & Formal definition of RE

  3. Agenda • Last week • Section 1.2 • NFA’s • Regular expression intro (Section 1.3) • Today • Continue regular expressions • This week • Sections 1.3 and 1.4

  4. Abuse of notation. These should be sets! RE inductive definition R is a regular expression if R is • a for some a   • ε •  • R1R2 where R1 and R2 are both regular expressions • R1R2 where R1 and R2 are both regular expressions • (R1*) where R1 is a regular expression

  5. RE’s and regular languages Theorem: A language is regular if and only if some regular expression describes it. • i.e., every regular expression has a corresponding DFA and vice versa

  6. RE’s and regular languages Lemma: If a language is described by a regular expression, then it is regular. • find an NFA corresponding to any regular expression • use inductive definition of RE’s

  7. a q1 q2 1. R=a for some a N = {{q1,q2},,,q1,{q2}} where (q1,a)={q2} and (r,x)= whenever r=q2 or x≠a

  8. q1 2. R=ε N = {{q1},,,q1,{q1}} where (q1,x)= for all x

  9. q1 3. R= N = {{q1},,,q1,} where (q1,x)= for all x

  10. Remaining constructions • R = R1R2 • R = R1R2 • R = R1* • These were all shown to be regular operators • We know we can construct NFA’s for R provided they exist for R1 and R2

  11. 0 R1 = 0 1 R2 = 1 R = Σ1 0 0 ε R3 = 01 ε 1 ε 1 ε ε 1 ε Example • R = 1 • R = (01)1

  12. 1 R1 = 1 0,1 R3 = * ε R = 1(0ε)* 0 R2 = 0ε ε 0 ε ε ε 0,1 1 ε ε ε ε ε ε Example2 • R = 1(0ε)*

  13. Equivalence of RE’s and DFA’s • We have seen that every RE has a corresponding NFA • Therefore, every RE has a corresponding DFA • I.e, every RE describes a regular language • We need to show that every regular language can be described by a RE • Begin by converting all DFA’s into GNFA’s • Generalized Non-deterministic Finite Automata

  14. GNFA’s • A GNFA is an NFA with the following properties: • The start state has transition arrows going to every other state, but no arrows coming in from any other state • There is exactly one accept state and there is an arrow from every other state to this state, but no arrows to any other state from the accept state • The start state is not the accept state

  15. GNFA’s (continued) • Except for the start and accept states, one arrow goes from every state to every other state and also from each state to itself • Instead of being labeled with symbols from the alphabet, transitions are labeled with regular expressions

  16. 01  0  1 0  10  Example GNFA

  17. Equivalence of DFA’s and RE’s • First show every DFA can be converted into a GNFA that accepts the same language • Then show that any GNFA has a corresponding RE that accepts the same language

  18. Converting a DFA into a GNFA • Add two new states • New start state with an ε jump to the original DFA’s start state • New accept state with an ε jump from each of the original DFA’s accept states • This new state will be the only accept state • All transition labels with multiple labels are relabeled with the union of the previous labels • All pairs of states without transitions get a transition labeled 

  19. 0 q2 1 qs qt q1 1 ε ε 0 q3 q4 0,1 0,1 Converting a DFA to a GNFA Add two new states

  20. qs qt ε ε Converting a DFA to a GNFA • All transition labels with multiple labels are relabeled with the union of the previous labels 0 q2 1 q1 1 0 q3 q4 0,1 01 0,1 01

  21. qs qt ε ε Converting a DFA to a GNFA • All pairs of states without transitions get a transition labeled  0 q2 1 q1 1 0 q3 q4 01 01

  22. qs qt ε ε Converting a DFA to a GNFA • The resulting state diagram is a GNFA • All GNFA properties are satisfied 0 q2 1 q1 1 0 q3 q4 01 01

  23. qs qt ε ε Converting a DFA to a GNFA • No step changed the strings accepted by the machine 0 q2 1 q1 1 0 q3 q4 01 01

  24. Converting a GNFA to a RE • If the GNFA has two states, then the label connecting the states is the RE • Otherwise, remove one state at a time without changing the language accepted by the machine until the GNFA has two states

  25. a12 q2 q1 a32 a13 q3 a33 Removing one state from a GNFA a12a13a33*a32 q1’ q2’

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