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CSE 311: Foundations of Computing

CSE 311: Foundations of Computing. Fall 2013 Lecture 24: NFAs, regular expressions, and NFA DFA. 0. 2. S0 [1]. S1 [0]. 0. 1. 2. 3. 3. 0. 1. 1. 0. S2 [1]. S3 [0]. 1. 2. 2. 3. 3. 1. 2. 2. 3. S4 [1]. S5 [0]. 0. 0. 1. 3. highlights. FSMs with output at states

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CSE 311: Foundations of Computing

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  1. CSE 311: Foundations of Computing Fall 2013 Lecture 24: NFAs, regular expressions, and NFADFA

  2. 0 2 S0[1] S1 [0] 0 1 2 3 3 0 1 1 0 S2[1] S3 [0] 1 2 2 3 3 1 2 2 3 S4[1] S5 [0] 0 0 1 3 highlights • FSMs with output at states • State minimization 0 2 S0[1] S1 [0] 0 1 2 3 0 1,3 3 1,2 0 S2[1] S3 [0] 1 ⇒ 2 3

  3. highlights Lemma: The language recognized by a DFA is the set of strings x that label some path from its start state to one of its final states 0 0 1 1 1 s0 s1 s2 s3 0 0,1

  4. nondeterministic finite automaton (NFA) • Graph with start state, final states, edges labeled by symbols (like DFA) but • Not required to have exactly 1 edge out of each state labeled by each symbol--- can have 0 or >1 • Also can have edges labeled by empty string  • Definition: x is in the language recognized by an NFA if and only if x labels a path from the start state to some final state 1 1 1 s0 s1 s2 s3 0,1 0,1

  5. three ways of thinking about NFAs • Outside observer: Is there a path labeled by x from the start state to some final state? • Perfect guesser: The NFA has input x and whenever there is a choice of what to do it magically guesses a good one (if one exists) • Parallel exploration: The NFA computation runs all possible computations on x step-by-step at the same time in parallel

  6. NFAs and regular expressions Theorem: For any set of strings (language) described by a regular expression, there is an NFA that recognizes . Proof idea: Structural induction based on the recursive definition of regular expressions... Note: One can also find a regular expression to describe the language recognized by any NFA but we won’t prove that fact

  7. regular expressions over  • Basis: • ,  are regular expressions • a is a regular expression for any a • Recursive step: • If A and B are regular expressions then so are: (A  B) (AB) A*

  8. basis • Case : • Case : • Case a:

  9. basis • Case : • Case : • Case a: a

  10. inductive hypothesis • Suppose that for some regular expressions and there exist NFAs and such that recognizes the language given by and recognizes the language given by

  11. inductive step Case(A  B):

  12. inductive step Case(A  B): λ λ

  13. inductive step Case(AB):

  14. inductive step Case(AB): λ λ

  15. inductive step CaseA*

  16. inductive step CaseA* λ λ NA λ

  17. build an NFA for (01 1)*0

  18. solution (01 1)*0 0  1     0     1

  19. NFAs and DFAs Every DFA is an NFA • DFAs have requirements that NFAs don’t have Can NFAs recognize more languages?

  20. NFAs and DFAs Every DFA is an NFA • DFAs have requirements that NFAs don’t have Can NFAs recognize more languages? No! Theorem: For every NFA there is a DFA that recognizes exactly the same language

  21. conversion of NFAs to a DFAs • Proof Idea: • The DFA keeps track of ALL the states that the part of the input string read so far can reach in the NFA • There will be one state in the DFA for each subset of states of the NFA that can be reached by some string

  22. conversion of NFAs to a DFAs New start state for DFA • The set of all states reachable from the start state of the NFA using only edges labeled  f   a,b,e,f a b  e DFA NFA

  23. conversion of NFAs to a DFAs For each state of the DFA corresponding to a set S of states of the NFA and each symbol s • Add an edge labeled s to state corresponding to T, the set of states of the NFA reached by starting from some state in S, then following one edge labeled by s, and then following some number of edges labeled by  • T will be  if no edges from S labeled s exist f d 1   1 g b 1 c b,e,f c,d,e,g  e 1 1

  24. conversion of NFAs to a DFAs Final states for the DFA • All states whose set contain some final state of the NFA c e a,b,c,e a b DFA NFA

  25. example: NFA to DFA a  1 0 b c 0 0,1 NFA DFA

  26. example: NFA to DFA a,b a  1 0 b c 0 0,1 NFA DFA

  27. example: NFA to DFA 0 a,b a 1  1 c 0 b c 0 0,1 NFA DFA

  28. example: NFA to DFA 0 a,b a 1  1 1 b c 0 b c 0 0 0,1 b,c NFA DFA

  29. example: NFA to DFA 0  a,b a 1 0 1  1 1 b c 0 b c 0 0 0,1 b,c NFA DFA

  30. example: NFA to DFA 0,1 0  a,b a 1 0 1  1 1 b c 0 b c 0 0 0,1 b,c NFA DFA

  31. example: NFA to DFA 0,1 0  a,b a 1 0 1  1 1 b c 0 1 b c 0 0 0,1 b,c a,b,c 0 NFA DFA

  32. example: NFA to DFA 0,1 0  a,b a 1 0 1  1 1 b c 0 1 b c 0 0 0 0,1 1 b,c a,b,c 0 NFA DFA

  33. exponential blow-up in simulating nondeterminism • In general the DFA might need a state for every subset of states of the NFA • Power set of the set of states of the NFA • n-state NFA yields DFA with at most 2n states • We saw an example where roughly 2n is necessary Is the nth char from the end a 1? • The famous “P=NP?” question asks whether a similar blow-up is always necessary to get rid of nondeterminism for polynomial-time algorithms

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