MC 306 Theory of Computation Tuesday, 10703

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MC 306 Theory of ComputationTuesday, 10/7/03

• Todays Class
• Review Regular expressions and the languages
  they determine (9/23/03 slides)
• Equivalence of regular expressions and DFAs/NFAs
• From regular expression to NFA (9/25/03 slides)
• From NFA to regular expression todays slides

• Assignments
• Exercise p. 62 2.22c
• Hand-in p. 62 2.22d

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Equivalence of NFAs/DFAs to Regular Expressions

• Theorem. Given an NFA or DFA M, there is a
  regular expression R such that L(R) L(M).
• By previous theorem, this means that NFAs/DFAs
  and regular expressions describe exactly the same
  class of languages.
• Thats why its reasonable to call DFA/NFA
  languages regular languages.

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Converting from NFA/DFA to Regular Expression

• Basic Idea Given an NFA M, how to get equivalent
  regular expression R
• 1. Initialization Modify M to have only one
  start state s with only outgoing arrows, and only
  one final state f with only incoming arrows. All
  other states have both incoming and outgoing
  arrows (or else we can delete them!).
• 2. State reduction Modify M to remove one state
  (not s or f), and get equivalent machine, but
• This machine will be permitted to use regular
  expressions on the arrows called a
  generalized NFA, or GNFA.
• 3. Repeat to delete all states except s and f.
• 4. Final GNFA has a single arrow, labeled with
  regular expression R L(R) L(M).

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Outline of conversion procedure
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Eliminating a state

• Each non-start, non-final state has
• In-arrows,
• Out-arrows
• May have loops
• Replace every combination of in-loop-out (or
  in-out, if no loops), with a single arrow (see
  top picture)
• In bottom picture, how many replacements do we
  have to make to eliminate state 2?

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Examples

• For each of the following DFA/NFAs M, find a
  regular expression R such that L(R) L(M).

Ex. 2
Ex. 1