CS 3240: Languages and Computation

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CS 3240 Languages and Computation

• Nondeterministic Finite Automata andEquivalence
  with Regular Expressions

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Review Regular Expressions

• Symbols and alphabet
• A symbol is a valid character in a language
• Alphabet is set of legal symbols
• Typically denoted as ?
• Metacharacters/metasymbols
• Defining reg-ex operations
• Escape character (\)
• Empty string ? and empty set ?
• Basic regular expressions
• Basic operations union, concatenation, repetition

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Review DFA

• A DFA is a five-tuple consisting of
• Alphabet ??
• A set of states Q
• A transition function d Q??? ? Q
• One start state q0
• One or more accepting states F ? Q
• A language accepted by a DFA is the set of
  strings such that DFA ends at an accepting state
  after processing the string

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Outline

• Nondeterministic Finite Automata (NFA)
• Equivalence between DFA and NFA
• Closure properties of regular languages
• Equivalence between regular expressions DFAs, and
  NFAs

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NFA

• Nondeterministic Finite Automaton
• Same input may produce multiple paths
• Allows transition with an empty string or
  transition from one state to different states
  given a character

q2
1
?
q1
q1
q2
q3
1
empty string transition
nondeterministic transition
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Example

• Accept strings containing either 101 or 11 as a
  substring

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Another Example

• Accept strings containing a 1 in the third
  position from the back

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How does NFA work?

• Start in start state
• If any ? transitions, clone a machine for each
• Read a symbol, clone a machine for each matching
  transition
• If a symbol is read and there is no way to exit
  from a state, then that machine dies
• At end of input if any machine accepts then accept

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Example Read 010110

• How would this NFA work?

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Read 010110
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Definition NFA

• A nondeterministic finite automaton (NFA) is
  defined by a 5-tuple, with
• Alphabet ??
• A set of states Q
• A transition function d Q ?Se? P(Q)
• One start state q0
• One or more accepting states F ? Q
• Notation Se S?e P(Q) is power set of Q
• What is the difference from DFA?

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Nondeterministic Transition

• The function d Q ?Se? P(Q) is the key
  difference!
• When reading symbol a while in state q, it may
  go to one of the states in d(q,a)?Q.
• e in Se allows empty-string transitions

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Languages

• Language accepted by a NFA is the set of strings
  such that NFA ends at an accepting state
• Each string of language is c1c2cn with ci?Se
  (possibly with e)
• States are qi ? d(qi-1,ci) for i1n
• qn is an accepting state

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DFA NFA

• Theorem For every language L accepted by an NFA,
  there is a DFA that accepts L.
• In other words, DFA and NFA are equivalent
  computational models.
• Proof idea When keeping track of a
  nondeterministic computation of an NFA N, use
  many fingers to point at the subset ? Q of
  states of N that can be reached on a given input
  string. We can simulate this computation with a
  DFA M with state space P(Q).

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Proof of DNANFA

• More formal proof Let A be the language
  recognized by the NFA N (Q,S,d,q0,F). Define
  the DFA M (Q,S,d,q0,F) by
• Q P(Q)
• d(R,a) q?Q q?d(r,a) for an r?R
• q0 q0
• F R?Q R contains an accept state of N
• Example

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Proof of DNANFA Contd

• This construction almost works, except for the
  e-transitions
• Solution Define- E(R) q q reachable from
  R using e steps - d(R,a) q?Q
  q?E(d(r,a)) for an r?R - q0 E(q0)
• DFA M accepts the same language as N.
• Therefore, a language is regular if and only if
  it is accepted by an NFA.

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NFA and Regular Expressions
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Regular Operations

• For languages L1 and L2, we define the following
  regular operations
• Union L1?L2 x x?L1 or x?L2
• Concatenation L1L2 xy x?L1 and y?L2
• Star L1 x1xk xj?L1 for all 1jk, k0
• Example L100,11, L20,1L1?L2
  0,1,00,11L1L2 000,001,110,111L1
  e,00,11,0000,0011,1100,1111,000000,
• Recall definition of regular expressions

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Closure under Union

• The regular languages are closed under the
  union operation. In other words, if L1 and L2 are
  regular languages, then so is L1 ? L2.
• Proof idea Construct a third automaton M3 such
  that L(M3) L1 ? L2.

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Union N N1 ? N2
N1
N2
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Closure under Concatenation

• The regular languages are closed under
  concatenation. In other words, if L1 and L2 are
  regular, then so is L1L2.
• Proof idea Construct a third automaton M3 such
  that L(M3) L1 L2.

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Concatenation N N1N2
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Closure under Star N N1
N1
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Regular Language Regular Expression

• As the names suggest, the following result holds
• Theorem A language is regular if and only if
  some regular expression describes it.
• Lemma If a language is described by a regular
  expression, then it is regular. (This follows
  from the closure properties of regular
  languages).
• Lemma If a language is regular, then it can be
  described by a regular expression.
• (This is harder and requires the notion of
  Generalized Nondeterministic Finite Automata
  (GNFA)).

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Regular Language ?Regular Expression

• If a language is regular, then it can be
  described by a regular expression.
• Proof Idea Use generalized nondeterministic
  finite automata where the labels on the
  transition arrows are regular expressions. For
  an R?R this GNFA recognizes L(R)
• Given a NFA/GNFA, repeatedly reduce it into a
  GNFA with a smaller number of states until only
  two states left.

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Example GNFA
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Example GNFA
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Definition Generalized NFA

• A Generalized nondeterministic finite automaton
  (GNFA) is defined by M(Q, S, d, qstart,
  qaccept) with
• Q finite set of states
• S the input alphabet
• qstart the start state
• qaccept the accept state
• d(Q\qaccept)?(Q\qstart) ? R the transition
  function
• R is the set of regular expressions over S

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DFA M ? GNFA M
Let M have k states Qq1,,qk
- Add two states qaccept and qstart
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Rip Internal state of GNFA

• If the GNFA M has more than 2 states, rip
  internal qrip to get equivalent GNFA M
• - Remove state qrip QQ\qrip
• - Change the transition function d to d(qi,qj)
  d(qi,qj) ? (d(qi,qrip)(d(qrip,qrip))d(qrip,qj))
  for every qi?Q\qaccept and qj?Q\qstart