Regular Languages. Kleene Theorem I. Proving Kleene Theorem. Kleene Theorem. Proving Kleene Theorem. Proving Kleene Theorem

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1 Regular Languages Kleene Theorem I Today we continue looking at our first class of languages: Regular languages Means of defining: Regular Expressions Machine for accepting: Finite Automata Kleene Theorem A language L over Σ is regular iff there exists an FA that accepts L.. If L is regular there exists an FA M such that L = L(M) 2. For any FA, M, L(M) is regular L(M), the language accepted by the FA can be expressed as a regular expression. Proving Kleene Theorem Approach Define 2 variants of the Finite Automata Nondeterministic Finite Automata (NFA) Nondeterministic Finite Automata with transitions ( -NFA) Prove that FA, NFA, and -NFA are equivalent w.r.t. the languages they accept For a regular expression, build a -NFA that accepts the same language For a DFA build a regular expression that describes the language accepted by the DFA. Proving Kleene Theorem We already showed the equivalence of DFA, NFA, and -NFA Left to do Given a RE, find a DFA that accepts the language described by the RE Actually find a -NFA Given a DFA, find an RE that describes the language accepted by the DFA Proving Kleene Theorem Tonight: Given a RE, find a DFA that accepts the language described by the RE Actually find a -NFA Wednesday: Given a RE, find a DFA that accepts the language described by the RE Actually find a -NFA

2 Theory Hall of Fame Steven Cole Kleene b. Hartford, Conn. PhD Princeton (934) Prof at U of Wisc at Madison ( ) Introduced Kleene Star op Defined regular expressions Anyone with a Theorem named after him/her gets in the THOF! Pt : RE -> DFA Since -NFA are equivalent to DFA w.r.t the class of languages they accept We can, given an RE, build an -NFA instead of an DFA that accepts the language described by the RE We can always then convert that -NFA to an equivalent DFA (using the algorithms presented last week) Regular Expression Recursive definition of regular languages / expression over Σ :. is a regular language and its regular expression is 2. {} is a regular language and is its regular expression 3. For each a Σ, {a} is a regular language and its regular expression is a Regular Expression 4. If L and L 2 are regular languages with regular expressions r and r 2 then -- L L 2 is a regular language with regular expression (r + r 2 ) -- L L 2 is a regular language with regular expression (r r 2 ) -- L * is a regular language with regular expression (r * ) Only languages obtainable by using rules -4 are regular languages. RE -> DFA We will build our -NFA by structural induction: Base case: Build an -NFA for, {}, and {a}, a Σ {} {a} a RE -> DFA Induction: Assume R and R 2 are regular expressions that describe languages L and L 2. Then, by the induction hypothesis, there exists -NFA, M and M 2 that accept L and L 2 Create -NFA that accept the languages described by: R + R 2 R R 2 R *

3 RE -> DFA Induction Hypothesis: L = L(M ) where M = (Q,Σ, q, δ, F ) L 2 = L(M 2 ) where M 2 = (Q 2,Σ, q 2, δ 2, F 2 ) Assume Q and Q 2 are disjoint Will build M u = (Q u,σ, q u, δ u,f u ) L(M u ) = L + L 2 M c = (Q c,σ, q c, δ c, F c ) L(M c ) = L L 2 M k = (Q k,σ, q k, δ k, F k ) L(M k ) = L * Using transitions, create a branch where the machine can either following one branch (representing one RE) or the other branch (representing the other RE) Start state of M Start state of M Start state of M 2 If a string is accepted by either of the existing Ms, it will be accepted by the new M. The set of accepting states of M will include each of the accepting states from M and M 2. M u = (Q u,σ, q u, δ u, F u ) Q u = Q Q 2 {q u } F u = F F 2 Transition function: δ u δ u (q u, ) = {q, q 2 } δ u (q u, a) = for all a Σ δ u (q, a) = δ (q, a) if q Q δ u (q, a) = δ 2 (q, a) if q Q 2 Build M to start at the start state of M and from any accepting state of M move directly to the start state of M 2 via a transition. -NFA for L -NFA for L 2

4 After being accepted by the first machine, a string will immediately be tested on the 2 nd machine The set of accepting states of the new M will be the same as that of the 2 nd machine. M c = (Q c,σ, q c, δ c, F c ) Q c = Q Q 2 Q c = q F c = F 2 Transition function δ c : δ c (q, a) = δ (q, a) if q Q δ c (q, a) = δ 2 (q, a) if q Q 2 For all q F, δ c (q, ) = δ (q, ) {q 2 } RE -> DFA: Kleene Star Create a new start state Go from new start state to original start state via a transition Go from any accepting state back to the new start state via a transition RE -> DFA: Kleene Star Make new start state the accepting state. Note that you can get from any excepting state to the new start state via a transition.

5 RE -> DFA: Kleene Star RE -> DFA: Kleene Star M k = (Q k,σ, q k, δ k, F k ) Q k = Q {q k } F k = {q k } Transition function δ k δ k (q, a) = δ (q, a) if q Q, δ k (q k, ) = {q } δ u (q u, a) = for all a Σ For all q A, δ k (q, ) = δ (q, ) {q k } Let s try an example Create an -NFA for the regular expression: (00 + ) * (0) * 0 (00 + ) * (0) *

6 (00 + ) * (0) * RE -> DFA: Summary What have we shown: Given a language L described by a regular expression, we can build an -NFA that accepts L Since -NFA are equivalent to DFAs, we can, if we wanted to, build an DFA to accept L. Kleene II DFA -> RE Problem Session Next Time Part of the proof is complete. Questions?