Nondeterminism Section 4 1 Deterministic FA required for

Nondeterminism [Section 4. 1] (Deterministic) FA required for every state q and every symbol of the alphabet to have exactly one arrow out of q labeled . What happens when we drop this requirement ? L = { w 2{a, b}* | w contains abba as a substring } Transition diagram (of this new type) :

Formal definition of an NFA [Section 4. 1] A nondeterministic finite automaton (NFA) is a 5 -tuple (Q, , q 0, A, ) where - Q is a finite set of states - is a finite alphabet (input symbols) - qo 2 Q is the initial state - A µ Q is a set of accepting states - : Q£ ___ is the transition function Give a formal definition of the NFA from the previous slide.

Formal definition of an NFA [Section 4. 1] Defining the computation of an NFA M=(Q, , q 0, A, ). Extended transition function * : Q£ * ____ : 1) For every q 2 Q, let *(q, ) = 2) For every q 2 Q, y 2 *, and 2 , let *(q, y ) = We say that a string x 2 * is accepted by M if ____. A string which is not accepted by M is rejected by M. The language accepted by M, denoted by L(M) is the set of all strings accepted by M.

NFA’s vs. FA’s [Section 4. 1] Is the following statement true ? For every FA M = (Q, , q 0, A, ) there exists an NFA M 1 = (Q 1, , q 1, A 1, 1) such that L(M) = L(M 1).

NFA’s vs. FA’s [Section 4. 1] Is the following statement true ? For every NFA M = (Q, , q 0, A, ) there exists an FA M 1 = (Q 1, , q 1, A 1, 1) such that L(M) = L(M 1). Example : a a a, b b

NFA’s vs. FA’s [Section 4. 1] Thm : For every NFA M = (Q, , q 0, A, ) there exists an FA M 1 = (Q 1, , q 1, A 1, 1) such that L(M) = L(M 1). Construction of M 1 : Proof that L(M) = L(M 1) :

NFA’s with Example : NFA accepting a*b*c*. [Section 4. 2]

Formal definition of an NFA- [Section 4. 2] A nondeterministic finite automaton with -transitions (NFA - ) is a 5 -tuple (Q, , q 0, A, ) where - Q is a finite set of states - is a finite alphabet (input symbols) - qo 2 Q is the initial state - A µ Q is a set of accepting states - : is the transition function Give a formal definition of the NFA- from the previous slide.

Formal definition of an NFA- [Section 4. 2] Defining the computation of an NFA- M=(Q, , q 0, A, ). Extended transition function * : ____ : 1) For every q 2 Q, let *(q, ) = 2) For every q 2 Q, y 2 *, and 2 , let *(q, y ) =

Formal definition of an NFA- [Section 4. 2] Defining the computation of an NFA- M=(Q, , q 0, A, ). Extended transition function * : Q£ * 2 Q : 1) For every q 2 Q, let *(q, ) = { q } 2) For every q 2 Q, y 2 *, and 2 , let *(q, y ) = -closure of a set S µ Q, denoted (S), is the set of all states reachable from S by a sequence of -transitions. Define (S) :

[Section 4. 2] Formal definition of an NFA- Defining the computation of an NFA- M=(Q, , q 0, A, ). Extended transition function * : Q£ * 2 Q : 1) For every q 2 Q, let *(q, ) = { q } 2) For every q 2 Q, y 2 *, and 2 , let *(q, y ) = [p 2 *(q, y) ( (p, )) We say that a string x 2 * is accepted by M if ____. A string which is not accepted by M is rejected by M. The language accepted by M, denoted by L(M) is the set of all strings accepted by M.

NFA- ’s vs. NFA’s [Section 4. 2] Is the following statement true ? For every NFA M = (Q, , q 0, A, ) there exists an NFA- M 1 = (Q 1, , q 1, A 1, 1) such that L(M) = L(M 1).

NFA- ’s vs. NFA’s [Section 4. 2] Is the following statement true ? For every NFA- M = (Q, , q 0, A, ) there exists an NFA M 1 = (Q 1, , q 1, A 1, 1) such that L(M) = L(M 1).

Proof of Kleene’s Thm [Section 4. 3] Kleene’s Thm: A language L over is regular iff there exists a finite automaton that accepts L. Part 1 : For every regular language there exists an NFA- accepting it.

Proof of Kleene’s Thm [Section 4. 3] Part 2 : Any language accepted by an FA is regular. Let M=({1, 2, 3, …, k}, , q 0, A, ) be an FA. We will show that for every p, q, r 2 Q, the following language is regular : L(p, q, r) = { x 2 * | *(p, x)=q and for every prefix y of x other than and x we have *(p, y) · r } Can we then conclude that L(M) is regular ?

Proof of Kleene’s Thm [Section 4. 3] Claim : Let M=({1, 2, 3, …, k}, , q 0, A, ) be an FA. For every p, q, r 2 Q, the following language is regular : L(p, q, r) = { x 2 * | *(p, x)=q and for every prefix y of x other than and x we have *(p, y) · r }