CSCI 4325 / 6339 Theory of Computation Zhixiang Chen

Chapter One Finite Automata

Outline Finite Automata Deterministic vs Nondeterministic finite automata, and their equivalence Finite Automata vs. regular expressions, and their equivalence Pumping lemma and non-regular languages

What Is a Finite Automaton? Input tape a b a b a a b b a … Input Tape Reading tape States, initial state, final states Moves Computation Accepting an input string Rejecting an input string q0 q5 q1 q4 q2 q3 Finite Control Reading head

Formal Definition A deterministic finite automaton is a quintuple M = (K, Σ, δ, s, F) where  K is a finite set of states;  Σ is a finite alphabet,  s  K is the initial state,  F  K is the set of final states,  δ is the transition function from K x Σ  K Explain K, Σ, δ, s, F.  The hardest one is δ.  What does δ(p, a) = q mean?

Configurations How to describe moves involved in a computation process?  Configurations Formal definition of configuration is very simple  A configuration is any element of K x Σ*. Given a configuration (q, w), q  K, w  Σ*,  What is its physical meaning?  For example, (q, aabbab)?

Yield Relation The yield relation We must understand that the yield relation describes exactly one step of a computation processing.

Because the yield relation is a relation, we can find a unique reflexive and transitive closure of. Denote the closure as Now, Given any two configurations (p, w) and (p’, w’),  If (p,w) (p’,w’)  then we say that (p,w) yields (p’,w’) stop! Think about the physical meaning of (p, w) yielding (p’, w’)!

The Language Accepted by a DFA The language accepted by an DFA M is  L(M) = { w: w  Σ*, (s, w) (q, e) for some q  F } Explain the definition.

Ex’s of DFA’s M = (K, Σ, δ, s, F)   δ is given in the following table Show computation form What is L(M)?

State Diagram State Diagram of a DFA M= (K, Σ, δ, s, F):  A directed graph:  for each state in K, draw a node  for each final state in F, draw a double circle.  Mark the initial node of the initial state >.  If δ(p, a) = q, then draw an edge from p to q and label it with a. Ex. The state diagram representation of a DFA is simple and intuitive. a a b b q o q 1

Design DFAs to Accept L1 = {w: w  {a,b}*, w has no three consecutive b’s} L2 = {w: w  {a,b}*, w has abab as a substring} L3 = {w: w  {a,b}*, w has no aa nor bb}

L1

L2 a b a,b b bb aa a q0q0 q1q1 q2q2 q3q3 q4q4 aa

L3 a a a b b b q1q1 q2q2 q3q3 q4q4 q5q5

Nondeterministic Finite Automata (NFA) Motivation of nondeterminism  Play chess, factor a number The key point is at each step, there are a fixed number of possible choices for the next. Some choices are good, some are bad. When we try to study the power of nondeterminism, we assume that a machine knows how to make a good choice.

NFA Examples Ex.Three NFAs accepting the same language defined by the regular expression  (ab v aba)* aa a b b q0q0 q1q1 q2q2 q3q3 q1q1 q0q0 a b ba q0q0 q1q1 q2q2 a a e b

Definition of NFA An NFA M = (K, ∑, ∆, s, F) where,  K is a finite set of states  ∑ is a finite alphabet  s is the initial state  F  K is the set of finite states  ∆ is the transition relation from K x (∑ ⋃ {e})  K Understand ∆ !  ∆ (p, a) may have more than one value  M is nondeterministic, because ∆ may have multiple values for a given pair of (p, a)

Configurations of NFAs Configuration  Pairs in K X ∑*  Meaning of a configuration? Yield Relation (q, w) (q’, w’) if w = aw’ for a Є ∑ ⋃ {e} and ∆ (q, a) = q’

The Reflexive and Transitive Closure Because the yield relation is a relation, we can find a unique reflexive and transitive closure of. Denote the closure as Now, Given any two configurations (p, w) and (p’, w’),  If (p,w) (p’,w’)  then we say that (p,w) yields (p’,w’) stop! Think about the physical meaning of (p, w) yielding (p’, w’)!

The Language Accepted by an NFA The language accepted by an NFA M is  L(M) = { w: w  *, (s, w) (q, e) for some q  F }

NFA Example a, b bb e b a

Design an NFA to accept  L = {w: not all symbols in  occured in w },  where  = {a, b, c, d } a,b,c a,b,d b,c,d a,c,d e e e e

Theorem. For each NFA, there is an equivalent DFA. Remark. Because each DFA has an equivalent NFA, the above result tells us that the power of NFA is the same as DFA.

Proof of the above Theorem Proof by Construction: Construct a DFA to simulate the given NDF Let the NFD M = ( k, , , s, F ) Wanted a DFA M’ = ( K’, , , s’, F’ )  Such that L( M ) = L( M’ ) Construction of M’   =   K’ = 2K  s’ = { q  K: (s, e) |---* (q, e ) }  F’ = { A  K’: A  F  }  For A  K, a   ( A, a ) = { q  K: ( p, a ) |---* ( q, e ) for some p  A }

Claim 1. M’ is deterministic by definition of  For any q  K, define E(q) = { p  K: ( q, e ) |---* ( p, e )} Claim 2. For any string w  *, p, q  K ( q, w ) |---* ( p, e )  ( E(q), w ) |---* ( P, e ) for some P  K’ such that p  P. Proof By induction on |w|  Basic step. For |w| = 0, w = e, we need to show: (q, e) |---* (p, e)  ( E(q), e )|---* (P, e) for some P  K’ such that p  P

Induction Hypothesis  Suppose the claim is true for all string w such that |w|  k Induction Step  We prove the claim is true for all string w such that |w| = k+ 1. Let w = ua for some a , u   * First prove: (q, w) |---* (p, e)  ( E(q), w ) |---* (P, e) for some P with p  P (q, w) |---* (p, e)   r1, r2 such that (q, w) |---* (r1, a) |---(r2, e) |---(p, e) apply induction hypothesis to (q, w) |---* (r1, a) Second we prove (E(q), w) |---*( P, e)  (q, w) |---*(p, e) for some P with p  P Again, let w = ua for some a , u  * (E(q), w) |---* (P, e)  (E(q), w) |---* (R1, a) |--- ( R2, e) |---* (P, e) Apply induction hypothesis to (E(q), w) |---* (P, e)  (E(q), w) |---* (R1, a)

Claim3: L(M) = L(M’)  By claim 2, for any w  *, (s, w) |---* (p, e), for p  F  (E(s), w) |---*, (P, e), for some P with p  P  Note E(s) = s’, P  F’

Equivalent DFA of NFA NFA DFA ae e b a b e e a a b b a,b a a

FA and Regular Expressions Theorem 1. The class of languages accepted by FA is closed under  (a) Union  (b) Concatenation  (c) Kleene star  (d) Complementation  (e) Intersection

Proof of Theorem 1 Proof.  (a) Union: Given Need to prove that there is a FA such that Idea for proving (a): by construction  Ideas for proving (b), (c) and (d): also by construction  Idea for proving (e):

Theorem 2. A language is regular if and only if it is accepted by a FA. Proof.  (The if-part) If a language is regular then it is accepted by a FA Recall that the class of regular languages is the smallest class of languages containing the empty set Ø and the singleton a, where a is a symbol, and closed under union, concatenation, and Kleene star. It is evident the empty set and all singletons are indeed accepted by FA; and by closure property of FA, every regular language is accepted by some FA.

(The only-if-part) If a language is accepted by a FA, then it is regular. Idea: Starting from the directed graph representation, find all paths from the initial state to the final states. For each path write a regular expression. Use Kleene star to deal with loops. Let be a FA. We shall construct a regular expression R such that. We shall represent L(M) as the union of many (but a finite number of) simple languages. Let. For i, j =1,…,n and k=0,…,n, we define R(i,j,k) as the set of all strings in that may drive M from state to state without passing through any intermediate state numbered k+1 or greater.

Note. By Math Induction, R(i,j,k) is regular for k=0,1,…,n. Hence, L(M) is regular.

The Pumping Theorem Theorem. Let be an infinite regular language. There is an integer such that any string with can be rewritten as such that and for each. Proof. Follows from the graph representation.

Irregular Languages w has equal number of a’s and b’s} The first two can be proved with the Pumping Theorem The third can be proved by contradiction and closure property of regular languages:  If L3 is regular, so is L1. But L1 is not

Algorithms for FA Theorem.  (a) There is an exponential time algorithm which, given an NFA, constructs an equivalent DFA.  (b) There is a polynomial time algorithm which, given a regular expression, construct an equivalent NFA  (c) There is an exponential time algorithm which, given an NFA, constructs an equivalent regular expression

 (d) There is a polynomial time algorithm which, given two DFAs, decides whether they are equivalent  (e) There is an exponential time algorithm which, given two NFAs, decided whether they are equivalent; similarly for equivalence of two regular expressions. For (d) and (e), note the following

More Algorithms Theorem. If L is a regular language, then there is an algorithm which, given tests whether it is in L in time. Proof. Construct a DFA to accept L.

More Algorithms Theorem. Let be an NFA. then there is an algorithm which, given tests whether it is in L(M) in time. Proof. Simulate the accepting process of M:  Step 0: Find E(s), which is s plus all states that are reachable with e-moves  Step 1: Start at E(s), find all states reachable passing the first letter of w plus their e-reachable states.  …  Step |w|: find the current state set as in Step 1 and decide whether a final state is in the current state set.  Estimate the time of the above process.

Conclusions NFA have the same computing power as DFA NFA have the same computing power as regular expressions Pumping Theorem vs. irregular languages Algorithms for  Constructing an equivalent NFA for a regular exp.  Constructing an equivalent regular exp. for an NFA  Constructing an equivalent DFA for an NFA  Deciding whether two FA are equivalent or not  Deciding whether a string is accepted by a DFA (or NFA) or not