Finite Automata. Theorems - Unit I SUSAN ELIAS. Professor Department of Computer Science & Engineering Sri Venkateswara College of Engineering

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1 Finite Automata Theorems - Unit I SUSAN ELIAS Professor Department of Computer Science & Engineering Sri Venkateswara College of Engineering September 17, 2012

2 Unit I - Guidelines Formal Definitions Definition of DFA Notations for DFA ˆδ for DFA Language of DFA Definition of NFA ˆδ for NFA Language of NFA Definition of ɛ-nfa ˆδ and Language for ɛ-nfa Theorems Problems

3 Unit I - Guidelines Formal Definitions Theorems NFA - DFA Theorems 2.11, 2.12 Theorem 2.22 Forms of Proofs - Chapter 1-1.2, 1.3, 1.4 Problems Design of DFA Problems Convert NFA, ɛ-nfa to DFA Problems

4 Introduction NFA - DFA To prove that NFA is equivalent to DFA ɛ-nfa - DFA To prove that ɛ-nfa is equivalent to DFA

5 Introduction Induction Principle : If we Prove S(i) and we prove that for all n i, S(n) implies S(n+1), then we may conclude S(n) for all n i.

6 Introduction IF-AND-ONLY-IF Statements A if and only if B A iff B A is equivalent to B

7 Introduction A if and only if B if part : if B then A only if part : if A then B

8 Introduction NFA - DFA Theorem 2.12 A language L is accepted by some DFA if and only if L is accepted by some NFA ɛ-nfa - DFA Theorem 2.22 A language L is accepted by some ɛ-nfa if and only if L is accepted by some DFA

9 NFA-DFA NFA - DFA Theorem 2.12 A language L is accepted by some DFA if and only if L is accepted by some NFA if part : Theorem Subset construction only if part : Trivial

10 NFA-DFA NFA - DFA - Theorem 2.11 If D = (Q D, Σ, δ D, {q 0 }, F D ) is a DFA constructed from NFA N = (Q N, Σ, δ N, q 0, F N ) by subset construction, then L(D) = L(N) Proof : Prove by induction on w that ˆ δ D ({q 0 }, w) = ˆ δ N (q 0, w) BASIS: Let w = 0 that is w = ɛ. By the Basis definition of ˆδ for DFA s and NFA s, both δˆ D ({q 0 }, ɛ) and δˆ N (q 0, ɛ) are {q 0 } INDUCTION: Let w be of length n + 1, assume the statement for length n Break w up as w = xa, where a is the first symbol of w. By the inductive hypothesis δˆ D ({q 0 }, x) = δˆ N (q 0, x) Let both these sets of N s states be {p 1, p 2,, p k }

11 NFA - DFA - Theorem 2.11 INDUCTION: The inductive part of the definition of ˆδ for NFA s says that δˆ N (q 0, w) = k i=1 δ N(p i, a) - (1) The subset construction on the other hand says that δ D ({p 1, p 2,, p k }, a) = k i=1 δ N(p i, a) - (2) Using Equation(2) and the fact that δˆ D ({q 0 }, x) = {p 1, p 2,, p k } in the inductive part of the definition of ˆδ for DFA s: δˆ D (q 0, w) = δ D ( δˆ D ({q 0 }, x), a) = δ D ({p 1, p 2,, p k }, a) = k i=1 δ N(p i, a) - (3) Thus Equations (1) and (3) demonstrate that δˆ D ({q 0 }, w) = δˆ N (q 0, w). When we observe that D and N both accept w if and only if δˆ D ({q 0 }, w) or δˆ N (q 0, w) respectively contain a state in F N the proof that L(D) = L(N) is complete

12 NFA - DFA - Theorem 2.12 NFA - DFA Theorem 2.12 A language L is accepted by some DFA if and only if L is accepted by some NFA if part : Theorem Subset construction only if part : Trivial - A transition diagram for a DFA can also be interpreted as the transition diagram for an NFA which happens to have exactly one choice of transition in any situation.

13 ɛ-nfa-dfa ɛ-nfa-dfa Theorem Theorem 2.22 A language L is accepted by some ɛ-nfa if and only if L is accepted by some DFA if part : Let L(D) be the language of a DFA Turn D into an ɛ NFA E by adding transitions δ(q, ɛ) = for all states q of D. Turn all the transitions of D on input symbol as δ E (q, a) = {p} Now transitions of E and D are the same only if part : Modified Subset Construction.

14 NFA-DFA - Theorem 2.22 Only-If part: Let E = (Q E, Σ, δ E, {q 0 }, F E ) be an ɛ NFA. Apply Modified Subset construction to produce DFA D = (Q D, Σ, δ D, {q 0 }, F D ). Need to show that L(D) = L(E) We prove that δˆ E (q 0, w) = δˆ D (q D, w) by induction on length of w BASIS: If w = 0, then w = ɛ It is known that δˆ E (q 0, ɛ) = ECLOSE(q 0 ) It is also known that q D = ECLOSE(q 0 ) It is known that for a DFA ˆδ(p, ɛ) = p for any state p So δˆ D (q D, ɛ) = ECLOSE(q 0 ) Thus is is proved that δˆ E (q 0, ɛ) = δˆ D (q D, ɛ) INDUCTION:

15 NFA - DFA - Theorem 2.22 INDUCTION: Suppose w = xa, where a is the final symbol of w and assume that the statement holds for x. That is δˆ E (q 0, x) = δˆ D (q 0, x) Let both these sets of states be {p 1, p 2,, p k } By the definition of ˆδ for ɛ-nfa we compute δˆ E (q 0, w) by 1 Let {r 1, r 2,, r m } be k i=1 δ E (p i, a) 2 Then δˆ E (q 0, w) = m j=1 ECLOSE(r j) Steps (1) and (2) are the same steps used in the Modified Subset Construction Thus δ ˆ D (q D, w) which is δ D ({p 1, p 2,, p k }, a) is the same set as δˆ E (q 0, w) Thus it is proved that δˆ E (q 0, w) = δˆ D (q D, w)