cs361-lecture02-1x1

Theory ofComputation

Deterministic Finite Automata

Barath RaghavanCS 361 Fall 2009Williams College

Tuesday, September 15, 2009

MISC

Anonymous feedbackHomework 1

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Claim:In any set of n hats, all are the same color.

“Proof” by induction:Base case: If n = 1, all hats are the same color.Inductive case: Assume that the claim holds for all

n ≤ k. Examine any set of n = k+1 hats. Remove one hat from the set; the remaining set has k hats and thus is of a single color. Replace the hat and remove a different hat from the set; the resulting set has k hats and thus is of a single color. The two sets overlap, thus they must be of the same color.

Therefore, all k+1 hats are the same color. ☐


Claim:In any set of n hats, all are the same color.

“Proof” by induction:Base case: If n = 1, all hats are the same color.Inductive case: Assume that the claim holds for all

n ≤ k. Examine any set of n = k+1 hats. Remove one hat from the set; the remaining set has k hats and thus is of a single color. Replace the hat and remove a different hat from the set; the resulting set has k hats and thus is of a single color. The two sets overlap, thus they must be of the same color.

Therefore, all k+1 hats are the same color. ☐


Deterministic

FiniteAutomata



“An organism or an automaton receives stimuli via its sensory receptor organs, and performs actions via its effector organs. To say that certain actions are a response to certain stimuli means, in the simplest case, that the actions are performed when and only when those stimuli occur... So we ask what kind of events are capable of being represented in the state of an automaton.”

-- Kleene, “Representation of events in nerve nets and finite automata” (1956)



1 2 3

1 2 32, 3 3

1, 21

Elevator Automaton


StatesAlphabetTransitionStart stateFinal states


Q set of states


Q set of states

Σ the alphabet


Q set of states

Σ the alphabet

δ : Q× Σ→ Q transitionfunction


Q set of states

Σ the alphabet


q0 ∈ Q start state


Q set of states

Σ the alphabet


q0 ∈ Q start state

F ⊆ Q set of final states


1 2 3

1 2 32, 3 3

1, 21


1 2 3

1 2 32, 3 3

1, 21

Q = { 1�, 2�, 3�}


1 2 3

1 2 32, 3 3

1, 21

Q = { 1�, 2�, 3�}Σ = {1, 2, 3}


1 2 3

1 2 32, 3 3

1, 21

Q = { 1�, 2�, 3�}Σ = {1, 2, 3}

q0 = 1� F = { 3�}Tuesday, September 15, 2009

1 2 3

1 2 32, 3 3

1, 21

Inputs

States

δ 1 2 31� 1� 2� 2�2� 1� 2� 3�3� 2� 2� 3�


What languages do these automata recognize?


A B0, 1

0, 1


L = {ε}

A B0, 1

0, 1


A B C D

0 0 0 0, 1

1 1 1


A B C D

0 0 0 0, 1

1 1 1

L = {w | w contains at least 3 1s}


?Tuesday, September 15, 2009

A B C D

0 1 0, 1

1

0

1 0

L = {w | w doesn’t contain 110}d)

L = {w | w contains at most two 1s}a)

L = {w | |w| < 3}b)

L = {w | |w| < 3 or contains at least two 1s}c)


A B C D

0 1 0, 1

1

0

1 0

L = {w | w doesn’t contain 110}


A B C D

0 1 0, 1

1

0

1 0



L = {w | |w| < 3}b)



A B C D

0 1 0, 1

1

0

1 0



L = {w | |w| < 3}b)



A B C D

0 1 0, 1

1

0

1 0



L = {w | |w| < 3}b)



A B C D

0 1 0, 1

1

0

1 0



L = {w | |w| < 3}b)



What automata recognize these

languages?


L = {w | |w| is odd and begins with 0 oris even and begins with 1}


L = {w | |w| is odd and begins with 0 oris even and begins with 1}

A B

C

0

1

0, 1

0, 1


L = {w | w = ε or every odd position of w is a 1}


L = {w | w = ε or every odd position of w is a 1}

A B

C

1

0, 10

0, 1



L = {w | w is not the string 11 or 111}


L = {w | w is not the string 11 or 111}

A B C D

E

1 1 1

0 00

0, 1

0, 1


Definition

1. A language L is regular if some DFA recognizes it.

2. The language L(A) recognized by a DFA A is regular.


L1 = {w | |w| is odd and begins with 0 oris even and begins with 1}

L2 = {w | w = ε or every odd position of w is a 1}

Can we build an automaton that accepts:

L3 = L1 ∪ L2


L1 = {w | |w| is odd and begins with 0 oris even and begins with 1}

L2 = {w | w = ε or every odd position of w is a 1}

L3 = L1 ∪ L2

A1 B1

C1

0

1

0, 1

0, 1

A2 B2

C2

1

0, 10

0, 1


How can we combine two DFAs?


Q = {(r1, r2) | r1 ∈ Q1 and r2 ∈ Q2}


Q = {(r1, r2) | r1 ∈ Q1 and r2 ∈ Q2}

Σ = Σ1 ∪Σ2


Q = {(r1, r2) | r1 ∈ Q1 and r2 ∈ Q2}

Σ = Σ1 ∪Σ2

δ((r1, r2), a) = (δ1(r1, a), δ2(r2, a))


Q = {(r1, r2) | r1 ∈ Q1 and r2 ∈ Q2}

Σ = Σ1 ∪Σ2

δ((r1, r2), a) = (δ1(r1, a), δ2(r2, a))

q0 = (q1, q2)


Q = {(r1, r2) | r1 ∈ Q1 and r2 ∈ Q2}

Σ = Σ1 ∪Σ2

δ((r1, r2), a) = (δ1(r1, a), δ2(r2, a))

q0 = (q1, q2)

F = {(r1, r2) | r1 ∈ F1 or r2 ∈ F2}


A1 B1

C1

0

1

0, 1

0, 1

A2 B2

C2

1

0, 10

0, 1


A1A2

B1A2 B1C2

C1B2 C1C2

0

1

10

0, 10, 1

0, 1


Regular languages are

closed under union.

if L1 and L2 are regular languagesthen L1 ∪ L2 is a regular language


Reading: Sipser 1.1



L = {w | w �= ε}

WHATAUTOMATONRECOGNIZES

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