1/20/2016

Sofya Raskhodnikova

Intro to Theory of Computation

LECTURE 4Last time:

• Equivalence of NFAs and DFAs

Today:

• Closure properties of regular

languages

• Equivalence of NFAs, DFAs

and regular expressions

Sofya Raskhodnikova; based on slides by Nick Hopper L4.1

Operations on languages

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Union: A B = { w | w A or w B }

Intersection: A B = { w | w A and w B }

Complement: A = { w | w A }

Reverse: AR = { w1 …wk | wk …w1 A }

Concatenation: A B = { vw | v A and w B }

Star: A* = { w1 …wk | k ≥ 0 and each wi A }

Regular

Union: A B = { w | w A or w B }

Concatenation: A B = { vw | v A and w B }

Star: A* = { w1 …wk | k ≥ 0 and each wi A }

L3.2

I-clicker problem (frequency: AC)

Let L be the set of words in English.

Then 𝑳 ∩ 𝑳𝑹 is

A. The set of English words in alphabetical order,

followed the same words in reverse alphabetical order.

B. {𝑤 ∣ 𝑤 is an English word or an English word written

backwords}.

C. {𝑤 ∣ 𝑤 is an English word that is a palindrom}.

D. None of the above.

Closure properties of the class of

regular languages

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

THEOREM. The class of regular languages

is closed under all 6 operations.

If A and B are regular, applying any of these

operation yields a regular language.

L3.4

Closure under reverse

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Theorem. The reverse of a regular language is also regular

Proof: Let L be a regular language and M be a DFA that

recognizes it. Construct an NFA 𝑴′ recognizing LR:

• Define 𝑴′ as M with the arrows

reversed.

• Make the start state of M be the

accept state in 𝑴′.

• Make a new start state that goes to all

accept states of M by ε-transitions.

ε ε ε

L3.5

New construction for A B

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

ε

ε

Construct an NFA M:

L(MA)=A

L(MB)=B

L3.6

Concatenation operation

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Concatenation: A B = { vw | v A and w B }

Theorem. If A and B are regular, A B is also regular.

Proof: Given DFAs M1 and M2, construct NFA by

connecting all accept states in M1 to the start state in M2.

ε

ε

L(M1)=A L(M2)=B

L3.7

Concatenation operation

1/21/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Concatenation: A B = { vw | v A and w B }

Theorem. If A and B are regular, A B is also regular.

Proof: Given DFAs M1 and M2, construct NFA by

connecting all accept states in M1 to the start state in M2.

• Make all states in M1 non-accepting.

ε

ε

L(M1)=A L(M2)=B

L3.8

Star operation

1/20/2016

Sofya Raskhodnikova; based on slides by Nick Hopper

Star: A* = { w1 …wk | k ≥ 0 and each wi A }

Theorem. If A is regular, A* is also regular.

L(M)=Aε

ε

ε

L3.9

The class of regular languages is

closed under

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Union: A B = { w | w A or w B }

Intersection: A B = { w | w A and w B }

Complement: A = { w | w A }

Reverse: AR = { w1 …wk | wk …w1 A }

Regular operations

Union: A B = { w | w A or w B }

Concatenation: A B = { vw | v A and w B }

Star: A* = { w1 …wk | k ≥ 0 and each wi A }

Other operations

L3.10

Regular expressions

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.11

• In a regular expression, we can use

– Constants: ε , , a set Σ, members of Σ.

– Regular operations: *, , ∪

• Examples:

– 0*1*

– (01)*

– 0*1*(ε 0)

• L(R) = the language regular expression R describes

= the set of all strings over the alphabet Σ={0,1}

= { w | w has a run of 0s followed by a run of 1s}

* ∪

Precedence

R2R1*(

EXAMPLE

R1*R2 R3 = ( ) )R3

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.12

{ w | w has exactly one character 1 }

0*10*

1)

2)

3)

Regular expressions: examples

{ w | w has length ≥ 3 and its 3rd symbol is 0 }

(01)(01) 0 (01)*

{ w | every odd position of w is a 1 }

(1(01))* (ε 1)

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.13

Regular expressions to NFAs

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Theorem. Every regular expression has an equivalent NFA.

Proof: Induction on the length of regular expression R.

Base case: length 1

Inductive step: follows from closure of the class of

regular languages under the regular operations.

L4.14

R =

R = ε

R =

I-clicker problem (frequency: AC)

What should the induction hypothesis be?

A. Suppose some regular expression of length 𝑘 can be

converted an NFA, for some 𝑘 ∈ ℕ.

B. Suppose all regular expressions of length 𝑘 can be

converted an NFA, for some 𝑘 ∈ ℕ.

C. Suppose all regular expressions of length at most k can

be converted an NFA, for some 𝑘 ∈ ℕ.

D. None of the above.

Regular expression to NFA

L4.161/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Transform (1(0 1))* to an NFA

1ε 1,0

ε

NFAs to regular expressions

1/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

Theorem. Every NFA has an equivalent regular expression.

Proof idea:

Transform NFA to a regular expression by removing states

and relabeling the arrows with regular expressions.

L4.17

0

1

001*0

Generalized NFAs

• Each transition is labeled with a regular

expression

• Unique and distinct start and accept states

• No transitions to the start state

• No transitions from the accept state

L4.181/20/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

R(qs,q) = a*b

R(qa,q) = Ø

R(q,qs) = Ø

G accepts w if it finds q0q1…qk, w1…wk:

wi is generated by R(qi,qi+1)

w = w1w2…wk

q0=qs, qk=qa

q

a b

aa*bqs qa

Generalized NFAs