1. Intro to Theory of Computation
LECTURE 4
Last time:
Equivalence of NFAs and DFAs
Today:
Closure properties of regular languages
Equivalence of NFAs, DFAs and regular expressions CS
464
Sofya Raskhodnikova
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

2. Operations on languages
Regular
Operations on languages
Complement: A = { w | w  A }
Union: A  B = { w | w  A or w  B }
Union: A  B = { w | w  A or w  B }
Intersection: A  B = { w | w  A and w  B }
Reverse: AR = { w1 …wk | wk …w1  A }
Concatenation: A  B = { vw | v  A and w  B }
Concatenation: A  B = { vw | v  A and w  B }
Star: A* = { w1 …wk | k ≥ 0 and each wi  A }
Star: A* = { w1 …wk | k ≥ 0 and each wi  A }
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

3. I-clicker problem (frequency: AC)
Let L be the set of words in English.
Then 𝑳 ∩𝑳 𝑹 is
The set of English words in alphabetical order, followed the same words in reverse alphabetical order.
{𝑤∣𝑤 is an English word or an English word written backwords}.
{𝑤∣𝑤 is an English word that is a palindrom}.
None of the above.
{madam, dad, but, tub, book}

4. Closure properties of the class of regular languages
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.
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

5. Sofya Raskhodnikova; based on slides by Nick Hopper
Closure under reverse
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
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

6. New construction for A  B
Construct an NFA M:
L(MA)=A ε ε
L(MB)=B
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

7. Concatenation operation
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
L(M1)=A
L(M2)=B ε
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

8. Concatenation operation
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
L(M1)=A
L(M2)=B ε
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

9. Sofya Raskhodnikova; based on slides by Nick Hopper
Star operation
Star: A* = { w1 …wk | k ≥ 0 and each wi  A }
Theorem. If A is regular, A* is also regular. ε ε
L(M)=A ε
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

10. The class of regular languages is closed under
Regular operations
Union: A  B = { w | w  A or w  B }
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
Complement: A = { w | w  A }
Intersection: A  B = { w | w  A and w  B }
Reverse: AR = { w1 …wk | wk …w1  A }
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

11. Sofya Raskhodnikova; based on slides by Nick Hopper
Regular expressions
In a regular expression, we can use
Constants: ε ,  , a set Σ, members of Σ.
Regular operations: *,  , ∪
Examples:
0*1*
(01)*
0*1*(ε 0)
L(R) = the language regular expression R describes
= { w | w has a run of 0s followed by a run of 1s}
= the set of all strings over the alphabet Σ={0,1}
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

12. Sofya Raskhodnikova; based on slides by Nick Hopper
Precedence
*  ∪
EXAMPLE
R1*R2  R3 = ( (
R1* ) R2 )  R3
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

13. Sofya Raskhodnikova; based on slides by Nick Hopper
Regular expressions: examples 1) 2) 3)
{ w | w has exactly one character 1 }
0*10*
{ w | w has length ≥ 3 and its 3rd symbol is 0 }
(01)(01) 0 (01)*
{ w | every odd position of w is a 1 }
(1(01))* (ε  1)
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

14. Regular expressions to NFAs
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.
R = ε
R =  
R = 
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

15. I-clicker problem (frequency: AC)
What should the induction hypothesis be?
Suppose some regular expression of length 𝑘 can be converted an NFA, for some 𝑘∈ℕ.
Suppose all regular expressions of length 𝑘 can be converted an NFA, for some 𝑘∈ℕ.
Suppose all regular expressions of length at most k can be converted an NFA, for some 𝑘∈ℕ.
None of the above.
{madam, dad, but, tub, book}

16. Regular expression to NFA
Transform (1(0  1))* to an NFA ε 1
1,0 ε
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

17. NFAs to regular expressions
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.
01*0 1
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

18. Sofya Raskhodnikova; based on slides by Nick Hopper
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
11/19/2018
Sofya Raskhodnikova; based on slides by Nick Hopper

19. G accepts w if it finds q0q1…qk, w1…wk: wi is generated by R(qi,qi+1)
Generalized NFAs
a  b
a*b a qs q qa
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
R(qs,q) = a*b
R(qa,q) = Ø
R(q,qs) = Ø