1. Lecture3 DFA vs. NFA, properties of RL
 2004 SDU

2. Regular operations
They are used to study the properties of regular languages.
Let A and B be languages, define the regular operations union, concatenation, and star as follows:
Example: A = {good, bad}, B = {boy, girl}
Then AB = {good, bad, boy, girl}; A  B ={goodboy, goodgirl, badboy, badgirl}, and
A*={, good, bad, goodgood, goodbad, badgood, badbad, goodgoodgood, goodgoodbad, goodbadgood, goodbadbad,…}
Union: AB = {x| x  A or x  B}
Concatenation: A  B = {xy | x  A and y  B}
Star: A* = {x1x2…xk | k  0 and each xi  A }
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3. Closure properties of Regular Languages
Regular languages are closed under:
Union   if A and B are regular languages, so is A  B
Concatenation  if A and B are regular languages, so is A  B
Star * if A is a regular language, so is A*.
 2004 SDU

4. General Construction for  and 
Let A1 and A2 be two regular languages, and M1 and M2 be the DFAs that recognize them, we want to construct a DFA M to recognize A1  A2.
Can M first simulate M1 and then simulate M2 on an given input?
Idea: Simulate two DFA's simultaneously.
Let M1 = (Q1, , 1, s1, F1) and M2 = (Q2, , 2, s2, F2)
Define:M = (Q, , , s, F) where:
Q = Q1  Q2
s = (s1, s2)
((q1, q2), ) = (1(q1, ), 2(q2, )), for each (q1, q2)Q, and each   
For Union, F = ? The proof?
Ans: (Q1  F2)  (F1  Q2)
For Intersection, F = ? The proof?
Ans: F1  F2
 2004 SDU

5. Example of union L1={w{0,1}*| w contains 01 as substring}
0,1 1 1 2 3
-redundant states
You can construct M in an inductive way and avoid redundancy!
1. (s1, s2)  Q
2. If (q1, q2)  Q, then
((q1,0), (q2,0))  Q
((q1,1), (q2,1))  Q 1
0,1 4 1 6 5 14 15 16 24 25 26 34 35 36 1
0,1
 2004 SDU

6. How about the other regular operations
The concatenation  ?
The star * ?
Can we use similar way?
To solve this problem, we introduce a new technique  non-determinism.
In deterministic computation, the next state is determined
In non-deterministic computation, several choices may exist
Page 48 for example, and 49 for clarity
 2004 SDU

7. An example of NFA
0,1 q1 1 q3 1 q4
0,1 q2
0,
What is the difference between DFA and NFA?
How does it compute on input ?
 2004 SDU

8. Formal definition of DFA
A DFA is a 5-tuple: M = (Q, , , s, F)
Where,
Q is finite set of states
 is finite input alphabet
s  Q is the initial state
F  Q is the set of final states
: Q   -> Q // note that this is a function
 2004 SDU

9. Nondeterministic Finite Automaton (NFA)
Generalization of DFA. Allows:
0 or more next states for the same (q, a).
Guessing
Transitions labeled by the empty string .
Changing state without reading input
Motivation: Flexibility, simplicity.
 2004 SDU

10. Formal definition of NFA
Notation:   =   {}.
An NFA is a 5-tuple: M = (Q, , , s, F) where:
Q - a finite set of states
 - a finite alphabet
s – the start state
F  Q – the set of final states
 : Q     P(Q), the power set of Q.
 2004 SDU

11. NFA -definition
Let N =(Q, , , q0, F) be an NFA and w a string over the alphabet .
We say N accepts w if we can write w as w = y1…ym ,where each yi
is a member of  and a sequence of states p0,…, pm exists with
three conditions:
p0=q0;
for every i, 1 i  m, pi(pi-1,yi);
pm F
Language recognized by an NFA N, denoted as L(N), is the set
of the strings accepted by N.
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12. Example of NFA   guess check Pr: How to understand some branch dies?
An NFA that accepts all strings
of 0’s the number of which
is either a multiple of 2 or a multiple
of 3.  
guess
check
Pr: How to understand some branch dies?
The formal description of this NFA is…?
The computing process on input 0000 is…?
more examples on pages 48, 51
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13. NFA vs. DFA Is NFA more powerful than DFA? Theorem: Proof Idea:
Ans: No.
Theorem:
For every NFA M there is an equivalent DFA M'
Proof Idea:
NFA is in a set of states at any point during reading a string.
DFA will use a state to keep track of this.
Important Assumption:
No transition labeled by epsilon.
(Will get rid of this assumption later.)
 2004 SDU

14. Equivalent DFA construction.
NFA N = (Q, , , s, F)
DFA M = (Q', , , s', F') where:
Q' = p(Q), power set of Q
s' = {s}
F' = {P  Q' | P  F is nonempty}
({q1,…,qm}, a) = (q1, a)  (q2, a)  ...  (qm, a)
i.e. find all the states that can be reached on a from all the NFA states in a DFA state.
 2004 SDU

15. How to handle epsilon transitions?
Define e-closure of state P, E(P) = {q|q can be reached from some p  P by traveling along 0 or more  arrows}.
Let: ({q1,…,qm}, a) = {qQ| q  E((q1, a))  E((q2, a))  ...  E((qm, a))
Let: s’=E(s)
Proof: at each step of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point.
See page56.
See page 57, example1.41
 2004 SDU

16. Construct DFA from NFA ………………………….w1 …………………….w2 … ……………………..wn
is an accepting path on w1…wn in NFA
is an accepting path on w1…wn in DFA
 2004 SDU

17. Construct from NFA to DFA (example)1  b a 3 a 2
a,b
{1,2} a b
{2}
a,b
{1} 
a,b a b a b b a a a
{2,3}
{1,2,3}
{3}
{1,3} b b
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18. Regular languages
Conclusion: A language is regular iff some NFA recognizes it.
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19. Construction for LºL‘ page 60
Let L=(Q1, , 1,s1,F1); L’=(Q2, , 2,s2,F2)
Define:
L’’ = (Q,,,s,F), where Q = Q1  Q2 s = s1 F = F2
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20. N’ N
N’’   
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21. L* Let L=(Q1, , 1, s1, F1) Define:
L* = (Q, , , s, F) Q = {s}  Q1 F = {s}  F1
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22. N* N   
 2004 SDU