1. NFA to DFA conversion and regular expressions
The Chinese University of Hong Kong
Fall 2010
CSCI 3130: Automata theory and formal languages
NFA to DFA conversion and regular expressions
Andrej Bogdanov

2. NFAs are as powerful as DFAs
Obviously, an NFA can do everything a DFA can do
How about the other way?
YES
If L is the language of some NFA, then it is also the language of some DFA.

3. Converting NFAs into DFAs
We will show a general way to convert every NFA into an equivalent DFA
Step 1: Simplify NFA by eliminating e-transitions
Step 2: Convert simplified NFA (without es)
We do this first

4. NFA to DFA conversion intuition
0, 1
NFA: 1 q0 q1 q2
DFA: 1
q0 or q1
q0 or q2 q0 1 1

5. NFA to DFA conversion intuition
0, 1
NFA: 1 q0 q1 q2
DFA: 1
{q0, q1}
{q0, q2} q0 1 1

6. General method NFA DFA states q0, q1, …, qn
Æ, {q0}, {q1}, {q0,q1}, …, {q0,…,qn}
one for each subset of states in the NFA
initial state q0
{q0}
transitions d
d’({qi1,…,qik}, a) =
d(qi1, a) ∪…∪ d(qik, a)
accepting states
F Í Q
F’ = {S: S contains some state in F}

7. Converting via the general method
0, 1
NFA: 1 q0 q1 q2
0, 1 1
DFA:
{q0, q1}
{q0, q2}
{q0, q1, q2}
{q1, q2}
{q0}
{q1}
{q2} Æ

8. Converting via the general method
After eliminating the dead states and transitions, we end up with the same picture 1 1
{q0}
{q0, q1} 1
0, 1 1 1 Æ
{q1}
{q0, q2}
{q0, q1, q2}
0, 1
{q2}
{q1, q2} 1

9. Why the method works
At the end, the DFA accepts when it is in a state that contains some accepting state of NFA
So the DFA accepts only when the NFA accepts too
After reading n symbols, the DFA is in state {qi1,…,qik} if and only if the NFA is in one of the states qi1,…,qik

10. Converting NFAs into DFAs
We will show a general way to convert every NFA into an equivalent DFA
Step 1: Simplify NFA by eliminating e-transitions
Step 2: Convert simplified NFA (without es)

11. Eliminating e-transitions
, 1 
eNFA: q0 q1 q2
Transition table of corresponding NFA:
inputs 1 q0
{q0, q1, q2}
{q1, q2} q1
{q0, q1, q2} Æ
states q2 Æ Æ
Accepting states of NFA:
q0, q1, q2

12. Eliminating e-transitions
, 1 
NFA: q0 q1 q2
0, 1
NFA without es: q0 q1 q2
0, 1

13. Eliminating e-transitions: General method
For every pair of states qi, qj and every symbol a Î S, replace every path like
For every accept state qf, make accepting all states connected to it via es: q4 qi q5 a e qi q5 q2 q0 qj e a qi a qi qj a e q9 q7 q3 qf

14. Eliminating e-transitions
, 1 
NFA: q0 q1 q2
, 1
NFA without es: q0 q1 q2
, 1

15. Regular expressions

16. String concatenation s = 011 t = 101 st = 011101 ts = 101011
ss =
sst =
s = a1…an
st = a1…anb1…bm
t = b1…bm

17. Operations on languages
The concatenation of languages L1 and L2 is
The n-th power of Ln is
The union of L1 and L2 is
L1L2 = {st: s  L1, t  L2}
Ln = {s1s2...sn: s1, s2, ..., sn  L}
L1  L2 = {s: s  L1 or s  L2}

18. Example L1 = {0, 01} L2 = {e, 1, 11, 111, …} any number of 1s L1L2
= {0, 01, 011, 0111, …}
 {01, 011, 0111, …}
= {0, 01, 011, 0111, …}
0 followed by any number of 1s
L12
= {00, 001, 010, 0101}
L22
= L2
L2n
= L2
(n ≥ 1)
L1  L2
= {0, 01, e, 1, 11, 111, ...}

19. Operations on languages
The star of L are all strings made up of zero or more chunks from L:
This is always infinite, and always contains e
Example: L1 = {01, 0}, L2 = {e, 1, 11, 111, …}. What is L1* and L2*?
L* = L0  L1  L2  …

20. Example L1 = {0, 01} L2 = {e, 1, 11, 111, …} any number of 1s L12
= {00, 001, 010, 0101}
L22
= L2
L2n
= L2
(n ≥ 1)
L1*:
is in L1*
is not in L1*
L2*
= L20  L21  L22  …
is not in L1*
= {e}  L21  L22  …
= L2
L1* are all strings that start
with 0 and do not contain
consecutive 1s
L2* = L2

21. Constructing languages with operations
Let’s say S = {0, 1}
We can construct languages by starting with simple ones, like {0}, {1} and combining them
{0}({0}{1})*
0(0+1)*
all strings that start with 0
({0}{1}*)({1}{0}*)
01*+10*
0 followed by any number of 1s, or 1 followed by any number of 0s

22. Regular expressions
A regular expression over S is an expression formed using the following rules:
The symbol Æ is a regular expression
The symbol e is a regular expression
For every a  S, the symbol a is a regular expression
If R and S are regular expressions, so are R+S, RS and R*.
A language is regular if it is represented by a regular expression

23. Examples S = {0, 1} 01* = 0(1*) = {0, 01, 011, 0111, …}
0 followed by any number of 1s
(01*)(01)
= {001, 0101, 01101, , …}
0 followed by any number of 1s and then 01

24. Examples 0+1 = {0, 1} strings of length 1 (0+1)*
any string
(0+1)*010
any string that ends in 010
(0+1)*01(0+1)*
any string that contatins the pattern 01

25. Examples ((0+1)(0+1))*+((0+1)(0+1)(0+1))*
all strings whose length is even or a mutliple of 3
= strings of length 0, 2, 3, 4, 6, 8, 9, 10, 12, ...
((0+1)(0+1))*
strings of even length
(0+1)(0+1)
strings of length 2
((0+1)(0+1)(0+1))*
strings of length a multiple of 3
(0+1)(0+1)(0+1)
strings of length 3

26. Examples ((0+1)(0+1)+(0+1)(0+1)(0+1))*
strings that can be broken in blocks,
where each block has length 2 or 3
(0+1)(0+1)+(0+1)(0+1)(0+1)
strings of length 2 or 3
(0+1)(0+1)
strings of length 2
(0+1)(0+1)(0+1)
strings of length 3

27. Examples ((0+1)(0+1)+(0+1)(0+1)(0+1))*
strings that can be broken in blocks,
where each block has length 2 or 3 e ✓ 1 ✗ 10 ✓
011 ✓
00110 ✓ ✓
this includes all strings except those of length 1
((0+1)(0+1)+(0+1)(0+1)(0+1))*
= all strings except 0 and 1

28. Examples (1+01+001)*(+0+00) ends in at most two 0s
there can be at most two 0s between consecutive 1s
there are never three consecutive 0s
Guess: ( )*(+0+00) = {x: x does not contain 000} e 00

29. Examples
Write a regular expression for all strings with two consecutive 0s.
S = {0, 1}
(anything) 00 (anything else)
(0+1)*00(0+1)*

30. Examples
Write a regular expression for all strings that do not contain two consecutive 0s.
S = {0, 1}
blocks ending in 1
last block
(1 + 01)
... at most one 0 in every block ending in 1
(e + 0)
... and at most one 0 in the last block
(1 + 01)*(e + 0)

31. Examples
Write a regular expression for all strings with an even number of 0s.
S = {0, 1}
even number of zeros = (two zeros)*
two zeros = 1*01*01*
(1*01*01*)*

32. Main theorem for regular languages
A language is regular if and only if it is the language of some DFA
NFA
regular expression
DFA
regular languages

33. Road map  
regular expression
NFA
NFA without e
DFA

34. Examples: regular expression → NFAq0 q1 q0 q1  q2 q3 q4 q5 1 M2
q0’
q1’  M2

35. Regular expressions
A regular expression over S is an expression formed using the following rules:
The symbol Æ is a regular expression
The symbol e is a regular expression
For every a  S, the symbol a is a regular expression
If R and S are regular expressions, so are R+S, RS and R*.

36. General method Æ e regular expr NFA q0 q0 a q0 q1 symbol a RS    q0MR MS q1

37. General method continued
regular expr
NFA MR  
R + S q0 q1   MS     R* q0 MR q1

38. Road map   
regular expression
NFA
NFA without e
DFA