1. CSC 3130: Automata theory and formal languages Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130 The Chinese University of Hong Kong NFA to DFA conversion and regular expressions Fall 2009

2. NFAs are as powerful as DFAs Obviously, an NFA can do everything a DFA can do But can it do more? Theorem If a language L is accepted by some NFA, then it is also accepted by some DFA. NO!

3. Proof of theorem We will show a general way to convert every NFA into an equivalent DFA Step 1: Simplify NFA by eliminating  -transitions Step 2: Convert simplified NFA (without  s) We do this first

4. NFA to DFA conversion intuition 1 0 0, 1 qq qq qq NFA: DFA: 1 qq q 0 or q 1 1 q 0 or q 2 1 0 0 0

5. NFA to DFA conversion intuition 1 0 0, 1 qq qq qq NFA: DFA: 1 qq {q 0, q 1 } 1 {q 0, q 2 } 1 0 0 0

6. General method NFADFA states q 0, q 1, …, q n  q 0 }, {q 1 }, {q 0,q 1 }, …, {q 0,…,q n } one for each subset of states in the NFA initial state q0q0 q0}q0} transitions  ’({q i1,…,q ik }, a) =  (q i1, a) ∪ … ∪  (q ik, a) accepting states F  Q F’ = {S: S contains some state in F}

7. 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 {q i1,…,q ik } if and only if the NFA is in one of the states q i1,…,q ik

8. Converting via the general method 1 0 0, 1 qq qq qq NFA: DFA: {q 0, q 1 } {q 0, q 2 }{q 0, q 1, q 2 } {q 1, q 2 } {q 0 } {q 1 } {q 2 }  0, 1 0 1 0 1 0 1 0 1 0 1 0 1

9. Converting via the general method {q 0, q 1 } {q 0, q 2 }{q 0, q 1, q 2 } {q 1, q 2 } {q 0 } {q 1 } {q 2 }  0, 1 0 1 0 1 0 1 0 1 0 1 0 1 After eliminating the dead states and transitions, we end up with the same picture

10. Proof of theorem We will show a general way to convert every NFA into an equivalent DFA Step 1: Simplify NFA by eliminating  -transitions Step 2: Convert simplified NFA (without  s)

11. Eliminating  -transitions q0q0 q1q1 q2q2 , 1 0 0   NFA: Transition table of corresponding NFA: states inputs 0 1 q0q0 q1q1 q2q2 {q 1, q 2 } {q 0, q 1, q 2 }    Accepting states of NFA: q 0, q 1, q 2

12. Eliminating  -transitions q0q0 q1q1 q2q2 , 1 0 0  NFA: NFA without  s: q0q0 q1q1 q2q2 0, 1 0 0 0 0

13. Eliminating  -transitions: General method For every state q i and every symbol a , replace every path out of q i like For every accept state q f, make accepting all states connected to it via  s: q4q4 qiqi q5q5 q2q2 q0q0 q5q5  a qiqi q5q5 a   qiqi q5q5 a qiqi a   q9q9 q7q7 q3q3 qfqf

14. Regular expressions

15. Operations on strings Given two strings s = a 1 …a n and t = b 1 …b m, we define their concatenation st = a 1 …a n b 1 …b m We define s n as the concatenation ss…s n times s = abb, t = cbast = abbcba s = 011s 3 = 011011011

16. Operations on languages The concatenation of languages L 1 and L 2 is Similarly, we write L n for LL…L ( n times) The union of languages L 1  L 2 is the set of all strings that are in L 1 or in L 2 Example: L 1 = {01, 0}, L 2 = { , 1, 11, 111, …}. What is L 1 L 2 and L 1  L 2 ? L 1 L 2 = {st: s  L 1, t  L 2 }

17. Operations on languages The star (Kleene closure) of L are all strings made up of zero or more chunks from L : –This is always infinite, and always contains  Example: L 1 = {01, 0}, L 2 = { , 1, 11, 111, …}. What is L 1 * and L 2 * ? L * = L 0  L 1  L 2  …

18. Constructing languages with operations Let’s fix an alphabet, say  = {0, 1} We can construct languages by starting with simple ones, like {0}, {1} and combining them {0}({0}  {1})* all strings that start with 0 ({0}{1}*)  ({1}{0}*) 0(0+1)* 01*+10*

19. Regular expressions A regular expression over  is an expression formed using the following rules: –The symbol  is a regular expression –The symbol  is a regular expression –For every a  , 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

20. Examples 01* (01*)(01) 0 followed by any number of 1 s = 0(1*)= {0, 01, 011, 0111, …} = {001, 0101, 01101, 011101, …} 0 followed by any number of 1 s and then 01  = {0, 1}

21. Examples (0+1)* = { , 0, 1, 00, 01, 10, 11, …} any string 0+1= {0, 1} (0+1)*01(0+1)* strings of length 1 any string that contatins the pattern 01 (0+1)*010 any string that ends in 010

22. Examples (0+1)(0+1) (0+1)(0+1)(0+1) strings of length 2 strings of length 3 ((0+1)(0+1))*+((0+1)(0+1)(0+1))* ((0+1)(0+1))* strings of even length ((0+1)(0+1)(0+1))* strings of length divisible by 3 all strings whose length is even or divisible by 3 = strings of length 0, 2, 3, 4, 6, 8, 9, 10, 12,...

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

24. 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 0011010011  1011010110 ✓ ✓✓✓✓✗ 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

25. Examples (1+01+001)*(  +0+00) ends in at most two 0 s there can be at most two 0 s between consecutive 1 s Guess: (1+01+001)*(  +0+00) = {x: x does not contain 000} there are never three consecutive 0 s 0110010110 001001000 

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

27. Examples Write a regular expression for all strings that do not contain two consecutive 0 s.  = {0, 1}... at most one 0 in every block ending in 1... and at most one 0 in the last block (1 + 01) (  + 0) (1 + 01)*(  + 0) 0110101101010 blocks ending in 1 last block

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

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

30. Road map NFA regular expression NFA without  DFA

31. M2M2 Examples: regular expression → NFA R 1 = 0 R 2 = 0 + 1 R 3 = (0 + 1)* q0q0 q1q1 0 q0q0 q1q1    q2q2 q3q3 0 q4q4 q5q5 1 q0’q0’q1’q1’  M2M2   

32. General method regular expr NFA  q0q0  q0q0 symbol a q0q0 q1q1 a RS q0q0 q1q1  MRMR MSMS 

33. General method continued regular expr NFA R + S q0q0 q1q1  MRMR MSMS   R*R* q0q0 q1q1  MRMR   

34. Road map NFA regular expression NFA without  DFA