1. Introduction to the Theory of Computation John Paxton Montana State University Summer 2003

2. Humor A busload of politicians were driving down a country road when, all of a sudden, the bus ran off the road and crashed into a tree in an old farmer's field. The old farmer, after seeing what had happened, went over to investigate. He then proceeded to dig a hole to bury the politicians. A few days later the local sheriff came out, saw the crashed bus, and asked the old farmer where all the politicians had gone. The old farmer said he had buried them. The sheriff asked the old farmer, "Were they all dead?" The old farmer replied, "Well, some of them said they weren't, but you know how them politicians lie."

3. 1.2 Nondeterminism In a nondeterministic machine, several choices might exist at a given point instead of only one.

4. Example 2 0,  0 1 q0q0 q1q1 q2q2

5. Example 3 Write an NDFA that recognizes the language consisting of all strings over {0, 1} containing a 1 in the third position from the end.

6. Example 3 10,1 q0q0 q1q1 q2q2 q3q3

7. Exercises Give a 3 state NFA that accepts the language: zero or more 0s, followed by zero or more ones, followed by zero or more 0s followed by a 0. Give a 1 state NFA that accepts the language {  }.

8. Formal Definition A nondeterministic finite automaton is a 5- tuple (Q, , , q 0, F) where 1.Q is a finite set called the states  is a finite set called the alphabet  : Q x    P(Q) is the transition function 4.q 0  Q is the start state 5.F  Q is the set of accept states

9. P(Q) The power set of states Consider Q = { q 0, q 1, q 2 } Then P(Q) = { {}, {q 0 }, {q 1 }, {q 2 }, {q 0, q 1 }, {q 0, q 2 }, {q 1, q 2 }, {q 0, q 1, q 2 } }

10. Example 3 Q = {q 0, q 1, q 2, q 3 }  = {0, 1} q 0 F = {q 3 }

11. Example 3,  01  q0q0 {q 0 }{q 0, q 1 }{} q1q1 {q 2 } {} q2q2 {q 3 } {} q3q3 {q 3 } {}

12. Acceptance Let N = (Q, , , q 0, F) be an NFA and w = w 1 w 2 …w n be a string over the alphabet . Then N “accepts” w if a sequence of states r 0 r 1 …r n exists in Q with the following three conditions: 1.r 0 = q 0 2.r i+1  (r i, w i+1 ) for 0 <= i <= n – 1 3.r n  F

13. Theorem Every nondeterministic finite automaton has an equivalent deterministic finite automaton.

14. Proof Part 1 Given a deterministic finite automaton, there is an equivalent nondeterministic finite automaton. Proof. Trivial!

15. Proof Part II. Given a nondeterministic finite automaton, there is an equivalent deterministic finite automaton. Proof. A bit harder …

16. Proof by Construction Let N = (Q, , , q 0, F) Construct M = (Q’, ,  ’, q 0 ’, F’) Q’ = P(Q) q 0 ’ = E [ {q 0 } ]

17. Proof by Construction F’ = {R  Q’ | R contains an accept state in N} For R  Q’ and a  , let  ’(R, a) = {q  Q | q  E[  (r, a)] for some r  R}

18. Convert Example 2 {q 0, q 1 } {q 1, q 2 } 0 {q 2 } 1 1

19. Exercise Convert the following NFA into an equivalent DFA.  a a a,b b q0q0 q1q1 q2q2

20. Theorems The class of regular languages is closed under the union operation. The class of regular languages is closed under the concatenation operation. The class of regular languages is closed under the star operation.

21. Closure Under Union Note: we already proved this once using DFAs. However, using NFAs, the proof is even easier so we will do it again!

22. Closure Under Union Let N 1 = (Q 1, ,  , q 1, F 1 ) Let N 2 = (Q 2, ,  , q 2, F 2 ) Construct N = (Q, , , q 0, F) Q = q 0 U Q 1 U Q 2

23. Closure Under Union F = F 1 U F 2  (q, a) =  1 (q, a) if q  Q 1  2 (q, a) if q  Q 2 {q 1, q 2 } if q = q 0 and a =  {} if q = q 0 and a <> 

24. Exercises Draw a picture that graphically displays how the preceding proof works. Draw an NFA that accepts the union of {w | w begins with a 1 and ends with a 0} and {w | w contains at least three 1s}

25. Closure Under Concatenation Let N 1 = (Q 1, ,  , q 1, F 1 ) Let N 2 = (Q 2, ,  , q 2, F 2 ) Construct N = (Q, , , q 0, F) Q = Q 1 U Q 2

26. Closure Under Concatenation q 0 = q 1 F = F 2  (q, a) =  1 (q, a) if q  Q 1 and !(q  F1)  1 (q, a) if q  F 1 and a <>   1 (q, a) U {q2} if q  F 1 and a =   2 (q, a) if q  Q 2

27. Exercises Draw a picture that graphically displays how the preceding proof works. Draw an NFA that accepts the concatenation of {w | w begins with a 1 and ends with a 0} and {w | w contains at least three 1s}

28. Closure Under Star Let N 1 = (Q 1, ,  , q 1, F 1 ) Construct N = (Q, , , q 0, F) Q = {q 0 } U Q 1 F = {q 0 } U F 1

29. Closure Under Star  (q, a) =  1 (q, a) if q  Q 1 and !(q  F 1 )  1(q, a) if q  F 1 and a <>   1(q, a) U {q 1 } if q  F 1 and a =  {q 1 } if q = q 0 and a =  {} if q = q 0 and a <> 

30. Exercises Draw a picture that graphically displays how the preceding proof works. Draw an NFA that accepts the star of {w | w begins with a 1 and ends with a 0}.