CSCI 2670 Introduction to Theory of Computing

1. CSCI 2670Introduction to Theory of Computing September 13

2. Agenda • Last class • Example DFA construction (on board) • Equivalence of RE’s and regular languages • GNFA’s • Today • More on equivalence of RE’s and regular languages • Non-regular languages? • Tomorrow • Dr. Canfield returns

3. RE’s and regular languages Theorem: A language is regular if and only if some regular expression describes it. • i.e., every regular expression has a corresponding DFA and vice versa

4. GNFA’s • To prove every DFA can be described using an RE, we introduce GNFA’s • A GNFA is an NFA with the following properties: • The start state has transition arrows going to every other state, but no arrows coming in from any other state • There is exactly one accept state and there is an arrow from every other state to this state, but no arrows to any other state from the accept state • The start state is not the accept state

5. GNFA’s (continued) • Except for the start and accept states, one arrow goes from every state to every other state and also from each state to itself • Instead of being labeled with symbols from the alphabet, transitions are labeled with regular expressions

6. Equivalence of DFA’s and RE’s • First show every DFA can be converted into a GNFA that accepts the same language • Then show that any GNFA has a corresponding RE that accepts the same language

7. 0 q2 1 qs qt q1 1 ε ε 0 q3 q4 0,1 0,1 Converting a DFA to a GNFA Add two new states

8. qs qt ε ε Converting a DFA to a GNFA • All transition labels with multiple labels are relabeled with the union of the previous labels 0 q2 1 q1 1 0 q3 q4 0,1 01 0,1 01

9. qs qt ε ε Converting a DFA to a GNFA • All pairs of states without transitions get a transition labeled  0 q2 1 q1 1 0 q3 q4 01 01

10. qs qt ε ε Converting a DFA to a GNFA • The resulting state diagram is a GNFA • All GNFA properties are satisfied 0 q2 1 q1 1 0 q3 q4 01 01

11. qs qt ε ε Converting a DFA to a GNFA • No step changed the strings accepted by the machine 0 q2 1 q1 1 0 q3 q4 01 01

12. Converting a GNFA to a RE • If the GNFA has two states, then the label connecting the states is the RE • Otherwise, remove one state at a time without changing the language accepted by the machine until the GNFA has two states

13. a12 q2 q1 a32 a13 q3 a33 Removing one state from a GNFA a12a13a33*a32 q1’ q2’

14. a12 a22 a11 q2 q1 a21 a13 a32 a23 a31 q3 a33 Accounting for loops a22a23a33*a32 a11a13a33*a31 a12a13a33*a32 q1’ q2’ a21a23a33*a31

15. Every DFA has a corresponding RE Proof: Let M be any DFA and let w be any string in Σ*. Convert M to G, a GNFA, then convert G to R, a regular expression, using methods shown in class. • Want to show wL(M) iff wL(R). • First show wL(M) iff wL(G). • Then show wL(G) iff wL(R).

16. wL(M) iff wL(G) • Let w = w1w2…wn, where each wi  Σ. • w  L(M) • iff there is a sequence of states q1, q2, …, qn+1 such that • q1 = q0 • qn+1F • qi+1 = (qi,wi) for each i = 1, 2, …, n • iff when w is read by G, the sequence of states qs, q1, q2, …, qn+1, qt would accept w • iff wL(G)

17. wL(G) iff wL(R) • Prove by induction on number of states in G Base case: If G has 2 states then clearly wL(G) iff wL(R). Induction step: Assume wL(G) iff wL(R) for every G with k-1 states. Prove w  L(G) iff wL(R) for every G with k states.

18. wL(R) if wL(G) Assume wL(G) and an accepting branch of the computation G enters on w is qs, q1, q2, …, qt. Let G’ be the GNFA that results from removing one of G’s states, qrip. There are two possibilities: Case 1: qrip is never entered in the computation of w. Then the same branch of computation exists in G’.

19. wL(R) if wL(G) Assume wL(G) and an accepting branch of the computation G enters on w is qs, q1, q2, …, qt. Let G’ be the GNFA that results from removing one of G’s states, qrip. There are two possibilities: Case 2: qrip is entered in the computation of w (bracketed byqi and qj). Then the new transition between qi and qj in G’ describes the computation that could be done on the computation of w through the branch qi, qrip, qj. So G’ accepts w. By induction wL(R).

20. wL(G) if wL(R) Assume wL(R). By induction hypothesis, wL(G’), the k-1 state GNFA resulting from removing one state from G. By construction, any computation in G’ can also be done in G – possibly going through an extra state qrip. Therefore, wL(G).

21. 1 1 0 q1 q2 0 Example

22. qs qt Example 1 1 0 q1 q2 0 ε ε Step 1: Add two new states

23. qs qt Example 1 101*0 1 0 q1 q2 0 ε ε 1*0 Step 2: Remove q1

24. qs qt Example 101*0 q2 ε 1*0 1*0(101*0)* Step 3: Remove q2

25. 1 1 0 q1 q2 0 Example So this DFA Is equivalent to the regular expression 1*0(101*0)*

26. Nonregular languages • So far, we have explored several ways to identify regular languages • DFA’s, NFA’s, GNFA’s, RE’s • There are many nonregular languages • {0n1n | n  0} • {101,101001,1010010001,…} • {w | w has the same number of 0s and 1s} • How can we tell if a language is not regular?

27. Property of regular languages • All regular languages can be generated by finite automata • States must be reused if the length of a string is greater than the number of states • If states are reused, there will be repetition

28. The pumping lemma Theorem: If A is a regular language, then there is a number p where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions • for each i  0, xyizis in A • |y| > 0, and • |xy|  p p is called the pumping length

29. Proof idea • Pumping length is equal to the number of states in the DFA whose language is A • p = |Q| • If A accepts a word w with |w| > p, then some state must be entered twice while processing w • Pigeonhole principle

30. y x z Proof idea • for each i  0, xyizis in A • |y| > 0, and • |xy|  p