CSCI 2670Introduction to Theory of

Computing

September 13

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

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

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:

1. The start state has transition arrows going to every other state, but no arrows coming in from any other state

2. 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

3. The start state is not the accept state

GNFA’s (continued)

4. Except for the start and accept states, one arrow goes from every state to every other state and also from each state to itself

5. Instead of being labeled with symbols from the alphabet, transitions are labeled with regular expressions

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

Converting a DFA to a GNFA

Add two new states

ε εqs qt

1

1

0

0

q1

q2

q3 q40,1

0,1

Converting a DFA to a GNFA

ε εqs qt

1

1

0

0

q1

q2

q3 q40,1

0,1

• All transition labels with multiple labels are relabeled with the union of the previous labels

01

01

Converting a DFA to a GNFA

ε εqs qt

1

1

0

0

q1

q2

q3 q4

• All pairs of states without transitions get a transition labeled

01

01

Converting a DFA to a GNFA

ε εqs qt

1

1

0

0

q1

q2

q3 q4

• The resulting state diagram is a GNFA– All GNFA properties are satisfied

01

01

Converting a DFA to a GNFA

ε εqs qt

1

1

0

0

q1

q2

q3 q4

• No step changed the strings accepted by the machine

01

01

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

Removing one state from a GNFA

q1’ q2’

a13

a32

a12

a33

q2q1

q3

a12a13a33*a3

2

Accounting for loops

a11a13a33*a3

1

q1’ q2’

a31

a13 a32

a23

a21

a12

a33

a22

q2q1

q3

a11

a22a23a33*a32

a21a23a33*a3

1

a12a13a33*a3

2

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 wL(M) iff wL(R).

First show wL(M) iff wL(G). Then show wL(G) iff wL(R).

wL(M) iff wL(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+1F• 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 wL(G)

wL(G) iff wL(R)

• Prove by induction on number of states in G

Base case: If G has 2 states then clearly wL(G) iff wL(R).

Induction step: Assume wL(G) iff wL(R) for every G with k-1 states. Prove w L(G) iff wL(R) for every G with k states.

wL(R) if wL(G)

Assume wL(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’.

wL(R) if wL(G)

Assume wL(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 wL(R).

wL(G) if wL(R)

Assume wL(R). By induction hypothesis, wL(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, wL(G).

Example

1

q1 q2

1

0

0

Example

1

q1 q2

1

0

0

qs qt

Step 1: Add two new states

ε ε

Example

1

q1 q2

1

0

0

qs qt

Step 2: Remove q1

ε

1*0

101*0

ε

Example

q2

qs qt

Step 3: Remove q2

1*0

101*0

ε

1*0(101*0)*

Example

1

q1 q2

1

0

0

So this DFA

Is equivalent to the regular expression 1*0(101*0)*

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?

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

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

1. for each i 0, xyiz is in A2. |y| > 0, and3. |xy| p

p is called the pumping length

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

Proof idea

1. for each i 0, xyiz is in A2. |y| > 0, and3. |xy| p

x

y

z