Note: Text Cross-References in Parenthesis Are for 2Nd Ed

Chapter 4 Properties of Regular Languages 2/14/03 ==> 10/1/03

Note: text cross-references in parenthesis are for 2nd ed.

NOTE: The “prime” symbol (as in L’) means complement

 Closure

If L1 and L2 are regular languages,

Then so are L1L2, L1L2, L1L2, L1’, and L*

 Regular languages are closed under reversal.

 Homomorphism:

Let  and  be two alphabets.

h:   * is a homomorphism: h(ai) = u  *,  ai  extending to strings we have:

w = a1a2...an

h(w) = h(a1)h(a2)...h(an)

 Homomorphic image:

If L is a language on , then homomorphic image of L is

h(L) = { h(w) : w  L }, h = homomorphism

 Example 4.2, 4.3

 Theorem 4.3

Let h be a homomorphism.

If L is a regular language, then its homomorphic image h(L) is also regular.

==> Regular languages are closed under homomorphisms.

CS373 Fall 2003 Chapter 4 N. Guydosh Page 1  Definition 4.2 Right Quotient

Let L1 and L2 be languages on same .

Then right quotient is defined as

L1/L2 = {x: xy  L1, for some y  L2}

==> “Set of all strings in L1 that have a suffix in L2 with this suffix removed”

n m Example: L1 = {a b : n  1, m  0}  {ba} m L2 = {b : m  1} n m L1/L2 = {a b : n  1, m  0}

aaabbbbb  L1  aaabb  L1/L2 bbb is suffix

==> aaab  L1/L2

==> aaa  L1/L2 Theorem 4.4

If L1 and L2 are regular languages over , then L1/L2 is also regular

==> Family of regular languages is closed under right quotient with a regular language.

See examples 4.4 and 4.5

 Elementary questions about regular languages (4.2) - or - Regular questions about elementary languages!

 Standard representation

A regular language is given in standard representation (SR) if, and only if, it is described by:

(1) Finite automaton, (2) Regular expression, or (3) Regular grammar

Informal descriptions and set notation are excluded.

CS373 Fall 2003 Chapter 4 N. Guydosh Page 2 Theorem 4.5

Given a standard representation of a regular language L on  and any w  *, then  an algorithm determining whether or not w  L.

==> Construct DFA for L and test w.

Theorem 4.6

 an algorithm for determining whether a regular language in standard representation form is empty, finite, or infinite.

==> Construct DFA. If simple path (no edge or vertex repeated) from q0 ==> qf then non-empty.

Question: Does any walk from q0 ==> qf implies  a simple path also? If q0 ==> qf path passes through vertex which is base of cycle then infinite

Theorem 4.7

Given two regular languages L1 and L2 in standard representation form,  an algorithm to determine if/(if not) L1 = L2.

==> Define regular language: L3 = (L1  L2’)  (L1’  L2) = L1  L2 ... “Exclusive OR”

Since L3 is a regular language,  DFA M accepting L3.

Use Theorem 4.6 to see if L3 is empty.

L3 =  if, and only if, L1 = L2. (Why?)

CS373 Fall 2003 Chapter 4 N. Guydosh Page 3 Identifying Non-Regular Languages (Sect 4.3)

 Pigeon hole principle (a target of denial)

 Put n objects into m boxes.

If n > m, then at least one box must have more than one item in it.

 Can prove L is non-regular language by assuming contrary ==> denying pigeon hole property.

Example 4.6 Prove L = {anbn : n  0} is non-regular

Assume contrary: (Assume L is regular language)

==>  DFA, M = (Q, {a,b}, , q0, F)

i Consider *(q0, a ) i > 0

==> infinite number of is and finite number of states.

n m ==>  q  Q and some n, m (n  m) such that *(q0, a ) = q = *(q0, a ) due to pigeon hole principle. ... q would be the first state to repeat for a sufficiently long string of a’s. But since, by assumption (contrary),

n n n M accepts a b , we have *(q, b ) = qf  F

[Uses: *(q,w1w2) = *(*(q,w1),w2)]

But also,

m n m n n *(q0,a b ) = *(*(q0,a ),b ) = *(q, b ) = qf  F

Contradiction since m  n and ambn should be rejected. QED

CS373 Fall 2003 Chapter 4 N. Guydosh Page 4  “Pumping Lemma” (another object of denial)

 Application of pigeon hole principle

Theorem 4.8

Let L be an infinite regular language.

Then, some integer m > 0 such that  w  L and |w|  m,

we can decompose w as

w = xyz

with |xy|  m

|y|  1

i such that wi = xy z, i  0 (pumped up w) is also in L.

Proof: L is regular language ==>  DFA M with states q0, q1, q2, ..., qn , (n+1 states)

Let w  L such that |w|  m = n + 1

M goes through the following sequence of states as it processes w:

q0, qi, qj, ..., qf

Since |w|  n + 1,

Number of states in walk = |w| + 1 > n + 1 See example below ... Number of states in walk > number of states

==> at least 1 repeated state (pigeon hole principle)

CS373 Fall 2003 Chapter 4 N. Guydosh Page 5 Example: w = abc

number of states = |w|+1 = 3+1. = 4. Sequence looks like:

q0 , qi , qj , ..., qr , ..., qr , ..., qf ------x y

 substrings x, y, z such that

*(q0 , x) = qr

*(qr , y) = qr

*(qr ,z) = qf

with |xy|  n + 1 and |y|  1 Note: Reason for |xy|  n + 1 =m: If qr is the first occurrence of a vertex which will be also the next repeated one, st Then it (1 qr) will occur no later than nth move since there are only n+1 states in going from q0 to qf . Since there can be no repetitions of states in the “y loop or segment”, and we need at least 1 move to complete the loop, it follows that number of moves in xy = |xy|  number of states = n+1 = m. Remember the example above relating number of moves (string length) with number of states.

nd Thus *(q0 , xz) = qf skip the loops, and |xy|  n + 1 = m … see p. 116 (120, 2 ed)

i and *(q0 , xy z) = qf i  0 any number of loops will work

xyiz = pumped string is accepted QED

CS373 Fall 2003 Chapter 4 N. Guydosh Page 6 Observation: - If assuming a language L is a regular language (accepted by DFA)

And pumping lemma is denied in the process, then L is not a regular language (contradiction)

- We are guaranteed existence of m as well as the decomposition xyz

We do not know what they are. - Cannot claim contradiction just because the pumping lemma is violated for some specific value of m or specific xyz … see middle p. 117 (bottom p. 120).

- But: Pumping Lemma holds  w  L, |w|  m and all i.

Thus If Pumping Lemma is violated for even one w or i, then L is not a regular language.

 Game Theory Approach to Applying Pumping Lemma

1. Opponent picks m

2. Given m, we pick w  L such that |w|  m (totally free to choose)

3. Opponent chooses decomposition xyz with |xy|  m and |y|  1. Opponent makes optimal choice.)

i 4. We try to pick some particular i such that the pumped string wi = xy z is not in L. … make no assumptions on what partition is – must guarantee violation for all possible partitions …

If we can do this, we win.

Else inconclusive (opponent wins).

CS373 Fall 2003 Chapter 4 N. Guydosh Page 7