chapter 4 properties of regular languages

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Chapter 4Chapter 4

Properties of Properties of

Regular LanguagesRegular Languages

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4.1:4.1: Closure Properties of RLs (1)Closure Properties of RLs (1)

A language A language LL over an alphabet over an alphabet is is regularregular if itif it has a regular set (expression) over

is accepted y DFA, NFA, or NFA-

is generated by regular grammar

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If If LL11 and and LL22 are regular languages, then are regular languages, then L1 L2 Union

= {w * : w L1 or w L2} L1 L2 Intersection

= {w * : w L1 and w L2} L1 L2 Concatenation

n languages: L1L2 …Ln = {w1w2…wn : wi Li} one languages: Ln = {w1w2…wn : wi L}

L* Kleene Star L*= L0 L1 L2 …

L1 Complement L = * - L = {w : w * and w L}

are also regular languagesare also regular languages

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Prove that RLs are closed under Prove that RLs are closed under complementationcomplementation.. If If LL is a regular language over is a regular language over , then , then LL = = ** - - LL is is

regular language.regular language. Proof:Proof:

If L is regular, there exists a DFA M recognizing L.

We can construct a DFA M’ for L by copying M to M’ except that all final states in M are changed to non-final, and all non-final states to final.

See next slide for a formal proofSee next slide for a formal proof

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Let Let MM11=(=(QQ, , , , , , qq00, , FF) a DFA that accepts ) a DFA that accepts LL((MM11)) Let Let MM22=(=(QQ, , , , , , qq00, , QQ--FF) a DFA that accepts ) a DFA that accepts LL((MM22)) Obviously both languages are regular languagesObviously both languages are regular languages Per definition 2.2: Per definition 2.2: LL((MM) = {) = {ww ** : : ((qq00, , w)w) FF} } The following are both trueThe following are both true

w * : (q0, w) F (q0, w) Q – F w * : (q0, w) Q - F (q0, w) F

Thus Thus LL((MM22) = ) = LL((MM11)) LL((MM11) is an arbitrary regular language and its ) is an arbitrary regular language and its

complement is also a regular language, therefore the complement is also a regular language, therefore the regular languages are closed under complementationregular languages are closed under complementation

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Are regular languages closed under Are regular languages closed under intersectionintersection?? If If LL11 and and LL22 are RLs, then are RLs, then LL11 LL22 is RL. is RL.

ProofProof:: Since RLs are closed under union and

complementation, they are also closed under intersection

L1, L2 are RLs, so L1, L2 are RLs, and L1 L2

is RL, so L1 L2 is RL.

Thus L1 L2 = L1 L2 is RL.

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Are regular languages closed under Are regular languages closed under differencedifference?? If If LL11 and and LL22 are RLs, is are RLs, is LL11 - - LL22 RL? Why? RL? Why?

L1 - L2 = L1 L2

Since RLs are closed under intersection and complementation, they are also closed under difference.

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Let Let LL, , MM and and NN are regular expressions, which of are regular expressions, which of the followings are correct?the followings are correct?

Commutative Rules:Commutative Rules: L M = M L LM = ML

Associative Rules:Associative Rules: (L M) N = L (M N) (LM)N = L(MN)

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Which of the followings are correct?Which of the followings are correct?

Left Distributive Rules: L(M N) = LM LN L (MN) = (L M)(L N)

Right Distributive Rules: (M N)L = ML NL (MN) L = (M L)(N L)

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Identities:Identities:

is the identity for union: L = L = L

is the identity for concatenation: L = L = L

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Other Rules for Kleene ClosureOther Rules for Kleene Closure

(L*)* = L*

L+ = LL* = L*L L* = L+ U {} * = {} * =

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4.3:4.3: Non-regular Languages (1)Non-regular Languages (1)

The language The language L = {an bn | n 0} is not regular There is NO DFA/NFA that accepts L

Theorem:Theorem: If If LL11 is RL and is RL and LL22 is not-regular, then is not-regular, then

The language L = L1 L2 is not necessarily regular

Proof:Proof: L1 = a*b* is regular L2 = {an bn | n 0} is not regular L1 L2 = L2 which is not regular

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The languageThe language L = {an bm | n, m 0 and n m} is not regular.

Proof: Proof: L is regular iff L a*b* is regular. But L a*b* = {an bn | n 0}, which we know

is not regular.

Then L is not regular.

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Many languages are non-regular:Many languages are non-regular: {anbn | n 0} {wwR} | w * } {an2 | n 0} {ap | p is prime} set of well-formed parentheses set of palindromes: {w = wR} …...

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We want to prove that We want to prove that LL = { = {aakkbbkk | | kk 0 0} is non-} is non-regular. Prove by contradiction:regular. Prove by contradiction:

Assume that there is a DFA M which recognizes L. Let n be the no. of states in M

Consider the acceptance of the input arbr where r = n:

q0 q1 q2 qr qr+1 q2ra a b b b…...…...

a a

arbr

path for accepting arbr

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Since r = n and M has only n states, there must be at least one state visited twice in the first r transitions. Let this state be visited at the ith and the jth steps, where j i.

By skipping the loop, ar-(j-i)br should also be accepted by M,

but this is contradictory since ar-(j-i)br L

q0

qi = qj

qf Fa path in M

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We want to prove that We want to prove that LL = { = {aakk22 | | kk 0} is non- 0} is non-

regular. Prove by contradiction:regular. Prove by contradiction:


M should also accept the string an2

q0 q1 q2 qn2a a

…...a a

path for accepting an2

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Since n2 n and M has only n states, there must be at least two equal states from q0 to qn2. Let them be qi and qj where j - i = m n.

By repeating the loop one more time, a(n2+m) is also accepted by M, which is a contradiction, since (n2+m) cannot be a square (the next square after n2 is (n+1)2 but n2+m (n+1)2)

q0

qi = qj

qn2 Fa path in M

A loop oflength m

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We want to prove that We want to prove that LL = { = {aapp | | pp is a prime} is is a prime} is non-regular. Prove by contradiction:non-regular. Prove by contradiction:


From Number Theory, we know that the no. of primes are infinite, so there exists a prime p n

q0 q1 q2 qpa a

…...a a

path for accepting ap

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Since p n and M has only n states, there must be at least two equal states from q0 to qp. Let them be qi and qj where j - i = m 0

By repeating the loop (p-m) times, a(p-m)m + (p-m) = a(p-m)(m+1) is also accepted by M, which is a contradiction since (p-m)(m+1) is not a prime, since it can be divided by (p-m) and (m+1) 1

q0

qi = qj

qp Fa path in M

A loop oflength m

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4.3:4.3: Pumping Lemma (1) Pumping Lemma (1)

If If LL is a regular language, there is a DFA which is a regular language, there is a DFA which recognizes recognizes LL. . All finite languages are regular

, {a}, {abaabb, b, baabbbbb, bb} So, only infinite languages are interesting for

this discussion.

What are non-regular languages?What are non-regular languages?

Pumping LemmaPumping Lemma A technique for proving a language L is NOT

regular

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4.3:4.3: Pumping Lemma (2)Pumping Lemma (2)

Any given DFA has a finite number of statesAny given DFA has a finite number of states Let Let kk = | = |QQ|| Consider any accepted input Consider any accepted input w = a1…an, such that |

w| = n kk DFA follows a path of states q0…qn:

This state numbering has nothing to do with state names in Q

May repeat states in this path

q0

a1 q1… qn-1 qn

a2 anan-1

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Since |Since |ww| | kk, path must contain at , path must contain at least one loopleast one loop Consider any one such loop (cycle) Break string & path into three parts: before,

during, & after loop

Given other inputs, could loop an arbitrary number Given other inputs, could loop an arbitrary number of timesof times (a1…ai)(ai+1…aj)*(aj+1…an) also accepted

q0

a1 qi=qj… qn

ai an

ai+1

…aj+1

…aj

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If language If language LL is regular, then for any is regular, then for any ww LL that is that is longer than |longer than |QQ|, there is a |, there is a substringsubstring of of ww that can be that can be repeated (repeated (pumpedpumped) any number of times with ) any number of times with resulting string remaining in the languageresulting string remaining in the language

Lemma:Lemma: Let G be the state diagram of a DFA with k states.

Any path of length k in G contains a cycle. Proof:Proof:

A path of length k contains k+1 nodes. Since there are only k nodes (states) in G, there must be a node, call it qi, that occurs in at least two positions in the path. The subpath from the first occurrence of qi to the second produces the desired cycle.

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Let Let GG be the state diagram of a DFA with be the state diagram of a DFA with kk states and let states and let pp be a path of length be a path of length kk..

The path p can be subdivided into subpaths u, v, and z where p = uvz

The length of uv is |uv| k, and v is a cycle

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Pumping Lemma Pumping Lemma Let Let LL be a regular language is accepted by a DFA be a regular language is accepted by a DFA

MM with with kk states. Let for ALL states. Let for ALL ww LL and | and |ww| | kk. . Then Then ww can be divided into three parts, can be divided into three parts, ww = = uuvvzz, , satisfying the following conditions:satisfying the following conditions:

|uv| k,

|v| 1, and

for all i 0, uviz L

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Proof: Let w L with |w| = n k where w(i) is the ith symbol of the string w.

Since n k and M has only k states, there must be at least one repeated state from q0 to qn

Let qi be the first such repeated state

q0 q1 q2 qnw(2) w(3) …...w(1) w(n)

q0

qi

qn Fa path in M

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Let u be the string obtained by traversing from q0 to qi, v be the string obtained by traversing the loop once (so |v| 1)

In the traversal from q0 to qi and then through the loop once back to qi, nothing except qi repeats, thus |uv| k

By traversing the loop 0 or more times, we obtain uviz for all i 0 and they should all be accepted by the DFA, i.e., in L.

q0

qi

qn Fa path in M

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How to use pumping lemma to prove some L isn’t regular? Lemma describes a property of regular

languages. Use proof by contradiction: If L doesn't have

this property, it is NOT regular

We cannot use the pumping lemma to prove a language is regular Since we have to check every string in L

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4.3: Pumping Lemma (10)4.3: Pumping Lemma (10)

We assume the given language L is regular Let n be the constant of pumping lemma Find a string w L where |w| n Divide w into 3 parts, w = uvz, where

|uv| n |v| 1

Claim that uviz are also in L for all i Find one i such that uviz LL Contradiction

Only one string Only one string ww is needed is needed do not need to consider all general cases choose w that can led to contradiction easily

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Example 1: L = {am bm | m 0} Assume L is regular, n is the # of DFA states Let w = anbn where |w| = 2n

n w = ap aj an-p-j bn where p + j n and j 1 w = u v z

u & v contain only a’s, since |uv| n Pick i = 2:

uv2z = ap aj aj an-p-j bn = an aj bn uv2z contains more a’s than b’s, since |v| 1 uv2z L

Contradiction and hence L is not regular

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Example 2:Example 2: L = {w | w {a, b}*, w has equal number of a’s

and b’s} w = an bn

Same argument as previous example

Choice of Choice of ww is important is important Different w’s may lead to different proofs, or

none being possible Choice of Choice of ii is important. is important.

Different i’s may lead to different proofs, or none being possible

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Example 3:Example 3: L = {(ab)i | i 0} L is regular: L = (ab)*

i.e., for any w, there is some split u, v, z that allows pumping

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Example 4: Prove that the language L of well-formed

parentheses is non-regular Assume L is regular and let n be the constant of

pumping lemma Let w = (n)n L, |w| = 2n n

we do not need to consider other possible cases like ()()

we only need to ensure |w| n Then w = (m (j (n-m-j )n where j 1 and m + j n

u v z

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By pumping lemma,By pumping lemma,

((mm((jjii((nn--mm--jj ))nn = = ((nn++jj((ii-1)-1) ))nn LL for all for all ii

When When ii = 0 = 0 ((nn--jj ))nn LL since since jj 11 nn--jj nn

ContradictionContradiction LL is not regular is not regular

We can choose We can choose i i = 2, = 2, ii = 3, ... = 3, ... but cannot choose i = 1

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To prove L is regular

1. Constructing a FSA (NFA or DFA), RE, or RG for L

2. Using closure property of regular sets Let ⊕ be a binary operation (such as union,

concatenation and intersection)L3 = L1 ⊕ L2 is regular, if both L1 and L2 are regular

Let ⊡ be an unary operation (such as complementation and Kleene star) L1 = ⊡ L2 is regular, if L2 is regular

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Example Example Prove that Prove that LL is regular, where is regular, where LL is the set of all is the set of all

strings with three consecutive 0's and strings with three consecutive 0's and = {0,1} = {0,1} Let L1 = (0 | 1)*, which is regular and

L2 = 000, which is also regular then L = L1 ． L2 ． L1 Hence L is regular due to the closure property

of regular sets under concatenation.

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Prove Prove LL is not regular is not regular 1. Using Pumping Lemma 2. Using closure property of regular sets

Let ⊕ be a binary operation (such as union, concatenation and intersection)if L3 = L1 ⊕ L2 and L1 is regular and L3 is not regular,

then L2 must be non-regular

Let ⊡ be an unary operation (such as complementation and Kleene star)if L1 = ⊡ L2 and L2 is not regular, then L1 is non-regular

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L = {w {a, b}*, w has equal number of a’s and b’s}

Consider L’ = L a*b* = {an bn | n 0}

By closure properties of RLs:if L is regular, then L’ is also regular

But since L’ not regular, so L can’t be regular

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Example Prove that L is not regular, where L is the set of all

strings over {a, b} such that the number of b is the double of the number of a. HW: Use pumping lemma to prove it.

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Common Mistakes Let L1 regular language, L3 and L4 are not regular. Is L regular?

L = L1 ⊕ L3

0n1n = 0*1* 0n1n

0*1* = 0*1* 0n1n

L = L3 ⊕ L4

0*1* = 0p1q 0q1p , where p q

L3 = L ⊕ L4

0n1n = 0*1* 0n1n

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Class DiscussionClass Discussion

Prove that L = {ww : w (0|1)*} is non-regular

Is L = {(01)6n : n 0} regular?

Is L = {(ab)n : n 0} regular?

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SummarySummary

We use the Pumping Lemma to prove a language is We use the Pumping Lemma to prove a language is not regularnot regular Note, does not work for all non-regular

languages, though

Choosing a good string Choosing a good string w w is first key stepis first key step

Choosing a good integer Choosing a good integer ii is second key step is second key step

Must apply argument to all legal Must apply argument to all legal uu, , vv, , zz

chapter 4 properties of regular languages

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