chapter 3 regular languages and regular grammars

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Chapter 3Chapter 3

Regular Languages andRegular Languages and

Regular GrammarsRegular Grammars

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3.1: Regular Expressions (1)3.1: Regular Expressions (1)

Regular Expression (RE): E is a regular expression over if E is one of:

a, where a If r and s are regular expressions (REs), then

the following expressions also regular: r | s (r or s) rs (r followed by s) r* (r repeated zero or more times)

Each RE has an equivalent regular language (RL)

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Regular Language (RL): L is a regular language over if L is one of:

empty set {} a set that contains empty

string a} where a

If R and S are regular languages (RL), then the following languages also regular: R S = {w | w R or w S} RS = {rs | r R and s S} R* = R0 R1 R2 R3 …

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Rules for Specifying Regular Expressions:Rules for Specifying Regular Expressions:

1. is a regular expression L =

2. If a is in , a is a regular expression L = {a}, the set containing the string a.

3. Let r and s be regular expressions with languages L(r) and L(s). Then

a. r | s is a RE L(r) L(s)

b. rs is a RE L(r) L(s)

c. r* is a RE (L(r))*

d. (r) is a RE L(r), extra parenthesis

precedence


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Examples:Examples: {0, 1}{00, 11} = {000, 011, 100, 111} {0} = {, 0, 00, 000, 0000, …} 11}* = {, 10, 1010, 101010, …,

01, 0101, 010101, …, 1001, 100101, 10010101, …, 0110, 011010, 01101010,

…} Notational shorthand:Notational shorthand:

L0 = Li = LLi-1

L+ = LL*

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Let Let LL be a language over {a, b}, each string in be a language over {a, b}, each string in LL contains the substring contains the substring bbbb L = {a, b}*{bb}{a, b}*

L is regular language (RL). Why?L is regular language (RL). Why? {a} and {b} are RLs {a, b} is RL {a, b}* is RL {b}{b} = {bb} is also RL Then L = {a, b}*{bb}{a, b}* is RL

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Let L be a language over {a, b}, each string in L begins and ends with an a contains at least one b L = {a}{a, b}*{b}{a, b}*{a} L = a*b*a

L is regular language (RL). Why? {a} and {b} are RLs {a, b}* is RL Then L = {a}{a, b}*{b}{a, b}*{a} is RL

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L = {a, b}*{bb}{a, b}* RE = (a|b)*bb(a|b)*

L = {a}{a, b}*{b}{a, b}*{a} RE = a(a|b)*b(a|b)*a

This RE = (a)|((b)*(c)) is equivalent to a|b*c

We say REs r and s are equivalent (r=s), iff r and s represent the same language Example: r = a|b, s = b|a r = s Why? Since L(r) = L(s) = {a, b}

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Let Let = {a, b}= {a, b}

RE a|b L = {a, b} RE (a|b)(a|b) L = {aa, ab, ba, bb} RE aa|ab|ba|bb same as above RE a* L = {, a , aa, aaa, …} RE (a|b)* L = set of all strings of

a’s and b’s including

RE (a*b*)* same as above RE a|a*b L = {a,b,ab,aab,aaab, …}

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Algebraic Properties of regular ExpressionsAlgebraic Properties of regular Expressions

AXIOM

r | s = s | r

r | (s | t) = (r | s) | t

(r s) t = r (s t)

r = r = r

r* = (r*)* = ( r | )+ = r+ |

r ( s | t ) = r s | r t( s | t ) r = s r | t r

r** = r*

r+ = r r*

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abcabc concatenation (“followed by”)

a | b | ca | b | c alternation (“or”)

** zero or more occurrences

++ one or more occurrences

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All strings of 1s and 0sAll strings of 1s and 0s(0 | 1)*

All strings of 1s and 0s beginning with a 1All strings of 1s and 0s beginning with a 11 (0 | 1)*

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All strings containing two or more All strings containing two or more 00s s (1|0)*0(1|0)*0(1|0)*

All strings containing an even number of All strings containing an even number of 00s s (1*01*01*)* | 1*

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All strings containing an All strings containing an eveneven number of number of 00s and s and even even number of number of 11ss( 0 0 | 1 1 )*(( 0 1 | 1 0 )( 0 0 | 1 1 )*( 0 1 | 1 0 )( 0 0 | 1 1 )*)*

All strings of alternating All strings of alternating 00s and s and 11ss( | 1 ) ( 0 1 )* ( | 0 )

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3.1: Regular Expressions (14)3.1: Regular Expressions (14) Strings over the alphabet {0, 1} with Strings over the alphabet {0, 1} with nono

consecutive 0'sconsecutive 0's (1 | 01 )* (0 | ) 1*(01+)* (0 | ) 1*(011*)* (0 | )

Strings over the alphabet {a, b} with Strings over the alphabet {a, b} with exactly three exactly three b'sb's a*ba*ba*ba*

Strings over the alphabet {a, b, c} containing (Strings over the alphabet {a, b, c} containing (at at least onceleast once) ) bcbc (a|b|c)*bc(a|b|c)*

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Strings over the alphabet {a, b} in which Strings over the alphabet {a, b} in which substrings substrings abab and and baba occur an unequal number of occur an unequal number of timestimes (a+b+)+ | (b+a+)+

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3.1: Regular Expressions (16)3.1: Regular Expressions (16) Describe the following in English:Describe the following in English:

(0|1)* all strings over {0, 1}

b*ab*ab*ab* all strings over {a, b} with exactly 3 a’s

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Examples of RE:Examples of RE:

01* {0, 01, 011, 0111, …..}

(01*)(01) {001, 0101, 01101, 011101, …..}

(0 | 1)* = {0, 1, 00, 01, 10, 11, …..} i.e., all strings of 0 and 1

(0 | 1)* 00 (0 | 1)* = {00, 1001, …..} i.e., all 0 and 1 strings containing a “00”

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More Examples of RE:More Examples of RE:

(1 | 10)* all strings starting with “1” and containing no “00”

(0 | 1)*011 all strings ending with “011”

0*1* all strings with no “0” after “1”

00*11* all strings with at least one “0” and one “1”, and no “0” after

“1”

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What languages do the following RE represent?What languages do the following RE represent?

(1 | 01 | 001)*( | 0 | 00)

((0 | 1)(0 | 1))* | ((0 | 1)(0 | 1)(0 | 1))*

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Home Study:Home Study:

Construct a RE over ={0,1} such that It does not contain any string with two consecutive “0”s It has no prefix with two or more “0”s than “1” nor two or

more “1”s than “0” The set of all strings ending with “00” The set of all strings with 3 consecutive 0’s The set of all strings beginning with “1”, which when

interpreted as a binary no., is divisible by 5 The set of all strings with a “1” at the 5th position from the

right The set of all strings not containing 101 as a sub-string

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Strings over the alphabet {Strings over the alphabet {((, , ))} where the } where the parentheses are balancedparentheses are balanced Not regular

L = {aL = {annbbnn | | nn 0} 0} L = ab, aabb, aaabbb, aaaabbbb, …} Not regular

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3.2:3.2: Connection Between RE & RL (1)Connection Between RE & RL (1)

A language A language LL is called is called regularregular if and only if there if and only if there exists some DFA M such that L = L(M).exists some DFA M such that L = L(M).

Since a DFA has an equivalent NFA, then Since a DFA has an equivalent NFA, then A language L is called regular if and only if

there exists some NFA N such that L = L(N).

If we have a RE If we have a RE rr, we can construct an NFA that , we can construct an NFA that accept L(accept L(rr).).

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2. For a in the regular expression, construct NFA

astartL = {a}

1. For in the regular expression, construct NFA

startL = {}

0. For in the regular expression, construct NFA

start L = { } =

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where i and f are new start / final states, and -moves are introduced from i to the old start states of Ms and Mt as well as from all of their final states to f.

3.(a) If s and t are regular expressions, Ms and Mt are their NFAs. s|t has NFA:

start i f

Ms

Mt

L = {L(Ms) L(Mt)}

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3.(b) If s and t are regular expressions, Ms, Mt their NFAs. st (concatenation) has NFA:

starti fMs Mt

Alternative:

overlap

Msstart i fMt

where i is the start state of Ms (or new under the alternative) and f is the final state of Mt (or new). Overlap maps final states of Ms to start state of Mt

L = {L(Ms)L(Mt)}

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fMsstart i

where : i is new start state and f is new final state

-move i to f (to accept null string)

-moves i to old start, old final(s) to f

-move old final to old start (WHY?)

3.(c) If s is a regular expressions and Ms its NFA,

s* (Kleene star) has NFA:L = {L(Ms)*}

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Build an Build an NFA-NFA- that accepts that accepts (a|b)(a|b)**babaaa

bb

baba

a|ba|b

astart

abstart q1

bstart

a

b

start

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Build an Build an NFA-NFA- that accepts that accepts (a|b)(a|b)**baba

(a|b)(a|b)**a

b

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Build an Build an NFA-NFA- that accepts that accepts (a|b)(a|b)**baba

ab

a

b

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r13

r12r5

r3 r11r4

r9

r10

r8r7

r6

r0

r1 r2

b

*c

a a

|

( )

b

|

*

c

(ab*c) | (a(b|c*))Decomposition for this regular expression:

What is the NFA? Let’s construct it !

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r3: a

r0: b

r2: c

b

r1:

r4 : r1 r2b

c

r5 : r3 r4

b

a c

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r11: a

r7: b

r6: c

c

r9 : r7 | r8

b

r10 : r9

c

r8:

c

r12 : r11 r10

b

a

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r13 : r5 | r12

b

a c

c

b

a

1

6543

8

2

10

9 12 13 14

11

15

7

16

17

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Let’s try Let’s try aa ( ( bb | | cc ) )**

1. 1. aa, , bb, & , & cc

2. 2. bb | | cc

3. ( 3. ( bb | | cc ) )**

S0 S1 a

S0 S1 b

S0 S1 c

S1 S2 b

S3 S4 c

S0 S5

S2 S3 b

S4 S5 c

S1 S6 S0 S7

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4. 4. aa ( ( bb | | cc ) )**

S0 S1 a

S4 S5 b

S6 S7 c

S3 S8 S2 S9

S0 S1 a

b | c

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Let : a

abb a*b+

3 patterns

NFA’s :

start

start

start

1

b

b

bb

a

a

a

2

3 4 5

87

6

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NFA for : a | abb | a*b+

0

b

b

bb

a

a

a

2

3 4 5

87

6

1

start

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Regular Expression to NFA-Regular Expression to NFA-

(a | ba)*a

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First Parsing StepFirst Parsing Step

concatenate

(a|ba)* a

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Second Parsing StepSecond Parsing Step

concatenate

* a

a|ba

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Third Parsing StepThird Parsing Step

concatenate

* a

|

a ba

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Fourth Parsing StepFourth Parsing Step

concatenate

* a

|

a concatenate

b a

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Identify Leaf NodesIdentify Leaf Nodes

concatenate

+ a

|

a concatenate

b a

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Convert Leaf NodesConvert Leaf Nodes

concatenate

*

|

concatenate

a

a

a

b

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Identify Convertible Node(s)Identify Convertible Node(s)

concatenate

*

|

concatenate

a

a

a

b

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Convert NodeConvert Node

concatenate

*

|

a

a ab

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Identify Convertible NodeIdentify Convertible Node

concatenate

*

|

a

a ab

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concatenate

* a

a

ab

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concatenate

* a

a

ab

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concatenate

a

a

ab

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concatenate

a

a

ab

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Convert Final NodeConvert Final Node

a

a

ab

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3.2:3.2: Expression Graphs (1)Expression Graphs (1)

NFA to RENFA to RE

If L is accepted by some NFA-, then L is represented by some regular expression

An expression graph is like a state diagram but it can have regular expressions as labels on arcs

An NFA- is an expression graph An expression graph can be reduced to one with

just two states If we reduce an NFA- in this way, the arc label

then corresponds to the regular expression representing it

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wik kwjij i

wji wik kj

wik kwjij i

wji (wii)* wik

kjwii

w

w*start

w1

(w1)*w2(w3 w4(w1)*w2)*start w2

w3

w4

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Merge Edges :

c

b

a

a | b | c

p

ac*b

ba c

Replace state by Edges

q0 q1

q0q1

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Let Let GG be the state diagram of a finite automata be the state diagram of a finite automata Let Let mm be the number of final states of be the number of final states of GG Make Make mm copies of copies of GG, each of which has one final , each of which has one final

state.state. Call these graphsCall these graphs G G11, , GG22, …, , …, GGmm For each For each GGtt

Repeat Do the steps in the previous slide

Until the only states in Gt are the start state and the single final state

Determine the RE of Gt The RE of The RE of GG is obtained by joining RE’s of each is obtained by joining RE’s of each

GGtt by by oror

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c3

c1 2

bG: start

b

c3

c1 2

bG1: start

b

c3

c1 2

bG2: start

b

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c3

c1 2

bG1: start

b

3cc

1

b b

1

b

b*

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c3

c1 2

bG2: start

b

3cc

1

b b

b*ccb*

RE for G b* | b*ccb*

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3.3 Regular Grammars (1)

A grammar is said to be right-linear if all production rules are of the form: A xB A x Where A, B NT, and x T*

A grammar is said to be left-linear if all production rules are of the form: A Bx A x Where A, B NT, and x T*

A regular grammar is one that either right-linear or left-linear, but not both

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The grammars G1 is right-linear G1=({S}, {a, b}, S, {S abS | a} L(G1) = {{ab}*a} r = (ab)*a

The grammars G2 is left-linear G2=({S, A, B}, {a, b}, S, { S Aab

A Aab |

B a } L(G2) = {a{ab}+} r = a(ab)+

Both grammars G1 and G2 are regular

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The grammars G is not regular G=({S, A, B}, {a, b}, S, { S A

A aB |

B Ab }

Every production rule is either in right-linear or left-linear from

The grammar itself is neither right-linear nor left-linear

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A language is said to be regular if it is generated by some regular grammar.

A regular language may be generated by both regular and non-regular grammars.

Derivations in regular grammars (right-linear) have a nice form There is at most one non-terminal present in a

sentential form and that non-terminal, if present, is the rightmost symbol in the string.

ab...cD ab...cdE D dE

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The grammars G1 & G2 generate the language a+b*

The grammar G1 is not a regular, because of the rule S AB does not have the specified form

G2 is a regular grammar (right-linear)

G1=({S,A,B}, {a,b}, S, {S AB A aA | aB bB | }

G2=({S,B}, {a,b}, S, {S aS | aB B bB | }

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The following grammars are regular grammars (right-linear)

G=({S,O}, {a,b}, S, {S aO | bO | O aS | bS }

G=({S,B}, {a,b}, S, {S aS | bB | B aB | bS}

G=({S,B,C}, {a,b,c}, S, {S bS | cS | aB | B aB | cS |

bC | C aB | bS | }

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Non-regular grammar S abSA |

A Aa |

Its equivalent regular grammar S aB |

B bS | bA A aA |

Both grammars generate the language | (ab)+a*

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The language a+b+ is generated by the grammar G G: S aS | aA

A bA | b And also accepted by NFA M

Derivation of aabb using CFG G S aS aaA aabA aabb

Accepting aabb using NFA M [S, aabb] ├ [S, abb] ├ [A, bb] ├ [A, b] ├ [Z, ]

S A Zba

a b

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The NFA M = (Q, , , S, F) can be constructed from the rules of a grammar G = (NT, , P, S) as follows: Q = NT {Z}, where Z NT if A a P

Q = NT otherwise

(A, a) = B whenever A aB P(A, a) = Z whenever A a P

F = {A | A P} {Z} if Z QF = {A | A P} otherwise

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The the grammar G generates and the NFA M accepts the language a*(a|b+) G: S aS | bB | a

B bB |

M

S B

Za

b

a b

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Given the following grammar G that generates the language over {a, b, c} that do not contain the substring abc G: S bS | cS | aB |

B aB | cS | bC | C aB | bS |

What is NFA M that accepts L(G)?

S B Cba

b, c a

ac

b

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The DFA M that accepts the language over {a, b} with even number of a’s and odd number of b’s is given

Give grammar G for M G: S aA | bB

A aS | bCB bS | aC | C aB | bA

S B

a

A Cb

a aa

b

b

b

chapter 3 regular languages and regular grammars

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