1 Chapter 3 Regular Languages

 2 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)

 3 3.1: Regular Expressions (2)   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 * = R 0  R 1  R 2  R 3  …

 4 Rules for Specifying Regular Expressions:  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 precedenceprecedence 3.1: Regular Expressions (3)

 5 3.1: Regular Expressions (4)  Examples:  {0, 1}{00, 11} = {000, 011, 100, 111}  {0}  = { , 0, 00, 000, 0000, …}       } * = { , 10, 1010, , …, 01, 0101, , …, 1001, , , …, 0110, , , …}  Notational shorthand:  L 0 =   L i = LL i-1  L + = LL *

 6 3.1: Regular Expressions (5)  Let L be a language over {a, b}, each string in L contains the substring bb  L = {a, b} * {bb}{a, b} *  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

 7 3.1: Regular Expressions (6)   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 is regular language (RL). Why?  {a} and {b} are RLs  {a, b} is RL  {a, b} * is RL  Then L = {a}{a, b} * {b}{a, b} * {a} is RL

 8 3.1: Regular Expressions (7)   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}

 9 3.1: Regular Expressions (8)  Let  = {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, …}

 : Regular Expressions (9)  Algebraic 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*

 : Regular Expressions (10) abc  concatenation (“followed by”) a | b | c  alternation (“or”)*  zero or more occurrences+  one or more occurrences

 : Regular Expressions (11)  All strings of 1s and 0s (0 | 1) *  All strings of 1s and 0s beginning with a 1 1 (0 | 1) *

 : Regular Expressions (12)  All strings containing two or more 0s (1|0) * 0(1|0) * 0(1|0) *  All strings containing an even number of 0s (1 * 01 * 01 * ) * | 1 *

 : Regular Expressions (13)  All strings containing an even number of 0s and even number of 1s Assume that ( 0 0 | 1 1 ) is X X* | (X* ( 0 1 | 1 0 ) X* ( 0 1 | 1 0 ) X*)* OR ( 0 0 | 1 1 ) * (( 0 1 | 1 0 )( 0 0 | 1 1 ) * ( 0 1 | 1 0 )( 0 0 | 1 1 ) * ) *  All strings of alternating 0s and 1s (  | 1 ) ( 0 1 ) * (  | 0 )

 : Regular Expressions (14)  Strings over the alphabet {0, 1} with no consecutive 0's  (1 | 01 ) * (0 |  )  1 * (01 + ) * (0 |  )  1 * (011 * ) * (0 |  )  Strings over the alphabet {a, b} with exactly three b's  a * ba * ba * ba *  Strings over the alphabet {a, b, c} containing (at least once) bc  (a|b|c) * bc(a|b|c) *

 : Regular Expressions (15)  Strings over the alphabet {a, b} in which substrings ab and ba occur an unequal number of times  (a + b + ) + | (b + a + ) +

 : Regular Expressions (16)  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

 : Regular Expressions (17)  Examples of RE:  01 *  {0, 01, 011, 0111, …..}  (01 * )(01)  {001, 0101, 01101, , …..}  (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”

 : Regular Expressions (18)  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”

 : Regular Expressions (19)  What languages do the following RE represent?  ((0 | 1)(0 | 1)) * | ((0 | 1)(0 | 1)(0 | 1)) *

 : Regular Expressions (20)  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

 :Connection Between RE & RL (1)  A language L is called regular if and only if there exists some DFA M such that L = L(M).  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 r, we can construct an NFA that accept L(r).

 :Connection Between RE & RL (2) 2. For a   in the regular expression, construct NFA a start L = {a} 1. For  in the regular expression, construct NFA  start L = {  } 0. For  in the regular expression, construct NFA start L = { } = 

 :Connection Between RE & RL (3) where i and f are new start / final states, and  -moves are introduced from i to the old start states of M s and M t as well as from all of their final states to f. 3.(a) If s and t are regular expressions, M s and M t are their NFAs. s|t has NFA: start  if  MsMs MtMt   L = {L(M s )  L(M t )}

 :Connection Between RE & RL (4) 3.(b) If s and t are regular expressions, M s, M t their NFAs. st (concatenation) has NFA: MsMs  start if MtMt  where i is the start state of M s (or new under the alternative) and f is the final state of M t (or new). Overlap maps final states of M s to start state of M t L = {L(M s )L(M t )}

 :Connection Between RE & RL (5) f MsMs  start 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 M s its NFA, s* (Kleene star) has NFA: L = {L(M s ) * }

 :Connection Between RE & RL (6)  Build an NFA-  that accepts (a|b) * ba abbaa|b a start a  b q1q1 b a b    

 :Connection Between RE & RL (7)  Build an NFA-  that accepts (a|b) * ba (a|b) * a b        

 :Connection Between RE & RL (8)  Build an NFA-  that accepts (a|b) * ba a  b  a b        

 :Connection Between RE & RL (9) r 13 r 12 r5r5 r3r3 r 11 r4r4 r9r9 r 10 r8r8 r7r7 r6r6 r0r0 r1r1 r2r2 b * c a a | ( ) b | * c (ab*c) | (a(b|c*)) Decomposition for this regular expression: What is the NFA? Let’s construct it !

 :Connection Between RE & RL (10) r3:r3: a r0:r0: b r2:r2: c b     r1:r1:r 4 : r 1 r 2 b     c r 5 : r 3 r 4 b     ac

 :Connection Between RE & RL (11) r 11 : a r7:r7: b r6:r6: c c     r 9 : r 7 | r 8   b r 10 : r 9 c     r8:r8: c     r 12 : r 11 r 10   b a

 :Connection Between RE & RL (12) r 13 : r 5 | r 12 b     ac c       b a   

 :Connection Between RE & RL (13) Let’s try a ( b | c ) * 1. a, b, & c 2. b | c 3. ( b | c ) * S0S0 S1S1 a S0S0 S1S1 b S0S0 S1S1 c S1S1 S2S2 b S3S3 S4S4 c S0S0 S5S5 S2S2 S3S3 b S4S4 S5S5 c S1S1 S6S6 S0S0 S7S7        

 :Connection Between RE & RL (14) 4. a ( b | c ) * S0S0 S1S1 a  S4S4 S5S5 b S6S6 S7S7 c S3S3 S8S8 S2S2 S9S9     S0S0 S1S1 a b | c 

 :Connection Between RE & RL (15) Let : a abb a*b + 3 patterns NFA’s : start 1 b b bb a a a

 :Connection Between RE & RL (16) NFA for : a | abb | a * b +    0 b b bb a a a start

 38 Regular Expression to NFA-  (a | ba) * a

 39 First Parsing Step concatenate (a|ba) * a

 40 Second Parsing Step concatenate *a a|ba

 41 Third Parsing Step concatenate *a | aba

 42 Fourth Parsing Step concatenate *a | a ba

 43 Identify Leaf Nodes concatenate +a | a ba

 44 Convert Leaf Nodes concatenate * | a a a b

 45 Identify Convertible Node(s) concatenate * | a a a b

 46 Convert Node concatenate * | a aab 

 47 Identify Convertible Node concatenate * | a aab 

 48 Convert Node concatenate * a a ab   

 49 Identify Convertible Node concatenate * a a ab   

 50 Convert Node concatenate a a ab      

 51 Identify Convertible Node concatenate a a ab      

 52 Convert Final Node a a ab       

 :Expression Graphs (1)  NFA 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

 :Expression Graphs (2) w ik k w ji ji w ji w ik kj  w ik k w ji ji w ji (w ii ) * w ik kj  w ii w  w * start w1w1   (w 1 ) * w 2 (w 3  w 4 (w 1 ) * w 2 ) * startw2w2 w3w3 w4w4

 :Expression Graphs (3)   Merge Edges : c b a a | b | c p ac * b b a c Replace state by Edges q0q0 q1q1 q0q0 q1q1

 :Expression Graphs (4)  Let G be the state diagram of a finite automata  Let m be the number of final states of G  Make m copies of G, each of which has one final state.  Call these graphs G 1, G 2, …, G m  For each G t  Repeat  Do the steps in the previous slide  Until the only states in G t are the start state and the single final state  Determine the RE of G t  The RE of G is obtained by joining RE’s of each G t by  or 

 :Expression Graphs (5) c 3 c 12 b G:start b c 3 c 12 b G 1 :start b c 3 c 12 b G 2 :start b

 :Expression Graphs (6) c 3 c 12 b G 1 :start b 3 cc 1 bb 1 b  b*

 :Expression Graphs (7) c 3 c 12 b G 2 :start b 3 cc 1 bb  b * ccb * RE for G  b * | b * ccb *