CMPU 240 Lnguge Theory nd Computtion Spring 2019 NFAs nd Regulr Expressions Lst clss: Introduced nondeterministic finite utomt with -trnsitions Tody: Prove n NFA- is no more powerful thn n NFA Introduce regulr expressions Fun: Assignment 1 is due Lte dedline is next clss Assignment 2 out Exm 1 in pprox. two weeks Recll NFA-, continued A lnguge L is clled regulr lnguge if there exists DFA D such tht L(D) = L. A nondeterministic finite utomton (NFA) is like DFA but cn hve missing trnsitions or multiple trnsitions defined on the sme input symbol. An NFA ccepts if ny possible series of choices leds to n ccepting stte. An NFA- my follow ny number of -trnsitions t ny point without consuming ny input.
Forml definition of NFA- A nondeterministic finite utomton with - trnsitions is five-tuple (Q, Σ, q 0, δ, F) where Q is finite set of sttes Σ is finite lphbet of symbols q 0 Q is the strt stte F Q is the set of ccepting sttes δ is the trnsition function from Q x (Σ {}) to P(Q) DFAs nd NFA-s -trnsitions re convenience, but they don t increse the power of FAs For ny NFA-, there is n equivlent DFA, i.e., one tht ccepts the sme lnguge. The construction is similr to the NFA-to-DFA construction Creting DFA from n NFA- 1. Compute the -closure for ech stte, i.e., the set of sttes rechble from tht stte following only - trnsitions. 2. The strt stte of the DFA is now the -closure of q 0, E({q 0 }) 3. Define δ for ech Σ nd ech -closed set S: If stte p S cn rech stte q on input (not!), then dd trnsition on input from S to E(q) 4. The set of finl sttes for the DFA now includes those sets tht contin t lest one ccepting stte of the NFA-. -closure exmple q 0 Find the set T = E({s}) T = {s} initil step T = {s, w} dd δ(s, ) T = {s, w, q 0 } dd δ(w, ) b b p r s w b t u v T = {s, w, q 0, p, t} dd δ(q 0, ) δ(p, ) = δ(t, ) =, so we re done: E({s}) = T = {s, w, q 0, p, t}
Exmple 1. Compute -closure for ll sttes: E({q}) = {q} E({r}) = {r, s} E({s}) = {r, s} 2. Strt stte = E({q}) = {q} 3. Compute δ for input nd set of sttes from step 1. δ({q}, 0) = E({s}) = {r, s} δ({q}, 1) = E({r}) = {r, s} δ({r, s}, 0) = E({q}) = {q} δ({r, s}, 1) = E({q}) = {q} 4. Finl sttes F D = {{r, s}} 0 q 1 r s 0 1 RESULTING DFA {q} 0, 1 0, 1 {r,s} Problem Convert this NFA- to DFA q 0 q 1 q 2 q 3 q 4 q 6 q 7 q 5 q 1 q 2 q 3 q 4 q 0 q 5 q 6 q 7 Step 1 E({q 0 }) = {q 0, q 1, q 6 } E({q 1 }) = {q 1 } E({q 2 }) = {q 2, q 3 } E({q 3 }) = {q 3 } E({q 4 }) = {q 4, q 5 } E({q 5 }) = {q 5 } E({q 6 }) = {q 6 } E({q 7 }) = {q 7, q 5 } Step 2 Strt = E({q 0 }) = {q 0, q 1, q 6 } Step 3 δ({q 0, q 1, q 6 }, ) = E({q 2, q 7 }) = {q 2, q 3, q 7, q 5 } δ({q 2, q 3, q 7, q 5 }, ) = E({q 4 }) = {q 4, q 5 } δ({q 4, q 5 }, ) = Step 4 Finl = { {q 2, q 3, q 7, q 5 }, {q 4, q 5 } } {q 0, q 1, q 6} {q 2, q 3, q 7, q 5} {q 4, q 5}
DFAs, NFAs, nd NFAs with -trnsitions ll describe sme clss of lnguges. Thus to show lnguge is regulr lnguge, you cn just build NFA- tht recognizes it, rther thn DFA. Mny times it is more convenient to build NFA- rther thn DFA, especilly if you wnt to keep trck of multiple possibilities. Regulr expressions We ve seen we cn show lnguge is regulr by constructing DFA for it constructing n NFA for it (with or without trnsitions) We cn lso show lnguge is regulr by constructing it out of other regulr lnguges using closure properties. Regulr expressions re concise nottion for describing how to ssemble lrger lnguge out of smller pieces.
A bottom-up pproch to the regulr lnguges: Strt with smll set of simple lnguges we know to be regulr Use closure properties to combine these to form more elborte lnguges Photogrph by Benjmin D. Eshm Opertors nd opernds Regulr expressions re built by combining three kinds of simple opernds: For ny symbol Σ, the regulr expression represents the lnguge {} The symbol is regulr expression representing the lnguge {} The symbol is regulr expression for the empty lnguge Opertion: Union If R 1 nd R 2 re regulr expressions, (R 1 R 2 ) is regulr expression for the union of the lnguges of R 1 nd R 2 : L((R 1 R 2 )) = L(R 1 ) L(R 2 ). These re combined using symbols for the regulr opertions union, conctention, nd Kleene str.
Opertion: Conctention When we conctente strings w nd x, we write wx or w x nd the result is ll the symbols from w followed by ll the symbols from x, e.g., w = wonder x = womn wx = wonderwomn This is like the + opertor in some progrmming lnguges, e.g., in Python, > w = 'wonder' > x = 'womn' > w+x 'wonderwomn' Opertion: Conctention, continued The conctention of two lnguges L 1 nd L 2 is the lnguge L 1 L 2 = {wx w L 1 nd x L 2 } E.g., consider the lnguges Noun = {Puppy, Rinbow, Whle, } Verb = {Hugs, Juggles, Loves, } Det = {A, The} The lnguge DetNounVerbDetNoun is {APuppyHugsTheWhle, TheRinbowJugglesTheRinbow, TheWhleLovesAPuppy, } Opertion: Conctention, continued Two views of L 1 L 2 : The set of ll strings tht cn be mde by conctenting string in L 1 with string in L 2. Opertion: Conctention, continued If R 1 nd R 2 nd regulr expressions, (R 1 R 2 ) is regulr expression for the conctention of the lnguges of R 1 nd R 2. The set of strings tht cn be split into two pieces: piece from L 1 followed by piece from L 2. Conceptully it s similr to the Crtesin product of two sets, only with strings.
Opertion: Kleene str We cn conctente lnguge with itself. Consider L = {, b} LL is the lnguge of strings formed by conctenting pirs of strings in L: {, b, b, bb} LLL is the set of strings formed by conctenting triples of strings in L: Etc. {, b, b, bb, b, bb, bb, bbb} Opertion: Kleene str, continued We cn define wht it mens to exponentite lnguge s follows: L 0 = {} Bse cse: Any string formed by conctenting zero strings together is the empty string. L n+1 = LL n Recursive cse: Conctenting n+1 strings together works by conctenting n strings, then conctenting one more. Opertion: Kleene str, continued An importnt opertion on lnguges is the Kleene closure, defined s L * = L 0 L 1 L 2 L 3 Intuitively, it gives ll possible wys of conctenting ny number of copies of strings in L together. Opertion: Kleene str, continued If R is regulr expression, (R * ) is regulr expression for the Kleene closure of the lnguge of R.
Forml definition of regulr expressions DEFINITION. R is regulr expression if R is 1 for some Σ 2 3 4 (R 1 R 2), where R 1 nd R 2 re regulr expressions 5 (R 1 R 2), where R 1 nd R 2 re regulr expressions 6 (R 1 *), where R 1 is regulr expression Bsis Recursive cses Every regulr expression rises by finite number of pplictions of these six rules. Order of opertions We cn omit prentheses (nd the conctention opertor) to mke regulr expressions more compct, but this mkes them mbiguous without defining precedence. 0 Prentheses (R) 1 Kleene str R * 2 Conctention R 1 R 2 or R 1 R 2 3 Union R 1 R 2 Empty strings, empty sets Do not confuse the regulr expressions: the lnguge contining only the empty string the lnguge contining no strings Identities: R = R R = R R is regulr expression if R is 1 for some Σ 2 3 4 (R 1 R 2 ), where R 1 nd R 2 re regulr expressions 5 (R 1 R 2 ), where R 1 nd R 2 re regulr expressions 6 (R 1 *), where R 1 is regulr expression Exmple: To prove (((b * )) ) is regulr expression over Σ = {, b}, show it cn be constructed ccording to the rules: 1 b is regulr by Rule 1 2 (b * ) is regulr by Rule 6 3 is regulr by Rule 1 4 ((b * )) is regulr by Rule 5 5 (((b * )) ) is regulr by Rule 4 pplied to expressions (4) nd (3)
Exmples L(001) = {001} L(0 10 * ) = {0, 1, 10, 100, 1000, } L( ((0(0 1)) * ) = the set of strings of 0s nd 1s, of even length, such tht every odd position hs 0 A few more exmples b * * b * (b) * Is this the sme s * b *? * b * * Is b in this? L = {x odd } = x(xx) * or (xx) * x but not x * xx * All strings of s nd bs of exctly length 3 L = {, b, b, bb, b, bb, bb, bbb} or ( b) ( b) ( b) or ( b) 3 RE for strings with n in them somewhere? ( b) * ( b) * RE for strings with t lest two s? b * b * ( b) * RE for strings with exctly two s? b * b * b * RE for strings with t lest one nd one b? Convenient shorthnds R + = RR * tht is, one or more strings from lnguge R conctented together. R k = conctention of k Rs. ( b) * ( b) * b( b) * ( b) * b( b) * ( b) * RE for strings tht end in b but do not contin? (b b) * (b b) = (b b) + RE for ll strings over {, b, c} hving no substring c? c * ( bc * ) *
Equlity of REs Two regulr expressions s nd t re equl if nd only if L(s) = L(t) Two regulr expressions cn look quite different yet describe the sme lnguge Exmple: Acknowledgments This lecture incorportes mteril from: Nncy Ide Keith Schwrz Michel Sipser Jeffrey Ullmn s = ( b) * nd t = (b * b) * *