Deterministic Finite Automata. Great Theoretical Ideas in Computer Science

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1 5-25 Gret Theoreticl Ides in Computer Science Let me show you mchine so simple tht you cn understnd it in less thn two minutes Deterministic Finite Automt Lecture 2 (Octoer 3, 28), The mchine ccepts string if the process ends in doule circle

2 Antomy of Deterministic Finite Automton sttes q, ccept sttes (F) q, q q 2 L(M) = All! strings of s nd s The Lnguge of Mchine M The mchine ccepts string if the strt stte process (q ) ends in doule circlesttes q 3 Antomy of Deterministic Finite Automton q, The lphet of finite utomton is the set where the symols come from: {,} The lnguge of finite utomton is the set of strings tht it ccepts q q 3 q 2 q q L(M) = { w w hs n even numer of s}

3 Nottion An lphet! is finite set (e.g.,! = {,}) M = (Q,!, ", q, F) where Q = {q, q, q 2, q 3 }! = {,} q $ Q is strt stte A string over! is finite-length sequence of elements of!. The set of ll strings over! is denoted y!*. For x string, x is the length of x q, F = {q, q 2 } % Q ccept sttes " : Q #! " Q trnsition function " q q q The unique string of length will e denoted y " nd will e clled the empty or null string A lnguge over! is set of strings over! q M q 3 q 2 q q 2 q 2 q 2 q 3 q 2 q 3 q q 2 A finite utomton is 5-tuple M = (Q,!, ", q, F) Q is the set of sttes! is the lphet " : Q #! # Q is the trnsition function q $ Q is the strt stte F % Q is the set of ccept sttes L(M) = the lnguge of mchine M = set of ll strings mchine M ccepts ABA The Automton Input String & Result Accept Reject Accept Accept

4 Wht mchine ccepts this lnguge? Wht is the lnguge ccepted y this mchine? L = ll strings in {,}* tht contin t lest one, L = ny string ending with Wht mchine ccepts this lnguge? L = strings with n odd numer of s nd ny numer of s Wht is the lnguge ccepted y this mchine?, L(M) = ny string with t lest two s

5 Wht mchine ccepts this lnguge? Build n utomton tht ccepts ll nd only those strings tht contin L = ny string with n nd,, q q q q Wht mchine ccepts this lnguge? L = strings with n even numer of pirs L = ll strings contining s consecutive sustring q q q q q Invrint: I m stte s exctly when s is the longest suffix of the input (so fr) forming prefix of., q

6 The Grep Prolem Input: Text T of length t, string S of length n Prolem: Does string S pper inside text T? Nïve method:, 2, 3, 4, 5,, t Cost: Roughly nt comprisons Rel-life Uses of DFAs Grep Coke Mchines Thermostts (fridge) Elevtors Trin Trck Switches Lexicl Anlyzers for Prsers Automt Solution Build mchine M tht ccepts ny string with S s consecutive sustring Feed the text to M Cost: t comprisons + time to uild M As luck would hve it, the Knuth, Morris, Prtt lgorithm uilds M quickly A lnguge is regulr if it is recognized y deterministic finite utomton L = { w w contins } is regulr L = { w w hs n even numer of s} is regulr

7 Union Theorem Given two lnguges, L nd L 2, define the union of L nd L 2 s L ' L 2 = { w w $ L or w $ L 2 } Ide: Run oth M nd M 2 t the sme time! Q = pirs of sttes, one from M nd one from M 2 = { (q, q 2 ) q $ Q nd q 2 $ Q 2 } = Q # Q 2 Theorem: The union of two regulr lnguges is lso regulr lnguge Theorem: The union of two regulr lnguges is lso regulr lnguge Theorem: The union of two regulr lnguges is lso regulr lnguge Proof Sketch: Let M = (Q,!, ", q, F ) e finite utomton for L 2 nd M 2 = (Q 2,!, " 2, q, F 2 ) e finite utomton for L 2 We wnt to construct finite utomton M = (Q,!, ", q, F) tht recognizes L = L ' L 2 q q p p

8 Automton for Union q,p q,p Theorem: The union of two regulr lnguges is lso regulr lnguge Corollry: Any finite lnguge is regulr q,p q,p Automton for Intersection q,p q,p q,p q,p The Regulr Opertions Union: A ' B = { w w $ A or w $ B } Intersection: A ( B = { w w $ A nd w $ B } Reverse: A R = { w w k w k w $ A } Negtion: A = { w w ) A } Conctention: A * B = { vw v $ A nd w $ B } Str: A* = { w w k k! nd ech w i $ A }

9 Regulr Lnguges Are Closed Under The Regulr Opertions We hve seen prt of the proof for Union. The proof for intersection is very similr. The proof for negtion is esy. Consider the lnguge L = { n n n > } i.e., unch of s followed y n equl numer of s No finite utomton ccepts this lnguge Cn you prove this? Are ll lnguges regulr? n n is not regulr. No mchine hs enough sttes to keep trck of the numer of s it might encounter

10 Tht is firly wek rgument Consider the following exmple M ccepts only the strings with n equl numer of s nd s! L = strings where the # of occurrences of the pttern is equl to the numer of occurrences of the pttern Cn t e regulr. No mchine hs enough sttes to keep trck of the numer of occurrences of Let me show you professionl strength proof tht n n is not regulr

11 Pigeonhole principle: Given n oxes nd m > n ojects, t lest one ox must contin more thn one oject Letterox principle: If the verge numer of letters per ox is x, then some ox will hve t lest x letters (similrly, some ox hs t most x) Advertisement You cn lern much more out these cretures in the FLAC course. Forml Lnguges, Automt, nd Computtion There is unique smllest utomton for ny regulr lnguge It cn e found y fst lgorithm. Theorem: L= { n n n > } is not regulr Proof (y contrdiction): Assume tht L is regulr Then there exists mchine M with k sttes tht ccepts L For ech + i + k, let S i e the stte M is in fter reding i,i,j + k such tht S i = S j, ut i - j M will do the sme thing on i i nd j i But vlid M must reject j i nd ccept i i Here s Wht You Need to Know Deterministic Finite Automt Definition Testing if they ccept string Building utomt Regulr Lnguges Definition Closed Under Union, Intersection, Negtion Using Pigeonhole Principle to show lnguge not regulr