11.1 Finite Automata. CS125 Lecture 11 Fall Motivation: TMs without a tape: maybe we can at least fully understand such a simple model?

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1 CS125 Lecture 11 Fll Finite Automt Motivtion: TMs without tpe: mybe we cn t lest fully understnd such simple model? Algorithms (e.g. string mtching) Computing with very limited memory Forml verifiction of distributed protocols, Hrdwre nd circuit design Exmple: Home Stereo P = power button (ON/OFF) S = source button (CD/Rdio/TV), only works when stereo is ON, but source remembered when stereo is OFF. Strts OFF, in CD mode. A computtionl problem: does given sequence of button presses w {P,S} leve the system with the rdio on? The Home Stereo DFA 11-1

2 Lecture Forml Definition of DFA A DFA M is 5-Tuple (Q,Σ,δ,q 0,F) Q : Finite set of sttes Σ : Alphbet δ : Trnsition function, Q x Σ Q q 0 : Strt stte, q 0 Q F : Accept (or finl) sttes, F Q If δ(p,σ) = q, then if M is in stte p nd reds symbol σ Σ then M enters stte q (while moving to next input symbol) Another Visuliztion b b b Input tpe Reding hed moves left to right, one squre t time Strt stte mrked with < Double-circled sttes re ccepting or finl Finite-stte control chnges stte depending on: current stte next symbol M ccepts string x if After strting M in the strt[initil] stte with hed on first squre, when ll of x hs been red, M winds up in finl stte.

3 Lecture Exmple Bounded Counting: A DFA tht recognizes {x : x hs n even # of s nd n odd # of b s} q 0 q 1 b b b b q 2 q 3 Trnsition function δ: b q 0 q 1 q 2 q 1 q 0 q 3 q 2 q 3 q 0 q 3 q 2 q 1. i.e. δ(q 0,) = q 1, etc. = strt stte = finl stte Q = {q 0,q 1,q 2,q 3 } Σ = {,b} F = {q 2 } Forml Definition of Computtion M = (Q,Σ,δ,q 0,F) ccepts w = w 1 w 2 w n Σ (where ech w i Σ) if there exist r 0,...,r n Q such tht 1. r 0 = q 0, 2. δ(r i,w i+1 ) = r i+1 for ech i = 0,...,n 1, nd 3. r n F. The lnguge recognized (or ccepted) by M, denoted L(M), is the set of ll strings ccepted by M.

4 Lecture Another Exmple, To Do On Your Own Pttern Recognition: A DFA tht ccepts { x : x hs b s substring}. Another Exmple, To Do On Your Own Pttern Recognition: A DFA tht ccepts { x : x hs bb s substring}. Using DFAs for Pttern Recognition Problem: given pttern w Σ of length m nd string x Σ of length n, decide whether w is substring of x. Algorithm: 1. Construct DFA M tht ccepts L w = {x Σ : w is substring of x}. Sttes re Q = {0,1,...,m}. Stte q represents: Trnsitions: δ(q,σ) = Time to construct M (nively): O(m 3 Σ ). 2. Run M on x. Time: O(n) The running time cn be improved to O(m +n), using n pproprite implicit representtion of the DFA. Widely used in prctice! (Look up the Knuth-Morris-Prtt lgorithm.)

5 Lecture Chrcterizing the Power of Finite Automt Def: A lnguge L Σ is regulr iff there is DFA M such tht L(M) = L. REG denotes the clss of regulr lnguges. The terminology regulr comes from n equivlent chrcteriztion in terms of regulr expressions (which we won t cover in lecture, but possibly will on problem set). Note tht REG TIME TM (n); it lso cn be shown tht REG CF. Unlike clsses ssocited with universl models (like TMs nd Word-RAMs), we hve firly complete understnding of the clss of regulr lnguges. In prticulr, Myhill-Nerode Theorem: A lnguge L Σ is regulr iff there re only finitely mny equivlence clsses under the following equivlence reltion L on Σ : x L y iff for ll strings z Σ, we hve xz L yz L. Moreover, the minimum number of sttes in DFA for L is exctly the number of equivlence clsses under L. (Exercises: refresh your memory on the definition of equivlence reltions nd equivlence clsses.) Proof:. Let M = (Q,Σ,δ,q 0,F) be DFA such tht L(M) = L. Note tht if x,y Σ drive M to the sme stte (strting from q 0 ), then for ll z Σ, xz nd yz drive M to the sme stte nd hence both re in L(M) = L or neither re in L(M). Thus x L y. Hence the number of equivlence clsses under L is t most Q.. Suppose L hs finitely mny equivlence clsses, where we write [x] L for the equivlence clss contining x. We construct DFA M = (Q,Σ,δ,q 0,F) s follows: Q is the set of equivlence clsses under L. q 0 = [ε] L. F = {[x] L : x L}. δ([x] L,σ) = [xσ] L. (Note tht this is well-defined: if x L y, then xσ L yσ, so the choice of the representtive x of the equivlence clss does not ffect the result.) By induction on x, it cn be shown tht running M on x leds to stte [x] L, nd hence we ccept exctly the strings in L.

6 Lecture Proving tht lnguges re nonregulr. strings tht re ll inequivlent under L. Some exmples follow: To show tht L is nonregulr, we only need to exhibit n infinite set of L = { n b n : n 0}. Exercise: prove tht ε,, 2, 3, 4,... re ll pirwise inequivlent under L. L = {w Σ : w = 2 n for some n 0}. Clim: ε,, 2, 3, 4,... re ll inequivlent under L. Suppose i L j for some i > j. Let k be ny power of 2 lrger thn i nd j. Then j k j L, so i k j L nd hence k + i j is power of 2. But 2k is the next lrger power of 2 fter k.. L = {w Σ : w = w R } (plindromes). Exercise: prove tht, 2 b, 3 b,... re pirwise inequivlent.