DFA (Deterministic Finite Automata) q a

Big pictur All lngugs Dcidl Turing mchins NP P Contxt-fr Contxt-fr grmmrs, push-down utomt Rgulr Automt, non-dtrministic utomt, rgulr xprssions

DFA (Dtrministic Finit Automt) 0 q 0 0 0 0 q

DFA (Dtrministic Finit Automt) 0 q 0 0 0 0 q Stts, this DFA hs 4 stts Trnsitions llld with lmnts of th lpht S = {0,}

DFA (Dtrministic Finit Automt) 0 q 0 0 Computtion on input w: Bgin in strt stt 0 q 0 0 q Rd input string in on-wy fshion Follow th rrows mtching input symols Whn input nds: ACCEPT if in ccpt stt REJECT if not

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 00

DFA (Dtrministic Finit Automt) lwys strt in strt stt q 0 0 0 0 q 0 Exmpl: Input string w = 00

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 00

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 00 ACCEPT cus nd in ccpt stt

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 00

DFA (Dtrministic Finit Automt) lwys strt in strt stt q 0 0 0 0 q 0 Exmpl: Input string w = 00

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 00

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 00 REJECT cus dos not nd in ccpt stt

DFA (Dtrministic Finit Automt) 0 q 0 0 0 q 0 Exmpl: Input string w = 0 ACCEPT w = 00 REJECT w = 00 ACCEPT w = 000 REJECT

Exmpl: 00 is n lmnt of L(M), 00 L(M) M := DFA (Dtrministic Finit Automt) q 0 0 M rcognizs lngug L(M) = { w : w strts with 0 nd nds with } L(M) is th lngug of strings cusing M to ccpt 0 0 0 q

Exmpl S = {0,} M := q 0 q 0 0 00 cuss M to ccpt, so 00 is in L(M) 00 L(M) 0 dos not cus M to ccpt, so 0 not in L(M), 00 L(M) 0000 L(M) 000 L(M) 0 L(M)

Exmpl M := q 0 q S = {0,} 0 0 L(M) = {w : w hs n vn numr of } Not: If thr is no, thn thr r zro, zro is n vn numr, so M should ccpt. Indd 0000000 L(M)

Exmpl S = {0,} M := 0 L(M) =?

Exmpl S = {0,} M := 0 L(M) = vry possil string ovr {0,} = {0,}*

Exmpl S = {0,} M := 0 0 q 0 0 L(M) =?

Exmpl S = {0,} M := 0 0 q 0 0 L(M) = ll strings ovr {0,} xcpt mpty string = {0,}* - { }

Exmpl S = {0,} M := 0 q 0 L(M) =? 0 0 0 0

Exmpl S = {0,} M := 0 q 0 0 0 0 0 L(M) = { w : w strts nd nds with sm symol } Mmory is ncodd in wht?

Exmpl S = {0,} M := 0 q 0 0 0 0 0 Rmmr 0 Rmmr L(M) = { w : w strts nd nds with sm symol } Mmory is ncodd in stts. DFA hv finit stts, so finit mmory

M := Convntion: q 0 0 W lrdy sw tht 0 0 L(M) = { w : w strts with 0 nd nds with } 0 q Th rrow q 0 lds to sink stt. If followd, M cn nvr ccpt

Convntion: M := q 0 0 0 0 q Don't nd to writ such rrows: If, from som stt, rd symol with no corrsponding rrow, imgin M gos into sink stt tht is not shown, nd REJECT. This mks picturs mor compct.

Anothr convntion: List multipl trnsition on sm rrow: 0,,2 Mns 2 0 This mks picturs mor compct.

Exmpl = {0,} M = 0, 0, L(M) =?

Exmpl = {0,} M = 0, 0, L(M) = 2 = {00,0,0,}

Exmpl from progrmming lngugs: Rcogniz strings rprsnting numrs: S = {0,,2,3,4,5,6,7,8,9, +, -,. } + - 0,...,9 0,...,9. 0,...,9 0,...,9 Not: 0,...,9 mns 0,,2,3,4,5,6,7,8,9: 0 trnsitions

Exmpl from progrmming lngugs: Rcogniz strings rprsnting numrs: S = {0,,2,3,4,5,6,7,8,9, +, -,. } + - 0,...,9 0,...,9. 0,...,9 Possily put sign (+, -) 0,...,9 Follow with ritrrily mny digits, ut t lst on Possily put dciml point Follow with ritrrily mny digits, possily non

Exmpl from progrmming lngugs: Rcogniz strings rprsnting numrs: S = {0,,2,3,4,5,6,7,8,9, +, -,. } + - 0,...,9 0,...,9. 0,...,9 Input w = 7 ACCEPT Input w = + REJECT Input w = -3.25 ACCEPT Input w = +2.35-. REJECT 0,...,9

Exmpl S = {0,} Wht out { w : w hs sm numr of 0 nd } Cn you dsign DFA tht rcognizs tht? It sms you nd infinit mmory W will prov ltr tht thr is no DFA tht rcognizs tht lngug!

Nxt: forml dfinition of DFA Usful to prov vrious proprtis of DFA Espcilly importnt to prov tht things CANNOT rcognizd y DFA. Usful to prctic mthmticl nottion

Stt digrm of DFA: On or mor stts Exctly on strt stt Som numr of ccpt stts Llld trnsitions xiting ch stt, for vry symol in S

Dfinition: A finit utomton (DFA) is 5-tupl (Q, S, d, q 0, F) whr Q is finit st of stts S is th input lpht d : Q X S Q is th trnsition function q 0 in Q is th strt stt F Q is th st of ccpt stts Q X S is th st of ordrd pirs (,) : Q, Exmpl {q,r,s}x{0,}={(q,0),(q,),(r,0),(r,),(s,0),(s,)}

q 0 q 0 0 Exmpl: ov DFA is 5-tupl (Q, S, d, q0, F) whr Q = { q 0, q } S = {0,} d(q0,0) =?

q 0 q 0 0 Exmpl: ov DFA is 5-tupl (Q, S, d, q0, F) whr Q = { q 0, q } S = {0,} d(q0,0) = q 0 d(q 0,) =?

q 0 q 0 0 Exmpl: ov DFA is 5-tupl (Q, S, d, q0, F) whr Q = { q 0, q } S = {0,} d(q0,0) = q 0 d(q 0,) = q d(q,0) = q d(q,) = q 0 q 0 in Q is th strt stt F =?

q 0 q 0 0 Exmpl: ov DFA is 5-tupl (Q, S, d, q0, F) whr Q = { q 0, q } S = {0,} d(q0,0) = q 0 d(q 0,) = q d(q,0) = q d(q,) = q 0 q 0 in Q is th strt stt F = { q0 } Q is th st of ccpt stts

Dfinition: A DFA (Q, S, d, q0, F) ccpts string w if w = w w 2 w k whr, i k, w i is in S (th k symols of w) Th squnc of k+ stts r 0, r,.., r k such tht: () r 0 = q 0, nd (2) r i+ = d(r i,w i+ ) 0 i < k hs r k in F (r i = stt DFA is in ftr rding i-th symol in w)

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0 w = 0 = w w 2 w 3 w = 0 w 2 = w 3 =

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0 w = 0 = w w 2 w 3 w = 0 w 2 = w 3 = W must show tht Th squnc of 3+=4 stts r 0, r, r 2, r 3 such tht: () r 0 = q 0 (2) r i+ = d(r i,w i+ ) 0 i < 3 hs r 3 in F

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0 w = 0 = w w 2 w 3 w = 0 w 2 = w 3 = r 0 = q 0 r :=?

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0 w = 0 = w w 2 w 3 w = 0 w 2 = w 3 = r 0 = q 0 r = d(r 0,w )=d(q 0,0 ) = q 0 r 2 :=?

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0 w = 0 = w w 2 w 3 w = 0 w 2 = w 3 = r 0 = q 0 r = d(r 0,w )=d(q 0,0 ) = q 0 r 2 = d(r,w 2 )=d(q 0, ) = q r 3 :=?

Exmpl q 0 q 0 0 Aov DFA (Q, S, d, q0, F) ccpts w = 0 w = 0 = w w 2 w 3 w = 0 w 2 = w 3 = r 0 = q 0 r = d(r 0,w )=d(q 0,0 ) = q 0 r 2 = d(r,w 2 )=d(q 0, ) = q r 3 = d(r 2,w 3 )=d(q, ) = q 0 r 3 = q 0 in F OK DONE!

Dfinition: For DFA M, w dnot y L(M) th st of strings ccptd y M: L(M) := { w : M ccpts w} W sy M ccpts or rcognizs th lngug L(M) Dfinition: A lngug L is rgulr if $ DFA M : L(M) = L

In th nxt lcturs w wnt to: Undrstnd powr of rgulr lngugs Dvlop ltrnt, compct nottion to spcify rgulr lngugs Exmpl: Unix commnd grp '\<c.*h\>' fil slcts ll words strting with c nd nding with h in fil

Undrstnd powr of rgulr lngugs: Suppos A, B r rgulr lngugs, wht out not A := { w : w is not in A } A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } Ar ths lngugs rgulr?

Undrstnd powr of rgulr lngugs: Suppos A, B r rgulr lngugs, wht out not A := { w : w is not in A } A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } Trminology: Ar rgulr lngugs closd undr not, U, o, *?

Thorm: If A is rgulr lngug, thn so is (not A)

Thorm: If A is rgulr lngug, thn so is (not A) Proof id:?????????? th st of ccpt stts

Thorm: If A is rgulr lngug, thn so is (not A) Proof id: Complmnt th st of ccpt stts Exmpl

Thorm: If A is rgulr lngug, thn so is (not A) Proof id: Complmnt th st of ccpt stts Exmpl: M := L(M) = q 0 q 0 0 { w : w hs vn numr of }

Thorm: If A is rgulr lngug, thn so is (not A) Proof id: Complmnt th st of ccpt stts Exmpl: M := q 0 q 0 0 M' := q 0 q 0 0 L(M) = { w : w hs vn numr of } L(M') = not L(M) = { w : w hs odd numr of }

Thorm: If A is rgulr lngug, thn so is (not A) Proof: Givn DFA M = (Q, S, d, q 0, F) such tht L(M) = A. Dfin DFA M' =?????????????????????????? This dfinition is th crtiv stp of this proof, th rst is (prhps complictd ut) mchnicl unwrpping dfinitions

Thorm: If A is rgulr lngug, thn so is (not A) Proof: Givn DFA M = (Q, S, d, q 0, F) such tht L(M) = A. Dfin DFA M' = (Q, S, d, q 0, F'), whr F' := not F. W nd to show L(M') = not L(M), tht is: for ny w,??????????????????????????

Thorm: If A is rgulr lngug, thn so is (not A) Proof: Givn DFA M = (Q, S, d, q 0, F) such tht L(M) = A. Dfin DFA M' = (Q, S, d, q 0, F'), whr F' := not F. W nd to show L(M') = not L(M), tht is: for ny w, M' ccpts w M dos not ccpt w. So lt w ny string of lngth k, nd considr th k+ stts r 0, r,.., r k from th dfinition of ccpt: () r 0 = q 0, nd (2) r i+ = d(r i,w i+ ) 0 i < k. How do w conclud?

Thorm: If A is rgulr lngug, thn so is (not A) Proof: Givn DFA M = (Q, S, d, q 0, F) such tht L(M) = A. Dfin DFA M' = (Q, S, d, q 0, F'), whr F' := not F. W nd to show L(M') = not L(M), tht is: for ny w, M' ccpts w M dos not ccpt w So lt w ny string of lngth k, nd considr th k+ stts r 0, r,.., r k from th dfinition of ccpt: () r 0 = q 0, nd (2) r i+ = d(r i,w i+ ) 0 i < k. Not tht r k in F' r k not in F, sinc F' = not F.

Wht is proof? A proof is n xplntion, writtn in English, of why somthing is tru. Evry sntnc must logiclly connctd to th prvious ons, oftn y so, hnc, sinc, tc. Your udinc is humn ing, NOT mchin.

Thorm: If A is rgulr lngug, thn so is (not A) Proof: DFA M = (Q, S, d, q 0, F) such tht L(M) = A. DFA M' = (Q, S, d, q 0, F'), whr F' := not F. L(M') = not L(M) M' ccpts w M dos not ccpt w k+ stts r 0, r,.., r k () r 0 = q 0, nd (2) r i+ = d(r i,w i+ ) 0 i < k. r k in F' r k not in F, F' = not F.

To know proof mns to know ll th pyrmid Wht is proof? Complmnt th st of ccpt stts Givn DFA M = (Q, S, d, q 0, F) such tht L(M) = A. Dfin DFA M' = (Q, S, d, q 0, F'), whr F' := not F. W nd to show L(M') = not L(M), tht is: for ny w, M' ccpts w M dos not ccpt w Considr th k+ stts r 0, r,.., r k such tht: () r 0 = q 0, nd (2) r i+ = d(r i,w i+ ) 0 i < k. Not tht r k in F' r k not in F, sinc F' = not F.

Exmpl = {0,} M = 0, 0, L(M) = 2 = {00,0,0,} Wht is DFA M' : L(M') = not 2 = ll strings xcpt thos of lngth 2?

Exmpl = {0,} M' = 0, 0, 0, 0, L(M') = not 2 = {0,}* - {00,0,0,} Do not forgt th convntion out th sink stt!

Suppos A, B r rgulr lngugs, wht out not A := { w : w is not in A } REGULAR A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i }

Thorm: If A, B r rgulr, thn so is A U B Proof id: Tk Crtsin product of stts In pir (q,q'), q trcks DFA for A, q' trcks DFA for B. Nxt w s n xmpl. In it w rvit with

Exmpl 0 M M A := B := c d 0 0 L(M L(M A ) = A =? B ) = B =?

Exmpl 0 M M A := B := c d 0 0 L(M A ) = A = L(M B ) = B = { w : w hs vn numr of } { w : w hs odd numr of 0} M AUB := How mny stts?

Exmpl 0 M M A := B := c d 0 0 L(M A ) = A = L(M B ) = B = { w : w hs vn numr of } { w : w hs odd numr of 0} M AUB :=,c,c L(M AUB ) = AUB = { w : w hs vn numr of, or odd numr of 0} 0,d,d 0

Thorm: If A, B r rgulr, thn so is A U B Proof: Givn DFA M A = (Q A,S, δ A,q A, F A ) such tht L(M) = A, DFA M B = (Q B,S, δ B,q B, F B ) such tht L(M) = B. Dfin DFA M = (Q, S, d, q 0, F), whr Q :=?

Thorm: If A, B r rgulr, thn so is A U B Proof: Givn DFA M A = (Q A,S, δ A,q A, F A ) such tht L(M) = A, DFA M B = (Q B,S, δ B,q B, F B ) such tht L(M) = B. Dfin DFA M = (Q, S, d, q 0, F), whr Q := Q A X Q B q 0 :=?

Thorm: If A, B r rgulr, thn so is A U B Proof: Givn DFA M A = (Q A,S, δ A,q A, F A ) such tht L(M) = A, DFA M B = (Q B,S, δ B,q B, F B ) such tht L(M) = B. Dfin DFA M = (Q, S, d, q 0, F), whr Q := Q A X Q B q 0 := (q A, q B ) F := {(q,q') Q : q F A or q' F B } δ( (q,q'), v) := (?,? )

Proof M ccpts w M A ccpts w or M B ccpts w Suppos tht M ccpts w of lngth k. From th dfinitions of ccpt nd M, th squnc (s 0, t 0 ) = q 0 = (q A, q B ), (s i+,t i+ )=d((s i,t i ),w i+ )=( δ A (s i,w i+ ), δ B (t i,w i+ ) 0 i< k hs (s k,t k ) F. By our dfinition of F, wht cn w sy out (s k,t k )?

Proof M ccpts w M A ccpts w or M B ccpts w W/out loss of gnrlity, ssum M A ccpts w, w =k. From th dfinition of M A ccpts w, th squnc r 0 := q A, r i+ := δ A (r i,w i+ ) 0 i < k, hs r k in?

Proof M ccpts w M A ccpts w or M B ccpts w W/out loss of gnrlity, ssum M A ccpts w, w =k. From th dfinition of M A ccpts w, th squnc r 0 := q A, r i+ := δ A (r i,w i+ ) 0 i < k, hs r k in F A. Dfin th squnc of k+ stts t 0 := q B, t i+ := δ B (t i,w i+ ) 0 i < k. M ccpts w cus th squnc?????????? (rcll stts in M r pirs)

Suppos A, B r rgulr lngugs, wht out not A := { w : w is not in A } REGULAR A U B := { w : w in A or w in B } REGULAR A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } Othr two r mor complictd! Pln: w introduc NFA prov tht NFA r quivlnt to DFA rprov A U B, prov A o B, A* rgulr, using NFA

Big pictur All lngugs Dcidl Turing mchins NP P Contxt-fr Contxt-fr grmmrs, push-down utomt Rgulr Automt, non-dtrministic utomt, rgulr xprssions

Non dtrministic finit utomt (NFA) DFA: givn stt nd input symol, uniqu choic for nxt stt, dtrministic: Nxt w llow multipl choics, non-dtrministic W lso llow -trnsitions: cn follow without rding nything

Exmpl of NFA q 0 q q, 2 Intuition of how it computs: Accpt string w if thr is wy to follow trnsitions tht nds in ccpt stt Trnsitions llld with symol in S = {,} must mtchd with input trnsitions cn followd without mtching

Exmpl of NFA q 0 q q, 2 Exmpl: Accpt (first follow -trnsition ) Accpt

ANOTHER Exmpl of NFA q 0 q q 3, q 2 Exmpl: Accpt (two ccpting pths, on Rjct uss th -trnsition) (two possil pths, ut nithr hs finl stt = q )

Dfinition: A non-dtrministic finit utomton (NFA) is 5-tupl (Q, S, d, q 0, F) whr Q is finit st of stts S is th input lpht d : Q X (S U {} ) Powrst(Q) q 0 in Q is th strt stt F Q is th st of ccpt stts Rcll: Powrst(Q) = st of ll susts of Q Exmpl: Powrst({,2}) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) = {q 0, q } d(q 0,) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) = {q 0, q } d(q 0,) = d(q,0) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) = {q 0, q } d(q 0,) = d(q,0) = d(q,) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) = {q 0, q } d(q 0,) = d(q,0) = d(q,) = d(q,) =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) = {q 0, q } d(q 0,) = d(q,0) = d(q,) = d(q,) = {q 0 } q 0 in Q is th strt stt F =?

0, q 0 q Exmpl: ov NFA is 5-tupl (Q, S, d, q0, F) Q = { q 0, q } S = {0,} d(q0,0) = {q 0 } d(q 0,) = {q 0, q } d(q 0,) = d(q,0) = d(q,) = d(q,) = {q 0 } q 0 in Q is th strt stt F = { q } Q is th st of ccpt stts

Dfinition: A NFA (Q, S, d, q0, F) ccpts string w if $ intgr k, k strings w = w w 2 w k (th symols of w, or ) w, w 2,, w k such tht whr i k, w i S U {} $ squnc of k+ stts r0, r,.., r k in Q such tht: r 0 = q 0 r i+ d(r i,w i+ ) 0 i < k r k is in F Diffrncs with DFA r in grn

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 =?

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 = q 0, r =?

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 = q 0, r = q, r 2 =? Trnsitions: r d(r 0,) = {q }

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 = q 0, r = q, r 2 = q 2, r 3 =? Trnsitions: r d(r 0,) = {q } r 2 d(r,) = {q,q 2 }

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 = q 0, r = q, r 2 = q 2, r 3 = q 0, r 4 =? Trnsitions: r d(r 0,) = {q } r 2 d(r,) = {q,q 2 } r 3 d(r 2,) = {q 0 }

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 = q 0, r = q, r 2 = q 2, r 3 = q 0, r 4 = q 2, r 5 =? Trnsitions: r d(r 0,) = {q } r 2 d(r,) = {q,q 2 } r 3 d(r 2,) = {q 0 } r 4 d(r 3,) = {q 2 }

Bck to first xmpl NFA: q 0 q q, 2 Accpts w = w =, w 2 =, w 3 =, w 4 =, w 5 = Accpting squnc of 5+ = 6 stts: r 0 = q 0, r = q, r 2 = q 2, r 3 = q 0, r 4 = q 2, r 5 = q 0 Trnsitions: r d(r 0,) = {q } r 2 d(r,) = {q,q 2 } r 3 d(r 2,) = {q 0 } r 4 d(r 3,) = {q 2 } r 5 d(r 4,) = {q 0 }

NFA r t lst s powrful s DFA, cus DFA r spcil cs of NFA Ar NFA mor powrful thn DFA? Surprisingly, thy r not: Thorm: For vry NFA N thr is DFA M : L(M) = L(N)

Thorm: For vry NFA N thr is DFA M : L(M) = L(N) Construction without trnsitions Givn NFA N (Q, S, d, q, F) Construct DFA M (Q', S, d', q', F') whr: Q' := Powrst(Q) q' = {q} F' = { S : S Q' nd S contins n lmnt of F} d'(s, ) := U s S d(s,) = { t : t d (s,) for som s S }

It rmins to dl with trnsitions Dfinition: Lt S st of stts. E(S) := { q : q cn rchd from som stt s in S trvling long 0 or mor trnsitions } W think of following trnsitions t ginning, or right ftr rding n input symol in S

Thorm: For vry NFA N thr is DFA M : L(M) = L(N) Construction including trnsitions Givn NFA N (Q, S, d, q, F) Construct DFA M (Q', S, d', q', F') whr: Q' := Powrst(Q) q' = E({q}) F' = { S : S Q' nd S contins n lmnt of F} d'(s, ) := E( U s S d(s,) ) = { t : t E( d (s,) ) for som s S }

Exmpl: NFA DFA convrsion NFA DFA 2 3, {,3} {3} {2} {2,3} {,2,3} Q DFA = Powrst(Q NFA ) = Powrst({,2,3}) = {,{},{2},{3},{,2}...} {} {,2}

Exmpl: NFA DFA convrsion NFA DFA 2 3, q DFA = E({q NFA }) = E({}) = {,3} {,3} {3} {2} {2,3} {,2,3} {} {,2}

Exmpl: NFA DFA convrsion NFA DFA 2 3, {,3} {3} {2} {2,3} {,2,3} F DFA = {S : S contins n lmnt of F NFA } {} {,2}

Exmpl: NFA DFA convrsion NFA DFA 2 3, {,3} {3} {2} {2,3} {,2,3} d DFA ({}, ) = E(d NFA (, )) = E( ) = {} {,2}

Exmpl: NFA DFA convrsion NFA DFA 2 3, {,3} {3} {2} {2,3} {,2,3} d DFA ({}, ) = E(d NFA (, )) = E({2}) = {2} {} {,2}

Exmpl: NFA DFA convrsion NFA DFA 2 3, {,3} {3} {2} {2,3} {,2,3} d DFA ({2}, ) = E(d NFA (2, )) = E({2,3}) = {2,3} {} {,2}

Exmpl: NFA DFA convrsion NFA DFA {,3} {3} 2 3, {2} {2,3} {,2,3} d DFA ({2}, ) = E(d NFA (2, )) = E({3}) = {3} {} {,2}

Exmpl: NFA DFA convrsion NFA DFA {,3} {3} 2 3, {2} {2,3} {,2,3} d DFA ({3}, ) = E(d NFA (3, )) = E({}) = {,3} {} {,2}

Exmpl: NFA DFA convrsion NFA DFA {,3} {3} 2 3, {2} {2,3} {,2,3} d DFA ({3}, ) = E(d NFA (3, )) = E( ) = {} {,2}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({2,3}, ) = E(d NFA (2,) U d NFA (3,)) = E({2,3} U {}) = {,2,3} {,3} {3} {2} {2,3} {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({2,3}, ) = E(d NFA (2,) U d NFA (3,)) = E({3} U ) = {3} {,3} {3} {2} {2,3} {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({,3}, ) = E(d NFA (,) U d NFA (3,)) = E( U {}) = {,3} {,3} {3} {2} {2,3} {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({,3}, ) = E(d NFA (,) U d NFA (3,)) = E({2} U ) = {2} {,3} {3} {2} {2,3} {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({,2}, ) = E(d NFA (,) U d NFA (2,)) = E( U {2,3}) = {2,3} {,3} {3} {2} {2,3} {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({,2}, ) = E(d NFA (,) U d NFA (2,)) = E({2} U {3}) = {2,3} {,3} {3} {2} {2,3}, {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({,2,3}, ) =E(d NFA (,) U d NFA (2,) U d NFA (3,)) =E( U {2,3} U {}) = {,2,3} {,3} {3} {2} {2,3}, {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA 2 3, d DFA ({,2,3}, ) =E(d NFA (,) U d NFA (2,) U d NFA (3,)) =E({2} U {3} U ) = {2,3} {,3} {3} {2} {2,3}, {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA, 2 3, d DFA (, ) = d DFA (, ) = {,3} {3} {2} {2,3}, {} {,2} {,2,3}

Exmpl: NFA DFA convrsion NFA DFA, 2 3, {,3} {3} {2} {2,3} {,2,3} W cn dlt th unrchl stts.

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 DFA {3} {} {,2,3} Q DFA = Powrst(Q NFA ) = Powrst({,2,3}) = {,{},{2},{3},{,2}...} {2,3} {2} {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 DFA {3} {} {,2,3} q DFA = E({q NFA }) = E({}) {2,3} {2} {,3} {,2} = {}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 DFA {3} {} {,2,3} F DFA = {S : S contins n lmnt of F NFA } {2,3} {2} {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 DFA {3} {} {,2,3} d DFA ({}, ) = E(d NFA (, )) = E( ) = {2,3} {2} {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 DFA {3} {} {,2,3} d DFA ({}, ) = E(d NFA (, )) {2,3} {2} {,3} {,2} = E({2,3}) = {,2,3}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 DFA {3} {} {,2,3} d DFA ({2}, ) = E(d NFA (2, )) {2,3} {2} {,3} {,2} = E({3}) = {,3}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({2}, ) NFA 3 = E(d NFA (2, )) = E( ) = {3} {} {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({3}, ) NFA 3 = E(d NFA (3, )) = E( ) = {3} {} {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({3}, ) NFA 3 = E(d NFA (3, )) = E( ) = {3} {}, {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({,2}, ) NFA = E(d NFA (,) U d NFA (2,)) = E( U {3}) = {,3} 3 {3} {}, {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({,2}, ) NFA = E(d NFA (,) U d NFA (2,)) = E({2,3} U ) = {,2,3} 3 {3} {}, {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({,3}, ) NFA = E(d NFA (,) U d NFA (3,)) = E( U ) = 3 {3} {}, {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({,3}, ) NFA = E(d NFA (,) U d NFA (3,)) = E({2,3} U ) = {,2,3} 3 {3} {}, {,2,3} {2} DFA {,3} {2,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({2,3}, ) NFA = E(d NFA (2,) U d NFA (3,)) = E({3} U ) = {,3} 3 {2,3} {3} {}, {,2,3} {2} DFA {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 d DFA ({2,3}, ) NFA = E(d NFA (2,) U d NFA (3,)) = E( U ) = 3 {2,3} {3} {}, {,2,3} {2} DFA {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 d DFA ({,2,3}, ) =E(d NFA (,) U d NFA (2,) U d NFA (3,)) =E( U {3} U ) = {,3} {2,3} {3} {}, {,2,3} {2} DFA {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 d DFA ({,2,3}, ) =E(d NFA (,) U d NFA (2,) U d NFA (3,)) =E({2,3} U U ) = {,2,3} {2,3} {3} {}, {,2,3} {2} DFA {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 d DFA (, ) = d DFA (, ) = {2,3} {3} {}, {,2,3} {2} DFA, {,3} {,2}

ANOTHER Exmpl: NFA DFA convrsion 2 NFA 3 W cn dlt th unrchl stts. DFA {}, {,2,3} {,3}

Summry: NFA nd DFA rcogniz th sm lngugs W now rturn to th qustion: Suppos A, B r rgulr lngugs, wht out not A := { w : w is not in A } REGULAR A U B := { w : w in A or w in B } REGULAR A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i }

Thorm: If A, B r rgulr lngugs, thn so is A U B := { w : w in A or w in B } Proof id: Givn DFA M A : L(M A ) = A, DFA M B : L(M B ) = B, Construct NFA N : L(N) = A U B N M A M B U =

Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q :=? M A N M U B =

M A Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q := {q} U Q A U Q B, F :=? N M U B =

Construction: M A Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: M U B = Q := {q} U Q A U Q B, F := F A U F B d(r,x) := { da (r,x) } if r in Q A nd x N d(r,x) :=? if r in QB nd x

Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q := {q} U Q A U Q B, F := F A U F B d(r,x) := { da (r,x) } if r in Q A nd x d(r,x) := { db (r,x) } if r in Q B nd x d(q,) :=? M A M U B = N

Exmpl Is L = {w in {0,}* : w is divisil y 3 OR w strts with } rgulr?

Exmpl Is L = {w in {0,}* : w is divisil y 3 OR w strts with } rgulr? OR is lik U, so try to writ L = L U L 2 whr L, L 2 r rgulr

Exmpl Is L = {w in {0,}* : w is divisil y 3 OR w strts with } rgulr? OR is lik U, so try to writ L = L U L 2 whr L, L 2 r rgulr L = {w : w is div. y 3} L 2 = {w : w strts with }

Exmpl Is L = {w in {0,}* : w is divisil y 3 OR w strts with } rgulr? OR is lik U, so try to writ L = L U L 2 whr L, L 2 r rgulr L = {w : w is div. y 3} L 2 = {w : w strts with } M = 0, 0, 0, L(M ) = L

Exmpl Is L = {w in {0,}* : w is divisil y 3 OR w strts with } rgulr? OR is lik U, so try to writ L = L U L 2 whr L, L 2 r rgulr L = {w : w is div. y 3} L 2 = {w : w strts with } M = 0, M 2 = 0, 0, 0, L(M ) = L L(M 2 ) = L 2

Exmpl Is L = {w in {0,}* : w is divisil y 3 OR w strts with } rgulr? OR is lik U, so try to writ L = L U L 2 whr L, L 2 r rgulr L = {w : w is div. y 3} L 2 = {w : w strts with } M = 0, L(M) = L(M 0, 0, ) U L(M 2 ) = L U L 2 0, = L L is rgulr.

W now rturn to th qustion: Suppos A, B r rgulr lngugs, thn not A := { w : w is not in A } REGULAR A U B := { w : w in A or w in B } REGULAR A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i }

Thorm: If A, B r rgulr lngugs, thn so is A o B := { w : w = xy for som x in A nd y in B }. Proof id: Givn DFAs M A, M B for A, B M A construct NFA N : L(N) = A o B. M ob = N

M A o M B = N Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q :=?

M A o M B = N Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q := Q A U Q B, q :=?

M A o M B = N Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q := Q A U Q B, q := q A, F :=?

M A o M B = N Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q := Q A U Q B, q := q A, F := F B d(r,x) :=? if r in QA nd x

M A o M B = N Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, DFA M B = (Q B, S, d B, q B, F B ) : L(M B ) = B, Construct NFA N = (Q, S, d, q, F) whr: Q := Q A U Q B, q := q A, F := F B d(r,x) := { da (r,x) } if r in Q A nd x d(r,) :=? if r in F A

Exmpl Is L = {w in {0,}* : w contins ftr 0} rgulr? Not: L = {0, 00000, 00, }

Exmpl Is L = {w in {0,}* : w contins ftr 0} rgulr? Lt L 0 = {w : w contins 0} L = {w : w contins }. Thn L = L 0 o L.

Exmpl Is L = {w in {0,}* : w contins ftr 0} rgulr? Lt L 0 = {w : w contins 0} L = {w : w contins }. Thn L = L 0 o L. M 0 = 0, 0 L(M 0 ) = L 0

Exmpl Is L = {w in {0,}* : w contins ftr 0} rgulr? Lt L 0 = {w : w contins 0} L = {w : w contins }. Thn L = L 0 o L. M 0 = 0, M = 0 0, 0 L(M 0 ) = L 0 L(M ) = L

Exmpl Is L = {w in {0,}* : w contins ftr 0} rgulr? Lt L 0 = {w : w contins 0} L = {w : w contins }. Thn L = L 0 o L. M = 0, 0 0, 0 L(M) = L(M 0 ) o L(M ) = L 0 o L = L L is rgulr.

W now rturn to th qustion: Suppos A, B r rgulr lngugs, thn not A := { w : w is not in A } REGULAR A U B := { w : w in A or w in B } REGULAR A o B := { w w 2 : w A nd w2 B } REGULAR A* := { w w 2 w k : k 0, w i in A for vry i }

Thorm: If A is rgulr lngug, thn so is A* := { w : w = w...w k, w i in A for i=,...,k } Proof id: Givn DFA M A : L(M A ) = A, Construct NFA N : L(N) = A* N M A * =

M A N * = Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, Construct NFA N = (Q, S, d, q, F) whr: Q :=?

M A N * = Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, Construct NFA N = (Q, S, d, q, F) whr: Q := {q} U Q A, F :=?

M A N * = Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, Construct NFA N = (Q, S, d, q, F) whr: Q := {q} U Q A, F := {q} U F A d(r,x) :=? if r in QA nd x

M A N * = Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, Construct NFA N = (Q, S, d, q, F) whr: Q := {q} U Q A, F := {q} U F A d(r,x) := { da (r,x) } if r in Q A nd x d(r,) :=? if r in {q} U FA

M A N * = Construction: Givn DFA MA = (Q A, S, d A, q A, F A ) : L(M A ) = A, Construct NFA N = (Q, S, d, q, F) whr: Q := {q} U Q A, F := {q} U F A d(r,x) := { da (r,x) } if r in Q A nd x d(r,) := { qa } if r in {q} U F A W hv L(N) = A*

Exmpl Is L = {w in {0,}* : w hs vn lngth} rgulr?

Exmpl Is L = {w in {0,}* : w hs vn lngth} rgulr? Lt L 0 = {w : w hs lngth = 2}. Thn L = L 0 *.

Exmpl Is L = {w in {0,}* : w hs vn lngth} rgulr? Lt L 0 = {w : w hs lngth = 2}. Thn L = L 0 *. M 0 = 0, 0, L(M 0 ) = L 0

Exmpl Is L = {w in {0,}* : w hs vn lngth} rgulr? Lt L 0 = {w : w hs lngth = 2}. Thn L = L 0 *. M = 0, 0, L(M) = L(M 0 )* = L 0 * = L L is rgulr.

W now rturn to th qustion: Suppos A, B r rgulr lngugs, thn not A := { w : w is not in A } A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } r ll rgulr!

W now rturn to th qustion: Suppos A, B r rgulr lngugs, thn not A := { w : w is not in A } A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } Wht out A B := { w : w in A nd w in B }?

W now rturn to th qustion: Suppos A, B r rgulr lngugs, thn not A := { w : w is not in A } A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } D Morgn's lws: A B = not ( (not A) U (not B) ) By ov, (not A) is rgulr, (not B) is rgulr, (not A) U (not B) is rgulr, not ( (not A) U (not B) ) = A B rgulr

W now rturn to th qustion: Suppos A, B r rgulr lngugs, thn not A := { w : w is not in A } A U B := { w : w in A or w in B } A o B := { w w 2 : w in A nd w 2 in B } A* := { w w 2 w k : k 0, w i in A for vry i } A B := { w : w in A nd w in B } r ll rgulr

Big pictur All lngugs Dcidl Turing mchins NP P Contxt-fr Contxt-fr grmmrs, push-down utomt Rgulr Automt, non-dtrministic utomt, rgulr xprssions

How to spcify rgulr lngug? Writ pictur complictd Writ down forml dfinition complictd d(q 0,0) = q 0, Us symols from S nd oprtions *, o, U good ({0} * U {}) o {00}

Rgulr xprssions: nything you cn writ with, ε, symols from S, nd oprtions *, o, U Convntions: Writ instd of {} Writ AB for A o B Writ for U So if = {,} thn = U Oprtion * hs prcdnc ovr o, nd o ovr U so U 0* mns U(0()*) Exmpl: 0, 0*, S*, S*00S*, (SS)*, 0 U 0

Dfinition Rgulr xprssions RE ovr r: Ø ε if in S R R' if R, R' r RE R U R' if R, R' r RE R* if R is RE

Dfinition Th lngug dscrid y RE: L(Ø) = Ø L( ε ) = { ε } L() = {} if in L(R R') = L(R) o L(R') L(R U R') = L(R) U L(R') L(R*) = L(R)*

Exmpl = {, } RE Lngug U? * ( U )* ** * * * * ( )* (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * ( U )* ** * * * * ( )* (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ** * * * * ( )* (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ** ll strings * * * * ( )* (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ll strings ** {w : w hs xctly on } * * * * ( )* (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ll strings ** {w : w hs xctly on } * * {w : w hs t lst on } * * ( )* (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ll strings ** {w : w hs xctly on } * * {w : w hs t lst on } * * ( )* {w : w contins th string } (***)* ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ll strings ** {w : w hs xctly on } * * {w : w hs t lst on } * * ( )* (***)* {w : w contins th string } {w : w hs vn lngth} ** Ø

Exmpl = {, } RE Lngug U {, } * {ε,,, } = { w : w hs only } ( U )* ll strings ** {w : w hs xctly on } * * {w : w hs t lst on } * * {w : w contins th string } ( )* {w : w hs vn lngth} (***)* {w : w contins vn numr of } ** Ø

Thorm: For vry RE R thr is NFA M: L(M) = L(R)

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M :=?

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M := R = M :=?

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M := R = M := R = M :=?

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M := R = M := R = M := R = R U R'?

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M := R = M := R = M := R = R U R' us construction for A U B sn rlir R = R o R'?

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M := R = M := R = M := R = R U R' us construction for A U B sn rlir R = R o R' us construction for A o B sn rlir R = R*?

Thorm: For vry RE R thr is NFA M: L(M) = L(R) Construction: R = M := R = M := R = M := R = R U R' us construction for A U B sn rlir R = R o R' us construction for A o B sn rlir R = R* us construction for A* sn rlir

Exmpl: RE NFA RE = ( U )*

Exmpl: RE NFA RE = ( U )* M = L(M )=L()

Exmpl: RE NFA RE = ( U )* M = M = L(M )=L() L(M )=L()

Exmpl: RE NFA RE = ( U )* M = L(M )=L()

Exmpl: RE NFA RE = ( U )* M = M = L(M )=L() L(M )=L()

Exmpl: RE NFA RE = ( U )* M U = L(M U )=L( U )

Exmpl: RE NFA RE = ( U )* M ( U )* = L(M ( U )* )=L(( U )*)=L(RE)

ANOTHER Exmpl: RE NFA RE =( U )*

ANOTHER Exmpl: RE NFA RE =( U )* M = L(M )=L()

ANOTHER Exmpl: RE NFA RE =( U )* M = M = L(M )=L() L(M )=L()

ANOTHER Exmpl: RE NFA RE =( U )* M U = L(M U )=L( U )

ANOTHER Exmpl: RE NFA RE =( U )* M U = M = L(M )=L() L(M U )=L( U )

ANOTHER Exmpl: RE NFA RE =( U )* M ( U ) = L(M ( U ) )=L(( U ))

ANOTHER Exmpl: RE NFA RE =( U )* M ( U ) = M = L(M )=L() L(M ( U ) )=L(( U ))

ANOTHER Exmpl: RE NFA RE =( U )* M ( U ) = M * = L(M * )=L(*) L(M ( U ) )=L(( U ))

ANOTHER Exmpl: RE NFA RE =( U )* M ( U )* = L(M ( U )* )=L(( U )*)=L(RE)

Rcp: Hr mns cn convrtd to W hv sn: RE NFA DFA Nxt w s: DFA RE In two stps: DFA Gnrlizd NFA RE

Gnrlizd NFA (GNFA) U * q 0 ** q Nondtrministic Trnsitions llld y RE Rd locks of input symols t tim

Gnrlizd NFA (GNFA) U * q 0 ** q Convntion: Uniqu finl stt Exctly on trnsition twn ch pir of stts xcpt nothing going into strt stt nothing going out of finl stt If rrow not shown in pictur, ll =

Dfinition: A gnrlizd finit utomton (GNFA) is 5-tupl (Q, S, d, q 0, q ) whr Q is finit st of stts S is th input lpht d : (Q - {q }) X (Q {q 0 }) Rgulr Exprssions q 0 in Q is th strt stt q in Q is th ccpt stt

Dfinition: GNFA (Q, S, d, q 0, q ) ccpts string w if intgr k, k strings w, w 2,, w k S* such tht w = w w 2 w k (divid w in k strings) $ squnc of k+ stts r0, r,.., r k in Q such tht: r 0 = q 0 w i+ L(d(r i,r i+ )) 0 i < k r k = q Diffrncs with NFA r in grn

Exmpl * * q 0 q q Accpts w = w =?

Exmpl * * q 0 q q Accpts w = w = w 2 =?

Exmpl * * q 0 q q Accpts w = w = w 2 = w 3 = r 0 =q 0 r =?

Exmpl * * q 0 q q Accpts w = w = w 2 = w 3 = r 0 =q 0 r =q r 2 =? w = L(d(r 0,r )) = L(d(q 0,q )) = L(*)

Exmpl * * q 0 q q Accpts w = w = w 2 = w 3 = r 0 =q 0 r =q r 2 =q r 3 =? w = L(d(r 0,r )) = L(d(q 0,q )) = L(*) w 2 = L(d(r,r 2 )) = L(d(q,q )) = L(*)

Exmpl * * q 0 q q Accpts w = w = w 2 = w 3 = r 0 =q 0 r =q r 2 =q r 3 = q w = L(d(r 0,r )) = L(d(q 0,q )) = L(*) w 2 = L(d(r,r 2 )) = L(d(q,q )) = L(*) w 3 = L(d(r 2,r 3 )) = L(d(q,q )) = L()

Thorm: DFA M GNFA N : L(N) = L(M) Construction: To nsur uniqu trnsition twn ch pir: U 0 0 To nsur uniqu finl stt, no trnsitions ingoing strt stt, no trnsitions outgoing finl stt:

Rpt until 2 stts rmin Thorm: GNFA N RE R : L(R) = L(N) Construction: If N hs 2 stts, thn N = thus R := S q 0 S q If N hs > 2 stts, limint som stt q r q 0, q : for vry ordrd pir q i, q j (possily qul) tht r connctd through q r R 2 R R R 2 *R 3 U R q R 4 i q 3 r q j q i q j R 4

Exmpl: DFA GNFA RE DFA q,c q 2

Exmpl: DFA GNFA RE GNFA q 0 q U c q 2 q

Exmpl: DFA GNFA RE q 0 q U c q 2 q Elimint q : r-drw GNFA with ll othr stts q 0 q 2 q

Exmpl: DFA GNFA RE q 0 q U c q 2 q Elimint q : find pth through q q 0 q 2 q

Exmpl: DFA GNFA RE Ø q 0 q U c q 2 q Elimint q : dd dg to nw GNFA Don't forgt: no rrow mns ll Ø q 0 * ( U c) U Ø q 2 q

Exmpl: DFA GNFA RE q 0 q U c q 2 q Elimint q : simplify RE on nw dg q 0 * ( U c) q 2 q

Exmpl: DFA GNFA RE q 0 q U c q 2 q Elimint q : if no mor pths through q, strt ovr q 0 * ( U c) q 2 q

Exmpl: DFA GNFA RE q 0 * ( U c) q 2 q Elimint q 2 : r-drw GNFA with ll othr stts q 0 q

Exmpl: DFA GNFA RE q 0 * ( U c) q 2 q Elimint q 2 : find pth through q 2 q 0 q

Exmpl: DFA GNFA RE q 0 * ( U c) q 2 q Elimint q 2 : dd dg to nw GNFA q 0 * ( U c) * U Ø q

Exmpl: DFA GNFA RE q 0 * ( U c) q 2 q Elimint q 2 : simplify RE on nw dg q 0 * ( U c) * q

Exmpl: DFA GNFA RE q 0 * ( U c) q 2 q Elimint q 2 : if no mor pths through q 2, strt ovr q 0 * ( U c) * q

Exmpl: DFA GNFA RE q 0 * ( U c) * q Only two stts rmin: RE = * ( U c) *

ANOTHER Exmpl: DFA GNFA RE DFA q c c q 2 q 3

ANOTHER Exmpl: DFA GNFA RE GNFA q 0 q c c q 2 q 3 q

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q r-drw GNFA with ll othr stts q 3 q q 0 q 2

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q find pth through q q 3 q q 0 q 2

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q dd dg to nw GNFA * U Ø q 3 q q 0 q 2

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q find nothr pth through q * U Ø q 3 q q 0 q 2

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q dd dg to nw GNFA * U Ø q 3 q q 0 *c U Ø q 2

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q find nothr pth through q * U Ø q 3 q q 0 *c U Ø q 2

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : dd dg to nw GNFA q c q 0 c q 2 q 3 * U Ø q q 3 don't forgt currnt q 2 q 3 dg! This tim is not Ø! q *c U Ø q 2 c* U

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q find nothr pth through q * U Ø q 3 q q 0 *c U Ø q 2 c* U

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : dd dg to q c c q 2 q 3 q don't forgt currnt q 2 q 2 dg! nw GNFA * U Ø q 3 q q 0 *c U Ø q 2 c* U c*c U

ANOTHER Exmpl: DFA GNFA RE q 0 Elimint q : q c c q 2 q 3 q whn no mor pths through q, strt ovr (nd simplify * q 3 q REs) q 0 *c c* U q 2 c*c U

ANOTHER Exmpl: DFA GNFA RE * q 3 q q 0 *c q 2 c* U Elimint q 2 : c*c U r-drw GNFA with ll othr stts q 0 * q 3 q

ANOTHER Exmpl: DFA GNFA RE * q 3 q q 0 *c q 2 c* U Elimint q 2 : c*c U find pth through q 2 q 0 * q 3 q

ANOTHER Exmpl: DFA GNFA RE * q 3 q q 0 *c q 2 c* U Elimint q 2 : c*c U dd dg to nw GNFA q 0 *c(c*c U )*(c* U ) U * q 3 q

ANOTHER Exmpl: DFA GNFA RE * q 3 q q 0 Elimint q 2 : *c whn no mor pths through q 2, strt ovr q 2 c* U c*c U q 0 *c(c*c U )*(c* U ) U * q 3 q

ANOTHER Exmpl: DFA GNFA RE q 0 *c(c*c U )*(c* U ) U * q 3 q Elimint q 3 : r-drw GNFA with ll othr stts q 0 q

ANOTHER Exmpl: DFA GNFA RE q 0 *c(c*c U )*(c* U ) U * q 3 q Elimint q 3 : find pth through q 3 Ø Ø don't forgt: no rrow mns Ø q 0 q

ANOTHER Exmpl: DFA GNFA RE q 0 *c(c*c U )*(c* U ) U * q 3 q Elimint q 3 : dd dg to nw GNFA Ø Ø (*c(c*c U )*(c* U ) U *) Ø* ε U Ø q 0 q

ANOTHER Exmpl: DFA GNFA RE q 0 *c(c*c U )*(c* U ) U * q 3 q Elimint q 3 : whn no mor pths through q 3, strt ovr (nd simplify REs) don't forgt: Ø*= ε q 0 *c(c*c U )*(c* U ) U * q

ANOTHER Exmpl: DFA GNFA RE q 0 *c(c*c U )*(c* U ) U * q Only two stts rmin: RE = *c(c*c U )*(c* U ) U *

Rcp: Hr mns cn convrtd to RE DFA NFA Any of th thr rcogniz xctly th rgulr lngugs (initilly dfind using DFA)

Ths convrsions r usd vry tim you ntr n RE, for xmpl for pttrn mtching using grp Th RE is convrtd to n NFA Thn th NFA is convrtd to DFA Th DFA rprsnttion is usd to pttrn-mtch Optimiztions hv n dvisd, ut this is still th gnrl pproch.

Wht lngug is NOT rgulr? Is { 0 n n : n 0 } = {ε, 0, 00, 000, } rgulr?

Pumping lmm: L rgulr lngug p 0 w L, w p x,y,z : w= xyz, y > 0, xy p i 0 : xy i z L Rcll y 0 =, y = y, y 2 = yy, y 3 = yyy,...

Pumping lmm: L rgulr lngug p 0 W will not s th proof. But hr's th id: p := Q for DFA rcognizing L w L, w p If w L, w p, thn during computtion 2 stts must th sm q Q y = portion of w tht rings ck to q cn rpt y nd still ccpt string x,y,z : w= xyz, y > 0, xy p i 0 : xy i z L

Pumping lmm: L rgulr lngug p 0 A w L, w p x,y,z : w= xyz, y > 0, xy p i 0 : xy i z L Usful to prov L NOT rgulr. Us contrpositiv: L rgulr lngug A sm s (not A) L not rgulr

Pumping lmm (contrpositiv) p 0 not A w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L L not rgulr To prov L not rgulr it is nough to prov not A Not A is th stuff in th ox.

Proving somthing lik l l l l l mns winning gm Thory is ll out winning gms!

Exmpl NAME THE BIGGEST NUMBER GAME Two plyrs: You, Advrsry. Ruls: First Advrsry sys numr. Thn You sy numr. You win if your numr is iggr. Cn you win this gm?

Exmpl NAME THE BIGGEST NUMBER GAME Two plyrs: You, Advrsry. Ruls: First Advrsry sys numr. Thn You sy numr. You win if your numr is iggr. You hv winning strtgy: if dvrsry sys x, you sy x+

Exmpl NAME THE BIGGEST NUMBER GAME Two plyrs: You, Advrsry. Ruls: First Advrsry sys numr. Thn You sy numr. You win if your numr is iggr., x y : y > x You hv winning strtgy: if dvrsry sys x, you sy x+ Clim is tru

Anothr xmpl: Thorm: NFA N DFA M : L(M) = L(N) W lrdy sw winning strtgy for this gm Wht is it?

Anothr xmpl: Thorm: NFA N DFA M : L(M) = L(N) W lrdy sw winning strtgy for this gm Th powr st construction.

Gms with mor movs: Chss, Chckrs, Tic-Tc-To You cn win if mov of th Advrsry mov You cn mk mov of th Advrsry mov You cn mk : You chckmt

Pumping lmm (contrpositiv) p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L Ruls of th gm: Advrsry picks p, You pick w L of lngth p, L not rgulr Advrsry dcomposs w in xyz, whr y > 0, xy p You pick i 0 Finlly, you win if xy i z L

Thorm: L := {0 n n : n 0} is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := 0 p p Advrsry movs x,y,z You mov i := 2 You must show xyyz L: p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L Sinc xy p nd w = xyz = 0 p p, y only hs 0 So xyyz = 0 p + y p Sinc y > 0, this is not of th form 0 n n DONE

Thorm: L := {w : w hs s mny 0 s } not rgulr Sm Proof: Us pumping lmm Advrsry movs p You mov w :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := {w : w hs s mny 0 s } not rgulr Sm Proof: Us pumping lmm Advrsry movs p You mov w := 0 p p Advrsry movs x,y,z You mov i :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := {w : w hs s mny 0 s } not rgulr Sm Proof: Us pumping lmm Advrsry movs p You mov w := 0 p p Advrsry movs x,y,z You mov i := 2 You must show xyyz L: p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L Sinc xy p nd w = xyz = 0 p p, y only hs 0 So xyyz =?

Thorm: L := {0 j k : j > k} is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := {0 j k : j > k} is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := 0 p+ p Advrsry movs x,y,z You mov i :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := {0 j k : j > k} is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := 0 p+ p Advrsry movs x,y,z You mov i := 0 You must show xz L: p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L Sinc xy p nd w = xyz = 0 p+ p, y only hs 0 So xz = 0 p + - y p Sinc y > 0, this is not of th form 0 j k with j > k

Thorm: L := {uu : u {0,}* } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := {uu : u {0,}* } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := 0 p 0 p Advrsry movs x,y,z You mov i :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := {uu : u {0,}* } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := 0 p 0 p Advrsry movs x,y,z You mov i := 2 You must show xyyz L: p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L Sinc xy p nd w = xyz = 0 p 0 p, y only hs 0 So xyyz = 0 p + y 0 p Sinc y > 0, first hlf of xyyz only 0, so xyyz L

Thorm: L := { n2 : n 0 } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := { n2 : n 0 } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := p2 Advrsry movs x,y,z You mov i :=? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := { n2 : n 0 } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := p2 Advrsry movs x,y,z You mov i := 2 You must show xyyz L: Sinc xy p, xyyz? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := { n2 : n 0 } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := p2 Advrsry movs x,y,z You mov i := 2 You must show xyyz L: Sinc xy p, xyyz p 2 + p < (p+) 2 Sinc y > 0, xyyz >? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := { n2 : n 0 } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := p2 Advrsry movs x,y,z You mov i := 2 You must show xyyz L: Sinc xy p, xyyz p 2 + p < (p+) 2 Sinc y > 0, xyyz > p 2 So xyyz cnnot wht? p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Thorm: L := { n2 : n 0 } is not rgulr Proof: Us pumping lmm Advrsry movs p You mov w := p2 Advrsry movs x,y,z You mov i := 2 You must show xyyz L: Sinc xy p, xyyz p 2 + p < (p+) 2 Sinc y > 0, xyyz > p 2 So xyyz cnnot squr. xyyz L p 0 w L, w p x,y,z : w = xyz, y > 0, xy p i 0 : xy i z L

Big pictur All lngugs Dcidl Turing mchins NP P Contxt-fr Contxt-fr grmmrs, push-down utomt Rgulr Automt, non-dtrministic utomt, rgulr xprssions