CS18) Lecture conversion between (Notes Regex continued pt 2) . and Automath recap . 2 Recap 5 algot Kleene's Start /NFA and convert it to T regex GBBBB By ... "BBUF state other than starts ②Choose accept and eliminate it 8) . * Y roin &B j do ③ If have parallel edyes & I endup in O G donors want to eliminate this chosen State , but need to add shortcuts # Onia of end up in either : of these situations 2020 Ans : & ↑ regex " To fro every donor to every Donor to receiver, state middle in receiver by passing Need to introduce as many shortcuts as there are donor and receiver pairs smart · Can choose way to eliminate states . 3 NFA 5 to Regex Conversion Example Example of converting NFA to regex : ↑ or 1 O Q · E - 20 - 0 00 Go000 ofYo O May need to Introduce 6 state - can less & shortcuts for one choose more elimina smartly choose incomings outgoing , arrow E on - d Yo in -00 of & O E > - nor do · Or objido paralleareduce T can ⑳ this - * EU 10 1 to combine parallel edyes ↳ ⑧ remaneedenorets outgoing be have ↓ 2 arrows union G o - incoming a us eliminate state this -or - Penae ( answer "1"in regex = union 5 4 RE to . NFA conversion opposite direction example : MenS Partition into implement blocks then Thorst connect · h O 0-0 mis yy Or need loop on regex blocks 01 luniOe) doesn't enforce order need add w/ new state E trans optio can'tadda better H t oppoto is o opt / L Did of structuron · optozoo ↑ 80 ↳ ↑ This is not the simplest machine but be very careful we simplifying O to Did ⑧ Mosmyrear , mon prop Examis z more were m 6 Lemma Pumping technique for how then given general (not all langs reg, prove some not reg, lang not reg = "pumping unmal) to prove 6 1 nonregular lang First . The dren - = 501 0011 , is number of Os andIs, only this is n = , finitely many 000 111 , : on p = As automaton visits some y only D accept 000 -..... Il =-. I w - p+/ an ones . accept state reads O, must state 9. more than once y part of z= ..... inside first loop after m because ( enter a state at least twice . part of w before first loop. w acupress string E X= = states states zeros · many OS t = w Must end up at wi0 number of p+ 1 · acceptina - DFF now of 3 bas Run # of states, can't keep frame seen so far . for L : Assume DFA · regular . nonregular Pf By contradiction - non LOL ame 00001111-.. ↑ : s d = 13 304 185A Ling ↓ while reading the OS 000000000 IIIIIIIIIIII sys X Y Z mor Since DFA accepts Xy2 it also accept m x2 contradiction be XZ is not in the Lang 6 . Lemma Pumping 2 ↑ free exam points, unlike closure questions, exam pumping lema is n Then, there is good for every longa · in X - 2, 2 · L Ce i XyztL for all i P - Xandy occupy earacters at mostP can Y = z 0, 1/ - - be can pumped L 11 strings are (xy)p # - t . · * E string non s . much easier empty PEIN Regular pump into string many times langs o patterns : in the xyz = L xyyyzt Xyyyyyyy can first p chars. z + XztL ↑ have can determine L what - pump y out least pr X # of states of the DFA a what about empty lang or or I string lang exist docent p is , a lang with · at finite strings , give me ur and fold what lang pis . Pumping Cemma proof If Do L: for DFA exam wi p · · DFA on length of string need to repeat is p symbols, some states = ---- use when pumping - 1 w/Lp . Does DFA capture all computational power ↳ /whim w accepts) After No, can't even check that #of OS is equal to out more y part accept state or first before cut can loop part -------inside--- = Languages lemma : at an w of = y appliesranular > can lang ? - be the size ## of states) . must end up X on =numberaes Run this only what should any string weL-s t Take practice pumpand DEA # of 1s . # notice Zine Context Free Grammars Y -------- even CFGS then : are not (check · conditions · #s then - twice some state at least while reading the of Y first O symbols) L for all is 0 furing machine (must visit (xy) = p repeated d 19 is repeated w = Xy z Xyiz a powerful enough (prev page) yE · · Lemma willmust statesa unvisited states when at Start 20 State and to take more need p symbols just at the start ↑ has full power Of computer More pumpingLemma on : contra pos 6 3 contra positive form = original . Pumping Lemma original Original If L is Fira contrapositive - form : - If fre not blue box then ( is non regular Negation of original blue box F => Xp L is nonrey . J WEL (w) =p V X/ y/ z : properties hold cant simultaneously . If first 3 fre last (4m) H 7i is false xyizL existential/universal . - Ex I love someone I love everyone . someone loves Everyone loves > EX - : : I love X I love X everyone I EXXY someone X -Y X loves X loves Y maritiesFront Y Example 6 4 . EX) for Pumping Lemma Examples L 204 = n20] : ↑ can't be implemented pickedyou for by Automata once someon p- you s define w Ia let p -string (V) in lang , at least choose i in control ⑳ P P string in than Is not cang anymore p chars Le28 : CS18) Last Lecture on Midterm 1 material Tool to ↳ prove langs are not pumpling lemma sufficiently long strings -2 shows all are regular - Lemma ~ Op no control over p. chosen (f) vint string explain why it can't be pumped while preserving 6 . membership in the lang alternate between F, 7 L = EW : W are as many game - is their loss Conly one winner) length regular contains zero sum your win contronit continued emply ! ur disadvantage 4 you to - Lis can choose ⑰we L 1w/2p 4 be pumpable. Prove nonregularity Pumping in Dx Os as 1s] Ty Let V = arbitrary Define W = 000 p T consider Al O II I ... -u .... p p F -anyt X/Y : 2 such that these properties hold: use or somewhere in between X choose and -p - + p - Xy2z@LEL nonney in E -pump string y - to i to result nonres in , i times string not language (will disrupt balance btween X(y) FX A ↓ from adversary E from "Exist" chooses this y written backwards ~= y outdoo C "All" if pump up or down string remains in language o fl - length at least p and in the lang + - L bad choice of string adversary . partitions string XY 2 to preventfrom pushing string outside work (can't prove because is XyizL Fi ↑ not is mistake choosing string then defining favorable . Doesn't #1 a X /Y , 2 How do you choose w ? lang Estrials error lang irregular) in control of the partition, you only pick w solution for Another L= GW : what equally many non regular Monday, - & If L reg was ↳ cangtang) reg lang, = but it's not = ↳ nin vey lang, so I must be non reg na03 "ow this is not nec can conclude t is : nonvey rec regular Os and 1s] L10 ON EXAM 0 = use Treg on ey &* orey EX) ( ~ = Con : n + m) -p + +P- & edit adversary may choose Y to be oneO but can't rely on this outside : adversary may of vey lang nonvey ? Alternate 3y + E down to pump = all regea push string out of lang #of Is z to pump - has numbergal > 80 + ↑ confusing ! nonney veg/reg he know is = canneveem eyed xy2 , from class reg nonnegut reg/nonreg up or one Divisie divisible zes case on on bu down, are equal - # of zeos make # of As to in exams w/o proof - pump 11 in string quisin V of, : chooseBotbetter po ! + p Xy] subset of ( lang? diff setup = can be Owilains = any statement proved class = non reg EX) 30 % / = : n > m3 ein lany Y W X need pump down &. to - of y) Xz has symbols : X2 would this bad w p-1 ones at most p-1 zeros . d . & L work if we w Adversary chooses y nonempty substring in first p noir onea remount good chose = me EX)L Slowin = : 11 3 ⑳on 2 im case y can en. be . XIL because doesn't start w/ 1 XzL because # of zeros < p Y unized wouldwork couldbeoutwo oomta not Y now to Picsing barely in Lang - = = Ein EE , 0 1 , 1 1111 , unary lang, string uses Exam Q Last III IIIIII III , .... 9 4 symbol (only one symbols same 3 langs of figure which reg , prove it gap between integers very large when #S get layer figure out Heuswill He prepare us > - square of sur long or non of closure pumping lemma) ↑ ceinus p2 is at So int n to be smallestent thats and a square p least p acceptable pl at most p (y)(p 1)2 ↓ < (xyz) + in soonesible p2 + 2p + 1 of be we Do both practiceams to be ↓ posted Xy 2 + = quee + p Hi ab it Capplication p2 Y examn of submitab rey prove im 3 bunch given consecutive squares, At least p chars : is the next squhre L . com make exams i EX) L = Ew = wR : weE0- 13 3 & * as X palindromes palindome ↓ are copies not of each other useless , anything 0000000000000 prev langs, strings accepted bols , w= not as standard pumping - doesn't make it ↓) noprovement to nonpalindrone w = 18000001 Occan's razor -try simplest ideas W Y = 00000000000000 adversary first palindne picks y symbols (always this --- xyyzz Grumpina Bumping copies more string is oddanyth "I" is - restriction) pumping down z in second half, not a palindrome first p from 00 XZ ↓ has less tuan p zeros then then ↑ hard to p more zes / explain , string may 1 or ↑ not palindrome may not have even/odd L ↑ need explanation even > always - EX) ~ L = 3ww w : + length 50- 13 ] * Y w =s string becomes 0000000 1 odd - 0 + 2025 So adversary can't Seitner is odd: zeros . NEED to have first half of OR remove all IX21 in length 0 Xy2fL or even XZ . 0 explanation and ends w/ starts second half ends W/ . 1 Ex) All strings of prime length L Sw / w/ prime = : come p = q = up w/ string at least p and length prime arbitrary s any prime . t . qIp . Need Y 0000000 = (xyy2) = a I ways to some this one * I much easier Nxy2 + lyl upqaddivan xyy ztL to pump up or down composite integer 2 + 121 - 9 -composite -ly1) · (not primes .6 5 True false ? or 1) closed under infinite unions Rey langs ↳ False Eoh n203 · are : infinitea = Y n C) = [buthourey 0 subsets of · ↳ regular languages are regular FALSE Soh n2032S * : w ↳ nonreg subsets · of reg nonregular languages are ↳ False because 0 50YV : n nonregular 203 · g · Nonney langs are closed under complement 6 True ↓ is regular #[ Thus, 1 is nonnegular is regular E) I nonvey II closed under union Nonregular langs ↳ False take any L that's nonney I is · are , L 1[ non · ↓ nonney rey 1) Nonvey FALSE = 20 langs : non Grey are or + LVT & nonney honey very = intersection & closed under 1 n203150/ rey nonrey so : n nonrey + m3 = "A nonvey lang & Tme is necessarily infinite Every finite large is of contraposive . reg 9 Lecture Mon . April 28 Cs181 Automata can't even ensure # the of Os and Is * -> Var not bound to 7 ( T CONTEXT-TE Automate to Turing Machines Notions Basic Example 2 = CFG C Ea, b3 # L V = its context GRAMMARS equivalent to automate but with one .I 7 are the same terminals states, instead of variables have - ES A, B) , stars - A S - B transition 7 A + a Ab - E A additional feature nonterminals/variables rules substitu tion rules B + bBa X : A derivation S - A eaAb > - Set of a Abb rure #1 rule #3 rule #3 > aaa Abbb - mie#3 > agabbb - rule#4 strings can derive from grammar languagethe grammar Lef The lang be derived can seatB] EX) set of G is the grammar of strings that . using G collapsing 2 mus to one Se A s + B A-aAblE B-6 Ba/E Lang ab : Sahb" : n = 03 UEbra" noS : ↓ anAbb Def A grammar is a tuple (V , E R S) where V = Set of vars = S Set of = (finite, non empty) terminals & set of R : , , rules start variable - 4 2 . EX) Reading CFGS STAA does - A terminate because can't eliminate all vars A - ↑ ↓ nothing changes not set of if we o removerl strings that can be derived 0 ↑ no strings : EX) Stabs/a ↳ regex In * (ab) a not required this to have Exam exam is is Ihr soin hours solid exams graded Mware midterm 25 % EX)s + aSb/E anbh ↑ : 30 points on exam Fou miss get = 0 n full 16 of exam, points built in curve- cantement Na FAIDFA EX)s + aSa/B B/Ea "b * * b an Ans : an beah : n / m2o 25 : points only collect EX) SeaSa/aBa In regex : - 65/b-b abrah : n , m2) Ex)S + ss/as/b + + ↓ (a b) Sess - sends with a 'S * SSS better E much wording M Saabla at s + SS - an 6 S for every SSS * b/chunce tab SSS - aS a SS > a SaaS S - a was was anabas ab abagabaab Saga SS > a S aaa SaS > a - EX S- AB * (aa) Ettals bBb/gt(bb) - * even # of As: as followed by even # of 6's (aa)"(bb)" or_ easier : * * (aa) (bb) - nimzo tonseed -only inna an ⑪ seelat/bT SET-aS0S round ↓ goana Answer : (EE) y delay terminating T does not terminate like by an extra S round * ⑪ arb FA CFGCanMement any nonaccept states dont generated mem break Aer : constructing CFGs EX)(av(avb)2 n 203 EEE : aa ... ... s - ElaSE 2 - alb string of even length EX) [(vb) "a (avb)r : n = 03 S + alES[ 2 - alb = Sahbu +- . aT/S nmSpeti S - El a Sb on cusing pocket analogy) * pla equal and Beb/E b's Exsambm n>m3 : much harder than prev s - aSB convention Beb/aX allows tothana ↓ as TeaT)aS ingramma ersive sa : alternationa of as 5-9Sb/Expase lang that we already know works salbm : Omeans alll confusing Se aSBBBIE Be bla E 3w : w = wRz * palindrome SeEsElE- - 2- alb does not wor , only s+ no coordination between & generates even length strings aSalb5b/El alb E) swiw contains 13 as seXaXaXaX X + &X)E 2 + alb Ex sw : w contains b aa Se XbaaX X + 2X/E 2 - alb ? &* Designing Programming Lang Using CFGs . programming langs All # · (av Identifiers = 4 .... are uz)(au var names (don't start / · program is CFGs· Uzu Oulv a Sa, b..... z, ↓ 0 . 1, 2, ..... 943 , + D ↑ any pos digit -112/3) OR - OR )9Edigit in D- D + Lealbic . - . /zletter we [L/ID) ↑ identifies ⑧ NeND/ ↑ numbers D + start (identifer W- quantity ↑ = Y E Clive a num content i wq or conditional WTWenWS/ devoid bybas meaning of their Eng if w wQ) test identifier optional C + I whitespace Qw' > Q identifer where & white space optional aw < wq var A + =w 2) W T + Qw = Gumber & All assignment: ww mandatory ↑ I - compared identifier E digit 10 any letterof alphabet · Ican be program mem (regex) nument quartity T symbol · -> stored one · ran number Q - IIN pos -. 93 = V- L quantity = Gall valid programs · # line breas CFG = ug) letters, not #s) - 2 ... assignment/conditional single statement statement types : · .... on numbers numbers = positive · based When wsweese was somigea statement
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