Formal Languages: Review

Transcription

1 Formal Languags: Rviw Alphabt: a finit st of symbols String: a finit squnc of symbols Languag: a st of strings String lngth: numbr of symbols in it String concatnation: w 1 w 2 Empty string: or ^ Languag concatnation: L 1 L 2 ={w 1 w 2 w 1 L 1, w 2 L 2 } String xponntiation: w k = ww w (k tims) Languag xponntiation: L k = LL L (k tims) LL = L 2 L k =LL k-1 L 0 ={} S={a,b} ababbaab L={a,aa,aaa, } aba =3 ab ba=abba "w w = w = w {1,2} {a,aa, } ={1a,2a,1aa,2aa, } a 3 =aaa {0,1} 32

2 Formal Languags: Rviw String rvrsal: w R Trivial languag: {} Empty languag: Ø All finit strings: S * L S * "L Thorm: S * is countabl, S * = Z Thorm: 2 S* is uncountabl. Thorm: S * contains no infinit strings. (aabc) R =cbaa Languag rvrsal: L R ={w R w L} {ab,cd} R ={ba,dc} Kln closur: L * = L 0 L 1 L 2 L 3... {a} * L + = L 1 L 2 L 3 L 4... {a} + Thorm: L + = LL * {} L=L {}=L Ø * ={} {a,aa,aaa, } dovtailing diagonalization finit strings in S i Thorm: (L * ) * =L * L * (L * ) * & (L * ) * L *

3 Finit Automata: Rviw Basic ida: a FA is a machin that changs stats whil procssing symbols, on at a tim. Finit st of stats: Q = {q 0, q 1, q 3,..., q k } Transition function: Initial stat: Final stats: d: Q S Q q 0 Q F Q Finit automaton is M=(Q, S, d, q 0, F) q 1 q i q 0 q k q j Ex: an FA that accpts all odd-lngth strings of zros: 0 q 0 q 1 0 M=({q 0,q 1 }, {0}, {((q 0,0),q 1 ), ((q 1,0),q 0 )}, q 0, {q 1 })

4 Finit Automata: Rviw FA opration: consum a string w S * on symbol at a tim whil changing stats Accptanc: nd up in a final stat Rjction: anything ls (including hang-up / crash) Ex: FA that accpts all strings of form abababab = (ab) * M a M=({q 0,q 1 }, {a,b}, {((q 0,a),q 1 ), ((q 1,b),q 0 )}, q 0, {q 0 }) b q 0 b q 2 q 1 a But M crashs on input string abba! Solution: add dad-nd stat to fully spcify M M a,b M =({q 0,q 1,q 2 }, {a,b}, {((q 0,a),q 1 ), ((q 1,b),q 0 ), ((q 0,b),q 2 ), ((q 1,b),q 2 ). ((q 2,a),q 2 ), ((q 2,b),q 2 ) }, q 0, {q 0 })

5 Finit Automata: Rviw Transition function d xtnds from symbols to strings: d:q S* Q d(q 0,wx) = d(d(q 0,w),x) whr d(q i,) = q i Languag of M is L(M)={w S* d(q 0,w) F} Dfinition: languag is rgular iff it is accptd by som FA. Thorm: Complmntation prsrvs rgularity. Proof: Invrt final and non-final stats in fully spcifid FA. b q 0 a b q 1 M q 1 q 2 a,b a b q 0 a M b q 2 a,b a L(M)=(ab) * L(M )= b(a+b) * + (a+b) * a + (a+b) * (aa+bb)(a+b) * M simulats M and dos th opposit!

6 Problm: dsign a DFA that accpts all strings ovr {a,b} whr any a s prcd any b s. Ida: skip ovr any contiguous a s, thn skip ovr any b s, and thn accpt iff th nd is rachd. a b a,b q 0 b q 1 a q 2 L = a*b* Q: What is th complmnt of L?

7 Problm: what is th complmnt of L = a*b*? Ida: writ a rgular xprssion and thn simplify. L = (a+b)*b + (a+b)*a + (a+b)* = (a+b)*b(a+b)*a(a+b)* = (a+b)*b + a(a+b)* = (a+b)*ba(a+b)* = a*b + a(a+b)* a b a,b q 0 b q 1 a q 2

8 Finit Automata: Rviw Thorm: Intrsction prsrvs rgularity. Proof: ( paralll simulation): Construct all supr-stats, on pr ach stat pair. Nw supr-transition function jumps among supr-stats, simulating old transition function Initial supr stat contains both old initial stats. Final supr stats contains pairs of old final stats. Rsulting DFA accpts sam languag as original NFA (but siz can b th product of two old sizs). Givn M 1 =(Q 1, S, d 1, q, F 1 ) and M 2 =(Q 2, S, d 2, q, F 2 ) construct M=(Q, S, d, q, F) Q = Q 1 Q 2 F = F 1 F 2 q=(q,q ) d :Q S Q d((q i,q j ),x) = (d 1 (q i,x),d 2 (q j,x))

9 Finit Automata: Rviw Thorm: Union prsrvs rgularity. Proof: D Morgan's law: L 1 L 2 = L 1 L 2 Or cross-product construction, i.., paralll simulation with F = (F 1 Q 2 ) (Q 1 F 2 ) Thorm: St diffrnc prsrvs rgularity. Proof: St idntity L 1 L 2 = L 1 L 2 Or cross-product construction, i.., paralll simulation with F = (F 1 (Q 2 F 2 )) Thorm: XOR prsrvs rgularity. Proof: St idntity L 1 L 2 = (L 1 L 2 ) (L 1 L 2 ) Or cross-product construction, i.., paralll simulation with F = (F 1 (Q 2 F 2 )) ((Q 1 F 1 ) F 2 ) Mta-Thorm: Idntity-basd proofs ar asir!

10 Finit Automata: Rviw Non-dtrminism: gnralizs dtrminism, whr many nxt movs ar allowd at ach stp: Old Nw d:q S Q d:2 Q S 2 Q Computation bcoms a tr. Accptanc: $ a path from root (start stat) to som laf (a final stat) Ex: non-dtrministically accpt all strings whr th 7 th symbol bfor th nd is a b : b a,b a,b a,b a,b a,b a,b q 0 q 1 q 2 q 3 q 4 q 5 q 6 q 7 a,b Input: ababbaaa Accpt!

11 Finit Automata: Rviw Thorm: Non-dtrminism in FAs dosn t incras powr. Proof: by simulation: Construct all supr-stats, on pr ach stat subst. Nw supr-transition function jumps among supr-stats, simulating old transition function Initial supr stat ar thos containing old initial stat. Final supr stats ar thos containing old final stats. Rsulting DFA accpts th sam languag as original NFA, but can hav xponntially mor stats. Q: Why dosn t this work for PDAs?

12 Finit Automata: Rviw Not: Powrst construction gnralizs th cross-product construction. Mor gnral constructions ar possibl. EC: Lt HALF(L)={v $ v,w S * ' v = w and vw L} Show that HALF prsrvs rgularity. A two way FA can mov its had backwards on th input: d:q S Q {lft,right} EC: Show that two-way FA ar not mor powrful than ordinary on-way FA. -transitions: q i q j q i q j On supr-stat! Thorm: -transitions don t incras FA rcognition powr. Proof: Simulat -transitions FA without using -transitions. i.., considr -transitions to b a form of non-dtrminism.

13 Th movi Nxt (2007) Basd on th scinc fiction story Th Goldn Man by Philip Dick Prmis: a man with th supr powr of non-dtrminism! At any givn momnt his rality branchs into multipl dirctions, and h can choos th branch that h prfrs! Transition function!

14 Top-10 Rasons to Study Non-dtrminism 1. Hlps us undrstand th ubiquitous concpt of paralllism / concurrncy; 2. Illuminats th structur of problms; 3. Can hlp sav tim & ffort by solving intractabl problms mor fficintly; 4. Enabls vast, dp, and gnral studis of compltnss thoris; 5. Hlps xplain why vrifying proofs & solutions sms to b asir than constructing thm;

15 Why Study Non-dtrminism? 6. Gav ris to nw and novl mathmatical approachs, proofs, and analyss; 7. Robustly dcoupls / abstracts complxity from undrlying computational modls; 8. Givs disciplind tchniqus for idntifying hardst problms / languags; 9. Forgd nw unifications btwn computr scinc, math & logic; 10. Non-dtrminism is intrsting fun, and cool!

16 Rgular Exprssions Rgular xprssions ar dfind rcursivly as follows: Ø mpty st q 0 {} trivial languag {x} " x S singlton languag q 0 q 0 x q 1 Inductivly, if R and S ar rgular xprssions, thn so ar: (R+S) union M 1 RS concatnation M 2 M 2 M 1 Compositions! R * Kln closur Exampls: aa(a+b) * bb (a+b) * b(a+b) * a(a+b) * Thorm: Any rgular xprssion is accptd by som FA. M

17 Rgular Exprssions A FA for a rgular xprssions can b built by composition: Ex: all strings ovr S={a,b} whr $ a b prcding an a (a+b) * b(a+b) * a(a+b) * = (a+b) * ba(a+b) * b a b a b a Why? b a b a b a b a b a b a b a b a b a Rmov prvious start/final stats

18 FA Minimization Ida: Equivalnt stats can b mrgd: b a a,b a,b b a b a b a b a b a b a b a a,b a,b b a a,b a,b

19 FA Minimization Thorm [Hopcroft 1971]: th numbr N of stats in a FA can b minimizd within tim O(N log N). Basd on arlir work [Huffman 1954] & [Moor 1956]. Conjctur: Minimizing th numbr of stats in a nondtrministic FA can not b don in polynomial tim. Thorm: Minimizing th numbr of stats in a pushdown automaton (or TM) is undcidabl. Projct ida: implmnt a finit automaton minimization tool. Try to dsign it to run rasonably fficintly. Considr also including: A rgular-xprssion-to-fa transformr, A non-dtrministic-to-dtrministic FA convrtr.

20 FAs and Rgular Exprssions Thorm: Any FA accpts a languag dnotd by som RE. Proof: Us gnralizd finit automata whr a transition can b a rgular xprssion (not just a symbol), and: Only 1 supr start stat and 1 (sparat) supr final stat. Each stat has transitions to all othr stats (including itslf), xcpt th supr start stat, with no incoming transitions, and th supr final stat, which has no outgoing transitions. M M Ø Ø Ø Ø Ø M Ø Ø Ø Ø Original FA M Gnralizd FA (GFA) M

21 FAs and Rgular Exprssions Now rduc th siz of th GFA by on stat at ach stp. A transformation stp is as follows: q i P q j q i P q j q i P + RS * T q j R S q T RS * T Such a transformation stp is always possibl, until th GFA has only two stats, th supr-start and supr-final stats: M P Labl of last rmaining transition is th rgular xprssion corrsponding to th languag of th original FA! Corollary: FAs and REs dnot th sam class of languags.

22 R+S = S+R Rgular Exprssions Idntitis R(ST) = (RS)T R(S+T) = RS+RT (R+S)T = RT+ST Ø * = * = R+Ø = Ø+R = R R = R = R (R * ) * = R * ( + R) * = R * (R * S * ) * = (R+S) * R+ R RØ R

23 Dcidabl Finit Automata Problms Df: A problm is dcidabl if $ an algorithm which can dtrmin (in finit tim) th corrct answr for any instanc. Givn a finit automata M 1 and M 2 : Q 1 : Is L(M 1 ) = Ø? Hint: graph rachability Q 2 : Is L(M 2 ) infinit? Hint: cycl dtction Q 3 : Is L(M 1 ) = L(M 2 )? Hint: considr L 1 -L 2 and L 2 -L 1 M M $? $? S * -{} Ø Ø

24 Rgular Exprssion Minimization Problm: find smallst quivalnt rgular xprssion Dcidabl (why?) Hard: PSPACE-complt Turing Machin Minimization Problm: find smallst quivalnt Turing machin Not dcidabl (why?) Not vn rcognizabl (why?)

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30 Contxt-Fr Grammars: Rviw Basic ida: st of production ruls inducs a languag Finit st of variabls: V = {V 1, V 2,..., V k } Finit st of trminals: T = {t 1, t 2,..., t j } Finit st of productions: P Start symbol: Productions: V i D whr V i V and D (V T)* Applying V i D to av i b yilds: a Db Not: productions do not dpnd on contxt - hnc th nam contxt fr! S

31 Contxt-Fr Grammars: Rviw Df: A languag is contxt-fr if it is accptd by som contxt-fr grammar. Thorm: All rgular languags ar contxt-fr. Thorm: Som contxt-fr languags ar not rgular. Ex: {0 n 1 n n > 0} Proof by pumping argumnt: long strings in a rgular languag contain a pumpabl substring. $ N ' "z L, z $ u,v,w S* ' z=uvw, uv, v 1, uv i w L " i Thorm: Som languags ar not contxt-fr. Ex: {0 n 1 n 2 n n > 0} Proof by pumping argumnt for CFL s.

32 Ambiguity: Rviw Df: A grammar is ambiguous if som string in its languag has two non-isomorphic drivations. Thorm: Som contxt-fr grammars ar ambiguous. Ex: G 1 : S SS a Drivation 1: S SS aa Drivation 2: S SS SSS aa Df: A contxt-fr languag is inhrntly ambiguous if vry contxt-fr grammar for it is ambiguous. Thorm: Som contxt-fr languags ar inhrntly ambiguous (i.., no non-ambiguous CFG xists). Ex: {a n b n c m d m m>0, n>0} {a n b m c n d m m>0, n>0}

33 Exampl: dsign a contxt-fr grammar for strings rprsnting all wll-balancd parnthsis. Ida: crat ruls for gnrating nsting & juxtaposition. G 1 : S SS (S) Ex: S SS (S)(S) ()() S (S) ((S)) (()) S (S) (SS)... (()((())())) Q: Is G 1 ambiguous? Anothr grammar: G 2 : S (S)S Q: Is L(G 1 ) = L(G 2 )? Q: Is G 2 ambiguous?

34 Exampl : dsign a contxt-fr grammar that gnrats all valid rgular xprssions. Ida: mbdd th RE ruls in a grammar. G: S a for ach a S L S (S) SS S* S+S S S* (S)* (S+S)* (a+b)* S SS SSSS abs*b aba*a Q: Is G ambiguous?

35 Pushdown Automata: Rviw Basic ida: a pushdown automaton is a finit automaton that can optionally writ to an unboundd stack. Finit st of stats: Q = {q 0, q 1, q 3,..., q k } q 1 Input alphabt: S Stack alphabt: G Transition function: d: Q (S {}) G 2 Q G* q i q j Initial stat: q 0 Q q 0 Final stats: F Q q k Pushdown automaton is M=(Q, S, G, d, q 0, F) Not: pushdown automata ar non-dtrministic!

36 Pushdown Automata: Rviw A pushdown automaton can us its stack as an unboundd but accss-controlld (last-in/first-out or LIFO) storag. A PDA accsss its stack using push and pop Stack & input alphabts may diffr. Input rad had only gos 1-way. Accptanc can b by final stat or by mpty-stack. Not: a PDA can b mad dtrministic by rstricting its transition function to uniqu nxt movs: d: Q (S {}) G Q G * M Input a b a stack

37 Pushdown Automata: Rviw Thorm: If a languag is accptd by som contxt-fr grammar, thn it is also accptd by som PDA. Thorm: If a languag is accptd by som PDA, thn it is also accptd by som contxt-fr grammar. Corrolary: A languag is contxt-fr iff it is also accptd by som pushdown automaton. I.E., contxt-fr grammars and PDAs hav quivalnt computation powr or xprssivnss capability.

38 Closur Proprtis of CFLs Thorm: Th contxt-fr languags ar closd undr union. Hint: Driv a nw grammar for th union. Thorm: Th CFLs ar closd undr Kln closur. Hint: Driv a nw grammar for th Kln closur. Thorm: Th CFLs ar closd undr with rgular langs. Hint: Simulat PDA and FA in paralll. Thorm: Th CFLs ar not closd undr intrsction. Hint: Find a countr xampl. Thorm: Th CFLs ar not closd undr complmntation. Hint: Us D Morgan s law.

39 Dcidabl PDA / CFG Problms Givn an arbitrary pushdown automata M (or CFG G) th following problms ar dcidabl (i.., hav algorithms): Q 1 : Is L(M) = Ø? Q 5 : Is L(G) = Ø? Q 2 : Is L(M) finit? Q 6 : Is L(G) finit? Q 3 : Is L(M) infinit? Q 7 : Is L(G) infinit? Q 4 : Is w L(M)? Q 8 : Is w L(G)?

40 Undcidabl PDA / CFG Problms Thorm: th following ar undcidabl (i.., thr xist no algorithms to answr ths qustions): Q: Is PDA M minimal? Q: Ar PDAs M 1 and M 2 quivalnt? Q: Is CFG G minimal? Q: Is CFG G ambiguous? Q: Is L(G 1 ) = L(G 2 )? Q: Is L(G 1 ) L(G 2 ) = Ø? Q: Is CFL L inhrntly ambiguous?

41 PDA Enhancmnts Thorm: 2-way PDAs ar mor powrful than 1-way PDAs. Hint: Find an xampl non-cfl accptd by a 2-way PDA. Thorm: 2-stack PDAs ar mor powrful than 1-stack PDAs. Hint: Find an xampl non-cfl accptd by a 2-stack PDA. Thorm: 1-quu PDAs ar mor powrful than 1-stack PDAs. Hint: Find an xampl non-cfl accptd by a 1-quu PDA. Thorm: 2-had PDAs ar mor powrful than 1-had PDAs. Hint: Find an xampl non-cfl accptd by a 2-had PDA. Thorm: Non-dtrminism incrass th powr of PDAs. Hint: Find a CFL not accptd by any dtrministic PDA.

42 Turing Machins: Rviw Basic ida: a Turing machin is a finit automaton that can optionally writ to an unboundd tap. Finit st of stats: Q = {q 0, q 1, q 3,..., q k } Tap alphabt: Blank symbol: Input alphabt: Transition function: Initial stat: Final stats: G b G S G {b} d: (Q F) G Q G {L,R} q 0 Q F Q Turing machin is M=(Q, G, b, S, d, q 0, F) q 1 q i q 0 q k q j

43 A Turing machin can us its tap as an unboundd storag but rads / writs only at had position. Initially th ntir tap is blank, xcpt th input portion Rad / writ had gos lft / right with ach transition A Turing machin is usually dtrministic Input string accptanc is by final stat(s) M Turing Machins: Rviw Input b b

44 Turing Machin Enhancmnts Largr alphabt: old: Σ={0,1} nw: Σ ={a,b,c,d} Ida: Encod largr alphabt using smallr on. Encoding xampl: a=00, b=01, c=10, d=11 b a d c a old: δ b nw: δ' 0 1

45 Turing Machin Enhancmnts Doubl-sidd infinit tap: Ida: Fold into a normal singl-sidd infinit tap old: δ L/R nw: δ' L/R L/R L/R R/L R/L

46 Turing Machin Enhancmnts Multipl hads: b b a b a b b a a Ida: Mark hads locations on tap and simulat bb b b A a b a b B b Aa a Modifid δ' procsss ach virtual had indpndntly: Each mov of δ is simulatd by a long scan & updat δ' updats & marks all virtual had positions

47 Turing Machin Enhancmnts Multipl taps: Ida: Intrlac multipl taps into a singl tap Modifid δ' procsss ach virtual tap indpndntly: Each mov of δ is simulatd by a long scan & updat δ' updats R/W had positions on all virtual taps

48 Turing Machin Enhancmnts Two-dimnsional tap: Ida: Flattn 2-D tap into a 1-D tap $ $ $ Modifid 1-D δ' simulats th original 2-D δ: Lft/right δ movs: δ' movs horizontally This is how compilrs implmnt 2D arrays! Up/down δ movs: δ' jumps btwn tap sctions

49 Turing Machin Enhancmnts Non-dtrminism: Ida: Paralll-simulat non-dtrministic thrads $ $ $ Modifid dtrministic δ' simulats th original ND δ: Each ND mov by δ spawns anothr indpndnt thrad All currnt thrads ar simulatd in paralll

50 Turing Machin Enhancmnts Combinations: ND H l l o W o r l d! 9 Π λ α τ ω ν Ida: Enhancmnts ar indpndnt (and commutativ with rspct to prsrving th languag rcognizd). Thorm: Combinations of nhancmnts do not incras th powr of Turing machins.

51 Turing -Rcognizabl vs. -Dcidabl w Input Accpt & halt Rjct & halt Df: A languag is Turing-dcidabl iff it is xactly th st of strings accptd by som always-halting TM. w Σ * = a b aa ab ba bb aaa aab aba abb baa bab bba bbbaaaa L(M) = { a, aa, aaa, aaaa } M(w) Not: M must always halt on vry input. Nvr runs forvr

52 Turing -Rcognizabl vs. -Dcidabl w Input Accpt & halt Rjct & halt Df: A languag is Turing-rcognizabl iff it is xactly th st of strings accptd by som Turing machin. w Σ * = a b aa ab ba bb aaa aab aba abb baa bab bba bbbaaaa L(M) = { a, aa, aaa, aaaa } M(w) Run forvr Not: M can run forvr on an input, which is implicitly a rjct (sinc it is not an accpt).

53 Rcognition vs. Enumration Df: Dcidabl mans Turing-dcidabl Rcognizabl mans Turing-rcognizabl Thorm: Evry dcidabl languag is also rcognizabl. Thorm: Som rcognizabl languags ar not dcidabl. Ex: Th halting problm is rcognizabl but not dcidabl. Not: Dcidability is a spcial cas of rcognizability. Not: It is asir to rcogniz than to dcid.

54 Famous Dcidrs A wrong dcision is bttr than indcision. I'm th dcidr, and I dcid what is bst.

55 Famous Dcidrs

56 Rcognition and Enumration Df: An numrator Turing machin for a languag L prints out prcisly all strings of L on its output tap. a $ a b $ b b a $ Not: Th ordr of numration may b arbitrary. Thorm: If a languag is dcidabl, it can b numratd in lxicographic ordr by som Turing machin. Thorm: If a languag can b numratd in lxicographic ordr by som TM, it is dcidabl.

57 Rcognition and Enumration Df: An numrator Turing machin for a languag L prints out prcisly all strings of L on its output tap. a $ a b $ b b a $ Not: Th ordr of numration may b arbitrary. Thorm: If a languag is rcognizabl, thn it can b numratd by som Turing machin. Thorm: If a languag can b numratd by som TM, thn it is rcognizabl.

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60 Dcidability w Input Accpt & halt Rjct & halt Nvr runs forvr Df: A languag is Turing-dcidabl iff it is xactly th st of strings accptd by som always-halting TM. Thorm: Th finit languags ar dcidabl. Thorm: Th rgular languags ar dcidabl. Thorm: Th contxt-fr languags ar dcidabl.

61 A Simpl Exampl Lt S = {x 3 + y 3 + z 3 x, y, z Z } Q: Is S infinit? A: Ys, sinc S contains all cubs. Q: Is S Turing-rcognizabl? A: Ys, sinc dovtailing TM can numrat S. Q: Is S Turing-dcidabl? A: Unknown! Q: Is 29 S? A: Ys, sinc =29 Q: Is 30 S? A: Ys, sinc ( ) 3 +( ) 3 +( ) 3 =30 Q: Is 33 S? A: Unknown! Thorm [Matiyasvich, 1970]: Hilbrt s 10 th problm (1900), namly of dtrmining whthr a givn Diophantin (i.., multi-variabl polynomial) quation has any intgr solutions, is not dcidabl.

62 Closur Proprtis of Dcidabl Languags Thorm: Th dcidabl languags ar closd undr union. Hint: us simulation. Thorm: Th dcidabl languags ar closd undr. Hint: us simulation. Thorm: Th dcidabl langs ar closd undr complmnt. Hint: simulat and ngat. Thorm: Th dcidabl langs ar closd undr concatnation. Hint: guss-factor string and simulat. Thorm: Th dcidabl langs ar closd undr Kln star. Hint: guss-factor string and simulat.

63 Closur Proprtis of Rcognizabl Languags Thorm: Th rcognizabl languags ar closd undr union. Hint: us simulation. Thorm: Th rcognizabl languags ar closd undr. Hint: us simulation. Thorm: Th rcognizabl langs ar not closd undr compl. Hint: rduction from halting problm. Thorm: Th rcognizabl langs ar closd undr concat. Hint: guss-factor string and simulat. Thorm: Th rcognizabl langs ar closd undr Kln star. Hint: guss-factor string and simulat.

64 Rducibilitis Df: A languag A is rducibl to a languag B if $ computabl function/map ƒ: * * whr "w w A ƒ(w) B A ƒ w B ƒ(w) Not: ƒ is calld a rduction of A to B Dnotation: A B Intuitivly, A is no hardr than B

65 Rducibilitis Df: A languag A is rducibl to a languag B if $ computabl function/map ƒ: * * whr "w w A ƒ(w) B A ƒ w B ƒ(w) Thorm: If A B and B is dcidabl thn A is dcidabl. Thorm: If A B and A is undcidabl thn B is undcidabl. Not: b vry carful about th mapping dirction!

66 Rduction Exampl 1 Df: Lt H b th halting problm for TMs running on w= Dos TM M halt on? H = { <M> * M() halts } Thorm: H is not dcidabl. Proof: Rduction from th Halting Problm H: Givn an arbitrary TM M and input w, construct nw TM M that if it ran on input x, it would: x Ignor x M 1. Ovrwrit x with th fixd w on tap; 2. Simulat M on th fixd input w; 3. Accpt M accpts w. Simulat M on w If M(w) halts thn Not: M halts on (and on any x *) M halts on w. A dcidr (oracl) for H can thus b usd to dcid H! Sinc H is undcidabl, H must b undcidabl also. halt Not: M is not run!

67 Rduction Exampl 2 Df: Lt L Ø b th mptynss problm for TMs Is L(M) mpty? L Ø = { <M> * L(M) = Ø } Thorm: L Ø is not dcidabl. Proof: Rduction from th Halting Problm H: Givn an arbitrary TM M and input w, construct nw TM M that if it ran on input x, it would: x Ignor x M 1. Ovrwrit x with th fixd w on tap; 2. Simulat M on th fixd input w; 3. Accpt M accpts w. Not: M halts on vry x * M halts on w. Simulat M on w If M(w) halts thn A dcidr (oracl) for L Ø can thus b usd to dcid H! Sinc H is undcidabl, L Ø must b undcidabl also. halt Not: M is not run!

68 Rduction Exampl 3 Df: Lt L rg b th rgularity problm for TMs Is L(M) rgular? L rg = { <M> * L(M) is rgular } Thorm: L rg is not dcidabl. Proof: Rduction from th Halting Problm H: Givn an arbitrary TM M and input w, construct nw TM M that if it ran on input x, it would: x 1. Accpt if x 0 n 1 n 2. Ovrwrit x with th fixd w on tap; 3. Simulat M on th fixd input w; 4. Accpt M accpts w. Not: L(M )= * M halts on w L(M )=0 n 1 n M dos not halt on w Accpt if x 0 n 1 n Ignor x M Simulat M on w If M(w) halts thn A dcidr (oracl) for L rg can thus b usd to dcid H! halt Not: M is not run!

69 Ric s Thorm Df: Lt a proprty P b a st of rcognizabl languags Ex: P 1 ={L L is a dcidabl languag} P 2 ={L L is a contxt-fr languag} P 3 ={L L = L * } P 4 ={{}} P 5 = Ø P 6 ={L L is a rcognizabl languag} L is said to hav proprty P iff L P Ex: (a+b) * has proprty P 1, P 2, P 3 & P 6 but not P 4 or P 5 {ww R } has proprty P 1, P 2, & P 6 but not P 3, P 4 or P 5 Df: A proprty is trivial iff it is mpty or it contains all rcognizabl languags.

70 Thorm: Th two trivial proprtis ar dcidabl. Proof: P non = Ø x Ric s Thorm Ignor x Say no Stop M non no P all ={L L is a rcognizabl languag} M non dcids P non x Ignor x Say ys Stop M all ys M all dcids P all Q: What othr proprtis (othr than P non and P all ) ar dcidabl? A: Non!

71 Ric s Thorm Thorm [Ric, 1951]: All non-trivial proprtis of th Turing-rcognizabl languags ar not dcidabl. Proof: Lt P b a non-trivial proprty. Without loss of gnrality assum Ø P, othrwis substitut P s complmnt for P in th rmaindr of this proof. Slct L P (not that L Ø sinc Ø P), and lt M L rcogniz L (i.., L(M L )=L Ø ). Assum (towards contradiction) that $ som TM M P which dcids proprty P: Not: x can b.g., a TM dscription. x Dos th languag dnotd by <x> hav proprty P? M P ys no

72 Rduction stratgy: us M p to solv th halting problm. Rcall that L P, and lt M L rcogniz L (i.., L(M L )=L Ø). Givn an arbitrary TM M & string w, construct M : x M w Ric s Thorm M halt start M L What is th languag of M? L(M ) is ithr Ø or L(M L )=L If M halts on w thn L(M )=L(M L )= L If M dos not halt on w thn L(M )= Ø sinc M L nvr starts => M halts on w iff L(M ) has proprty P Oracl M P can dtrmin if L(M ) has proprty P, and thrby solv th halting problm, a contradiction! ys ys Dos th languag dnotd by <x> hav proprty P? M P ys no

73 Ric s Thorm Corollary: Th following qustions ar not dcidabl: givn a TM, is its languag L: Empty? Finit? Infinit? Co-finit? Rgular? Contxt-fr? Inhrntly ambiguous? Dcidabl? L= *? L contains an odd string? L contains a palindrom? L = {Hllo, World}? L is NP-complt? L is in PSPACE? Warning: Ric s thorm applis to proprtis (i.., sts of languags), not (dirctly to) TM s or othr objct typs!

74 Contxt-Snsitiv Grammars Problm: dsign a contxt-snsitiv grammar to accpt th (non-contxt-fr) languag {1 n $1 2n n 1} Ida: gnrat n 1 s to th lft & to th right of $; thn doubl n tims th # of 1 s on th right. S 1ND1E /* Bas cas; E marks nd-of-string */ N 1ND $ /* Loop: n 1 s and n D s; nd with $ */ D1 11D /* Each D doubls th 1 s on right */ DE E /* Th E cancls out th D s */ E ε /* Procss nds whn th E vanishs */

75 Exampl: Gnrating strings in {1 n $1 2n n 1} S 1ND1E D1 11D E ε N 1ND $ DE E S 1ND1E 11NDD1E 11ND11DE 111NDD11DE 111ND11D1DE 111N11D1D1DE 111N11D1D1E 111$11D1D1E 111$1111DD1E 111$1111D11DE 111$111111D1DE 111$ DDE 111$ DE 111$ E 111$ ε = 1 3 $1 8 = 1 3 $1 23

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