Computability and computational complexity

Transcription

1 Computability ad computatioal complexity Lecture 4: Uiversal Turig machies. Udecidability Io Petre Computer Sciece, Åbo Akademi Uiversity Fall toukokuu

2 Cotet Uiversal Turig machies Udecidability 21. toukokuu

3 UNIVERSAL TURING MACHINES 21. toukokuu

4 Uiversal Turig machies Ay TM has a fixed program, solvig a sigle problem Questio: Is there a programmable Turig machie? A uiversal Turig machie U takes as iput the descriptio of aother Turig machie M ad a iput x for M, ad the simulates M o x: U(M,x)=M(x) Assume that the alphabet ad the states of all Turig machies are itegers ={1,2,, } K={ +1,, + K, + K +1,, + K +6}, where the last 6 symbols stad for the directioal istructios, for h, yes, ad o the iitial state is +1 all umbers are represeted i biary, each with eough leadig 0s so that they are of the same legth 21. toukokuu

5 Uiversal Turig machies Descriptio of a UTM start with K ad i biary a descriptio of d follows: ( (,s), (p,r,d) ) the add a separator symbol the add a descriptio of x, with each symbol of x as a biary umber, separated with aother separator symbol o tape S 1 U keeps a descriptio of M o tape S 2 it keeps the curret cofiguratio of M U simulates M Sca S 2 to fid a iteger correspodig to a state Search S 1 for a rule of d matchig the curret state ad the scaed symbol Apply the rule 21. toukokuu

6 Smallest uiversal Turig machie 1962, M.Misky: 7-state, 4-symbol UTM; i short (7,4) Later proposals: (15,2), (9,3), (6,4), (5,5), (4,6), (3,9), (2,18) Smallest kow UTM (i terms of descriptioal complexity) is by Yurii Rogozhi: (4,6)-machie usig oly 22 istructios 21. toukokuu

7 Towards udecidability ON RE LANGUAGES 21. toukokuu

8 Recall: acceptig vs. decidig Let LÌ(S-{#})* A Turig machie decides L iff for every xî (S-{#})*, if xîl, the M(x)= yes if xïl, the M(x)= o I this case, we say that L is a recursive laguage A Turig machie accepts L iff for every xî (S-{#})*, xîl iff M(x)= yes if xïl, M might ot termiate o x we could ask that M(x)= for all xïl I this case we say that L is a recursively eumerable laguage Note: ay recursive laguage is recursively eumerable 21. toukokuu

9 Recursively eumerable The laguage E(M) eumerated by a Turig machie M: E(M)={x: (s,>,1)"*(,y#x#,1), for some x,y} I other words: all strigs x such that at some poit durig the operatio of M, there is a time whe the tape eds with #x# 21. toukokuu

10 L is recursively eumerable iff there is a TM M such that L=E(M) This machie E eumerates the strigs i laguage L This machies M accepts laguage L: for ay iput strig x, wait to see whether it is eumerated by E; if it comes out, the M accepts x This machie M accepts laguage L This machies E eumerates the strigs i L: for each i>0, simulate M up to i steps o each of the first i strigs (lexicographic order); output ay strig that gets accepted 21. toukokuu

11 Strigs i lexicographic order w 1 w 2 w 3 w 4 w 5 1 Number of steps toukokuu

12 No recursively eumerable laguages Recursively eumerable laguages Recursive laguages 21. toukokuu

13 Fiite Ifiite coutable Ifiite ucoutable Alphabet! All strigs over! All sets of strigs (laguages) over! 21. toukokuu

14 Fiite Ifiite coutable Ifiite ucoutable S Powerset 2 S of S S is coutable: S={s 1, s 2, s 3, } Elemets of the powerset: {s 5, s 19, s 34 }, {s 1, s 729, s 1234, s 2459, s } Each of them ca be writte as a ifiite seuece of 0/1: 1 o positio i iff s i is i that set Assume that the powerset is coutable: L 1, L 2, L 3, L 4, Cosider the followig elemet D of the powerset: Iclude s i i D iff s i is ot i L i, for all i>0 Observe that D is ot i our eumeratio If D=L k, the the k-th bit should be its ow complemet!!! Powerset Ecodig s 1 s 2 s 3 L L L toukokuu

15 No-RE laguages Follow-up from the previous slide: there are (far) more laguages tha Turig machies Ucoutable umber of laguages vs. coutable umber of TM Coclusio: there exist o-recursively eumerable laguages (caot be described by algorithms) No recursively eumerable laguages Recursively eumerable laguages Recursive laguages 21. toukokuu

16 What about udecidable laguages Questio 1: are there udecidable laguages? I other words, laguages that are RE, but ot recursive Questio 2: are there also atural udecidable problems? Aswer: YES!!! No recursively eumerable laguages Recursively eumerable laguages Recursive laguages 21. toukokuu

17 UNDECIDABILITY 21. toukokuu

18 Haltig problem HALTING. Give the descriptio of Turig machie M ad a iput x, will M halt o x? We prove that HALTING is a udecidable problem We prove that H is ot a recursive laguage, i.e., there is o Turig machie to decide it We prove also that H is recursively eumerable Deote H={M;x: M(x)¹ } the set of YES istaces of HALTING 21. toukokuu

19 Haltig problem H={M;x: M(x)¹ } Propositio. H is recursively eumerable. Proof. Cosider a uiversal Turig machie U ad modify it so that wheever it halts, it does so i a acceptig state. Let U be the ew machie. Note ow that U accepts H: If M,xÎH, the M(x)¹ ad so, U(M,x)¹, i.e., U (M,x)= yes If M,xÏH, the M(x)= ad so, U(M,x)=, i.e., U (M,x)= 21. toukokuu

20 HALTING is udecidable Theorem. H is ot recursive. I other words: HALTING is udecidable Proof: Assume that H is recursive: there is a TM M H decidig H. This meas that for ay TM M ad iput x, we ca decide if M halts o x. Techiue of the proof: diagoalizatio. Ituitive idea: All iput strigs s 1 s 2 s 3 All Turig machies M M M toukokuu

21 Formal proof Costruct a TM D as follows: O iput M, D first simulates M H o iput M;M util it is about to halt If M H is about to accept, the D eters ito a ifiite loop If M H is about to reject, the D halts i a acceptig state I other words: D(M)= if M H (M;M)= yes ad D(M)= yes otherwise Questio: What is D(D)? I other words, does machie D halt o iput D? If D(D)=, the M H (D;D)= yes, i.e., D;DÎH, i.e., D halts o iput D, i.e. D(D)¹ ; cotradictio If D(D)¹, the M H (D;D)¹ yes, i.e., D;DÏH, i.e., D does ot halt o iput D, i.e., D(D)= ; cotradictio Coclusio: there is o TM decidig H 21. toukokuu

22 More udecidability results May other properties ca be show to be udecidable usig the techiue of reductio ad our proof of HALTING beig udecidable HALTING Trasformatio computable by a TM The A must be udecidable PROBLEM A If A were decidable, the HALTING would also be decidable decide M o x by decidig whether t(m,x)îa 21. toukokuu

23 More udecidability Propositio. The followig laguages are ot recursive: {M: M halts o all iputs} {M;x: there is y such that M(x)=y} {M;x: the computatio of M o x uses all states of M} {M;x;y: M(x)=y} I other words: it is udecidable whether: (i) M halts o all iputs, (ii) M gives a output o x; Proof. Oly show here the first property, usig the reductio techiue Let M be a TM ad x a strig Costruct TM M that o a iput y works as follows: if y=x, the it simulates M o x, otherwise it halts Clearly, M halts o x iff M halts o all iputs 21. toukokuu

24 Ay o-trivial property of TM is udecidable! Very geeral result: Rice s theorem Deote by L(M) the laguage accepted by a Turig machie M Rice s theorem. Suppose that C is a proper, o-empty subset of the set of all recursively eumerable laguages. The the followig problem is udecidable: Give a Turig machie M, is L(M)ÎC? Proof. We ca assume ÆÏC (otherwise repeat the argumet below for the complemet of C, also a proper, o-empty subset of RE laguages). Also, there is LÎC, accepted by TM M L Cosider a arbitrary TM M ad a strig x. Costruct a TM M x such that M halts o x iff L(M x )ÎC M x o iput y: first it simulates M o x ad if it halts, the it cotiues by simulatig M L o y Note: L(M x )=L if M halts o x ad it is Æ otherwise I other words, sice ÆÏC: L(M x )ÎC iff M halts o x 21. toukokuu

25 More o recursive laguages Propositio. If L is recursive, the so is its complemet. Proof: make all acceptig states be rejectig states ad the other way aroud Propositio. L is recursive iff both L ad its complemet are recursively eumerable Proof. Direct implicatio clear by the previous result Reverse implicatio Let L be accepted by a TM M ad its complemet by a TM M Build a TM decidig L: O a secod tape make a copy of the iput Simulate o the two tapes machies M ad M o iput x Oe step of oe machie, followed by oe step of the other machie x must be i either L or L evetually oe of the machies will accept it Give aswer yes if it was accepted o the first tape, o if it was accepted o the secod tape Coseuece: the complemet of HALTING is ot recursively eumerable The set of Turig machies M ad iputs x for which M does ot halt o x 21. toukokuu

26 Learig objectives Uderstad the cocept of udecidability Ability to prove udecidability results through reductios 21. toukokuu