6.045J/18.400J: Automata, Computability and Complexity. Quiz 2: Solutions. Please write your name in the upper corner of each page.

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1 6045J/18400J: Automt, Computbility nd Complexity Mrh 30, 2005 Quiz 2: Solutions Prof Nny Lynh Vinod Vikuntnthn Plese write your nme in the upper orner of eh pge Problem Sore Totl Q2-1

2 Problem 1: True or Flse (20 points) Full redit will be given for orret nswers If you inlude justifition for your nswers, you my obtin prtil redit for inorret nswers 1 True or Flse: There exists Turing mhine tht enumertes set of (enodings of) deider Turing mhines, suh tht inludes Turing mhines tht deide infinitely mny different deidble lnguges True It is esy to enumerte set of Turing mhines suh tht for eh, reognizes the lnguge 2 True or Flse: There exists Turing mhine tht enumertes set of (enodings of) deider Turing mhines, suh tht inludes t lest one Turing mhine tht deides eh deidble lnguge Flse Disprove by digonlizing (Rell tht this ws Problem 2 of homework 5) 3 True or Flse: There exists Turing mhine tht enumertes n infinite set of (enodings of) deider Turing mhines, suh tht every mhine tht outputs is miniml (Here miniml mens tht there is no other smller deider Turing mhine tht deides the sme lnguge) Flse This is n pplition of Reursion Theorem Consider mhine tht first gets its own desription (vi Reursion Theorem), nd proeeds to run the enumertor till outputs mhine tht hs longer desription thn Then it ontinues to simulte, nd therefore is not miniml This is ontrdition 4 True or Flse: Rie s Theorem immeditely implies tht "! # is Turing mhine nd $ &%')(+* (, is undeidble True This is non-trivil property of lnguges A mhine tht deides the empty lnguge - hs the property, but mhine / tht deides 0 ( does not

3 5 True or Flse: If 1 is deidble lnguge nd must be Turing-reognizble is Turing reognizble lnguge, then 2134 Flse Does not hold if 5 0 ( nd is : is 6798 ;, whih is not reognizble 6 True or Flse: If 1 is Turing reognizble lnguge nd be Turing-reognizble is deidble lnguge, then 5< must True is, in prtiulr, reognizble Then, this follows from the losure of Turing-reognizble lnguges under ontention (this ws gin homework problem!!) 7 True or Flse: A three-dimensionl Turing mhine is like n ordinry Turing mhine exept tht its tpe storge onsists of three dimensionl tpe, where eh tpe ell is unit ube In one step, the single tpe hed n move north, south, est, west, up, or down The lss of lnguges reognized by three-dimensionl Turing mhines is extly the Turing-reognizble lnguges True One n simulte this by single-tpe mhine 8 True or Flse: The lss of lnguges reognized by three-stk mhines is extly the Turing reognizble lnguges True Three stks is equivlent to two stks whih is s powerful s Turing mhines

4 = 0 S Problem 2: (25 points) Consider the following forml desription of Turing Mhine, where >@?BAC> AC> AD>@EBAD>FHGIGIJKHL<AD> MCJNCJOGPLQ, RA *+, TA@*BA<V Assume tht ny unspeified trnsitions go to >@MQJNCJ GPL 0 R 1 R 0 1 L L 0,R L 1,L q 0 q q q R R q ept 1 (5 points) Write out the epting omputtion history of on input T*, in the form given in lss nd in Sipser s book Desribe the behvior represented in this omputtion history in words >@?1T* W> * X*Y>& W> * > E ZW W>H?[Z Z\> FHGIGIJKHL

5 > * > * > * 2 (5 points) Wht lnguge does reognize? (Give preise definition) B]9*] #@^[_ R 3 (10 points) Give the set of tiles for the modified Post Correspondene Problem, for this prtiulr mhine nd input T* Indite whih is the initil tile We hve strted things off by listing the tiles needed for ompleting the mth from the point where n epting stte is enountered You must define the initil tile nd the tiles needed to represent ll the moves The initil tile : b >@?T* d Tiles for the right moves : >H?1 e> > f g> >2* *Y> > E5 W>H? >H?2 h> FHGIGIJPK<L Tiles for the left moves : B> B> >@E B> EV >@E B B> E * >@E R* *> *, * > >@E+*, * >@EY >@E+* * > E * >@E+*j* e> h e> > EYi e> EY > EYd e> E * > EYW* The lphbet tiles: The len-up tiles : > FHGIGIJPK<L * > FHGIGIJKHL, > F@GIGIJKHL > FHGIGIJKHL *,,, * g> F@GIGIJKHL *Y> FHGIGIJKHL, > F@GIGIJKHL > FHGIGIJKHL O > F@GIGIJKHL, d k l,, l h> FHGIGIJKHL > FHGIGIJKHL, > F@GIGIJKHL > FHGIGIJKHL,

6 * > 4 (5 points) Write the epting omputtion history you wrote for prt (b) twie, one bove the other, nd mrk the boundries of the MPCP tiles involved in the mth You my skip the prt involved in terminting the omputtion just mrk the tiles up to the first ourrene of the ept stte m >? R* e> >@EVh >? e> * * d > E e>@? > * * > b l *> * >H?, e> FHGIGIJPK<L k d

7 Problem 3: (10 points) Suppose we hve Turing Mhine tht, on eh input string n, either hlts with n output string on its tpe or loops forever Desribe briefly how to onstut n enumertor Turing mhine tht enumertes the outputs produed by on ll the inputs This is simple dove-til onstrution The enumertor does the following: In the first phse, it runs the mhine on the string o for * step In the seond phse, it runs on o for p steps, nd on for * step More generlly, in the q Lsr phse, it runs on the Lsr string (in the lexiogrphi order) for qt3u steps This ensures tht if outputs something on input n, it does so in finite mount of time, nd eventully outputs it too

8 ;! Problem 4: (20 points) Let wvxxy{z} d "!@#~ ept ny strings of odd length epts ll strings of even length nd does not 1 (2 points) Does Rie s Theorem pply to vxxyuz} m? Why or why not? If it does pply, then wht does this imply bout EVENODD? Rie s theorem does pply to vxxyuz} m, beuse this is lerly lnguge property nd it is non-trivil This is beuse mhine tht epts extly the even length strings hs this property, nd mhine 4 tht epts nothing, does not This mens tht vy{z} m is undeidble 2 (9 points) Is EVENODD Turing-reognizble? Prove your nswer You my use ny results proved in lss or in Sipser s book, but if you do, then ite the results expliitly No, wvxxy{z} d is not Turing reognizble! We show this by reduing 6 ; 798 to vxy{z} m Given AO, we onstrut mhine!5!& suh tht AO if nd only if vxy{z} m first heks if its input is of even length If so, it epts Else, it runs on nd epts if epts, nd rejets if rejets If " indeed epts, then 0 ( nd therefore, "Vƒ vxy{z} m On the other hnd, if does not ept, then is preisely the set of ll even length strings Therefore, " vxy{z} m

9 ! 3 (9 points) Is the omplement of EVENODD Turing-reognizble? Agin, prove your nswer Agin, you my use ny results proved in lss or in the book, but ite them expliitly No, the omplement of vxxy{z} m is not Turing reognizble We show this by reduing 6 ; 798 to the omplement of vxxyuz} m, or equivlently, to vxy{z} m!!5 Given AD, we onstrut mhine suh tht AO 6$798 "!& if nd only if wvxxy{z} d first heks if its input is of odd length If so, it rejets Else, it runs on nd epts if epts, nd rejets if rejets If " indeed epts, then is preisely the set of ll even-length strings nd therefore, vxy{z} m On the other hnd, if does not ept, then - Therefore, "Vƒ vxy{z} m

10 !!! Problem 5: (15 points) Let be the following lnguge of Turing mhine desriptions: "!& j is Turing mhine with input lphbet RA *+ nd epts every string onsisting of just zeros (it my ept other strings) Prove tht is undeidble using the Reursion Theorem Do this by filling in the following proof outline: Suppose for the ske of ontrdition tht is deidble Let be the deider for Define Turing mhine : : On input do: Obtin the desription of itself This is possible beuse of the Reursion theorem Run on input s If If epts then rejet rejets then ept If epts ll strings onsisting of only zeros, then epts!& beuse is deider for nd s, But this implies tht, rejets ll strings (nd in prtiulr rejets ll strings of the form ( ), whih is impossible On the other hnd, if does not ept ll strings onsisting of only zeros, then!xƒ rejets, beuse is deider for, nd But this implies tht, epts ll strings, (nd in prtiulr epts ll strings of the form ( ), whih is lso impossible Therefore, we hve ontrdition, nd nnot be deidble

11 ^ Problem 6: (10 points) Consider new kind of mhine, q -Queue Mhine A q -Queue Mhine hs the sme generl struture s q -Counter Mhine or q -Stk Mhine However, it hs q queues for storge insted of q ounters or stks Initilly, eh queue > is empty It supports the following opertions (Let S be the lphbet of queue symbols): 1 > ˆ9 ˆ : tkes symbol in S, nd dds it to the end of queue > 2 Š > ˆ ˆ9 : removes the symbol t the front of queue >, if > is nonempty If > is empty, this opertion does nothing 3 tœt IŽ : boolen, whih returns 1 if queue > is urrently empty, 0 otherwise Briefly outline n rgument tht the eptne problem for p -Queue-Mhines is undeidble We do this by showing tht q -queue mhine n simulte q -ounter mhine This, in prtiulr, shows tht the eptne problem for q -ounter mhine n be redued to the eptne problem for q -queue mhine Sine we know tht the eptne problem for p -ounter mhine is undeidble, the eptne problem for p -queue mhine is undeidble, too We need to show how to simulte the inrement, derement nd iszero funtions on ounter, using queue inrement: is implemented by enqueueing symbol on the queue derement: is implemented by dequeueing symbol from the queue iszero: is implemented by invoking the funtion empty on the queue, nd returning true if nd only if empty returns true It is trivil to see tht this orretly implements the funtionlity of ounter Sine the eptne problem for p -ounter mhine is undeidble, we onlude tht the eptne problem for p -queue mhine is undeidble too

12 Srth Work