This lecture covers Chapter 8 of HMU: Properties of CFLs


 Herbert Harrell
 10 months ago
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1 This lecture covers Chpter 8 of HMU: Properties of CFLs Turing Mchine Extensions of Turing Mchines Restrictions of Turing Mchines Additionl Reding: Chpter 8 of HMU.
2 Turing Mchine: Informl Definition B B B B c b b b B B An tpe extending infinitely in both sides A reding hed tht cn edit tpe, move right or left. A finite control. A string is ccepted if finite control reches finl/ccepting stte 2 / 22
3 Turing Mchine: Forml Definition A Turing mchine M = (Q, Σ, Γ, δ, q 0, B, F ) such tht: Q: finite set of sttes Σ: finite set of input symbols Γ: finite set of tpe symbols such tht Σ Γ δ: trnsition function. δ is prtil function over Q Γ, where the first component is viewed s the present stte, nd the second is viewed s the tpe symbol red. If δ(q, X ) is defined, then Tpe symbol Next Stte Reding hed direction to move next Present stte (q;x) =(q 0 ;Y;D) The symbol replcing X B Γ \ Σ is the blnk symbol. All but finite number of tpe symbols re Bs. q 0: the initil stte of the TM. F : the set of finl/ccepting sttes fo the TM. Hed lwys moves to the left or right. Being sttionry is not n option. The Turing Mchine is deterministic. 3 / 22
4 Describing TMs Turing mchines cn be defined by describing δ using trnsition tble. They cn lso be defined using trnsition digrms (with lbels ppropritely ltered) A TM tht ccepts ny binry string tht does not contin 111 0=0! 0=0! 1=1! q 1 q 0 B=B! 1=1! B=B! 0=0! B=B! q f q 3 1=1! q 2 4 / 22
5 Instntneous Descriptions of TMs An instntneous description (or configurtion) of TM is complete description of the system tht enbles one to determine the trjectory of the TM s it opertes. The instntneous description or configurtion or ID of TM contins 3 prts: () The (finite, nontrivil) portion of tpe to the left of the reding hed; (b) the stte tht the TM is presently in; nd (c) the (finite, nontrivil) portion of the tpe to the right of the reding hed. Stte, Tpe contents, Reding hed loction 1 pple i pple q B B X 1 X 2 X 3 X i X B B ID segment to the strict left stte segment from the hed onwrds z } { z} { z } { X 1 X i 1 q X i X hed i Blnks z } { q B B X 1 X 2 X 3 X B B segment to the strict left z } { stte X 1 X B i 1 z} { q hed z i Blnks } { q B B X 1 X 2 X 3 X B B hed stte z} { q segment from the hed onwrds z } { B i X 1 X 5 / 22
6 Moves of TM Just s in the cse of PDA, we use to indicte single move of TM M, nd M M to indicte zero or finite number of moves of TM. Present ID Trnsition Next ID X 1 X i 1 qx i X (1 <i< ) X 1 X B i 1 q qb i X 1 :::X (q;x i )=(q 0 ;Y;R) (q;x i )=(q 0 ;Y;L) (q;b) =(q 0 ;Y;R) (q;b) =(q 0 ;Y;L) (q;b) =(q 0 ;Y;R) (q;b) =(q 0 ;Y;L) X 1 X i 1 Yq 0 X i+1 X X 1 X i 2 q 0 X i 1 YX i+1 X { X1 { { X 1 X B i 1 Yq 0 X 1 X 1 q 0 X Y i =1 X B i 2 q 0 BY i > 1 Yq 0 X 2 X i =0 Yq 0 B i 1 X 1 X i>0 q 0 BY X 2 X i =0 q 0 BY B i 1 X 1 X i>0 6 / 22
7 Lnguge ccepted by TM A string w is in the lnguge ccepted by TM M iff q 0w αpβ for some p F. M Another notion of cceptnce tht is common is to require TM to hlt (i.e., no further trnsitions re possible). It is lwys possible to design TM such tht the TM hlts when it reches finl stte without chnging the lnguge the TM ccepts. However, we cnnot require (ll) TMs to hlt for ll inputs. A lnguge L is sid to be recursively enumerble if it is ccepted by some TM. A lnguge L is sid to be recursive if both L nd L c re recursively enumerble. Regulr Contextfree Recursive Recursively Enumerble (RE) 7 / 22
8 Extensions of TMs Extensions of TMs 8 / 22
9 Extensions of TMs MultipleTrck TMs Multipletrck TM There re k trcks, ech hving symbols written on them. The mchine cn only red symbols from ech tpe corresponding to one loction, i.e., ll symbols in column t ny one time. A ktrck TM with tpe lphbet Γ hs the sme lngugecceptnce power s TM with tpe lphbet Γ k. X 1 X 2. X k 9 / 22
10 Extensions of TMs Multitpe TMs Multipletpe TM There re k tpes, ech hving symbols written on them. The mchine cn ech tpe independently, i.e., the symbols red from ech tpe need not correspond to the sme loction After red of ech tpes, ech reding hed cn move independently to the right, left, or sty sttionry. X 1 X k X / 22
11 Extensions of TMs Multitpe TMs Theorem Every lnguge tht is ccepted by multitpe TM is lso recursively enumerble (i.e., ccepted by some stndrd TM). Proof of Theorem Let L be ccepted by ktpe TM M. We ll devise 2ktrck TM M tht ccepts L. Every even tpe of M hs the sme lphbet s tht of the ktpe TM. The 2i th trck of M contins exctly the sme contents s the i th tpe of M. Every odd trck hs n lphbet {B,, }, nd contins single or ; the 2i 1 th trck of M contins or t the loction where the i th reding hed of M is locted. M M / 22
12 Extensions of TMs Multitpe TMs Proof of Theorem The stte of M hs 3 components: () the stte of M; (b) the number of s to its strict left; nd (c) vector of length k with ech component tking vlue in Γ {?}. Ech move of M corresponds to multiple moves of M. Ech move of M corresponds of sweep of the tpe from the loction of the leftmost dgger to tht of the rightmost dgger nd bck performing the chnges to trcks tht M would do to its corresponding tpes. At the beginning of the sweep, the hed of M is t loction where the leftmost is nd the stte of M is (q, 0, [?,,?]). The hed moves to the right uncovering s nd the corresponding trck symbols. The right sweep ends when the second component is k M Stte: (q ; 0; [0; 1; 1]) Stte: q 12 / 22 M
13 Extensions of TMs Multitpe TMs Proof of Theorem At this stge, M knows the input symbols M will hve red, nd knows wht ctions to tke. It then sweeps left mking pproprite chnges to the trcks (just like M does to its tpe) ech time is encountered. M lso moves the s ccordingly nd lters it to to indicte tht it hs processed this trck. The left sweep ends when the second component is zero. At this time, M would hve completed moving the s nd the trck contents; they ll now mtch those of M. The TM then sweeps right nd returns reverting ech bck to. M then moves the stte to (q, 0, [?,,?]) nd strt the next sweep if q is not finl stte. Note tht M mimics M nd hence the lnguges ccepted re identicl. 13 / 22
14 Extensions of TMs Multitpe TMs The running time of TM M with input w is the number of moves M mkes before it hlts. (If it does not, the running time is ). The time complexity T M : {0, 1,...} {0, 1,...} of TM M is defined s follows: T M (n) := mximum running time of M for n input w of length n symbols. Theorem The time tken for M in Theorem to process n moves of M is O(n 2 ). Outline of Proof of Theorem After n moves of M, ny two heds of M cn be t most 2n loctions prt. Ech sweep then requires 8n moves of M. Ech trck updte requires finite number of moves. Totlly, to updte the trcks, Θ(k) time steps re needed. Loction of tpe heds n 0 n!!!! 0 Moves of M n 2n prt 14 / 22
15 Extensions of TMs Nondeterministic TMs Nondeterministic TM: δ(q, X ) is set of triples representing possible moves. Theorem For every nondeterministic TM N, there is TM M such tht L(M) = L(N). Outline of Proof of Theorem ID 1 (N does BredthFirst explortion of IDs of M) ID 2;1 ID 2;2 ID 2;k ID 3;1 ID 3;2 ID 3;3 ID 3;4 ID 3; Tpe 1 ID 1 ID 1 ID 2;1 ID 2;2 ID 2;k (If M does not hlt t ID1) (If M does not hlt t ID1 nd ID2;1) ID 1 ID 2;1 ID 2;2 ID 2;k ID 3;1 ID 3;2 (If M does not hlt t ID1, ID2;1 nd ID2;2) ID 1 ID 2;1 ID 2;2 ID 2;k ID 3;1 ID 3;2 ID 3;3 ID 3;4 15 / 22
16 Extensions of TMs Outline of Proof of Theorem We cn devise 2tpe TM M tht simultes N. M first replces the content of the first tpe by followed by the ID tht N is initilly in, which is then followed by specil symbol, which serves s ID seprtor. (M uses the second tpe s scrtch tpe to enble this opertion). If the ID corresponds to finl stte, N hlts (s would M). If not, M then identifies ll possible choices for the next IDs for N nd enters ech one of them followed by t the right end of it s first tpe. (Agin, M uses the second tpe s scrtch tpe to enble this opertion) M then serches for to the right of, chnges the to (to signify tht it is processing the succeeding ID), nd processes tht ID in the similr wy. M hlts t n ID it iff M would t tht ID. 16 / 22
17 Restrictions of TMs Restrictions of TMs 17 / 22
18 Restrictions of TMs TM Semiinfinite Tpe Theorem Every recursively enumerble lnguge is lso ccepted by TM with semiinfinite tpe. Outline of Proof of Theorem Given TM M tht ccepts lnguge L, construct twotrck TM M s follows. The first nd second trcks of M re the R nd L semiinfinite prts of the tpe of M. First, write specil symbol, sy t the leftmost prt of the second trck; this indictes to M tht left move is not to be ttempted t this loction. At ny time, M keeps trck of whether M is to the right or left of its strt loction. If M is to the strict right of its strt loction, M mimics M on the first trck. If M is to the strict left of its strt loction, M mimics M on second trck, but with the hed directions reversed. M detects the strt by the symbol. It cn be formlly shown tht M ccepts string iff M ccepts it. M L $ R B B 2 L $ R b b b M b B B 1 2 L $ R R $ L 18 / 22
19 Restrictions of TMs Multistck Mchines A multistck mchne is PDA with severl independent stcks (i.e., one cn be popping symbol, while the other is writing symbol). Theorem Every recursively enumerble lnguge is ccepted by twostck PDA Outline of Proof of Theorem TM PDA PDA PDA B b B b b b S S 1 2 S b 3 b R semiinfinite portion of TM s tpe Strict L semiinfinite portion of TM s tpe indictes the end of the stck content (to prevent PDA from hlting) If TM moves right chnging tpe symbol X to Y nd stte from q to q, PDA moves from stte q to q popping X from left stck nd pushing Y to the right stck. 19 / 22
20 Restrictions of TMs Counter Mchines A counter mchine is multistck mchine whose stck lphbet contins two symbols: Z 0 (stck end mrker) nd X Z 0 is initilly in the stck. Z 0 my be replced by X i Z 0 for ny i 0 X my be replced by X i for ny i 0. A counter mchine effectively stores nonnegtive number. X X X Z 0 X X X X X Z 0 20 / 22
21 Restrictions of TMs Counter Mchines Theorem Every recursively enumerble lnguge is ccepted by threecounter mchine Outline of Proof of Theorem We know twostck PDA cn simulte ny TM. We ll show tht 3counter mchine cn simulte ny PDA. WLOG, let the stck lphbet of Γ = {0, 1,..., r}. Suppose the first stck contins Y 1(top),..., Y k. Then the first counter stores Y 1 + ry r k 1 Y k. Similrly for the second stck. The third counter is used to chnge the two stck contents. Popping the top symbol stck (sy A) = finding quotient when Y 1 + ry r k 1 Y k is divided by r. pop r X s from stck A, nd push single X on the third stck. Repet until ll X s re exhusted on the stck where popping is performed. Now empty stck A nd copy the third stck contents onto stck A. Chnge Y 1 to some Y 1 requires dding or subtrcting, which is done by popping or pushing the corresponding number of X s. 21 / 22
22 Restrictions of TMs Counter Mchines Outline of Proof of Theorem pushing symbol Z onto stck (sy A) = compute rc + Z where C is the number presently stored in the stck A. pop one X from stck A, nd push r X s on the third stck. Finlly push Z X s onto the third stck. Now empty stck A nd copy the third stck contents onto stck A. Since the bove three re the only opertions needed to simulte TM on twostck PDA, we cn stimulte 2stck PDA nd hence TM using 3counter mchine. Theorem Every recursively enumerble lnguge is ccepted by twocounter mchine Outline of Proof of Theorem The key ide: simulte three counters using one, nd use the other for mnipultions. The first counter stores 2 i 3 j 5 k where i, j, k re the contents of the 3counter mchine. Updtes to the stck involve: () divide by 2,3, or 5; (b) multiply by 2,3, or 5; or (c) identify if i or j or k is zero (check divisibility). Ech opertion cn be esily seen to be done with spre counter. 22 / 22
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