Note on ETM
Skip ahead to section 2.2 or 2.3 and show that ETM is not recognizable. Then by the theorem that. “a language is decidable iff it is both recognizable and
Reductions
This in turn implies that ETM is undecidable. At least one of them must be not recognizable. ETM is recognizable. ETM is not recognizable.
Homework 8 Solutions
Consider the emptiness problem for Turing machines: ETM = { ?M?
Homework 10 Solutions
However ETM is Turing-recognizable (HW 8
Tutorial 5
If both L and L are recognizable then L is decidable. As we know that ATM HALTTM
5.4 No. For example consider A = {0 1
But ¬ETM is Turing recognizable and ¬ATM is not
10 Reducibility
HALTTM is Turing-recognizable since it can be recognized by TM U. HALTTM is not Turing-decidable. ETM = {< M >
CS 341: Foundations of CS II Marvin K. Nakayama Computer
Since ETM is undecidable (Theorem 5.2) EQTM must be undecidable. • We'll see later that EQTM is not Turing-recognizable not co-Turing-recognizable
Problem 1 Problem 2
Show that the language ETM = {?M?
COMPSCI 501: Formal Language Theory Reducibility so far
4 Mar 2019 But ETM is Turing-recognizable (why?) which would mean ATM recognizable (false). ? Mapping reductions may not exist!
[PDF] Note on ETM
Skip ahead to section 2 2 or 2 3 and show that ETM is not recognizable Then by the theorem that “a language is decidable iff it is both recognizable and
[PDF] Homework 10 Solutions
However ETM is Turing-recognizable (HW 8 problem 4) and ATM is not Turing-recognizable (Corollary 4 23) contradicting Theorem 5 22 3 Consider the language
[PDF] CS 341: Foundations of CS II Marvin K Nakayama Computer
Rice's Theorem: any nontrivial property of the language of a TM is undecidable • ETM is not Turing-recognizable • EQTM is neither Turing-recognizable nor co-
[PDF] Problem 1 - JHU CS
Show that the language ETM = {?M?M is a Turing machines and L(M) = ? } is not Turing-recognizable Proof: We provide two different proofs Proof 1: We
[PDF] Tutorial 5
As we know that ATM HALTTM ETM are recognizable their complements cannot be rec- ognizable because then the languages would be decidable and we know that
[PDF] 10 Reducibility
HALTTM is Turing-recognizable since it can be recognized by TM U HALTTM is not Turing-decidable Proof: We will reduce ATM to HALTTM Assume TM R decides
[PDF] 1 Reducability
HALTTM = {?Mw? M is a TM that halts on input w} is undecidable If A ?m B and B is recognizable then A is recognizable Corollary 5 23
[PDF] Reducibility Decidable vs Undecidable vs Recognizable ?
26 oct 2020 · Theorem: Every nontrivial property about recognizable languages (of Turing machines) is undecidable • The proof is a generalization of the
[PDF] reducibility
29 juil 2013 · We mentioned that ETM is co-TM recognizable We will prove next that ETM is undecidable Intuition: You cannot solve this problem UNLESS
[PDF] Reductions
HALTTM is undecidable since ATM is undecidable This in turn implies that ETM is undecidable At least one of them must be not recognizable
Is ETM Turing-recognizable?
ETM is not Turing-recognizable. Rice's Theorem: Every nontrivial property of the Turing-recognizable languages is undecidable.Is ETM undecidable proof?
Note that ?M1? ? ETM ?? L(M1) = ? ?? M accepts w ?? ?M, w? ? ATM. But then TM S decides ATM, which is undecidable. Therefore, TM R cannot exist, so ETM is undecidable.Is EQTM Turing-recognizable?
EQTM is not recognizable.- ATM = {?M,w?M is a Turing machine and M accepts w} However, unlike ADFA and ACFG, ATM is not decidable. But ATM is Turing-recognizable.
Reductions
LetAandBbe languages over the same alphabet .
Analgorithmthat transforms:
Strings inAto strings inB.
Strings not inAto strings not inB.
Bis decidable)Ais decidable.
Ais undecidable)Bis undecidable.
One way to show a problemBto be undecidable is to reduce an undecidable problemAtoB. 1Reductions
A Turing machineMcomputesa functionfif:
Mhalts on all inputs.
On inputxit writesf(x) on the tape and halts.
Such a functionfis called acomputablefunction.
Examples: increment, addition, multiplication, shift. Any algorithm with output is computing a function. 2Reductions
LetAandBbe languages over .
AisreducibletoBif and only if:
there exists a computable functionf: !such that for allw2; w2A,f(w)2B.Notation:AmB.
FACT:AmB,AmB.
w2A,f(w)2Bis equivalent tow62A,f(w)62B. 3Reductions
To Re-iterate:
1.Construction:f(w) fromwby an algorithm.
2.Correctness:w2A,f(w)2B.
4An Example involving DFAs
EQDFA=fA;BjA;Bare DFAs andL(A) =L(B)g.
EDFA=fAjAis a DFA andL(A) =;g.
Areduction machineon inputA;Btwo DFAs:
1. Constructs the DFAA0such thatL(A0) =L(A).
2. Constructs the DFAB0such thatL(B0) =L(B).
3. Constructs the DFAM1such thatL(M1) =L(A)\L(B0).
4. Constructs the DFAM2such thatL(M2) =L(A0)\L(B).
5. Constructs the DFACsuch thatL(C) =L(M1)[L(M2).
6. OutputsC.
Correctness:
SupposeL(A) =L(B). Then,L(C) =;.
SupposeL(A)6=L(B). Then,L(C)6=;.
That is,EQDFAmEDFA.
5An Example involving CFGs
ALLCFG=fGjGis a CFG andL(G) = g.
EQCFG=fG;HjG;Hare CFGs andL(G) =L(H)g.
Areduction machineon inputGa context-free grammar with alphabet :1. Constructs a CFGHwith rules of the formS0 aS0j, for alla2.
2. Outputs (G;H).
L(H) = .
Correctness:
SupposeGgenerates all strings in . Then,L(G) =L(H). SupposeGdoes not generate some string in . ThenL(G)6=L(H).That is, ALLCFGmEQCFG.
6The Halting problem
HALTTM=fM;wjMis a DTM andMhalts onwg.
The reduction machine outputs a DTM that loops wheneverMreaches the rejecting state.On inputM;w:
1. Constructs the following machineM0:
Read inputx.
SimulateMonx.
IfMaccepts, halt and accept.
IfMhalts and rejects, enter a loop.
2. OutputsM0;w.
That is,ATMmHALTTM.
HALTTMis undecidable sinceATMis undecidable.
7The Halting problem
Three machines:
A reduction machine that is a DTM.
Input to the reduction machine isM;w, whereMis a DTM. Output of the reduction machine isM0;w, whereM0is a DTM. 8The Halting problem
The output DTMM0has the inputMhard-coded into it.
LetM= (Q;;;;qs;qa;qr). What isM0?
M0= (Q0;;;0;qs;qa;qr0):
Q0=Q[ fql;qr0g.
Dene0as follows:
For all statesp, for all statesq6=qr, for all symbolsa;b, for allD2 fL;R;Sg: if(p;a) = (q;b;D) then0(p;a) = (q;b;D). For all statesp, for all symbolsa;b, for allD2 fL;R;Sg: if(p;a) = (qr;b;D) then0(p;a) = (ql;b;S). For all symbolsa, include the transition0(ql;a) = (ql;a;S). M0is constructed fromMby the reduction machine.
9ETM=fMjMis a TM andL(M)6=;g
A reduction machine on inputM;w:
1. Constructs the following machineM0:
Read inputx.
Ifx6=wthen reject.
Ifx=w, SimulateMonw.
IfMaccepts, halt and accept.
2. OutputsM0.
Correctness:
IfMacceptswthenL(M0) is not empty.
IfMdoes not acceptwthenL(M0) is empty.
That is,ATMmETM.
10ETM=fMjMis a TM andL(M)6=;g
ATMmETM.
This implies thatETMis undecidable.
This in turn implies thatETMis undecidable.
At least one of them must be not recognizable.ETMis recognizable. ETMis not recognizable.
11EQTM=fM1;M2jM1andM2are TMs andL(M1) =L(M2)g
EQTMis undecidable.
IsEQTMrecognizable?
See the text book.
12Language is Context-free
CONTEXTFREETM=fMjMis a TM andL(M) is contextfreeg
A reduction machine on inputM;w:
1. Constructs the following machineM0:
Read inputx.
Ifxhas the form 0n1n2nthen halt and accept.
Ifxis not of this form, simulateMonw.
IfMaccepts, halt and accept.
2. OutputsM0.
Correctness
L(M0) is ifMacceptsw.
L(M0) is not context-free ifMdoes not acceptw.
ATMmCONTEXTFREETM.
CONTEXTFREETMis undecidable sinceATMis undecidable. 13 Language is Not RegularREGULARTM=fMjMis a TM andL(M) is NOT regulargA reduction machine on inputM;w:
1. Constructs the following machineM0:
Read inputx.
Ifxis not of the form 0n1nthen halt and reject.
Ifxis of this form, simulateMonw.
IfMaccepts, halt and accept.
2. OutputsM0.
Correctness:
L(M0) is;ifMdoes not acceptw.
L(M0) is not regular ifMacceptsw.
A TMmREGULARTM.REGULARTMis undecidable sinceATMis undecidable. 14Not Recognizable
A TMmREGULARTM=)ATMmREGULARTM.REGULARTMis not recognizable. ATMmREGULARTM=)ATMmREGULARTM
REGULAR
TMis not recognizable.
15Properties of Reductions
1.AmBandBmC=)AmC.
2.AmB=)AmB
3.AmBandBis decidable =)Ais decidable.
4. LetAbe recognizable. Then,AmAif and only ifAandAare decidable.
5.AmBandBis recognizable =)Ais recognizable.
16quotesdbs_dbs14.pdfusesText_20[PDF] is fog a colloid
[PDF] is form 8332 necessary
[PDF] is france constitution written or unwritten
[PDF] is full fat milk homogeneous or heterogeneous
[PDF] is glucose a crystalloid
[PDF] is green a primary color
[PDF] is hamburg a rich city
[PDF] is hand sanitizer dangerous goods
[PDF] is handset leasing profitable for telecom companies
[PDF] is high alkalinity in drinking water bad
[PDF] is interest income taxable in singapore
[PDF] is iphone packaging recyclable
[PDF] is iron a pure substance
[PDF] is iron filings a mixture