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# Halt tm is undecidable

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is undecidable! The halting problem <M, w> S’s solver R’s solver convert S to R accept reject M accept reject w Let us use S = A TM qAssume that M R solves R = HALT TM qSolve S using M R by constructing M S using M R qA contradiction since S is not solvable! Map the solution of R to a solution to S M accept w or not Since this problem is undecidable, does that mean there isn't one algorithm that works for every instance, or does it even mean there is an instance for which we cannot arrive at an answer? i.e. is there a program you can write for which it would make sense to say: "this program doesn't halt or not halt, it's undecidable". A Turing machine may halt for other reasons as well. We say that it "accepts" an input if the entire input is read (or otherwise properly processed). If the TM doesn't reach the accept state (due to bad/non-accepting input), it will halt on a "fail" state -- but it can still halt. As long as it does actually halt then the TM works as designed. • This is an undecidable problem. Theorem 5.1 HALT TM = { ¢M, w ²| M is a TM and halts on input w } is Turing-recognizable, but not decidable. Proof. U of Theorem E is easy to convert into a TM that simulates the computation of M on input w and accepts if and only if the computation being simulated halts. In computability theory and computational complexity theory, an undecidable problem is a decision problem for which it is proved to be impossible to construct an algorithm that always leads to a correct yes-or-no answer. The halting problem is an example: it can be proven that there is no algorithm that correctly determines whether arbitrary programs eventually halt when run.

HALT TM • HALT TM ={<M,w>| M is a TM and halts on input w} – It is undecidable • We need to reduce A TM to HALT TM, where A TM already proven to be undecidable – Can use HALT TM to solve A TM • Proof by contradiction – Assume HALT TM is decidable and show this implies A TM is decidable 10/17/19 Theory of Computation - Fall'19 Halt TM by reducing A TM to Halt TM. Let Halt TM = {hM,wi: M is a TM and M halts on input w}. Theorem Halt TM is undecidable. We assume that Halt TM is decidable and use that assumption to show that A TM is decidable, a contradiction. The key idea is to show that A TM is reducible to Halt TM. Assume that we have a TM R that decides Halt TM ... The Totality Problem is Undecidable The halting problem can be used to show that other problems are undecidable. Totality Problem: A function (or program) F is said to be total if F(x) is defined for all x (or similarly, if F(x) halts for all x). Determining whether or not a function F is total is undecidable. If xhas the form 0n1 n2 then halt and accept. If xis not of this form, simulate Mon w. If Maccepts, halt and accept. 2. Outputs M0. Correctness L(M0) is if Maccepts w. L(M0) is not context-free if Mdoes not accept w. A TM m CONTEXT FREE TM. CONTEXT FREE TM is undecidable since A TM is undecidable. 13 • This is an undecidable problem. Theorem 5.1 HALT TM = { ¢M, w ²| M is a TM and halts on input w } is Turing-recognizable, but not decidable. Proof. U of Theorem E is easy to convert into a TM that simulates the computation of M on input w and accepts if and only if the computation being simulated halts.

15-0: Universal TM •Turing Machines are “Hard Wired” •Addition machine only adds •0 n1 2n machine only determines if a string is in the language 0n1n2 •Have seen one “ProgrammableTM” •Random Access Computer TM 15-1: Universal TM •We can create a “Universal Turing Machine”

To be clear, we just showed how from any undecidable language to construct another one which is a counterexample for the claim. If accepts, accept. If rejects, reject. Use diagonalization argument like that for HALT TM. You may Because this is impossible, TM H must not exist, so ATM is undecidable. To be clear, we just showed how from any undecidable language to construct another one which is a counterexample for the claim. If accepts, accept. If rejects, reject. Use diagonalization argument like that for HALT TM. You may Because this is impossible, TM H must not exist, so ATM is undecidable. know that ATM is unsolvable. The key is thus how to reduce ATM to HALTTM. Assume that R is the TM that decides HALTTM, i.e., for any given M and w, R will tell us if M halts on w. We will construct another TM S, that decides if M accepts w : For any given M and w, we run R ﬁrst, if R says M does not halt on w, S rejects hM,wi, since then there is

COMPSCI350: Computability, Complexity and Quantum Computing11 / 105 Solving problems by testing the halting problem 2 A conjecture is nitely refutableif verifying a nite number of A language is Turing-decidable if it halits in an accepting state for every input in the language, and halts in a rejecting state for every other input. Intuitively, for recognizability we allow our TM to run forever on inputs that are not in the language, while for decidability we require that the TM halt on every input.

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May 06, 2019 · This article introduces undecidable languages by using reduction and computation history. In addition, Linear Bounded Automaton (LBA) is also introduced. Thm. HALT TM. is undecidable.Proof by contradiction. Assume that HALT. TM is decidable, and some TM M halt. decides it.Construct a TM M. ATM that decides A TM:But is A. TM is undecidable, a contradiction. So the assumption is false and HALT TM. is undecidable.

Feb 04, 2018 · TOC: Undecidability of the Halting Problem Topics discussed: This lecture shows how can we prove the Undecidability of the Halting problem. Contribute: http:... If input TM p will not halt on i, then halt halts and outputs “false”. So, we can view halt as a function that takes parameters p and i and produces the output “true” or “false” depending on whether or not TM p halts on input i. Based on halt, we can trivially construct a new TM, which we can call trouble, which will But the latter language is undecidable by Rice’s theorem. This contradiction shows that the language in question is undecidable. It is recognizable, because a recognizer could simply run Mon and accept whenever Mhalted. Because the language is undecidable and recognizable, it cannot be co-recognizable. C.D. For brevity, call the language A.

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steps. Halt rejecting if ever it makes a move left (right), and halt accepting at the end of the simulation otherwise. By construction this is a halting TM. Moreover, when we are waiting for a move left (right) by the simulated M, this machine is steadily moving right (left). It may make at most lsuch steps before meeting a It's not showing that halt() "becomes" undecidable. It's showing that if we think we have a TM that solves the halting problem, then we can always break it. So, it must be the case that we don't have a TM that solves the halting problem. \$\endgroup\$ – Mark Reitblatt Dec 16 '10 at 3:44

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Sep 28, 2014 · I think it is recursive enumerable by definition because if TM take a input string and it weather it test a string and say that Yes its present in LANG but if there is string and which TM unable to say weather string is in given language or TM goes in infinite loop i.e. never halt then correspondnce class of language is called Recursive enumerable. Apr 07, 2020 · To prove a certain language is undecidable, we can come up with a reduction that reduce a known undecidable language (A) to our target (B). This means that, we can solve B, by running A is its subroutine(sub-function). Some popular undecidable problems: The acceptance problem for Turing machines is undecidable, know that ATM is unsolvable. The key is thus how to reduce ATM to HALTTM. Assume that R is the TM that decides HALTTM, i.e., for any given M and w, R will tell us if M halts on w. We will construct another TM S, that decides if M accepts w : For any given M and w, we run R ﬁrst, if R says M does not halt on w, S rejects hM,wi, since then there is

A language is Turing-decidable if it halits in an accepting state for every input in the language, and halts in a rejecting state for every other input. Intuitively, for recognizability we allow our TM to run forever on inputs that are not in the language, while for decidability we require that the TM halt on every input.

The M recognizes the complement of L. On the other hand, suppose L′ is Turing recognized by M′ and co-Turing recognized by M′′. Then let D be the machines which on input w simulates each of M′ and M′′ ﬁrst for 1 step, then for 2 steps, etc. If M′ ever accepts the D accepts and if M′′ ever accepts then D rejects. know that ATM is unsolvable. The key is thus how to reduce ATM to HALTTM. Assume that R is the TM that decides HALTTM, i.e., for any given M and w, R will tell us if M halts on w. We will construct another TM S, that decides if M accepts w : For any given M and w, we run R ﬁrst, if R says M does not halt on w, S rejects hM,wi, since then there is A proof which uses this general statement to prove that a particular problem is undecidable is a proof by reduction – in your case, you prove the undecidability \$\mathit{HALT}_{\mathit{TM}}\$ by reducing a problem you already know is undecidable to \$\mathit{HALT}_{\mathit{TM}}\$.

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Reducibility & Undecidable Problems Overview. Last time we talked about decidable problems relating to finite automata and context-free grammars. To be clear, we just showed how from any undecidable language to construct another one which is a counterexample for the claim. If accepts, accept. If rejects, reject. Use diagonalization argument like that for HALT TM. You may Because this is impossible, TM H must not exist, so ATM is undecidable. – Given the above definition, no final state of a TM need to have any transitions. Henceforth, this is our assumption. – If x is NOT in L(M) then M may enter an infinite loop, or halt in a non-final state. – Some TMs halt on ALL inputs, while others may not. In either case the language defined by TM is still well defined. To be clear, we just showed how from any undecidable language to construct another one which is a counterexample for the claim. If accepts, accept. If rejects, reject. Use diagonalization argument like that for HALT TM. You may Because this is impossible, TM H must not exist, so ATM is undecidable.

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A linear bounded automaton (LBA) is a TM that is permitted to use only the portion of the tape containing the input (i e cannot move the head beyond the input)the input (i.e., cannot move the head beyond the input). Lemma Let Mbe an LBA with qstates and gsymbols in the tape alphabet.

Dec 16, 2016 · Halt on every input means it is a halting TM. Which halts on infinite no. of string. Now, complement of these it is "TM halt on empty string". That is RE. • So, either M or M' will halt and accept • If M halts and accepts then halt and accept • If M' halts and accepts then halt and reject The TM that runs M and M' simultaneously always halts and accepts or rejects so it decides L and L is recursive. If L is recursively enumerable and L is not recursive then L' is not recursively enumerable.

HALTtm = {<M, w> | M is a TM and M halts on input w} use undecidability of Atm to prove the undecidability of the halting problem (recognizable) Answer better than no answer whether it is good answer or bad answer If H is halting TM for HALT_{TM}, U is halting TM for A_{TM} because A_{TM} is undecidable, HALT_{TM} is undecidable recognizable That means there will be a Turing machine that says 'yes' if the input is P2 but may or may not halt for the input which is not in P2. As we know that one can convert an instance of w in P1 to an instance x in P2. Then apply a TM to check whether x is in P2. If x is accepted that also means w is accepted. Why? HALT and the (Complement of HALT) are both undecidable and disjoint, and because of disjointness L1 ∪ Halt and L2 ∪ Halt are both also undecidable. 2. We say a function f:N −→ N is total computable if there is TM M which start with an natural number n on its tape and computes f(n) on its tape (and halts).

To be clear, we just showed how from any undecidable language to construct another one which is a counterexample for the claim. If accepts, accept. If rejects, reject. Use diagonalization argument like that for HALT TM. You may Because this is impossible, TM H must not exist, so ATM is undecidable. Thm. HALT TM. is undecidable.Proof by contradiction. Assume that HALT. TM is decidable, and some TM M halt. decides it.Construct a TM M. ATM that decides A TM:But is A. TM is undecidable, a contradiction. So the assumption is false and HALT TM. is undecidable. IEvery nite language is decidable: For example, by a TM that has all the strings in the language \hard-coded" into it. IWe just saw some example algorithms all of which terminate in a nite number of steps, and output yes or no (accept or reject). i.e., They decide the corresponding languages. M is a tm and m accepts at least two different strings (source: on YouTube) M is a tm and m accepts at least two different strings ...

COMPSCI350: Computability, Complexity and Quantum Computing11 / 105 Solving problems by testing the halting problem 2 A conjecture is nitely refutableif verifying a nite number of

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Can you be a nurse with a felony in texasSince this problem is undecidable, does that mean there isn't one algorithm that works for every instance, or does it even mean there is an instance for which we cannot arrive at an answer? i.e. is there a program you can write for which it would make sense to say: "this program doesn't halt or not halt, it's undecidable". know that ATM is unsolvable. The key is thus how to reduce ATM to HALTTM. Assume that R is the TM that decides HALTTM, i.e., for any given M and w, R will tell us if M halts on w. We will construct another TM S, that decides if M accepts w : For any given M and w, we run R ﬁrst, if R says M does not halt on w, S rejects hM,wi, since then there is • This is an undecidable problem. Theorem 5.1 HALT TM = { ¢M, w ²| M is a TM and halts on input w } is Turing-recognizable, but not decidable. Proof. U of Theorem E is easy to convert into a TM that simulates the computation of M on input w and accepts if and only if the computation being simulated halts. Thm. HALT TM. is undecidable.Proof by contradiction. Assume that HALT. TM is decidable, and some TM M halt. decides it.Construct a TM M. ATM that decides A TM:But is A. TM is undecidable, a contradiction. So the assumption is false and HALT TM. is undecidable.

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If H returns NO, then halt. The following is the block diagram of an ‘Inverted halting machine’ − Further, a machine (HM) 2 which input itself is constructed as follows − If (HM) 2 halts on input, loop forever. Else, halt. Here, we have got a contradiction. Hence, the halting problem is undecidable. Sep 28, 2014 · I think it is recursive enumerable by definition because if TM take a input string and it weather it test a string and say that Yes its present in LANG but if there is string and which TM unable to say weather string is in given language or TM goes in infinite loop i.e. never halt then correspondnce class of language is called Recursive enumerable.

TM Compute Decider Reduction f(M,w) Mˆ,q,w HALT TM YES NO Given the reduction, if is decidable, then is decidable STATE TM HALT TM A contradiction! since is undecidable Halting Problem Decider HALT TM Another undecidable problem. • ETM = {⟨M⟩⃒M is a TM and L(M) = ∅} is undecidable. • Proof is again by reduction from A TM: we suppose TM R decides E TM and use it to define a TM that decides A TM as follows: ‣Check that input has form ⟨M,w⟩; if not, reject.

We can understand Undecidable Problems intuitively by considering Fermat’s Theorem, a popular Undecidable Problem which states that no three positive integers a, b and c for any n>2 can ever satisfy the equation: a^n + b^n = c^n. If we feed this problem to a Turing machine to find such a solution which gives a contradiction then a Turing ... HALT TM is Undecidable. HALT TM = {<M, w> | M is a TM and M halts on input w} Prove that HALT TM is undecidable: Assume that TM R is a DECIDER for HALT TM. Construct TM S to decide A TM, as follows: "On input <M, w>, an encoding of a TM M and a string w: Run TM R on input <M, w> to see if M halts on w

HALT TM • HALT TM ={<M,w>| M is a TM and halts on input w} – It is undecidable • We need to reduce A TM to HALT TM, where A TM already proven to be undecidable – Can use HALT TM to solve A TM • Proof by contradiction – Assume HALT TM is decidable and show this implies A TM is decidable 10/17/19 Theory of Computation - Fall'19 • So, either M or M' will halt and accept • If M halts and accepts then halt and accept • If M' halts and accepts then halt and reject The TM that runs M and M' simultaneously always halts and accepts or rejects so it decides L and L is recursive. If L is recursively enumerable and L is not recursive then L' is not recursively enumerable.