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Questions about problems which cannot be solved by any Turing machine.

4
votes
Let $A_w=\{\langle\,M\,\rangle\mid L(M)=\{w\}\}$ and $\mathrm{HALT}=\{(\langle\,M\,\rangle, w)\mid M\text{ halts on }w\}$. It's known that the complement, $\overline{\mathrm{HALT}}$, is unrecognizable …
answered Dec 23 '15 by Rick Decker
1
vote
$A$ is recognizable by assumption. Since $\overline{A}=B\cup C$ is recognizable (recognizable languages are closed under union) we have that $A$ and $\overline{A}$ are recognizable, and hence $A$ must …
answered Feb 24 '17 by Rick Decker
4
votes
Hint: The language $\{\langle\,M\,\rangle\mid L(M)\text{ is unrecognizable}\}$ is empty (why?) and the empty language is decidable (why?).
answered Feb 3 '16 by Rick Decker
4
votes
In answer to your edited question, If we accept the definition $L(M)$ as the set of all strings a given Turing Machine $M$ can accept, when I ask can we construct it, in the simplest way I suppose …
answered Sep 15 '14 by Rick Decker
2
votes
$RE$ and $co-RE$ are sets of sets of strings. $H_{TM}$ and $\overline{A_{TM}}$ are sets of strings. You correctly have that $H_{TM}\in RE$ and $\overline{A_{TM}}\in co-RE$. However, you cannot then co …
answered Jun 21 '15 by Rick Decker
2
votes
Your intuition is correct: your $L$ is unrecognizable. Here's a hint on how to proceed. If we define $A_\text{TM}= \{(\langle M\rangle,w)\mid M \text{ accepts }w\}$, then it's known that its compleme …
answered Jul 11 '16 by Rick Decker
1
vote
Since the problem didn't restrict $x$ or $y$, let $x$ be any string and let $y=x$. Then $M$ either accepts $x$ or it doesn't. If $M$ accepts $x$ (and so accepts $y$) then $\langle M, x, y\rangle\in L$ …
answered Mar 3 '15 by Rick Decker
10
votes
Major edit of my original: A naive reading of your question seems to be, let $P$ be the problem $P=$ Given a language, $L$, is it decidable? Then you ask Is $P$ decidable? As D.W. and D …
answered May 6 '15 by Rick Decker
2
votes
Assuming you have a decider $R$ for DISJOINT, you could use this to make a decider D for $E_\text{TM} = \{\langle M\rangle\mid L(M)=\emptyset\}$ as follows: D(<M>) = return R(<M>, <A>) where $A$ …
answered Nov 20 '14 by Rick Decker
1
vote
The language $E=\{\langle\,M\,\rangle\mid M \text{accepts no string}\}$ is known to be not recursively enumerable. Consider a reduction $E\le L$ defined by $$ \langle\,M\,\rangle\mapsto (\langle\,M\,\ …
answered Feb 2 '16 by Rick Decker
1
vote
You're in luck here: since every transition of your TM moves the head right, the state of this machine will never depend on the previous contents of the tape. This means that your TM will act as if it …
answered May 17 '14 by Rick Decker
1
vote
Let's look just at $M_2$ for the moment. If $M$ accepts $w$, then $M$ will accept strings of the form $0^n1^n$ in step (1) and everything else in step (2). In other words If $M$ accepts $w$, then the …
answered Apr 27 '17 by Rick Decker
2
votes
The problem is that your proposed $D$ either isn't a TM or doesn't lead to a contradiction. Suppose we did as you suggested, defining a machine $D$ that required as input a pair $(\langle\,M\,\rangle, …
answered Jan 2 '16 by Rick Decker
1
vote
To expand on Stephen's last remark, a classic non-constructive proof (though not about graphs) is the conjecture For every two irrational numbers $r, s$ the value of $r^s$ is irrational. This is …
answered Jun 10 '14 by Rick Decker
4
votes
We have the language (commonly called $A_\text{TM}$): $$ A_\text{TM}=\{(\langle M\rangle, w)\mid M\text{ is a TM and $M$ accepts $w$}\} $$ and we want to show that this language is undecidable. We pr …
answered Dec 10 '16 by Rick Decker

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