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I want to show that

$\qquad L = \{\langle M \rangle \mid \text{TM $M$ will write a $z$ to the tape at some point for some input}\}$

is undecidable.

I'm really not sure how to show this is undecidable. I know that a language is undecidable if M would not be able to determine if 'z' is or is not in the language.

Raphael
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susangel15
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    @Raphael Being a beginner myself, I am unable to see how the link is able to provide answer to this question. Am I missing something ? – advocateofnone Nov 08 '15 at 00:07
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    As a hint you can do this proof almost exactly like the halting problem. Roughly you just have to replace "loop" with "print z" – Jake Nov 08 '15 at 00:13
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    @sasha The reference question gives techniques for proving claims such as this, at least one of which works. (Hint: reduce from the Halting problem. Examples can be found via [tag:computability+reductions].) – Raphael Nov 08 '15 at 08:00
  • Whether 'z' is in the language has little to do with whether $M$ writes a 'z' to its tape: it might be that writing a 'z' to the tape is part of the intermediate calculations $M$ uses when deciding whether to accept an input string. – Rick Decker Nov 08 '15 at 18:18
  • @sasha. You're quite right. While many of our reference questions are quite good, they aren't as helpful as they could be for visitors to this site, especially newcomers, who are looking for a nudge in the right direction. Voting to close a question because there is an extended tutorial available that talks about some general solutions is no more helpful than if a student in my class came to me saying "I don't quite understand how to do problem 3 on the homework" and I responded "Read the text. Now get out of here." – Rick Decker Nov 08 '15 at 19:59

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