I'm trying to prove that the following language is undecidable:$$ \{ \langle M, w \rangle ~|~ M \text{ is a TM where its head moves left a finite number of times on } w \} $$

But I'm having a bit of trouble. I know I have to do some type of reduction, but I'm not really sure what. Can a Turing Machine be simulated by one which only moves left? if so, then I think I could show $A_{TM}$ reduces to it. Any hints would be appreciated.


Hint: Reduce from the halting problem. Given a Turing machine $M$ and an input $w$, you want to come up with a new pair $\langle M',w \rangle$ such that $M$ halts on $w$ iff $M'$ makes a finite number of left moves on $w'$. Now if $M$ halts on $w$ then in particular it makes a finite number of left moves, but the other direction is not true. Think of a way of ensuring that if $M$ makes an infinite number of moves in any direction, then $M'$ makes an infinite number of left moves. (You might want to add some spurious moves.)

  • $\begingroup$ I've been working on this, but still don't quite have it. Why should $M'$ make an infinite number of left moves if $M$ makes infinite moves in any direction? Shouldn't $M'$ halt if $M$ makes and infinite number of right moves? $\endgroup$ Nov 11 '13 at 20:00
  • $\begingroup$ and what do you mean by spurious moves? I've never heard that phrase before. $\endgroup$ Nov 11 '13 at 20:01
  • $\begingroup$ It's not a technical phrase. It just means moves which are necessary for the computation, but can help with ensuring that the machine makes infinitely many left moves. $\endgroup$ Nov 11 '13 at 20:02
  • $\begingroup$ I've been thinking about using using $M'$ to create another TM $M''$ to check if $M$ makes a finite number of right moves on $w$. Then the halting problem could be solved by checking if $M'$ and $M''$ accept (the contradiction). Am I on the right track? I really appreciate all the help. $\endgroup$ Nov 11 '13 at 23:08
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    $\begingroup$ I can't follow you. If you have a concrete suggestion, you can try to prove that it works. If you can't seem to prove that it works, perhaps it doesn't. $\endgroup$ Nov 11 '13 at 23:45

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