Is $L=\{<M> | L(M) \ is \ finite\} $ decidable ? M is a TM.

I think its relative simple to proof with the theorem of rice. But I am interested in a solution which does not use the Rice theorem.

This my try : Let f(<m,w>) be a function which works in the following way :

  1. Run w on M
  2. If M accepts Construct a TM Mwhich accepts only the word w and return M
  3. If M rejects Construct a TM Mwhich accepts everything. Return M

So if m is in $A_{TM}= \{<M,w>|M \ accepts \ w\}$ we know that f(<m,w>) is in L. If m is not in A then we know that f(<m,w>) does accept every word and therefore infinity words. So f(<m,w>) not in L.

Is this a correct mapping reduction ?


The function you defined is not a reduction at all - it may not even stop!

The problem is running $m$ on $w$: can you be sure $m$ wont be stuck in an infinite loop on $w$? you can't.

You can define a proper reduction as follows: (on input $<m,w>$)

Create the machine $M_{m,w}$that does the following algorithm, and return in: (on input $s$)

  1. Emulate $m$ on $w$ for $|s|$ steps. If $m$ halted in that time, reject $s$. Otherwise, accept $s$.

I will leave it for you to prove this is a proper reduction from $H_{TM}$ to $L$ (its a good exercise!)

  • $\begingroup$ I dont understand your solution. I think if I edit the third step from my solution it should be fine, right ? I would edit step 3 to this: If m loops or reject construct a TM Mwhich accepts everything. Return M $\endgroup$ – Frank Jul 17 '20 at 11:14
  • $\begingroup$ You can't know whether something loops or not - thats the halting problem, and as we all know - its not decidable! $\endgroup$ – nir shahar Jul 17 '20 at 11:15
  • $\begingroup$ Could maybe provide the solution to your given answer ? I am not able to solve it. $\endgroup$ – Frank Jul 22 '20 at 9:05

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