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Let a language A = {(M,w) : M is a TM and w is a string such that w is accepted by M in an even number of steps}.

How can I prove that this is undecidable? I have considered trying to build the ATM proof assuming this language. Language A can cover acceptance under even steps but how would I cover acceptance under uneven steps? Language A would reject in that case. How would you go about trying to build the proof via this path?

Another thing I wonder is if Rice's theorem covers this language. Would this be considered a functional property? Could I use it to prove the undecidability of this language?

Thanks

Raphael
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Ricardo Ferreira da Silva
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2 Answers2

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This is a bit of a nit-picky question, since you really need to pay attention to the parity of the number of steps you TM does.

We'll show that the problem is undecidable by a reduction from $A_{TM}$. There are many ways to construct such a reduction. In my opinion, the following is the least technical.

Given input $(M,w)$ for $A_{TM}$, we will construct a new output $(M',\epsilon)$, such that $M$ accepts $w$ iff $M'$ accepts $\epsilon$ within an even number of steps.

We construct $M'$ as follows: given input $x$, if $x\neq \epsilon$, then $M'$ rejects.

Otherwise, if $x=\epsilon$, $M'$ works as follows: simulates the run of $M$ on $w$. If $M$ accepts, it clears the tape by going to the rightmost non-blank cell, and then writes blanks all the way to the left. It then simulates $M$ again, erases the tape again in exactly the same manner, and accepts.

If $M$ rejects, then $M'$ rejects as well.

The idea here is that if $M$ accepts $w$, then the operation of $M'$ on $\epsilon$ is to simulate $M$ on $w$ twice. Therefore, it will take an even number of steps.

Proving the correctness and computability of this procedure is not hard.

Shaull
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In short, if you could decide membership in the language $A$, then you could decide whether a Turing machine accepts a word. Because TM acceptance is undecidable, this proves that a decider for $A$ can't exist.

Here's a machine that uses a subroutine for $A$ in order to solve the acceptance problem. Given a machine and word $\langle M, w\rangle$, design a new machine $M^\prime$. The new machine is exactly the same as $M$, except that it has an extra step in the beginning where it does nothing. (e.g. $M^\prime$ has a new starting state $q_0^\prime$. Initially, the machine reads the tape, prints nothing, doesn't move, and transitions into the start state $q_0$ of $M$.)

Now, if $M$ rejects $w$ then so does $M^\prime$. Otherwise, they both accept $w$, and $M^\prime$ takes one more step than $M$— so one of the two machines takes an even number of steps to accept.

So, if we could decide whether a TM accepts a word in an even number of steps, we could decide whether $M$ accepts $w$. Just test $\langle M, w\rangle$ and $\langle M^\prime, w\rangle$ and return ACCEPT if either one takes an even number of steps to accept, or else return REJECT.

Rice's theorem doesn't apply because "accepting in an even number of steps" is not a property that languages have, but a property that specific Turing machines have. (Two Turing machines can have the same language, but one of them might belong to $A$ and the other one not.) Rice's theorem applies just to languages of the form

$$\{\langle M\rangle : M\text{ is a TM and }L(M)\text{ has property }p \}$$

user326210
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