This question is part of the undecidable lecture by Jeff Erickson.

$$\{\langle M\rangle\mid M \text{ accepts at least one string }w\text{ in space }|w|^2\}$$

We should prove that this language is undecidable. So what i tried is to create a new TM $M'$ where $M'$ moves $|w|^2$ steps to the right and than tries to accept $w$. So in my opinion that would solve the Accept Problem which we know is undecidable.

My problem with this is that we don't actually get $w$ as an input we just ask if there is a word that accepts after $|w|^2$ steps. And i don't really know how to handle this.


1 Answer 1


Let $\text{AnySpaceSquare}$ denote the language in the question.

Let us reduce the well-known undecidable language $\text{Accept} :=\{\langle M, w\rangle\mid\text{Turing machine }M\text{ accepts }w\}$ to $\text{AnySpaceSquare}$.

Suppose a TM $M$ and a word $w$ is given. Construct TM $M'$ so that upon any input, it will simulate $M$ on input $w$, ignoring the input completely.

  1. Upon any input $v$, $M'$ erases that input.
  2. $M'$ writes $w$ on the tape. (So, $M'$ should have $w$ hard-encoded in its states and transitions.) Up to now, $M'$ should use at most $\max(|v|,|w|)+1$ spaces.
  3. $M'$ runs as if it were $M$.
  4. If the simulated $M$ accepts/rejects/runs forever/whatever, so does $M'$ respectively.

Let us check the map $\langle M, w\rangle\to \langle M'\rangle$.

  • If $\langle M, w\rangle\not\in \text{Accept}$, $M'$ never accepts.
    So $\langle M'\rangle\not\in\text{AnySpaceSquare}$.
  • Otherwise, $\langle M, w\rangle\in \text{Accept}$.
    Let us say $M$ accepts $w$ in $n$ steps. These $n$ steps use at most $n+1$ spaces. Pick any word $v$ that is long enough, $M'$ should accept $v$ within $|v|^2$ spaces. In fact, any word at least $n+|w|+1$ long would be fine.
    So, $\langle M'\rangle\in\text{AnySpaceSquare}$.

Hence, if we could decide $\text{AnySpaceSquare}$, we could also decide $\text{Accept}$.

Exercise. $ \{\langle M\rangle\mid M \text{ accepts at least one string }w\text{ in time }|w|^2\}$ is undecidable.


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