[fnrslvr]

This might get you away from the literal statement of Rice's theorem, but we tend to ask questions about our programs' behaviour across all possible inputs, and these questions tend to remain impossible to solve in general.

For example, finding out whether a time-bounded Turing machine outputs 0 all inputs is still undecidable. (coRE-complete, in fact. That's better in some sense than the Pi_2-completeness of the same problem for Turing machines without the time bound, but since neither admits a solution in the form of a computer program, a lot of people would consider the difference an academic curiosity only.)

[a1369209993]

> For example, finding out whether a time-bounded Turing machine outputs 0 all inputs is still undecidable.

No it's not; call the time bound N; the TM can only inspect N symbols, so there are 2^N possible inputs; a exhaustive check takes O(N*2^N) operations.

[fnrslvr]

> call the time bound N

Call the time bound c|x|^k, where x is the input and c and k are constants.

I see what point you're making, and I agree that you get yourself a nice, merely coNP-complete problem if you're willing to cut off the space of inputs at a point by enforcing a constant time bound or whatever. More than anything, this kind of argument makes me shudder at just how hard the very worst stuff in NP must be.

[Dylan16807]

When people talk about "limited time and/or memory" for a runtime they're almost certainly talking about something constant, or a small polynomial of the input size.

But yes, it should reduce to NP territory. If you actually had a nondeterministic CPU, then for most use cases you could fully evaluate these machines in seconds.

[a1369209993]

> limited time [to] a small polynomial of the input size.

This is backwards; I meant that if you have any time bound whatsoever, the number of inputs you can inspect is limited to (a architecture-specific finite multiple of) the number of operations you can execute (because inspecting a input is such a operation).

[Dylan16807]

I wasn't talking about your argument that the calculation is finite. I was talking about how N is chosen in the first place.

[a1369209993]

> to cut off the space of inputs at a point by enforcing a constant time bound

That is literally the problem statement you gave. I was pointing out that it is very much not undecidable.

It might be (almost certainly is, in this case) computationally intractable both in the general case and in practice, but it's still decidable.

[Dylan16807]

> For example, finding out whether a time-bounded Turing machine outputs 0 all inputs is still undecidable.

Is that a problem? I mean, if you're installing a timeout then haven't you already decided that the answer is "no"?

[fnrslvr]

If the time bound is being put in place in the hopes of being able to automatically analyze the behaviour of the code, then no.

I singled out "outputting 0" on every input because it's among the simplest possible specs you could try try to check with the kind of "very powerful semantic analysis" tool the top-level post was hoping for. In practice you probably want to check a more complex spec -- you want your code to solve a specific problem on whatever input it receives -- but if the "always output 0" spec is impossible to automatically check, then what hope do more relevant specs have?