Relations between deciding languages and computing functions in advice machines


I’m trying to understand implications of translating between functions and languages for P/Poly complexity. I’m not sure whether the following all makes sense. Giving it my best shot given my current understanding of the concepts. (I have a project in which I want to discuss Hava Siegelmann’s analog recurrent neural nets, which recognize languages in P/Poly, but I’d like to understand and be able to explain to others implications this has for computing functions.)

Suppose I want to use an advice Turing $ T_1$ machine to calculate a function from binary strings to binary strings $ f: {0,1}* \rightarrow {0,1}*$ . $ T_1$ will be a machine that can compute $ f$ in polynomial time given advice that is polynomial-size on the length of arguments $ s$ to $ f$ , i.e. $ f$ is in P/Poly. (Can I say this? I have seen P/Poly defined only for languages, but not for functions with arbitrary (natural number) values.)

Next suppose I want to treat $ f$ as defining a language $ L(f)$ , by encoding its arguments and corresponding values into strings, where $ L(f) = \{\langle s,f(s)\rangle\}$ and $ \langle\cdot,\cdot\rangle$ encodes $ s$ and $ f(s)$ into a single string.

For an advice machine $ T_2$ that decides this language, the inputs are of length $ n = |\langle s,f(s)\rangle|$ , so the relevant advice for such an input will be the advice for $ n$ .


Question 1: If $ T_1$ can return the result $ f(s)$ in polynomial time, must there be a machine $ T_2$ that decides $ \{\langle s,f(s)\rangle\}$ in polynomial time? I think the answer is yes. $ T_2$ can extract $ s$ from $ \{\langle s,f(s)\rangle\}$ , and then use $ T_1$ to calculate $ f(s)$ , and then encode $ s$ with $ f(s)$ and compare it with the original encoded string. Is that correct?


Question 2 (my real question): If we are given a machine $ T_2$ that can decide $ \{\langle s,f(s)\rangle\}$ in polynomial time, must there be a way to embed $ T_2$ in a machine $ T_3$ so that $ T_3$ can return $ f(s)$ in polynomial time?

I suppose that if $ T_2$ must include $ T_1$ , then the answer is of course yes. $ T_3$ just uses the capabilities of $ T_1$ embedded in $ T_2$ to calculate $ f(s)$ . But what if $ T_2$ decides $ L(f)$ some other way? Is that possible?

If we are given $ s$ , we know its length, but not the length of $ f(s)$ . So in order to use $ T_2$ to find $ f(s)$ , it seems there must be a sequential search through all strings $ s_f = \{\langle s,r\rangle\}$ for arbitrary $ r$ . (I’m assuming that $ f(s)$ is unbounded, but that $ f$ has a value for every $ s$ . So the search can take an arbitrary length of time, but $ f(s)$ will ultimately be found.)

One thought I have is that the search for a string $ s_f$ that encodes $ s$ with $ f(s)$ has time complexity that depends on the length of the result $ f(s)$ (plus $ |s|$ , but that would be swamped when $ f(s)$ is long).

So now the time complexity does not have to do with the length of the input, but only the length of $ f(s)$ . Maybe $ L(f)$ is in P/Poly if $ f$ is in P? (Still confused here.)

Thinking about these questions in terms of Boolean circuits has not helped.