# The limited predictor problem

post by cousin_it · 2012-03-21T00:15:26.176Z · score: 10 (11 votes) · LW · GW · Legacy · 8 comments## Contents

1. Solving the problem when the agent has a halting oracle 2. Failing to solve the problem when N is algorithmically random 3. Solving the problem when N is large but has a short definition 4. A broader view None 8 comments

This post requires some knowledge of logic, computability theory, and K-complexity. Much of the credit goes to Wei Dai. The four sections of the post can be read almost independently.

The limited predictor problem (LPP) is a version of Newcomb's Problem where the predictor has limited computing resources. To predict the agent's action, the predictor simulates the agent for N steps. If the agent doesn't finish in N steps, the predictor assumes that the agent will two-box. LPP is similar to the ASP problem, but with simulation instead of theorem proving.

#### 1. Solving the problem when the agent has a halting oracle

Consider the agent defined in "A model of UDT with a halting oracle", and a predictor that can run the agent's code step-by-step, with oracle calls and all. Turns out that this agent solves LPP correctly if N is high enough. To understand why, note that the agent offloads all interesting work to oracles that return instantly, so the agent's own runtime is provably bounded. If that bound is below N, the agent's oracle will prove that the predictor predicts the agent correctly, so the agent will one-box.

#### 2. Failing to solve the problem when N is algorithmically random

Consider a setting without oracles, with only Turing-computable programs. Maybe the agent should successively search for proofs somehow?

Unfortunately you can't solve most LPPs this way, for a simple but surprising reason. Assume that the predictor's time limit N is a large and algorithmically random number. Then the predictor's source code is >log(N) bits long, because N must be defined in the source code. Then any proof about the world program must also have length >log(N), because the proof needs to at least quote the world program itself. Finding a proof by exhaustive search takes exponential time, so the agent will need >N steps. But the predictor simulates the agent for only N steps. Whoops!

#### 3. Solving the problem when N is large but has a short definition

As usual, let U be the world program that returns a utility value, and A be the agent program that returns an action and has access to the world's source code. Consider the following algorithm for A:

- From L=1 to infinity, search for proofs up to length L of the form "if A()=a and runtime(A)<g(L), then U()=u", where g(L) is an upper bound on runtime(A) if A stops the search at length L. Upon finding at least one proof for each possible a, go to step 2.
- Search for proofs up to length f(L) of the form "if runtime(A)<g(L), then A()≠a", where f(L) is some suitably fast-growing function like 10^L. If such a proof is found, return a.
- If we're still here, return the best a found on step 1.

This algorithm is very similar to the one described in "A model of UDT without proof limits", but with the added complication that A is aware of its own runtime via the function g(L). By an analogous argument, A will find the "intended" proof that the predictor predicts A correctly if runtime(A) is small enough, as long as the "intended" proof exists and isn't too long relative to the predictor's time limit N. More concretely, A will solve all instances of LPP in which N is larger than g(L), where L is the length of the "intended" proof. For example, if f(L)=10^L, then g(L) is doubly exponential, so A will successfully solve LPPs where the predictor's source code defines N using triple exponentials or some more compact notation.

#### 4. A broader view

TDT and UDT were originally designed for solving "decision-determined" problems. The agent figures out how the resulting utility logically depends on the agent's action, then returns the action with the highest utility, thus making the premise true.

But a cleverly coded decision program can also control other facts about itself. For example, the program may figure out how the resulting utility depends on the program's return value *and running time*, then choose the best return value *and choose how long to keep running*, thus making both premises true. This idea is a natural extension of quining (you carefully write a program that can correctly judge its own runtime so far) and can be generalized to memory consumption and other properties of programs.

With enough cleverness we could write a program that would sometimes decide to waste time, or run for an even number of clock cycles, etc. We did not need so much cleverness in this post because LPP lies in a smaller class that we may call "LPP-like problems", where utility depends only on the agent's return value and runtime, and the dependence on runtime is monotonous - it never hurts to return the same value earlier. That class also includes all the usual decision-determined problems like Newcomb's Problem, and our A also fares well on those.

I was surprised to find so many new ideas by digging into such a trivial-looking problem as LPP. This makes me suspect that advanced problems like ASP may conceal even more riches, if only we have enough patience to approach them properly...

## 8 comments

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In section 2, you say:

Unfortunately you can't solve most LPPs this way [...]

By solving most LPPs, do you mean writing a general-purpose agent program that correctly maximizes its utility function under most LPPs? I tried to write a program to see if I could show a counterexample, but got stuck when it came to defining what exactly a solution would consist of.

Does the agent get to know N? Can we place a lower bound on N to allow the agent to time to parse the problem and become aware of its actions? Otherwise, wouldn't low N values force failure for any non-trivial agent?

Can we place a lower bound on N to allow the agent to time to parse the problem and become aware of its actions?

Yes, of course that's allowed. I started out thinking that I could write a program that solved LPP for all N above a certain bound, and then found the obstacle described in section 2.

Could you explain a little more why explicitly representing the running time is helpful for the LPP? Is it true that the agent without the "runtime(A)<g(L)" terms fails to solve the problem, while this one succeeds? I don't understand how it is possible. The running time bound would still be there in any proof of "if A()=a then U()=u", except it would be implicit...

I guess the question is whether any proof of "if A()=a then U()=u" will be found at all. (The formulation of LPP makes it easier to prove something like "if A()=a and runtime(A)<N, then U()=u".) But maybe you're right and such a proof will be found. It's a little tricky...

Hmm. It seems to me, the following holds:

(agent proves "if A()=a then U()=u") => "if A()=a and runtime(A)<N then U()=u"

(agent proves "if A()=a and runtime(A)<N then U()=u") => "if A()=a then U()=u"

where the right parts are logically implied, but not necessarily proved by the agent, although in almost all cases they would be, the proof steps from one to the other being so small...

In any case, the idea of logically controlling the non-obvious properties of computation is splendid, thanks for writing it up!

Thanks for the compliment ;-) but I still don't completely understand your comment. If you mean the version of A that doesn't refer to runtime(A) explicitly, what's the short proof of the second statement?

"if A()=a and runtime(A)<N then U()=u" can only be proved by an agent that represents runtime() explicitly, which means it would probably find (relatively short) general proofs of the form
(max proof length) runtime < some function of L,

and

agent proves X => length of proof of X < max proof length.

Although, now that I think of it more, this does not help, as in this setting the agent may easily prove X but not "agent proves X"... You're right, the proof of the second statement is not obviously short.

Then the predictor's source code is >log(N) bits long, because N must be defined in the source code.

N = 3^^^3 ?