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Loebian cooperation in the tiling agents problem
post by Vladimir Slepnev 53 days ago | Alex Mennen, Vadim Kosoy, Abram Demski, Patrick LaVictoire and Stuart Armstrong like this | 4 comments

The tiling agents problem is about formalizing how AIs can create successor AIs that are at least as smart. Here’s a toy model I came up with, which is similar to Benya’s old model but simpler. A computer program X is asked one of two questions:

  • Would you like some chocolate?

  • Here’s the source code of another program Y. Do you accept it as your successor?


Program X will be called a “winner” if is says “yes” to chocolate, every program Y accepted as successor by X also says “yes” to chocolate, every program Z accepted as successor by Y also says “yes” to chocolate, and so on. Otherwise X will be called a “loser”.

Let’s also assume that X has access to a provability oracle for Peano arithmetic (PA), to avoid dealing with logical uncertainty.

Here’s some simple winning programs:

  1. Always say “yes” to chocolate. Refuse to accept any successor.

  2. Always say “yes” to chocolate. Accept Y as successor iff Y has the same source code as X.

  3. Always say “yes” to chocolate. Accept Y as successor iff PA proves that Y is a winner.

Unfortunately (1) is too restrictive, (2) is too sensitive to implementation details, and (3) refuses to accept itself as successor for Goedelian reasons. How do we write a more general winning program?

Note that all programs above are analogous to strategies in the Prisoner’s Dilemma with visible source code: (1) is defection, (2) is quining cooperation, and (3) is simulating the opponent. But in the PD we also have Loebian cooperation! We could try applying it to our problem naively, like this:

  1. Always say “yes” to chocolate. Accept Y as successor iff PA proves that “X is a winner” implies “Y is a winner”.

At first glance that program looks nice. It accepts itself as successor, and also accepts all programs that PA proves to be winners. Unfortunately it also accepts all losers, because the sentence “X accepts Y” is provable for Loebian reasons.

But not all is lost! Here’s a better way to apply Loebian cooperation to our problem:

  1. Always say “yes” to chocolate. Accept Y as successor iff PA proves that Y can’t lose in fewer steps than X.

(Losing in one step means saying “no” to chocolate, losing in two steps means accepting a program that says “no” to chocolate, and so on. Our program X will accept a program Y iff the following is provable in PA: for all n such that X can’t lose in n steps, Y also can’t lose in n steps.)

Proof that X is a winner: X can’t lose in one step by construction. Therefore Y can’t lose in one step, as long as PA is sound. Therefore X can’t lose in two steps. Therefore Y can’t lose in two steps, as long as PA is sound. And so on, QED.

It’s easy to see that X accepts itself as successor, as well as modifications of itself, and also accepts all programs that PA proves to be winners. That seems nice. And the condition “Y can’t lose in fewer steps than X” also makes sense philosophically, because it means “X doesn’t move toward losing”, which gives hope that it can be generalized.

Some followup questions: I don’t know how this approach will work on more complicated problems, whether it’s related to Benya’s parametric polymorphism, and whether the analogy between self-modification and cooperation will turn out to be fruitful (even if they are difficult for the same reasons, which seems obvious now).

What do you think?



by Vladimir Slepnev 47 days ago | Vadim Kosoy likes this | link

Wei has proposed this program by email:

  1. Always say “yes” to chocolate. Accept Y as successor iff PA proves that Y always says “yes” to chocolate and Y accepts a subset of successors that X accepts.

Figuring out the relationship between 5 and 6 (do they accept each other, are they equivalent) is a really fun exercise. So far I’ve been able to find a program that’s accepted by 5 but not 6. Won’t spoil people the pleasure of figuring it out :-)

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by Stuart Armstrong 50 days ago | link

I’m not sure this would work, and it might be tied to ambiguity about what “steps” mean.

Consider:

Y: Run X to completion. Then say “no” to chocolate.

Then PA proves that Y doesn’t lose in less steps than X (since X doesn’t do anything in more than N steps while Y runs N+1 steps before taking action), yet it’s clear that Y loses.

I think it’s because “lose in n steps” is not clear.

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by Vladimir Slepnev 50 days ago | Stuart Armstrong likes this | link

It doesn’t mean computation steps. Losing in 1 step means you say “no” to chocolate, losing in 2 steps means you accept some program that says “no” to chocolate, and so on. Sorry, I thought that was the obvious interpretation, I’ll edit the post to make it clear.

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by Stuart Armstrong 50 days ago | link

Ah, thanks! That seems more sensible.

reply



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