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:
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:
Always say “yes” to chocolate. Refuse to accept any successor.
Always say “yes” to chocolate. Accept Y as successor iff Y has the same source code as X.
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:
- 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:
- 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?