I don’t understand this answer and cousin_it doesn’t either (I just asked him).
The predictor is supposed to one-box, but the agent isn’t supposed to infer (within M steps) that the predictor one-boxes (just as in finite diagonal step, the objective is to make a decision unpredictable, not impossible).
If the predictor outputs “one-box”, then the agent must prove this if it enumerates all proofs within a certain length because there is a proof-by-simulation of this fact, of length proportional to the predictor’s run time, right? I don’t see how the diagonal step can prevent the agent from finding this proof, unless it makes the predictor not output “one-box”.
The agent doesn’t use enough steps to simulate the predictor, it decides early (because it finds a proof that predictor conditionally one-boxes early), which is also what might allow the predictor to conditionally predict agent’s one-boxing within predictor’s limited computational resources. The M steps where the agent protects the predictor from being unconditionally predictable is a very small number here, compared to agent’s potential capability.
I don’t understand this answer and cousin_it doesn’t either (I just asked him).
If the predictor outputs “one-box”, then the agent must prove this if it enumerates all proofs within a certain length because there is a proof-by-simulation of this fact, of length proportional to the predictor’s run time, right? I don’t see how the diagonal step can prevent the agent from finding this proof, unless it makes the predictor not output “one-box”.
The agent doesn’t use enough steps to simulate the predictor, it decides early (because it finds a proof that predictor conditionally one-boxes early), which is also what might allow the predictor to conditionally predict agent’s one-boxing within predictor’s limited computational resources. The M steps where the agent protects the predictor from being unconditionally predictable is a very small number here, compared to agent’s potential capability.