In the alternative algorithm for the five-and-ten problem, why should we use the first proof that we find? How about this algorithm:

A2 :=
  Spend some time t searching for proofs of sentences of the form
  "A2() = a → U() = x"
  for a ∈ {5, 10}, x ∈ {0, 5, 10}.
  For each found proof and corresponding pair (a, x):
    if x > x*:
      a* := a
      x* := x
  Return x*

If this one searches long enough (depending on how complicated U is), it will return 10, even if the non-spurious proofs are longer than the spurious ones.

I guess then it would have to prove that it will find a proof with x > 0 within t. This is difficult.

Decision Theory

by abramdemski, Scott Garrabrant 1 min read31st Oct 201837 comments

101

Ω 24


Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.

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