>> Correct me if wrong, but isn't the halting
problem only
>> undecidable when the length of the program is unbounded? Wouldn't the AI assign non-zero >> probability to a machine that solved the halting problem for >> programs up to size S? (S is the number of stars in the sky, grains of sand, >> atoms in the universe, etc...) As an aside, this would actually be my best guess as to >> what was really going on if I were presented with such a box (and I'm not >> even programmed with UD+ASSA, AFAIK). Any sufficiently advanced >> technology is indistinguishable form magic (but not actual magic) and all that ;->... >> >> Moshe The AI would assign approximately 2^-S to this
probability. A human being would intuitively assign a significantly greater a
priori probability, especially for larger values of S. As S goes
to infinity, the AI's probability would converge to 0, whereas the human's
would converge to some positive constant.
Why 2^-S? Being able to solve the halting problem
for programs up to size S is equivalent to knowing the first S bits of the
halting probability (Chaitin's Omega). Since Omega is incompressible by a
Turing machine, the length of the shortest algorithmic description of the first
S bits of Omega is just S (plus a small constant). See http://www.umcs.maine.edu/~chaitin/xxx.pdf.
Here's another way to see why the AI's method of
induction does not capture our intuitive notion. Supposed we've determined
empirically that the black box can solve the halting problem for programs up to
some S. No matter how large S is, the AI would still only assign a probability
of 2^-100 to the black box being able to solve halting problems for
programs of size S+100. |

- Re: is induction unformalizable? Wei Dai
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- RE: is induction unformalizable? "Hal Finney"
- Re: is induction unformalizable? Bruno Marchal

- Re: is induction unformalizable? "Hal Finney"
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- Re: is induction unformalizable? "Hal Finney"
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