On 14 Feb 2012, at 04:00, Stephen P. King wrote:

On 2/13/2012 5:54 PM, acw wrote:On 2/12/2012 17:29, Stephen P. King wrote:Hi Folks,I would like to bring the following to your attention. I thinkthat wedo need to revisit this problem. http://lesswrong.com/lw/19d/the_anthropic_trilemma/ The Anthropic Trilemma <http://lesswrong.com/lw/19d/the_anthropic_trilemma/> snipI gave a tentative (and likely wrong) possible solution to it inanother thread. The trillema is much lessened if one considers arelative measure on histories (chains of OMs) and their length.That is, if a branch has more OMs, it should be more likely.The first horn doesn't apply because you'd have to keep the copiesrunning indefinitely (merging won't work).The second horn, I'm not so sure if it's avoided: COMP-immortalityimplies potentially infinite histories (although mergers may makethem finite), which makes formalizing my idea not trivial.The third horn only applies to ASSA, not RSSA (implicit in COMP).The fourth horn is acceptable to me, we can't really deny Boltzmannbrains, but they shouldn't be that important as the experienceisn't spatially located anyway(MGA). The white rabbit problem ismore of a worry in COMP than this horn.The fifth horn is interesting, but also the most difficult tosolve: it would require deriving local physics from COMP.My solution doesn't really solve the first horn though, it justmakes it more difficult: if you do happen to make 3^^^3 copies ofyourself in the future and they live very different and long lives,that might make it more likely that you end up with a continuationin such a future, however making copies and merging them shortlyafterwards won't work.Hi ACW,This solution only will work for finite and very special versionsof infinite sets. For the infinities like that of the Integers, itwill not work because any proper subset of the infinite set isidentical to the complete set as we can demonstrated with a one-to-one map between the odd integers and the integers.

`You should not confuse bijection (set isomorphism) and equality. Also,`

`measure exists on infinite discrete sets, by weakening the sigma-`

`additivity constraints. And then, finally, the measure problem bears`

`on infinite extension of computations, and they are 2^aleph_0.`

Remember the one line UD program: For all i, j,k compute the kth first steps of phi_i(j).

`We can describe a computation a sequence phi_i(j)^0,`

`phi_i(j)^1, .... , phi_i(j)^k.`

`That set is enumerable, but the set of all sequences going through`

`equivalent 1p-steps is not enumerable, and you can define a measure by`

`just using the normal distribution in a manner similar to the`

`dovetailing on the reals. This has just to be corrected to take into`

`account the constraints of self-reference, which seems to be the`

`origin of an arithmetical quantization, negative amplitude of`

`probability, etc.`

Given that the number of computations that a universal TM can runis at least the countable infinity of the integers, we cannot use acomparison procedure to define the measure.

`You confuse the computations made by the UD, and observed by an`

`outsider, and the infinite computations going through your actual 1p-`

`state. Those includes all the dummies dovetailing on the reals, and`

`cannot be enumerable.`

`Think about the iterated self-duplication. It leads to the usual`

`Gaussian.`

(Maybe this is one of the reasons many very smart people have tried,unsuccessfully, to ban infinite sets...)

`Not al all. The infinite set have been introduced to make the measure`

`problem more easy, even for problem handling finite objects when they`

`are very numerous.`

`Mathematical logic explains that finite and enumerable is more complex`

`than the continuum, which existence is basically motivated by`

`searching to simplify the problem. For example, Fermat on the reals is`

`trivial. Not so on non negative integers.`

Bruno http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.