As I understand it the UD generates all possible programs and as it
generates each one it runs one step of it before generating the next.
Does that not mean that eventually it will generate the program which
is generating what we understand to be some observer moments for us at
this particular time. This is where I was thinking of the foliation
bit - each hypersurface is a snapshot in time of the universe as
experienced by me.  This being said would that not mean they would
necessarily be in order or are you thinking that some other program.
could generate by chance a perfectly good observer moment that was out
of sync?


Best

Nick

On Jan 5, 2:09 pm, Stathis Papaioannou <stath...@gmail.com> wrote:
> 2010/1/6 Quentin Anciaux <allco...@gmail.com>:
>
>
>
>
>
> >> > It seems to me that it depends if the computation is iterative or not... 
> >> > in
> >> > other words, to compute step N you must have computed step N-1 before 
> >> > that.
>
> >> > If you can directly compute step N without computing prior step, 
> >> > S2/S1/S3 is
> >> > possible. If not you had necessarily computed step S1 before S2, only by
> >> > doing a replay of a previously done computation you could do it :
>
> >> > - first generate S1/S2/S3 in order and save each intermediate result, 
> >> > then
> >> > you can do
> >> > - S2 (taking the previously intermediate result of S1), S1 then S3 
> >> > (taking
> >> > S2 result).
>
> >> > But running the same thing more times add a priori nothing. If the 
> >> > process
> >> > is iterative then "in order" computation win the measure battle (because 
> >> > any
> >> > out of order one require a genuine in order computation before).
>
> >> Another way to compute S2 without using S1 would be to run the UD.
>
> > Yes but the UD will generate infinitely more often the in order S1/S2/S3
> > than out of order... with what you are saying I don't even understand
> > what is a computation if not a rules ordered sequential state order.
>
> A UD running on an actual computer for a finite time *could* generate
> S2 before S1. There is nothing in the experience of S to indicate
> which was generated first, even though if he had to guess with no
> other information he is more likely to be right if he guesses he is
> being generated sequentially.
>
> --
> Stathis Papaioannou- Hide quoted text -
>
> - Show quoted text -

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