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


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