Abram, This is a very interesting function. I have spent a lot of time thinking about it. However, I do not believe that does, in any way, prove or indicate that Solomonoff Induction is convergent. I want to discuss the function but I need to take more time to study some stuff and to work various details out. (Although I have thought a lot about it, I am writing this under a sense of deadline, so it may not be well composed.)
My argument was that Solomonoff's conjecture, which was based (as far as I can tell) on 'all possible programs', was fundamentally flawed because the idea of 'all possible programs' is not a programmable definition. All possible programs is a domain, not a class of programs that can be feasibly defined in the form of an algorithm that could 'reach' all the programs. The domain of all possible programs is trans-infinite just as the domain of irrational numbers are. Why do I believe this? Because if we imagine that infinite algorithms are computable, then we could compute irrational numbers. That is, there are programs that, given infinite resources, could compute irrational numbers. We can use the binomial theorem, for example to compute the square root of 2. And we can use trial and error methods to compute the nth root of any number. So that proves that there are infinite irrational numbers that can be computed by algorithms that run for infinity. So what does this have to do with Solomonoff's conjecture of all possible programs? Well, if I could prove that any individual irrational number could be computed (with programs that ran through infinity) then I might be able to prove that there are trans-infinite programs. If I could prove that some trans-infinite subset of irrational numbers could be computed then I might be able to prove that 'all possible programs' is a trans-infinite class. Now Abram said that since his sum, based on runtime and program length, is convergent it can prove that Solomonoff Induction is convergent. Even assuming that his convergent sum method could be fixed up a little, I suspect that this time-length bound is misleading. Since a Turing Machine allows for erasures this means that a program could last longer than his time parameter and still produce an output string that matches the given string. And if 'all possible programs' is a trans-infinite class then there are programs that you are going to miss. Your encoding method will miss trans-infinite programs (unless you have trans-cended the trans-infinite.) However, I want to study the function and some other ideas related to this kind of thing a little more. It is an interesting function. Unfortunately I also have to get back to other-worldly things. Jim Bromer On Mon, Jul 26, 2010 at 2:54 PM, Abram Demski <[email protected]> wrote: > Jim, > > I'll argue that solomonoff probabilities are in fact like Pi, that is, > computable in the limit. > > I still do not understand why you think these combinations are necessary. > It is not necessary to make some sort of ordering of the sum to get it to > converge: ordering only matters for infinite sums which include negative > numbers. (Perhaps that's where you're getting the idea?) > > Here's my proof, rewritten from an earlier post, using the properties of > infinite sums of non-negative numbers. > > (preliminaries) > > Define the computation as follows: we start with a string S which we want > to know the Solomonoff probability of. We are given a time-limit T. We start > with P=0, where P is a real number with precision 2*log_4(T) or more. We use > some binary encoding for programs which (unlike normal programming > languages) does not contain syntactically invalid programs, but still will > (of course) contain infinite loops and so on. We run program "0" and "1" for > T/4 each, "00", "01", "10" and "11" for T/16 each, and in general run each > program of length N for floor[T/(4^N)] until T/(4^N) is less than 1. Each > time we run a program and the result is S, we add 1/(4^N) to P. > > (assertion) > > P converges to some value as T is increased. > > (proof) > > If every single program were to output S, then T would converge to 1/4 + > 1/4 + 1/16 + 1/16 + 1/16 + 1/16 + ... that is, 2*(1/(4^1)) + 4*(1/(4^2)) + > 8*(1/(4^3)) + ... which comes to 1/2 + 1/4 + 1/8 + 1/16 + .. i.e. 1/(2^1) + > 1/(2^2) + 1/(2^3) + ... ; it is well-known that this sequence converges to > 1. Thus 1 is an upper bound for P: it could only get that high if every > single program were to output S. > > 0 is a lower bound, since we start there and never subtract anything. In > fact, every time we add more to P, we have a new lower bound: P will never > go below a number once it reaches it. The sum can only increase. Infinite > sums with this property must either converge to a finite number, or go to > infinity. However, we already know that 1 is an upper bound for P; so, it > cannot go to infinity. Hence, it must converge. > > --Abram > > On Mon, Jul 26, 2010 at 9:14 AM, Jim Bromer <[email protected]> wrote: > >> As far as I can tell right now, my theories that Solomonoff Induction >> is trans-infinite were wrong. Now that I realize that the mathematics do >> not support these conjectures, I have to acknowledge that I would not be >> able to prove or even offer a sketch of a proof of my theories. Although I >> did not use rigourous mathematics while I have tried to make an assessment >> of the Solomonoff method, the first principle of rigourous mathematics is to >> acknowledge that the mathematics does not support your supposition when they >> don't. >> >> Solomonoff Induction isn't a mathematical theory because the desired >> results are not computable. As I mentioned before, there are a great many >> functions that are not computable but which are useful and important because >> they tend toward a limit which can be seen in with a reasonable number of >> calculations using the methods available. Pi is one such function. (I am >> presuming that pi would require an infinite expansion which seems right.) >> >> I have explained, and I think it is a correct explanation, that there is >> no way that you could make an apriori computation of all possible >> combinations taken from infinite values. So you could not even come up with >> a theoretical construct that could take account of that level of >> complexity. It is true that you could come up with a theoretical >> computational method that could take account of any finite number of values, >> and that is what we are talking about when we talk about the infinite, but >> in this context it only points to a theoretical paradox. Your theoretical >> solution could not take the final step of computing a probability for a >> string until it had run through the infinite combinations and this is >> impossible. The same problem does not occur for pi because the function >> that produces it tends toward a limit. >> >> The reason I thought Solomonoff Induction was trans-infinite was because I >> thought it used the Bayesian notion to compute the probability using all >> possible programs that produced a particular substring following a given >> prefix. Now I understand that the desired function is the computation of >> only the probability of a particular string (not all possible substrings >> that are identical to the string) following a given prefix. I want to study >> the method a little further during the next few weeks, but I just wanted to >> make it clear that, as far as now understand the program, that I do not >> think that it is trans-infinite. >> >> Jim Bromer >> *agi* | Archives <https://www.listbox.com/member/archive/303/=now> >> <https://www.listbox.com/member/archive/rss/303/> | >> Modify<https://www.listbox.com/member/?&;>Your Subscription >> <http://www.listbox.com/> >> > > > > -- > Abram Demski > http://lo-tho.blogspot.com/ > http://groups.google.com/group/one-logic > *agi* | Archives <https://www.listbox.com/member/archive/303/=now> > <https://www.listbox.com/member/archive/rss/303/> | > Modify<https://www.listbox.com/member/?&;>Your Subscription > <http://www.listbox.com/> > ------------------------------------------- agi Archives: https://www.listbox.com/member/archive/303/=now RSS Feed: https://www.listbox.com/member/archive/rss/303/ Modify Your Subscription: https://www.listbox.com/member/?member_id=8660244&id_secret=8660244-6e7fb59c Powered by Listbox: http://www.listbox.com
