Re: The Rapidly-Accelerating Computer
Hal Finney wrote: >It's not clear to me what is wrong with letting a standard TM operate for >a transfinite number of steps. We usually give the TM an infinite tape, >and we feel free to imagine the infinite tape initialized with an infinite >number of symbols. Only with oracle machine. >That in itself implies, in a sense, an infinite amount >of time or effort to initialize the TM. As Wei Dai says that would be equivalent to the UTM with an oracle for the halting problem. (at least with the right symbol on the tape). Note that there are still unsolvable problems for such an UTM. There is an infinite lattice of degrees of unsolvability. A transfinite alpha-machine with alpha stritly bigger than omega could do much more than an omega machine. >Why would we hesitate to accept >the infinite stream of 1's but not the infinite stream of 0's? Because the zero are supposed to be there by default. To put the "ones" you need omega times. Wei Dai wrote: >I'll point out that this model >actually makes it clear that normal observer + RAC is not equivalent to >CSO + normal computer. A CSO + a never stopping computer can emulate the halting problem in the sense that he can guess that a non-stopping program will never stop ... until he changes his mind (after seeing a program which halts). In this way he will always be wrong, but he will be less and less wrong with time. In the limit he will be correct. It is ambiguous if the limit can be said to exist in our case. (There is a "well know" theorem in Recursion theory explaining this: the Schoenfield modulus lemma). Bruno
Re: The Rapidly-Accelerating Computer
On Fri, Oct 13, 2000 at 08:25:39PM -0700, [EMAIL PROTECTED] wrote: > I'm not sure it would be zero. The program for the CSO is not > particularly complex compared to other observer programs. If you have the > program for a constant-speed observer then you only need to simulate the > program, inserting ever increasing delays between simulated clock cycles. > > Now all you have to do is wait infinity+1 ticks of the UTM and you will > have your CSO at subjective time 1, and the program to create him was > not particularly long or unlikely. Did you really have in mind something like the following? for (i=0; i This sounds correct; it's hard to imagine a problem which takes an > infinite amount of computation to solve, but whose solutions could be > tested in finite time. Is this a theorem of computational theory? If the solutions can be tested in finite time, then you can solve the problem by testing all possible solutions, and this process would halt in a finite amount of time. (I'm assuming that "no solution" counts as a solution.) > On the other hand there might be theoretical reasons to believe in the > RAC; for example, if the laws of physics appear to be such as to allow > for infinitely fast computation, then it might be that we believe in > the RAC due to our understanding of the details of its construction. > It's like our belief today in the correctness of large proofs that can > only be verified by computer. It boils down to how to define the measure of observer moments. If you define it with a standard UTM, then nothing can convince you that RACs exist. If the laws of physics appear to allow infinitely fast computation, you'll just assume that you don't have a complete understanding of those laws.
Re: The Rapidly-Accelerating Computer and Black Holes
George Levy writes: > An RAC/CSO may be physically realizable using a black hole. Just fly near a > black hole and drop a computer into it, preferably one that runs Microsoft > software. As it falls down you'll observe a CSO. Or leave the computer behind > and YOU take the plunge. As you get nearer the BH the computer will look like > an RAC. Of course this doesn't take into account quantum effects. This doesn't quite work. From the outside, you do observe the computer effectively slow to a halt (actually it stops emitting light altogether) as it falls in, but after all there's no trick to making computers that halt. What you want is to *be* the CSO, so that you see an infinite amount of computation elsewhere. In the case of an infalling observer, he does not see the universe go through an infinite amount of time. The problem is that he passes the event horizon at the speed of light, doppler shifting the rest of the universe and slowing its observed speed. The net result is that he only sees a finite amount of events pass in the outside world as he falls in. If you could build a solid shell around the event horizon and lower yourself arbitrarily close to it (not falling in), you could be a CSO, but you can't do that. No material is strong enough to let you get arbitrarily close to the event horizon. So you cannot see an infinite amount of computation done outside, unless you have an infinitely strong material (the observer would also have to be infinitely small). Hal
Re: The Rapidly-Accelerating Computer and Black Holes
An RAC/CSO may be physically realizable using a black hole. Just fly near a black hole and drop a computer into it, preferably one that runs Microsoft software. As it falls down you'll observe a CSO. Or leave the computer behind and YOU take the plunge. As you get nearer the BH the computer will look like an RAC. Of course this doesn't take into account quantum effects. If you are really "dying" to know the answers to a problem that only an RAC could do, then you could try this little experiment. Good luck, send me a post card. (Unfortunately, it will never reach me, and I won't get it) George Levy
Re: The Rapidly-Accelerating Computer
On Fri, Oct 13, 2000 at 10:57:57PM +0200, Saibal Mitra wrote: > Yes, this is inherent in the construction of the CSO. s(t) has to be > smaller than 1 for any finite t. But what about transfinite values for t? > Since transfinite numbers can be described in a mathematical consistent way, > it is possible to define a mathematical model of the CSO surviving beyond > s(t) = 1 . Following Tegmark one should thus believe that the CSO will > experience this moment. But from Tegmark it's not possible to derive a measure for such an observer moment. If you follow my suggestion that the measure of an observer moment is equal to its universal a priori probability (which is defined as the probability that a universal Turing machine with a uniformly random input tape will produce that output), then the measure of the CSO at subjective time 1 is zero (unless you count observers having hallucinations that they are the CSO at subjective time 1). If the measure is defined in terms of an automata with access to an oracle for the halting problem, then the measure of the CSO would presumably be much larger. I don't really know how to justify using the universal Turing machine instead of any other automata. Suppose someone said he had access to a RAC. How could he prove it to you? The most he could do is show that he could solve any problem that you can verify the solution of, but unless you already had access to a RAC, all such problems can be solved in finite time on a regular Turing machine. For an observer that starts out with finite computing resources, there is no way to verify that anything purported to be a RAC actually is one, so there is no possible empirical reason for him not to believe that RACs don't exist.
Re: The Rapidly-Accelerating Computer
Wei writes: > Even if a continually slowing observer (CSO) could exist, it's > relationship to a normal computer would not be the same as that of a > normal observer to a RAC. To a normal observer, there is some finite > subjective time in the future when the RAC will have gone through an > infinite number of clock cycles, but to the CSO there is no finite > subjective for when a normal computer will have gone through an infinite > number of clock cycles. This is obvious when you consider that any finite > subjective time for the CSO is also a finite objective time. Doesn't this assume that objective time is discrete? With continuous objective time there is no objective fact of the matter about whether a given interval is finite or infinite. There are algebraic transformations like t' = 1/t which turn finite times into infinite, and vice versa. In the example above with the RAC, you have a time t measured by an ordinary observer and a time t' measured by the RAC. Then in the CSO case you have the time t measured by the CSO and time t' measured by an ordinary computer. The relationship between t and t' seems to be the same in each case. The only difference is in terms of "objective time", but it's not clear that is uniquely defined (or even meaningful at all). Hal
Re: The Rapidly-Accelerating Computer
Wei Dai wrote: > On Thu, Sep 14, 2000 at 01:13:35PM -0400, [EMAIL PROTECTED] wrote: > > It may be impossible to construct such a machine in our universe, but can we > > achieve the same results by slowing down the consciousness of the observer > > observing a conventional computer? In other words, each observer's clock > > cycle (assuming a computer model for the observer) would double in duration > > in relation to the computer clock. Could there be such an observer in our > > universe? I suspect that there can't be because the construction of the > > observer's clock woud require smaller and smaller energy packets in the > > presence of constant background noise. > > Even if a continually slowing observer (CSO) could exist, it's > relationship to a normal computer would not be the same as that of a > normal observer to a RAC. To a normal observer, there is some finite > subjective time in the future when the RAC will have gone through an > infinite number of clock cycles, but to the CSO there is no finite > subjective for when a normal computer will have gone through an infinite > number of clock cycles. This is obvious when you consider that any finite > subjective time for the CSO is also a finite objective time. > > But if as a function of objective time t the subjective time s(t) is such that Lim t ---> Infinity s(t) = 1 then after a subjective time of 1 second an infinite amount of objective time will have passed, unless you assume that the CSO can only exist at times s(t) < 1 . Saibal
RE: The Rapidly-Accelerating Computer
It is meaninless fr5om an objective point of view, but from a 'classical universe perspective' it has meaning. There is no objective meaning other that 'everything exists'. > -Original Message- > From: [EMAIL PROTECTED] [SMTP:[EMAIL PROTECTED]] > Sent: Friday, September 15, 2000 5:06 PM > To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; > [EMAIL PROTECTED] > Subject: RE: The Rapidly-Accelerating Computer > > James writes: > > There is no distinction. No observer-moments are related. No > observations > > are related to events. But of course, all observer-moments exists, and > all > > events exist, so you could argue that all observations are accurate. > > So in this formulation, the question of whether a RAC could exist is > meaningless? That doesn't sound like a very useful approach. > > Hal
RE: The Rapidly-Accelerating Computer
James writes: > There is no distinction. No observer-moments are related. No observations > are related to events. But of course, all observer-moments exists, and all > events exist, so you could argue that all observations are accurate. So in this formulation, the question of whether a RAC could exist is meaningless? That doesn't sound like a very useful approach. Hal
RE: The Rapidly-Accelerating Computer
James writes: > (And note that there are no 'observers' (sets of observer-moments in time > with an objective identity across them), only monad-like observer-moments. > So this observer-moment, which is all you are, survives 'forever'. I don't > suppose anyone will ever take this seriously this but I can't help saying > it whenever I get the chance.) So how would you analyze the question of whether there can exist a universe with an RAC, and an observer who sees the result of its infinite calculation? How would you define the situation in terms of observer-moments? Hal
