X-Message-Number: 18505 Date: Thu, 07 Feb 2002 14:25:58 +0100 From: Henri Kluytmans <> Subject: Re: Simulation, qubits Yvan Bozzonetti wrote : >A classical computer using a 813 bits word could indeed have only one planck >cube full. On the other hand, a quantum computer works on a superposition of >all possible words or address. so all cubles are present here. A quantum computer of n qubits can designate the same limited number of states as a conventional computer with a register of n bits. The advantage of a quantum computer is that it can operate on all possible combinations of those n bits in one single operational step! >At the end of a computation run, the interference process let only the >address in the "on" state. These define all places at all time not empty in >the simulated universe. If you have a quantum computer of 813 qubits, and you let it collapse, you will get a number of 813 bits. And could you explain to me how that number of 813 bits would tell me which of the 2^813 Planck cubes would be in the "on" state, and which ones would be "off" ? Or, to be precise, because the time dimension is to be eliminated, (you're looking at "all time") : how can a number of 813 bits tell me which of 10^183 (~2^608) cubes are in the "on" state ???? As far as I can see, that is not possible. ---- I asked : >>Please explain to me why precisely 2^813 out of all 2^(2^813) >>combinations remain ??? Yvan Bozzonetti replied : >2^(2^813) combine each planck's cube with each other: this is time travel and >muliuniverse, a case explicitely discarded before. Yes I know. That's why I also mentioned the number of Planck cubes at any point in time ( =10^183 ) which is still much larger than 813. >2^813 addresses (words) is >all can be hold in the classical universe. Nope, 2^813 addresses is the representation of a universe where, for example, only ONE Planck cube can be "on" in the complete universe at any point in time ... This is what seems to be your assumption about the universe. >If you go beyond, you have more than one word for each planck's cube, >if you do that, you have a sub-space of >a kind or another, something that is beyond the classical case. Nope. (See above.) Rate This Message: http://www.cryonet.org/cgi-bin/rate.cgi?msg=18505