X-Message-Number: 18501
From: 
Date: Wed, 6 Feb 2002 11:16:16 EST
Subject: Re: CryoNet #18488 - #18497

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From: Henri Kluytmans <>:


> >To simulate the Universe (ten billion light years in radius for ten billion 
> >years) you need 10^244 Planck's cubes ( One Planck's unit for each 
> dimension 
> >in space and time). This could be run by a quantum computer with a 813 
> bits 
> >word in a single computation cycle.
> 
> I could see this to be the case if, for example, only ONE Planck cube out 
> of all 10^244 could be in the "on" state simultaneously for each overall
> state (and all the other Planck cubes would then be in the "off" state). 
> 
> But this doesn't seem to be the case.

Assume each cube has an address. Because there are 2^813 cube, an address is 
a number 813 bits long, that is, a word in the computer. This word (address) 
in full (in state) if it is present, it is empty (out state) if it is not 
present in the list of all possible address.

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.

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. 

>And what would be that state you will be looking for ?
>
>I.e. what would be the state(s) you want to let the quantum computer 
>collapse into?

See above. 
I don't think the final state(s) must be colapsed to the classical domain at 
the end, better to link it with another quantum state.
 
>For example, to factor a product of two primes with a combined size 
>of 1000 bits. You will need at least a 1000 qubits. You will use 
>the right quantum logic to let the qubits collapse into that single 
>state that contains the two primes. There are 2^1000 possible 
>combinations, but there is only 1 possible collapsed state.
>The quantum logic will determine the collapsed state.

True, by the way, your example demonstrates that some simple to think about 
problems are larger than a full classical simulation of the observable 
universe.

>Please explain to me why precisely 2^813 out of all 2^(2^813) 
>combinations remain ???

2^(2^813) combine each planck's cube with each other: this is time travel and 
muliuniverse, a case explicitely discarded before. 2^813 addresses (words) is 
all can be hold in the classical 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.

Yvan Bozzonetti.


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