X-Message-Number: 18622
Date: Fri, 22 Feb 2002 12:36:27 +0100
From: Henri Kluytmans <>
Subject: Re: Simulation

-->Hi,

Sorry, for the late reply, but i'm a little busy. It could be 
that further replies from me will take just as long. :(

I wrote :

>>If you have a quantum computer of 813 qubits, and you let it collapse, 
>>you will get a number of 813 bits. 

Yvan Bozzonetti replied :

>The problem here is the way you think a quantum computer must be used.
>Clearly, you are wired on the factorization problem: here we start with all 
>possible combinations and at the end only one is selected. Furthermore, the 
>entire computation is collapsed to the classical domain (if there was no 
>collapse, we would have at least two states: the "answer state" and the
empty 
>one). 

I guess, with the empty one, you mean the superposition of all possible 
combinations at the start ?

>Here, in the Universe simulation, no classical collapse is intended, the 
>"result" remains in the quantum state as said before (it may be linked to 
>another computation round ). 

Indeed, one has the option to not let the complete system collapse and 
still leave part of the qubits in a quantum state. But if part of the 
result remains in a quantum state then usually this is, because you 
will use this part for further quantum computations.

A far as I know you need to collapse to normal bits in the end of 
the process to read out the result, you cannot read out the 
contents of a quantum bit without destroying the superposition.

However in an earlier posting you wrote :

"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."

Theoretically it is possible the assign the address of every 
(2^608) planck cube to any combination in the 2^608 possible 
combination of a system of 608 qubits. But to perform the 
simulation itself you will need at least 2^608 qubits. 
Furthermore to read out the final result you will need 
2^608 qubits that you will let collapse into normal bits.

(I've left out the time dimension.)

>So there is not a single word selected, what 
>remains is a set of words, each one is an elementary cube  address., 
>precisely the addresses in the "on" state.

And how do you read them out ??????

You would still need to perform more quantum logic, and you would 
still need 2^608 qubits to transfer result into, and then collapse 
those qubits, to be able to read out the bit content.

And how do you perform the simulation itself without more qubits?


Cheers,
>Hkl

Rate This Message: http://www.cryonet.org/cgi-bin/rate.cgi?msg=18622