The problem is, bit-reversible does not mean the simulation is exact


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送交者: wasguru 于 2009-01-07, 14:50:17:

回答: Maybe not for you but evidently a lot of people (wink 008) 由 james_hussein_bond 于 2009-01-07, 14:37:02:

Besides roundoff errors, there are still truncation errors, which is inevitable in any finite difference algorithm. You can make a MD simulation bit-reversible, but that does not mean there is no numerical noise.

Actually, I doubt that there is really no roundoff error. Bit-reversible does not even mean the roundoff error is zero.

Just think about this: One common computation in MD simulations is taking square roots. Square roots are generally irrational. I doubt there is any way to exactly represent square roots in computer (except symbolically, which is not feasible in MD). Therefore there are alway round-off errors.





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