Thank you very much for the reply. But I'm still confused.
When only 3rd atom index is negative, idumi = -1 and iduml = 1
so that kdiv=(2+idumi+iduml)/4 = (2 -1 +1)/4 = 1/2. Thus "kdiv" is not zero.
I would like to know what I am still misunderstanding.
<-----Original Message----->
> From: "David A. Case" <case.scripps.edu>
> To: amber.scripps.edu
> Cc:
> Sent: 2006-10-21 03:42
> Subject: Re: AMBER: Question about sander code
On Sat, Oct 21, 2006, Kijeong Kwac wrote:
> kdiv becomes zero if the 3^rd and 4^th atom indexes are negative.
^^^ or
kdiv becomes zero if the either the third or fourth indices (or both) are
negative.
> In this situation I think that, when only 3^rd atom index is negative in the
> prmtop file, the value of the variable fmn(ic0) should be zero in the
> line 802 so that fmuln be zero.
This is not true (see above). In the scenario you present, fmuln will be zero
by virtue of the fact that kdiv is zero.
[I know I already sent this information to you off the list, and I gather what
I said earlier was not helpful to you. But I'm repeating it here so that
others can build on my explanation (or explain why it is wrong).]
...dac
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Received on Sun Oct 22 2006 - 06:07:29 PDT
