I had some misunderstanding in this argument.
Now I know how it works.
Thanks.
<-----Original Message----->
> From: "Kijeong Kwac" <pine994.naver.com>
> To: amber.scripps.edu
> Cc:
> Sent: 2006-10-21 07:41
> Subject: RE: Re: Re: AMBER: Question about sander code
Thank you for your reply.
It is clear that "kdiv" is not zero when only 3rd atom index is negative.
2 - 1 + 1 = 2, so (2 - 1 + 1) / 4 = 2/4 = 1/2, not zero.
I think there should be some other way to make the variable "fmuln" to become zero.
<-----Original Message----->
> From: "David A. Case" <case.scripps.edu>
> To: amber.scripps.edu
> Cc:
> Sent: 2006-10-21 06:50
> Subject: Re: Re: AMBER: Question about sander code
On Sat, Oct 21, 2006, Kijeong Kwac wrote:
> 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.
This is integer arithmetic: (2 -1 +1)/4 = 0.
...dac
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Received on Sun Oct 22 2006 - 06:07:30 PDT
