<div dir="ltr">Thank you very much. I need to build a toy system to test it, but sounds just about right. I might even hard-code the square of threshold to save another flop, following your philosophy. A flop is a flop :)<div>
<br></div><div>Again, thank you. </div><div><br></div><div>Murat</div></div><div class="gmail_extra"><br><br><div class="gmail_quote">On Thu, Sep 26, 2013 at 12:10 AM, Bogdan Costescu <span dir="ltr"><<a href="mailto:bcostescu@gmail.com" target="_blank">bcostescu@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div class="im">On Wed, Sep 25, 2013 at 5:29 PM, MURAT OZTURK <<a href="mailto:murozturk@ku.edu.tr">murozturk@ku.edu.tr</a>> wrote:<br>
> Is it OK to intorduce the conditional into pdihs(), and bypass even calling<br>
> dopdihs() if r_kj ( as reported by calling dih_angle() ) is less then<br>
> threshold?<br>
<br>
</div>Yes. It makes sense to abort calculating as soon as you have enough<br>
information to make the decision. You can place your test immediately<br>
after calling dih_angle(), as it gives you r_jk. The code could look<br>
like (based on your own proposal):<br>
<br>
/* outside of the loop */<br>
threshold2 = threshold * threshold;<br>
<br>
for (i = 0; (i < nbonds); )<br>
{<br>
...<br>
phi = dih_angle(...)<br>
midDist2 = iprod(r_jk, r_jk);<br>
if (midDist2 > threshold2)<br>
continue;<br>
...<br>
}<br>
<br>
Please note the comparison done on the squares of distances instead of<br>
distances, saving some CPU cycles.<br>
<br>
Good luck!<br>
<span class="HOEnZb"><font color="#888888">Bogdan<br>
</font></span><div class="HOEnZb"><div class="h5">--<br>
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