[cctbxbb] 26-fold symm expansion of asu?

Mustyakimov, Marat mustyakimovm at ornl.gov
Wed Jul 23 07:01:13 PDT 2014

The fast way to generate symmetry equivalent ASU:
Apply symmetry operation to unit vectors: ex = SymOp*(1,0,0),  ey=SymOp*(0,1,0) and ez=SymOp*(0,0,1);
Then every map point of the symmetry related asu is obtained from the original one:
   new_ijk = i*ex + j*ey + k*ez  where (i,j,k) is the grid point in the original ASU.

Marat Mustyakimov
P.O. Box 2008, Bldg. 8600, MS-6475
Oak Ridge National Lab
Oak Ridge TN 37831  USA
e-mail: mustyakimovm at ornl.gov<mailto:mustyakimovm at ornl.gov>
e-mail: marat.mustyakimov at gmail.com<mailto:marat.mustyakimov at gmail.com>
Tel: +1-865-253-3924

On Jul 22, 2014, at 9:58 PM, Pavel Afonine <pafonine at lbl.gov<mailto:pafonine at lbl.gov>> wrote:

Hi Alastair,

I think I can write a piece of code using cctbx that would do what you want and probably that would be what you call "slow". Could you please explain some more what it is for and why, and may be give some context? Perhaps that helps me to see if we have something similar already available..

For instance we have very efficient code that computes maps in ASU, and also that expands map in ASU into the whole P1 box. This is something Marat wrote recently and is very efficient. May be you can reuse that or at least harvest ideas?

I'm copying to Marat in case he has comments.


On 7/21/14, 2:14 PM, Alastair Fyfe wrote:
I'd be grateful for any pointers to clipper or cctbx code that addresses the following: given a "brick" of map grid coordinates that enclose the asu, obtain the 26 symmetry-related "bricks" which contact it. Calculating this by successively generating symmetry mates and testing seems slow; I was wondering whether there was a better way.
thank you,
Alastair Fyfe
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