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: [email protected]mailto:[email protected] e-mail: [email protected]mailto:[email protected] Tel: +1-865-253-3924 On Jul 22, 2014, at 9:58 PM, Pavel Afonine mailto:[email protected]> 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. Pavel On 7/21/14, 2:14 PM, Alastair Fyfe wrote: Hi, 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 UCSC _______________________________________________ cctbxbb mailing list [email protected]mailto:[email protected] http://phenix-online.org/mailman/listinfo/cctbxbb