The full-matrix solution

Zhang, K. Y. J.; Cowtan, K. D.; Main, P.

doi:10.1107/97809553602060000687

International
Tables for
Crystallography
Volume F
Crystallography of biological macromolecules
Edited by M. G. Rossmann and E. Arnold

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International Tables for Crystallography (2006). Vol. F. ch. 15.1, pp. 322-323 | 1 | 2 |

Section 15.1.5.2.2. The full-matrix solution

K. Y. J. Zhang,^a K. D. Cowtan^b ^* and P. Main^c

^a Division of Basic Sciences, Fred Hutchinson Cancer Research Center, 1100 Fairview Ave N., Seattle, WA 90109, USA,^bDepartment of Chemistry, University of York, York YO1 5DD, England, and ^cDepartment of Physics, University of York, York YO1 5DD, England
Correspondence e-mail: cowtan+email@ysbl.york.ac.uk

15.1.5.2.2. The full-matrix solution

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The equations to be solved for the electron-density shifts, $[\delta\rho\left({\bf x}\right)]$ , are from the Jacobian of equation (15.1.5.2), $[\cases{(2V/N){\textstyle\sum\limits_{\bf y}} \rho\left({\bf y}\right)\psi \left({\bf x} - {\bf y}\right) - \delta\rho\left({\bf x}\right) = \Delta\rho\left({\bf x}\right)\cr \delta\rho\left({\bf x}\right) = \Delta H\left({\bf x}\right)\hfill\cr}, \eqno(15.1.5.19)]$ where $[\Delta\rho\left({\bf x}\right)]$ is the residual to Sayre's equation, $[\Delta\rho({\bf x}) = \rho({\bf x}) - (V/N){\textstyle\sum\limits_{\bf y}} \rho^{2} ({\bf y})\psi ({\bf x} - {\bf y}), \eqno(15.1.5.20)]$ and $[\Delta H\left({\bf x}\right)]$ is the residual to the linear density-modification equations, $[\Delta H \left({\bf x}\right) = H ({\bf x}) - \rho({\bf x}). \eqno(15.1.5.21)]$ Starting from a trial solution of $[\delta \rho_{0} \left({\bf x}\right) = {\bf 0}]$ , the initial residual vector is $[\eqalignno{{\bf r}_{0} \left({\bf x}\right) &= (2/V)\rho\left({\bf x}\right){\textstyle\sum\limits_{\bf h}} {\theta \left(\overline{\bf h}\right)} \Delta F\left({\bf h}\right)\exp \left(- 2\pi i{\bf hx}\right) &\cr &\quad- \Delta\rho\left({\bf x}\right) + \Delta H\left({\bf x}\right),&(15.1.5.22)}]$ where $[\eqalignno{\Delta F\left({\bf h}\right) &= F\left({\bf h}\right) - \theta \left({\bf h}\right)G\left({\bf h}\right), &(15.1.5.23)\cr G\left({\bf h}\right) &= (V/N){\textstyle\sum\limits_{\bf y}} \rho^{2} \left({\bf y}\right) \exp \left(2\pi i{\bf hy}\right) &(15.1.5.24)}%(15.1.5.24)]$ and $[\Delta\rho\left({\bf x}\right) = (1/V){\textstyle\sum\limits_{\bf h}} \Delta F\left({\bf h}\right) \exp \left(-2\pi i{\bf hx}\right). \eqno(15.1.5.25)]$ Thus, only three FFTs are required to calculate the initial residual. The residual of Sayre's equation is given in equation (15.1.5.23).

The calculation of $[{\bf q}_{k}]$ in equation (15.1.5.14) is achieved in a similar manner using FFTs, $[\eqalignno{{\bf q}_{k} = {\bf Jp}_{k} &= \left\{{{(1/V){\textstyle\sum\nolimits_{\bf h}} \left[2a\left({\bf h}\right)\theta \left({\bf h}\right) - b\left({\bf h}\right)\right]\exp \left(-2\pi i{\bf hx}\right)} \over {p_{k} \left({\bf x}\right)}}\right\}\cr &= \left[{{Q_{k} \left({\bf x}\right)} \over {p_{k} \left({\bf x}\right)}}\right], &(15.1.5.26)}]$ where the vector is partitioned as shown above, and $[\eqalignno{a\left({\bf h}\right) &= (V/N){\textstyle\sum\limits_{\bf y}} \rho\left({\bf y}\right) p_{k} \left({\bf y}\right)\exp \left(2\pi i{\bf hy}\right), &(15.1.5.27)\cr b\left({\bf h}\right) &= (V/N){\textstyle\sum\limits_{\bf y}}\; p_{k} \left({\bf y}\right)\exp \left(2\pi i{\bf hy}\right). &(15.1.5.28)}%(15.1.5.28)]$

Similarly, vector $[{\bf s}_{k}]$ in equation (15.1.5.16) is obtained from $[\displaylines{{\bf s}_{k} = {\bf J}^{T} {\bf q}_{k} = (2/V)\rho\left({\bf x}\right){\textstyle\sum\limits_{\bf h}} \theta \left(\overline{\bf h}\right) \left[ 2a\left({\bf h}\right)\theta \left({\bf h}\right) - b\left({\bf h}\right)\right]\exp \left(-2\pi i{\bf hx}\right)\hfill\cr \qquad- Q_{k} \left({\bf x}\right) + p_{k} \left({\bf x}\right),\hfill (15.1.5.29)}]$ where $[Q_{k}\left({\bf x}\right)]$ is defined in equation (15.1.5.26).

The remaining calculations in equations (15.1.5.12), (15.1.5.13), (15.1.5.15), (15.1.5.17) and (15.1.5.18) require either the inner product of a pair of vectors or a linear combination of vectors, both of which are very quick to calculate. Each iteration of the conjugate gradient requires four FFTs, as described in equations (15.1.5.26 –15.1.5.29).

References

International Tables for Crystallography (2006). Vol. F. ch. 15.1, pp. 322-323