Hi Mario, There is no guaranteed way. On the Mac or on Linux you may be able to stop the computation by sending Chimera's Python interpreter an INT signal (terminal interrupt) [I don't know of any such option on Windows]. Open a terminal window (on Mac: /Applications/Utilities/ Terminal.app). Find Chimera's Python process via: ps augxww | grep python You'll see something along the lines of: pett 86576 98.7 13.4 1218088 562256 ?? R 11:58AM 2:17.97 /Users/pett/Applications/Chimera-daily-x11.app/Contents/ Resources/bin/python2.5 /Users/pett/Applications/Chimera-daily-x11.app/ Contents/Resources/share/__main__.py The first number (86576) is the process ID of Chimera's Python interpreter. Send it an interrupt signal by: kill -INT 86576 If all goes well, Chimera will bring up an error dialog saying something about KeyboardInterrupt. The Match->Align computation will have stopped with the structures in the same position as they were at the end of the last iteration, the last alignment window will be up, and the Reply Log will be available. On Linux, if you started Chimera yourself from the command line you may be able to just hit control-C (or whatever your interrupt character is) instead of the rigamarole above. --Eric On Oct 27, 2009, at 11:46 AM, Mario Borgnia wrote:
Hi Eric,
Thanks for the fast response. My other question is how to terminate the process in progress without compromising the output of previous iterations. Thanks!!!
Mario
Eric Pettersen wrote:
Hi Mario, The iterations are independent. So you could iterate twice, save a session, and then iterate further later. You could use the 'Save' button on the Reply Log to save info as needed.
--Eric
P.S. Sorry it's slow. It's all in Python and given how complex it is it's unlikely it's going to be recoded in C++ anytime soon.
Eric Pettersen
UCSF Computer Graphics Lab
On Oct 27, 2009, at 7:47 AM, Mario Borgnia wrote:
Hi, I launched a massive match-align procedure with five iterations without knowing that it would take so long to complete. Is it possible to stop it at iteration 2 without loosing the information computed to that point? Also, is it possible to continue from that point on in a subsequent round? Thanks!
Mario