David, thank you for you help! It works great and is really fast! Because programmers are lazy using the STL is preferred :)
Glen, I think the STL function works almost the same way as you suggestion. Good work! Christian, thank you for your suggestion. If it's still too slow for the users (although I don't think so) I'll use some of them. It's great to get so much help in a very short time! Joachim > C++ has a standard algorithm called set_intersection - found in the header > <algorithm>. Given two sorted sequences, it will output the items that are > found in both sequences. This function is very efficient, and with use of > it, I think your problem should become trivial. Documentation for this > algorithm can be found at http://www.sgi.com/tech/stl/set_intersection.html > > Of course, to use it, your sequences will need to have standards-conforming > iterators. If you need any help using the function, let me know. > > -David. > > ----- Original Message ----- > From: "Joachim Ansorg" <[EMAIL PROTECTED]> > To: <[EMAIL PROTECTED]> > Sent: Tuesday, January 07, 2003 12:46 AM > Subject: [sword-devel] Question for Optimization / speed experts > > > Hi! > > Some questiosn for optimizations experts :) > > > > I'm working on a piece of code in BibleTime which gives me headaches. > > BibleTime supports displaying different lexicons side by side, it offers > > only > > > the keys which are available in all of the chosen lexicons. > > > > The function which creates the list of entries which are available in > > all modules is ways too slow and too CPU consuming. > > I implemented it the following way: > > > > 1. Sort the modules by number of entries (from small to large) > > 2. Use the modules with the fewest entries and go through all of it's > > entries: > > 3. Is the entry in all other modules? Add the entry to a list of good > > entries it it is, otherwise continue with the next entry of the smallest > > list > > > and compare it with all other chosen modules. > > > > Assume the user selected BDB together with StrongsHebrew. Each of the > > modules > > > has around 10000 entries. So the outer loop would have 10000 iterations, > > the > > > inner loop would have another 10000 iterations. So we get > > 10.000*10.000=100.000.000 = one hunred million iterations. This is ways > > too > > > slow and ways too much. > > This needs on my 1.2GHz Atlon with 512 MB Ram more then five seconds. > > > > If there are three, four or even five modules side by side we get much > > more > > > loops. > > > > Are the other ways to do such a task? Since the entry lists are sorted by > > the > > > alaphabet this would be a good point to start. But I don't know how to do > > it > > > best. > > > > I'd be really glad for help with this case! > > > > Thank you! > > Joachim > > -- > > Joachim Ansorg > > www.bibletime.info > > www.ansorgs.de -- Joachim Ansorg www.bibletime.info www.ansorgs.de
