[sword-devel] Question for Optimization / speed experts
Joachim Ansorg
sword-devel@crosswire.org
Tue, 7 Jan 2003 03:39:05 +0100
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" <joachim@ansorgs.de>
> To: <sword-devel@crosswire.org>
> 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