The Dijkstra algorithm tries to find the path between two nodes which have
the lowest cost (where cost could be f.ex. a property on each relationship).
So that algorithm calculates costs and sums them for each path it tries out.
Check out http://en.wikipedia.org/wiki/Dijkstra's_algorithm for more
2010/6/23 Suruchi Deodhar deodharsuru...@gmail.com
Hi Mattias,
Thanks for your answers.
Sorry for the confusion wrt the java execution command. I saved CLASSPATH
in .bashrc file and ran following command.
/usr/java/default/bin/java -Xmx2048M BuildGraph
In response to your question
Do
2010/6/23 Atle Prange atle.pra...@gmail.com
Hm, i'll have to fix that...
Any thoughts on a Trie implementation? Would it be able to compete?
I have no idea on performance or what would be the best approach. I though
your alphabet-relationship-types approach sounded quite interesting. Or as a
I guess i have to brush the dust of my tree knowledge then...
-atle
On Thu, 2010-06-24 at 09:43 +0200, Mattias Persson wrote:
2010/6/23 Atle Prange atle.pra...@gmail.com
Hm, i'll have to fix that...
Any thoughts on a Trie implementation? Would it be able to compete?
I have no idea
Hi all!
I'm creating a web-based admin interface for Neo4j. Something that can
handle configuration, data browsing, backups and so on. I've been hacking on
some Neo4j things related to the web previously, for instance the new
UI at try.neo4j.org
http://try.neo4j.org/http://try.neo4j.org and the
I agree, lets keep it immutable, then its safe to keep the expander around.
Two suggestions:
Your could provide a reusable expander:
Expander e = Expanders.outgoing().along(FRIEND).and(ENEMY);
for(Node n:e.expand(node)){
...
}
or a onetime expander:
for(Node n :
I guess I would have just added a clear() method to the expansion. But of
course I understand that using mutable expansions requires that the user
actually understand the implications, which goes against the simplicity of
the DSL.
On Thu, Jun 24, 2010 at 12:23 AM, Tobias Ivarsson
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