using neo4j get spanning tree from graph with loops - neo4j

I want to find the spanning tree from graph with loops. I cannot use regular bfs traversal here. so I check the allsimplepaths java function api, It seems find loop between two nodes. right now i select a random root, but don't know the end points. so i just want to get the spanning tree from graph while the it has many loops maybe. so it should convert to DAG and then give the tree structures. The graph may have more than one spanning tree.
how to do this? can allsimplepaths applied here?

Look at TraversalDescription with an appropriate uniqueness (NODE_GLOBAL) and Path-Expanders that follow the interesting relationships.

Related

how new traverse api works

nowadaya i m learning new traverse api of neo4j and i followed the link below
http://neo4j.com/docs/stable/tutorial-traversal-java-api.html
so now i know how to use uniqueness,evaluater etc.
that is i know how to change beahviours of the api.
but the thing i want to know is that how exactly it traverse.
for example im trying to find neighbours of a node.
does neo4j use index to find this?
does neo4j keep a hash to find neighbours?
more specifically, when i write the following code for example.
TraversalDescription desc = database.traversalDescription().breadthFirst().evaluator( Evaluators.toDepth( 3) );
node =database.getNodeById(4601410);
Traverser traverser = desc.traverse(node);
in my description i used breadthFirst. So it means that when i give node to traverse, the code should find the first neighbours. So how the api finds the first neighbours is the thing i want to know. Is there a pointer to neighbours in node? So when i say traverse until to depth 3 it finds the first neighbours and then take the neighbours as node in a recursive function and so on? So if we say to depth 10 then it can be slow?
so what i want exactly is how i can change the natural behaviour of the api to traverse?
Simplified, Neo4j stores records representing nodes and relationships a.s.o. in its store. Every node is represented by a node record on disk, that record contains a pointer (direct offset into relationship store) for the first relationship (neighbour if you will). Relationship records link to each other, so getting all neighbours for a node will read the node record, its relationship pointer to that relationship record and continue following those forward pointers until the end of that chain. Does that answer your question?
TraversalDescription features a concept of PathExpander - that is the component deciding which relationships will be used for the next step. Use TraversalDescription.expand() for this.
You can either use your own implementation for PathExpander or use one of the predefined methods in PathExpanders.
If you just want your traversal follow specific relationship types you can use TraversalDescription.relationships() to specify those.

Cypher: Ordering Nodes on Same Level by Property on Relationship

I am new to Neo4j and currently playing with this tree structure:
The numbers in the yellow boxes are a property named order on the relationship CHILD_OF.
My goal was
a) to manage the sorting order of nodes at the same level through this property rather than through directed relationships (like e.g. LEFT, RIGHT or IS_NEXT_SIBLING, etc.).
b) being able to use plain integers instead of complete paths for the order property (i.e. not maintaining sth. like 0001.0001.0002).
I can't however find the right hint on how or if it is possible to recursively query the graph so that it keeps returning the nodes depth-first but for the sorting at each level consider the order property on the relationship.
I expect that if it is possible it might include matching the complete path iterating over it with the collection utilities of Cypher, but I am not even close enough to post some good starting point.
Question
What I'd expect from answers to this question is not necessarily a solution, but a hint on whether this is a bad approach that would perform badly anyways. In terms of Cypher I am interested if there is a practical solution to this.
I have a general idea on how I would tackle it as a Neo4j server plugin with the Java traversal or core api (which doesn't mean that it would perform well, but that's another topic), so this question really targets the design and Cypher aspect.
This might work:
match path = (n:Root {id:9})-[:CHILD_OF*]->(m)
WITH path, extract(r in rels(path) | r.order) as orders
ORDER BY orders
if it complains about sorting arrays then computing a number where each digit (or two digits) are your order and order by that number
match path = (n:Root {id:9})-[:CHILD_OF*]->(m)
WITH path, reduce(a=1, r in rels(path) | a*10+r.order) as orders
ORDER BY orders

Topological sort of DAG

I have a dag (Tree) in which the directed edges are only of three kinds :
Left to right (siblings)
Child to Parent
Parent to a child
Specifically , the problem is to evaluate an attribute parse tree, but it doesn't matter what the specific problem is.
Sort of :
What traversal is guaranteed to give a topological sort of the nodes ?
I think inorder will fail but some places it is suggested that inorder is the way to go. I know reverse post order woeks on general DAGS but I think there must be a simpler traversal for my case.
Since your graph is a DAG, and you therefore have no back-edges, you can use depth-first search to traverse your graph and add nodes to your sorted list in the order in which they come off the DFS stack.

low speed on cypher query

I use query
"START a=node("+str(node1)+"),
b =node("+str(node2)+")
MATCH p=shortestPath(a-[:cooperate*..200]-b)
RETURN length(p)"
to see the path between a and b. I have many nodes, so when i run the query, sometimes it runs fast and sometimes run slowly.I use neo4j 1.9 community. Can anyone helps?
Query time is proportional to the amount of the graph searched. Your query allows for very deep searches, up to depth 200. If a. and b. are very close, you'll not search much of the graph, and the query will return very fast. If a. and b. are separated by 200 edges, you will search a very large swathe of graph (perhaps the whole graph?), which for a large graph will be much slower.
Is the graph changing between the two queries, is it possible these two nodes end up in different places in relation to eachother between the queries? For example if you generate some random data to populate the graph?

Neo4j - get all articulation vertices

using Neo4j, I would like to get all the articulation vertices (vertices/nodes that when removed, splits the graph in more connected components) from my graph.
Is there an easy way to do it (without completely re-implementing DFS)?
Alternatively, is there a possibility to do a traversal with the exclusion of a certain node? (and its relationships) (I have a fairly small number of nodes, using neo4j embedded so optimal O() is not critical)
you could exclude nodes by not continuing past them, e.g. with the Traversal Framework, see http://docs.neo4j.org/chunked/snapshot/tutorials-java-embedded-traversal.html#_new_traversal_framework. Also, you could implement your own RelationshipExpander that will not expand relationships to your node to avoid in a traversal, see http://components.neo4j.org/neo4j/1.5.M01/apidocs/org/neo4j/graphdb/RelationshipExpander.html
HTH
/peter

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