Is there any way to access rpl neighbors of each node at the root of rpl network? I know nodes do not send neghbors to the root. But, is it possible to access each node's neighbor table as a global variable only in the simulation?
No, it's not possible. Simulations also don't work like that - it is not possible to see the variables of another node from the perspective of a node's C code.
The best you can do is to implement periodic sending of the nodes' neighbor tables to the root.
Related
When I say Stateful Node, I mean a node that carries ‘state info,’ such as the path that leads to this node. E.g. R1 is a node, and
state1: link coming from path 1
state2: link coming from path 2
Is there any way I could create such a node in Neo4j? While traversing such a node, I expect it to behave like this:
if state 1, and input is x, then [:has] node1
if state one and input is y, then stop
if state two and input is z, then [: has] node 2.
I want to convert node R1 to a stateful node so that it keeps the information mentioned above. Does Neo4J support such nodes? If so, could you guide me to a resource? Also, does the cipher query support the ‘stateful’ approach so I can set the state according to the path from which R1 is produced?
In the Neo4j architecture, a relationship is a doubly linked-list that stores pointers to the start and end nodes.
It sounds like what you're looking to do is create nodes that store that same information for all relationships that touch it, and then have behavior based on how the graph reaches them.
This is more akin to logic control, and Cypher handles that through filters on relationship type, node labels, and properties.
However, you can always set properties of nodes based on queries. For example:
MATCH (:AUTH_T)-[:HAS]->(n:R1)
SET R1.reached_by = "HAS"
Then you could do something with that in the future, like if you want to know if node n was reached by another method.
I am playing around with some graph theory algorithms in neo4j. I am trying to find the minimum spanning tree (mst) within my network. I synthetically created a network of 10 000 people. Each person has 12 relationship types each one linking him back to the other 9999 and each relationship with its own weight assigned.
The problem I have however is the fact that according to the definition the results must be a tree spanning over the ENTIRE network. The neo4j function however only returns a very small sub-graph (only about 12 nodes) of the entire network.
The code I am using looks like this:
MATCH (a:Name {Name:"Dillon Snow"})
CALL algo.mst(a,"Weight",{stats:true})
YIELD loadMillis, computeMillis, writeMillis, weightSum, weightMin, weightMax, relationshipCount
RETURN loadMillis, computeMillis, writeMillis, weightSum, weightMin, weightMax, relationshipCount
What can I change to get the function to return the mst spreading through the entire network
algo.mst.* has not been adapted to the matured Neo4j-Graph-Algorithms-CoreAPI in its current release (3.2.5.2/3.3.0.0 # Dec 2017) which might lead to unexpected results. But there is a pull request in the pipe, you can expect some changes in the next release.
Anyway.. The procedure should add a new relationship-type (default mst) to your nodes. In a connected graph each node should be connected as well while a disconnected graph leads to connections only between the nodes of this particular connected component (from your startNode).
If i understand you right you have multiple relationship types and more then one of them between a pair of nodes? E.g. Node A is connected to Node B with several relations, each of them with a different type and property value. This is a problem. In general the Graph-Algorithms-API does not support multible releationships. Each pair of nodes can only have one connection per direction. Although you can import multible types the core-api itself has no idea of the underlying type. If multible relationships between a pair of nodes get imported usualy the last one wins. This has been mentioned in the documentation ;)
To overcome this limitation you could replace your relationship types with some kind of artificial nodes. When traversing over the result tree the occurence of one of those nodes would indicate the original relationship.
I used spatial features of Neo4j. I faced following problem in that. I used Node API to create a node and add that node to Spatial Layer.
Que 1 :- In Neo4j Browser client when i retrive all node it shows me two node with same node no. (One is real node and other is added to spatial layer). I think second node is for indexing purpose. Need clerification on this. I am wrong or it's bug or it's normal behaviour.
Que 2 :- If I update latitude and longitude of real node then second node in spatial layer is not updated with new value of co-ordinates. How can i update both node?
Any Help will be very helpful for me.
Thanks.
Neo4j spatial adds a second node within the geo R-Tree that represents the location of your domain node.
You have to remove it from the index and re-add it again.
As #Michael pointed correctly, it's a normal behavior as it's a representation of location of your node.
To remove a node from an index, you can delete the node, using "internal" id of the indexed node. Please check my answer here, for the detailed steps on how to do that. After that you can easily re-add the node to index.
A singly connected graph is a directed graph which has at most 1 path from u to v ∀ u,v.
I have thought of the following solution:
Run DFS from any vertex.
Now run DFS again but this time starting from the vertices in order of decreasing finish time. Run this DFS only for vertices which are not visited in some previous DFS. If we find a cross edge in the same component or a forward edge, then it is not Singly connected.
If all vertices are finished and no such cross of forward edges, then singly connected.
O(V+E)
Is this right? Or is there a better solution.
Update : atmost 1 simple path.
A graph is not singly connected if one of the two following conditions satisfies:
In the same component, when you do the DFS, you get a road from a vertex to another vertex that has already finished it's search (when it is marked BLACK)
When a node points to >=2 vertices from another component, if the 2 vertices have a connection then it is not singly connected. But this would require you to keep a depth-first forest.
A singly connected component is any directed graph belonging to the same entity.
It may not necessarily be a DAG and can contain a mixture of cycles.
Every node has atleast some link(in-coming or out-going) with atleast one node for every node in the same component.
All we need to do is to check whether such a link exists for the same component.
Singly Connected Component could be computed as follows:
Convert the graph into its undirected equivalent
Run DFS and set the common leader of each node
Run an iteration over all nodes.
If all the nodes have the same common leader, the undirected version of the graph is singly connected.
Else, it contains of multiple singly connected subgraphs represented by their corresponding leaders.
Is this right?
No, it's not right. Considering the following graph which is not singly connected. The first component comes from a dfs beginning with vertex b and the second component comes from a dfs beginning with vertex a.
The right one:
Do the DFS, the graph is singly connected if all of the three following conditions satisfies:
no foward edges
no cross edges in the same component
there is no more than 1 cross edges between any two of components
In neo4j should all nodes connect to node 0 so that you can create a traversal that spans across all objects? Is that a performance problem when you get to large datasets? If so, how many nodes is too much? Is it ok not to have nodes connect to node 0 if I don't see a use case for it now, assuming I use indexes for finding specific nodes?
There is no need or requirement to connect everything to the root node. Indexes work great in finding starting points for your traversal. If you have say less then 5000 nodes connected to a starting node (like the root node), then a relationship scan is cheaper than an index lookup.
To judge what is better, you need to know a bit more about the domain.