Neo4j Creating relationship between nodes that already map to a relationship - graph

Currently, I have a query in Neo4j that returns all the nodes in my graph that are pointed to by multiple nodes.
The query to return these nodes (in picture) looks like this:
MATCH (n)-[r:CLINICAL_SIGNIFICANCE]->()
WITH n, count(r) as rel_cnt
WHERE rel_cnt > 1
MATCH (c)-[r:PROTEIN_CHANGE]->(n)
return c, n
Is there a way to loop through the nodes labeled as c (blue nodes) and if they point to the same node labeled as n (yellow nodes), create a relationship between the nodes labeled as c (blue nodes)?

Related

Finding an edge by unique ID or name

It is possible to search a subgraph and node by their unique names;
n = agnode(g, "myUniqueNodeName", FALSE);
h = agsubg(g, "myUniqueSubgrahName", FALSE);
Likewise, is there a way to search edges in a strict directed graph by their unique names?
e = agedge (g, u, v, "e28", FALSE);
Documentation indicates that:
The 'name' of an edge (more correctly, identifier) is treated as a
unique indentifier for edges between a particular node pair. That is,
there can only be at most one edge with name e28 between any given u
and v, but there can be many other edges between other nodes.
It seems that there must be a list of edges that can be searched by name. Otherwise, a separate (ID -> edge) map will need to be maintained separately.

Get node descendants in a tree graph

I have a directed graph (grafopri1fase1) the graph has no loops and it has a tree structure (not binary tree).
I have an array of nodes (meterdiretti) that i have extracted from the graph (grafopri1fase1) matching a condition.
I would like to know starting from each node of Meterdiretti how many nodes are under each node of Meterdiretti.
The result I would like to have is a Matrix with the following format
first column------------ second column
meterdiretti[1] -------- total amount of nodes reachable starting from meterdiretti[1]
meterdiretti[2] -------- total amount of nodes reachable starting from meterdiretti[2]
....
meterdiretti[n] ----------total amount of nodes reachable starting from meterdiretti[n]
Take a punt at what you want - it would be good if you could add a reproducible example to your question.
I think what you want is to count the descendents of a node. You can do this with neighborhood.size and mode="out" argument.
library(igraph)
# create a random graph
g <- graph.tree(17, children = 2)
plot(g, layout=layout.reingold.tilford)
# test on a single node
neighborhood.size( g, vcount(g), "1", "out") - 1
# [1] 16
# apply over a few nodes
neighborhood.size( g, vcount(g), c(1,4,7), "out") - 1
[1] 16 4 2

Query ArangoDB general-graph for common neighbors with more than 2 start vertices?

Consider the following example graph:
 
Given the vertices A, B and C (creators), how to figure out their common neighbors?(projects all 3 participated in)For two vertices, I could simply use GRAPH_COMMON_NEIGHBORS("myGraph", A, B), but what if I want to query for 3 or more? Expected result: 1 and 2.
Given the same vertices, how can I make it return common neighbors with no other connections?(creators exclusively participated in a project, no additional edges allowed)?Expected result: 1, because 2 has an edge coming from D, which isn't one of the starting vertices.
You can simply pass the same set of vertices as both parameters for common neighbors. Then repack the result in a better format for AQL to compute the intersection:
let res = (
let nodes = ["a/A","a/B","a/C"]
for n in GRAPH_COMMON_NEIGHBORS("g",nodes , nodes)
for f in VALUES(n)
return VALUES(f)
)
return CALL("intersection", res[0])

Find length of longest trail in directed unweighted graph

I have a directed, unweighted, possibly cyclic graph that can contain loops and multiple duplicate edges (i.e. two edges from node 1 to node 2).
I would now like to find the length of the longest trail in this graph, i.e. the longest path that:
- uses no edge twice (but if there are multiple edges from node 1 to node 2, it can use every one of them)
- possibly visits nodes several time (i.e. it does not have to be a simple path)
In particular, is this problem NP-hard? I know that the longest simple path is NP-hard (reducing Hamiltonian Path to it) and the longest trail with edge reusal is in P (Bellman ford with weight -1 on every edge). However, with this problem, I am not quite sure and I could not find good information on it.
Although I am not completely sure, I think that this problem is NP-hard. As I understand, your question arises due to multiple edges between nodes. The graphs that has multiple edges between same nodes can be expanded to larger graphs with no multiple edges between them. Thus, a graph with multiple edges between same nodes has no difference than a graph without multiple edges.
Let me walkthrough a simple example to explain:
Let there be a graph with 3 nodes (A,B,C) and 5 edges between them (A to B, A to B, B to A, B to C, C to A)
This graph can be expanded and shown with 5 nodes and 7 edges.
Lets expand the node A to 3 different nodes (A1, A2, A3). When we adjust the edges according to previous edges, there exists 7 edges(A1 to B, A2 to B, B to A3, B to C, C to A1, C to A2, C to A3)
As a result, now we have a graph without multiple edges and can be evaluated with the help of Hamiltonian and Bellman Ford.
Hope I've at least cleared the problem a bit.

Finding Subgraph in a Graph

I have a graph shown here. Just to node that nodes B_0, B_1, belong to node of type B, C_0, C_1. C_2, C_3 belong to node of type C and so on.
Now, I want to find multiple subgraphs, which could satify criteria like defined by this example -
Criteria -
subgraph contains 1 node of type A, 1 node of type B, 1 node of type C, one node of type D.
subgraph has one edge from node of type A to one node of type B, one edge connecting type B and type C and one node connecting type C and type D.
subgraph contains one edge from type A going out of subgraph to type B node, one edge
from type B to type C node outside, one edge from type D to type E outside.
Now this description should give result as -
subgraph :: A_0, B_0, C_1, D_1
subgraph :: A_0, B_0, C_0, D_0
subgraph :: A_0, B_1, C_2, D_2
subgraph :: A_0, B_1, C_3, D_3
I want to know, if there is any algorithm, by which I can find such sub-graphs?
I tried to figure out an algorithm by making all possible combinations. However, this would be exponential to number of nodes in subgraph. Thus, I want to know if there exists an efficient way to calculate it. Or if there exists a problem of similar nature in Graph Theory?
You can start by visiting all nodes of type A. For each A node, look at all nodes connected to it which are of type B. From there look all nodes of type C and so on, keeping track of the nodes you've visited from the last A node. Then whenever you reach a subnode that completes the criteria of your search, you add all the list of nodes from the A node until the point where you are. Essentially you're doing a depth-first search where you keep traversing into the graph as long as a node meets the criteria of what should follow and backtrack whenever there's no more valid nodes (ie. which would produce a valid subgraph) going out from your current node.

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