找到ArangoDB中节点数的交叉节点？

Question

I have a number of nodes connected through intermediate node of other type. Like on picture There are can be multiple middle nodes. I need to find all the middle nodes for a given number of nodes and sort it by number of links between my initial nodes. In my example given A, B, C, D it should return node E (4 links) folowing node F (3 links). Is this possible? If not may be it can be done using multiple requests? I was thinking about using SHORTEST_PATH function but seems it can only find path between nodes from the same collection?

Answer 1

Very nice question, it challenged the AQL part of my brain ;) Good news: it is totally possible with only one query utilizing GRAPH_COMMON_NEIGHBORS and a portion of math. Common neighbors will count for how many of your selected vertices a cross is the connecting component (taking into account ordering AEB is different from BEA) using combinatorics we end up having a*(a-1)=c many combinations, where c is comupted. We use p/q formula to identify a (the number of connected vertices given in your set).

If the type of vertex is encoded in an attribute of the vertex object the resulting AQL looks like this:

FOR x in ( ( let nodes = ["nodes/A","nodes/B","nodes/C","nodes/D"] for n in GRAPH_COMMON_NEIGHBORS("myGraph",nodes , nodes) for f in VALUES(n) for s in VALUES(f) for candidate in s filter candidate.type == "cross" collect crosses = candidate._key into counter return {crosses: crosses, connections: 0.5 + SQRT(0.25 + LENGTH(counter))} ) ) sort x.connections DESC return x

If you put the crosses in a different collection and filter by collection name the query will even get more efficient, we do not need to open any vertices that are not of type cross at all. FOR x in ( ( let nodes = ["nodes/A","nodes/B","nodes/C","nodes/D"] for n in GRAPH_COMMON_NEIGHBORS("myGraph",nodes, nodes, {"vertexCollectionRestriction": "crosses"}, {"vertexCollectionRestriction": "crosses"}) for f in VALUES(n) for s in VALUES(f) for candidate in s collect crosses = candidate._key into counter return {crosses: crosses, connections: 0.5 + SQRT(0.25 + LENGTH(counter))} ) ) sort x.connections DESC return x

Both queries will yield the result on your dataset:

[ { "crosses": "E", "connections": 4 }, { "crosses": "F", "connections": 3 } ]