Gremlin：如何在有向无环图中有效地找到“根”？

Question

I'm trying to write a gremlin query that efficiently solves the "confluent rivers" problem (for a lack of a better name, maybe there is one in graph theory?). Here's an example:

The task: given one of the root nodes, deliver a map containing the IDs of downstream nodes as keys, with all of their river root IDs (ie the end nodes reached by going all paths upwards again from the current node) as values.

For instance, in the example graph above, and for root node 0, the result should be:

{
 "0": ["0"],
 "1": ["0", "4"],
 "2": ["0", "5", "8"],
 "3": ["0", "4", "5", "8"],
 "6": ["0", "4"]
}

I'm particularly worried here about walking paths multiple times. For example, after calculating the roots of "2", I'd like to reuse that result for computing the roots of its downstream node "3".

Any clues how that might work for a large directed acyclic graph?

Answer 1

Given your diagram we can create the following graph.

g.addV('0').as('0').
  addV('1').as('1').
  addV('2').as('2').
  addV('3').as('3').
  addV('4').as('4').
  addV('5').as('5').
  addV('6').as('6').
  addV('7').as('7').
  addV('8').as('8').
  addE('link').from('0').to('1').
  addE('link').from('0').to('2').
  addE('link').from('1').to('6').
  addE('link').from('1').to('3').
  addE('link').from('2').to('3').
  addE('link').from('4').to('1').
  addE('link').from('5').to('7').
  addE('link').from('7').to('2').
  addE('link').from('8').to('7').iterate()  

The query below starts at '0' and finds all the leaf nodes and then works backwards to find all the roots. The output does not include the starting node ('0') but if necessary the query can be tweaked to include that.

gremlin>  g.V().hasLabel('0').
......1>        repeat(out()).emit().
......2>        until(__.not(out())).dedup().
......3>        group().
......4>          by(label()).
......5>          by(repeat(__.in('link')).
......6>             until(__.not(__.in('link'))).
......7>             label().dedup().
......8>             fold())

==>[1:[0,4],2:[0,8,5],3:[0,8,4,5],6:[0,4]]       

If ordering is important the query can be updated to order the lists.

UPDATED

Adding an extra example that also includes '0' as a key in the results.

gremlin>  g.V().hasLabel('0').
......1>        emit().repeat(out()).
......2>        until(__.not(out())).dedup().
......3>        group().
......4>          by(label()).
......5>          by(coalesce(
......6>              repeat(__.in('link')).
......7>              until(__.not(__.in('link'))).
......8>              label().dedup().
......9>              fold()))

==>[0:[],1:[0,4],2:[0,5,8],3:[0,4,5,8],6:[0,4]]