Gremlin simple path query, to get path based on first edge encountered property

Question

Sample Graph - actual graph Image See here

Code to generate Vertex

        graph.addV("organization")
            .property("name", "CITI")
            .property("type", "ORG")
            .property(T.id, "1")
            .property("orgName", "CITI")
            .iterate();
        graph.addV("component")
            .property("name", "comop1")
            .property("type", "Physical")
            .property("orgName", "CITI")
            .property("app", "APP1")
            .property(T.id, "4013496")
            .iterate();
        graph.addV("component")
            .property("name", "comp2")
            .property("app", "APP2")
            .property("orgName", "ORG")
            .property("type", "System")
            .property(T.id, "2105820")
            .iterate();
        graph.addV("component")
            .property("name", "comp3")
            .property("app", "APP2")
            .property("orgName", "CITI")
            .property("type", "Logical")
            .property(T.id, "2105830")
            .iterate();
        graph.addV("component")
            .property("name", "comp4")
            .property("app", "APP2")
            .property("orgName", "CITI")
            .property("type", "Logical")
            .property(T.id, "2100982")
            .iterate();
        graph.addV("component")
            .property("name", "comp5")
            .property("app", "APP3")
            .property("orgName", "CITI")
            .property("type", "Logical")
            .property(T.id, "4007086")
            .iterate();
        graph.addV("component")
            .property("name", "comp6")
            .property("app", "APP3")
            .property(T.id, "4007087")
            .property("orgName", "CITI")
            .property("type", "Logical")
            .iterate();
        graph.addV("component")
            .property("name", "comp7")
            .property("app", "APP3")
            .property("orgName", "CITI")
            .property("type", "Logical")
            .property(T.id, "4003585")
            .iterate();
        graph.addV("component")
            .property("name", "comp8")
            .property("app", "APP3")
            .property("orgName", "CITI")
            .property("type", "Logical")
            .property(T.id, "4003586")
            .iterate();
        
        graph.addV("organization")
            .property("name", "BOFA")
            .property("orgName", "BOFA")
            .property("type", "Logical")
            .property(T.id, "2")
            .iterate();
        graph.addV("organization")
            .property("name", "JPMC")
            .property("orgName", "JPMC")
            .property("type", "Logical")
            .property(T.id, "3")
            .iterate();

Code to generate Edges

        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge1").from(__.V("1")).to(__.V("4013496")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge2").from(__.V("1")).to(__.V("4013496")).iterate();
        
        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge3").from(__.V("4013496")).to(__.V("2105820")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge4").from(__.V("4013496")).to(__.V("2105820")).iterate();

        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge5").from(__.V("2105820")).to(__.V("2105830")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge6").from(__.V("2105820")).to(__.V("2105830")).iterate();

        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge7").from(__.V("2105830")).to(__.V("2100982")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge8").from(__.V("2105830")).to(__.V("2100982")).iterate();

        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge9").from(__.V("2100982")).to(__.V("4007086")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge10").from(__.V("2100982")).to(__.V("4007087")).iterate();

        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge11").from(__.V("4007086")).to(__.V("4003585")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge12").from(__.V("4007087")).to(__.V("4003586")).iterate();

        graph.addE("commercialService").property("name", "CS1").property(T.id, "edge13").from(__.V("4003585")).to(__.V("2")).iterate();
        graph.addE("commercialService").property("name", "CS2").property(T.id, "edge14").from(__.V("4003586")).to(__.V("3")).iterate();

I have this sample graph, initially 2 edges are coming until they separate out, E1,E2...E14 are the edge ID's and CS1 and CS2 are "name" property of edge.(See Image attached above "Sample Graph")

I am trying to get simple path using the below query

This is a java gremlin query

graph.V("1").
      repeat(outE().otherV().simplePath()).
      until(outE().count().is(0)).
      dedup().
      group().
        by("name").
        by(path()).
      next();

This gives me result as Map<Object, Object>, where key as JPMC and BOFA and 2 different path's as map value.

path[v[1], e[edge1][1-commercialService->4013496], v[4013496], e[edge4][4013496-commercialService->2105820], v[2105820], e[edge6][2105820-commercialService->2105830], v[2105830], e[edge7][2105830-commercialService->2100982], v[2100982], e[edge10][2100982-commercialService->4007087], v[4007087], e[edge12][4007087-commercialService->4003586], v[4003586], e[edge14][4003586-commercialService->3], v[3]]

But when iterate over this path in Java and try to find the edge property "name", I am getting value as CS1 and CS2. It seems when graph is preparing the path it doesn't matter which edge is used to reach the next node.

Where as I am looking for something where we can get the path grouped by "name" property of the edge, like below

path[v[1], e[edge1][1-commercialService->4013496], v[4013496], e[edge3][4013496-commercialService->2105820], v[2105820], e[edge5][2105820-commercialService->2105830], v[2105830], e[edge7][2105830-commercialService->2100982], v[2100982], e[edge9][2100982-commercialService->4007087], v[4007087], e[edge11][4007087-commercialService->4003586], v[4003586], e[edge13][4003586-commercialService->3], v[2]]

2nd Solution tried

graph.V(orgId).repeat(outE().order().by("name").otherV().simplePath()).until(outE().count().is(0)).dedup().path().toList();

This time it is always traversing through single Edge, till we reach the common node. Output :

path[v[1], e[edge1][1-commercialService->4013496], v[4013496], e[edge3][4013496-commercialService->2105820], v[2105820], e[edge5][2105820-commercialService->2105830], v[2105830], e[edge7][2105830-commercialService->2100982], v[2100982], e[edge9][2100982-commercialService->4007086], v[4007086], e[edge11][4007086-commercialService->4003585], v[4003585], e[edge13][4003585-commercialService->2], v[2]]

path[v[1], e[edge1][1-commercialService->4013496], v[4013496], e[edge3][4013496-commercialService->2105820], v[2105820], e[edge5][2105820-commercialService->2105830], v[2105830], e[edge7][2105830-commercialService->2100982], v[2100982], e[edge10][2100982-commercialService->4007087], v[4007087], e[edge12][4007087-commercialService->4003586], v[4003586], e[edge14][4003586-commercialService->3], v[3]]

There is also a way to pass on the "name" property value in the query itself to follow particular path. But I don't have that value with me to pass on. Instead I am thinking if we can some how use the "name" property from the very first edge we encounter in the path ?
Also is there any way to get all the properties of vertex/edge populated when we fetch the path ?

Tried below query as well graph.V(orgId).repeat(outE().otherV().simplePath()).until(outE().count().is(0)).dedup().path().toList() — Parvesh Kumar, Jul 19 '22 at 11:33
To your second question, while not always a good idea as it returns a lot of data (potentially), you can do `path().by(valueMap())`. For the first question I'm not 100% sure what you want to have the query return. Do you only want to follow edges with the same label? — Kelvin Lawrence, Jul 19 '22 at 12:39
As a side note, while graph diagrams are nice, adding the `addV` and `addE` steps that produce the sample graph to the question will help people give you tested answers. See here for an example of creating a sample graph https://stackoverflow.com/questions/72869520/how-to-find-common-vertex-count-and-sort-the-result-in-gremlin — Kelvin Lawrence, Jul 19 '22 at 12:45
@KelvinLawrence yes we can use valueMap, but it gives me everything as plain Object, due to which I can not identify whether its an Vertex or Edge. So I moved to other way where we can fetch more details above the vertex or edge — Parvesh Kumar, Jul 20 '22 at 10:31
@KelvinLawrence : Added the code to create Vertex/Edges and yes you are right I want to follow the particular edge based on the "name" property of the Edge, but I can not pass that name in has("name", "CS1 or CS2") as I don't have that information while querying the graph. I was thinking some thing where I can use the name property from the first edge we encounter for rest of the traversal. So I am still getting 2 paths but each path will use the same edge. — Parvesh Kumar, Jul 20 '22 at 10:35
Thanks for updating with the sample graph - I will take a look. — Kelvin Lawrence, Jul 26 '22 at 14:25
Will there only ever be 2 parallel edges maximum between the initial sets of vertices in the graph/diagram staring from `V[1]` ? The path from `V[1]` has quite an aggressive fan out as at each depth there are 2x more possible paths to that spot. — Kelvin Lawrence, Jul 26 '22 at 22:10
@KelvinLawrence As of now I believe we could have only 2 parallel edges, don't have clear picture of exact data right now. — Parvesh Kumar, Jul 28 '22 at 10:52
Sorry, it took me a while to get back to this. I have added an answer. — Kelvin Lawrence, Aug 07 '22 at 16:33

Kelvin Lawrence · Accepted Answer · 2022-08-07T17:50:49.863

I initially tried using sack to write the query but I ran into an unexpected issue. I then decided to use as labelling. This resulted with the query below:

g.V('1').
  outE().as('e').
  values('name').as('n').
  select('e').
  inV().
  repeat(
    outE().where(eq('n')).by('name').by().inV()).
  until(not(out())).
  path().from('e').
    by('name').
    by(label)

This yields

1   path[CS1, component, CS1, component, CS1, component, CS1, component, CS1, component, CS1, component, CS1, organization]
2   path[CS2, component, CS2, component, CS2, component, CS2, component, CS2, component, CS2, component, CS2, organization]

You can of course label the path however you wish.

g.V('1').
  outE().as('e').
  values('name').as('n').
  select('e').
  inV().
  repeat(
    outE().where(eq('n')).by('name').by().inV()).
  until(not(out())).
  path().from('e').
    by('name')

Which yields:

1   path[CS1, comop1, CS1, comp2, CS1, comp3, CS1, comp4, CS1, comp5, CS1, comp7, CS1, BOFA]
2   path[CS2, comop1, CS2, comp2, CS2, comp3, CS2, comp4, CS2, comp6, CS2, comp8, CS2, JPMC]

Using this approach, if you want the starting vertex in the path result, we have to jiggle things a bit so that the value (edge name) we calculated before the repeat starts is not itself part of the path result. In the prior examples the from step just started the path with the first edge. The version below includes the initial vertex.

g.V('1').as('v1').
  outE().values('name').as('n').
  select('v1').as('start').
  repeat(
    outE().where(eq('n')).by('name').by().inV()).
  until(not(out())).
  path().from('start').
    by('name')

and we now get:

1   path[CITI, CS1, comop1, CS1, comp2, CS1, comp3, CS1, comp4, CS1, comp5, CS1, comp7, CS1, BOFA]
2   path[CITI, CS2, comop1, CS2, comp2, CS2, comp3, CS2, comp4, CS2, comp6, CS2, comp8, CS2, JPMC]

Here is a visual representation of the final query. Hopefully this matches your image that we started the discussion with.

this works great when we are only looking at outgoing edges from a given vertex, is it possible to track bothE(), I tried but getting timeout exception where as I have the until condition specified — Parvesh Kumar, Nov 30 '22 at 08:41

Gremlin simple path query, to get path based on first edge encountered property

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