Ping!
Any ideas on this.
Thanks
Pieter
On 23/01/2017 19:00, pieter-gmail wrote:
> Hi,
>
> Ran some more tests, seems all is not well.
>
> So the previous example always works (on TinkerPop and Neo4j) because
> bothE is not seen as two queries but rather one and all is well.
>
> Running the same test with a union instead has the same issue as Sqlg.
>
> @Test
> public void testLazy() {
> final TinkerGraph graph = TinkerGraph.open();
> final Vertex a1 = graph.addVertex(T.label, "A");
> final Vertex b1 = graph.addVertex(T.label, "B");
> final Vertex c1 = graph.addVertex(T.label, "C");
> a1.addEdge("ab", b1);
> a1.addEdge("ac", c1);
>
> AtomicInteger count = new AtomicInteger(0);
> graph.traversal().V(a1).union(outE(), inE()).forEachRemaining(edge -> {
> a1.addEdge("ab", b1); #1
> c1.addEdge("ac", a1); #2
> count.getAndIncrement();
> });
> Assert.assertEquals(2, count.get());
> }
>
> If only #1 is executed the test passes as the outE() is first traversed
> and the subsequent inE() is unchanged.
> However is #2 is executed by the time inE() executes its sees 2 new in
> edges and the count == 4.
>
> I tested on Neo4j and it has the same behavior.
>
> As far as I can tell all is actually behaving as expected with lazy
> iteration assumed.
>
> The unspecified semantics is with bothE(), both sides done immediately
> like Neo4j and TinkerPop, or out and in done lazily as with Sqlg.
> If lazily which side first as if affects the semantics.
>
> Another caveat is if barrier steps are injected and then the semantics
> is not the same as without the barrier step.
>
> Cheers
> Pieter
>
> On 23/01/2017 14:38, Stephen Mallette wrote:
>> I'd say TinkerGraph demonstrates the expected behavior - I guess we don't
>> have a test that enforces that?
>>
>> On Sat, Jan 21, 2017 at 3:23 PM, pieter-gmail <pieter.mar...@gmail.com>
>> wrote:
>>
>>> Ok, thanks.
>>>
>>> Its tricky as it messes with the laziness and visibility of the queries.
>>> Can really think of a solution though.
>>>
>>> Cheers
>>> Pieter
>>>
>>> On 21/01/2017 21:14, Daniel Kuppitz wrote:
>>>> Hi Pieter,
>>>>
>>>> forEachRemaining iterates over the result and hence I would expect the
>>>> result to be 2. Otherwise you would / should end up with an endless loop.
>>>> However, the same behavior is seen when you replace forEachRemaining
>>> with a
>>>> sideEffect lambda step. Not sure what the expected behavior is / should
>>> be,
>>>> but I think I prefer the way it's done now.
>>>>
>>>> Cheers,
>>>> Daniel
>>>>
>>>>
>>>> On Sat, Jan 21, 2017 at 7:30 PM, pieter-gmail <pieter.mar...@gmail.com>
>>>> wrote:
>>>>
>>>>> Hi,
>>>>>
>>>>> I need to clarify the expected semantics of gremlin's lazy iteration
>>>>> semantics.
>>>>> The following gremlin is what highlighted it.
>>>>>
>>>>> ```
>>>>> @Test
>>>>> public void testLazy() {
>>>>> final TinkerGraph graph = TinkerGraph.open();
>>>>> final Vertex a1 = graph.addVertex(T.label, "A");
>>>>> final Vertex b1 = graph.addVertex(T.label, "B");
>>>>> final Vertex c1 = graph.addVertex(T.label, "C");
>>>>> a1.addEdge("ab", b1);
>>>>> a1.addEdge("ac", c1);
>>>>>
>>>>> AtomicInteger count = new AtomicInteger(0);
>>>>> graph.traversal().V(a1).bothE().forEachRemaining(edge -> {
>>>>> a1.addEdge("ab", b1);
>>>>> //a1.addEdge("ac", c1);
>>>>> count.getAndIncrement();
>>>>> });
>>>>> Assert.assertEquals(2, count.get());
>>>>> }
>>>>>
>>>>> ```
>>>>>
>>>>> On TinkerGraph the count is always 2.
>>>>>
>>>>> On Sqlg's Postgresql dialect the `bothE` first traverses the 'ac' edges
>>>>> adds in a 'ab' edge and then traverses the 'ab' edges and ends up with a
>>>>> count of 3.
>>>>>
>>>>> On Sqlg's HSQLDB and H2 dialects the `bothE` first traverses the 'ab'
>>>>> edges adds in a 'ab' edge then traverses the 'ac' edges and ends up with
>>>>> a count of 2.
>>>>>
>>>>> So for Sqlg the added edge will be seen by subsequent traversals due to
>>>>> the lazy nature but the order affects the result.
>>>>>
>>>>> TinkerGraph seems to get both 'ab' and 'ac' edges upfront and does not
>>>>> subsequently see the added edge. A bit like it has a hidden barrier step
>>>>> before the `forEachRemaining`.
>>>>>
>>>>> What is the expected semantics in this situation?
>>>>> Is a traversal that traverses an element that has been modified earlier
>>>>> in the same traversal suppose to see the change?
>>>>>
>>>>> Thanks
>>>>> Pieter
>>>>>
>>>>>
