Unexpected Behavior: Parallel Execution of Asymmetric Graph

[jasontjahjono]

I have this graph:

   B -----
 /         \
A           E 
 \         /
   C --- D

where B takes 5 seconds to run, C takes 2 seconds, and D takes 2 seconds. (i used timeout inside the nodes)
I expect the graph to run in 5 seconds, because I thought C and D can run in parallel while B is running (2+2 < 5 seconds).

However, in practice, it takes 7 seconds. B and C runs in parallel, but C waits for B to finish before moving into D.

Here's how I built the graph:

const workflow = new StateGraph(GraphState)
  .addNode('A', nodeA)
  .addNode('B', nodeB)
  .addNode('C', nodeC)
  .addNode('D', nodeD)
  .addNode('E', nodeE)
  .addEdge(START, 'A')
  .addEdge('A', 'B')
  .addEdge('A', 'C')
  .addEdge('C', 'D')
  .addEdge(['B', 'D'], 'E')
  .addEdge('E', END);

Here's the result:

NODE: In Node A (6ms elapsed)
NODE: In Node B (8ms elapsed)
NODE: In Node C (8ms elapsed)
NODE: In Node D (5016ms elapsed)
NODE: In Node E (7023ms elapsed)
RESULT { messages: [ 'A', 'B', 'C', 'D', 'E' ], startTime: 1735541316193 }

How do I make sure D does not wait for B to finish? I assumed this could be solved using subgraphs but I want to check if there's a more elegant solution than building a subgraph.

[jacoblee93]

Hey @jasontjahjono, I believe we need #720 to land - you could then return a Send from node C.

Otherwise I don't think there's currently a way to get around the fact that nodes are processed one step at a time.

For now, subgraph is the best way to go. Will leave this open.

CC @nfcampos in case I'm missing something