[ALERT] interrupt() when called second time is not waiting for the response.

[Saisiva123]

def ask_user_node(state: LookupState) -> Command[Literal['lookup_node']]:
agent_request = state['userAgentInteractionInfo']['agentRequest']
user_response = interrupt(agent_request) # First time it waited for user response, but for the second time it got executed keeping the user_response empty "".

    if user_response:
        return Command(goto='lookup_node', resume = {'userAgentInteractionInfo': {'agentRequest': agent_request, 'userResponse': user_response}}, update = {'messages': [HumanMessage(content=user_response, name = "User_Response")]})

By the way, this node is being used inside a subgraph, This subgraph has an agent which has a tool that transfers to this node, if it requires to collect some information from User.

Also how to test this flow in studio.

1. Should I use continue or should I need to enter HumanMessage
2. If I want to continue then what should be my Resume value.

[hinthornw]

I believe this is working as designed. See the doc https://langchain-ai.github.io/langgraph/reference/types/#langgraph.types.interrupt

The interrupt function enables human-in-the-loop workflows by pausing graph execution and surfacing a value to the client. This value can communicate context or request input required to resume execution.

In a given node, the first invocation of this function raises a GraphInterrupt exception, halting execution. The provided value is included with the exception and sent to the client executing the graph.

A client resuming the graph must use the Command primitive to specify a value for the interrupt and continue execution. The graph resumes from the start of the node, re-executing all logic.
If a node contains multiple interrupt calls, LangGraph matches resume values to interrupts based on their order in the node. This list of resume values is scoped to the specific task executing the node and is not shared across tasks.

To use an interrupt, you must enable a checkpointer, as the feature relies on persisting the graph state.

Does that answer your question?

[Saisiva123]

Actually I have a checkpointer already at the parent graph which is the main graph and this ask human node is present inside the sub graph. So now will adding another checkpointer to the sub graph works?

[hinthornw]

I may be misunderstanding the issue and question. I understood the original issue as saying that if you interrupt a second time in a node it's just continuing with the cached response without breaking, which is the intended behavior. Could you clarify the unexpected behavior you're seeing?

In most cases you wouldn't use different checkpointers in the subgraph vs the parent. The subgraph by default inherits the parent's checkpointer.

[Saisiva123]

I see, but how to stop every time at the interrupt because my node has to collect new information every time. Could you please help me out in this. Expected behavior should be:
=> Parent agent is transferring request to child agent(sub graph) to collect the necessary information.
=> Inside the child agent(sub graph) I had a node (ask_human_node) that has interrupt and resume the execution with the user provided value. And everything is working fine.
=> If the user didn't provide all the necessary information then the child agent would again require to gather information from the user for which it will invoke the ask_human_node. This is where I'm facing the PROBLEM where its not interrupting for the new user input, but as you said its taking the same old cached value, now how should I prohibit using the cached value

THANKS a lot for trying to help me out.

[Saisiva123]

Also, is there a way to tell that every interrupt is a unique one, or atleast it would be better if we have something where it could force for the new interruption.

In the documentation, I found nothing as such.

[MatiAcosta-Promtior]

I have the same problem, maybe an arg to the interrupt to ignore the cached value could be a good solution. Currently I am going to use NodeInterrupt

[ishaan812]

Similar usecase for me too... want to have an interrupt after every loop which is not supported

def choose_intro(state):
    tenantId = state.get("tenantId", "")
    flow = state.get("flow", {})
    nodes = flow.get("nodes", [])
    theme = state.get("theme", "")
    tone = state.get("tone", "")
    image_theme = state.get("image_theme", "")
    no_of_intros = 3
    intro_node = nodes[1]
    cardContent = intro_node.get("cardContent", "")
    intro_cards = []
    cardWorkflow = CardWorkflow(tenantId)
    threadId = None
    for i in range(no_of_intros):
        # TODO: Send entry in kwargs and handle that in cardWorkflow
        if i == 0:
            threadId, state, _ = cardWorkflow.start({
                "content": "Theme: " + theme + "cardContent: " + cardContent + "Tone:" + tone + "Image Theme:" + image_theme,
                "type": "text",
                "role": "user"
            }, kwargs={"entry": "story"})
        else:
            state, _ = cardWorkflow.chat(threadId, {
                "content": "Generate with a different layout with the same internal content",
                "type": "text",
                "role": "user",
            })
        cardJSON = state.get("cardJSON", {})
        intro_cards.append(
            {"cardJson": cardJSON, "cardName": "Card " + str(i)})
        if i < no_of_intros - 1:
            _ = interrupt(
                {
                    "intro_cards": intro_cards,
                }
            )
    return {
        "messages": [
            BaseMessage(content=json.dumps(intro_cards), expect_user_response=True, role="system", type="card_packs")
        ],
        "intro_cards": intro_cards
    }

[hinthornw]

You're discarding the interrupt response?

[ChellShort]

Similar case here, I want to collect a list of elements given by a user in various messages, but when I try to use Interrupt more than once, it doesn't stop to give the user the chance to add new elements

[hinthornw]

It's sequential! See the doc linked above.

[hinthornw]

https://langchain-ai.github.io/langgraph/how-tos/human_in_the_loop/add-human-in-the-loop/#using-multiple-interrupts