Mark tool parameter as ignored

[nmccann]

With the recent introduction of kwargs flowing through to tools, it is now possible to have an ai_function that does this:

@ai_function
def search_tool(query: str, **kwargs: Any) -> str:
   deterministic_filter = kwargs.get("deterministic_filter")

   return f"Performed query {query} with filter {deterministic_filter}"

And call it like so:

# Set up agent that uses the search_tool

result = await agent.run("How fast is a cheetah?", deterministic_filter="animals")

With this, the agent's LLM will see that search_tool expects a query parameter that should be a str, and it will populate that parameter. The deterministic_filter will end up propagating through to the kwargs and ensure that the filter is always the same, and not subject to the whims of the LLM.

This kwargs approach is nice, but has downsides. For one, additional work is needed to extract the value from kwargs before it can be used (and potentially requires casting it to the desired type). Without looking at the implementation, it's unclear what kwargs are applicable for the search_tool, or what type they should be.

An alternative would be:

@ai_function
def search_tool(query: str, deterministic_filter: str, **kwargs: Any) -> str:
   return f"Performed query {query} with filter {deterministic_filter}"

In my testing the end result is the same, because the value from kwargs takes priority over the LLM's generated value. However this isn't ideal as the LLM potentially wastes tokens on parameters it shouldn't worry about, and creates ambiguity about whether the value of the parameter is deterministic or not.

My idea would be to leverage some of Pydantic's capabilities (maybe combining SkipJSONSchema with Annotated) to denote a parameter as ignored for the purposes of the tool's representation to the LLM.

I thought SemanticKernel had something similar, but the closest thing I was able to find with some quick searching was this which doesn't seem to have been implemented.

Edit: In my testing, the second example actually uses the LLM's generated value if you omit **kwargs: Any from the function definition. I'm not sure why that would be the case.