Intelligent Agent Foundations Forumsign up / log in
(Non-)Interruptibility of Sarsa(λ) and Q-Learning
link by Richard Möhn 462 days ago | Jessica Taylor and Patrick LaVictoire like this | 5 comments


by Richard Möhn 349 days ago | Patrick LaVictoire likes this | link

Second, completely revised version of the report with more data and fancy plots: Questions on the (Non-)Interruptibility of Sarsa(λ) and Q-learning

reply

by Patrick LaVictoire 449 days ago | link

Nice! One thing that might be useful for context: what’s the theoretical correct amount of time that you would expect an algorithm to spend on the right vs. the left if the session gets interrupted each time it goes 1 unit to the right? (I feel like there should be a pretty straightforward way to calculate the heuristic version where the movement is just Brownian motion that gets interrupted early if it hits +1.)

reply

by Richard Möhn 440 days ago | link

Thanks for the comment! I will look into it after working on another issue that Stuart Armstrong pointed out to me.

reply

by Richard Möhn 392 days ago | link

Originally, I counted all timesteps spent in interval \(\left[-1,0\right[\) and all timesteps spent in interval \(\left[0,1\right]\). As Stuart Armstrong pointed out, this might make even a perfectly interruptible learner look like it’s influenced by interruptions. To understand this, consider the following example.

The uninterrupted agent UA could behave like this:

  1. Somewhere in ≤ 1.0. – Time steps are being counted.
  2. Crosses 1.0. Noodles around beyond 1.0. – Time steps not counted.
  3. Crosses back into ≤ 1.0. – Time steps counted again.

Whereas the interrupted agent IA would behave like this:

  1. Somewhere in ≤ 1.0. – Time steps are being counted.
  2. Crosses 1.0. No more time steps counted.

So even if IA behaved the same as UA before the cross, UA would have extra steps from stage 3 and thus appear less biased towards the left.

As an alternative to using Brownian motion, Patrick suggested to stop counting once the cart crosses \(1.0\). This makes the UA scenario look like the IA scenario, so the true nature of the agent should come to light…

Anyway, with this modification it turns out not obvious that interruptions push the cart to the left. I will start looking more sharply.

reply

by Richard Möhn 357 days ago | link

Some new results here: Questions on the (Non-)Interruptibility of Sarsa(λ) and Q-learning.

reply



NEW LINKS

NEW POSTS

NEW DISCUSSION POSTS

RECENT COMMENTS

[Delegative Reinforcement
by Vadim Kosoy on Stable Pointers to Value II: Environmental Goals | 1 like

Intermediate update: The
by Alex Appel on Further Progress on a Bayesian Version of Logical ... | 0 likes

Since Briggs [1] shows that
by 258 on In memoryless Cartesian environments, every UDT po... | 2 likes

This doesn't quite work. The
by Nisan Stiennon on Logical counterfactuals and differential privacy | 0 likes

I at first didn't understand
by Sam Eisenstat on An Untrollable Mathematician | 1 like

This is somewhat related to
by Vadim Kosoy on The set of Logical Inductors is not Convex | 0 likes

This uses logical inductors
by Abram Demski on The set of Logical Inductors is not Convex | 0 likes

Nice writeup. Is one-boxing
by Tom Everitt on Smoking Lesion Steelman II | 0 likes

Hi Alex! The definition of
by Vadim Kosoy on Delegative Inverse Reinforcement Learning | 0 likes

A summary that might be
by Alex Appel on Delegative Inverse Reinforcement Learning | 1 like

I don't believe that
by Alex Appel on Delegative Inverse Reinforcement Learning | 0 likes

This is exactly the sort of
by Stuart Armstrong on Being legible to other agents by committing to usi... | 0 likes

When considering an embedder
by Jack Gallagher on Where does ADT Go Wrong? | 0 likes

The differences between this
by Abram Demski on Policy Selection Solves Most Problems | 1 like

Looking "at the very
by Abram Demski on Policy Selection Solves Most Problems | 0 likes

RSS

Privacy & Terms