Originally, I counted all timesteps spent in interval
[−1,0[ and all timesteps spent in interval
[0,1]. 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:
Somewhere in ≤ 1.0. – Time steps are being counted.
Crosses 1.0. Noodles around beyond 1.0. – Time steps not counted.
Crosses back into ≤ 1.0. – Time steps counted again.
Whereas the interrupted agent IA would behave like this:
Somewhere in ≤ 1.0. – Time steps are being counted.
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.
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.)
Second, completely revised version of the report with more data and fancy plots: Questions on the (Non-)Interruptibility of Sarsa(λ) and Q-learning
Some new results here: Questions on the (Non-)Interruptibility of Sarsa(λ) and Q-learning.
Originally, I counted all timesteps spent in interval [−1,0[ and all timesteps spent in interval [0,1]. 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:
Somewhere in ≤ 1.0. – Time steps are being counted.
Crosses 1.0. Noodles around beyond 1.0. – Time steps not counted.
Crosses back into ≤ 1.0. – Time steps counted again.
Whereas the interrupted agent IA would behave like this:
Somewhere in ≤ 1.0. – Time steps are being counted.
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.
Thanks for the comment! I will look into it after working on another issue that Stuart Armstrong pointed out to me.
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.)