Use deep reinforcement learning with recursive actions?

Question

Can we use recursive actions in deep reinforcement learning ? If yes how ?

For example, in a reassignment problem, for a task i the neural networks returns a node m, and then there will be some condition : If the node m is busy (If there's already a task j assigned to that node), then the task j will be reassigned to another node and so on until a vacant node will be returned (We suppose that there's always at least one vacant node).

Answer 1

Your description doesn't seem so much like a "recurrent action" as a "failed to progress" situation.

The simplest ways to deal with possibility of a blocked/failed action are:

• Leave state and time step unchanged and pick next best action, without any new policy evaluations. This is effectively the approach used by many turn-based game playing RL, such as AlphaGo, although the rejection of non-valid actions is done using the game engine and not technically tried repeatedly in the environment, that does not make a big difference in all cases.

Or

• Update the time step, receive a reward (maybe zero) and try again on next time step. The state information might usefully include blocked actions (or they can be applied programmatically as in the first option). You would typically use this approach if there was a real cost to attempting the action (i.e. a negative reward for losing time) and/or it is possible from the state information to predict which actions could be blocked, and you want to include these factors in the agent's behaviour.

The second option is the approach you might use in a maze solver, if it could only sense walls by bumping in to them.

If the blocked actions are completely unpredictable from the agent's observations, then it doesn't matter too much which approach you use. However, if there is a cost for failing to act, then the second approach would give you more accurate action values in case you are using a value-based approach or need to assess the performance of your agent numerically.