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R. Brafman and M. Tennenholtz Presented by Daniel Rasmussen
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Exploration versus Exploitation Should we follow our best known path for a guaranteed reward, or take an unknown path for a chance at a greater reward? Explicit policies do not work well in multiagent settings R-Max uses an implicit exploration/exploitation policy How does it do this? Why should we believe it works?
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Background R-Max algorithm Proof of near-optimality Variations on the algorithm Conclusions
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A set of (normal form) games Each game has a set of possible actions and rewards associated with those actions as usual We add a probabilistic transfer function that maps the current game and the players’ actions onto the next game
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(2,2)(3, 1) (2, 3)(1, 1) (4,2)(1, 1) (2, 2)(1, 0) (0,2)(5, 1) (2, 4)(3, 1) (4,3)(3, 3) (2, 3)(1, 1) (2,0)(3, 0) (2, 1)(1, 1) (5,2)(0, 1) (4, 3)(5, 1) 0.7 0.3 0.2 0.4
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History: the set of all past states, the actions taken in those states, and the rewards received (plus the current state) (S x A 2 x R) t x S Policy: a mapping from the set of histories to the set of possible actions in the current state P: (S x A 2 x R) t x S A, for all t >= 0 Value of a policy: expected average reward if we follow the policy for T steps Denoted U(s, p a, p o, T) U(p a ) = min po min s lim T..∞ U(s, p a, p o, T)
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ε-return mixing time: how many steps do we need to take before the expected average reward is within ε of the stable average Min T such that U(s, p a, p o, T) > U(p a ) – ε for all s and p o Optimal policy: the policy with the highest U(p a ) We will always parameterize this in ε and T (i.e. the policy with the highest U(p a ) whose ε-return mixing time is T)
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The agent always knows what state it is in After each stage the agent knows what actions were taken and what rewards were received We know the maximum possible reward, R max We know the ε-return mixing time of any policy
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When faced with the choice between a known and unknown reward, always try the unknown reward
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Maintain an internal model of the stochastic game Calculate an optimal policy according to model and carry it out Update model based on observations Calculate a new optimal policy and repeat
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Input N: number of games k: number of actions in each game ε: the error bound δ: the probability of failure R max : the maximum reward value T: the ε-return mixing time of an optimal policy
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Initializing the internal model Create states {G 1...G n } to represent the stages in the stochastic game Create a fictitious game G 0 Initialize all rewards to (R max, 0) Set all transfer functions to point to G 0 Associate a boolean known/unknown variable with each entry in each game, initialized to unknown Associate a list of states reached with each entry, which is initially empty
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(2,2)(3, 1) (2, 3)(1, 1) (4,2)(1, 1) (2, 2)(1, 0) (0,2)(5, 1) (2, 4)(3, 1) (4,3)(3, 3) (2, 3)(1, 1) (2,0)(3, 0) (2, 1)(1, 1) (5,2)(0, 1) (4, 3)(5, 1) 0.7 0.3 0.2 0.4 G1G1 G2G2 G3G3 G4G4 G5G5 G6G6
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R max,0 unknown {} unknown {} unknown {} unknown {} G1G1 G2G2 G3G3 G4G4 G5G5 G6G6 G0G0
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Repeat Compute an optimal policy for T steps based on the current internal model Execute that policy for T steps After each step: ▪ If an entry was visited for the first time, update the rewards based on observations ▪ Update the list of states reached from that entry ▪ If the list of states reached now contains c+1 elements ▪ mark that entry as known ▪ update the transition function ▪ compute a new policy
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R max,0 unknown {} unknown {} unknown {} unknown {} G1G1 G2G2 G3G3 G4G4 G5G5 G6G6 G0G0
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2,2R max,0 unknown {(G 4,1)} unknown {} unknown {} unknown {} G1G1 G2G2 G3G3 G4G4 G5G5 G6G6 G0G0
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2,2R max,0 3, 3 R max,0 unknown {(G 4,1)} unknown {} unknown {} unknown {} G1G1 G2G2 G3G3 G4G4 G5G5 G6G6 G0G0 unknown {(G 1,1)}
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2,2R max,0 3, 3 R max,0 unknown {(G 4,1)(G 2,1)} unknown {} unknown {} unknown {} G1G1 G2G2 G3G3 G4G4 G5G5 G6G6 G0G0 unknown {(G 1, 1)}
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2,2R max,0 3, 3 R max,0 unknown {} unknown {} unknown {} G1G1 G2G2 G3G3 G4G4 G5G5 G6G6 G0G0 unknown {(G 1, 1)} 0.5 known {(G 4,1)(G 2,1)}
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Goal: prove that if the agent follows the R-Max algorithm for T steps the average reward will be within ε of the optimum Outline of proof: After T steps the agent will have either obtained near-optimum average reward or it will have learned something new There are a polynomial number of parameters to learn, therefore the agent can completely learn its model in polynomial time The adversary can block learning, but if it does so then the agent will obtain near-optimum reward Either way the agent wins!
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Lemma 1: If the agent’s model is a sufficiently close approximation of the true game, then an optimal policy in the model will be near- optimal in the game We guarantee that the model is sufficiently close by waiting until we have c + 1 samples before marking an entry known
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Lemma 2: the difference between the expected reward based on the model and the actual reward will be less than the exploration probability times R max |V model – V actual | < eR max large |V model – V actual | large exploration probability
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Combining Lemma 1 and 2 In unknown states the agent has an unrealistically high expectation of reward (R max ), so Lemma 2 tells us that the probability of exploration is high In known states Lemma 1 tells us that we will obtain near-optimal reward We will always be in either a known or unknown state, therefore we will always explore with high probability or obtain near-optimal reward If we explore for long enough then (almost) all states will be known, and we are guaranteed near-optimal reward
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Remove assumption that we know the ε- return mixing time of an optimal policy, T Remove assumption that we know R max Simplified version for repeated games
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R-Max is a model-based reinforcement learning algorithm that is guaranteed to converge on the near-optimal average reward in polynomial time Works on zero-sum stochastic games, MDPs, and repeated games The authors provide a formal justification for optimism under uncertainty heuristic Guarantee that the agent either obtains near-optimal reward or learns efficiently
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For complicated games N, k, and T are all likely to be high, so polynomial time will still not be computationally feasible We do not consider how the agent’s behaviour might impact the adversary’s
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