1. Policy Evaluation & Policy Iteration S&B: Sec 4.1, 4.3; 6.5

2. The Bellman equation The final recursive equation is known as the Bellman equation: Unique soln to this eqn gives value of a fixed policy π when operating in a known MDP M = 〈 S, A,T,R 〉 When state/action spaces are discrete, can think of V and R as vectors and T π as matrix, and get matrix eqn:

3. Exercise Solve the matrix Bellman equation (i.e., find V ): I formulated the Bellman equations for “state- based” rewards: R(s) Formulate & solve the B.E. for “state-action” rewards ( R(s,a) ) and “state-action-state” rewards ( R(s,a,s’) )

4. Policy values in practice “Robot” navigation in a grid maze

5. Policy values in practice Optimal policy, π*

6. Policy values in practice Value function for optimal policy, V*

7. A harder “maze”...

8. Optimal policy, π*

9. A harder “maze”... Value function for optimal policy, V*

10. A harder “maze”... Value function for optimal policy, V*

11. Still more complex...

12. Optimal policy, π*

13. Still more complex... Value function for optimal policy, V*

14. Still more complex... Value function for optimal policy, V*

15. Planning: finding π* So we know how to evaluate a single policy, π How do you find the best policy? Remember: still assuming that we know M = 〈 S, A,T,R 〉

16. Planning: finding π* So we know how to evaluate a single policy, π How do you find the best policy? Remember: still assuming that we know M = 〈 S, A,T,R 〉 Non-solution: iterate through all possible π, evaluating each one; keep best

17. Policy iteration & friends Many different solutions available. All exploit some characteristics of MDPs: For infinite-horizon discounted reward in a discrete, finite MDP, there exists at least one optimal, stationary policy (may exist more than one equivalent policy) The Bellman equation expresses recursive structure of an optimal policy Leads to a series of closely related policy solutions: policy iteration, value iteration, generalized policy iteration, etc.

18. The policy iteration alg. Function: policy_iteration Input: MDP M = 〈 S, A,T,R 〉, discount γ Output: optimal policy π* ; opt. value func. V* Initialization: choose π 0 arbitrarily Repeat { V i =eval_policy( M, π i, γ) // from Bellman eqn π i+1 =local_update_policy( π i, V i ) } Until ( π i+1 ==π i ) Function: π’ =local_update_policy( π, V ) for i=1..| S | { π’(s i ) =argmax a ∈ A {sum j ( T(s i,a,s j )*V(s j ) )} }

19. Why does this work? 2 explanations: Theoretical: The local update w.r.t. the policy value is a contractive mapping, ergo a fixed point exists and will be reached See, “contraction mapping”, “Banach fixed- point theorem”, etc. http://math.arizona.edu/~restrepo/475A/Notes/sourcea/n ode22.html http://planetmath.org/encyclopedia/BanachFixedPointTh eorem.html Contracts w.r.t. the Bellman Error:

20. Why does this work? The intuitive explanation It’s doing a dynamic-programming “backup” of reward from reward “sources” At every step, the policy is locally updated to take advantage of new information about reward that is propagated back by the evaluation step Value “propagates away” from sources and the policy is able to say “hey! there’s reward over there! I can get some of that action if I change a bit!”

21. P.I. in action PolicyValue Iteration 0

22. P.I. in action PolicyValue Iteration 1

23. P.I. in action PolicyValue Iteration 2

24. P.I. in action PolicyValue Iteration 3

25. P.I. in action PolicyValue Iteration 4

26. P.I. in action PolicyValue Iteration 5

27. P.I. in action PolicyValue Iteration 6: done

28. Properties & Variants Policy iteration Known to converge (provable) Observed to converge exponentially quickly # iterations is O(ln(| S |)) Empirical observation; strongly believed but no proof (yet) O(| S | 3 ) time per iteration (policy evaluation) Other methods possible Linear program (poly time soln exists) Value iteration Generalized policy iter. (often best in practice)

29. Q : A key operative Critical step in policy iteration π’(s i ) =argmax a ∈ A {sum j ( T(s i,a,s j )*V(s j ) )} Asks “What happens if I ignore π for just one step, and do a instead (and then resume doing π thereafter)?” Often used operation. Gets a special name: Definition: the Q function, is: Policy iter says: “Figure out Q, act greedily according to Q, then update Q and repeat, until you can’t do any better...”

30. What to do with Q Can think of Q as a big table: one entry for each state/action pair “If I’m in state s and take action a, this is my expected discounted reward...” A “one-step” exploration: “In state s, if I deviate from my policy π for one timestep, then keep doing π, is my life better or worse?” Can get V and π from Q :