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100 things I know
 

100 things I know

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Anexo a reporte avances proyecto "Más allá del aprendizaje por refuerzo con recompensas promedio"

Anexo a reporte avances proyecto "Más allá del aprendizaje por refuerzo con recompensas promedio"

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    100 things I know 100 things I know Presentation Transcript

    • 100 things I know. Part I of III Reinaldo Uribe M Mar. 4, 2012
    • SMDP Problem Description. 1. In a Markov Decision Process, a (learning) agent is embedded in an envionment and takes actions that affect that environment. States: s ∈ S. Actions: a ∈ As ; A = s∈S As . (Stationary) system dynamics: transition from s to s after taking a, with probability a Pss = p(s |s, a) Rewards: Ra . Def. r(s, a) = E Ra | s, a ss ss At time t, the agent is in state st , takes action at , transitions to state st+1 and observes reinforcement rt+1 with expectation r(st , at ).
    • SMDP Problem Description. 2. Policies, value and optimal policies. An element π of the policy space Π indicates what action, π(s), to take at each state. The value of a policy from a given state, v π (s) is the expected cumulative reward received starting in s and following π: ∞ v π (s) = E γ t r(st , π(st )) | s0 = s, π t=0 0 < γ ≤ 1 is a discount factor. An optimal policy, π ∗ has maximum value at every state: π ∗ (s) ∈ argmax v π (s) ∀s π∈Π π∗ v ∗ (s) = v (s) ≥ v π (s) ∀π ∈ Π
    • SMDP Problem Description. 3. Discount Makes infinite-horizon value bounded if rewards are bounded. Ostensibly makes rewards received sooner more desirable than those received later. But, exponential terms make analysis awkward and harder... ... and γ has unexpected, undesirable effects, as shown in Uribe et al. 2011 Therefore, hereon γ = 1. See section Discount, at the end, for discussion.
    • SMDP Problem Description. 4. Average reward models. A more natural long term measure of optimality exists for such cyclical tasks, based on maximizing the average reward per action. Mahadevan 1996 n−1 1 ρπ (s) = lim E r(st , π(st )) | s0 = 0, π n→∞ n t=0 Optimal policy: ρ∗ (s) ≥ ρπ (s) ∀s, π ∈ Π Remark: All actions equally costly.
    • SMDP Problem Description 5. Semi-Markov Decision Process: usual approach, transition times. Agent is in state st and takes action π(st ) at decision epoch t. After an average of Nt units of time, the sistem evolves to state st+1 and the agent observes rt+1 with expectation r(st , π(st )). In general, Nt (st , at , st+1 ). Gain (of a policy at a state): n−1 π E t=0 r(st , π(st )) | s0 = s, π ρ (s) = lim n→∞ n−1 E t=0 Nt | s0 = s, π Optimizing gain still maximizes average reward per action, but actions are no longer equally weighted. (Unless all Nt = 1)
    • SMDP Problem Description 6.a Semi-Markov Decision Process: explicit action costs. Taking an action takes time, costs money, or consumes energy. (Or any combination thereof) Either way, real valued cost kt+1 not necessarily related to process rewards. Cost can depend on a, s and (less common in practice) s . Generally, actions have positive cost. We simply require all policies to have positive expected cost. Wlog the magnitude of the smallest nonzero average action cost is forced to be unity: |k(a, s)| ≥ 1 ∀k(a, s) = 0
    • SMDP Problem Description 6.b Semi-Markov Decision Process: explicit action costs. Cost of a policy from a state: n−1 cπ (s) = lim E k(st , π(st )) | s0 = s, π n→∞ t=0 So cπ (s) > 0 ∀π ∈ Π, s. Nt = k(st , π(st )). Only their definition/interpretation changes. Gain v π (s)/n ρπ (s) = cπ (s)/n
    • SMDP Problem Description 7. Optimality of π ∗ : π ∗ ∈ Π with gain n−1 E t=0 r(st , π(st )) | s0 = s, π ∗ ∗ v π (s) π∗ ∗ ρ (s) = ρ (s) = lim = π∗ n→∞ n−1 = s, π ∗ c (s) E t=0 k(st , π(st )) | s0 is optimal if ρ∗ (s) ≥ ρπ (s) ∀s, π ∈ Π, as it was in ARRL. Notice that the optimal policy doesn’t necessarily maximize v π or minimize cπ . Only optimizes their ratio.
    • SMDP Problem Description 8. Policies in ARRL and SMDPs are evaluated using the average-adjusted sum of rewards: n−1 H π (s) = lim E (r(st , π(st )) − ρπ (s)) | s0 = s, π n→∞ t=0 Puterman 1994, Abounadi et al. 2001, Ghavamzadeh & Mahadevan 2007 This signals the existence of bias optimal policies that, while gain optimal, also maximize the transitory rewards received before entering recurrence. We are interested in gain-optimal policies only. (It is hard enough...)
    • SMDP Problem Description 9. The Unichain Property A process is unichain if every policy has a single, unique recurrent class. I.e. if for every policy, all recurrent states communicate between them. All methods rely on the unichain property. (Because, if it holds:) ρπ (s) is constant for all s. ρπ (s) = ρπ Gain and value expressions simplify. (See next) However, deciding if a problem is unichain is NP-Hard. Tsitsiklis 2003
    • SMDP Problem Description 10. Unichain property under recurrent states. Feinberg & Yang, 2010 A state is recurrent if it belongs to a recurrent class of every policy. A recurrent state can be found, or proven not to exist, in polynomial time. If a recurrent state exists, determining whether the unichain property holds can be done in polynomial time. (We are not going to actually do it–it requires knowledge of the system dynamics–but good to know!) Recurrent states seem useful. In fact, existence of a recurrent state is more critical to our purposes that the unichain property. Both will be required in principle for our methods/analysis, until their necessity is furher qualified in section Unichain Considerations below.
    • Intermission
    • Generic Learning Algorithm 11. The relevant expressions under our assumptions simplify, losing dependence on s0 The following Bellman equation holds for average-adjusted state value: H π (s) = r(s, π(s)) − k(s, π(s))(ρπ ) + Eπ H π (s ) (1) Ghavamzadeh & Mahadevan 2007 Reinforcement Learning methods exploit Eq. (1), running the process and substituting: State for state-action pair value. Expected for obseved reward and cost. ρπ for an estimate. H π (s ) for its current estimate.
    • Generic Learning Algorithm 12. Algorithm 1 Generic SMDP solver Initialize repeat forever Act Do RL to find value of current π Usually 1-step Q-learning Update ρ.
    • Generic Learning Algorithm 13. Model-based state value update: H t+1 (st ) ← max r(st , a) + Ea H t (st+1 ) a Ea emphasizes that expected value of next state depends on action chosen/taken. Model free state-action pair value update: Qt+1 (st , at ) ← (1 − γt ) Qt (st , at )+ γt rt+1 − ρt ct+1 + max Qt (st+1 , a) a In ARRL, ct = 1 ∀t
    • Generic Learning Algorithm 14.a Table of algorithms. ARRL Algorithm Gain update t AAC r(si , π i (si )) i=0 Jalali and Ferguson 1989 ρt+1 ← t+1 t+1 R–Learning ρ ← (1 − α)ρt + Schwartz 1993 α rt+1 + max Qt (st+1 , a) − max Qt (st , a) a a H–Learning ρt+1 ← (1−αt )ρt +αt rt+1 − H t (st ) + H t (st+1 ) αt Tadepalli and Ok 1998 αt+1 ← αt + 1 SSP Q-Learning ρt+1 ← ρt + αt min Qt (ˆ, a) s Abounadi et al. 2001 a t HAR r(si , π i (si )) i=0 Ghavamzadeh and Mahadevan 2007 ρt+1 ← t+1
    • Generic Learning Algorithm 14.b Table of algorithms. SMDPRL Algorithm Gain update SMART t Das et al. 1999 r(si , π i (si )) i=0 ρt+1 ← t MAX-Q c(si , π i (si )) Ghavamzadeh and Mahadevan 2001 i=0
    • SSP Q-Learning 15. Stochastic Shortest Path Q-Learning Most interesting. ARRL If unichain and exists s recurrent (Assumption 2.1 ): ˆ SSP Q-learning is based on the observation that the average cost under any stationary policy is simply the ratio of expected total cost and expected time between two successive visits to the reference state [ˆ] s Thus, they propose (after Bertsekas 1998) making the process episodic, splitting s into the (unique) initial and terminal ˆ states. If the Assumption holds, termination has probability 1. Only the value/cost of the initial state are important. Optimal solution “can be shown to happen” when H(ˆ) = 0. s (See next section)
    • SSP Q-Learning 16. SSPQ ρ update. ρt+1 ← ρt + αt min Qt (ˆ, a), s a where 2 αt → ∞; αt < ∞. t t But it is hard to prove boundedness of {ρt }, so suggested instead ρt+1 ← Γ ρt + αt min Qt (ˆ, a) , s a with Γ(·) a projection to [−K, K] and ρ∗ ∈ (−K, K).
    • A Critique 17. Complexity. Unknown. While RL is PAC. 18. Convergence. Not always guaranteed (ex. R-Learning). When proven, asymptotic: convergence to the optimal policy/value if all state-action pairs are visited infinite times. Usually proven depending on decaying learning rates, which make learning even slower.
    • A Critique 19. Convergence of ρ updates. ... while the second “slow” iteration gradually guides [ρt ] to the desired value. Abounadi et al. 2001 It is the slow one! Must be so for sufficient approximation of current policy value for improvement. Initially biased towards (likely poor) observed returns at the start. A long time must probably pass following the optimal policy for ρ to converge to actual value.
    • Our method 20. Favours an understanding of the −ρ term, either alone in ARRL or as a factor of costs in SMDPs, not so much as an approximation to average rewards but as a punishment for taking actions, which must be made “worth it” by the rewards. I.e. nudging. Exploits the splitting of SSP Q-Learning, in order to focus on the value/cost of a single state, s. ˆ Thus, also assumes the existence of a recurrent state, and that the unichain policy holds. (For the time being) Attempts to ensure an accelerated convergence of ρ updates. In a context in which certain, efficient convergence can be easily introduced.
    • Intermission
    • Fractional programming 21. So, ‘Bertsekas splitting’ of s into initial sI and terminal sT . ˆ Then, from sI Any policy π ∈ Π has an expected return until termination v π (sI ), and an expected cost until termination cπ (sI ). v π (sI ) The ARRL problem, then, becomes max π π∈Π c (sI ) Lemma v π (sI ) argmax = argmax v π (s) + ρ∗ (−cπ (s)) π∈Π cπ (sI ) π∈Π For ρ∗ such that max v π (s) + ρ(−cπ (s)) = 0 π∈Π
    • Fractional programming 22. Implications. Assume the gain, ρ∗ is known. Then, the nonlinear SMDP problem reduces to RL. Which is better studied, well understood, simpler, and for which sophisticated, efficient algorithms exist. If we only use (r − ρ∗ c)(s, a, s ). Problem: ρ∗ is usually not known.
    • Nudging 23. Idea: Separate reinforcement learning (leave it to the pros) from updating ρ. Thus, value-learning becomes method-free. We can use any old RL method. Gain update is actually the most critical step. Punish too little, and the agent will not care about hurrying, only collecting reward. Punish too much, and the agent will only care about finishing already. In that sense, (r − ρc) is like picking fruit inside a maze.
    • Nudging 24. The problem reduces to a sequence of RL problems. For a sequence of (temporarily fixed) ρk Some of the methods already provide an indication of the sign of ρ updates. We just don’t hurry to update ρ after taking a single action. Plus the method comes armed with a termination condition: As soon as H k (sI ) = 0 then π k = π ∗ .
    • Nudging 25. Algorithm 2 Nudged SSP Learning Initialize repeat Set reward scheme to (r − cρ) Solve by any RL method. Update ρ From current H π (sI ) until H π (sI ) = 0
    • w − l space 26. D We will propose a method for updating ρ and show that it minimizes uncertainty between steps. For that, we will use a transformation that extends the work of our CIG paper. But First. Let D be a bound on the magnitude of unnudged reward D ≥ lim sup{H π (sI ) | ρ = 0} π∈Π D ≤ lim inf {H π (sI ) | ρ = 0} π∈Π Observe interval (−D, D) bounds ρ∗ but the upper bound is tight only in ARRL if all of D reward is received in a single step from sI .
    • w − l space 27. All policies π ∈ Π, from (that is, at) sI have: real expected value |v π (sI )| ≤ D. positive cost cπ (sI ) ≥ 1 28.a w − l transformation: D+v π (sI ) D−v π (sI ) w= 2cπ (sI ) l= 2cπ (sI )
    • w − l space 28.b w − l plane. D l w D
    • w − l space 29. Properties: w, l ≥ 0 w, l ≤ D D w+l = ≤D cπ (s I) v π (sI ) = D ⇒ l=0 v π (sI ) = −D ⇒ w=0 lim (w, l) = (0, 0) cπ (sI )→∞ 30. Inverse transformation: π π v π (sI ) = D wπ −lπ w +l cπ (sI ) = D wπ1 π +l
    • Intermission
    • w − l space 31. Value. w π − lπ v π (sI ) = D D w π + lπ Level sets are lines. w–axis, expected D. l–axis, expected −D. w = l, expected 0. 5D −D −0. l Optimization → fanning from l = 0. 0 Not convex, but splits the 0.5D space. So optimizers are vertices of D w D convex hull of policies.
    • w − l space 32. Cost. D 1 cπ (sI ) = D wπ+ lπ Level sets are lnes with slope −1. w + l = D, expected cost 1. l Cost decreases with distance 1 to the origin. Cost optimizers (both max 2 and min) also vertices. 4 8 w D
    • w − l space 33. The origin. Policies of infinite expected cost. Mean the problem is not unichain or sI is not recurrent. And are troublesome for optimizing value. So under our assumptions, the origin does not belong to the space
    • Nudged value in the w − l space 34. SMDP problem in w − l. π π v π (sI ) D wπ −lπ argmax π = argmax w 1 +l = argmax wπ − lπ π∈Π c (sI π∈Π D wπ +lπ π∈Π D /2 D − l ● ● ● ●● ● ● ● ● ● ●● ●● ● ● ● ● ● ●● ● ● ●●●●● ● ● ● ● ● ●●●●● ●●●●● ● ● ● ● ●● ● ● ●● ● ● ● ●● ● ● ● ● ●●● ● ●●● ● ● ● ● ● ●●●●● ●●● ●● ● ●● ● ● ● ● ● ●●● ● ●●●●●●●● ●●● ●● ●● ●● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ●● ●● ●● ● ●● ●● ●●●●●●●●● ● ● ● ● ● ● ●●●● ● ● ● ●●●●● ● ●●●● ● ● ● ● ● ●●● ●●●● ●●●●●●●●●●●●●●●● ● ● ● ● ●●●●● ●●●●●●●● ●●●● ● ● ●●●●●●● ●●●●●●●●●●●● ● ●●●● ● ● ●●●●●● ●● ● ●●● ● ● ● ● ●●● ● ● ● ● ●●●● ●● ●●● ● ● ●● ● ● ● ●● ●●●●●●●●●●●●●●●●●● ● ●● ● ● ●●●●●●●●●●●●●●●●●●● ● ● ● ●●● ●●●● ●●●● ● ● ● ● ● ● ● ●●● ●●●●●●●●● ● ●● ● ● ● ●● ● ● ● ● ●●●●●●●●●● ●●●●●●● ●●● ●● ● ● ●●●●●●●●●●●●●●●● ● ● ● ● ●●● ●● ●●●●●●●●●●●●●●●●●●●●●● ●●● ● ●●● ●●●● ●●●●● ●● ●●●● ● ● ● ● ● ● ●●●● ● ● ● ●● ●●●●●●●●●●●●●●●●●●●●●●● ● ● ● ● ● ● ●●●●●●●●●●●●●●●●●●●●● ● ● ●●●● ●●●● ●●●●●●● ●● ● ● ●●●●●●●●●●●●●● ●●●●● ● ●● ●●●●●●●●●●●●●●● ● ● ● ● ●●●●● ●●●●●●●●●●●●●●●●● ● ● ● ●● ● ● ●●●● 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    • Nudged value in the w − l space 35. Nudged value, for some ρ. argmax v π (sI ) − ρcπ (sI ) π∈Π w π − lπ 1 = argmax D π + lπ − ρD π π∈Π w w + lπ w π − lπ − ρ = argmax D π∈Π w π + lπ
    • Nudged value in the w − l space 36. Nudged value level sets ˆ (For a set ρ and all policies π with a given h) ˆ ˆ D−h π D lπ = ˆ wˆ − ρ ˆ D+h D+hˆ Lines! ˆ Slope depends only on h (i.e., not on ρ)
    • Nudged value in the w − l space 37. Pencil of lines ˆ ˇ For a set ρ, any two h and h level set lines have intersection: ρ ρ ,− 2 2 Pencil of lines with that vertex.
    • Nudged value in the w − l space D 38. Zero nudged value. D−0 π D lπ = ˆ wˆ − ρ D+0 D+0 l lπ = w π − ρ ˆ ˆ −D Unity slope. 0 Negative values above, positive below. w D D  ρ   ρ    If whole cloud above w = l, some  ,−     2 2  negative nudging is the optimizer. (Encouragement)
    • Nudged value in the w − l space D l ● ● ● ●● ● ● ● ● ●● ●● ● ●● ● ● ●● ●● ● ● ● ● ● ● ●●●● ● ● ● ● ● ● ●●● ● ● ● ● ● ● ● ● ●●●●●●●●● ●●●●● ● ● ● ●● ● ● ● ● ● ●● ● ● ● ● ● ● ●●●●●●●●●●●●●●● ● ● ● ●●●●●●●●● ●●●●● ● ● ● ●●●●●●●●●●●● ● ●● ● ●●●●●●●●●● ●●●● ● ●● ●● ● ●●●● ●● ● ●● ● ●●●●●●●●●●●●●●●●●● ● ●● ● ● ●●●●●●●●● ● ●● ●●● ●●●●●●●●●● ● ● ●●● ●●●●●●●●●●●●● ●● ● ●● ●●● ● ●● ● ● ●●●●●●●●●●●●●●●●●● ● ● ● ●● ●● ●●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●● ● ● ● ●● ●●●●●● ●● ●●●●●●●●●●●●●●●●● ● ●● ● ●●● ● ●●●●●●●●●●●●●●●●●●●●●●● ●●●●● ●●● ●● ●●●●●●●●●●● ● ●● ●● ●●●●●●●●●●●●●●●● ● ●●● ● ● ●●●●●●●●●●●● ● ●● ●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●● ●●● ●● ● ●● ● ● ● ●●● ● ●●●●●●●●●●●●●●●●●●●●●●● ●● ●●●●●●●●●●●●●● ● ●● ●● ● ●● ●●●●●●●●●●●●●●● ● ●●●● ● ● ●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●●●●●● ●●●●●●●●●●● ● ● ●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●● ● ● ●●●● ●●●● ● ●●●●●●●●●●●●●●● ●● ● ●● ●●●●●●●●●●●●●●●●●●●● ● ●●●●● ●●●● ●● ● ● ● ●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●● ● ● ●●● ● ●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●● ●●● ● ●● ●●●●●●●●●●●●●●●●●●●●● ●●●●●●●●●●●●●●● ● ● ● ● ●● ●● ●● ●●●●●●● ● ●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●●●●● ●●●●●●●●●●●●●●●●●●● ●● ●●●●●●●●●●●●●● ● ● ● ●●●● ●●● ● ●● ● ●● ● ●●●●●●●●●●●●●●●●●●●●●● ● ● ● ● ●●●●●●●● ● ● ● ● ●●●●●●●●●●●●●●●●●●●● ● ● ●●● ●●●● ●●●●●●●●●●●●● ●● ● ● ●●●●●●●●●●●●●●●●● ●● ●● ●●●●●●●●●● ●● ●●●●●●●●●●●●●●●●●●●●●● ● ●●● ●●●●●●●●● ●● ● ● ●●●●●● ●●●● ● ● ●●●●●●●●●●●●●●●●● ● ● ●●● ●●● ● ●●●●●●●●●●●●●●●●●● ● ● ●● ●●●●●●●●●●● ●●●●●●●●●●●●●●●●●●●●●●● ●● ● ● ●●●●●●● ●●●●● ● ●●●●●●●●●●●●●●●● ●●● ●●●●●●●●●●● ●●●● ● ● ● ●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●●●● ● ● ● ● ●●●●● ● ●● ● ●●●●●●●●●●●●●●●●● ●●● ●●● ● ●●●●●●●●●● ● ● ● ●● ●●● ●●● ●● ● ●●● ● ● ●● ●●●●●●●●●●●●●●●●●●●● ●●● ●● ●●●●●●●●●●●●●●●●●● ●● ● ●●●●● ●● ●● ●● ● ●●●●●●●●●●●●● ●●●● ● ● ●●●●●●●●●●●●●●●●●●●● ●●●●●●●●●●●●●●●●●●●●●● ●●● ●●●●●●●●●●●●●●●●● ● ● ● ●●●●●●●●●●●●●●●●●●●●● ● ● ● ●● ●●●●●●●●●●●●●●●●●●●●●●●● ●● ●● ●●●●●●●●●●●●● ● ●●●●●●●●●●●● ●● ● ●● ●●●● ●● ● ● ●●●●●●●●●●●● ● ● ● ●● ● ● ●●●●●●●●●●●●●●●● ●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●● ● ● ●● ●●●●●●●●●●●●●●●●●●●●●● ● ● ●● ●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●● ● ●● ●●● ●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●●●●●● ●● ●● ●●●●●●●●●●●●●●●●●● ● ●●●●● ● ●●●●●●● ● ●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●● ● ● ● ●●●●●●●●●●●●●●●●● ● ● ● ●● ● ● ●●●●●●● ●●● ●● ● ●● ●●●●●● ●●●● ●● ● ●● ● ● ● ●● ●● ● ●●●●●●●● ●●●● ● ● ●●●●●●●●●●●●●●●●●● ●●●●●●●●●●●● ● ● ● ● ● ● ●●●●●●●●●●●●●●●●●●●●● ● ●● ●●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●● ● ● ● ● ●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●●●● ● ●● ●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●● ● ● ● ● ●● ●●● ● ● ●● ●●●●●●●●●●●●●● ●● ● ● ●●●●●●●● ● ● ● ● ●●●●●●●●● ● ● ●●●●● ●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●●●●●● ●●●●●●●●●●●●●●●●●●●● ● ● ● ● ● ●● ● ●● ●●●●● ●● ●●●●●●●●●●● ● ●●●● ●●●●●●●●●●● ●●●●●●●●●●●●●●●●●● ● ● ●● ●●●●●●●● ● ●●●●●●●●●●● ● ● ●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●● ● ●● ●● ● ● ● ● ●●●●● ● ●● ● ● ●●●●●●●●●●●●●●●●●●●●● ● ●● ● ●● ● ●●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●● ● ● ●●●●●●●● ●●●● ● ●●●● ● ● ● ●●●●●●●●●●● ●● ● ●●●●●●●●●●●●● ●●● ●●●●●●●●●●●●●●● ●●●●●●●●●●●●●●●●●●●●●●● ●● ● ●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●● ● ● ●●●●●●●●●●●●●●●●●●●●●● ● ●● ●● ● ●● ●●● ●● ●● ●●●●● ● ● ● ●●●●●●●●●●●●●●● ●● ● ●●● ●● ●●●●●● ●●●● ● ● ●● ●●●●●●●●●●●●●● ●● ● ● ●● ● ● ●●● ● ● ● ●● ●●●●●●●●● ●● ● ●●●●●●●●●●●●●●●●●● ● ● ●●●●●●●● ●●●●● ● ● ●● ● ● ●●●●●● ●● ● ●●●●●●●●●●●●●●● ● ● ●●●●● ● ● ● ●●●● ●●●●●●●●●● ● ●● ●●● ● ●● ● ● ● ●●●●●●●●●●●●●●●● ● ● ●●●●●● ●●●● ●● ● ●●●●●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●●●●●● ● ● ●●● ● ● ●●● ●●● ●●●● ● ● ● ●●●●●●●●●●●● ● ● ● ● ●●●●●●●●●●●●●●●●● ●● ●●●●●●● ●●● ● ● ● ●●●●●●●●●●●● ●● ● ●● ●●●●●●●●●●●●●● ●●● ●● ●● ● ● ●●●● ●● ● ● ●●●●● ●●●●●●● ●●●●●●●●●●●● ● ● ● ●●●●●●●●●●●●●●●●●●●●●● ● ● ● ●●●●●●●●●●●●● ●● ● ● ● ●●●●●●●●●●●●●●●●● ●● ● ●●●●●●● ● ● ● ● ●●● ● ● ● ● ●●●●●●●●●●●●●●●●●● ● ●●●●●●●●●●●●● ● ● ●●●●●● ● ●●● ● ● ●● ● ● ●● ● ●● ● ●●●●●●●●●●●●●●● ●●● ● ● ●●● ● ● ● ● ●●● ●● ● ●●● ●● ● ● ●● ● ●●●●●●●●● ●●● ●● ● ●● ●● ●●●●●●● ●●● ●●● ● ● ●●●● ●●● ●● ●● ● ● ● ● ● ● ● ●● ●● ● ● ●●●●● ● ● ● ●● ● ● ●●● ●●●●●●●● ● ● ● ● ● ●● ● ●● ●●●●● ●●● ●●● ●● ● ● ●●●●● ● ● ● ● ● ●● ● ● ●●●●● ● ●●●●● ● ● ● ●●●●●●●● ● ●●● ●●● ● ● ●● ●● ● ● ● ●●●●● ● ● ●●●●● ●●●●●● ●● ● ● ●● ● ● ●●●● ● ● ● ●● ● ●●●●● ● ● ●● ●●●●●●● ●●● ● ● ● ●●● ●● ● ●●●● ● ●●●●● ●● ● ● ● ●● ● ● ● ● ● ●● ●● ●● ● ● ●● ● ●●● ●●●●● ●●● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ●● ●● ●● ● ● ●●● ● ● ● ● ● ●● ● ● w D
    • Nudged value in the w − l space 40. Initial bounds on ρ∗ . −D ≤ ρ∗ ≤ D (Duh! but nice geometry)
    • Enclosing triangle 42. Nomenclature D 41. Definition. ABC such that: ABC ⊂ w − l space. (w∗ , l∗ ) ∈ ABC. l B ● Slope of AB segment, unity. mβ mα wA ≤ wB 1 ● C wA ≤ wC A● mγ w D
    • Enclosing Triangle 43. (New) bounds on ρ. Def. Slope mζ projection of point X(wX , lX ) to w = −l line. mζ wX − lX Xζ = mζ + 1 Bounds: Aα = Bα ≤ ρ∗ ≤ Cα wA − lA = wB − lB ≤ ρ∗ ≤ wC − lC 44. So, collinearity (of A, B and C) implies optimality. (Even if there are multiple optima)
    • Right and left uncertainty 45. Iterating inside an enclosing triangle. 1 Set ρ to some value within the bounds ˆ (wA − lA ≤ ρ ≤ wC − lC ). ˆ 2 Solve problem with rewards (r − ρc). ˆ 46. Optimality. If h(sI ) = 0 Done! Optimal policy found for current problem solves SMDP and termination condition has been met.
    • Right and left uncertainty 47.a If h(sI ) > 0 Right uncertainty. l B ● ● S T ● ● C A● w y1
    • Right and left uncertainty 47.b Right uncertainty. Derivation: y1 = Sα − Tα 1 = ((1 − mβ )wS − (1 − mγ )wT − (mγ − mβ )wC ) 2 Maximization: ∗ 2s ab(ρ/2 − Cβ )(ρ/2 − Cγ ) + a(ρ/2 − Cβ ) + b(ρ/2 − Cγ ) y1 = c s = sign(mβ − mγ ) a = (1 − mγ )(mβ + 1) b = (1 − mβ )(mγ + 1) c = (b − a) = 2(mγ − mβ )
    • Right and left uncertainty 48.a If h(sI ) < 0 Left uncertainty. l B ● ● ● C A● R ● y2 w
    • Right and left uncertainty 48. Left uncertainty. Is maximum where expected. (When value level set crosses B) y2 = Rα − Qα ∗ (ρ/2 − Bα ) y2 = (Bα − Bγ ) (ρ/2 − Bγ )
    • Right and left uncertainty 49. Fundamental lemma. As ρ grows, maximal right uncertainty is monotonically ˆ decreasing and maximal left uncertainty is monotonically increasing, and both are non-negative with minimum 0.
    • Optimal nudging 50. Find ρ (between the bounds, obviously) such that the ˆ maximum resulting uncertainty, either left or right, is min. Since both are monotonic and have minimum 0, min max when maximum left and right uncertainties are equal. Remark: bear in mind this (↑) is the worst case. It can terminate immediately. ρ is gain, but neither biased towards observations (initial or ˆ otherwise), nor slowly updated. Optimal nudging is “optimal” in the sense that with this update the maximum uncertainty range of resulting ρ values is minimum.
    • Optimal nudging 51. Enclosing triangle into enclosing triangle. 52. Strictly smaller (both area and, importantly, resulting uncertainty)
    • Obtaining an initial enclosing triangle 53. Setting ρ(0) = 0 and solving. Maximizes reward irrespective of cost. (Usual RL problem) Can be interpreted geometrically as fanning from the w axis to find the policy with w, l coordinates that subtends the smallest angle. The resulting optimizer maps to a point somewhere along a line with intercept at the origin. 54. Optimum of the SMDP problem above but not behind that line. Else, contradiction.
    • Obtaining an initial enclosing triangle 56. Either way, after iter. 0, uncertainty reduces in at least half.
    • Conic intersection 57. Maximum right uncertainty is a conic!    c −(b + a) −Cα c r ∗  −(b + a) ∗ r y1 1 c (Cβ a + Cγ b)   y1  = 0 2 −Cα c (Cβ a + Cγ b) Cα c 1 58. Maximum left uncertainty is a conic!    0 1 (Bγ − Cγ ) r ∗ ∗ r y2 1  1 0 −Bγ   y2  = 0 (Bγ − Cγ ) −Bγ −2Bα (Bγ − Cγ ) 1
    • Conic intersection 59. Intersecting them is easy. 60. And cheap. (Requires in principle constant time and simple matrix operations) 61. So plug it in!
    • Termination Criteria 62. We want to reduce uncertainty to ε. Because it is a good idea. (Right?) So there’s your termination condition right there. 63. Alternatively, stop when |h(k) (sI )| < . 64. In any case, if the same policy remains optimal and the sign of its nudged value changes between iterations, stop: It is the optimal solution of the SMDP problem.
    • Finding D 65. A quick and dirty method: 1 Maximize cost (or episode length, all costs equal 1). 2 Multiply by largest unsigned reinforcement. 66. So, at most one RL problem more. 67. If D is estimated too large, wider initial bounds and longer computation, but ok. 68. If D is estimated too small (by other methods, of course), points outside the triangle in w − l space. (But where?)
    • Recurring state + unichain considerations 69. Feinberg and Yang: Deciding whether the unichain condition holds can be done in polynomial time if a recurring state exists. 70. Existence of a recurring state is common in practice. 71. (Future work) It can maybe be induced using ε–MDPs. (Maybe). 72. At least one case in which no unichain is no problem: games. Certainty of positive policies. Non-positive chains. 73. Happens! (See experiments)
    • Complexity 74. Discounted RL is PAC (–efficient). 75. In problem size parameters (|S|, |A|) and 1/γ. 76. Episodic undiscounted RL is also PAC. (Following similar arguments, but slightly more intricate derivations) 77. So we call a PAC (–efficient) method a number of times.
    • Complexity 78. Most worstest case foreverest when choosing ρ(k) is not reducing uncertainty. 79. Reducing it in half is a better bound for our method. 80. ... and it is a tight bound... 81. ... in cases that are nearly optimal from the outset. 82. So, at worst, log 1 calls to PAC: ε PAC!
    • Complexity 83. Whoops, we proved complexity! That’s a first for SMDP (or ARRL, for that matter). 84. And we inherit convergence from invoked RL, so there’s also that.
    • Tipically much faster 85. Worst case happens when we are ”already there. 86. Otherwise, depends, but certainly better. 87. Multi-iteration reduction in uncertainty way better than 0.5· , because it accumulates geometrically. 88. Empirical complexity better than the already very good upper bound.
    • Bibliography I S. Mahadevan. Average reward reinforcement learning: Foundations, algorithms, and empirical results. Machine Learning, 22(1):159–195, 1996. Reinaldo Uribe, Fernando Lozano, Katsunari Shibata, and Charles Anderson. Discount and speed/execution tradeoffs in markov decision process games. In Computational Intelligence and Games (CIG), 2011 IEEE Conference on, pages 79–86. IEEE, 2011.