4/21

1. Progression corresponds to finding a single path in the transition graph What about regression? 4/21 Make-up class on this Friday No class on next Tuesday Don't print hidden slides

2. Interpreting progression and regression in the transition graph • In the transition graph (corresponding to the atomic model) • progression search corresponds to finding a single path • Regression search corresponds to simultaneously starting from multiple states (all of which satisfy the goal conditions), and effectively searching in parallel until one of the paths reaches the initial state • Alternately, you can see regression as searching in the space of sets of states, with the termination condition being that any of the states is an initial state. • ..In contrast, planning with an incomplete state is also a search in the space of belief states (remember the vaccum world), except the termination condition requires that every state in the belief state is a goal state.

3. Conditional effects don’t change the progression much at all Why? (because the state in which the operator is being applied is known. So you know whether or not the conditional effect actually happens) Handling conditional effects in regression planning introduces “secondary” preconditions Consider regressing goals {P,Q} over an action A with two conditional effects: R=>P; J=>~Q What happens if A has two more effects: U=> P; N=>~Q Handling Conditional Effects

5. ~Have(cake) eaten(cake) Have(cake) eaten(cake) bake Eat ~Have(cake) eaten(cake) No-op Eat Have(cake) ~eaten(cake) Have(cake) ~eaten(cake) No-op No-op Graph has leveled off, when the prop list has not changed from the previous iteration Have(cake) ~eaten(cake) Don’t look at curved lines for now… The note that the graph has leveled off now since the last two Prop lists are the same (we could actually have stopped at the Previous level since we already have all possible literals by step 2)

6. h-A h-B Pick-A Pick-B ~cl-A ~cl-B ~he onT-A onT-A onT-B onT-B cl-A cl-A cl-B cl-B he he

7. h-A on-A-B St-A-B on-B-A h-B Pick-A h-A h-B Pick-B ~cl-A ~cl-A ~cl-B ~cl-B St-B-A ~he ~he onT-A onT-A Ptdn-A onT-A onT-B onT-B onT-B Ptdn-B cl-A cl-A cl-A Pick-A cl-B cl-B cl-B Pick-B he he he

8. How do we use reachability heuristics for regression? Progression Regression

9. Neither hlev nor hsum work well always P1 P0 A0 True cost of {p1…p100} is 1 (needs just one action reach) Hlev says the cost is 1 Hsum says the cost is 100 Hlev better than Hsum q q B1 p1 B2 p2 B3 p3 p99 B99 p100 B100 P1 P0 A0 q q True cost of {p1…p100} is 100 (needs 100 actions to reach) Hlev says the cost is 1 Hsum says the cost is 100 Hsum better than Hlev p1 p2 B* p3 p99 Hrelax will get it correct both times.. p100

10. Relaxed plan for our blocks example h-A on-A-B St-A-B on-B-A h-B Pick-A h-A h-B Pick-B ~ cl -A ~ cl -A ~ cl -B ~ cl -B St-B-A ~he ~he onT -A onT -A Ptdn -A onT -A onT -B onT -B onT -B Ptdn -B cl -A cl -A cl -A Pick-A cl -B cl -B cl -B Pick-B he he he

11. “Relaxed plan” • Suppose you want to find a relaxed plan for supporting literals g1…gm on a k-length PG. You do it this way: • Start at kth level. Pick an action for supporting each gi (the actions don’t have to be distinct—one can support more than one goal). Let the actions chosen be {a1…aj} • Take the union of preconditions of a1…aj. Let these be the set p1…pv. • Repeat the steps 1 and 2 for p1…pv—continue until you reach init prop list. • The plan is called “relaxed” because you are assuming that sets of actions can be done together without negative interactions. Optimal relaxed plan is still NP-hard No backtracking needed!

12. h-sum; h-lev; h-relax • Given a set of literals {l1…lk} • H-lev is the earliest level in which all of them are present • H-sum is the sum of the earliest level in which each of them are present • H-relax is the length of the plan to support the literals • H-lev is lower than or equal to h-relax • H-sum is larger than or equal to H-lev • H-lev is admissible • H-relax is not admissible unless you find optimal relaxed plan • Which is NP-Hard..

13. q t r 5 6 p q r 6 7 p q r 5 6 r q p 5 6 r q p s t 5 6 7 p q r s t 5 6 7 p q r s t 6 7 8 q t r s p 5 6 7 ops oqs ors ops opq oqt orq opq oqr orp opr opr oqt oqt otp oqt o78 o67 o67 o67 o56 o67 o56 o56 1 3 3 5 )=5 h( 1 5 Planning Graphs for heuristics • Construct planning graph(s) at each search node • Extract relaxed plan to achieve goal for heuristic 1 1 1 o12 o12 q 5 2 2 G oG o23 3 3 3 o34 o34 4 4 o45 r 5 5 q5 p 5 G G G oG oG oG opq 3 3 3 o34 o34 opr 4 4 p5 r5 o45 p 6 5 5 5 o56 o56 o56 6 6 o67 7 p6 1 1 1 o12 o12 2 2 G oG o23 3 5 5 5 o56 o56 6 6 o67 7

14. What if actions have non-uniform costs?

15. Challenges in Cost Propagation

17. Planning PSP MDPs • In addition to actions having costs, we might have goals with rewards, with the understanding that if you achieve a goal, you get the corresponding reward • So now, the objective of planning is to find a plan that has the highest net benefit measured as the difference between the cumulative reward for the goals achieved and the cumulative cost of the actions used • This problem, called partial satisfaction planning, is both easy (since an “empty” plan is a solution, just not a very good one) and hard (since now the “quality of the plan” in terms of its net benefit is more important) • It is possible to extend the planning graph heuristics to this problem • On top of this, we might also want to say that rewards are not limited to just goals achieved in the final state, but can also be gathered for visiting certain good states on the way • Such goals are called “trajectory constraints” • Even further, we can consider a scenario where the actions are stochastic • By this time it is not even clear that a sequence of actions is an adequate form for the solution. We need to understand it first at the atomic level—and we shall do so. By the way, this problem is called Markov Decision Process. [MDPs can be done at propositional and relational level, but we won’t discuss that in this class]. • If our masochism continues unabated, we can also now say that in addition to actions being stochastic, we have partial observability • This will lead to a generalization of MDP called POMDP (Partially Observable MDP); we won’t cover this in this course. • ..but as long as we are naming things, if we consider actions with durations, we get Semi-MDPs; if we consider other agents, we get decentralized-MDPs (and in each case we can have PO versions..)

18. Repeat [can generalize to have action costs C(a,s)] If Mij matrix is not known a priori, then we have a reinforcement learning scenario..

19. What does a solution to an MDP look like? • The solution should tell the optimal action to do in each state (called a “Policy”) • Policy is a function from states to actions (* see finite horizon case below*) • Not a sequence of actions anymore • Needed because of the non-deterministic actions • If there are |S| states and |A| actions that we can do at each state, then there are |A||S| policies • How do we get the best policy? • Pick the policy that gives the maximal expected reward • For each policy p • Simulate the policy (take actions suggested by the policy) to get behavior traces • Evaluate the behavior traces • Take the average value of the behavior traces. We will concentrate on infinite horizon problems (infinite horizon doesn’t necessarily mean that that all behavior traces are infinite. They could be finite and end in a sink state)

20. Optimal Policies depend on rewards.. Repeat - - - -

21. Make-up class: Tomorrow (Friday) 10:30—11:45 in DCDC Conference Room 175 [pass Bisonwitches, turn right] 4/23 If you are twenty and not a liberal, you are heartless. If you are sixty and not a conservative, you are mindless. --Winston Churchill But why is Rao putting this here? He better not be hinting that the campus republicans are heartless or geriatric..

22. If you are twenty and not a liberal, you are heartless If you are sixty and not a conservative, you are mindless --Churchill Horizon & Policy • We said policy is a function from states to actions.. but we sort of lied. • Best policy is non-stationary, i.e.,depends on how long the agent has to “live” – which is called “horizon” • More generally, a policy is a mapping from <state, time-to-death>  <action> • So, if we have a horizon of k, then we will have k policies • If the horizon is infinite, then policies must all be the same.. (So infinite horizon case is easy!)

23. What does a solution to an MDP look like? • The solution should tell the optimal action to do in each state (called a “Policy”) • Policy is a function from states to actions (* see finite horizon case below*) • Not a sequence of actions anymore • Needed because of the non-deterministic actions • If there are |S| states and |A| actions that we can do at each state, then there are |A||S| policies • How do we get the best policy? • Pick the policy that gives the maximal expected reward • For each policy p • Simulate the policy (take actions suggested by the policy) to get behavior traces • Evaluate the behavior traces • Take the average value of the behavior traces. We will concentrate on infinite horizon problems (infinite horizon doesn’t necessarily mean that that all behavior traces are infinite. They could be finite and end in a sink state)

24. If you are twenty and not a liberal, you are heartless If you are sixty and not a conservative, you are mindless --Churchill Horizon & Policy • How long should behavior traces be? • Each trace is no longer than k (Finite Horizon case) • Policy will be horizon-dependent (optimal action depends not just on what state you are in, but how far is your horizon) • Eg: Financial portfolio advice for yuppies vs. retirees. • No limit on the size of the trace (Infinite horizon case) • Policy is not horizon dependent We will concentrate on infinite horizon problems (infinite horizon doesn’t necessarily mean that that all behavior traces are infinite. They could be finite and end in a sink state)

25. How to evaluate a policy? • Step 1: Define utility of a sequence of states in terms of their rewards • Assume “stationarity” of preferences • If you prefer future f1 to f2 starting tomorrow, you should prefer them the same way even if they start today • Then, only two reasonable ways to define Utility of a sequence of states • U(s1, s2 sn) = n R(si) • U(s1, s2 sn) = n°i R(si) (0 ·°· 1) • Maximum utility bounded from above by Rmax/(1 - °) • Step 2: Utility of a policy ¼ is the expected utility of the behaviors exhibited by an agent following it. E [ 1t=0°t R(st) | ¼ ] • Step 3: Optimal policy ¼* is the one that maximizes the expectation: argmax¼ E [ 1t=0°t R(st) | ¼ ] • Since there are only A|s| different policies, you can evaluate them all in finite time (Haa haa..)

26. How to handle unbounded state sequences? • If we don’t have a horizon, then we can have potentially infinitely long state sequences. Three ways to handle them • Use discounted reward model ( ith state in the sequence contributes only °i R(si) • Assume that the policy is proper (i.e., each sequence terminates into an absorbing state with non-zero probability). • Consider “average reward per-step”

27. Utility of a State • The (long term) utility of a state s with respect to a policy \pi is the expected value of all state sequences starting with s • U¼(s) = E [ 1t=0°t R(st) | ¼ , s0 =s ] • The true utility of a state s is just its utility w.r.t optimal policy U(s) =U¼*(s) • Thus, U and ¼* are closely related • ¼*(s) = argmaxas’ Mass’ U(s’) • As are utilities of neighboring states • U(s) = R(s) + ° argmaxas’ Mass’ U(s’) Bellman Eqn

28. Optimal Utility depends on Optimal Policy If you go to Tiger Hill near Darjeeling, and only look towards the direction the Sun is rising, you may not understand what the brouhaha is all about; but if you look the other side, you see this enchanting view of Kanchanjunga

29. Think of these as h*() values… Called value function U* Think of these as related to h* values Repeat U* is the maximal expected utility (value) assuming optimal policy

30. Bellman Equations as a basis for computing optimal policy • Qn: Is there a simpler way than having to evaluate |A||S| policies? • Yes… • The Optimal Value and Optimal Policy are related by the Bellman Equations • U(s) = R(s) + ° argmaxas’ Mass’ U(s’) • ¼*(s) = argmaxas’ Mass’ U(s’) • The equations can be solved exactly through • “value iteration” (iteratively compute U and then compute ¼*) • “policy iteration” ( iterate over policies) • Or solve approximately through “real-time dynamic programming”

31. .8 .1 .1 U(i) = R(i) + ° maxj Maij U(j) + °

32. Updates can be done synchronously OR asynchronously --convergence guaranteed as long as each state updated infinitely often Why are values coming down first? Why are some states reaching optimal value faster? .8 .1 .1

33. Value Iteration Demo • http://www.cs.ubc.ca/spider/poole/demos/mdp/vi.html • Things to note • The way the values change (states far from absorbing states may first reduce and then increase their values) • The convergence speed difference between Policy and value

34. Terminating Value Iteration • The basic idea is to terminate the value iteration when the values have “converged” (i.e., not changing much from iteration to iteration) • Set a threshold e and stop when the change across two consecutive iterations is less than e • There is a minor problem since value is a vector • We can bound the maximum change that is allowed in any of the dimensions between two successive iterations by e • Max norm ||.|| of a vector is the maximal value among all its dimensions. We are basically terminating when ||Ui – Ui+1|| < e

35. 4/28 (held on 4/24) Policy Iteration Real-time Dynamic Programming Min-max Search Alpha-beta pruning

36. Policies converge earlier than values • There are finite number of policies but infinite number of value functions. • So entire regions of value vector are mapped to a specific policy • So policies may be converging faster than values.Search in the space of policies • Given a utility vector Ui we can compute the greedy policy pui • The policy loss of pui is ||Upui-U*|| • (max norm difference of two vectors is the maximum amount by which they differ on any dimension) P4 P3 V(S2) U* P2 P1 V(S1) Consider an MDP with 2 states and 2 actions

37. n linear equations with n unknowns. We can either solve the linear eqns exactly, or solve them approximately by running the value iteration a few times (the update wont have the “max” operation)

38. Bellman equations when actions have costs • The model discussed in class ignores action costs and only thinks of state rewards • C(s,a) is the cost of doing action a in state s • Assume costs are just negative rewards.. • The Bellman equation then becomes U(s) = R(s) + ° maxa [ -C(s,a) + s’ R(s’) Mass’ ] • Notice that the only difference is that -C(s,a) is now inside the maximization • With this model, we can talk about “partial satisfaction” planning problems where • Actions have costs; goals have utilities and the optimal plan may not satisfy all goals.

39. Real Time Dynamic Programming • Value and Policy iteration are the bed-rock methods for solving MDPs. Both give optimality guarantees • Both of them tend to be very inefficient for large (several thousand state) MDPs (Polynomial in |S|  ) • Many ideas are used to improve the efficiency while giving up optimality guarantees • E.g. Consider the part of the policy for more likely states (envelope extension method) • Interleave “search” and “execution” (Real Time Dynamic Programming) • Do limited-depth analysis based on reachability to find the value of a state (and there by the best action you should be doing—which is the action that is sending you the best value) • The values of the leaf nodes are set to be their immediate rewards • Alternatively some admissible estimate of the value function (h*) • If all the leaf nodes are terminal nodes, then the backed up value will be true optimal value. Otherwise, it is an approximation… RTDP For leaf nodes, can use R(s) or some heuristic value h(s)

40. The expected value computation is fine if you are maximizing “expected” return If you are --if you are risk-averse? (and think “nature” is out to get you) V2= min(V3,V4) What if you see this as a game? If you are perpetual optimist then V2= max(V3,V4) Min-Max! If you have deterministic actions then RTDP becomes RTA* (if you use h(.) to evaluate leaves

41. Claude Shannon (finite look-ahead) Chaturanga, India (~550AD) (Proto-Chess) Von Neuman (Min-Max theorem) Donald Knuth (a-b analysis) John McCarthy (a-b pruning)

42. Game Playing (Adversarial Search) • Perfect play • Do minmax on the complete game tree • Alpha-Beta pruning (a neat idea that is the bane of many a CSE471 student) • Resource limits • Do limited depth lookahead • Apply evaluation functions at the leaf nodes • Do minmax • Miscellaneous • Games of Chance • Status of computer games..

43. Kriegspiel (blind-fold chess) Snakes-and-ladders is perfect information with chance think of the utter boringness of deterministic snakes and ladders Not that the normal snakes-and-ladders has any real scope for showing your thinking power (your only action is dictated by the dice—so the dice can play it as a solitaire—at most they need your hand..).

45. Searching Tic Tac Toe using Minmax A game is considered Solved if it can be shown that the MAX player has a winning (or at least Non-losing) Strategy This means that the backed-up Value in the Full min-max Tree is +ve

46. <= 2 <= 2 <= 5 <= 14 Cut 2 14 5 2 • Whenever a node gets its “true” value, its parent’s bound gets updated • When all children of a node have been evaluated (or a cut off occurs below that node), the current bound of that node is its true value • Two types of cutoffs: • If a min node n has bound <=k, and a max ancestor of n, say m, has a bound >=j, then cutoff occurs as long as j >=k • If a max node n has bound >=k, and a min ancestor of n, say m, has a bound <=j, then cutoff occurs as long as j<=k

47. Another alpha-beta example Project 2 assigned

48. (order nodes in terms of their static eval values) Click for an animation of Alpha-beta search in action on Tic-Tac-Toe

50. 4/24 class ended here