Machine Learning: Symbol-based
10d
10.0 Introduction
10.1 A Framework for
Symbol-based Learning 10.2 Version Space Search 10.3 The ID3 Decision Tree
Induction Algorithm 10.4 Inductive Bias and
Learnability
10.5 Knowledge and Learning 10.6 Unsupervised Learning 10.7 Reinforcement Learning 10.8 Epilogue and
References 10.9 Exercises
Additional references for the slides:
Thomas Dean, James Allen, and Yiannis Aloimonos,
Reinforcement Learning
• A form of learning where the agent can
explore and learn through interaction with the environment
• The agent learns a policy which is a mapping from states to actions. The policy tells what the best move is in a particular state.
• It is a general methodology: planning,
decision making, search can all be viewed as some form of the reinforcement learning.
Tic-tac-toe: a different approach
• Recall the minimax approach:
The agent knows its current state. Generates a two layer search tree taking into account all the possible moves for itself and the opponent. Backs up values from the leaf nodes and takes the best move assuming that the opponent will also do so.
• An alternative is to directly start playing with an opponent (does not have to be perfect,
but could as well be). Assume no prior knowledge or lookahead. Assign “values” to states: 1 is win 0 is loss or draw
0.5 is anything else
Notice that 0.5 is arbitrary, it cannot differentiate
between good moves and bad moves. So, the learner has no guidance initially.
It engages in playing. When the game ends, if it is a win, the value 1 will be
propagated backwards. If it is a draw or a loss, the
value 0 is propagated backwards. Eventually,
earlier states will be labeled to reflect their “true” value.
After several plays, the learner will learn the best
Issues in generalizing this approach
• How will the state values be initialized or propagated backwards?
• What if there is no end to the game (infinite horizon)?
• This is an optimization problem which
suggests that it is hard. How can an optimal policy be learned?
A simple robot domain
0 1
3 2
The robot is in one of the states: 0, 1, 2, 3. Each one represents an office, the offices are connected in a ring.
Three actions are available:
+ moves to the “next”
state
- moves to the “previous”
state
@ remains at the same state
+ +
+ +
@ @
@
@
-
-
- -
The robot domain (cont’d)
• The robot can observe the label of the state it is in and perform any action corresponding to an arc leading out of its current state.
• We assume that there is a clock governing the passage of time, and that at each tick of the
clock the robot has to perform an action.
• The environment is deterministic, there is a unique state resulting from any initial state and action.
• Each state has a reward:
10 for state 3, 0 for the others.
The reinforcement learning problem
• Given information about the environment
States
Actions
State-transition function (or diagram)
• Output a policy p: states → actions, i.e., find the best action to execute at each state
• Assumes that the state is completely
observable (the agent always knows which state it is in)
Compare three policies
a. Every state is mapped to @
The value of this policy is 0, because the robot will never get to office 3.
b. Every state is mapped to + policy 0
The value of this policy is , because the robot will end up in office 3 infinitely often.
c. Every state is except 3 is mapped to +, 3 is mapped to @ policy 1
The value of this policy is also , because the robot will end up (stay) in office 3 infinitely often.
Compare three policies
So, it is easy to rule case a out, but how can we show that policy 1 is better than policy 0?
One way would be to compute the average reward per tick:
POLICY 1
The average reward per tick for state 0 is 10.
POLICY 0
The average reward per tick for state 0 is 10/4.
Another way would be to assign higher values for immediate rewards and apply a discount to future rewards.
Discounted cumulative reward
Assume that the robot associates a higher value with more immediate rewards and therefore discounts future rewards.
The discount rate () is a number between 0 and 1 used to discount future rewards.
The discounted cumulative reward for a
particular state with respect to a given policy is the sum for n from 0 to infinity of n times the reward associated with the state reached after the n-th tick of the clock.
POLICY 1
The discounted cumulative POLICY 0
The discounted cumulative
Discounted cumulative reward (cont’d)
Take = 0.5
For state 0 with respect to policy 0:
0.50 x 0 + 0.51 x 0 + 0.52 x 0 + 0.53 x 10 + 0.54 x 0 + 0.55 x 0 + 0.56 x 0 + 0.57 x 10 + …
= 1.25 + 0.078 + … = 1.33 in the limit For state 0 with respect to policy 1:
0.50 x 0 + 0.51 x 0 + 0.52 x 0 + 0.53 x 10 +
0.54 x 10 + 0.55 x 10 + 0.56 x 10 + 0.57 x 10 + …
= 2.5 in the limit
Discounted cumulative reward (cont’d)
Let j be a state,
R(j) be the reward for ending up in state j,
be a fixed policy,
(j) be the action dictated by in state j,
f(j,a) be the next state given the robot starts in state j and performs action a,
Vi(j) be the estimated value of state j with
respect to the policy after the i-th iteration of the algorithm
Using a dynamic programming algorithm, one can obtain a good estimate of V, the value function for policy as i .
A dynamic programming algorithm to compute values for states for a policy
1. For each j, set V0(j) to 0.
2. Set i to 0.
3. For each j, set Vi+1 (j) to R(j) + Vi( f(j,) ) ).
4. Set i to i + 1.
5. If i is equal to the maximum number of iterations,
then return Vi otherwise, return to step 3.
Values of states for policy 0
• initialize
V(0) = 0
V(1) = 0
V(2) = 0
V(3) = 0
• iteration 0
For office 0: R(0) + V(1) = 0 + 0.5 x 0 = 0
For office 1: R(1) + V(2) = 0 + 0.5 x 0 = 0
For office 2: R(2) + V(3) = 0 + 0.5 x 0 = 0
For office 3: R(3) + V(1) = 10 + 0.5 x 0 = 10
(iteration 0 essentially initializes values of states to their immediate rewards)
Values of states for policy 0 (cont’d)
• iteration 0 V(0) = V(1) = V(2) = 0 V(3)=10
• iteration 1
For office 0: R(0) + V(1) = 0 + 0.5 x 0 = 0
For office 1: R(1) + V(2) = 0 + 0.5 x 0 = 0
For office 2: R(2) + V(3) = 0 + 0.5 x 10 = 5
For office 3: R(3) + V(0) = 10 + 0.5 x 0 = 10
• iteration 2
For office 0: R(0) + V(1) = 0 + 0.5 x 0 = 0
For office 1: R(1) + V(2) = 0 + 0.5 x 5 = 2.5
For office 2: R(2) + V(3) = 0 + 0.5 x 10 = 5
Values of states for policy 0 (cont’d)
• iteration 2 V(0) = 0 V(1) = 2.5 V(2) = 5 V(3) = 10
• iteration 3
For office 0: R(0) + V(1) = 0 + 0.5 x 2.5 = 1.25
For office 1: R(1) + V(2) = 0 + 0.5 x 5 = 2.5
For office 2: R(2) + V(3) = 0 + 0.5 x 10 = 5
For office 3: R(3) + V(0) = 10 + 0.5 x 0 = 10
• iteration 4
For office 0: R(0) + V(1) = 0 + 0.5 x 2.5 = 1.25
For office 1: R(1) + V(2) = 0 + 0.5 x 5 = 2.5
For office 2: R(2) + V(3) = 0 + 0.5 x 10 = 5
For office 3: R(3) + V(1) = 10 + 0.5 x 1.25 = 10.625
Values of states for policy 1
• initialize
V(0) = 0
V(1) = 0
V(2) = 0
V(3) = 0
• iteration 0
For office 0: R(0) + V(1) = 0 + 0.5 x 0 = 0
For office 1: R(1) + V(2) = 0 + 0.5 x 0 = 0
For office 2: R(2) + V(3) = 0 + 0.5 x 0 = 0
For office 3: R(3) + V(3) = 10 + 0.5 x 0 = 10
Values of states for policy 1 (cont’d)
• iteration 0 V(0) = V(1) = V(2) = 0 V(3)=15
• iteration 1
For office 0: R(0) + V(1) = 0 + 0.5 x 0 = 0
For office 1: R(1) + V(2) = 0 + 0.5 x 0 = 0
For office 2: R(2) + V(3) = 0 + 0.5 x 10 = 5
For office 3: R(3) + V(3) = 10 + 0.5 x 10 = 15
• iteration 2
For office 0: R(0) + V(1) = 0 + 0.5 x 0 = 0
For office 1: R(1) + V(2) = 0 + 0.5 x 5 = 2.5
For office 2: R(2) + V(3) = 0 + 0.5 x 15 = 7.5
For office 3: R(3) + V(3) = 10 + 0.5 x 15 = 17.5
Values of states for policy 1 (cont’d)
• iteration 2 V(0) = 0 V(1) = 2.5 V(2) = 7.5 V(3) = 17.5
• iteration 3
For office 0: R(0) + V(1) = 0 + 0.5 x 2.5 = 1.25
For office 1: R(1) + V(2) = 0 + 0.5 x 7.5 = 3.75
For office 2: R(2) + V(3) = 0 + 0.5 x 17.5 = 8.75
For office 3: R(3) + V(3) = 10 + 0.5 x 17.5 = 18.75
• iteration 4
For office 0: R(0) + V(1) = 0 + 0.5 x 3.75 = 1.875
For office 1: R(1) + V(2) = 0 + 0.5 x 8.75 = 4.375
For office 2: R(2) + V(3) = 0 + 0.5 x 18.75 = 9.375
For office 3: R(3) + V(3) = 10 + 0.5 x 18.75 = 19.375
Compare policies
• Policy 0 after iteration 4
For office 0: R(0) + V(1) = 0 + 0.5 x 2.5 = 1.25
For office 1: R(1) + V(2) = 0 + 0.5 x 5 = 2.5
For office 2: R(2) + V(3) = 0 + 0.5 x 10 = 5
For office 3: R(3) + V(1) = 10 + 0.5 x 1.25 = 10.625
• Policy 1 after iteration 4
For office 0: R(0) + V(1) = 0 + 0.5 x 3.75 = 1.875
For office 1: R(1) + V(2) = 0 + 0.5 x 8.75 = 4.375
For office 2: R(2) + V(3) = 0 + 0.5 x 18.75 = 9.375
For office 3: R(3) + V(3) = 10 + 0.5 x 18.75 = 19.375
• Policy 1 is better because each state has
Temporal credit assignment problem
• It is the problem of assigning credit or blame to the actions in a sequence of actions where
feedback is available only at the end of the sequence.
• When you lose a game of chess or checkers, the blame for your loss cannot necessarily be
attributed to the last move you made, or even the next-to-the-last move.
• Dynamic programming solves the temporal credit assignment problem by propagating rewards
backwards to earlier states and hence to actions earlier in the sequence of actions determined by a
Computing an optimal policy
Given a method for estimating the value of states with respect to a fixed policy, it is
possible to find an optimal policy. We would like to maximize the discounted cumulative reward.
Policy iteration [Howard, 1960] is an algorithm that uses the algorithm for computing the value of a state as a subroutine.
Policy iteration algorithm
1. Let 0 be an arbitrary policy.
2. Set i to 0.
3. Compute V0 (j) for each j.
4. Compute a new policy i+1 so that i+1 (j) is the action a maximizing R(j) + Vi( f(j,) ) .
5. If i+1 = i , then return i; otherwise, set i to i + 1, and go to step 3.
Policy iteration algorithm (cont’d)
A policy is said to be the optimal policy if there is no other policy ’ and state j such that V’ (j) > V (j) and for all k j V’ (j) > V (j) .
The policy iteration algorithm is guaranteed to terminate in a finite number of steps with an optimal policy.
Comments on reinforcement learning
• A general model where an agent can learn to function in dynamic environments
• The agent can learn while interacting with the environment
• No prior knowledge except the (probabilistic) transitions is assumed
• Can be generalized to stochastic domains (an action might have several different probabilistic consequences, i.e., the state-transition function is not deterministic)
• Can also be generalized to domains where the
Famous example: TD-Gammon
(Tosauro, 1995)• Learns to play Backgammon
• Immediate reward:
+100 if win -100 if lose
0 for all other states
• Trained by playing 1.5 million games against itself (several weeks)
• Now approximately equal to best human player (won World Cup of Backgammon in 1992; among top 3 since 1995)
• Predecessor: NeuroGammon [Tesauro and Sejnowski, 1989] learned from examples of labeled moves (very tedious for human expert)
Other examples
• Robot learning to dock on battery charger
• Pole balancing
• Elevator dispatching [Crites and Barto, 1995]: better than industry standard
• Inventory management [Van Roy et. Al]:
10-15% improvement over industry standards
• Job-shop scheduling for NASA space missions [Zhang and Dietterich, 1997]
• Dynamic channel assignment in cellular phones [Singh and Bertsekas, 1994]
Common characteristics
• delayed reward
• opportunity for active exploration
• possibility that state only partially observable
• possible need to learn multiple tasks with same sensors/effectors
• there may not be an adequate teacher
