;; Reinforcement Learning : Exploration vs Exploitation : Multi-Armed Bandits ;; I'm reading the excellent: ;; Reinforcement Learning: An Introduction ;; by Richard S. Sutton and Andrew G. Barto ;; The book's website, on which is available a complete pdf, is here: ;; http://www.incompleteideas.net/book/the-book.html ;; In Chapter 2, they introduce multi-armed bandits as a simplified model problem ;; On the basis that you don't understand anything you can't explain to a computer, I thought I'd code it up: ;; Here is a 2 armed bandit (defn bandit [action] (case action :arms? [:right :left] :right (if (< (rand) 0.5) 4 0) :left (if (< (rand) 0.2) 5 0) :oops!!)) ;; We can ask it how many arms it's got, and what they're called (bandit :arms?) ; [:right :left] ;; And we can pull those arms. Rewards are variable. (bandit :right) ; 4 ; 4 ; 4 ; 0 ; 0 ; 0 ; 0 (bandit :left) ; 5 ; 0 ; 0 ; 0 ; 5 ; 0 ; 5 ; 0 ;; Once we pull an arm, we'll have an action/reward pair (bandit :right) ; 4 ;; the pair would be: [:right 4] ;; Here's a function that yanks an arm at random, and gives us such a pair (defn random-yank [bandit] (let [a (rand-nth (bandit :arms?))] [a (bandit a)])) (random-yank bandit) ; [:left 0] (random-yank bandit) ; [:right 4] ;; And a utility function to take the average of a sequence. We need to be able to provide a default value if the sequence is empty. (defn average ([seq default] (if (empty? seq) default (/ (reduce + seq) (count seq)))) ([seq] (average seq 0))) ;; with some examples (average [1 2 3 4 5]) ; 3 (average (list) 10) ; 10 (average (list 1) 2) ; 1 (average [] 100) ; 100 ;; If we just pull arms at random we get an average reward of about 1.5 (float (average (map second (repeatedly 1000 #(random-yank bandit))))) ; 1.49 ;; Since we can see the code for this particular bandit, we know that ;; the expected value of pulling the right arm is 2 (a half-chance of ;; a reward of 4) and the expected reward for the left arm is 0.2*5 = 1 ;; So if we were seeking to maximize reward, we'd probably be best to pull the right arm all the time. (float (average (map bandit (repeat 10000 :right)))) ; 1.9912 (float (average (map bandit (repeat 10000 :left )))) ; 0.985 ;; The interesting question is, if we don't know how the bandit works, how should we design an algorithm that gets the most reward? ;; (Or at least do better than yanking arms at random!) ;; One thing our algorithm is going to have to do is keep some state to record what happens. ;; Let's start by recording the results of all pulls to date: ;; At first, we know nothing, so we can set up a table to represent that we know nothing (defn initial-state [bandit] (into {} (for [k (bandit :arms?)] [k (list)]))) ;; We haven't pulled either arm yet (initial-state bandit) ; {:right (), :left ()} ;; When we get a new action reward/pair, we'll add the result to our state (defn update-state [state [action reward]] (update-in state [action] #(conj % reward))) ;; here are some examples of using update-state (update-state {:right (), :left ()} [:right 2]) ; {:right (2), :left ()} (reduce update-state {:right (), :left ()} [[:right 2] [:left 3] [:right 4] [:right 5]]) ; {:right (5 4 2), :left (3)} ;; here's how we can use it to record the result of ten random yanks (reduce update-state (initial-state bandit) (repeatedly 10 #(random-yank bandit))) ; {:right (4 4 0 0 0), :left (0 0 0 0 5)} ;; Once we actually have some data, we can make estimates of the expected rewards ;; mapvals applies a function to every value in a map, returning a new map with the same keys (defn mapvals [m f] (into {} (for [[k v] m] [k (f v)]))) ;; examples (mapvals {} inc) ; {} (mapvals {:a 1} inc) ; {:a 2} (mapvals {:a 1, :b 2} inc) ; {:a 2, :b 3} (mapvals {:a 1, :b 2, :c 3} #(* % %)) ; {:a 1, :b 4, :c 9} ;; In the book, Q_t(a) is the current estimate (at time t) ;; We'll use as our estimate of the value of an action the average value seen so far, or zero if we have no information (defn Q [state] (mapvals state #(average % 0))) ;; examples (Q '{:right (5 4 2), :left (3)}) ; {:right 11/3, :left 3} (Q '{:right (5 4 2), :left ()}) ; {:right 11/3, :left 0} (Q (initial-state bandit)) ; {:right 0, :left 0} (Q (update-state (initial-state bandit) (random-yank bandit))) ; {:right 0, :left 2} ;; let's check that we get roughly what we expect in the long run (Q (reduce update-state (initial-state bandit) (repeatedly 10000 #(random-yank bandit)))) ; {:right 9832/5015, :left 1027/997} ;; If we have estimates of the value of each arm, then a good way to ;; use them is to pull the arm with the highest estimate. ;; This is called 'exploitation', as opposed to 'exploration', which ;; is when you try things you think may be suboptimal in order to get ;; information ;; The 'greedy' action is the one with the highest expected value. Of ;; course there may be more than one greedy action especially at first. ;; To help with this, another utility function: ;; max-keys finds the keys with the highest value in a map, and returns a list with just these keys and values (defn max-keys [m] (let [slist (reverse (sort-by second m)) [_ max] (first slist)] (take-while #(= (second %) max) slist))) ;; examples (max-keys {}) ; () (max-keys {1 0}) ; ([1 0]) (max-keys {1 0, 2 0}) ; ([2 0] [1 0]) (max-keys {1 0, 2 1}) ; ([2 1]) (max-keys {1 0, 2 1, 3 -1 , 4 -3, 5 2, 6 2}) ; ([6 2] [5 2]) ;; if there is a tie for the greedy action, we can choose at random between the candidates ;; And so we can go from estimates to greedy action like this: (defn greedy-action [estimates] (first (rand-nth (max-keys estimates)))) ;; examples (greedy-action '{:right 10, :left 3}) ; :right (greedy-action '{:right 10, :left 3 :centre 20}) ; :centre (greedy-action '{:right 10, :left 3 :centre 3}) ; :right (greedy-action '{:right 3, :left 3 :centre 3}) ; :right (greedy-action (Q '{:right (5 4 2), :left (3)})) ; :right (greedy-action (Q '{:right (), :left (3)})) ; :left (greedy-action (Q (initial-state bandit))) ; :left ;; after a lot of random pulls, the greedy action should reliably be the one with the highest expected payoff (greedy-action (Q (reduce update-state (initial-state bandit) (repeatedly 10000 #(random-yank bandit))))) ; :right ;; OK, so we have our stage set, a way of recording what's happened, and some helpful functions defined. ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;; Our first try at a learning algorithm will be 'by hand', as it were. ;; We'll always make the 'greedy' choice. ;; At first, we have no records to go on (initial-state bandit) ; {:right (), :left ()} ;; expected values for both levers are therefore zero (Q (initial-state bandit)) ; {:right 0, :left 0} ;; so the greedy action will get chosen at random (greedy-action (Q (initial-state bandit))) ; :left ;; in this case, we've chosen :left, and the bandit's response is (bandit :left) ; 0 ;; we record it (update-state (initial-state bandit) [:left 0]) ; {:right (), :left (0)} ;; and we have a new state '{:right (), :left (0)} ;; new estimates (Q '{:right (), :left (0)}) ; {:right 0, :left 0} ;; and again, we choose at random (greedy-action (Q '{:right (), :left (0)})) ; :left ;; the bandit is not feeling very generous (bandit :left) ; 0 (update-state '{:right (), :left (0)} [:left 0]) ; {:right (), :left (0 0)} ;; new state: '{:right (), :left (0 0)} ;; new estimates (Q '{:right (), :left (0 0)}) ; {:right 0, :left 0} ;; this time we choose :right (greedy-action (Q '{:right (), :left (0 0)})) ; :right ;; and the bandit pays out! (bandit :right) ; 4 (update-state '{:right (), :left (0 0)} [:right 4]) ; {:right (4), :left (0 0)} ;; the greedy action will be :right now, because we have evidence that right is better. (greedy-action (Q '{:right (4), :left (0 0)})) ; :right ;; You get the idea...... ;; Let's automate that.... ;; Given a state and a bandit, we decide an action and the bandit ;; responds, producing an action/reward pair, and a new state (defn greedy-algorithm [bandit state] (let [action (greedy-action (Q state)) reward (bandit action)] [[action reward] (update-state state [action reward])])) (greedy-algorithm bandit (initial-state bandit)) ; [[:left 0] {:right (), :left (0)}] ;; To get something we can iterate: (defn step [[[a r] state]] (greedy-algorithm bandit state)) (iterate step [ [:dummy :dummy] (initial-state bandit)]) ;; ([[:dummy :dummy] {:right (), :left ()}] ;; [[:left 5] {:right (), :left (5)}] ;; [[:left 0] {:right (), :left (0 5)}] ;; [[:left 0] {:right (), :left (0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 0 0 0 0 0 0 5)}] ;; [[:left 5] {:right (), :left (5 0 0 0 0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 5 0 0 0 0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 5 0 0 0 0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 5 0 0 0 0 0 0 0 0 0 0 5)}] ;; [[:left 0] {:right (), :left (0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 5)}] ;; In this case, the greedy algorithm happens to get a payout on its ;; first try, and decides that it will pull that arm for ever. It ;; never even tries the other arm. ;; Try again: (iterate step [ [:dummy :dummy] (initial-state bandit)]) ;;([[:dummy :dummy] {:right (), :left ()}] ;; [[:right 0] {:right (0), :left ()}] ;; [[:right 0] {:right (0 0), :left ()}] ;; [[:left 0] {:right (0 0), :left (0)}] ;; [[:right 4] {:right (4 0 0), :left (0)}] ;; [[:right 4] {:right (4 4 0 0), :left (0)}] ;; [[:right 4] {:right (4 4 4 0 0), :left (0)}] ;; [[:right 4] {:right (4 4 4 4 0 0), :left (0)}] ;; [[:right 4] {:right (4 4 4 4 4 0 0), :left (0)}] ;; [[:right 4] {:right (4 4 4 4 4 4 0 0), :left (0)}] ;; [[:right 0] {:right (0 4 4 4 4 4 4 0 0), :left (0)}] ;; [[:right 0] {:right (0 0 4 4 4 4 4 4 0 0), :left (0)}] ;; [[:right 4] {:right (4 0 0 4 4 4 4 4 4 0 0), :left (0)}] ;; [[:right 0] {:right (0 4 0 0 4 4 4 4 4 4 0 0), :left (0)}] ;; [[:right 4] {:right (4 0 4 0 0 4 4 4 4 4 4 0 0), :left (0)}] ;; [[:right 0] {:right (0 4 0 4 0 0 4 4 4 4 4 4 0 0), :left (0)}] ;; In this case, it tried the right arm a couple of times, then had a ;; go with the left arm, then went back to the right arm, won a ;; payout, and then got hung up on pulling the right arm repeatedly. ;; We've got a couple of problems here! ;; First is that the algorithm has clearly got into a state where it ;; always pulls the left arm (in the first case), and the right ;; arm (in the second case). ;; It can't be doing the right thing in both cases. ;; Secondly the state is growing linearly, as the algorithm remembers ;; all previous results. That's giving us algorithmic complexity ;; problems and the calculation will get slower and slower, and ;; eventually run out of memory.

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## Thursday, December 13, 2018

### Reinforcement Learning : Exploration vs Exploitation : Multi-Armed Bandits

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