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[SOLVED] Coms w4701: artificial intelligence homework 3 problem 1: mcts practice (12 points) the partial game tree below was discussed in class on the topic of monte carlo tree search

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The partial game tree below was discussed in class on the topic of Monte Carlo tree search. Each
node shows the win rate: number of playout wins / total number of playouts from that node. The
leaf node labeled 0/0 was just expanded in the middle of a MCTS iteration.(a) Suppose that a rollout is performed and the player corresponding to the purple nodes (root,
third layer, and newly expanded leaf) wins. Sketch a copy of the tree and indicate all win rates
after backpropagation, whether updated or not.(b) Using the new win rates from your tree in (a) above and the exploration parameter α = 1,
compute the UCT values of each of the nodes in the second layer of the tree (immediate children
of the root node). Which of these three nodes is traversed by the selection policy in the next
MCTS iteration?(c) We will refer to the child node you found above as n. Suppose that the player corresponding to
the second layer (orange) always loses in simulation. After how many iterations of MCTS will a
child node of the root different from n be selected? (You may use a program like WolframAlpha
to solve any nonlinear equations.)(d) Starting once again from your tree in (a), solve for the minimum value of α for which a child
node of the root different from n be selected in the next MCTS iteration. Explain why we need
a higher, not lower, α value in order to select one of the other two nodes.We will model a mini-blackjack game as a MDP. The goal is to draw cards from a deck containing
2s, 3s, and 4s (with replacement) and stop with a card sum as close to 6 as possible without going
over. The possible card sums form the states: 0, 2, 3, 4, 5, 6, “done”. The last state is terminal
and has no associated actions. From all other states, one action is to draw a card and advance
to a new state according to the new card sum, with “done” representing card sums of 7 and 8.
Alternatively, one may stop and receive reward equal to the current card sum, also advancing to
“done” afterward.(a) Draw a state transition diagram of this MDP. The diagram should be a graph with seven
nodes, one for each state. Draw edges that represent transitions between states due to the
draw action only; you may omit transitions due to the stop action. Write the transition
probabilities adjacent to each edge.(b) Based on the given information and without solving any equations, what are the optimal actions
and values of states 5 and 6? You may assume that V
∗
(done) = 0. Then using γ = 0.9, solve
for the optimal actions and values of states 4, 3, 2, and 0 (you should do so in that order).Briefly explain why dynamic programming is not required for this particular problem.(c) Find the largest value of γ that would possibly lead to a different optimal action in state 3
(compared to those above) but leave all others the same. Is there any nonzero value of γ that
would yield a different optimal action for state 0? Why or why not?Let’s revisit the mini-blackjack game but from the perspective of dynamic programming. You will
be thinking about both value iteration and policy iteration at the same time. Assume γ = 0.9.(a) Let’s initialize the time-limited state values: V0(s) = 0 for all s (we will ignore “done”). Find
the state values of V1 after one round of value iteration. You do not need to write out every
calculation if you can briefly explain how you infer the new values.(b) Coincidentally, V0 = 0 are also the values for the (suboptimal) policy π0(s) = draw for all
s. If we were to run policy iteration starting from π0, what would be the new policy π1 after
performing policy improvement? Choose the draw action in the case of ties.(c) Perform a second round of value iteration to find the values V2. Have the values converged?
(d) Perform a second round of policy iteration to find the policy π2. Has the policy converged?Let’s now study the mini-blackjack game from the perspective of using reinforcement learning
given a set of game episodes and transitions. This time, assume γ = 1. We initialize all values and
Q-values to 0 and observe the following episodes of state-action sequences:
• 0, draw, 3, draw, done (reward = 0)
• 0, draw, 2, draw, 4, draw, done (reward = 0)
• 0, draw, 4, draw, 6, stop, done (reward = 6)
• 0, draw, 3, draw, 5, stop, done (reward = 5)
• 0, draw, 2, draw, 5, stop, done (reward = 5)(a) Suppose that the above episodes were generated by following a fixed policy. According to Monte
Carlo prediction, what are the values of the six states other than the “done” state? Explain
whether the order in which we see these episodes affects the estimated state values.(b) Suppose we use temporal-difference learning with α = 0.8 instead. Write out each of the
updates for which a state value is changed. Again explain whether the order in which we see
these episodes affects the estimated state values.(c) Now suppose we had generated the above transitions using Q-learning, starting with all Q-values
initialized to zero. Which Q values are nonzero after all episodes are complete? Assuming that
draw is the default “exploit” action in the case of equal Q values, which transitions would be
considered exploratory?We will now extend mini-blackjack from 6 to 21 and add in a few other twists. We have 22 numbered states, one for each possible card sum from 0 to 21, in addition to the “done” terminal state.
From each non-terminal state we can either stop and receive reward equal to the current card sum,
or we can draw an additional card.When drawing a card from the deck, any of the standard set of cards may show up, but the jack,
queen, and king cards are treated as having value equal to 10 (aces will just be treated as 1s). We
will still be drawing cards with replacement, so the probability of drawing a card with a value 1
through 9 is 1
13 each, while the probability of obtaining a 10 value is 4
13 . Lastly, we will add in a
constant living reward that is received with every draw action.Given this information, we can thus model the problem as a Markov decision process and solve
for the optimal policy and value functions. You will be implementing several of the dynamic
programming and reinforcement learning algorithms in the provided blackjack Python file.5.1: Value Iteration (12 points)
Implement value iteration to compute the optimal values for each of the 22 non-terminal states.
The first argument is the initial value function V0 stored in a 1D NumPy array, with the index
corresponding to the state. The other arguments are the living reward, discount factor, and stopping
threshold. As discussed in class, your stopping criterion should be based on the maximum value
change between successive iterations. Your value updates should be synchronous; only use Vi to
compute Vi+1. When finished, your function should return the converged values array V
∗
.5.2: Policy Extraction (10 points)
An agent would typically care more for a policy than a value function. Implement value to policy,
which takes in a set of values, living reward, and discount factor to compute the best policy
associated with the provided values. Return the policy as a NumPy array with 22 entries (one for
each state), with value 0 representing stop and value 1 representing draw.5.3: DP Analysis (9 points)
Now that you can generate optimal policies and values, we can study the impact of living reward
and discount factor on the problem. For each of the following experiments, you can simply use an
initial set of values all equal to 0.(a) Compute and plot the values V
∗ and policy π
∗
for living reward lr = 0 and γ = 1. You should
see that V
∗
consists of three continuous “segments”. Briefly explain why the discontinuities
(which show up as “dips”) between the segments exist, referring to the optimal policy found
and game rules.(b) Experiment with decreasing the discount factor, e.g. in 0.1 decrements. For sufficiently low
values of γ you should see that the three segments of V
∗ merge into two. Show the plots for
an instance of this effect and report the γ value you used. Briefly explain the changes that you
observe in V
∗ and π
∗
.(c) Reset γ to 1 and experiment with changing the living reward, e.g. using intervals of 1 or 2.
Which segments of V
∗
shift for negative living rewards or slightly positive living rewards? In
which direction do these values shift in each case? How does π
∗
change as these state values
shift? Find approximate thresholds of the living reward in which π
∗ becomes stop in all states,
and alternatively draw in all states.5.4: Temporal-Difference Learning (12 points)
You will now investigate using reinforcement learning, and in particular the Q learning algorithm,
to learn the optimal policy and values for our blackjack game purely through self-play. We will need
to keep track of Q values; we can use a 22×2 NumPy array Q, so that we have Q[s,a]= Q(s, a). a=0
corresponds to the stop action and a=1 corresponds to the draw action. To allow for exploration,
we will use ε-greedy action selection.The Qlearn function takes in an initial array of Q values, living reward lr, discount factor γ,
learning rate α, exploration rate ε, and the number of transitions N. In addition to updating Q,
your procedure should keep track of the states, actions, and rewards seen in a N × 3 array record.After initializing the arrays and initial state, the procedure within the simulation loop is as follows:
• Use the ε-greedy method to determine whether we explore or exploit.
• If stopping, the reward is r = s, where s is the current state. If drawing, the reward is
r = lr; also call the draw() function so that you can compute the successor state s
′
.• Update the appropriate Q value, as well as record with the current state, action, and reward.
You should use 0 for the “Q value” of the “done” state.
• Reset s to 0 if we either took the stop action or s
′ > 21, else set s = s
′
.
After finishing N transitions, return Q and record.5.5: RL Analysis (9 points)
After you implement Qlearn, you should ensure that it is working correctly by comparing the
learned state values (the maximum values in each row of Q) with those returned by value iteration.
A suitable default set of learning parameters would be to have a low α (e.g., 0.1), low ε (e.g., 0.1)
and high N (e.g., 50000).Once you are ready, call the provided RL analysis function (no additional code needed) and answer
the following questions using the plots that you see. Your Qlearn must return the two arrays as
specified above in order for this to work properly.(a) The first plot shows the number of times each state is visited when running Qlearn with lr = 0,
γ = 1, α = ε = 0.1, and N ranging from 0 to 50k in 10k intervals. Explain how the value of N
is important to ensuring that all states are visited sufficiently. You should also note that the
two most visited states are the same regardless of N; explain why these two states attract so
much attention.(b) The second plot shows the cumulative reward received for a game using the same parameters
above, except with N = 10000 and ε ranging from 0 to 1 in 0.2 intervals. Explain what you
see with the ε = 0 curve and why we need exploration to learn. Looking at the other curves,
how does increasing exploration affect the overall rewards received and why?(c) The third plot shows the estimated state values using the same parameters above, except with
ε = 0.1 and α ranging from 0.1 to 1. While the set of curves may be hard to read, each should
bear some resemblance to the true state values V
∗
. What is a problem that arises when α
is too low, particularly with less visited states? What is a problem that arises when α is too
high? Try to compare the stability or smoothness of the different curves.Submission
You should have one PDF document containing your solutions for problems 1-4, as well as the
information and plots for 5.3 and 5.5. You should also have a completed Python file implementing
problem 5; make sure that all provided function headers and the filename are unchanged.Submit
the document and .py code file to the respective assignment bins on Gradescope. For full credit,
you must tag your pages for each given problem on the former.