Tutorial 1: Reinforcement Learning
Tutorial 1: Reinforcement Learning
Kexin Ren & Younesse Kaddar (Lecturer: Nicolas Perrin)
1. Markov Decision Problems
Our problem is a Markov Decision Problems (MDP) which is described by a tuple $(𝒳, 𝒰, 𝒫_0, 𝒫,r,γ)$ where $𝒳$ is the statespace, $𝒰$ the actionspace, $𝒫_0$ the distribution of the initial state, $𝒫$ the transition function, $r$ the reward function, and $γ ∈ [0, 1]$ a parameter called the discount factor.
The state in which the robot starts, denoted $x_0$, is drawn from the initial state distribution $𝒫_0$. We will choose it such that the robot always starts in state $0$. Then, given the state $x_t$ at time $t$ and the action $u_t$ at that same time, the next state $x_{t+1}$ depends only on $x_t$ and $u_t$ according to the transition function $𝒫$:
If the robot tries to get out of the grid (for example if it uses action $N$ in state $0$), it will stay in his current state. The robot also receives a reward $r_t$ according to the reward function $r(x_t, u_t)$, which depends on the state $x_t$ and the action $u_t$. We choose to give a reward $1$ when the robot is in state $15$ and does the action $NoOp$, and $0$ for any other stateaction pair.
Questions
1. Define self.gamma
We may set $γ$ to be equal to $0.95 ∈ [0, 1]$, for instance.
class mdp():
def __init__(self):
# [...]
self.gamma = 0.95
2. In __init__(self)
, define P0
, the distribution of the initial state.
As the robot always starts in state $0$: $𝒫_0$ is equal to the Dirac measure $δ_0$
class mdp():
def __init__(self):
# [...]
self.P0 = np.zeros(self.nX)
self.P0[0] = 1
# [...]
3. Write down the transition probabilities for each triplet $𝒫(\text{state}, \text{action}, \text{next state})$ for the example above
class mdp():
def __init__(self):
# [...]
self.P = np.empty((self.nX,self.nU,self.nX))
self.P[:,N,:]= np.array([
[1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0]])
self.P[:,S,:] = self.P[::1,N,::1]
self.P[:,E,:] = np.array([
[0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0],
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0]])
self.P[:,W,:] = self.P[::1,E,::1]
self.P[:,NoOp,:]= np.eye(self.nX)
# [...]
4. Define the reward function r, stored in the matrix self.r
As reward is equal to $1$ if the robot is in state $15$ and does the action $NoOp$, and $0$ otherwise:
class mdp():
def __init__(self):
# [...]
self.r = np.zeros((self.nX, self.nU))
self.r[15, NoOp] = 1
2. Dynamic Programming
The overall goal is to maximize
A policy is a mapping $π : X ⟶ U$ that assigns, to each state $x$, an action $u ≝ π(x)$ that the robot should execute whenever in state $x$.
The state value of a policy $π$ is the total discounted reward that the robot expects to receive when starting from a given state $x$ and following policy $π$:
The stateaction value of a policy $π$ is the total discounted reward that the robot expects to receive when starting from a given state $x$, taking the action $u$ and then following policy $π$:
The optimal policy is the policy $π^\ast$ such that for every state:
In particular, we have:
2.1 Value Iteration
It stems from the above that:
out of which we get a recursive algorithm to compute $Q^\ast$:
1. Set the proper ranges in the for i
and for j
loops.
i
ranges over the statesj
ranges over the actions
therefore:
def VI(self):
Q = np.zeros((self.nX,self.nU))
# [...]
for i in range(self.nX):
for j in range(self.nU):
# [...]
2. Complete the assignment Q[i,j] =
Based on the equation $(8)$:
Q[i,j] = self.r[i,j] + self.gamma * np.sum(self.P[i,j,:] * Qmax)
As a result, the VI
method becomes:
def VI(self):
Q = np.zeros((self.nX, self.nU))
pol = N*np.ones((self.nX, 1))
quitt = False
iterr = 0
while quitt==False:
iterr += 1
Qold = Q.copy()
Qmax = Qold.max(axis=1)
for i in range(self.nX):
for j in range(self.nU):
Q[i,j] = self.r[i,j] + self.gamma * np.sum(self.P[i,j,:] * Qmax)
Qmax = Q.max(axis=1)
pol = np.argmax(Q,axis=1)
if (np.linalg.norm(QQold))==0:
quitt = True
return [Q,pol]
and by running python run.py
, the following policy figure is displayed:
which makes sense, since the state $15$ (denoted by $16$ on the figure) is the most attractive with its reward of $1$ (whereas the other states have a zero reward).
3. Add a second goal: modify the reward function to give $0.9$ when the robot is in state $5$ and does not move. Explain when and why the robot chooses to go to this new goal instead of the first one. Explain what parameter is responsible of this behaviour.
The discount factor $γ$ determines a tradeoff between exploration (exploring farther states) and exploitation/greediness (exploiting the rewards of the nearby one). As it happens: the smaller the parameter $γ$, the more the robot tends to exploit the closest state associated with a (strictly) positive reward (even if there might be a state farther on which a given action leads to a bigger reward).
Here, there are two different optimal policies, depending on the value of $γ$:
Value of $γ$  Optimal Policy 

Greedier policy: $γ < γ_0 ≝ \sqrt{0.9}$  
More exploratory policy: $γ ≥ γ_0$ 
When it comes to the greedier policy: for states close to the state $5$ (denoted by $6$ on the pictures), the robot tends to head to the state $5$, even if its reward ($=0.9$) is inferior the reward ($=1$) of the state $15$ (denoted by $16$ on the pictures)
On the contrary, with the more exploratory policy: apart from the states $0, 1$ and $4$ (which are one step away from the state $5$), the robot favors the state $15$, i.e. the longterm bigger reward over the smaller yet closer (for the states $2, 6, 8$ and $9$) reward of the state $5$.
4. Change self.P
to implement stochastic (nondeterministic) transitions. Use comments in the code to describe the transitions you chose and the results you observed while running VI
on the new transitions you defined.
NB: in this question, for convenience
 we’ll call a state out of which the transition function is nondeterministic a stochastic state
 we’ll use the old reward function, which amounts to zero everywhere except on state $15$ when the robot doesn’t move (in which case it is set to be $1$)
To implement stochastic transitions, we use the first state and the seventh state as examples, with the following code (having nondeterministic transitions for every state overcomplicate things: one state with nondeterministic transitions is enough to have a good idea of what’s going on):
pos = np.random.rand(5)
n = 0 # n = 0 for 1st state, n = 6 for 7th state
for i in range(self.nU):
self.P[n,i,np.where(self.P[n,i,:]==1)]=pos[i]/sum(pos)
We found that, when the first step (at state $0$) is stochastic, the result is the same as the deterministic one (as shown in Figure 2.1.4.1) which makes sense because all routes will actually weigh the same under these conditions; when the 7th state (state $6$) is stochastic, no neighboring state of state $6$ will have the agent (via the optimal policy) choose to go through state $6$ (as shown in Figure 2.1.4.2), because there is a possibility of going “backwards” (i.e. not towards to the most attractive state (due to its reward), state $15$) on state $6$, which will reduce the $Q$value of other states heading for state $6$.
NB: in the figures, the states range from $1$ to $16$
On the whole, with a stochastic transition function: the less “stable” (i.e. likely to lead to a given state for a given action) a stochastic state is, the more it is likely to be averted, if it is on the way to a highreward state.
2.2. Policy Iteration
By definition of the state value function of a given policy $π$, we have:
But as $𝒳$ is finite: by setting $\textbf{V}_π$ (resp. $\textbf{R}_π$) to be the vectormatrix $(V^π(x))_{x ∈ 𝒳}$ (resp. $(r(x, π(x)))_{x ∈ 𝒳}$), and
it comes that
which yields another algorithm to compute the optimal policy, along with
1. Fill the code PI(self)
to implement policy iteration, and test it. Compare the converge speed of VI
and PI
.
Using the previous equations, it comes that:
def PI(self):
Q = np.empty((self.nX, self.nU))
pol = N*np.ones(self.nX, dtype=np.int16)
R = np.zeros(self.nX)
P = np.zeros((self.nX, self.nX))
quitt = False
iterr = 0
while quitt==False:
iterr += 1
for i in range(self.nX):
R[i] = self.r[i,pol[i]]
for j in range(self.nX):
P[i,j] = self.P[i,pol[i],j]
V = np.dot(np.linalg.inv(Iself.gamma*P), R)
for i in range(self.nX):
for j in range(self.nU):
Q[i,j] = self.r[i,j] + self.gamma * np.sum(self.P[i,j,:] * V)
print(Q[i,j])
pol_old = pol.copy()
pol = np.argmax(Q, axis=1)
if np.array_equal(pol, pol_old):
quitt = True
return [Q, pol]
We compared the convergences of VI
and PI
and found that, VI
algorithm converges after 667 iterations while the PI
algorithm converges after 3 iterations, supporting the fact that PI
is more efficient than VI (gamma is set to 0.95
).
In a terminal:
python mtimeit s'import mdp' 'mdp.mdp().PI()'
>> 100 loops, best of 3: 2.66 msec per loop
python mtimeit s'import mdp' 'mdp.mdp().VI()'
>> 10 loops, best of 3: 250 msec per loop
which suggests that
PI
is approximately 100 times faster thanVI
3. Reinforcement Learning: Modelfree approaches
3.1. Temporal Difference Learning (TDlearning)
Based on $⊛$ and $⊛⊛$, we get:
as a result of which we define the temporal difference errors (TD errors):
3.2. TD(0)
At time $t$, we denote by
 $V^{(t})$ the estimation of the state value function
 $x_t$ the current state
 $δ_t ≝ δ_{V^{(t)}}(x_t)$
 $α$ the learning rate
From the $δ$rule, the following update of the state value function can be derived:
1. Open the file mdp.py
, and implement a function called MDPStep
. This function returns a next state and a reward given a state and an action. Hint: To draw a number according to a vector of probabilities, you can use the function discreteProb
.
def MDPStep(self,x,u):
# This function executes a step on the MDP M given current state x and action u.
# It returns a next state y and a reward r
y = self.discreteProb(self.P[x,u,:]) # y is sampled according to the distribution self.P[x,u,:]
r = self.r[x,u] # r is be the reward of the transition
return [y,r]
2. Fillin the missing lines in the TD
function, which implements $TD(0)$. To obtain samples from MDP, you will use the MDPStep
function
def TD(self,pol):
V = np.zeros((self.nX,1))
nbIter = 100000
alpha = 0.1
for i in range(nbIter):
x = np.floor(self.nX*np.random.random()).astype(int)
[y, r] = self.MDPStep(x, pol[x])
V[x] += alpha * (r + self.gamma * V[y]  V[x])
return V
By

modifying
run.py
so that one computes:[Q,pol] = m.PI() V = m.TD(pol)

and adding the line
print(np.argsort(V,axis=0))
in the
TD
method, just before the return, to sort the states according to their estimated value (in increasing order)
it appears that the estimated values are sorted as follows:
for

the deterministic transition function and the reward function of question 2.1.3. (the reward is null everywhere except at state $5$ (where it amounts to $0.9$) and state $15$ (where it amounts to $1$) when the robot doesn’t move)

the following policy ($γ < \sqrt{0.9}$):
which makes perfect sense, intuitively.
3. Write a function compare
that takes in input a state value function $V$, a policy $π$, a stateaction value function $Q$, and returns True
if and only if $V$ and $Q$ are consistent with respect to $π$ up to some precision, i.e. if $∀x ∈ 𝒳, V^π(x) = Q^π(x, π(x)) ± ε$.
We are asked to program a function which test if $∀x ∈ 𝒳, \vert V^π(x)  Q^π(x, π(x)) \vert ≤ ε_{max}$, for a threshold value $ε_{max}$: i.e. if the infinity norm of the difference of the vectors is smaller than $ε_{max}$.
def compare(self,V,Q,pol,eps=0.0001):
Q_pol = np.array([[Q[x, pol[x]]] for x in range(V.size)])
return np.linalg.norm(VQ_pol, ord=np.inf)<eps
4. Use the compare
function to verify that $TD(0)$ converges towards the proper value function, using it on the policy returned by VI
or PI
.
[Q1,pol1] = m.VI()
[Q2,pol2] = m.PI()
V1 = m.TD(pol1)
V2 = m.TD(pol2)
print(m.compare(V1,Q1,pol1))
print(m.compare(V2,Q2,pol2))
yields True
and True
, so $TD(0)$ does converge towards the proper value function.
3.3. QLearning
In accordance with the softmaxpolicy:
we update the stateaction value function as follows:
1. Implement the function softmax
that returns the softmax policy.
def softmax(self,Q,x,tau):
# Returns a softmax probability distribution over actions
# Inputs :
#  Q : a Qfunction reprensented as a nX times nU matrix
#  x : the state for which we want the softmax distribution
#  tau : temperature parameter of the softmax distribution
# Output :
#  p : probabilty of each action according to the softmax distribution
# (column vector of length nU)
p = np.exp(Q[x,:]/tau)
return p/np.sum(p)
2. Fillin the missing lines to implement the QLearning
function.
def QLearning(self,tau):
# This function computes the optimal statevalue function and the corresponding policy using QLearning
# Initialize the stateaction value function
Q = np.zeros((self.nX,self.nU))
# Run learning cycle
nbIter = 100000
alpha = 0.01
for i in range(nbIter):
# Draw a random state
x = np.floor(self.nX*np.random.random()).astype(int)
# Draw an action using a softmax policy
u = self.discreteProb(self.softmax(Q,x,tau))
# Perform a step of the MDP
[y,r] = self.MDPStep(x, u)
# Update the stateaction value function with QLearning
Q[x,u] += alpha * (r + self.gamma * np.max(Q[y,:])  Q[x,u])
# Compute the corresponding policy
Qmax = Q.max(axis=1)
pol = np.argmax(Q,axis=1)
return [Qmax,pol]
3. Run $Q$Learning several times. What do you observe?
NB: in this question, the numbering of the states is the displayed one by plotPolicy
(i.e. the states range from $1$ to $16$).
We found that the policy is stochastic, except that for the rightmost column (i.e. states $13, 14$ and $15$), bottom row (i.e. states $4, 8, 12$) and of course, state $16$, the policy is fixed. The states in the rightmost column always lead to the South
action, towards the state $16$, while the states in the bottom row always lead to the East
action, towards the state $16$.
This makes sense because for the rightmost column and the bottom row, there is only one choice of action (or direction) for them which will go towards the state $16$, and the robot will learn the only right direction when softmax is applied. Unlike states in these two lines, other states have actions associated to them in a nondeterministic way, as a stochastic strategy is used in the algorithm.
4. Compare the statevalue function and the policy computed using $Q$Learning with the ones you obtained with VI
and PI
.
NB: in this question, for convenience, the old reward function is referred to (which amounts to zero everywhere except on state $16$ when the robot doesn’t move (in which case it is set to be $1$))
Regarding the policy:

on the one hand,
VI
andPI
are deterministic algorithms, always ouputting the optimal (up to the precision ofnp.linalg.norm(QQold) == 0
andnp.array_equal(pol, pol_old)
) policy: we exploration/exploitation tradeoff is dealt with by the discount factor $γ$ 
on the other hand, $Q$Learning generates a policy (converging to the optimal one as
nbIter
increases) nondeterministically: on top of the discount factor $γ$, there is another layer which has an impact on the exploration/exploitation tradoff: the softmax policy used to draw the very samples that enable us to update the $Q$function
Regarding the statevalue function:

the results may differ for every trial of $Q$Learning depending on the stochastic experience of the robot, while the results remain the same each time for
VI
andPI
(which are deterministic). 
moreover, the statevalue matrices of
VI
andPI
have coefficients that don’t vary “too much”: they range over smaller values (usually $13$  $20$), and the actions towards the highreward state $16$ lead to a higher value, as expected. On the contrary, the statevalue matix of $Q$Learning usually presents (with respect to its coefficients) more discrepancies: with disparity in value magnitude and sometimes no apparent correlation between the value and direction towards the highreward state $16$.
4. ModelBased Reinforcement Learning (MBRL)
RealTime Dynamic Programming:
We update the estimate of the transition probabilty $\hat{𝒫}^{(t)}$ as follows:
where $N_t(x,u)$ is the number of visits to the pair $(x,u)$ before (and including) time $t$.
As for the reward function estimate:
Those are used in the following updating rule:
1. Fillin RTDP
(RealTime Dynamic Programming) function
def RTDP(self):
Q = np.zeros((self.nX,self.nU))
hatP = np.ones((self.nX,self.nU,self.nX))/self.nX
N = np.ones((self.nX,self.nU))
nbIter = 10000
for iterr in range(nbIter):
# Draw a random pair of state and action
x = np.floor(self.nX*np.random.random()).astype(int)
u = np.floor(self.nU*np.random.random()).astype(int)
# One step of the MDP for this stateaction pair
[y,r] = self.MDPStep(x,u)
# Compute the estimate of the transition probabilities
hatP[x,u,:] *= (1  1/N[x, u])
hatP[x,u,:] += (np.arange(self.nX) == y).astype(int)/N[x, u]
# Updating rule for the stateaction value function
Qmax = Q.max(axis=1)
Q[x,u] = r + self.gamma * np.sum(hatP[x,u,:]*Qmax)
N[x,u] += 1
Qmax = Q.max(axis=1)
pol = np.argmax(Q,axis=1)
return [Qmax,pol]
2. Suppose now that the reward is stochastic: modify MDPStep
by adding a Gaussian noise of standard deviation $0.1$ to the reward. What happens if you set the standard deviation to $1.0$?
def MDPStep(self,x,u,sigma=0.1):
# This function executes a step on the MDP M given current state x and action u.
# It returns a next state y and a reward r
y = self.discreteProb(self.P[x,u,:]) # y is sampled according to the distribution self.P[x,u,:]
r = self.r[x,u] + sigma * np.random.randn() # r is be the reward of the transition
return [y,r]
Noisy rewards have an effect on how good a job the computed policy is doing at maximizing the overall cumulated rewards (we may call that the efficiency of the policy). The smaller the standard deviation of the noise, the more efficient the computed policy becomes, since the rewards that were encountered during RTDP tend to be closer and closer to their regular/actual values. It all hinges on how the standard deviation of the noise compares to the order of magnitude of the mean rewards.
With a small (compared to how large the mean rewards can get) noise standard deviation (e.g. $0.1 < 1 = \text{reward of state } 16$), the policy becomes just a little less efficient. However, with a big noise standard deviation value (e.g. $1$, which is as large as the mean rewards can get), the robot is completely confused: for instance, it is difficult to draw any conclusion when receiving a reward of $1$ on a given sample: it may mean that the current stateaction pair has
 a mean reward of $0$ (nonattractive state) with a noise of $+1$
 or a mean reward of $1$ (highly attractive state) with a noise of $0$
As a result, one can barely figure out any reliable solution.
3. Implement RTDP2
, a variant of RTDP that handles this stochastic reward by computing the model $\hat{r}$ of the mean reward for each state and action (like in equation $(17)$ for $\hat{P}$).
The mean reward $\hat{r}$ is computed as follows (with the same notations as before):
def RTDP2(self):
Q = np.zeros((self.nX,self.nU))
hatP = np.ones((self.nX,self.nU,self.nX))/self.nX
hatR = np.zeros((self.nX,self.nU))
N = np.ones((self.nX,self.nU))
nbIter = 10000
for iterr in range(nbIter):
# Draw a random pair of state and action
x = np.floor(self.nX*np.random.random()).astype(int)
u = np.floor(self.nU*np.random.random()).astype(int)
# One step of the MDP for this stateaction pair
[y,r] = self.MDPStep(x,u,sigma=0.1)
# Compute the estimate of the transition probabilities
hatP[x,u,:] *= (1  1/N[x, u])
hatP[x,u,:] += (np.arange(self.nX) == y).astype(int)/N[x, u]
# Compute the estimate of the reward
hatR[x,u] = ((N[x, u]1)*hatR[x,u] + r)/N[x, u]
# Updating rule for the stateaction value function
Qmax = Q.max(axis=1)
Q[x,u] = hatR[x,u] + self.gamma * np.sum(hatP[x,u,:]*Qmax)
N[x,u] += 1
Qmax =Q.max(axis=1)
pol = np.argmax(Q,axis=1)
return [Qmax,pol]
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