In learning about MDP
's I am having trouble with value iteration
. Conceptually this example is very simple and makes sense:
If you have a 6
sided dice, and you roll a 4
or a 5
or a 6
you keep that amount in $
but if you roll a 1
or a 2
or a 3
you loose your bankroll and end the game.
In the beginning you have $0
so the choice between rolling and not rolling is:
k = 1
If I roll : 1/6*0 + 1/6*0 + 1/6*0 + 1/6*4 + 1/6*5 + 1/6*6 = 2.5
I I don't roll : 0
since 2.5 > 0 I should roll
k = 2:
If I roll and get a 4:
If I roll again: 4 + 1/6*(-4) + 1/6*(-4) + 1/6*(-4) + 1/6*4 + 1/6*5 + 1/6*6 = 4.5
If I don't roll: 4
since 4.5 is greater than 4 I should roll
If I roll and get a 5:
If I roll again: 5 + 1/6*(-5) + 1/6*(-5) + 1/6*(-5) + 1/6*4 + 1/6*5 + 1/6*6 = 5
If I don't roll: 5
Since the difference is 0 I should not roll
If I roll and get a 6:
If I roll again: 6 + 1/6*(-6) + 1/6*(-5) + 1/6*(-5) + 1/6*4 + 1/6*5 + 1/6*6 = 5.5
If I don't roll: 6
Since the difference is -0.5 I should not roll
What I am having trouble with is converting that into python code. Not because I am not good with python, but maybe my understanding of the pseudocode is wrong. Even though the Bellman equation does make sense to me.
I borrowed
the Berkley code for value iteration
and modified it to:
isBadSide = [1,1,1,0,0,0]
def R(s):
if isBadSide[s-1]:
return -s
return s
def T(s, a, N):
return [(1./N, s)]
def value_iteration(N, epsilon=0.001):
"Solving an MDP by value iteration. [Fig. 17.4]"
U1 = dict([(s, 0) for s in range(1, N+1)])
while True:
U = U1.copy()
delta = 0
for s in range(1, N+1):
U1[s] = R(s) + max([sum([p * U[s1] for (p, s1) in T(s, a, N)])
for a in ('s', 'g',)])
delta = max(delta, abs(U1[s] - U[s]))
if delta < epsilon:
return U
print(value_iteration(6))
# {1: -1.1998456790123457, 2: -2.3996913580246915, 3: -3.599537037037037, 4: 4.799382716049383, 5: 5.999228395061729, 6: 7.199074074074074}
Which is the wrong answer. Where is the bug in this code? Or is it an issue of my understanding of the algorithm?
Let B
be your current balance.
If you choose to roll, the expected reward is 2.5 - B * 0.5
.
If you choose not to roll, the expected reward is 0
.
So, the policy is this: If B < 5
, roll. Otherwise, don't.
And the expected reward on each step when following that policy is V = max(0, 2.5 - B * 0.5)
.
Now, if you want to express it in terms of the Bellman equation, you need to incorporate the balance into the state.
Let the state <Balance, GameIsOver>
consist of the current balance and the flag that defines whether the game is over.
stop
:
<B, false>
into <B, true>
roll
:
<B, false>
into <0, true>
with the
probability 1/2
<B, false>
into <B + 4, false>
with the
probability 1/6
<B, false>
into <B + 5, false>
with the
probability 1/6
<B, false>
into <B + 6, false>
with the
probability 1/6
<B1, true>
into <B2, false>
Using the notation from here:
π(<B, false>) = "roll", if B < 5
π(<B, false>) = "stop", if B >= 5
V(<B, false>) = 2.5 - B * 0.5, if B < 5
V(<B, false>) = 0, if B >= 5
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