### Markov decision process

This week's methods are all built to solve __M__arkov __D__ecision __P__rocesses. In the broadest sense, an MDP is defined by how it changes states and how rewards are computed.

State transition is defined by $P(s' |s,a)$ - how likely are you to end at state $s'$ if you take action $a$ from state $s$. Now there's more than one way to define rewards, but we'll use $r(s,a,s')$ function for convenience.

_This notebook is inspired by the awesome_ [CS294](https://github.com/berkeleydeeprlcourse/homework/blob/36a0b58261acde756abd55306fbe63df226bf62b/hw2/HW2.ipynb) _by Berkeley_

For starters, let's define a simple MDP from this picture:




```python
import sys, os
if 'google.colab' in sys.modules and not os.path.exists('.setup_complete'):
    !wget -q https://raw.githubusercontent.com/yandexdataschool/Practical_RL/master/setup_colab.sh -O- | bash
    !wget -q https://raw.githubusercontent.com/yandexdataschool/Practical_RL/master/week02_value_based/mdp.py
    !touch .setup_complete

# This code creates a virtual display to draw game images on.
# It will have no effect if your machine has a monitor.
if type(os.environ.get("DISPLAY")) is not str or len(os.environ.get("DISPLAY")) == 0:
    !bash ../xvfb start
    os.environ['DISPLAY'] = ':1'
```


```python
transition_probs = {
    's0': {
        'a0': {'s0': 0.5, 's2': 0.5},
        'a1': {'s2': 1}
    },
    's1': {
        'a0': {'s0': 0.7, 's1': 0.1, 's2': 0.2},
        'a1': {'s1': 0.95, 's2': 0.05}
    },
    's2': {
        'a0': {'s0': 0.4, 's2': 0.6},
        'a1': {'s0': 0.3, 's1': 0.3, 's2': 0.4}
    }
}
rewards = {
    's1': {'a0': {'s0': +5}},
    's2': {'a1': {'s0': -1}}
}

from mdp import MDP
mdp = MDP(transition_probs, rewards, initial_state='s0')
```

We can now use MDP just as any other gym environment:


```python
print('initial state =', mdp.reset())
next_state, reward, done, info = mdp.step('a1')
print('next_state = %s, reward = %s, done = %s' % (next_state, reward, done))
```

    initial state = s0
    next_state = s2, reward = 0.0, done = False


but it also has other methods that you'll need for Value Iteration


```python
print("mdp.get_all_states =", mdp.get_all_states())
print("mdp.get_possible_actions('s1') = ", mdp.get_possible_actions('s1'))
print("mdp.get_next_states('s1', 'a0') = ", mdp.get_next_states('s1', 'a0'))
print("mdp.get_reward('s1', 'a0', 's0') = ", mdp.get_reward('s1', 'a0', 's0'))
print("mdp.get_transition_prob('s1', 'a0', 's0') = ", mdp.get_transition_prob('s1', 'a0', 's0'))
```

    mdp.get_all_states = ('s0', 's1', 's2')
    mdp.get_possible_actions('s1') =  ('a0', 'a1')
    mdp.get_next_states('s1', 'a0') =  {'s0': 0.7, 's1': 0.1, 's2': 0.2}
    mdp.get_reward('s1', 'a0', 's0') =  5
    mdp.get_transition_prob('s1', 'a0', 's0') =  0.7


### Optional: Visualizing MDPs

You can also visualize any MDP with the drawing fuction donated by [neer201](https://github.com/neer201).

You have to install graphviz for system and for python. 

1. * For ubuntu just run: `sudo apt-get install graphviz` 
   * For OSX: `brew install graphviz`
2. `pip install graphviz`
3. restart the notebook

__Note:__ Installing graphviz on some OS (esp. Windows) may be tricky. However, you can ignore this part alltogether and use the standart vizualization.


```python
from mdp import has_graphviz
from IPython.display import display
print("Graphviz available:", has_graphviz)
```

    Graphviz available: True



```python
if has_graphviz:
    from mdp import plot_graph, plot_graph_with_state_values, plot_graph_optimal_strategy_and_state_values
    display(plot_graph(mdp, graph_size="50,50"))
```


    

    


### Value Iteration

Now let's build something to solve this MDP. The simplest algorithm so far is __V__alue __I__teration

Here's the pseudo-code for VI:

---

`1.` Initialize $V^{(0)}(s)=0$, for all $s$

`2.` For $i=0, 1, 2, \dots$
 
`3.` $ \quad V_{(i+1)}(s) = \max_a \sum_{s'} P(s' | s,a) \cdot [ r(s,a,s') + \gamma V_{i}(s')]$, for all $s$

---

First, let's write a function to compute the state-action value function $Q^{\pi}$, defined as follows

$$Q_i(s, a) = \sum_{s'} P(s' | s,a) \cdot [ r(s,a,s') + \gamma V_{i}(s')]$$



```python
def get_action_value(mdp, state_values, state, action, gamma):
    """ Computes Q(s,a) as in formula above """

    v = 0
    for next_state in mdp.get_next_states(state, action):
        v += (mdp.get_transition_prob(state, action, next_state)
              * (mdp.get_reward(state, action, next_state) + gamma * state_values[next_state]))

    return v
```


```python
import numpy as np
test_Vs = {s: i for i, s in enumerate(sorted(mdp.get_all_states()))}
assert np.isclose(get_action_value(mdp, test_Vs, 's2', 'a1', 0.9), 0.69)
assert np.isclose(get_action_value(mdp, test_Vs, 's1', 'a0', 0.9), 3.95)
```

Using $Q(s,a)$ we can now define the "next" V(s) for value iteration.
 $$V_{(i+1)}(s) = \max_a \sum_{s'} P(s' | s,a) \cdot [ r(s,a,s') + \gamma V_{i}(s')] = \max_a Q_i(s,a)$$


```python
def get_new_state_value(mdp, state_values, state, gamma):
    """ Computes next V(s) as in formula above. Please do not change state_values in process. """
    if mdp.is_terminal(state):
        return 0

    v = -1e30
    action = None
    for possible_action in mdp.get_possible_actions(state):
        cur_v = get_action_value(mdp, state_values, state, possible_action, gamma)
        if cur_v > v:
            v = cur_v
            action = possible_action
    
    return v
```


```python
test_Vs_copy = dict(test_Vs)
assert np.isclose(get_new_state_value(mdp, test_Vs, 's0', 0.9), 1.8)
assert np.isclose(get_new_state_value(mdp, test_Vs, 's2', 0.9), 1.08)
assert np.isclose(get_new_state_value(mdp, {'s0': -1e10, 's1': 0, 's2': -2e10}, 's0', 0.9), -13500000000.0), \
    "Please ensure that you handle negative Q-values of arbitrary magnitude correctly"
assert test_Vs == test_Vs_copy, "Please do not change state_values in get_new_state_value"
```

Finally, let's combine everything we wrote into a working value iteration algo.


```python
# parameters
gamma = 0.9            # discount for MDP
num_iter = 100         # maximum iterations, excluding initialization
# stop VI if new values are this close to old values (or closer)
min_difference = 0.001

# initialize V(s)
state_values = {s: 0 for s in mdp.get_all_states()}

if has_graphviz:
    display(plot_graph_with_state_values(mdp, state_values))

for i in range(num_iter):

    # Compute new state values using the functions you defined above.
    # It must be a dict {state : float V_new(state)}
    new_state_values = {s: get_new_state_value(mdp, state_values, s, gamma) for s in state_values.keys()}

    assert isinstance(new_state_values, dict)

    # Compute difference
    diff = max(abs(new_state_values[s] - state_values[s])
               for s in mdp.get_all_states())
    print("iter %4i   |   diff: %6.5f   |   " % (i, diff), end="")
    print('   '.join("V(%s) = %.3f" % (s, v) for s, v in state_values.items()))
    state_values = new_state_values

    if diff < min_difference:
        print("Terminated")
        break
```


    

    


    iter    0   |   diff: 3.50000   |   V(s0) = 0.000   V(s1) = 0.000   V(s2) = 0.000
    iter    1   |   diff: 0.64500   |   V(s0) = 0.000   V(s1) = 3.500   V(s2) = 0.000
    iter    2   |   diff: 0.58050   |   V(s0) = 0.000   V(s1) = 3.815   V(s2) = 0.645
    iter    3   |   diff: 0.43582   |   V(s0) = 0.581   V(s1) = 3.959   V(s2) = 0.962
    iter    4   |   diff: 0.30634   |   V(s0) = 0.866   V(s1) = 4.395   V(s2) = 1.272
    iter    5   |   diff: 0.27571   |   V(s0) = 1.145   V(s1) = 4.670   V(s2) = 1.579
    iter    6   |   diff: 0.24347   |   V(s0) = 1.421   V(s1) = 4.926   V(s2) = 1.838
    iter    7   |   diff: 0.21419   |   V(s0) = 1.655   V(s1) = 5.169   V(s2) = 2.075
    iter    8   |   diff: 0.19277   |   V(s0) = 1.868   V(s1) = 5.381   V(s2) = 2.290
    iter    9   |   diff: 0.17327   |   V(s0) = 2.061   V(s1) = 5.573   V(s2) = 2.481
    iter   10   |   diff: 0.15569   |   V(s0) = 2.233   V(s1) = 5.746   V(s2) = 2.654
    iter   11   |   diff: 0.14012   |   V(s0) = 2.389   V(s1) = 5.902   V(s2) = 2.810
    iter   12   |   diff: 0.12610   |   V(s0) = 2.529   V(s1) = 6.042   V(s2) = 2.950
    iter   13   |   diff: 0.11348   |   V(s0) = 2.655   V(s1) = 6.168   V(s2) = 3.076
    iter   14   |   diff: 0.10213   |   V(s0) = 2.769   V(s1) = 6.282   V(s2) = 3.190
    iter   15   |   diff: 0.09192   |   V(s0) = 2.871   V(s1) = 6.384   V(s2) = 3.292
    iter   16   |   diff: 0.08272   |   V(s0) = 2.963   V(s1) = 6.476   V(s2) = 3.384
    iter   17   |   diff: 0.07445   |   V(s0) = 3.045   V(s1) = 6.558   V(s2) = 3.467
    iter   18   |   diff: 0.06701   |   V(s0) = 3.120   V(s1) = 6.633   V(s2) = 3.541
    iter   19   |   diff: 0.06031   |   V(s0) = 3.187   V(s1) = 6.700   V(s2) = 3.608
    iter   20   |   diff: 0.05428   |   V(s0) = 3.247   V(s1) = 6.760   V(s2) = 3.668
    iter   21   |   diff: 0.04885   |   V(s0) = 3.301   V(s1) = 6.814   V(s2) = 3.723
    iter   22   |   diff: 0.04396   |   V(s0) = 3.350   V(s1) = 6.863   V(s2) = 3.771
    iter   23   |   diff: 0.03957   |   V(s0) = 3.394   V(s1) = 6.907   V(s2) = 3.815
    iter   24   |   diff: 0.03561   |   V(s0) = 3.434   V(s1) = 6.947   V(s2) = 3.855
    iter   25   |   diff: 0.03205   |   V(s0) = 3.469   V(s1) = 6.982   V(s2) = 3.891
    iter   26   |   diff: 0.02884   |   V(s0) = 3.502   V(s1) = 7.014   V(s2) = 3.923
    iter   27   |   diff: 0.02596   |   V(s0) = 3.530   V(s1) = 7.043   V(s2) = 3.951
    iter   28   |   diff: 0.02336   |   V(s0) = 3.556   V(s1) = 7.069   V(s2) = 3.977
    iter   29   |   diff: 0.02103   |   V(s0) = 3.580   V(s1) = 7.093   V(s2) = 4.001
    iter   30   |   diff: 0.01892   |   V(s0) = 3.601   V(s1) = 7.114   V(s2) = 4.022
    iter   31   |   diff: 0.01703   |   V(s0) = 3.620   V(s1) = 7.133   V(s2) = 4.041
    iter   32   |   diff: 0.01533   |   V(s0) = 3.637   V(s1) = 7.150   V(s2) = 4.058
    iter   33   |   diff: 0.01380   |   V(s0) = 3.652   V(s1) = 7.165   V(s2) = 4.073
    iter   34   |   diff: 0.01242   |   V(s0) = 3.666   V(s1) = 7.179   V(s2) = 4.087
    iter   35   |   diff: 0.01117   |   V(s0) = 3.678   V(s1) = 7.191   V(s2) = 4.099
    iter   36   |   diff: 0.01006   |   V(s0) = 3.689   V(s1) = 7.202   V(s2) = 4.110
    iter   37   |   diff: 0.00905   |   V(s0) = 3.699   V(s1) = 7.212   V(s2) = 4.121
    iter   38   |   diff: 0.00815   |   V(s0) = 3.708   V(s1) = 7.221   V(s2) = 4.130
    iter   39   |   diff: 0.00733   |   V(s0) = 3.717   V(s1) = 7.230   V(s2) = 4.138
    iter   40   |   diff: 0.00660   |   V(s0) = 3.724   V(s1) = 7.237   V(s2) = 4.145
    iter   41   |   diff: 0.00594   |   V(s0) = 3.731   V(s1) = 7.244   V(s2) = 4.152
    iter   42   |   diff: 0.00534   |   V(s0) = 3.736   V(s1) = 7.249   V(s2) = 4.158
    iter   43   |   diff: 0.00481   |   V(s0) = 3.742   V(s1) = 7.255   V(s2) = 4.163
    iter   44   |   diff: 0.00433   |   V(s0) = 3.747   V(s1) = 7.260   V(s2) = 4.168
    iter   45   |   diff: 0.00390   |   V(s0) = 3.751   V(s1) = 7.264   V(s2) = 4.172
    iter   46   |   diff: 0.00351   |   V(s0) = 3.755   V(s1) = 7.268   V(s2) = 4.176
    iter   47   |   diff: 0.00316   |   V(s0) = 3.758   V(s1) = 7.271   V(s2) = 4.179
    iter   48   |   diff: 0.00284   |   V(s0) = 3.762   V(s1) = 7.275   V(s2) = 4.183
    iter   49   |   diff: 0.00256   |   V(s0) = 3.764   V(s1) = 7.277   V(s2) = 4.185
    iter   50   |   diff: 0.00230   |   V(s0) = 3.767   V(s1) = 7.280   V(s2) = 4.188
    iter   51   |   diff: 0.00207   |   V(s0) = 3.769   V(s1) = 7.282   V(s2) = 4.190
    iter   52   |   diff: 0.00186   |   V(s0) = 3.771   V(s1) = 7.284   V(s2) = 4.192
    iter   53   |   diff: 0.00168   |   V(s0) = 3.773   V(s1) = 7.286   V(s2) = 4.194
    iter   54   |   diff: 0.00151   |   V(s0) = 3.775   V(s1) = 7.288   V(s2) = 4.196
    iter   55   |   diff: 0.00136   |   V(s0) = 3.776   V(s1) = 7.289   V(s2) = 4.197
    iter   56   |   diff: 0.00122   |   V(s0) = 3.778   V(s1) = 7.291   V(s2) = 4.199
    iter   57   |   diff: 0.00110   |   V(s0) = 3.779   V(s1) = 7.292   V(s2) = 4.200
    iter   58   |   diff: 0.00099   |   V(s0) = 3.780   V(s1) = 7.293   V(s2) = 4.201
    Terminated



```python
if has_graphviz:
    display(plot_graph_with_state_values(mdp, state_values))
```


    

    



```python
print("Final state values:", state_values)

assert abs(state_values['s0'] - 3.781) < 0.01
assert abs(state_values['s1'] - 7.294) < 0.01
assert abs(state_values['s2'] - 4.202) < 0.01
```

    Final state values: {'s0': 3.7810348735476405, 's1': 7.294006423867229, 's2': 4.202140275227048}


Now let's use those $V^{*}(s)$ to find optimal actions in each state

 $$\pi^*(s) = argmax_a \sum_{s'} P(s' | s,a) \cdot [ r(s,a,s') + \gamma V_{i}(s')] = argmax_a Q_i(s,a)$$
 
The only difference vs V(s) is that here we take not max but argmax: find action such with maximum Q(s,a).


```python
def get_optimal_action(mdp, state_values, state, gamma=0.9):
    """ Finds optimal action using formula above. """
    if mdp.is_terminal(state):
        return None

    v = -1e30
    action = None
    for possible_action in mdp.get_possible_actions(state):
        cur_v = get_action_value(mdp, state_values, state, possible_action, gamma)
        if cur_v > v:
            v = cur_v
            action = possible_action

    return action
```


```python
assert get_optimal_action(mdp, state_values, 's0', gamma) == 'a1'
assert get_optimal_action(mdp, state_values, 's1', gamma) == 'a0'
assert get_optimal_action(mdp, state_values, 's2', gamma) == 'a1'

assert get_optimal_action(mdp, {'s0': -1e10, 's1': 0, 's2': -2e10}, 's0', 0.9) == 'a0', \
    "Please ensure that you handle negative Q-values of arbitrary magnitude correctly"
assert get_optimal_action(mdp, {'s0': -2e10, 's1': 0, 's2': -1e10}, 's0', 0.9) == 'a1', \
    "Please ensure that you handle negative Q-values of arbitrary magnitude correctly"
```


```python
if has_graphviz:
    display(plot_graph_optimal_strategy_and_state_values(mdp, state_values, get_action_value))
```


    

    



```python
# Measure agent's average reward

s = mdp.reset()
rewards = []
for _ in range(10000):
    s, r, done, _ = mdp.step(get_optimal_action(mdp, state_values, s, gamma))
    rewards.append(r)

print("average reward: ", np.mean(rewards))

assert(0.40 < np.mean(rewards) < 0.55)
```

    average reward:  0.442


### Frozen lake


```python
from mdp import FrozenLakeEnv
mdp = FrozenLakeEnv(slip_chance=0)

mdp.render()
```

    *FFF
    FHFH
    FFFH
    HFFG
    



```python
def value_iteration(mdp, state_values=None, gamma=0.9, num_iter=1000, min_difference=1e-5, verbose=True):
    """ performs num_iter value iteration steps starting from state_values. Same as before but in a function """
    state_values = state_values or {s: 0 for s in mdp.get_all_states()}
    it = 0
    for i in range(num_iter):

        # Compute new state values using the functions you defined above. It must be a dict {state : new_V(state)}
        new_state_values = {s: get_new_state_value(mdp, state_values, s, gamma) for s in state_values.keys()}

        assert isinstance(new_state_values, dict)

        # Compute difference
        diff = max(abs(new_state_values[s] - state_values[s])
                   for s in mdp.get_all_states())

        if verbose:
            print("iter %4i   |   diff: %6.5f   |   V(start): %.3f " %
                  (i, diff, new_state_values[mdp._initial_state]))

        state_values = new_state_values
        it += 1
        if diff < min_difference:
            break

    return state_values, it
```


```python
state_values, _ = value_iteration(mdp)
```

    iter    0   |   diff: 1.00000   |   V(start): 0.000 
    iter    1   |   diff: 0.90000   |   V(start): 0.000 
    iter    2   |   diff: 0.81000   |   V(start): 0.000 
    iter    3   |   diff: 0.72900   |   V(start): 0.000 
    iter    4   |   diff: 0.65610   |   V(start): 0.000 
    iter    5   |   diff: 0.59049   |   V(start): 0.590 
    iter    6   |   diff: 0.00000   |   V(start): 0.590 



```python
s = mdp.reset()
mdp.render()
for t in range(100):
    a = get_optimal_action(mdp, state_values, s, gamma)
    print(a, end='\n\n')
    s, r, done, _ = mdp.step(a)
    mdp.render()
    if done:
        break
```

    *FFF
    FHFH
    FFFH
    HFFG
    
    down
    
    SFFF
    *HFH
    FFFH
    HFFG
    
    down
    
    SFFF
    FHFH
    *FFH
    HFFG
    
    right
    
    SFFF
    FHFH
    F*FH
    HFFG
    
    down
    
    SFFF
    FHFH
    FFFH
    H*FG
    
    right
    
    SFFF
    FHFH
    FFFH
    HF*G
    
    right
    
    SFFF
    FHFH
    FFFH
    HFF*
    


### Let's visualize!

It's usually interesting to see what your algorithm actually learned under the hood. To do so, we'll plot state value functions and optimal actions at each VI step.


```python
import matplotlib.pyplot as plt
%matplotlib inline


def draw_policy(mdp, state_values):
    plt.figure(figsize=(3, 3))
    h, w = mdp.desc.shape
    states = sorted(mdp.get_all_states())
    V = np.array([state_values[s] for s in states])
    Pi = {s: get_optimal_action(mdp, state_values, s, gamma) for s in states}
    plt.imshow(V.reshape(w, h), cmap='gray', interpolation='none', clim=(0, 1))
    ax = plt.gca()
    ax.set_xticks(np.arange(h)-.5)
    ax.set_yticks(np.arange(w)-.5)
    ax.set_xticklabels([])
    ax.set_yticklabels([])
    Y, X = np.mgrid[0:4, 0:4]
    a2uv = {'left': (-1, 0), 'down': (0, -1), 'right': (1, 0), 'up': (0, 1)}
    for y in range(h):
        for x in range(w):
            plt.text(x, y, str(mdp.desc[y, x].item()),
                     color='g', size=12,  verticalalignment='center',
                     horizontalalignment='center', fontweight='bold')
            a = Pi[y, x]
            if a is None:
                continue
            u, v = a2uv[a]
            plt.arrow(x, y, u*.3, -v*.3, color='m',
                      head_width=0.1, head_length=0.1)
    plt.grid(color='b', lw=2, ls='-')
    plt.show()
```


```python
state_values = {s: 0 for s in mdp.get_all_states()}

for i in range(10):
    print("after iteration %i" % i)
    state_values, _ = value_iteration(mdp, state_values, num_iter=1)
    draw_policy(mdp, state_values)
# please ignore iter 0 at each step
```


```python
from IPython.display import clear_output
from time import sleep
mdp = FrozenLakeEnv(map_name='8x8', slip_chance=0.1)
state_values = {s: 0 for s in mdp.get_all_states()}

for i in range(30):
    clear_output(True)
    print("after iteration %i" % i)
    state_values, _ = value_iteration(mdp, state_values, num_iter=1)
    draw_policy(mdp, state_values)
    sleep(0.5)
# please ignore iter 0 at each step
```

Massive tests


```python
mdp = FrozenLakeEnv(slip_chance=0)
state_values, _ = value_iteration(mdp)

total_rewards = []
for game_i in range(1000):
    s = mdp.reset()
    rewards = []
    for t in range(100):
        s, r, done, _ = mdp.step(
            get_optimal_action(mdp, state_values, s, gamma))
        rewards.append(r)
        if done:
            break
    total_rewards.append(np.sum(rewards))

print("average reward: ", np.mean(total_rewards))
assert(1.0 <= np.mean(total_rewards) <= 1.0)
print("Well done!")
```

    iter    0   |   diff: 1.00000   |   V(start): 0.000 
    iter    1   |   diff: 0.90000   |   V(start): 0.000 
    iter    2   |   diff: 0.81000   |   V(start): 0.000 
    iter    3   |   diff: 0.72900   |   V(start): 0.000 
    iter    4   |   diff: 0.65610   |   V(start): 0.000 
    iter    5   |   diff: 0.59049   |   V(start): 0.590 
    iter    6   |   diff: 0.00000   |   V(start): 0.590 
    average reward:  1.0
    Well done!



```python
# Measure agent's average reward
mdp = FrozenLakeEnv(slip_chance=0.1)
state_values, _ = value_iteration(mdp)

total_rewards = []
for game_i in range(1000):
    s = mdp.reset()
    rewards = []
    for t in range(100):
        s, r, done, _ = mdp.step(
            get_optimal_action(mdp, state_values, s, gamma))
        rewards.append(r)
        if done:
            break
    total_rewards.append(np.sum(rewards))

print("average reward: ", np.mean(total_rewards))
assert(0.8 <= np.mean(total_rewards) <= 0.95)
print("Well done!")
```

    iter    0   |   diff: 0.90000   |   V(start): 0.000 
    iter    1   |   diff: 0.72900   |   V(start): 0.000 
    iter    2   |   diff: 0.62330   |   V(start): 0.000 
    iter    3   |   diff: 0.50487   |   V(start): 0.000 
    iter    4   |   diff: 0.40894   |   V(start): 0.000 
    iter    5   |   diff: 0.34868   |   V(start): 0.349 
    iter    6   |   diff: 0.06529   |   V(start): 0.410 
    iter    7   |   diff: 0.05832   |   V(start): 0.468 
    iter    8   |   diff: 0.01139   |   V(start): 0.480 
    iter    9   |   diff: 0.00764   |   V(start): 0.487 
    iter   10   |   diff: 0.00164   |   V(start): 0.489 
    iter   11   |   diff: 0.00094   |   V(start): 0.490 
    iter   12   |   diff: 0.00022   |   V(start): 0.490 
    iter   13   |   diff: 0.00011   |   V(start): 0.490 
    iter   14   |   diff: 0.00003   |   V(start): 0.490 
    iter   15   |   diff: 0.00001   |   V(start): 0.490 
    iter   16   |   diff: 0.00000   |   V(start): 0.490 
    average reward:  0.873
    Well done!



```python
# Measure agent's average reward
mdp = FrozenLakeEnv(slip_chance=0.25)
state_values, _ = value_iteration(mdp)

total_rewards = []
for game_i in range(1000):
    s = mdp.reset()
    rewards = []
    for t in range(100):
        s, r, done, _ = mdp.step(
            get_optimal_action(mdp, state_values, s, gamma))
        rewards.append(r)
        if done:
            break
    total_rewards.append(np.sum(rewards))

print("average reward: ", np.mean(total_rewards))
assert(0.6 <= np.mean(total_rewards) <= 0.7)
print("Well done!")
```

    iter    0   |   diff: 0.75000   |   V(start): 0.000 
    iter    1   |   diff: 0.50625   |   V(start): 0.000 
    iter    2   |   diff: 0.39867   |   V(start): 0.000 
    iter    3   |   diff: 0.26910   |   V(start): 0.000 
    iter    4   |   diff: 0.18164   |   V(start): 0.000 
    iter    5   |   diff: 0.14013   |   V(start): 0.140 
    iter    6   |   diff: 0.07028   |   V(start): 0.199 
    iter    7   |   diff: 0.06030   |   V(start): 0.260 
    iter    8   |   diff: 0.02594   |   V(start): 0.285 
    iter    9   |   diff: 0.01918   |   V(start): 0.305 
    iter   10   |   diff: 0.00858   |   V(start): 0.313 
    iter   11   |   diff: 0.00560   |   V(start): 0.319 
    iter   12   |   diff: 0.00260   |   V(start): 0.321 
    iter   13   |   diff: 0.00159   |   V(start): 0.323 
    iter   14   |   diff: 0.00076   |   V(start): 0.324 
    iter   15   |   diff: 0.00045   |   V(start): 0.324 
    iter   16   |   diff: 0.00022   |   V(start): 0.324 
    iter   17   |   diff: 0.00012   |   V(start): 0.325 
    iter   18   |   diff: 0.00006   |   V(start): 0.325 
    iter   19   |   diff: 0.00003   |   V(start): 0.325 
    iter   20   |   diff: 0.00002   |   V(start): 0.325 
    iter   21   |   diff: 0.00001   |   V(start): 0.325 
    average reward:  0.638
    Well done!



```python
# Measure agent's average reward
mdp = FrozenLakeEnv(slip_chance=0.2, map_name='8x8')
state_values, _ = value_iteration(mdp)

total_rewards = []
for game_i in range(1000):
    s = mdp.reset()
    rewards = []
    for t in range(100):
        s, r, done, _ = mdp.step(
            get_optimal_action(mdp, state_values, s, gamma))
        rewards.append(r)
        if done:
            break
    total_rewards.append(np.sum(rewards))

print("average reward: ", np.mean(total_rewards))
assert(0.6 <= np.mean(total_rewards) <= 0.8)
print("Well done!")
```

    iter    0   |   diff: 0.80000   |   V(start): 0.000 
    iter    1   |   diff: 0.57600   |   V(start): 0.000 
    iter    2   |   diff: 0.41472   |   V(start): 0.000 
    iter    3   |   diff: 0.29860   |   V(start): 0.000 
    iter    4   |   diff: 0.24186   |   V(start): 0.000 
    iter    5   |   diff: 0.19349   |   V(start): 0.000 
    iter    6   |   diff: 0.15325   |   V(start): 0.000 
    iter    7   |   diff: 0.12288   |   V(start): 0.000 
    iter    8   |   diff: 0.09930   |   V(start): 0.000 
    iter    9   |   diff: 0.08037   |   V(start): 0.000 
    iter   10   |   diff: 0.06426   |   V(start): 0.000 
    iter   11   |   diff: 0.05129   |   V(start): 0.000 
    iter   12   |   diff: 0.04330   |   V(start): 0.000 
    iter   13   |   diff: 0.03802   |   V(start): 0.033 
    iter   14   |   diff: 0.03332   |   V(start): 0.058 
    iter   15   |   diff: 0.02910   |   V(start): 0.087 
    iter   16   |   diff: 0.01855   |   V(start): 0.106 
    iter   17   |   diff: 0.01403   |   V(start): 0.120 
    iter   18   |   diff: 0.00810   |   V(start): 0.128 
    iter   19   |   diff: 0.00555   |   V(start): 0.133 
    iter   20   |   diff: 0.00321   |   V(start): 0.137 
    iter   21   |   diff: 0.00247   |   V(start): 0.138 
    iter   22   |   diff: 0.00147   |   V(start): 0.139 
    iter   23   |   diff: 0.00104   |   V(start): 0.140 
    iter   24   |   diff: 0.00058   |   V(start): 0.140 
    iter   25   |   diff: 0.00036   |   V(start): 0.141 
    iter   26   |   diff: 0.00024   |   V(start): 0.141 
    iter   27   |   diff: 0.00018   |   V(start): 0.141 
    iter   28   |   diff: 0.00012   |   V(start): 0.141 
    iter   29   |   diff: 0.00007   |   V(start): 0.141 
    iter   30   |   diff: 0.00004   |   V(start): 0.141 
    iter   31   |   diff: 0.00003   |   V(start): 0.141 
    iter   32   |   diff: 0.00001   |   V(start): 0.141 
    iter   33   |   diff: 0.00001   |   V(start): 0.141 
    average reward:  0.756
    Well done!


# HW Part 1: Value iteration convergence

### Find an MDP for which value iteration takes long to converge  (0.5 pts)

When we ran value iteration on the small frozen lake problem, the last iteration where an action changed was iteration 6--i.e., value iteration computed the optimal policy at iteration 6. Are there any guarantees regarding how many iterations it'll take value iteration to compute the optimal policy? There are no such guarantees without additional assumptions--we can construct the MDP in such a way that the greedy policy will change after arbitrarily many iterations.

Your task: define an MDP with at most 3 states and 2 actions, such that when you run value iteration, the optimal action changes at iteration >= 50. Use discount=0.95. (However, note that the discount doesn't matter here--you can construct an appropriate MDP with any discount.)

Note: value function must change at least once after iteration >=50, not necessarily change on every iteration till >=50.


```python
transition_probs = {
    's0': {
        'a0': {'s0': 0.999, 's1': 0.001}
    },
    's1': {
        'a0': {'s1': 1},
        'a1': {'s1': 0.98435, 's2': 0.01565}  # very carefully trial and error :)
    },
    's2': {
        'a0': {'s2': 1}
    }
}
rewards = {
    's0': {'a0': {'s0': -1, 's1': +1e3}},
    's1': {'a0': {'s1': +1e3}, 'a1': {'s1': -5e2, 's2': +1e4}},
    's2': {'a0': {'s2': +1e4}}
}

gamma = 0.95

from mdp import MDP
from numpy import random
mdp_bonus = MDP(transition_probs, rewards, initial_state='s0')
# Feel free to change the initial_state
```


```python
if has_graphviz:
    display(plot_graph(mdp_bonus, graph_size="50,50"))
```


    

    



```python
state_values = {s: 0 for s in mdp_bonus.get_all_states()}
policy = np.array([get_optimal_action(mdp_bonus, state_values, state, gamma)
                   for state in sorted(mdp_bonus.get_all_states())])

for i in range(100):
    print("after iteration %i" % i)
    state_values, _ = value_iteration(mdp_bonus, state_values, num_iter=1)

    print(policy)
    new_policy = np.array([get_optimal_action(mdp_bonus, state_values, state, gamma)
                           for state in sorted(mdp_bonus.get_all_states())])

    n_changes = (policy != new_policy).sum()
    print("N actions changed = %i \n" % n_changes)
    policy = new_policy

# please ignore iter 0 at each step
```

    after iteration 0
    iter    0   |   diff: 10000.00000   |   V(start): 0.001 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 1
    iter    0   |   diff: 9000.00000   |   V(start): 0.902 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 2
    iter    0   |   diff: 8100.00000   |   V(start): 2.522 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 3
    iter    0   |   diff: 7290.00000   |   V(start): 4.707 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 4
    iter    0   |   diff: 6561.00000   |   V(start): 7.329 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 5
    iter    0   |   diff: 5904.90000   |   V(start): 10.276 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 6
    iter    0   |   diff: 5314.41000   |   V(start): 13.457 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 7
    iter    0   |   diff: 4782.96900   |   V(start): 16.795 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 8
    iter    0   |   diff: 4304.67210   |   V(start): 20.228 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 9
    iter    0   |   diff: 3874.20489   |   V(start): 23.701 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 10
    iter    0   |   diff: 3486.78440   |   V(start): 27.172 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 11
    iter    0   |   diff: 3138.10596   |   V(start): 30.607 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 12
    iter    0   |   diff: 2824.29536   |   V(start): 33.978 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 13
    iter    0   |   diff: 2541.86583   |   V(start): 37.263 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 14
    iter    0   |   diff: 2287.67925   |   V(start): 40.445 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 15
    iter    0   |   diff: 2058.91132   |   V(start): 43.512 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 16
    iter    0   |   diff: 1853.02019   |   V(start): 46.455 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 17
    iter    0   |   diff: 1667.71817   |   V(start): 49.268 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 18
    iter    0   |   diff: 1500.94635   |   V(start): 51.947 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 19
    iter    0   |   diff: 1350.85172   |   V(start): 54.491 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 20
    iter    0   |   diff: 1215.76655   |   V(start): 56.899 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 21
    iter    0   |   diff: 1094.18989   |   V(start): 59.175 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 22
    iter    0   |   diff: 984.77090   |   V(start): 61.319 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 23
    iter    0   |   diff: 886.29381   |   V(start): 63.335 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 24
    iter    0   |   diff: 797.66443   |   V(start): 65.227 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 25
    iter    0   |   diff: 717.89799   |   V(start): 67.001 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 26
    iter    0   |   diff: 646.10819   |   V(start): 68.660 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 27
    iter    0   |   diff: 581.49737   |   V(start): 70.210 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 28
    iter    0   |   diff: 523.34763   |   V(start): 71.656 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 29
    iter    0   |   diff: 471.01287   |   V(start): 73.003 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 30
    iter    0   |   diff: 423.91158   |   V(start): 74.256 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 31
    iter    0   |   diff: 381.52042   |   V(start): 75.421 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 32
    iter    0   |   diff: 343.36838   |   V(start): 76.503 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 33
    iter    0   |   diff: 309.03154   |   V(start): 77.507 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 34
    iter    0   |   diff: 278.12839   |   V(start): 78.437 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 35
    iter    0   |   diff: 250.31555   |   V(start): 79.299 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 36
    iter    0   |   diff: 225.28400   |   V(start): 80.096 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 37
    iter    0   |   diff: 202.75560   |   V(start): 80.833 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 38
    iter    0   |   diff: 182.48004   |   V(start): 81.513 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 39
    iter    0   |   diff: 164.23203   |   V(start): 82.142 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 40
    iter    0   |   diff: 147.80883   |   V(start): 82.722 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 41
    iter    0   |   diff: 133.02795   |   V(start): 83.256 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 42
    iter    0   |   diff: 119.72515   |   V(start): 83.749 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 43
    iter    0   |   diff: 107.75264   |   V(start): 84.203 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 44
    iter    0   |   diff: 96.97737   |   V(start): 84.620 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 45
    iter    0   |   diff: 87.27964   |   V(start): 85.005 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 46
    iter    0   |   diff: 78.55167   |   V(start): 85.358 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 47
    iter    0   |   diff: 70.69650   |   V(start): 85.683 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 48
    iter    0   |   diff: 63.62685   |   V(start): 85.981 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 49
    iter    0   |   diff: 57.26417   |   V(start): 86.255 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 50
    iter    0   |   diff: 51.53775   |   V(start): 86.507 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 51
    iter    0   |   diff: 46.38398   |   V(start): 86.737 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 52
    iter    0   |   diff: 41.74558   |   V(start): 86.949 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 53
    iter    0   |   diff: 37.57102   |   V(start): 87.143 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 54
    iter    0   |   diff: 33.81392   |   V(start): 87.321 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 55
    iter    0   |   diff: 30.43253   |   V(start): 87.484 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 56
    iter    0   |   diff: 27.38927   |   V(start): 87.633 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 57
    iter    0   |   diff: 24.65035   |   V(start): 87.770 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 58
    iter    0   |   diff: 22.18531   |   V(start): 87.895 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 59
    iter    0   |   diff: 19.96678   |   V(start): 88.009 
    ['a0' 'a0' 'a0']
    N actions changed = 0 
    
    after iteration 60
    iter    0   |   diff: 17.97010   |   V(start): 88.114 
    ['a0' 'a0' 'a0']
    N actions changed = 1 
    
    after iteration 61
    iter    0   |   diff: 16.17309   |   V(start): 88.210 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 62
    iter    0   |   diff: 14.55578   |   V(start): 88.297 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 63
    iter    0   |   diff: 13.10021   |   V(start): 88.377 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 64
    iter    0   |   diff: 11.79018   |   V(start): 88.450 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 65
    iter    0   |   diff: 10.61117   |   V(start): 88.517 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 66
    iter    0   |   diff: 9.55005   |   V(start): 88.578 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 67
    iter    0   |   diff: 8.59504   |   V(start): 88.634 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 68
    iter    0   |   diff: 7.73554   |   V(start): 88.685 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 69
    iter    0   |   diff: 6.96199   |   V(start): 88.731 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 70
    iter    0   |   diff: 6.26579   |   V(start): 88.774 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 71
    iter    0   |   diff: 5.63921   |   V(start): 88.812 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 72
    iter    0   |   diff: 5.07529   |   V(start): 88.848 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 73
    iter    0   |   diff: 4.56776   |   V(start): 88.880 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 74
    iter    0   |   diff: 4.11098   |   V(start): 88.909 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 75
    iter    0   |   diff: 3.69988   |   V(start): 88.936 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 76
    iter    0   |   diff: 3.32990   |   V(start): 88.960 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 77
    iter    0   |   diff: 2.99691   |   V(start): 88.982 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 78
    iter    0   |   diff: 2.69722   |   V(start): 89.003 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 79
    iter    0   |   diff: 2.42749   |   V(start): 89.021 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 80
    iter    0   |   diff: 2.18475   |   V(start): 89.038 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 81
    iter    0   |   diff: 1.96627   |   V(start): 89.053 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 82
    iter    0   |   diff: 1.76964   |   V(start): 89.067 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 83
    iter    0   |   diff: 1.59268   |   V(start): 89.080 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 84
    iter    0   |   diff: 1.43341   |   V(start): 89.091 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 85
    iter    0   |   diff: 1.29007   |   V(start): 89.102 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 86
    iter    0   |   diff: 1.16106   |   V(start): 89.111 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 87
    iter    0   |   diff: 1.04496   |   V(start): 89.120 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 88
    iter    0   |   diff: 0.94046   |   V(start): 89.128 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 89
    iter    0   |   diff: 0.84641   |   V(start): 89.135 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 90
    iter    0   |   diff: 0.76177   |   V(start): 89.142 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 91
    iter    0   |   diff: 0.68560   |   V(start): 89.148 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 92
    iter    0   |   diff: 0.61704   |   V(start): 89.153 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 93
    iter    0   |   diff: 0.55533   |   V(start): 89.158 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 94
    iter    0   |   diff: 0.49980   |   V(start): 89.163 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 95
    iter    0   |   diff: 0.44982   |   V(start): 89.167 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 96
    iter    0   |   diff: 0.40484   |   V(start): 89.170 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 97
    iter    0   |   diff: 0.36435   |   V(start): 89.174 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 98
    iter    0   |   diff: 0.32792   |   V(start): 89.177 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    
    after iteration 99
    iter    0   |   diff: 0.29513   |   V(start): 89.180 
    ['a0' 'a1' 'a0']
    N actions changed = 0 
    


### Value iteration convervence proof (0.5 pts)
**Note:** Assume that $\mathcal{S}, \mathcal{A}$ are finite.

Update of value function in value iteration can be rewritten in a form of Bellman operator:

$$(TV)(s) = \max_{a \in \mathcal{A}}\mathbb{E}\left[ r_{t+1} + \gamma V(s_{t+1}) | s_t = s, a_t = a\right]$$

Value iteration algorithm with Bellman operator:

---
&nbsp;&nbsp; Initialize $V_0$

&nbsp;&nbsp; **for** $k = 0,1,2,...$ **do**

&nbsp;&nbsp;&nbsp;&nbsp; $V_{k+1} \leftarrow TV_k$

&nbsp;&nbsp;**end for**

---

In [lecture](https://docs.google.com/presentation/d/1lz2oIUTvd2MHWKEQSH8hquS66oe4MZ_eRvVViZs2uuE/edit#slide=id.g4fd6bae29e_2_4) we established contraction property of bellman operator:

$$
||TV - TU||_{\infty} \le \gamma ||V - U||_{\infty}
$$

For all $V, U$

Using contraction property of Bellman operator, Banach fixed-point theorem and Bellman equations prove that value function converges to $V^*$ in value iterateion$

**Proof.**

Consider $||V_{k+1} - V^*||_{\infty}$. As $V^*$ is a fixed point of operator $T$, $TV^* = V^*$.
$$\begin{align}
||V_{k+1} - V^*|| = ||TV_k - TV^*|| \le&\ \gamma ||V_k - V^*|| = \gamma^{k+1}||V_0-V^*|| \implies \\
0 \le \lim\limits_{k \to +\infty} ||V_{k+1} - V^*|| \le&\ \lim\limits_{k \to +\infty} \gamma^{k+1}||V_0-V^*|| = [0 < \gamma < 1] = 0 \implies \\
\lim\limits_{k \to +\infty} ||V_{k+1} - V^*|| =&\ 0
\end{align}$$

**Q.E.D.**

### Bonus. Asynchronious value iteration (2 pts)

Consider the following algorithm:

---

Initialize $V_0$

**for** $k = 0,1,2,...$ **do**

&nbsp;&nbsp;&nbsp;&nbsp; Select some state $s_k \in \mathcal{S}$    

&nbsp;&nbsp;&nbsp;&nbsp; $V(s_k) := (TV)(s_k)$

**end for**

---


Note that unlike common value iteration, here we update only a single state at a time.

**Homework.** Prove the following proposition:

If for all $s \in \mathcal{S}$, $s$ appears in the sequence $(s_0, s_1, ...)$ infinitely often, then $V$ converges to $V*$

**Proof.**

Consider vector $\Delta^{t+1} = |V_{t+1} - V^*| = [|V_{t+1}(s_i) - V^*(s_i)|]_{i = \overline{1,|\mathcal{S}|}}$.

Let's split sequence $S$ of $s_i$-s into all subsequences with same index $S_i$. It's guaranteed that all of them are infinite, that is $|S_i| \to +\infty$. Then, we take a look at $||V_{t+1}-V^*||_{\infty}$. It's obvious that $||V_{t+1}-V^*||_{\infty} \le ||V_{t+1}-V^*||_1 = \sum\limits_{i=1}^{|\mathcal{S}|} \Delta_i^{t+1}$.

Taking a limit on both sides:
$$0 \le \lim\limits_{t \to +\infty} ||V_{t+1}-V^*||_{\infty} \le \lim\limits_{t \to +\infty}\sum\limits_{i=1}^{|\mathcal{S}|} \Delta_i^{t+1} = \sum\limits_{i=1}^{|\mathcal{S}|}\lim\limits_{t \to +\infty} \Delta_i^{t+1}$$

Last equality is valid for finite $|\mathcal{S}|$ iff $\lim\limits_{t \to +\infty} \Delta_i^{t+1}$ exists for all $i$. But
$$\lim\limits_{t \to +\infty} \Delta_i^{t+1} = \lim\limits_{t \to +\infty} |V_{t+1}(s_i) - V^*(s_i)|$$

Introducing $S_i(t) = \sum\limits_{j=1}^{t} \mathcal{I}(S[j] = s_i)$ where $\mathcal{I}$ is the indicator function. Then, in terms of $S_i(t)$ $s_k$ appearing infinitely many times in $S$ means that $S_i(t) \to +\infty$ as $t \to +\infty$. And limit above can be expressed in terms of $S_i(t)$ as well!

$$\lim\limits_{t \to +\infty} |V_{t+1}(s_i) - V^*(s_i)| \le \lim\limits_{t \to +\infty} \gamma^{S_i(t)} |V_0(s_i) - V^*(s_i)| = 0$$

Substituting this gives us:
$$0 \le \lim\limits_{t \to +\infty} ||V_{t+1}-V^*||_{\infty} \le \sum\limits_{i=1}^{|\mathcal{S}|}\lim\limits_{t \to +\infty} \gamma^{S_i(t)} |V_0(s_i) - V^*(s_i)| = [0 < \gamma < 1] = 0$$

Then $\lim\limits_{t \to +\infty} ||V_{t+1}-V^*||_{\infty} = 0$. **Q.E.D.**

# HW Part 2: Policy iteration

## Policy iteration implementateion (2 pts)

Let's implement exact policy iteration (PI), which has the following pseudocode:

---
Initialize $\pi_0$   `// random or fixed action`

For $n=0, 1, 2, \dots$
- Compute the state-value function $V^{\pi_{n}}$
- Using $V^{\pi_{n}}$, compute the state-action-value function $Q^{\pi_{n}}$
- Compute new policy $\pi_{n+1}(s) = \operatorname*{argmax}_a Q^{\pi_{n}}(s,a)$
---

Unlike VI, policy iteration has to maintain a policy - chosen actions from all states - and estimate $V^{\pi_{n}}$ based on this policy. It only changes policy once values converged.


Below are a few helpers that you may or may not use in your implementation.


```python
transition_probs = {
    's0': {
        'a0': {'s0': 0.5, 's2': 0.5},
        'a1': {'s2': 1}
    },
    's1': {
        'a0': {'s0': 0.7, 's1': 0.1, 's2': 0.2},
        'a1': {'s1': 0.95, 's2': 0.05}
    },
    's2': {
        'a0': {'s0': 0.4, 's1': 0.6},
        'a1': {'s0': 0.3, 's1': 0.3, 's2': 0.4}
    }
}
rewards = {
    's1': {'a0': {'s0': +5}},
    's2': {'a1': {'s0': -1}}
}

from mdp import MDP
mdp = MDP(transition_probs, rewards, initial_state='s0')
```

Let's write a function called `compute_vpi` that computes the state-value function $V^{\pi}$ for an arbitrary policy $\pi$.

Unlike VI, this time you must find the exact solution, not just a single iteration.

Recall that $V^{\pi}$ satisfies the following linear equation:
$$V^{\pi}(s) = \sum_{s'} P(s,\pi(s),s')[ R(s,\pi(s),s') + \gamma V^{\pi}(s')]$$

You'll have to solve a linear system in your code. (Find an exact solution, e.g., with `np.linalg.solve`.)


```python
def compute_vpi(mdp, policy, gamma):
    """
    Computes V^pi(s) FOR ALL STATES under given policy.
    :param policy: a dict of currently chosen actions {s : a}
    :returns: a dict {state : V^pi(state) for all states}
    """
    states = mdp.get_all_states()
    state_to_idx = dict(zip(states, np.arange(len(states))))
    system = np.diag(np.ones(len(states)))
    rhs = np.zeros(len(states))
    for (i, state) in enumerate(states):
        policy_action = policy[state]
        if policy_action:  # if state has no actions, policy_action will be an empty string
            for next_state in mdp.get_next_states(state, policy_action):
                p = mdp.get_transition_prob(state, policy_action, next_state)
                r = mdp.get_reward(state, policy_action, next_state)
                rhs[i] += p * r
                system[i][state_to_idx[next_state]] -= gamma * p
            
    vpi = np.linalg.solve(system, rhs).tolist()
    return dict(zip(states, vpi))
```


```python
test_policy = {s: np.random.choice(
    mdp.get_possible_actions(s)) for s in mdp.get_all_states()}
new_vpi = compute_vpi(mdp, test_policy, gamma)

print(new_vpi)

assert type(
    new_vpi) is dict, "compute_vpi must return a dict {state : V^pi(state) for all states}"
```

    {'s0': 17.245615232375616, 's1': 20.350781409950176, 's2': 18.153279191974335}


Once we've got new state values, it's time to update our policy.


```python
def compute_new_policy(mdp, vpi, gamma):
    """
    Computes new policy as argmax of state values
    :param vpi: a dict {state : V^pi(state) for all states}
    :returns: a dict {state : optimal action for all states}
    """
    states = mdp.get_all_states()
    q_pi = lambda state, action: np.sum([mdp.get_transition_prob(state, action, next_state)
                                         * (mdp.get_reward(state, action, next_state) + gamma * vpi[next_state])
                                         for next_state in mdp.get_next_states(state, action)])
    policy = {}
    for state in states:
        actions = mdp.get_possible_actions(state)
        if actions:
            policy[state] = actions[np.argmax([q_pi(state, action) for action in actions])]
        else:
            policy[state] = ''
            
    return policy
```


```python
new_policy = compute_new_policy(mdp, new_vpi, gamma)

print(new_policy)

assert type(
    new_policy) is dict, "compute_new_policy must return a dict {state : optimal action for all states}"
```

    {'s0': 'a1', 's1': 'a0', 's2': 'a0'}


__Main loop__


```python
def policy_iteration(mdp, policy=None, gamma=0.9, num_iter=1000, min_difference=1e-5, iter_callback=None):
    """ 
    Run the policy iteration loop for num_iter iterations or till difference between V(s) is below min_difference.
    If policy is not given, initialize it at random.
    """
    def get_random_policy():
        policy = {}
        states = mdp.get_all_states()
        for state in states:
            actions = mdp.get_possible_actions(state)
            if actions:
                policy[state] = actions[np.random.randint(0, len(actions))]
            else:
                policy[state] = ''
        return policy
    
    def diff(vpi, new_vpi):
        vpi_np = np.array(list(vpi.values()))
        new_vpi_np = np.array(list(new_vpi.values()))
        return np.linalg.norm(vpi_np - new_vpi_np, ord=np.inf)
    
    if policy is None:
        policy = get_random_policy()
    state_values = compute_vpi(mdp, policy, gamma)
    it = 0
    for i in range(num_iter):
        new_policy = compute_new_policy(mdp, state_values, gamma)
        new_state_values = compute_vpi(mdp, new_policy, gamma)
        d = diff(state_values, new_state_values)
        policy_is_optimal = d < min_difference
        state_values = new_state_values
        policy = new_policy
        it += 1
        if iter_callback:
            iter_callback(it=it, diff=d, policy=policy, vpi=state_values)
        if policy_is_optimal:
            break

    return state_values, policy, it
```

__Your PI Results__


```python
gamma = 0.95
```


```python
def play_vi_game(mdp, gamma=0.9, verbose=True):
    if verbose:
        print('Value iteration log')
    state_values, total_iters = value_iteration(mdp, gamma=gamma, verbose=verbose)

    total_rewards = []
    for game_i in range(1000):
        s = mdp.reset()
        rewards = []
        for t in range(100):
            s, r, done, _ = mdp.step(
                get_optimal_action(mdp, state_values, s, gamma))
            rewards.append(r)
            if done:
                break
        total_rewards.append(np.sum(rewards))
    return np.mean(total_rewards), total_iters

def play_pi_game(mdp, gamma=0.9, verbose=True):
    pi_callback = lambda it, diff, policy, vpi: 1
    if verbose:
        print('Policy iteration log')
        pi_callback = lambda it, diff, policy, vpi:\
            print(f'iter {it:4d}   |   diff: {diff:.5f}   |   V(start): {vpi[(0,0)]:.3f}')
    vpi, policy, total_iters = policy_iteration(mdp, gamma=gamma, iter_callback=pi_callback)
    
    total_rewards = []
    for game_i in range(1000):||
        s = mdp.reset()
        rewards = []
        for t in range(100):
            s, r, done, _ = mdp.step(policy[s])
            rewards.append(r)
            if done:
                break
        total_rewards.append(np.sum(rewards))
    return np.mean(total_rewards), total_iters
```


```python
from tqdm import tqdm

def print_mdp_comparison(mdp, mdp_name, gamma=0.9, verbose=False):
    print('###############################################')
    print(f'Playing MDP {mdp_name}')
    print('###############################################')
    
    vi_history = []
    vi_iters = []
    pi_history = []
    pi_iters = []
    
    for i in tqdm(range(25)):
        vi_r, vi_it = play_vi_game(mdp, gamma, verbose)
        pi_r, pi_it = play_pi_game(mdp, gamma, verbose)
        vi_history.append(vi_r)
        vi_iters.append(vi_it)
        pi_history.append(pi_r)
        pi_iters.append(pi_it)

    print('Average VI reward stats')
    print(f'Mean: {np.mean(vi_history)}, std: {np.std(vi_history)}')
    print(f'Mean iter: {np.mean(vi_iters)}')
    print('Average PI reward stats')
    print(f'Mean: {np.mean(pi_history)}, std: {np.std(pi_history)}')
    print(f'Mean iter: {np.mean(pi_iters)}')
    
MDPs = [mdp_bonus,
        FrozenLakeEnv(slip_chance=0.25),
        FrozenLakeEnv(slip_chance=0.2, map_name='8x8')]
MDP_names = ['Bonus 1',
             'Small lake',
             'Big lake'
             ]

for (mdp, mdp_name) in zip(MDPs, MDP_names):
    print_mdp_comparison(mdp, mdp_name, gamma)
```

      0%|          | 0/25 [00:00<?, ?it/s]

    ###############################################
    Playing MDP Bonus 1
    ###############################################


    100%|██████████| 25/25 [02:24<00:00,  5.80s/it]
      0%|          | 0/25 [00:00<?, ?it/s]

    Average VI reward stats
    Mean: 16500.61112, std: 2816.309295078682
    Mean iter: 406.0
    Average PI reward stats
    Mean: 16856.87312, std: 2260.2912720156983
    Mean iter: 1.52
    ###############################################
    Playing MDP Small lake
    ###############################################


    100%|██████████| 25/25 [00:14<00:00,  1.68it/s]
      0%|          | 0/25 [00:00<?, ?it/s]

    Average VI reward stats
    Mean: 0.6535599999999999, std: 0.013896992480389428
    Mean iter: 24.0
    Average PI reward stats
    Mean: 0.6488, std: 0.013401492454200775
    Mean iter: 4.48
    ###############################################
    Playing MDP Big lake
    ###############################################


    100%|██████████| 25/25 [00:38<00:00,  1.54s/it]

    Average VI reward stats
    Mean: 0.77256, std: 0.015365103318884661
    Mean iter: 42.0
    Average PI reward stats
    Mean: 0.7697200000000001, std: 0.010148970391128363
    Mean iter: 6.56


    


**Tricky MDP:**

For tricky MDP policy iteration achieves better values in only a few iterations. Value iteration has difficulties converging because the probability of getting into very profitable state is rather low and discounting takes a lot of time to have an effect.

**Frozen lake envs:**

Stats above suggest that policy iteration converges to something slightly worse than optimal value iteration policy. Even though we calculate average reward estimate based on 25 runs of both algorithms, it still may differ quite significantly between runs. If we actually run some statistical test, I guess it'll tell us that they actually have the same average reward. This might be the case because environments are quite straight-forward and both algorithms converge to globally optimal policy. Apart from that, PI seems to be more stable than VI overall (its std is lower than that of VI).

## Policy iteration convergence (3 pts)

**Note:** Assume that $\mathcal{S}, \mathcal{A}$ are finite.

We can define another Bellman operator:

$$(T_{\pi}V)(s) = \mathbb{E}_{r, s'|s, a = \pi(s)}\left[r + \gamma V(s')\right]$$

And rewrite policy iteration algorithm in operator form:


---

Initialize $\pi_0$

**for** $k = 0,1,2,...$ **do**

&nbsp;&nbsp;&nbsp;&nbsp; Solve $V_k = T_{\pi_k}V_k$   

&nbsp;&nbsp;&nbsp;&nbsp; Select $\pi_{k+1}$ s.t. $T_{\pi_{k+1}}V_k = TV_k$ 

**end for**

---

To prove convergence of the algorithm we need to prove two properties: contraction an monotonicity.

#### Monotonicity (0.5 pts)

For all $V, U$ if $V(s) \le U(s)$   $\forall s \in \mathcal{S}$ then $(T_\pi V)(s) \le (T_\pi U)(s)$   $\forall s \in  \mathcal{S}$

**Proof.**

$$\begin{align}
V(s) \le U(s) \implies V(s) - U(s) \le&\ 0 \ \forall s \in \mathcal{S} \\
T_\pi V(s) - T_\pi U(s) =&\ \mathbb{E}_{r, s'|s, a = \pi(s)}\left[r + \gamma V(s')\right] - \mathbb{E}_{r, s'|s, a = \pi(s)}\left[r + \gamma U(s')\right] \\
=&\ \mathbb{E}_{r, s'|s, a = \pi(s)}\left[\gamma(V(s') - U(s'))\right] \\
=&\ \sum\limits_{s'} \underbrace{P(r, s' | s, a = \pi(s))\gamma}_{\ge 0}\underbrace{(V(s') - U(s'))}_{\le 0} \le 0
\end{align}
$$

**Q.E.D.**

#### Contraction (1 pts)

$$
||T_\pi V - T_\pi U||_{\infty} \le \gamma ||V - U||_{\infty}
$$

for all $V, U$.

**Proof.**

$$\begin{align}
||T_\pi V - T_\pi U||_{\infty} =&\ \max\limits_s \left|\sum\limits_{s'} P(r, s' | s, a = \pi(s))\gamma(V(s') - U(s'))\right| \le \\
\le&\ \max\limits_s \sum\limits_{s'} \gamma\left|P(r, s' | s, a = \pi(s))(V(s') - U(s'))\right| \le \\
\le&\ \max\limits_s \sum\limits_{s'} \gamma\max\limits_{s'}|V(s') - U(s')|P(r, s' | s, a = \pi(s)) = \\
=&\ \gamma||V - U||_{\infty}\max\limits_s \underbrace{\sum\limits_{s'} P(r, s' | s, a = \pi(s))}_{=1} = \\
=&\ \gamma||V - U||_{\infty}
\end{align}$$

**Q.E.D.**

#### Convergence (1.5 pts)

Prove that there exists iteration $k_0$ such that $\pi_k = \pi^*$ for all $k \ge k_0$

**Proof.**

It's easy to see that $V_k \le V_{k+1} \le V^*$. Using monotonicity, we conclude that $T_{\pi_{k+1}} V_k \le T_{\pi_{k+1}}V_{k+1} = V_{k+1}$. Now, consider

$$|V_{k+1}-V^*| = V^* - V_{k+1} \le [-V_{k+1} \le -T_{\pi_{k+1}}V_k] \le V^* - T_{\pi_{k+1}}V_k = T_{\pi_{k+1}}V^* - T_{\pi_{k+1}}V_k = |T_{\pi_{k+1}}V^* - T_{\pi_{k+1}}V_k|$$

Then
$$||V_{k+1}-V^*||_{\infty} \le ||T_{\pi_{k+1}}V^* - T_{\pi_{k+1}}V_k||_{\infty} \le \gamma||V_k-V^*||_{\infty} \le \gamma^k||V_0-V^*||_{\infty}$$

Taking limit on both sides, we prove that $V_{k+1}$ converges to $V^*$ as $k$ approaches infinity. Using $\epsilon-\delta$ formalism, this means that
$$\forall \epsilon > 0 \ \exists k_0(\epsilon): \forall k \ge k_0(\epsilon) \ ||V_k-V^*||_{\infty} \le \epsilon$$

Because $\mathcal{S}$, $\mathcal{A}$ are finite, there exists a finite sequence $E = ||V_{\pi_i} - V^*||_{\infty},  i=\overline{1,|\mathcal{S}||\mathcal{A}|}$. Denote $\epsilon_0 = \min\{E_i | E_i \ne 0\}$ (this minimum always exists because sequence is finite) and take $\epsilon = \frac{\epsilon_0}{2}$. The only possible value for $||V_k-V^*||_{\infty}$ is then $0$ (optimal policy always exists and $||V_{\pi^*} - V^*||_{\infty} = 0$ because $V_{\pi^*} = V^*$). And due to definition of the limit, there has to exist some $k_0$ such that for all $k \ge k_0$ $||V_k-V^*||_{\infty} = 0$. Then $\pi_k = \pi^*$. **Q.E.D.**


```python

```

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