mdp linear algebra approach cannot stop

[zdarktknight]

This is an excellent example. However, when I tried the linear algebra approach in the mdp post, the while loop cannot stop.

[mpatacchiola]

Hi @zdarktknight

Which Numpy method are you using? It would be useful if you post your code here.

[zdarktknight]

Thank you very much for your reply. I am using python 3.6 with numpy version 1.15.1. Here is my code:

import numpy as np
def return_expected_action(u, T, v):
    actions_array = np.zeros(4)
    for action in range(4):
         #Expected utility of doing a in state s, according to T and u.
         actions_array[action] = np.sum(np.multiply(u, np.dot(v, T[:,:,action])))
    return np.argmax(actions_array)

def return_policy_evaluation_linalg(p, r, T, gamma):
    u = np.zeros(12)
    for s in range(12):
        if not np.isnan(p[s]):
            action = int(p[s])
            u[s] = np.linalg.solve(np.identity(12) - gamma*T[:,:,action], r)[s]
    return u

def linalg():
    gamma = 0.999
    epsilon = 0.0001
    iteration = 0
    T = np.load("T.npy")
    
    p = np.random.randint(0, 4, size=(12)).astype(np.float32)
    p[5] = np.NaN
    p[3] = p[7] = -1
    
    #Utility vectors
    u = np.array([0.0, 0.0, 0.0,  0.0,
                  0.0, 0.0, 0.0,  0.0,
                  0.0, 0.0, 0.0,  0.0])
    #Reward vector
    r = np.array([-0.04, -0.04, -0.04,  +1.0,
                  -0.04,   0.0, -0.04,  -1.0,
                  -0.04, -0.04, -0.04, -0.04])
    while True:
        iteration = iteration + 1
        
        # 1. Policy evaluation
        u_old = u.copy()
        u = return_policy_evaluation_linalg(p, r, T, gamma)     # old Utility, old actions

        # Stoping acriteria
        unchanged = True

        for s in range(12):
            if not np.isnan(p[s]) and not p[s] == -1:
                v = np.zeros((1,12))
                v[0,s] = 1.0

                # Policy improvement
                a = return_expected_action(u, T, v)
                if a != p[s]:                                  # update action
                    p[s] = a
                    unchanged = False
        
        if unchanged:
            break
             
        # Numerical problem ??? cannot stop with this criteria
        if iteration == 1000: 
            print('stop by iteration limit')
            break

    print("=================== FINAL RESULT ==================")
    print("Iterations: " + str(iteration))
    print("Gamma: " + str(gamma))
    print("Epsilon: " + str(epsilon))
    print("===================================================")
    print(u[0:4])
    print(u[4:8])
    print(u[8:12])
    print("===================================================")
    print_policy(p, shape=(3,4))
    print("===================================================")


def print_policy(p, shape):
 counter = 0
    policy_string = ""
    for row in range(shape[0]):
        for col in range(shape[1]):
            if(p[counter] == -1): policy_string += " *  "            
            elif(p[counter] == 0): policy_string += " ^  "
            elif(p[counter] == 1): policy_string += " <  "
            elif(p[counter] == 2): policy_string += " v  "           
            elif(p[counter] == 3): policy_string += " >  "
            elif(np.isnan(p[counter])): policy_string += " #  "
            counter += 1
        policy_string += '\n'
    print(policy_string)

if __name__ == "__main__":
    linalg()

[MurtAlsa]

Hello @mpatacchiola This repo is excellent work! I learned a lot from this repo. I was facing the same issue when I ran the algebraic approach. I was hoping if there's any valid justification? I believe there's an issue with the stopping criteria as is.

[MurtAlsa]

To recall: I believe the code is implemented well! However, for the algebraic approach, when I looked at the pseudocode mentioned in the book ( ch.17 - Figure 17.7), it's not the same as you have it in your code.

At line 186 in your code it shows if a != p[s]: but in the book it's if a > p[s]:

Best,