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I am trying to learn tabular Q learning by using a table of states and actions (i.e. no neural networks). I was trying it out on the FrozenLake environment. It's a very simple environment, where the task is to reach a G starting from a source S avoiding holes H and just following the frozen path which is F. The $4 \times 4$ FrozenLake grid looks like this

SFFF
FHFH
FFFH
HFFG

I am working with the slippery version, where the agent, if it takes a step, has an equal probability of either going in the direction it intends or slipping sideways perpendicular to the original direction (if that position is in the grid). Holes are terminal states and a goal is a terminal state.

Now I first tried value iteration which converges to the following set of values for the states

[0.0688909  0.06141457 0.07440976 0.05580732 0.09185454 0. 0.11220821 0.         0.14543635 0.24749695 0.29961759 0. 0.         0.3799359  0.63902015 0.        ]

I also coded policy iteration, and it also gives me the same result. So I am pretty confident that this value function is correct.

Now, I tried to code the Q learning algorithm, here is my code for the Q learning algorithm

def get_action(Q_table, state, epsilon):
    """
    Uses e-greedy to policy to return an action corresponding to state
    
    Args:
        Q_table: numpy array containing the q values
        state: current state
        epsilon: value of epsilon in epsilon greedy strategy
        env: OpenAI gym environment 
    """
    return env.action_space.sample() if np.random.random() < epsilon else np.argmax(Q_table[state]) 


def tabular_Q_learning(env):
    """
    Returns the optimal policy by using tabular Q learning
    
    Args:
        env: OpenAI gym environment
        
    Returns:
        (policy, Q function, V function) 
    """
    
    # initialize the Q table
    # 
    # Implementation detail: 
    # A numpy array of |x| * |a| values
    
    Q_table = np.zeros((env.nS, env.nA))
    
    # hyperparameters
    epsilon = 0.9
    episodes = 500000
    lr = 0.81

    
    for _ in tqdm_notebook(range(episodes)):
        # initialize the state
        state = env.reset()
        
        if episodes / 1000 > 21:
            epsilon = 0.1
        
        t = 0
        while True: # for each step of the episode
            # env.render()
            # print(observation)
         
            # choose a from s using policy derived from Q 
            action = get_action(Q_table, state, epsilon) 
            
            # take action a, observe r, s_dash
            s_dash, r, done, info = env.step(action)
            
            # Q table update 
            Q_table[state][action] += lr * (r + gamma * np.max(Q_table[s_dash]) - Q_table[state][action])
            
            state = s_dash
            
            t += 1
            
            if done:
                # print("Episode finished after {} timesteps".format(t+1))
                break
        # print(Q_table)
    
    policy = np.argmax(Q_table, axis=1)
    V = np.max(Q_table, axis=1)
    
    return policy, Q_table, V

I tried running it and it converges to a different set of values which is following [0.26426802 0.03656142 0.12557195 0.03075882 0.35018374 0. 0.02584052 0. 0.37657211 0.59209091 0.15439031 0. 0. 0.60367728 0.79768863 0. ]

I am not getting, what is going wrong. The implementation of Q learning is pretty straightforward. I checked my code, it seems right.

Any pointers would be helpful.

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1 Answer 1

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I was able to solve the problem.

The main issue for non-convergence was that I was not decaying the learning rate appropriately. I put a decay rate of $-0.00005$ on the learning rate lr, and subsequently Q-Learning also converged to the same value as value iteration.

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    $\begingroup$ You may also have been able to work with a lower fixed learning rate of e.g. 0.01 - that should be low enough to converge to an optimal policy, and result in close enough to true Q values that you could easily see the similarity with the dynamic programming approaches. $\endgroup$ Oct 15, 2019 at 8:49
  • $\begingroup$ Feel free to accept your own answer by clicking on the checkmark sign. $\endgroup$
    – nbro
    Nov 1, 2020 at 16:51

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