My question is about using the SARSA algorithm in reinforcement learning for an undiscounted, continuing (non-episodic) problem (can it be used for such a problem?)
I have been studying the textbook by Sutton and Barto, and they show how to modify the Q-learning algorithm so that it can be used for undiscounted problems. They refer to the new algorithm (for undiscounted problems) as R-learning, in Chapter 6.7. Basically, in R-learning, the update rule for Q(s,a) on each iteration is:
Q(s,a) = Q(s,a) + alpha * [r - rho + max_a{Q(s',a)} - Q(s,a)]
Here, rho is updated on each iteration only if a greedy action is chosen at state s. The update rule for rho is:
rho = rho + beta * [r - rho + max_a{Q(s',a)} - max_a{Q(s,a)}]
(Here, alpha and beta are learning parameters.) Now, my question is to do with SARSA, rather than Q-learning. I want to modify the SARSA algorithm so that it is suitable for average reward (undiscounted) problems, in the same way that the Q-learning was modified to be used for average reward problems (I don't know if this is possible?). However, in the literature I cannot find an explanation of exactly how SARSA should be modified for an average reward problem.
Here is my guess for how SARSA should be used in an undiscounted problem. I would guess that the update rule should be:
Q(s,a) = Q(s,a) + alpha * [r - rho + Q(s',a') - Q(s,a)],
where a' is the action actually chosen at state s. This seems fairly obvious. But how should rho be updated? My guess is that since SARSA is an on-policy algorithm, rho should always be updated on each iteration - regardless of whether or not a greedy action is chosen at s - and the update rule should simply be:
rho = rho + beta * [r - rho + Q(s',a') - Q(s,a)].
Could somebody tell me if this is correct? Or should rho still be updated based on the optimal actions at the states s and s'?
First of all, the problem is that an undiscounted non-episodic task is an ill conditioned problem, because the expected reward is divergent (unless rewards have some property which makes them diminishing in the future).
EDIT: I'm sorry, I've look-up the referenced chapter in the book, and noticed that indeed the R-learning IS a method to tackle undiscounted non-episodic tasks.
AD REM: I think that the idea behind updating rho in such manner is to estimate the average reward of the current policy. Therefore I'm guessing that even though SARSA is an on-policy method, you should update rho only if greedy action has been chosen. And that's because if you want to accurately estimate the average reward of the current policy, you should take into account only events that would occur when you would be following this policy. Events that occur as a result of exploration steps do not represent an accurate sample of "what's this policy is worth". That's of course only an intuitive argument - I don't have any experience with R-learning nor did I analysed this issue formally.
Related
I'm learning how add the max-min ant system into my current ant system. From what I've read the trial pheromone is initialized tMax, tMax is calculated by,
tMax = 1 / best tour length
But how exactly would it be possible to initialize the trail pheromone to tMax if it depends on a tour which doesn't yet exist?
tMin also depends on tMax which also makes it impossible to initialize without a best solution.
In MMAS, all edges are initialised to tauMax, but the definition of tauMax is slightly different from what you've stated above:
tauMax <-- 1 / (rho * bestTourSoFarLength),
where rho is the evaporation rate (typically set to 0.5), and where focus on the tour length is best so far (see below). tauMax is repeatedly updated during algorithm execution, each time the incumbent best tour (so far) is updated.
For initialisation, an initial feasible tour is constructed heuristically. Typically, the nearest neighbour tour for a random starting city is used.
Recall that in the context of stochastic optimisation methods such as Ant Colony Optimization (ACO)/MMAS, we can generally not prove optimality of the best incumbent solution (tour) at algorithm termination (we know from practice, however, that ACO/MMAS does perform well on some set of problems, most notably variations of the travelling salesman problem (TSP)). Hence, in contexts such as these, the term "best solution" non-rigorously denote varied meanings from varied authors; "best so far", "best at algorithm termination" and so on, so be aware of that when reading literature on the subject.
Finally, as a note, tauMin depends---as you've noted---on tauMax, but is typically not updated after initialisation. When imposing pheromone limits, the important "dynamic" part is the monotonically decreasing tauMax, whereas pheromones for most edges will land at the constant tauMin eventually, due to evaporation. A suitable value of tauMin is given by the hideous expression (based on empirical data)
tauMin = tauMax*(1-(0.05)^(1/n))/((n/2-1)*(0.05)^(1/n)).
I understood how belief states are updated in POMDP. But in Policy and Value function section, in http://en.wikipedia.org/wiki/Partially_observable_Markov_decision_process I could not figure out how to calculate value of V*(T(b,a,o)) for finding optimal value function V*(b). I have read a lot of resources on the internet but none explain how to calculate this clearly. Can some one provide me with a mathematically solved example with all the calculations or provide me with a mathematically clear explanation.
You should check out this tutorial on POMDPs:
http://cs.brown.edu/research/ai/pomdp/tutorial/index.html
It includes a section about Value Iteration, which can be used to find an optimal policy/value function.
I try to use the same notation in this answer as Wikipedia.
First I repeat the Value Function as stated on Wikipedia:
V*(b) is the value function with the belief b as parameter. b contains the probability of all states s, which sum up to 1:
r(b,a) is the reward for belief b and action a which has to be calculated using the belief over each state given the original reward function R(s,a): the reward for being in state s and having done action a.
We can also write the function O in terms of states instead of belief b:
this is the probability of having observation o given a belief b and action a. Note that O and T are probability functions.
Finally the function τ(b,a,o) gives the new belief state b'=τ(b,a,o) given the previous belief b, action a and observation o. Per state we can calculate the new probability:
Now the new belief b' can be used to calculate iteratively: V(τ(b,a,o)).
The optimal value function can be approached by using for example Value Iteration which applies dynamic programming. Then the function is iteratively updated until the difference is smaller then a small value ε.
There is a lot more information on POMDPs, for example:
Sebastian Thrun, Wolfram Burgard, and Dieter Fox. 2005. Probabilistic Robotics (Intelligent Robotics and Autonomous Agents). The MIT Press.
A brief introduction to reinforcement learning
A POMDP Tutorial
Reinforcement Learning and Markov Decision Processes
I'm playing around with Neural Networks trying to understand the best practices for designing their architecture based on the kind of problem you need to solve.
I generated a very simple data set composed of a single convex region as you can see below:
Everything works fine when I use an architecture with L = 1, or L = 2 hidden layers (plus the output layer), but as soon as I add a third hidden layer (L = 3) my performance drops down to slightly better than chance.
I know that the more complexity you add to a network (number of weights and parameters to learn) the more you tend to go towards over-fitting your data, but I believe this is not the nature of my problem for two reasons:
my performance on the Training set is also around 60% (whereas over-fitting typically means you have a very low training error and high test error),
and I have a very large amount of data examples (don't look at the figure that's only a toy figure I uplaoded).
Can anybody help me understand why adding an extra hidden layer gives
me this drop in performances on such a simple task?
Here is an image of my performance as a function of the number of layers used:
ADDED PART DUE TO COMMENTS:
I am using a sigmoid functions assuming values between 0 and 1, L(s) = 1 / 1 + exp(-s)
I am using early stopping (after 40000 iterations of backprop) as a criteria to stop the learning. I know it is not the best way to stop but I thought that it would ok for such a simple classification task, if you believe this is the main reason I'm not converging I I might implement some better criteria.
At least on the surface of it, this appears to be a case of the so-called "vanishing gradient" problem.
Activation functions
Your neurons activate according to the logistic sigmoid function, f(x) = 1 / (1 + e^-x) :
This activation function is used frequently because it has several nice properties. One of these nice properties is that the derivative of f(x) is expressible computationally using the value of the function itself, as f'(x) = f(x)(1 - f(x)). This function has a nonzero value for x near zero, but quickly goes to zero as |x| gets large :
Gradient descent
In a feedforward neural network with logistic activations, the error is typically propagated backwards through the network using the first derivative as a learning signal. The usual update for a weight in your network is proportional to the error attributable to that weight times the current weight value times the derivative of the logistic function.
delta_w(w) ~= w * f'(err(w)) * err(w)
As the product of three potentially very small values, the first derivative in such networks can become small very rapidly if the weights in the network fall outside the "middle" regime of the logistic function's derivative. In addition, this rapidly vanishing derivative becomes exacerbated by adding more layers, because the error in a layer gets "split up" and partitioned out to each unit in the layer. This, in turn, further reduces the gradient in layers below that.
In networks with more than, say, two hidden layers, this can become a serious problem for training the network, since the first-order gradient information will lead you to believe that the weights cannot usefully change.
However, there are some solutions that can help ! The ones I can think of involve changing your learning method to use something more sophisticated than first-order gradient descent, generally incorporating some second-order derivative information.
Momentum
The simplest solution to approximate using some second-order information is to include a momentum term in your network parameter updates. Instead of updating parameters using :
w_new = w_old - learning_rate * delta_w(w_old)
incorporate a momentum term :
w_dir_new = mu * w_dir_old - learning_rate * delta_w(w_old)
w_new = w_old + w_dir_new
Intuitively, you want to use information from past derivatives to help determine whether you want to follow the new derivative entirely (which you can do by setting mu = 0), or to keep going in the direction you were heading on the previous update, tempered by the new gradient information (by setting mu > 0).
You can actually get even better than this by using "Nesterov's Accelerated Gradient" :
w_dir_new = mu * w_dir_old - learning_rate * delta_w(w_old + mu * w_dir_old)
w_new = w_old + w_dir_new
I think the idea here is that instead of computing the derivative at the "old" parameter value w, compute it at what would be the "new" setting for w if you went ahead and moved there according to a standard momentum term. Read more in a neural-networks context here (PDF).
Hessian-Free
The textbook way to incorporate second-order gradient information into your neural network training algorithm is to use Newton's Method to compute the first and second order derivatives of your objective function with respect to the parameters. However, the second order derivative, called the Hessian matrix, is often extremely large and prohibitively expensive to compute.
Instead of computing the entire Hessian, some clever research in the past few years has indicated a way to compute just the values of the Hessian in a particular search direction. You can then use this process to identify a better parameter update than just the first-order gradient.
You can learn more about this by reading through a research paper (PDF) or looking at a sample implementation.
Others
There are many other optimization methods that could be useful for this task -- conjugate gradient (PDF -- definitely worth a read), Levenberg-Marquardt (PDF), L-BFGS -- but from what I've seen in the research literature, momentum and Hessian-free methods seem to be the most common ones.
Because the number of iterations of training required for convergence increases as you add complexity to a neural network, holding the length of training constant while adding layers to a neural network will certainly result in you eventually observing a drop like this. To figure out whether that is the explanation for this particular observation, try increasing the number of iterations of training that you're using and see if it improves. Using a more intelligent stopping criterion is also a good option, but a simple increase in the cut-off will give you answers faster.
As far as my understanding is, it's possible to replace a look-up-table for Q-values (state-action-pair-evaluation) by a neural network for estimating these state-action pairs. I programmed a small library, which is able to propagate and backpropagate through a self-built neural network for learning wanted target-values for a certain in-out-put.
So I also found this site while googling, and googling through the whole web (as it felt for me): http://www.cs.indiana.edu/~gasser/Salsa/nn.html where the Q-learning combined with a neural network is shortly explained.
For each action, there's an extra output neuron, and the activation-value of one of these output-"units" tells me, the estimated Q-value. (One question: Is the activation value the same as the "output" of the neuron or something different?)
I used the standard sigmoid-function as activation-function, so the range of the function-values x is
0<x<1
So I thought, my target value should always be from 0.0 to 1.0 -> Question: Is that point of my understanding correct? Or did I missunderstand something about that?
If yes, there comes following problem:
The equation for calculating the target-reward / new Q-value is:
q(s,a) = q(s,a) + learningrate * (reward + discountfactor * q'(s,a) - q(s,a))
So how do I perform this equation to get the right target for the neural network, if targets should be from 0.0 to 1.0?!
How do I calculate good reward-values? Is moving toward the aim more worth it, than going away from it? (more +reward when nearing the aim than -reward for bigger distance to aim?)
I think there are some missunderstandings of mine. I hope, you can help me to answer that questions. Thank you very much!
Using a neural-network to store q-value is a good extension of table lookup. This makes it possible to use q-learning when the state space is continuous.
input layer ......
|/ \ | \|
output layer a1 a2 a3
0.1 0.2 0.9
Suppose you have 3 actions available. Above shows the outputs from the neural network using current state and learned weights. So you know a3 is the best action to go with.
Now the questions you have:
One question: Is the activation value the same as the "output" of the neuron or something different?
Yes, I think so. In the referred link, the author said:
Some of the units may also be designated output units; their activations represent the network's response.
So I thought, my target value should always be from 0.0 to 1.0 -> Question: Is that point of my understanding correct? Or did I missunderstand something about that?
If you choose sigmoid as your activation function, for sure you output will be from 0.0 to 1.0. There are different choices of activation functions, e.g., here. Sigmoid is one of the most popular choices though. I think the output value being from 0.0 to 1.0 is not a problem here. if at current time, you have only two available actions, Q(s,a1) = 0.1, Q(s,a2) = 0.9, you know that action a2 is much better than a1 with respective to q-value.
So how do I perform this equation to get the right target for the neural network, if targets should be from 0.0 to 1.0?! How do I calculate good reward-values?
I am not sure for this, but you can try to clamp the new target q-value to be between 0.0 and 1.0, i.e.,
q(s,a) = min(max(0.0, q(s,a) + learningrate * (reward + discountfactor * q'(s,a) - q(s,a))), 1.0)
Try to do some experiments for finding a proper reward value.
Is moving toward the aim more worth it, than going away from it? (more +reward when nearing the aim than -reward for bigger distance to aim?)
Normally you should give more reward when it's close to the aim if you use the classical update equation, so that the new q-value gets increased.
Although I know that SARSA is on-policy while Q-learning is off-policy, when looking at their formulas it's hard (to me) to see any difference between these two algorithms.
According to the book Reinforcement Learning: An Introduction (by Sutton and Barto). In the SARSA algorithm, given a policy, the corresponding action-value function Q (in the state s and action a, at timestep t), i.e. Q(st, at), can be updated as follows
Q(st, at) = Q(st, at) + α*(rt + γ*Q(st+1, at+1) - Q(st, at))
On the other hand, the update step for the Q-learning algorithm is the following
Q(st, at) = Q(st, at) + α*(rt + γ*maxa Q(st+1, a) - Q(st, at))
which can also be written as
Q(st, at) = (1 - α) * Q(st, at) + α * (rt + γ*maxa Q(st+1, a))
where γ (gamma) is the discount factor and rt is the reward received from the environment at timestep t.
Is the difference between these two algorithms the fact that SARSA only looks up the next policy value while Q-learning looks up the next maximum policy value?
TLDR (and my own answer)
Thanks to all those answering this question since I first asked it. I've made a github repo playing with Q-Learning and empirically understood what the difference is. It all amounts to how you select your next best action, which from an algorithmic standpoint can be a mean, max or best action depending on how you chose to implement it.
The other main difference is when this selection is happening (e.g., online vs offline) and how/why that affects learning. If you are reading this in 2019 and are more of a hands-on person, playing with a RL toy problem is probably the best way to understand the differences.
One last important note is that both Suton & Barto as well as Wikipedia often have mixed, confusing or wrong formulaic representations with regards to the next state best/max action and reward:
r(t+1)
is in fact
r(t)
When I was learning this part, I found it very confusing too, so I put together the two pseudo-codes from R.Sutton and A.G.Barto hoping to make the difference clearer.
Blue boxes highlight the part where the two algorithms actually differ. Numbers highlight the more detailed difference to be explained later.
TL;NR:
| | SARSA | Q-learning |
|:-----------:|:-----:|:----------:|
| Choosing A' | π | π |
| Updating Q | π | μ |
where π is a ε-greedy policy (e.g. ε > 0 with exploration), and μ is a greedy policy (e.g. ε == 0, NO exploration).
Given that Q-learning is using different policies for choosing next action A' and updating Q. In other words, it is trying to evaluate π while following another policy μ, so it's an off-policy algorithm.
In contrast, SARSA uses π all the time, hence it is an on-policy algorithm.
More detailed explanation:
The most important difference between the two is how Q is updated after each action. SARSA uses the Q' following a ε-greedy policy exactly, as A' is drawn from it. In contrast, Q-learning uses the maximum Q' over all possible actions for the next step. This makes it look like following a greedy policy with ε=0, i.e. NO exploration in this part.
However, when actually taking an action, Q-learning still uses the action taken from a ε-greedy policy. This is why "Choose A ..." is inside the repeat loop.
Following the loop logic in Q-learning, A' is still from the ε-greedy policy.
Yes, this is the only difference. On-policy SARSA learns action values relative to the policy it follows, while off-policy Q-Learning does it relative to the greedy policy. Under some common conditions, they both converge to the real value function, but at different rates. Q-Learning tends to converge a little slower, but has the capabilitiy to continue learning while changing policies. Also, Q-Learning is not guaranteed to converge when combined with linear approximation.
In practical terms, under the ε-greedy policy, Q-Learning computes the difference between Q(s,a) and the maximum action value, while SARSA computes the difference between Q(s,a) and the weighted sum of the average action value and the maximum:
Q-Learning: Q(st+1,at+1) = maxaQ(st+1,a)
SARSA: Q(st+1,at+1) = ε·meanaQ(st+1,a) + (1-ε)·maxaQ(st+1,a)
What is the difference mathematically?
As is already described in most other answers, the difference between the two updates mathematically is indeed that, when updating the Q-value for a state-action pair (St, At):
Sarsa uses the behaviour policy (meaning, the policy used by the agent to generate experience in the environment, which is typically epsilon-greedy) to select an additional action At+1, and then uses Q(St+1, At+1) (discounted by gamma) as expected future returns in the computation of the update target.
Q-learning does not use the behaviour policy to select an additional action At+1. Instead, it estimates the expected future returns in the update rule as maxA Q(St+1, A). The max operator used here can be viewed as "following" the completely greedy policy. The agent is not actually following the greedy policy though; it only says, in the update rule, "suppose that I would start following the greedy policy from now on, what would my expected future returns be then?".
What does this mean intuitively?
As mentioned in other answers, the difference described above means, using technical terminology, that Sarsa is an on-policy learning algorithm, and Q-learning is an off-policy learning algorithm.
In the limit (given an infinite amount of time to generate experience and learn), and under some additional assumptions, this means that Sarsa and Q-learning converge to different solutions / "optimal" policies:
Sarsa will converge to a solution that is optimal under the assumption that we keep following the same policy that was used to generate the experience. This will often be a policy with some element of (rather "stupid") randomness, like epsilon-greedy, because otherwise we are unable to guarantee that we'll converge to anything at all.
Q-Learning will converge to a solution that is optimal under the assumption that, after generating experience and training, we switch over to the greedy policy.
When to use which algorithm?
An algorithm like Sarsa is typically preferable in situations where we care about the agent's performance during the process of learning / generating experience. Consider, for example, that the agent is an expensive robot that will break if it falls down a cliff. We'd rather not have it fall down too often during the learning process, because it is expensive. Therefore, we care about its performance during the learning process. However, we also know that we need it to act randomly sometimes (e.g. epsilon-greedy). This means that it is highly dangerous for the robot to be walking alongside the cliff, because it may decide to act randomly (with probability epsilon) and fall down. So, we'd prefer it to quickly learn that it's dangerous to be close to the cliff; even if a greedy policy would be able to walk right alongside it without falling, we know that we're following an epsilon-greedy policy with randomness, and we care about optimizing our performance given that we know that we'll be stupid sometimes. This is a situation where Sarsa would be preferable.
An algorithm like Q-learning would be preferable in situations where we do not care about the agent's performance during the training process, but we just want it to learn an optimal greedy policy that we'll switch to eventually. Consider, for example, that we play a few practice games (where we don't mind losing due to randomness sometimes), and afterwards play an important tournament (where we'll stop learning and switch over from epsilon-greedy to the greedy policy). This is where Q-learning would be better.
There's an index mistake in your formula for Q-Learning.
Page 148 of Sutton and Barto's.
Q(st,at) <-- Q(st,at) + alpha * [r(t+1) + gamma * max Q(st+1,a) -
Q(st,at) ]
The typo is in the argument of the max:
the indexes are st+1 and a,
while in your question they are st+1 and at+1 (these are correct for SARSA).
Hope this helps a bit.
In Q-Learning
This is your:
Q-Learning: Q(St,At) = Q(St,At) + a [ R(t+1) + discount * max Q(St+1,At) - Q(St,At) ]
should be changed to
Q-Learning: Q(St,At) = Q(St,At) + a [ R(t+1) + discount * max Q(St+1,a) - Q(St,At) ]
As you said, you have to find the maximum Q-value for the update eq. by changing the a, Then you will have a new Q(St,At). CAREFULLY, the a that give you the maximum Q-value is not the next action. At this stage, you only know the next state (St+1), and before going to next round, you want to update the St by the St+1 (St <-- St+1).
For each loop;
choose At from the St using the Q-value
take At and observe Rt+1 and St+1
Update Q-value using the eq.
St <-- St+1
Until St is terminal
The only difference between SARSA and Qlearning is that SARSA takes the next action based on the current policy while qlearning takes the action with maximum utility of next state
I didn't read any book just I see the implication of them
q learning just focus on the (action grid)
SARSA learning just focus on the (state to state) and observe the action list of s and s' and then update the (state to state grid)
Both SARSA and Q-learnig agents follow e-greedy policy to interact with environment.
SARSA agent updates its Q-function using the next timestep Q-value with whatever action the policy provides(mostly still greedy, but random action also accepted). The policy being executed and the policy being updated towards are the same.
Q-learning agent updates its Q-function with only the action brings the maximum next state Q-value(total greedy with respect to the policy). The policy being executed and the policy being updated towards are different.
Hence, SARSA is on-policy, Q-learning is off-policy.