# A2C Continuous for Pendulum-v0 working implementation, negation for loss and entropy calculation

very good implementation of A2C continuous for Pendulum-v0

Code has snippet to stop execution when mean of last 10 or 20 is higher than -20 but the results look like:

episode: 706   score: [-13.13392661]
episode: 707   score: [-12.91221984]
episode: 708   score: [-50.38036647]
episode: 709   score: [-74.58410041]
episode: 710   score: [-138.1596521]
episode: 711   score: [-87.3867222]
episode: 712   score: [-63.28444052]
episode: 713   score: [-0.37368592]
episode: 714   score: [-13.28473712]
episode: 715   score: [-117.78089523]
episode: 716   score: [-25.65207563]
episode: 717   score: [-0.36829411]
episode: 718   score: [-50.81750735]
episode: 719   score: [-0.33565775]
episode: 720   score: [-0.47168285]
episode: 721   score: [-0.35240929]
episode: 722   score: [-0.40577252]
episode: 723   score: [-0.37114168]
episode: 724   score: [-25.73963544]
episode: 725   score: [-37.70957794]


Even with the reward/10 line, still pretty good. However, I don't understand these lines regarding negation of loss and why the entropy equation looks different from what I saw in Packt Publishing Deep Reinforcement Learning Hands-On per picture below: The code:

 def actor_optimizer(self):
#placeholders for actions and advantages parameters coming in
action = K.placeholder(shape=(None, 1))

# mu = K.placeholder(shape=(None, self.action_size))
# sigma_sq = K.placeholder(shape=(None, self.action_size))

mu, sigma_sq = self.actor.output

#defined a custom loss using PDF formula, K.exp is element-wise exponential
pdf = 1. / K.sqrt(2. * np.pi * sigma_sq) * K.exp(-K.square(action - mu) / (2. * sigma_sq))
#log pdf why?
log_pdf = K.log(pdf + K.epsilon())
#entropy looks different from log(sqrt(2 * pi * e * sigma_sq))
#Sum of the values in a tensor, alongside the specified axis.
entropy = K.sum(0.5 * (K.log(2. * np.pi * sigma_sq) + 1.))

exp_v = K.sum(exp_v + 0.01 * entropy)
#loss is a negation
actor_loss = -exp_v

#adjust params with custom train function
#return custom train function
return train


Again, the entropy equation coded was this: entropy = K.sum(0.5 * (K.log(2. * np.pi * sigma_sq) + 1.)) which looks different from what's given in the textbook photo above.

Also, why is the loss a negation? actor_loss = -exp_v?

Is it negated because it is gradient ascent rather than gradient descent of the objective function for a policy gradient?

They are the same after simple algebraic manipulations. The entropy of a single variable Gaussian distribution with pdf $$p(x|\mu, \sigma)$$ is \begin{align} \mathcal{H}(p) & = ln(\sqrt{2 \pi e \sigma^2}) \\ & = 0.5 \cdot ln(2 \pi e \sigma^2) \\ & = 0.5 \cdot (ln(2 \pi \sigma^2) + 1) \end{align} In policy gradient, we assume these $$\sigma$$s are isotropic. Thus the total entropy is the sum above $$\mathcal{H}$$s.
Yes. In policy gradient you would like to maximize the likelihood log_pdf of getting higher episode returns advantages. At the same time we prefer a policy of high entropy. Deep learning frameworks usually only provide optimizer for minimizing the loss. Therefore you negate exp_v as loss.