This implements paper Prioritized experience replay, using a binary segment tree.

```
13import random
14
15import numpy as np
```

Prioritized experience replay samples important transitions more frequently. The transitions are prioritized by the Temporal Difference error (td error), $\delta$.

We sample transition $i$ with probability, where $\alpha$ is a hyper-parameter that determines how much prioritization is used, with $\alpha = 0$ corresponding to uniform case. $p_i$ is the priority.

We use proportional prioritization $p_i = |\delta_i| + \epsilon$ where $\delta_i$ is the temporal difference for transition $i$.

We correct the bias introduced by prioritized replay using importance-sampling (IS) weights in the loss function. This fully compensates when $\beta = 1$. We normalize weights by $\frac{1}{\max_i w_i}$ for stability. Unbiased nature is most important towards the convergence at end of training. Therefore we increase $\beta$ towards end of training.

We use a binary segment tree to efficiently calculate $\sum_k^i p_k^\alpha$, the cumulative probability, which is needed to sample. We also use a binary segment tree to find $\min p_i^\alpha$, which is needed for $\frac{1}{\max_i w_i}$. We can also use a min-heap for this. Binary Segment Tree lets us calculate these in $\mathcal{O}(\log n)$ time, which is way more efficient that the naive $\mathcal{O}(n)$ approach.

This is how a binary segment tree works for sum; it is similar for minimum. Let $x_i$ be the list of $N$ values we want to represent. Let $b_{i,j}$ be the $j^{\mathop{th}}$ node of the $i^{\mathop{th}}$ row in the binary tree. That is two children of node $b_{i,j}$ are $b_{i+1,2j}$ and $b_{i+1,2j + 1}$.

The leaf nodes on row $D = \left\lceil {1 + \log_2 N} \right\rceil$ will have values of $x$. Every node keeps the sum of the two child nodes. That is, the root node keeps the sum of the entire array of values. The left and right children of the root node keep the sum of the first half of the array and the sum of the second half of the array, respectively. And so on…

Number of nodes in row $i$, This is equal to the sum of nodes in all rows above $i$. So we can use a single array $a$ to store the tree, where,

Then child nodes of $a_i$ are $a_{2i}$ and $a_{2i + 1}$. That is,

This way of maintaining binary trees is very easy to program.
*Note that we are indexing starting from 1*.

We use the same structure to compute the minimum.

`18class ReplayBuffer:`

`88 def __init__(self, capacity, alpha):`

We use a power of $2$ for capacity because it simplifies the code and debugging

`93 self.capacity = capacity`

$\alpha$

`95 self.alpha = alpha`

Maintain segment binary trees to take sum and find minimum over a range

```
98 self.priority_sum = [0 for _ in range(2 * self.capacity)]
99 self.priority_min = [float('inf') for _ in range(2 * self.capacity)]
```

Current max priority, $p$, to be assigned to new transitions

`102 self.max_priority = 1.`

Arrays for buffer

```
105 self.data = {
106 'obs': np.zeros(shape=(capacity, 4, 84, 84), dtype=np.uint8),
107 'action': np.zeros(shape=capacity, dtype=np.int32),
108 'reward': np.zeros(shape=capacity, dtype=np.float32),
109 'next_obs': np.zeros(shape=(capacity, 4, 84, 84), dtype=np.uint8),
110 'done': np.zeros(shape=capacity, dtype=np.bool)
111 }
```

We use cyclic buffers to store data, and `next_idx`

keeps the index of the next empty
slot

`114 self.next_idx = 0`

Size of the buffer

`117 self.size = 0`

`119 def add(self, obs, action, reward, next_obs, done):`

Get next available slot

`125 idx = self.next_idx`

store in the queue

```
128 self.data['obs'][idx] = obs
129 self.data['action'][idx] = action
130 self.data['reward'][idx] = reward
131 self.data['next_obs'][idx] = next_obs
132 self.data['done'][idx] = done
```

Increment next available slot

`135 self.next_idx = (idx + 1) % self.capacity`

Calculate the size

`137 self.size = min(self.capacity, self.size + 1)`

$p_i^\alpha$, new samples get `max_priority`

`140 priority_alpha = self.max_priority ** self.alpha`

Update the two segment trees for sum and minimum

```
142 self._set_priority_min(idx, priority_alpha)
143 self._set_priority_sum(idx, priority_alpha)
```

`145 def _set_priority_min(self, idx, priority_alpha):`

Leaf of the binary tree

```
151 idx += self.capacity
152 self.priority_min[idx] = priority_alpha
```

Update tree, by traversing along ancestors. Continue until the root of the tree.

`156 while idx >= 2:`

Get the index of the parent node

`158 idx //= 2`

Value of the parent node is the minimum of it’s two children

`160 self.priority_min[idx] = min(self.priority_min[2 * idx], self.priority_min[2 * idx + 1])`

`162 def _set_priority_sum(self, idx, priority):`

Leaf of the binary tree

`168 idx += self.capacity`

Set the priority at the leaf

`170 self.priority_sum[idx] = priority`

Update tree, by traversing along ancestors. Continue until the root of the tree.

`174 while idx >= 2:`

Get the index of the parent node

`176 idx //= 2`

Value of the parent node is the sum of it’s two children

`178 self.priority_sum[idx] = self.priority_sum[2 * idx] + self.priority_sum[2 * idx + 1]`

`180 def _sum(self):`

The root node keeps the sum of all values

`186 return self.priority_sum[1]`

`188 def _min(self):`

The root node keeps the minimum of all values

`194 return self.priority_min[1]`

`196 def find_prefix_sum_idx(self, prefix_sum):`

Start from the root

```
202 idx = 1
203 while idx < self.capacity:
```

If the sum of the left branch is higher than required sum

`205 if self.priority_sum[idx * 2] > prefix_sum:`

Go to left branch of the tree

```
207 idx = 2 * idx
208 else:
```

Otherwise go to right branch and reduce the sum of left branch from required sum

```
211 prefix_sum -= self.priority_sum[idx * 2]
212 idx = 2 * idx + 1
```

We are at the leaf node. Subtract the capacity by the index in the tree to get the index of actual value

`216 return idx - self.capacity`

`218 def sample(self, batch_size, beta):`

Initialize samples

```
224 samples = {
225 'weights': np.zeros(shape=batch_size, dtype=np.float32),
226 'indexes': np.zeros(shape=batch_size, dtype=np.int32)
227 }
```

Get sample indexes

```
230 for i in range(batch_size):
231 p = random.random() * self._sum()
232 idx = self.find_prefix_sum_idx(p)
233 samples['indexes'][i] = idx
```

$\min_i P(i) = \frac{\min_i p_i^\alpha}{\sum_k p_k^\alpha}$

`236 prob_min = self._min() / self._sum()`

$\max_i w_i = \bigg(\frac{1}{N} \frac{1}{\min_i P(i)}\bigg)^\beta$

```
238 max_weight = (prob_min * self.size) ** (-beta)
239
240 for i in range(batch_size):
241 idx = samples['indexes'][i]
```

$P(i) = \frac{p_i^\alpha}{\sum_k p_k^\alpha}$

`243 prob = self.priority_sum[idx + self.capacity] / self._sum()`

$w_i = \bigg(\frac{1}{N} \frac{1}{P(i)}\bigg)^\beta$

`245 weight = (prob * self.size) ** (-beta)`

Normalize by $\frac{1}{\max_i w_i}$, which also cancels off the $\frac{1}{N}$ term

`248 samples['weights'][i] = weight / max_weight`

Get samples data

```
251 for k, v in self.data.items():
252 samples[k] = v[samples['indexes']]
253
254 return samples
```

`256 def update_priorities(self, indexes, priorities):`

`261 for idx, priority in zip(indexes, priorities):`

Set current max priority

`263 self.max_priority = max(self.max_priority, priority)`

Calculate $p_i^\alpha$

`266 priority_alpha = priority ** self.alpha`

Update the trees

```
268 self._set_priority_min(idx, priority_alpha)
269 self._set_priority_sum(idx, priority_alpha)
```

`271 def is_full(self):`

`275 return self.capacity == self.size`