强化学习读书笔记 - 06~07 - 时序差分学习(Temporal-Difference Learning)

强化学习读书笔记 - 06~07 - 时序差分学习(Temporal-Difference Learning)

学习笔记：
Reinforcement Learning: An Introduction, Richard S. Sutton and Andrew G. Barto c 2014, 2015, 2016

数学符号看不懂的，先看看这里：

时序差分学习简话

时序差分学习结合了动态规划和蒙特卡洛方法，是强化学习的核心思想。

时序差分这个词不好理解。改为当时差分学习比较形象一些 - 表示通过当前的差分数据来学习。

蒙特卡洛的方法是模拟（或者经历）一段情节，在情节结束后，根据情节上各个状态的价值，来估计状态价值。
时序差分学习是模拟（或者经历）一段情节，每行动一步（或者几步），根据新状态的价值，然后估计执行前的状态价值。
可以认为蒙特卡洛的方法是最大步数的时序差分学习。
本章只考虑单步的时序差分学习。多步的时序差分学习在下一章讲解。

数学表示
根据我们已经知道的知识：如果可以计算出策略价值（\(\pi\)状态价值\(v_{\pi}(s)\)，或者行动价值\(q_{\pi(s, a)}\)），就可以优化策略。
在蒙特卡洛方法中，计算策略的价值，需要完成一个情节(episode)，通过情节的目标价值\(G_t\)来计算状态的价值。其公式：
Formula MonteCarlo

\[V(S_t) \gets V(S_t) + \alpha \delta_t \\ \delta_t = [G_t - V(S_t)] \\ where \\ \delta_t \text{ - Monte Carlo error} \\ \alpha \text{ - learning step size} \]

时序差分的思想是通过下一个状态的价值计算状态的价值，形成一个迭代公式（又）：
Formula TD(0)

\[V(S_t) \gets V(S_t) + \alpha \delta_t \\ \delta_t = [R_{t+1} + \gamma\ V(S_{t+1} - V(S_t)] \\ where \\ \delta_t \text{ - TD error} \\ \alpha \text{ - learning step size} \\ \gamma \text{ - reward discount rate} \]

注：书上提出TD error并不精确，而Monte Carlo error是精确地。需要了解，在此并不拗述。

时序差分学习方法

本章介绍的是时序差分学习的单步学习方法。多步学习方法在下一章介绍。

  • 策略状态价值\(v_{\pi}\)的时序差分学习方法(单步\多步)
  • 策略行动价值\(q_{\pi}\)的on-policy时序差分学习方法: Sarsa(单步\多步)
  • 策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法: Q-learning(单步)
  • Double Q-learning(单步)
  • 策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法(带importance sampling): Sarsa(多步)
  • 策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法(不带importance sampling): Tree Backup Algorithm(多步)
  • 策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法: \(Q(\sigma)\)(多步)

策略状态价值\(v_{\pi}\)的时序差分学习方法

单步时序差分学习方法TD(0)

  • 流程图
Reinforcement Learning - TD0 Reinforcement Learning - TD0 s S s_1 S' s->s_1  A   R v V(S) s->v v_1 V(S') s_1->v_1 v->v_1 R
  • 算法描述

Initialize \(V(s)\) arbitrarily \(\forall s \in \mathcal{S}^+\)
Repeat (for each episode):
  Initialize \(\mathcal{S}\)
  Repeat (for each step of episode):
   \(A \gets\) action given by \(\pi\) for \(S\)
   Take action \(A\), observe \(R, S'\)
   \(V(S) \gets V(S) + \alpha [R + \gamma V(S') - V(S)]\)
   \(S \gets S'\)
  Until S is terminal

多步时序差分学习方法

  • 流程图
Reinforcement Learning - TD n Reinforcement Learning - TD n s S s_1 ... s->s_1  A0   Rk v V(S) s->v s_2 Sn s_1->s_2  An-1   Rn v_1 V(...) s_1->v_1 v_2 V(Sn) s_2->v_2 v->v_1 Rk v_1->v_2 Rn
  • 算法描述

Input: the policy \(\pi\) to be evaluated
Initialize \(V(s)\) arbitrarily \(\forall s \in \mathcal{S}\)
Parameters: step size \(\alpha \in (0, 1]\), a positive integer \(n\)
All store and access operations (for \(S_t\) and \(R_t\)) can take their index mod \(n\)

Repeat (for each episode):
  Initialize and store \(S_0 \ne terminal\)
  \(T \gets \infty\)
  For \(t = 0,1,2,\cdots\):
   If \(t < T\), then:
    Take an action according to \(\pi(\dot \ | S_t)\)
    Observe and store the next reward as \(R_{t+1}\) and the next state as \(S_{t+1}\)
    If \(S_{t+1}\) is terminal, then \(T \gets t+1\)
   $\tau \gets t - n + 1 \ $ (\(\tau\) is the time whose state's estimate is being updated)
   If \(\tau \ge 0\):
    \(G \gets \sum_{i = \tau + 1}^{min(\tau + n, T)} \gamma^{i-\tau-1}R_i\)
    if \(\tau + n \le T\) then: \(G \gets G + \gamma^{n}V(S_{\tau + n}) \qquad \qquad (G_{\tau}^{(n)})\)
    \(V(S_{\tau}) \gets V(S_{\tau}) + \alpha [G - V(S_{\tau})]\)
  Until \(\tau = T - 1\)

这里要理解\(V(S_0)\)是由\(V(S_0), V(S_1), \dots, V(S_n)\)计算所得；\(V(S_1)\)是由\(V(S_1), V(S_1), \dots, V(S_{n+1})\)。

策略行动价值\(q_{\pi}\)的on-policy时序差分学习方法: Sarsa

单步时序差分学习方法

  • 流程图
Reinforcement Learning - TD Sarsa Reinforcement Learning - TD Sarsa s S s_1 S' s->s_1  A   R q Q(S, A) s->q q_1 Q(S', A') s_1->q_1 q->q_1 R
  • 算法描述

Initialize \(Q(s, a), \forall s \in \mathcal{S}, a \in \mathcal{A}(s)\) arbitrarily, and \(Q(terminal, \dot \ ) = 0\)
Repeat (for each episode):
  Initialize \(\mathcal{S}\)
  Choose \(A\) from \(S\) using policy derived from \(Q\) (e.g. \(\epsilon-greedy\))
  Repeat (for each step of episode):
   Take action \(A\), observe \(R, S'\)
   Choose \(A'\) from \(S'\) using policy derived from \(Q\) (e.g. \(\epsilon-greedy\))
   \(Q(S, A) \gets Q(S, A) + \alpha [R + \gamma Q(S', A') - Q(S, A)]\)
   \(S \gets S'; A \gets A';\)
  Until S is terminal

多步时序差分学习方法

  • 流程图
Reinforcement Learning - TD Sarsa Reinforcement Learning - TD Sarsa s S s_1 ... s->s_1  A   Rk q Q(S, A) s->q s_2 Sn s_1->s_2 An-1 Rn q_1 Q(...) s_1->q_1 q_2 Q(Sn, An) s_2->q_2 q->q_1 Rk q_1->q_2 Rn
  • 算法描述

Initialize \(Q(s, a)\) arbitrarily \(\forall s \in \mathcal{S}^, \forall a in \mathcal{A}\)
Initialize \(\pi\) to be \(\epsilon\)-greedy with respect to Q, or to a fixed given policy
Parameters: step size \(\alpha \in (0, 1]\),
  small \(\epsilon > 0\)
  a positive integer \(n\)
All store and access operations (for \(S_t\) and \(R_t\)) can take their index mod \(n\)

Repeat (for each episode):
  Initialize and store \(S_0 \ne terminal\)
  Select and store an action \(A_0 \sim \pi(\dot \ | S_0)\)
  \(T \gets \infty\)
  For \(t = 0,1,2,\cdots\):
   If \(t < T\), then:
    Take an action \(A_t\)
    Observe and store the next reward as \(R_{t+1}\) and the next state as \(S_{t+1}\)
    If \(S_{t+1}\) is terminal, then:
     \(T \gets t+1\)
    Else:
     Select and store an action \(A_{t+1} \sim \pi(\dot \ | S_{t+1})\)
   $\tau \gets t - n + 1 \ $ (\(\tau\) is the time whose state's estimate is being updated)
   If \(\tau \ge 0\):
    \(G \gets \sum_{i = \tau + 1}^{min(\tau + n, T)} \gamma^{i-\tau-1}R_i\)
    if \(\tau + n \le T\) then: \(G \gets G + \gamma^{n} Q(S_{\tau + n}, A_{\tau + n}) \qquad \qquad (G_{\tau}^{(n)})\)
    \(Q(S_{\tau}, A_{\tau}) \gets Q(S_{\tau}, A_{\tau}) + \alpha [G - Q(S_{\tau}, A_{\tau})]\)
    If {\pi} is being learned, then ensure that \(\pi(\dot \ | S_{\tau})\) is \(\epsilon\)-greedy wrt Q
  Until \(\tau = T - 1\)

策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法: Q-learning

Q-learning 算法（Watkins, 1989）是一个突破性的算法。这里利用了这个公式进行off-policy学习。

\[Q(S_t, A_t) \gets Q(S_t, A_t) + \alpha [R_{t+1} + \gamma \underset{a}{max} \ Q(S_{t+1}, a) - Q(S_t, A_t)] \]

单步时序差分学习方法

  • 算法描述

Initialize \(Q(s, a), \forall s \in \mathcal{S}, a \in \mathcal{A}(s)\) arbitrarily, and \(Q(terminal, \dot \ ) = 0\)
Repeat (for each episode):
  Initialize \(\mathcal{S}\)
  Choose \(A\) from \(S\) using policy derived from \(Q\) (e.g. \(\epsilon-greedy\))
  Repeat (for each step of episode):
   Take action \(A\), observe \(R, S'\)
   \(Q(S, A) \gets Q(S, A) + \alpha [R + \gamma \underset{a}{max} \ Q(S‘, a) - Q(S, A)]\)
   \(S \gets S';\)
  Until S is terminal

  • Q-learning使用了max，会引起一个最大化偏差(Maximization Bias)问题。
    具体说明，请看书上的Example 6.7。**
    使用Double Q-learning可以消除这个问题。

Double Q-learning

单步时序差分学习方法

Initialize \(Q_1(s, a)\) and \(Q_2(s, a), \forall s \in \mathcal{S}, a \in \mathcal{A}(s)\) arbitrarily
Initialize \(Q_1(terminal, \dot \ ) = Q_2(terminal, \dot \ ) = 0\)
Repeat (for each episode):
  Initialize \(\mathcal{S}\)
  Repeat (for each step of episode):
   Choose \(A\) from \(S\) using policy derived from \(Q_1\) and \(Q_2\) (e.g. \(\epsilon-greedy\))
   Take action \(A\), observe \(R, S'\)
   With 0.5 probability:
    \(Q_1(S, A) \gets Q_1(S, A) + \alpha [R + \gamma Q_2(S', \underset{a}{argmax} \ Q_1(S', a)) - Q_1(S, A)]\)
   Else:
    \(Q_2(S, A) \gets Q_2(S, A) + \alpha [R + \gamma Q_1(S', \underset{a}{argmax} \ Q_2(S', a)) - Q_2(S, A)]\)
   \(S \gets S';\)
  Until S is terminal

策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法(by importance sampling): Sarsa

考虑到重要样本，把\(\rho\)带入到Sarsa算法中，形成一个off-policy的方法。
\(\rho\) - 重要样本比率(importance sampling ratio)

\[\rho \gets \prod_{i = \tau + 1}^{min(\tau + n - 1, T -1 )} \frac{\pi(A_t|S_t)}{\mu(A_t|S_t)} \qquad \qquad (\rho_{\tau+n}^{(\tau+1)}) \]

多步时序差分学习方法

  • 算法描述

Input: behavior policy \mu such that \(\mu(a|s) > 0，\forall s \in \mathcal{S}, a \in \mathcal{A}\)
Initialize \(Q(s，a)\) arbitrarily \(\forall s \in \mathcal{S}^, \forall a in \mathcal{A}\)
Initialize \(\pi\) to be \(\epsilon\)-greedy with respect to Q, or to a fixed given policy
Parameters: step size \(\alpha \in (0, 1]\),
  small \(\epsilon > 0\)
  a positive integer \(n\)
All store and access operations (for \(S_t\) and \(R_t\)) can take their index mod \(n\)

Repeat (for each episode):
  Initialize and store \(S_0 \ne terminal\)
  Select and store an action \(A_0 \sim \mu(\dot \ | S_0)\)
  \(T \gets \infty\)
  For \(t = 0,1,2,\cdots\):
   If \(t < T\), then:
    Take an action \(A_t\)
    Observe and store the next reward as \(R_{t+1}\) and the next state as \(S_{t+1}\)
    If \(S_{t+1}\) is terminal, then:
     \(T \gets t+1\)
    Else:
     Select and store an action \(A_{t+1} \sim \pi(\dot \ | S_{t+1})\)
   $\tau \gets t - n + 1 \ $ (\(\tau\) is the time whose state's estimate is being updated)
   If \(\tau \ge 0\):
    \(\rho \gets \prod_{i = \tau + 1}^{min(\tau + n - 1, T -1 )} \frac{\pi(A_t|S_t)}{\mu(A_t|S_t)} \qquad \qquad (\rho_{\tau+n}^{(\tau+1)})\)
    \(G \gets \sum_{i = \tau + 1}^{min(\tau + n, T)} \gamma^{i-\tau-1}R_i\)
    if \(\tau + n \le T\) then: \(G \gets G + \gamma^{n} Q(S_{\tau + n}, A_{\tau + n}) \qquad \qquad (G_{\tau}^{(n)})\)
    \(Q(S_{\tau}, A_{\tau}) \gets Q(S_{\tau}, A_{\tau}) + \alpha \rho [G - Q(S_{\tau}, A_{\tau})]\)
    If {\pi} is being learned, then ensure that \(\pi(\dot \ | S_{\tau})\) is \(\epsilon\)-greedy wrt Q
  Until \(\tau = T - 1\)

Expected Sarsa

  • 流程图
Reinforcement Learning - TD Expected Sarsa Reinforcement Learning - TD Expected Sarsa s S s_1 ... s->s_1  A   Rk q Q(S, A) s->q s_2 Sn s_1->s_2 An-1 Rn q_1 Q(...) s_1->q_1 q_2 sum(pi(a|Sn) * Q(Sn, a)) s_2->q_2 q->q_1 Rk q_1->q_2 Rn
* 算法描述 略。

策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法(不带importance sampling): Tree Backup Algorithm

Tree Backup Algorithm的思想是每步都求行动价值的期望值。
求行动价值的期望值意味着对所有可能的行动\(a\)都评估一次。

多步时序差分学习方法

  • 流程图
Reinforcement Learning - TD Tree Backup Reinforcement Learning - TD Tree Backup s S s_1 ... s->s_1  A   Rk q Q(S, A) s->q s_2 Sn s_1->s_2 An-1 Rn q_1 sum(pi(a|...) * Q(..., a)) s_1->q_1 q_2 sum(pi(a|Sn) * Q(Sn, a)) s_2->q_2 q->q_1 Rk q_1->q_2 Rn
  • 算法描述

Initialize \(Q(s，a)\) arbitrarily \(\forall s \in \mathcal{S}^, \forall a in \mathcal{A}\)
Initialize \(\pi\) to be \(\epsilon\)-greedy with respect to Q, or to a fixed given policy
Parameters: step size \(\alpha \in (0, 1]\),
  small \(\epsilon > 0\)
  a positive integer \(n\)
All store and access operations (for \(S_t\) and \(R_t\)) can take their index mod \(n\)

Repeat (for each episode):
  Initialize and store \(S_0 \ne terminal\)
  Select and store an action \(A_0 \sim \pi(\dot \ | S_0)\)
  \(Q_0 \gets Q(S_0, A_0)\)
  \(T \gets \infty\)
  For \(t = 0,1,2,\cdots\):
   If \(t < T\), then:
    Take an action \(A_t\)
    Observe and store the next reward as \(R_{t+1}\) and the next state as \(S_{t+1}\)
    If \(S_{t+1}\) is terminal, then:
     \(T \gets t+1\)
     \(\delta_t \gets R - Q_t\)
    Else:
     \(\delta_t \gets R + \gamma \sum_a \pi(a|S_{t+1})Q(S_{t+1},a) - Q_t\)
     Select arbitrarily and store an action as \(A_{t+1}\)
     \(Q_{t+1} \gets Q(S_{t+1},A_{t+1})\)
     \(\pi_{t+1} \gets \pi(S_{t+1},A_{t+1})\)
   $\tau \gets t - n + 1 \ $ (\(\tau\) is the time whose state's estimate is being updated)
   If \(\tau \ge 0\):
    \(E \gets 1\)
    \(G \gets Q_{\tau}\)
    For \(k=\tau, \dots, min(\tau + n - 1, T - 1):\)
     \(G \gets\ G + E \delta_k\)
     \(E \gets\ \gamma E \pi_{k+1}\)
    \(Q(S_{\tau}, A_{\tau}) \gets Q(S_{\tau}, A_{\tau}) + \alpha [G - Q(S_{\tau}, A_{\tau})]\)
    If {\pi} is being learned, then ensure that \(\pi(a | S_{\tau})\) is \(\epsilon\)-greedy wrt \(Q(S_{\tau},\dot \ )\)
  Until \(\tau = T - 1\)

策略行动价值\(q_{\pi}\)的off-policy时序差分学习方法: \(Q(\sigma)\)

\(Q(\sigma)\)结合了Sarsa(importance sampling), Expected Sarsa, Tree Backup算法，并考虑了重要样本。
当\(\sigma = 1\)时，使用了重要样本的Sarsa算法。
当\(\sigma = 0\)时，使用了Tree Backup的行动期望值算法。

多步时序差分学习方法

  • 流程图
Reinforcement Learning - TD Q(sigma) Reinforcement Learning - TD Q(sigma) s S s_1 ... s->s_1  A   R. q Q(S, A) s->q s_2 ... s_1->s_2 A. R. q_1 Q(...) s_1->q_1 sigma = 1 s_3 ... s_2->s_3 A. R. q_2 sum(pi(a|...) * Q(...,a)) s_2->q_2 sigma = 0 s_4 Sn s_3->s_4 An-1 Rn q_3 Q(...) s_3->q_3 sigma = 1 q_4 sum(pi(a|Sn) * Q(Sn,a)) s_4->q_4 sigma = 0 q->q_1 R. q_1->q_2 R. q_2->q_3 R. q_3->q_4 Rn
  • 算法描述

Input: behavior policy \mu such that \(\mu(a|s) > 0，\forall s \in \mathcal{S}, a \in \mathcal{A}\)
Initialize \(Q(s，a)\) arbitrarily \forall s \in \mathcal{S}^, \forall a in \mathcal{A}$
Initialize \(\pi\) to be \(\epsilon\)-greedy with respect to Q, or to a fixed given policy
Parameters: step size \(\alpha \in (0, 1]\),
  small \(\epsilon > 0\)
  a positive integer \(n\)
All store and access operations (for \(S_t\) and \(R_t\)) can take their index mod \(n\)

Repeat (for each episode):
  Initialize and store \(S_0 \ne terminal\)
  Select and store an action \(A_0 \sim \mu(\dot \ | S_0)\)
  \(Q_0 \gets Q(S_0, A_0)\)
  \(T \gets \infty\)
  For \(t = 0,1,2,\cdots\):
   If \(t < T\), then:
    Take an action \(A_t\)
    Observe and store the next reward as \(R_{t+1}\) and the next state as \(S_{t+1}\)
    If \(S_{t+1}\) is terminal, then:
     \(T \gets t+1\)
     \(\delta_t \gets R - Q_t\)
    Else:
     Select and store an action as \(A_{t+1} \sim \mu(\dot \ |S_{t+1})\)
     Select and store \(\sigma_{t+1})\)
     \(Q_{t+1} \gets Q(S_{t+1},A_{t+1})\)
     \(\delta_t \gets R + \gamma \sigma_{t+1} Q_{t+1} + \gamma (1 - \sigma_{t+1})\sum_a \pi(a|S_{t+1})Q(S_{t+1},a) - Q_t\)
     \(\pi_{t+1} \gets \pi(S_{t+1},A_{t+1})\)
     \(\rho_{t+1} \gets \frac{\pi(A_{t+1}|S_{t+1})}{\mu(A_{t+1}|S_{t+1})}\)
   $\tau \gets t - n + 1 \ $ (\(\tau\) is the time whose state's estimate is being updated)
   If \(\tau \ge 0\):
    \(\rho \gets 1\)
    \(E \gets 1\)
    \(G \gets Q_{\tau}\)
    For \(k=\tau, \dots, min(\tau + n - 1, T - 1):\)
     \(G \gets\ G + E \delta_k\)
     \(E \gets\ \gamma E [(1 - \sigma_{k+1})\pi_{k+1} + \sigma_{k+1}]\)
     \(\rho \gets\ \rho(1 - \sigma_{k} + \sigma_{k}\tau_{k})\)
    \(Q(S_{\tau}, A_{\tau}) \gets Q(S_{\tau}, A_{\tau}) + \alpha \rho [G - Q(S_{\tau}, A_{\tau})]\)
    If \({\pi}\) is being learned, then ensure that \(\pi(a | S_{\tau})\) is \(\epsilon\)-greedy wrt \(Q(S_{\tau},\dot \ )\)
  Until \(\tau = T - 1\)

总结

时序差分学习方法的限制：学习步数内，可获得奖赏信息。
比如，国际象棋的每一步，是否可以计算出一个奖赏信息？如果使用蒙特卡洛方法，模拟到游戏结束，肯定是可以获得一个奖赏结果的。

参照

posted @ 2017-03-09 15:23  SNYang  阅读(12046)  评论(0)  编辑  收藏  举报