diff --git a/Notebooks/C1.ipynb b/Notebooks/C1.ipynb
new file mode 100644
index 0000000..b5952a3
--- /dev/null
+++ b/Notebooks/C1.ipynb
@@ -0,0 +1,285 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "f6b663c9",
+   "metadata": {},
+   "source": [
+    "# 第 1 章：强化学习基本概念 (Basic Concepts)\n",
+    "本笔记本旨在复现 3x3 网格世界中的基础马尔可夫决策过程 (MDP) 元素。"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "6771d846",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "状态空间大小: 9\n",
+      "动作空间大小: 5\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 定义状态空间 (9个格子)\n",
+    "states = [\"s1\", \"s2\", \"s3\", \"s4\", \"s5\", \"s6\", \"s7\", \"s8\", \"s9\"]\n",
+    "\n",
+    "# 定义动作空间 (5种动作)\n",
+    "actions = [\"a1\", \"a2\", \"a3\", \"a4\", \"a5\"] # 分别代表：上、右、下、左、原地\n",
+    "\n",
+    "print(f\"状态空间大小: {len(states)}\")\n",
+    "print(f\"动作空间大小: {len(actions)}\")"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "20ed6dce",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "在 s1 采取动作 a2，下一个状态是: s2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 构建状态转移字典，格式：(当前状态, 动作): 下一个状态\n",
+    "transitions = {}\n",
+    "\n",
+    "# s1 的转移\n",
+    "transitions[(\"s1\", \"a1\")] = \"s1\" # 向上撞墙，反弹回 s1\n",
+    "transitions[(\"s1\", \"a2\")] = \"s2\" # 向右进入 s2\n",
+    "transitions[(\"s1\", \"a3\")] = \"s4\" # 向下进入 s4\n",
+    "transitions[(\"s1\", \"a4\")] = \"s1\" # 向左撞墙，反弹回 s1\n",
+    "transitions[(\"s1\", \"a5\")] = \"s1\" # 原地不动\n",
+    "\n",
+    "# s2 的转移\n",
+    "transitions[(\"s2\", \"a1\")] = \"s2\" # 向上撞墙，反弹回 s2\n",
+    "transitions[(\"s2\", \"a2\")] = \"s3\" # 向右进入 s3\n",
+    "transitions[(\"s2\", \"a3\")] = \"s5\" # 向下进入 s5\n",
+    "transitions[(\"s2\", \"a4\")] = \"s1\" # 向左进入 s1\n",
+    "transitions[(\"s2\", \"a5\")] = \"s2\" # 原地不动\n",
+    "\n",
+    "# s3 的转移\n",
+    "transitions[(\"s3\", \"a1\")] = \"s3\" # 向上撞墙\n",
+    "transitions[(\"s3\", \"a2\")] = \"s3\" # 向右撞墙\n",
+    "transitions[(\"s3\", \"a3\")] = \"s6\" # 向下进入 s6\n",
+    "transitions[(\"s3\", \"a4\")] = \"s2\" # 向左进入 s2\n",
+    "transitions[(\"s3\", \"a5\")] = \"s3\" # 原地不动\n",
+    "\n",
+    "# s4 的转移\n",
+    "transitions[(\"s4\", \"a1\")] = \"s1\" # 向上进入 s1\n",
+    "transitions[(\"s4\", \"a2\")] = \"s5\" # 向右进入 s5\n",
+    "transitions[(\"s4\", \"a3\")] = \"s7\" # 向下进入 s7\n",
+    "transitions[(\"s4\", \"a4\")] = \"s4\" # 向左撞墙\n",
+    "transitions[(\"s4\", \"a5\")] = \"s4\" # 原地不动\n",
+    "\n",
+    "# s5 的转移\n",
+    "transitions[(\"s5\", \"a1\")] = \"s2\" # 向上进入 s2\n",
+    "transitions[(\"s5\", \"a2\")] = \"s6\" # 向右进入 s6\n",
+    "transitions[(\"s5\", \"a3\")] = \"s8\" # 向下进入 s8\n",
+    "transitions[(\"s5\", \"a4\")] = \"s4\" # 向左进入 s4\n",
+    "transitions[(\"s5\", \"a5\")] = \"s5\" # 原地不动\n",
+    "\n",
+    "# s6 的转移\n",
+    "transitions[(\"s6\", \"a1\")] = \"s3\" # 向上进入 s3\n",
+    "transitions[(\"s6\", \"a2\")] = \"s6\" # 向右撞墙\n",
+    "transitions[(\"s6\", \"a3\")] = \"s9\" # 向下进入 s9\n",
+    "transitions[(\"s6\", \"a4\")] = \"s5\" # 向左进入 s5\n",
+    "transitions[(\"s6\", \"a5\")] = \"s6\" # 原地不动\n",
+    "\n",
+    "# s7 的转移\n",
+    "transitions[(\"s7\", \"a1\")] = \"s4\" # 向上进入 s4\n",
+    "transitions[(\"s7\", \"a2\")] = \"s8\" # 向右进入 s8\n",
+    "transitions[(\"s7\", \"a3\")] = \"s7\" # 向下撞墙\n",
+    "transitions[(\"s7\", \"a4\")] = \"s7\" # 向左撞墙\n",
+    "transitions[(\"s7\", \"a5\")] = \"s7\" # 原地不动\n",
+    "\n",
+    "# s8 的转移\n",
+    "transitions[(\"s8\", \"a1\")] = \"s5\" # 向上进入 s5\n",
+    "transitions[(\"s8\", \"a2\")] = \"s9\" # 向右进入 s9\n",
+    "transitions[(\"s8\", \"a3\")] = \"s8\" # 向下撞墙\n",
+    "transitions[(\"s8\", \"a4\")] = \"s7\" # 向左进入 s7\n",
+    "transitions[(\"s8\", \"a5\")] = \"s8\" # 原地不动\n",
+    "\n",
+    "# s9 的转移\n",
+    "transitions[(\"s9\", \"a1\")] = \"s6\" # 向上进入 s6\n",
+    "transitions[(\"s9\", \"a2\")] = \"s9\" # 向右撞墙\n",
+    "transitions[(\"s9\", \"a3\")] = \"s9\" # 向下撞墙\n",
+    "transitions[(\"s9\", \"a4\")] = \"s8\" # 向左进入 s8\n",
+    "transitions[(\"s9\", \"a5\")] = \"s9\" # 原地不动\n",
+    "\n",
+    "def get_next_state(state, action):\n",
+    "    return transitions.get((state, action), state) # 如果没定义，默认原地不动\n",
+    "\n",
+    "print(f\"在 s1 采取动作 a2，下一个状态是: {get_next_state('s1', 'a2')}\")"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "788f4de0",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "s1撞墙奖励: -1\n",
+      "进入目标s9奖励: 1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 定义奖励规则\n",
+    "def get_reward(state, action, next_state):\n",
+    "    # 如果撞墙（当前状态和下一个状态一样，且不是原地动作）\n",
+    "    if state == next_state and action != \"a5\":\n",
+    "        return -1\n",
+    "    # 如果进入目标状态 s9\n",
+    "    elif next_state == \"s9\":\n",
+    "        return 1\n",
+    "    # 如果进入禁区 s6 或 s7 (根据书中图 1.2)\n",
+    "    elif next_state in [\"s6\", \"s7\"]:\n",
+    "        return -1\n",
+    "    # 其他情况\n",
+    "    else:\n",
+    "        return 0\n",
+    "\n",
+    "# 测试一下\n",
+    "test_r1 = get_reward(\"s1\", \"a1\", \"s1\")\n",
+    "test_r2 = get_reward(\"s8\", \"a2\", \"s9\")\n",
+    "print(f\"s1撞墙奖励: {test_r1}\") # 应该输出 -1\n",
+    "print(f\"进入目标s9奖励: {test_r2}\") # 应该输出 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "id": "c64e665e",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "--- 运行轨迹并计算回报 ---\n",
+      "Step 0: s1 --a2--> s2, Reward: 0\n",
+      "Step 1: s2 --a3--> s5, Reward: 0\n",
+      "Step 2: s5 --a3--> s8, Reward: 0\n",
+      "Step 3: s8 --a2--> s9, Reward: 1\n",
+      "最终折扣回报: 0.7290000000000001\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 定义一条书中的轨迹 (图 1.6 左侧的好策略轨迹)\n",
+    "# 格式：[(状态, 动作), ...]\n",
+    "good_trajectory = [\n",
+    "    (\"s1\", \"a2\"), (\"s2\", \"a3\"), (\"s5\", \"a3\"), (\"s8\", \"a2\")\n",
+    "]\n",
+    "\n",
+    "# 计算回报的函数\n",
+    "def calculate_discounted_return(trajectory, gamma=0.9):\n",
+    "    total_return = 0.0\n",
+    "    \n",
+    "    for t, (state, action) in enumerate(trajectory):\n",
+    "        next_state = get_next_state(state, action)\n",
+    "        reward = get_reward(state, action, next_state)\n",
+    "        \n",
+    "        # 计算折扣回报\n",
+    "        total_return += (gamma ** t) * reward\n",
+    "        print(f\"Step {t}: {state} --{action}--> {next_state}, Reward: {reward}\")\n",
+    "        \n",
+    "    return total_return\n",
+    "\n",
+    "print(\"--- 运行轨迹并计算回报 ---\")\n",
+    "final_return = calculate_discounted_return(good_trajectory, gamma=0.9)\n",
+    "print(f\"最终折扣回报: {final_return}\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "5de54db1",
+   "metadata": {},
+   "source": [
+    "接下来来有一些随机策略的case"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "id": "6b05c85f",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "在状态 s1 进行 10 次动作选择测试：\n",
+      "[np.str_('a2'), np.str_('a2'), np.str_('a2'), np.str_('a2'), np.str_('a3'), np.str_('a3'), np.str_('a3'), np.str_('a3'), np.str_('a2'), np.str_('a3')]\n",
+      "向右(a2)的次数: 5, 向下(a3)的次数: 5\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "# 定义一个随机策略字典\n",
+    "# 格式：状态: {动作: 概率, 动作: 概率}\n",
+    "stochastic_policy = {\n",
+    "    \"s1\": {\"a2\": 0.5, \"a3\": 0.5}, # 图 1.5 中的设定\n",
+    "    \"s2\": {\"a3\": 1.0},            # 其他状态假设是确定性的 100%\n",
+    "    \"s5\": {\"a3\": 1.0},\n",
+    "    \"s8\": {\"a2\": 1.0}\n",
+    "}\n",
+    "\n",
+    "def choose_action(state, policy):\n",
+    "    # 如果策略里有这个状态的动作概率分布\n",
+    "    if state in policy:\n",
+    "        action_probs = policy[state]\n",
+    "        actions = list(action_probs.keys())\n",
+    "        probs = list(action_probs.values())\n",
+    "        # 根据概率分布随机选择一个动作\n",
+    "        chosen_action = np.random.choice(actions, p=probs)\n",
+    "        return chosen_action\n",
+    "    else:\n",
+    "        return \"a5\" # 默认原地不动\n",
+    "\n",
+    "# 测试随机策略：让智能体在 s1 连续做 10 次决定\n",
+    "print(\"在状态 s1 进行 10 次动作选择测试：\")\n",
+    "choices = [choose_action(\"s1\", stochastic_policy) for _ in range(10)]\n",
+    "print(choices)\n",
+    "print(f\"向右(a2)的次数: {choices.count('a2')}, 向下(a3)的次数: {choices.count('a3')}\")"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "GymRL",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.13.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
diff --git a/Notebooks/C2.ipynb b/Notebooks/C2.ipynb
new file mode 100644
index 0000000..2a88605
--- /dev/null
+++ b/Notebooks/C2.ipynb
@@ -0,0 +1,184 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "31a4e03b",
+   "metadata": {},
+   "source": [
+    "# 第 2 章：状态价值与贝尔曼方程\n",
+    "贝尔曼方程描述了所有状态价值之间的关系。\n",
+    "我们将使用**矩阵-向量形式** 来表示贝尔曼方程： $$v_{\\pi} = r_{\\pi} + \\gamma P_{\\pi}v_{\\pi}$$\n",
+    "以及使用**迭代法** 来求解它： $$v_{k+1} = r_{\\pi} + \\gamma P_{\\pi}v_{k}$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "8e67777b",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "--- 贝尔曼方程迭代求解过程 ---\n",
+      "第 1 次迭代: v(s1)=-0.5000, v(s2)=1.0000, v(s3)=1.0000, v(s4)=1.0000\n",
+      "第 2 次迭代: v(s1)=0.4000, v(s2)=1.9000, v(s3)=1.9000, v(s4)=1.9000\n",
+      "第 3 次迭代: v(s1)=1.2100, v(s2)=2.7100, v(s3)=2.7100, v(s4)=2.7100\n",
+      "第 4 次迭代: v(s1)=1.9390, v(s2)=3.4390, v(s3)=3.4390, v(s4)=3.4390\n",
+      "第 5 次迭代: v(s1)=2.5951, v(s2)=4.0951, v(s3)=4.0951, v(s4)=4.0951\n",
+      "第 100 次迭代: v(s1)=8.4997, v(s2)=9.9997, v(s3)=9.9997, v(s4)=9.9997\n",
+      "\n",
+      "最终收敛的状态价值: [ 8.5 10.  10.  10. ]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "# 1. 设定折扣因子\n",
+    "gamma = 0.9\n",
+    "\n",
+    "# 2. 定义书本图 2.5 中的期望奖励向量 r_pi (Reward vector)\n",
+    "# s1 的期望奖励是 0.5*0 + 0.5*(-1) = -0.5\n",
+    "r_pi = np.array([-0.5, 1.0, 1.0, 1.0])\n",
+    "\n",
+    "# 3. 定义状态转移矩阵 P_pi (Transition matrix)\n",
+    "# 每一行代表 s1, s2, s3, s4; 每一列代表转移到 s1, s2, s3, s4 的概率\n",
+    "P_pi = np.array([\n",
+    "    [0.0, 0.5, 0.5, 0.0],  # s1 有 0.5 概率到 s2, 0.5 概率到 s3\n",
+    "    [0.0, 0.0, 0.0, 1.0],  # s2 有 1.0 概率到 s4\n",
+    "    [0.0, 0.0, 0.0, 1.0],  # s3 有 1.0 概率到 s4\n",
+    "    [0.0, 0.0, 0.0, 1.0]   # s4 有 1.0 概率停留在 s4\n",
+    "])\n",
+    "\n",
+    "# 4. 迭代法求解状态价值 (Iterative solution)\n",
+    "v_values = np.zeros(4) # 初始时，假设所有状态价值为 0\n",
+    "iterations = 100\n",
+    "\n",
+    "print(\"--- 贝尔曼方程迭代求解过程 ---\")\n",
+    "for k in range(iterations):\n",
+    "    # 核心公式：v_{k+1} = r_pi + gamma * P_pi * v_k\n",
+    "    v_next = r_pi + gamma * P_pi.dot(v_values)\n",
+    "    \n",
+    "    if k < 5 or k == iterations - 1:\n",
+    "        print(f\"第 {k+1} 次迭代: v(s1)={v_next[0]:.4f}, v(s2)={v_next[1]:.4f}, v(s3)={v_next[2]:.4f}, v(s4)={v_next[3]:.4f}\")\n",
+    "        \n",
+    "    v_values = v_next\n",
+    "\n",
+    "print(\"\\n最终收敛的状态价值:\", np.round(v_values, 2))\n",
+    "# 你可以将这个输出结果与书中 2.5 节手算的 8.5, 10, 10, 10 进行对比！"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "2086b1ab",
+   "metadata": {},
+   "source": [
+    "# 2.8 从状态价值到动作价值 (Action Value)\n",
+    "动作价值 $q_\\pi(s,a)$ 评估的是在特定状态下采取特定动作的好坏。\n",
+    "\n",
+    "它由两部分组成：\n",
+    "1. **即时奖励的期望**：采取动作 $a$ 后立刻得到的奖励。\n",
+    "2. **未来奖励的期望**：进入下一个状态 $s'$ 后，未来的状态价值 $\\gamma v_\\pi(s')$。\n",
+    "\n",
+    "数学公式为：\n",
+    "$$q_{\\pi}(s,a) = \\sum_{r\\in\\mathcal{R}}p(r|s,a)r + \\gamma\\sum_{s'\\in\\mathcal{S}}p(s'|s,a)v_{\\pi}(s')$$\n",
+    "\n",
+    "状态价值其实就是所有可能动作价值的加权平均：\n",
+    "$$v_{\\pi}(s) = \\sum_{a\\in\\mathcal{A}}\\pi(a|s)q_{\\pi}(s,a)$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "a3e76dd6",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "--- 计算在状态 s1 下，所有可能动作的价值 (Q-values) ---\n",
+      "策略内的动作: q(s1, a2向右) = -1 + 0.9 * v(s2) = 8.0\n",
+      "策略内的动作: q(s1, a3向下) = 0 + 0.9 * v(s3) = 9.0\n",
+      "\n",
+      "策略外的动作: q(s1, a1向上) = -1 + 0.9 * v(s1) = 6.65\n",
+      "策略外的动作: q(s1, a4向左) = -1 + 0.9 * v(s1) = 6.65\n",
+      "策略外的动作: q(s1, a5原地) =  0 + 0.9 * v(s1) = 7.65\n",
+      "\n",
+      "--- 验证公式：状态价值是动作价值的加权平均 ---\n",
+      "根据 Q 值反推的 v(s1) = 0.5 * 8.0 + 0.5 * 9.0 = 8.5\n",
+      "这与我们之前迭代出来的 v(s1) = 8.5 完全一致！\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "# 1. 继承之前的已知数据\n",
+    "gamma = 0.9\n",
+    "# 假设这是我们在 Cell 8 算出来的最终状态价值\n",
+    "v_values = {\"s1\": 8.5, \"s2\": 10.0, \"s3\": 10.0, \"s4\": 10.0}\n",
+    "\n",
+    "# 2. 定义动作价值计算函数 q(s, a)\n",
+    "def calculate_q_value(state, action, reward, next_state, gamma, v_values):\n",
+    "    # q(s,a) = 立即奖励 + gamma * 下一个状态的价值\n",
+    "    return reward + gamma * v_values[next_state]\n",
+    "\n",
+    "print(\"--- 计算在状态 s1 下，所有可能动作的价值 (Q-values) ---\")\n",
+    "\n",
+    "# --- 策略会选择的动作 (向右 a2, 向下 a3) ---\n",
+    "# a2: 向右进入 s2，得奖励 -1\n",
+    "q_s1_a2 = calculate_q_value(\"s1\", \"a2\", -1, \"s2\", gamma, v_values)\n",
+    "print(f\"策略内的动作: q(s1, a2向右) = -1 + 0.9 * v(s2) = {q_s1_a2}\") # 预期为 8.0\n",
+    "\n",
+    "# a3: 向下进入 s3，得奖励 0\n",
+    "q_s1_a3 = calculate_q_value(\"s1\", \"a3\", 0, \"s3\", gamma, v_values)\n",
+    "print(f\"策略内的动作: q(s1, a3向下) = 0 + 0.9 * v(s3) = {q_s1_a3}\") # 预期为 9.0\n",
+    "\n",
+    "\n",
+    "# --- 策略不会选择的动作 (向上 a1, 向左 a4, 原地 a5) ---\n",
+    "# a1: 向上撞墙，弹回 s1，得奖励 -1\n",
+    "q_s1_a1 = calculate_q_value(\"s1\", \"a1\", -1, \"s1\", gamma, v_values)\n",
+    "print(f\"\\n策略外的动作: q(s1, a1向上) = -1 + 0.9 * v(s1) = {q_s1_a1:.2f}\") # 预期为 6.65\n",
+    "\n",
+    "# a4: 向左撞墙，弹回 s1，得奖励 -1\n",
+    "q_s1_a4 = calculate_q_value(\"s1\", \"a4\", -1, \"s1\", gamma, v_values)\n",
+    "print(f\"策略外的动作: q(s1, a4向左) = -1 + 0.9 * v(s1) = {q_s1_a4:.2f}\") # 预期为 6.65\n",
+    "\n",
+    "# a5: 原地不动，停在 s1，得奖励 0\n",
+    "q_s1_a5 = calculate_q_value(\"s1\", \"a5\", 0, \"s1\", gamma, v_values)\n",
+    "print(f\"策略外的动作: q(s1, a5原地) =  0 + 0.9 * v(s1) = {q_s1_a5:.2f}\") # 预期为 7.65\n",
+    "\n",
+    "\n",
+    "print(\"\\n--- 验证公式：状态价值是动作价值的加权平均 ---\")\n",
+    "# 我们的策略是: 0.5概率选a2, 0.5概率选a3\n",
+    "v_s1_calculated = 0.5 * q_s1_a2 + 0.5 * q_s1_a3\n",
+    "print(f\"根据 Q 值反推的 v(s1) = 0.5 * {q_s1_a2} + 0.5 * {q_s1_a3} = {v_s1_calculated}\")\n",
+    "print(f\"这与我们之前迭代出来的 v(s1) = {v_values['s1']} 完全一致！\")"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "GymRL",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.13.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
diff --git a/Notebooks/C3.ipynb b/Notebooks/C3.ipynb
new file mode 100644
index 0000000..4371e73
--- /dev/null
+++ b/Notebooks/C3.ipynb
@@ -0,0 +1,166 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "745510b1",
+   "metadata": {},
+   "source": [
+    "# 第 3 章：最优状态价值与贝尔曼最优方程\n",
+    "\n",
+    "## 1. 什么是最优策略？\n",
+    "如果一个策略 $\\pi^*$ 在所有状态下的价值，都大于或等于其他任何策略的价值，即对于所有 $s \\in \\mathcal{S}$，都有 $v_{\\pi^*}(s) \\ge v_\\pi(s)$，那么 $\\pi^*$ 就是最优策略。\n",
+    "最优策略的状态价值，被称为**最优状态价值 (Optimal State Value)** $v^*$。\n",
+    "\n",
+    "## 2. 贝尔曼最优方程 (BOE)\n",
+    "之前我们算的是某个固定策略的价值。现在我们要算的是“最优”价值。\n",
+    "BOE 的核心思想是：在某个状态下的最优价值，等于**所有可能动作中，动作价值 $q(s,a)$ 的最大值**。\n",
+    "\n",
+    "数学表达为：\n",
+    "$$v(s) = \\max_{a \\in \\mathcal{A}} q(s,a)$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "f31bd6cc",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "--- 策略改进演示 ---\n",
+      "动作 a1 的价值: 6.2\n",
+      "动作 a2 的价值: 8.0\n",
+      "动作 a3 的价值: 9.0\n",
+      "动作 a4 的价值: 6.2\n",
+      "动作 a5 的价值: 7.2\n",
+      "\n",
+      "=> 最佳动作是 'a3'，其价值为 9.0。\n",
+      "结论：只要我们在每个状态都选择 Q 值最大的动作，就能得到一个更好的、甚至是最优的策略！\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "# 假设我们在状态 s1 算出了 5 个动作的 Q 值 (参考书中 3.1 节的计算)\n",
+    "# a1: 向上, a2: 向右, a3: 向下, a4: 向左, a5: 原地不动\n",
+    "q_values_s1 = {\n",
+    "    \"a1\": 6.2,  # 撞墙，-1 + 0.9*v(s1)\n",
+    "    \"a2\": 8.0,  # 原策略向右，进禁区，-1 + 0.9*v(s2)\n",
+    "    \"a3\": 9.0,  # 向下避开禁区，0 + 0.9*v(s3)\n",
+    "    \"a4\": 6.2,  # 撞墙，-1 + 0.9*v(s1)\n",
+    "    \"a5\": 7.2   # 原地不动，0 + 0.9*v(s1)\n",
+    "}\n",
+    "\n",
+    "# 寻找使 Q 值最大的动作 (贪心策略 Greedy Policy)\n",
+    "best_action = max(q_values_s1, key=q_values_s1.get)\n",
+    "max_q_value = q_values_s1[best_action]\n",
+    "\n",
+    "print(\"--- 策略改进演示 ---\")\n",
+    "for action, q_val in q_values_s1.items():\n",
+    "    print(f\"动作 {action} 的价值: {q_val}\")\n",
+    "\n",
+    "print(f\"\\n=> 最佳动作是 '{best_action}'，其价值为 {max_q_value}。\")\n",
+    "print(\"结论：只要我们在每个状态都选择 Q 值最大的动作，就能得到一个更好的、甚至是最优的策略！\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "618e235d",
+   "metadata": {},
+   "source": [
+    "## 3. 求解 BOE：压缩映射与迭代法\n",
+    "BOE 看起来很难解，因为它方程两边都有未知数 $v$。但数学上的**压缩映射定理 (Contraction Mapping Theorem)** 告诉我们三个好消息：\n",
+    "1. **存在性**：BOE 一定有解。\n",
+    "2. **唯一性**：最优状态价值 $v^*$ 的解是唯一的。\n",
+    "3. **算法**：可以通过不断迭代 $v_{k+1} = \\max_\\pi(r_\\pi + \\gamma P_\\pi v_k)$ 来逼近最优解，这个过程叫**价值迭代 (Value Iteration)**=。\n",
+    "\n",
+    "## 4. 影响最优策略的因素\n",
+    "* **折扣因子 $\\gamma$ 的魔力**：\n",
+    "  * $\\gamma$ 较大 (如 0.9)：智能体目光长远，为了以最快速度拿到目标大奖，甚至愿意冒险穿过有惩罚的禁区。\n",
+    "  * $\\gamma$ 较小 (如 0.5 或更小)：智能体变得短视且保守，宁愿绕远路也不敢冒险。\n",
+    "  * 此外，$\\gamma$ 天生就不鼓励“无意义的绕路”，因为绕路会让未来的奖励因折扣变得更小。\n",
+    "* **奖励的仿射变换不变性**：如果把所有的奖励都乘以一个正数，或者都加上一个固定的常数，算出来的最优策略**完全不会改变**。"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "41d7c569",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "--- 情景 1：每走一步没有任何惩罚 (即时奖励为 0) ---\n",
+      "直达路径(A)的回报: 9.0\n",
+      "绕路路径(B)的回报: 7.290000000000001\n",
+      "结论：哪怕每步惩罚是 0，因为有 gamma 的折现，智能体依然知道直达路径更好！\n",
+      "\n",
+      "--- 情景 2：给每一步都增加 -1 的惩罚 (仿射变换) ---\n",
+      "增加惩罚后，直达路径(A)的回报: 8.00\n",
+      "增加惩罚后，绕路路径(B)的回报: 4.58\n",
+      "结论：增加统一惩罚后，直达路径(A)依然大于绕路路径(B)。相对好坏关系不变，最优策略不变。\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 验证：假设两条路径，路径A直达终点(2步)，路径B绕路(4步)\n",
+    "gamma = 0.9\n",
+    "target_reward = 10.0\n",
+    "\n",
+    "print(\"--- 情景 1：每走一步没有任何惩罚 (即时奖励为 0) ---\")\n",
+    "# 路径A: 0 + 0.9 * 10\n",
+    "return_path_A = 0 + (gamma ** 1) * target_reward \n",
+    "# 路径B: 0 + 0.9*0 + 0.9^2*0 + 0.9^3 * 10\n",
+    "return_path_B = 0 + 0 + 0 + (gamma ** 3) * target_reward\n",
+    "\n",
+    "print(f\"直达路径(A)的回报: {return_path_A}\")\n",
+    "print(f\"绕路路径(B)的回报: {return_path_B}\")\n",
+    "print(\"结论：哪怕每步惩罚是 0，因为有 gamma 的折现，智能体依然知道直达路径更好！\\n\")\n",
+    "\n",
+    "print(\"--- 情景 2：给每一步都增加 -1 的惩罚 (仿射变换) ---\")\n",
+    "# 路径A: -1 + 0.9 * (10 - 1) -> 简化理解为每步 -1，最后一步拿到目标\n",
+    "return_path_A_penalty = -1 + (gamma ** 1) * target_reward\n",
+    "# 路径B: 绕远路\n",
+    "return_path_B_penalty = -1 + gamma*(-1) + (gamma**2)*(-1) + (gamma**3)*target_reward\n",
+    "\n",
+    "print(f\"增加惩罚后，直达路径(A)的回报: {return_path_A_penalty:.2f}\")\n",
+    "print(f\"增加惩罚后，绕路路径(B)的回报: {return_path_B_penalty:.2f}\")\n",
+    "print(\"结论：增加统一惩罚后，直达路径(A)依然大于绕路路径(B)。相对好坏关系不变，最优策略不变。\")"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "413fb515",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "GymRL",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.13.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
diff --git a/Notebooks/C4.ipynb b/Notebooks/C4.ipynb
new file mode 100644
index 0000000..5d97fdb
--- /dev/null
+++ b/Notebooks/C4.ipynb
@@ -0,0 +1,230 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "13468a42",
+   "metadata": {},
+   "source": [
+    "# 第 4 章：价值迭代与策略迭代 (Value Iteration and Policy Iteration)\n",
+    "\n",
+    "本章介绍的算法属于**动态规划 (Dynamic Programming)**，因为它们需要提前知道环境的完整模型（状态转移概率和奖励）。\n",
+    "\n",
+    "## 1. 价值迭代 (Value Iteration)\n",
+    "价值迭代是直接求解第三章“贝尔曼最优方程 (BOE)”的迭代算法。\n",
+    "它在每次迭代中包含两步：\n",
+    "* **策略更新 (Policy Update)**：算出 Q 值，找到当前最贪心（收益最大）的动作。\n",
+    "* **价值更新 (Value Update)**：把状态的价值直接更新为最大的 Q 值：$v_{k+1}(s) = \\max_a q_k(s,a)$。\n",
+    "* **注意**：价值迭代中间产生的 $v_k$ 仅仅是计算过程中的临时数值，它们并不代表任何特定策略的真实状态价值！\n",
+    "\n",
+    "## 2. 策略迭代 (Policy Iteration)\n",
+    "策略迭代的过程更像是一个稳扎稳打的“评估-改进”循环。\n",
+    "* **策略评估 (Policy Evaluation)**：对当前策略，死磕到底，算出它真实的、收敛的状态价值 $v_{\\pi_k}$（这通常需要内部再套一个循环）。\n",
+    "* **策略改进 (Policy Improvement)**：根据算出来的真实价值，选择 Q 值最大的动作，更新策略。这保证了新策略一定比老策略更好（或者一样好）。"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "9ae8b2ba",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "# --- 1. 定义 2x2 环境 ---\n",
+    "states = [\"s1\", \"s2\", \"s3\", \"s4\"]\n",
+    "actions = [\"up\", \"right\", \"down\", \"left\", \"stay\"]\n",
+    "gamma = 0.9\n",
+    "\n",
+    "# 状态转移规则 (依据 Table 4.1 逆向推导的简单网格规律)\n",
+    "def get_transition(state, action):\n",
+    "    # s4 是目标(吸收态)，到了就停在原地\n",
+    "    if state == \"s4\": return \"s4\"\n",
+    "    \n",
+    "    if state == \"s1\":\n",
+    "        if action == \"right\": return \"s2\"\n",
+    "        if action == \"down\": return \"s3\"\n",
+    "        if action == \"stay\": return \"s1\"\n",
+    "        return \"s1\" # 撞墙反弹\n",
+    "    elif state == \"s2\":\n",
+    "        if action == \"left\": return \"s1\"\n",
+    "        if action == \"down\": return \"s4\"\n",
+    "        if action == \"stay\": return \"s2\"\n",
+    "        return \"s2\" # 撞墙反弹\n",
+    "    elif state == \"s3\":\n",
+    "        if action == \"up\": return \"s1\"\n",
+    "        if action == \"right\": return \"s4\"\n",
+    "        if action == \"stay\": return \"s3\"\n",
+    "        return \"s3\" # 撞墙反弹\n",
+    "    return state\n",
+    "\n",
+    "# 奖励规则 (依据 Table 4.1 提取)\n",
+    "def get_reward(state, action, next_state):\n",
+    "    if state == \"s4\": return 1 # 目标奖励\n",
+    "    \n",
+    "    # 判断是否撞墙 (尝试移动但留在原地)\n",
+    "    if state == next_state and action != \"stay\":\n",
+    "        return -1\n",
+    "    \n",
+    "    # 正常移动的奖励\n",
+    "    if next_state == \"s2\": return -1 # 禁区\n",
+    "    if next_state == \"s4\": return 1  # 目标\n",
+    "    return 0 # 其他移动 (书中这题为 0)\n",
+    "\n",
+    "# 辅助函数：计算 Q 值 (这里假设转移是 100% 确定性的)\n",
+    "def compute_q_value(state, action, v_values):\n",
+    "    next_s = get_transition(state, action)\n",
+    "    reward = get_reward(state, action, next_s)\n",
+    "    return reward + gamma * v_values[states.index(next_s)]"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "id": "32dc5456",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "=== 开始价值迭代 (Value Iteration) ===\n",
+      "迭代 1: V值 = [0. 1. 1. 1.], 策略 = ['down', 'down', 'right', 'up']\n",
+      "迭代 2: V值 = [0.9 1.9 1.9 1.9], 策略 = ['down', 'down', 'right', 'up']\n",
+      "迭代 3: V值 = [1.71 2.71 2.71 2.71], 策略 = ['down', 'down', 'right', 'up']\n",
+      "迭代 4: V值 = [2.44 3.44 3.44 3.44], 策略 = ['down', 'down', 'right', 'up']\n",
+      "迭代 5: V值 = [3.1 4.1 4.1 4.1], 策略 = ['down', 'down', 'right', 'up']\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(\"=== 开始价值迭代 (Value Iteration) ===\")\n",
+    "\n",
+    "# 初始化 V 值为 0\n",
+    "V_vi = np.zeros(len(states))\n",
+    "policy_vi = [\"stay\"] * len(states) \n",
+    "iterations = 5\n",
+    "\n",
+    "for k in range(iterations):\n",
+    "    new_V = np.zeros(len(states))\n",
+    "    \n",
+    "    # 遍历所有状态\n",
+    "    for i, s in enumerate(states):\n",
+    "        q_values = []\n",
+    "        # 遍历所有动作，计算 Q 值\n",
+    "        for a in actions:\n",
+    "            q = compute_q_value(s, a, V_vi)\n",
+    "            q_values.append(q)\n",
+    "            \n",
+    "        # 核心：价值更新 (直接取最大的 Q 值)\n",
+    "        best_q = max(q_values)\n",
+    "        new_V[i] = best_q\n",
+    "        \n",
+    "        # 核心：策略更新 (记录最大 Q 值对应的动作)\n",
+    "        best_action_idx = np.argmax(q_values)\n",
+    "        policy_vi[i] = actions[best_action_idx]\n",
+    "        \n",
+    "    print(f\"迭代 {k+1}: V值 = {np.round(new_V, 2)}, 策略 = {policy_vi}\")\n",
+    "    \n",
+    "    # 如果价值不再变化，说明收敛了\n",
+    "    if np.max(np.abs(new_V - V_vi)) < 1e-5:\n",
+    "        print(f\"-> 价值迭代在第 {k+1} 步提前收敛！\")\n",
+    "        V_vi = new_V\n",
+    "        break\n",
+    "        \n",
+    "    V_vi = new_V"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "id": "9e6c5d0d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "=== 开始策略迭代 (Policy Iteration) ===\n",
+      "\n",
+      "第 1 轮主迭代，当前策略: ['stay', 'stay', 'stay', 'stay']\n",
+      "  评估完成，当前策略的真实 V值 = [  0. -10.   0.  10.]\n",
+      "\n",
+      "第 2 轮主迭代，当前策略: ['down', 'down', 'right', 'up']\n",
+      "  评估完成，当前策略的真实 V值 = [ 9. 10. 10. 10.]\n",
+      "-> 策略不再改变，策略迭代收敛！最优策略已找到。\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(\"=== 开始策略迭代 (Policy Iteration) ===\")\n",
+    "\n",
+    "V_pi = np.zeros(len(states))\n",
+    "# 初始给一个极差的策略：全部原地不动 (类似书中图 4.3 的烂策略)\n",
+    "current_policy = [\"stay\", \"stay\", \"stay\", \"stay\"] \n",
+    "\n",
+    "for k in range(5):\n",
+    "    print(f\"\\n第 {k+1} 轮主迭代，当前策略: {current_policy}\")\n",
+    "    \n",
+    "    # --- 步骤 1: 策略评估 (Policy Evaluation) ---\n",
+    "    # 死磕到底，一直循环直到 V 值收敛，算出当前策略的真实价值\n",
+    "    while True:\n",
+    "        new_V = np.zeros(len(states))\n",
+    "        for i, s in enumerate(states):\n",
+    "            action = current_policy[i] # 只看当前策略指定的动作\n",
+    "            new_V[i] = compute_q_value(s, action, V_pi)\n",
+    "            \n",
+    "        if np.max(np.abs(new_V - V_pi)) < 1e-5:\n",
+    "            break\n",
+    "        V_pi = new_V\n",
+    "    print(f\"  评估完成，当前策略的真实 V值 = {np.round(V_pi, 2)}\")\n",
+    "    \n",
+    "    # --- 步骤 2: 策略改进 (Policy Improvement) ---\n",
+    "    policy_stable = True\n",
+    "    for i, s in enumerate(states):\n",
+    "        old_action = current_policy[i]\n",
+    "        \n",
+    "        # 看看有没有更好的动作\n",
+    "        q_values = [compute_q_value(s, a, V_pi) for a in actions]\n",
+    "        best_action = actions[np.argmax(q_values)]\n",
+    "        \n",
+    "        current_policy[i] = best_action\n",
+    "        if old_action != best_action:\n",
+    "            policy_stable = False # 策略发生了改变\n",
+    "            \n",
+    "    if policy_stable:\n",
+    "        print(\"-> 策略不再改变，策略迭代收敛！最优策略已找到。\")\n",
+    "        break"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "5282df09",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "GymRL",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.13.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}