{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "intro001",
   "metadata": {},
   "source": [
    "# 第8章｜校园生活证据问答助手\n",
    "\n",
    "> **核心问题：** 如果模型并不知道本校的菜单和制度，它为什么仍可能给出一段流畅的答案？\n",
    "\n",
    "本 Notebook 用 6 道固定题拆开观察检索增强生成（RAG）的完整链路：**资料分块 → 文本表示 → 相似度检索 → 阈值判断 → 证据拼接 → 基于证据回答 → 引用核查**。\n",
    "\n",
    "你最终要完成一个离线可运行的最小 RAG，并亲自解释：\n",
    "\n",
    "1. 检索器是否找到了正确原文；\n",
    "2. 生成器是否真的依据原文回答；\n",
    "3. 资料中没有答案时，系统为什么应当拒答；\n",
    "4. 只改变 `TOP_K` 后，结果为何发生变化。\n",
    "\n",
    "这不是“点一下看效果”的演示。所有核心代码都能查看，实验结论必须由你根据输出和原文作出。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "goals001",
   "metadata": {},
   "source": [
    "## 学习目标与成果\n",
    "\n",
    "完成本项目后，你应该能够：\n",
    "\n",
    "- 用自己的话说明 RAG 的“检索、拼接、生成”三步；\n",
    "- 读懂字符 n-gram TF-IDF、余弦相似度、Top-K 和相似度阈值在代码中的作用；\n",
    "- 区分“检索命中”和“回答有依据”两级验收；\n",
    "- 比较无检索猜测与有检索回答，但不把一次演示冒充正式模型评测；\n",
    "- 只修改一个主变量，保留可复现的实验记录；\n",
    "- 说明本案例为什么优先考虑 RAG，而不是把动态制度写入模型参数。\n",
    "\n",
    "交付物包括：运行后的 Notebook、`TOP_K=1` 与 `TOP_K=3` 对照、至少 5 条引用核查、知识库数据卡和 AI 协同记录。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "predict1",
   "metadata": {},
   "source": [
    "## 0. 运行前先预测\n",
    "\n",
    "先打开 `chapters/ch08-rag/student/experiment_log.md`，在运行代码之前写下你的判断：\n",
    "\n",
    "1. Q6“学校游泳馆几点开门？”在资料库没有答案时，普通聊天模型可能怎样回答？\n",
    "2. `TOP_K=1` 改成 `TOP_K=3` 后，检索到的资料更多，回答一定会更好吗？\n",
    "3. 如果检索器找错了原文，但回答碰巧正确，系统能否通过工程验收？为什么？\n",
    "\n",
    "使用 Cline 时，可以让它先解释将要读取和修改的文件，再等待你批准。不要让 AI 代写实验结论。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "setup001",
   "metadata": {},
   "source": [
    "## 1. 找到本章资源并加载完整实现\n",
    "\n",
    "Notebook 可能从 `teaching_resources/` 或 `notebooks/` 启动。下面的代码向上寻找资源根目录，再加载兼容入口 `scripts_optional/ch08/minimal_rag.py`。\n",
    "\n",
    "该脚本是本章完整、可测试的参考实现，没有使用 LangChain 等大型编排框架。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "setup002",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "from pathlib import Path\n",
    "import importlib.util\n",
    "import inspect\n",
    "import json\n",
    "import re\n",
    "import sys\n",
    "\n",
    "start = Path.cwd().resolve()\n",
    "RESOURCE_ROOT = next(\n",
    "    path for path in [start, *start.parents]\n",
    "    if (path / \"scripts_optional/ch08/minimal_rag.py\").exists()\n",
    ")\n",
    "SCRIPT = RESOURCE_ROOT / \"scripts_optional/ch08/minimal_rag.py\"\n",
    "CHAPTER_DIR = RESOURCE_ROOT / \"chapters/ch08-rag\"\n",
    "\n",
    "spec = importlib.util.spec_from_file_location(\"ch08_minimal_rag\", SCRIPT)\n",
    "rag = importlib.util.module_from_spec(spec)\n",
    "sys.modules[spec.name] = rag\n",
    "spec.loader.exec_module(rag)\n",
    "\n",
    "print(\"资源根目录：\", RESOURCE_ROOT)\n",
    "print(\"完整算法文件：\", SCRIPT.relative_to(RESOURCE_ROOT))\n",
    "print(\"离线检索器：\", rag.CharNgramTfidfProvider.name)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "data0001",
   "metadata": {},
   "source": [
    "## 2. 先看资料：知识库质量决定回答上限\n",
    "\n",
    "本章使用三套课程虚构资料：\n",
    "\n",
    "- 食堂菜单；\n",
    "- 图书馆制度；\n",
    "- 社团章程。\n",
    "\n",
    "每个 Markdown 二级标题对应一个语义完整的知识片段，标题开头是稳定段落编号。例如 `LIBRARY-02` 不只是显示标签，它还必须能够返回 `library.md` 中的真实原文。\n",
    "\n",
    "三份资料不代表任何真实学校，也不包含个人信息。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "data0002",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "chunks = rag.load_markdown_chunks(CHAPTER_DIR / \"data\")\n",
    "\n",
    "print(f\"共读取 {len(chunks)} 个知识片段：\")\n",
    "for chunk in chunks:\n",
    "    print(f\"{chunk.id:12}｜{chunk.title:8}｜来源文件：{chunk.source}\")\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "data0003",
   "metadata": {},
   "source": [
    "### 检查一段原文\n",
    "\n",
    "先不要提问。直接查看 `LIBRARY-02`，确认“最多借几本书”这个事实在资料中怎样表述。后面引用核查必须回到这里，而不是只相信模型给出的编号。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "data0004",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "chunk_by_id = {chunk.id: chunk for chunk in chunks}\n",
    "sample = chunk_by_id[\"LIBRARY-02\"]\n",
    "\n",
    "print(\"段落编号：\", sample.id)\n",
    "print(\"段落标题：\", sample.title)\n",
    "print(\"来源文件：\", sample.source)\n",
    "print(\"原文：\", sample.text)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "repr0001",
   "metadata": {},
   "source": [
    "## 3. 文本怎样变成可以比较的数字\n",
    "\n",
    "离线基线不调用网络模型，而是使用字符 n-gram TF-IDF：\n",
    "\n",
    "1. 把中文文本拆成连续的 1～3 字符片段；\n",
    "2. 计算片段在当前文档中的突出程度（TF）；\n",
    "3. 降低所有文档都常见的片段权重（IDF）；\n",
    "4. 把结果组成向量并单位化；\n",
    "5. 查询向量与资料向量做点积，得到余弦相似度。\n",
    "\n",
    "这个方法便于阅读和复现，但主要依赖文字重叠，**不等同于语义 embedding**。它是断网时用于观察 RAG 机制的确定性基线。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "repr0002",
   "metadata": {},
   "source": [
    "### 查看完整文本表示代码\n",
    "\n",
    "下面不是伪代码，而是从本章参考实现中直接读取实际运行的函数和类。阅读时重点寻找：\n",
    "\n",
    "- n-gram 是在哪里生成的；\n",
    "- TF 与 IDF 在哪里相乘；\n",
    "- 向量在哪里单位化。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "repr0003",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "print(inspect.getsource(rag.char_ngrams))\n",
    "print(inspect.getsource(rag.CharNgramTfidfProvider))\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "repr0004",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "provider = rag.CharNgramTfidfProvider()\n",
    "document_texts = [f\"{chunk.title}。{chunk.text}\" for chunk in chunks]\n",
    "provider.fit(document_texts)\n",
    "document_vectors = provider.encode(document_texts)\n",
    "\n",
    "print(\"知识库向量形状：\", document_vectors.shape)\n",
    "print(\"词表大小：\", len(provider.vocabulary))\n",
    "print(\"第1段向量长度：\", round(float((document_vectors[0] @ document_vectors[0]) ** 0.5), 3))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "search01",
   "metadata": {},
   "source": [
    "## 4. 检索：排序结果不等于可靠证据\n",
    "\n",
    "检索器总能排出“最相近”的 Top-K 段，即使所有资料都不能回答问题。因此本项目把过程分成两步：\n",
    "\n",
    "- **候选排序：** 取相似度最高的 Top-K；\n",
    "- **证据接受：** 只有分数达到固定阈值的候选才能进入生成提示词。\n",
    "\n",
    "相似度分数不是“答案正确概率”。它只能表示：在当前知识库、当前文本表示方法下，这两段文字有多接近。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "search02",
   "metadata": {},
   "source": [
    "### 查看完整检索与回答代码\n",
    "\n",
    "重点阅读 `retrieve()`、`accepted_evidence()`、`build_prompt()` 和 `answer()`。你会看到检索与生成是两个明确分开的环节。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "search03",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "print(inspect.getsource(rag.MinimalRAG.retrieve))\n",
    "print(inspect.getsource(rag.MinimalRAG.accepted_evidence))\n",
    "print(inspect.getsource(rag.MinimalRAG.build_prompt))\n",
    "print(inspect.getsource(rag.MinimalRAG.mock_generate))\n",
    "print(inspect.getsource(rag.MinimalRAG.answer))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "search04",
   "metadata": {},
   "source": [
    "### 运行一次检索\n",
    "\n",
    "先用 Q2 观察：正确段落应当是 `LIBRARY-02`。默认取 2 个候选，但不代表两个候选都会成为证据。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "search05",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "assistant = rag.MinimalRAG(chunks, min_score=rag.MIN_SCORE)\n",
    "question = \"借阅者一次最多借几本书？\"\n",
    "hits = assistant.retrieve(question, top_k=2)\n",
    "\n",
    "for hit in hits:\n",
    "    decision = \"通过阈值\" if hit.score >= assistant.min_score else \"未采用\"\n",
    "    print(f\"{hit.id}｜score={hit.score:.3f}｜{decision}｜{hit.title}\")\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "prompt01",
   "metadata": {},
   "source": [
    "## 5. 拼接与生成：只让模型看到已接受证据\n",
    "\n",
    "`build_prompt()` 把通过阈值的资料和问题组合起来，并明确要求：\n",
    "\n",
    "- 只能依据给定资料；\n",
    "- 没有证据时固定回答“资料中未提到该信息。”；\n",
    "- 每个事实附段落编号。\n",
    "\n",
    "离线 mock 生成器只摘录证据中的句子并附编号。它不是大语言模型，也不追求自然对话；它用于建立全班断网时仍可复现、可核查的最低基线。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "prompt02",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "result = assistant.answer(question, top_k=2)\n",
    "\n",
    "print(\"发送给生成器的提示词：\")\n",
    "print(result[\"prompt\"])\n",
    "print(\"\\n离线回答：\")\n",
    "print(result[\"answer\"])\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "model001",
   "metadata": {},
   "source": [
    "### 可选：接入经教师校验的 Qwen 模型\n",
    "\n",
    "参考实现预留了两个小接口：\n",
    "\n",
    "```python\n",
    "embedding = rag.CallableEmbeddingProvider(my_embed_function, \"approved-qwen-embedding\")\n",
    "assistant = rag.MinimalRAG(chunks, provider=embedding)\n",
    "result = assistant.answer(\n",
    "    question,\n",
    "    generation_provider=my_qwen_chat_function,\n",
    ")\n",
    "```\n",
    "\n",
    "- `my_embed_function(list[str])` 为每段文字返回一个等长向量；\n",
    "- `my_qwen_chat_function(prompt)` 接收已经拼好的证据提示词并返回文本；\n",
    "- API Key 只能从环境变量或学校配置读取，不能写入 Notebook；\n",
    "- 不得向云端上传真实个人资料或未经授权的校务文档；\n",
    "- 接入真实模型后，必须重新运行 6 道固定题和两级验收。\n",
    "\n",
    "网页端 Qwen、DeepSeek 可以用于讨论提示词和核验思路，但手工复制的回答也要记录平台显示的模型名称、日期和完整提示词。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "fixed001",
   "metadata": {},
   "source": [
    "## 6. 六道固定题：形成无检索与有检索对照\n",
    "\n",
    "Q1～Q5 在资料内有明确答案，Q6 是库外题。\n",
    "\n",
    "“无检索回答”使用明确标注的课程模拟猜测，只用于说明没有证据仍强行作答的风险，**不是对 Qwen、DeepSeek 或其他模型的正式评测结果**。“有检索回答”则来自上面的实际离线程序。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "fixed002",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "questions = rag.load_questions(CHAPTER_DIR / \"data/questions.json\")\n",
    "baseline_rows = rag.run_fixed_questions(top_k=2)\n",
    "\n",
    "for item, row in zip(questions, baseline_rows):\n",
    "    print(f\"\\n{item['id']}｜{item['question']}\")\n",
    "    print(\"无检索模拟：\", rag.naive_no_retrieval_answer(item[\"id\"]))\n",
    "    print(\"有检索回答：\", row[\"answer\"])\n",
    "    print(\"通过阈值证据：\", row[\"accepted_ids\"] or \"无（应拒答）\")\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "gate0001",
   "metadata": {},
   "source": [
    "## 7. 两级验收：不要把错误藏进一个总分\n",
    "\n",
    "### 第一级：检索层\n",
    "\n",
    "- 库内题：期望段落是否进入“通过阈值证据”？\n",
    "- 库外题：是否没有任何候选通过阈值？\n",
    "\n",
    "### 第二级：生成层\n",
    "\n",
    "- 回答中的引用编号是否都属于已接受证据？\n",
    "- 库外题是否明确拒答？\n",
    "- 回答是否添加了原文没有的事实？\n",
    "\n",
    "检索错误但回答碰巧正确，仍然不能通过；检索正确但生成器加入资料外内容，同样不能通过。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "gate0002",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "for row in baseline_rows:\n",
    "    retrieval_label = \"通过\" if row[\"retrieval_pass\"] else \"未通过\"\n",
    "    grounding_label = \"通过\" if row[\"grounding_pass\"] else \"未通过\"\n",
    "    print(\n",
    "        f\"{row['id']}｜检索层={retrieval_label}｜\"\n",
    "        f\"生成层={grounding_label}｜证据={row['accepted_ids'] or '无，执行拒答'}\"\n",
    "    )\n",
    "\n",
    "assert all(row[\"retrieval_pass\"] for row in baseline_rows)\n",
    "assert all(row[\"grounding_pass\"] for row in baseline_rows)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "outside1",
   "metadata": {},
   "source": [
    "### 单独观察库外题\n",
    "\n",
    "Q6 与图书馆“开馆时间”有部分文字重叠，因此检索器仍会把图书馆段落排在前面。但分数没有达到阈值，它不能进入证据区，系统必须拒答。\n",
    "\n",
    "这个例子说明：**Top-K 排名靠前不等于资料真正能回答问题。**\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "outside2",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "outside = assistant.answer(\"学校游泳馆几点开门？\", top_k=2)\n",
    "\n",
    "for hit in outside[\"hits\"]:\n",
    "    decision = \"通过阈值\" if hit[\"id\"] in outside[\"accepted_ids\"] else \"未采用\"\n",
    "    print(f\"{hit['id']}｜score={hit['score']:.3f}｜{decision}\")\n",
    "print(\"最终回答：\", outside[\"answer\"])\n",
    "\n",
    "assert outside[\"accepted_ids\"] == []\n",
    "assert outside[\"answer\"] == rag.REFUSAL\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "student1",
   "metadata": {},
   "source": [
    "## 8. 学生主实验：只改变 TOP_K\n",
    "\n",
    "这是本章唯一必改变量。\n",
    "\n",
    "1. 先令 `STUDENT_TOP_K = 1`，运行并记录；\n",
    "2. 在实验日志写下改成 3 后的预测；\n",
    "3. 再令 `STUDENT_TOP_K = 3`，重新运行；\n",
    "4. 保持分块、表示法、阈值和六道题完全不变。\n",
    "\n",
    "如果同时修改多个变量，就无法判断结果究竟由什么引起。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "student2",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "# TODO（学生必改）：先用 1 完成记录，再改成 3。\n",
    "STUDENT_TOP_K = 1\n",
    "\n",
    "student_rows = rag.run_fixed_questions(top_k=STUDENT_TOP_K)\n",
    "for row in student_rows:\n",
    "    candidate_ids = [hit[\"id\"] for hit in row[\"hits\"]]\n",
    "    print(\n",
    "        f\"{row['id']}｜候选={candidate_ids}｜\"\n",
    "        f\"通过阈值={row['accepted_ids'] or '无'}｜回答={row['answer']}\"\n",
    "    )\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "compare1",
   "metadata": {},
   "source": [
    "### 用同一函数并排查看 1 与 3\n",
    "\n",
    "程序只整理客观输出，不会替你完成“为什么”的解释。你要特别检查：Top-K 增大后，是否出现更多同主题但不能回答当前问题的候选。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "compare2",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "def compact_results(top_k):\n",
    "    rows = rag.run_fixed_questions(top_k=top_k)\n",
    "    return {\n",
    "        row[\"id\"]: {\n",
    "            \"candidates\": [hit[\"id\"] for hit in row[\"hits\"]],\n",
    "            \"accepted\": row[\"accepted_ids\"],\n",
    "            \"retrieval_pass\": row[\"retrieval_pass\"],\n",
    "            \"grounding_pass\": row[\"grounding_pass\"],\n",
    "        }\n",
    "        for row in rows\n",
    "    }\n",
    "\n",
    "comparison = {\n",
    "    \"TOP_K=1\": compact_results(1),\n",
    "    \"TOP_K=3\": compact_results(3),\n",
    "}\n",
    "print(json.dumps(comparison, ensure_ascii=False, indent=2))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "cite0001",
   "metadata": {},
   "source": [
    "## 9. 引用核查：编号必须回到真实 Markdown 原文\n",
    "\n",
    "至少抽查 5 条回答。在实验记录中把每条引用标为：\n",
    "\n",
    "- A：原文完全支持；\n",
    "- B：原文部分支持，但回答有简化、合并或改写；\n",
    "- C：原文没有该事实，或者编号不存在。\n",
    "\n",
    "下面的代码只帮助定位原文，等级和处理意见必须由你本人判断。\n"
   ]
  },
  {
   "cell_type": "code",
   "id": "cite0002",
   "metadata": {},
   "execution_count": null,
   "outputs": [],
   "source": [
    "for row in student_rows[:5]:\n",
    "    print(f\"\\n{row['id']}｜回答：{row['answer']}\")\n",
    "    for source_id in row[\"accepted_ids\"]:\n",
    "        source = chunk_by_id[source_id]\n",
    "        print(f\"  {source_id}（{source.source}）：{source.text}\")\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "datacard",
   "metadata": {},
   "source": [
    "## 10. 新增资料之前先写数据卡\n",
    "\n",
    "在教师允许下，可以复制一份知识库到个人工作目录，新增一个有唯一编号的课程自编段落和一道验证题。不要直接覆盖教师核准基准。\n",
    "\n",
    "数据卡至少记录：\n",
    "\n",
    "- 资料名称与来源；\n",
    "- 是课程自编、公开许可还是得到授权；\n",
    "- 版本日期和更新责任人；\n",
    "- 是否包含姓名、学号、联系方式等个人信息；\n",
    "- 可以回答什么、不能回答什么；\n",
    "- 一道验证题及期望段落编号。\n",
    "\n",
    "新增资料后要重新建立索引并运行全部固定题，确认新资料没有让旧题退化。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "ragfine",
   "metadata": {},
   "source": [
    "## 11. 为什么本项目优先 RAG，而不是微调\n",
    "\n",
    "食堂菜单、借阅制度和社团章程都可能更新，回答还要求回到原文核验。本项目因此优先更新外部知识库和检索索引：\n",
    "\n",
    "- **RAG：** 模型参数不变，回答前检索可更新资料；\n",
    "- **微调：** 调整模型参数，更适合相对稳定的任务方式、格式、术语或表达风格；\n",
    "- **组合使用：** 复杂系统可以用微调建立行为能力，再用 RAG 提供动态事实。\n",
    "\n",
    "这不是“RAG 永远优于微调”的结论，而是根据更新频率、可追溯要求、数据成熟度和预算做出的工程选择。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "summary1",
   "metadata": {},
   "source": [
    "## 12. 完成检查与结论\n",
    "\n",
    "提交前确认：\n",
    "\n",
    "- [ ] 我在运行前写了预测；\n",
    "- [ ] Notebook 已从第一格运行到最后一格；\n",
    "- [ ] `TOP_K=1` 与 `TOP_K=3` 各运行一次，其他变量不变；\n",
    "- [ ] 6 道题分别完成检索层和生成层验收；\n",
    "- [ ] Q6 没有证据时明确拒答；\n",
    "- [ ] 至少 5 条引用回到 Markdown 原文核查；\n",
    "- [ ] 我保存了“原始意图—AI建议—人工选择—实际改动—验证”记录；\n",
    "- [ ] 新增资料有数据卡，不含未经许可的个人信息；\n",
    "- [ ] 我能用不超过 200 字说明：RAG 改善了什么，仍不能保证什么。\n",
    "\n",
    "**工程底线：** RAG 可以降低无依据生成风险，但不能自动修正错误资料，也不能保证检索、生成和引用永不出错。可靠系统必须保留拒答、日志、测试和人工核验。\n"
   ]
  }
 ],
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