linear_regression_project.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {},
   "outputs": [],
   "source": [
    "# 任意选一个你喜欢的整数，这能帮你得到稳定的结果\n",
    "seed = 9078"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 欢迎来到线性回归项目\n",
    "\n",
    "若项目中的题目有困难没完成也没关系，我们鼓励你带着问题提交项目，评审人会给予你诸多帮助。\n",
    "\n",
    "所有选做题都可以不做，不影响项目通过。如果你做了，那么项目评审会帮你批改，也会因为选做部分做错而判定为不通过。\n",
    "\n",
    "其中非代码题可以提交手写后扫描的 pdf 文件，或使用 Latex 在文档中直接回答。"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 1 矩阵运算\n",
    "\n",
    "## 1.1 创建一个 4*4 的单位矩阵"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "metadata": {},
   "outputs": [],
   "source": [
    "# 这个项目设计来帮你熟悉 python list 和线性代数\n",
    "# 你不能调用任何NumPy以及相关的科学计算库来完成作业\n",
    "\n",
    "\n",
    "# 本项目要求矩阵统一使用二维列表表示，如下：\n",
    "A = [[1,2,3], \n",
    "     [2,3,3], \n",
    "     [1,2,5]]\n",
    "\n",
    "B = [[1,2,3,5], \n",
    "     [2,3,3,5], \n",
    "     [1,2,5,1]]\n",
    "\n",
    "# 向量也用二维列表表示\n",
    "C = [[1],\n",
    "     [2],\n",
    "     [3]]\n",
    "\n",
    "#TODO 创建一个 4*4 单位矩阵\n",
    "I = [[1,0,0,0],\n",
    "    [0,1,0,0],\n",
    "    [0,0,1,0],\n",
    "    [0,0,0,1]]\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.2 返回矩阵的行数和列数"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO 返回矩阵的行数和列数\n",
    "def shape(M):\n",
    "    row_num = len(M)\n",
    "    col_num = len(M[0]) if row_num >= 1 else 0\n",
    "    return row_num,col_num"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.001s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 shape 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_shape"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.3 每个元素四舍五入到特定小数数位"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO 每个元素四舍五入到特定小数数位\n",
    "# 直接修改参数矩阵，无返回值\n",
    "def matxRound(M, decPts=4):\n",
    "    rowNum, colNum = shape(M)\n",
    "    for r in range(rowNum):\n",
    "        for c in range(colNum):\n",
    "            M[r][c] = round(M[r][c], decPts)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.066s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 matxRound 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_matxRound"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.4 计算矩阵的转置"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 58,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO 计算矩阵的转置\n",
    "def transpose(M):\n",
    "    #谢谢你的建议，这种写法更简洁！！\n",
    "    #这里用到了python函数调用时，传参数的一个语法糖－－分解参数\n",
    "    #zip(*M)中的 *M ， 其中M是一个二维iterable，会被分解多个list，一个list是一行\n",
    "    #这多个list又被zip拉链在一起，就使得每一列成为一个list\n",
    "    return [list(col) for col in zip(*M)]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.008s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 transpose 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_transpose"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.5 计算矩阵乘法 AB"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 60,
   "metadata": {
    "scrolled": true
   },
   "outputs": [],
   "source": [
    "# TODO 计算矩阵乘法 AB，如果无法相乘则raise ValueError\n",
    "def matxMultiply(A, B):\n",
    "    aRowNum,aColNum = shape(A)\n",
    "    bRowNum,bColNum = shape(B)\n",
    "#     print aRowNum, aColNum\n",
    "#     print bRowNum, bColNum\n",
    "    if (aColNum!=bRowNum and aRowNum!=bColNum) or aColNum==0 or bRowNum==0 or aRowNum==0 or bColNum==0:\n",
    "        raise ValueError('cannot mulitply the matrixs or dimension is/are 0!')\n",
    "    transformedB = transpose(B)\n",
    "    newM = []\n",
    "    for aRow in A:\n",
    "        newRow = [sum([ab[0]*ab[1] for ab in zip(aRow, bCol)]) for bCol in transformedB]\n",
    "        newM.append(newRow)\n",
    "    return newM"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 61,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.057s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 matxMultiply 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_matxMultiply"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "\n",
    "# 2 Gaussign Jordan 消元法\n",
    "\n",
    "## 2.1 构造增广矩阵\n",
    "\n",
    "$ A = \\begin{bmatrix}\n",
    "    a_{11}    & a_{12} & ... & a_{1n}\\\\\n",
    "    a_{21}    & a_{22} & ... & a_{2n}\\\\\n",
    "    a_{31}    & a_{22} & ... & a_{3n}\\\\\n",
    "    ...    & ... & ... & ...\\\\\n",
    "    a_{n1}    & a_{n2} & ... & a_{nn}\\\\\n",
    "\\end{bmatrix} , b = \\begin{bmatrix}\n",
    "    b_{1}  \\\\\n",
    "    b_{2}  \\\\\n",
    "    b_{3}  \\\\\n",
    "    ...    \\\\\n",
    "    b_{n}  \\\\\n",
    "\\end{bmatrix}$\n",
    "\n",
    "返回 $ Ab = \\begin{bmatrix}\n",
    "    a_{11}    & a_{12} & ... & a_{1n} & b_{1}\\\\\n",
    "    a_{21}    & a_{22} & ... & a_{2n} & b_{2}\\\\\n",
    "    a_{31}    & a_{22} & ... & a_{3n} & b_{3}\\\\\n",
    "    ...    & ... & ... & ...& ...\\\\\n",
    "    a_{n1}    & a_{n2} & ... & a_{nn} & b_{n} \\end{bmatrix}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 62,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO 构造增广矩阵，假设A，b行数相同\n",
    "def augmentMatrix(A, b):\n",
    "    aRowNum,aColNum = shape(A)\n",
    "    bRowNum,bColNum = shape(b)\n",
    "    if aRowNum!=bRowNum or aRowNum==0 or aColNum==0 or bRowNum==0 or bColNum!=1:\n",
    "        raise ValueError('bad formed matrix in augmentMatrix!')\n",
    "    #更简洁的写法！！\n",
    "    return [ra+rb for ra,rb in zip(A,b)]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 63,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.004s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 augmentMatrix 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_augmentMatrix"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 2.2 初等行变换\n",
    "- 交换两行\n",
    "- 把某行乘以一个非零常数\n",
    "- 把某行加上另一行的若干倍："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 64,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO r1 <---> r2\n",
    "# 直接修改参数矩阵，无返回值\n",
    "def swapRows(M, r1, r2):\n",
    "    M[r1],M[r2] = M[r2],M[r1]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 65,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.002s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 swapRows 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_swapRows"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 66,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO r1 <--- r1 * scale\n",
    "# scale为0是非法输入，要求 raise ValueError\n",
    "# 直接修改参数矩阵，无返回值\n",
    "def scaleRow(M, r, scale):\n",
    "    if scale==0:\n",
    "        raise ValueError(\"(scaleRow) scale should not be 0!\")\n",
    "    M[r] = [i*scale for i in M[r]]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 67,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.003s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 scaleRow 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_scaleRow"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 68,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO r1 <--- r1 + r2*scale\n",
    "# 直接修改参数矩阵，无返回值\n",
    "def addScaledRow(M, r1, r2, scale):\n",
    "    M[r1] = [t[0]+t[1]*scale for t in zip(M[r1], M[r2])]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 69,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 0.002s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 addScaledRow 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_addScaledRow"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 2.3  Gaussian Jordan 消元法求解 Ax = b"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 2.3.1 算法\n",
    "\n",
    "步骤1 检查A，b是否行数相同\n",
    "\n",
    "步骤2 构造增广矩阵Ab\n",
    "\n",
    "步骤3 逐列转换Ab为化简行阶梯形矩阵 [中文维基链接](https://zh.wikipedia.org/wiki/%E9%98%B6%E6%A2%AF%E5%BD%A2%E7%9F%A9%E9%98%B5#.E5.8C.96.E7.AE.80.E5.90.8E.E7.9A.84-.7Bzh-hans:.E8.A1.8C.3B_zh-hant:.E5.88.97.3B.7D-.E9.98.B6.E6.A2.AF.E5.BD.A2.E7.9F.A9.E9.98.B5)\n",
    "    \n",
    "    对于Ab的每一列（最后一列除外）\n",
    "        当前列为列c\n",
    "        寻找列c中 对角线以及对角线以下所有元素（行 c~N）的绝对值的最大值\n",
    "        如果绝对值最大值为0\n",
    "            那么A为奇异矩阵，返回None (你可以在选做问题2.4中证明为什么这里A一定是奇异矩阵)\n",
    "        否则\n",
    "            使用第一个行变换，将绝对值最大值所在行交换到对角线元素所在行（行c） \n",
    "            使用第二个行变换，将列c的对角线元素缩放为1\n",
    "            多次使用第三个行变换，将列c的其他元素消为0\n",
    "            \n",
    "步骤4 返回Ab的最后一列\n",
    "\n",
    "**注：** 我们并没有按照常规方法先把矩阵转化为行阶梯形矩阵，再转换为化简行阶梯形矩阵，而是一步到位。如果你熟悉常规方法的话，可以思考一下两者的等价性。"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 2.3.2 算法推演\n",
    "\n",
    "为了充分了解Gaussian Jordan消元法的计算流程，请根据Gaussian Jordan消元法，分别手动推演矩阵A为***可逆矩阵***，矩阵A为***奇异矩阵***两种情况。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 70,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "   8,   3,  -5,  -9 ||  1  \n",
      "  -1,   7,   7,   8 ||  1  \n",
      "  -5,   7,   6,  -7 ||  1  \n",
      "   1,  -2,  -5,   2 ||  1  \n"
     ]
    }
   ],
   "source": [
    "# 不要修改这里！\n",
    "from helper import *\n",
    "\n",
    "A = generateMatrix(4,seed,singular=False)\n",
    "b = np.ones(shape=(4,1)) # it doesn't matter\n",
    "Ab = augmentMatrix(A.tolist(),b.tolist()) # please make sure you already correct implement augmentMatrix\n",
    "printInMatrixFormat(Ab,padding=4,truncating=0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "请按照算法的步骤3，逐步推演***可逆矩阵***的变换。\n",
    "\n",
    "在下面列出每一次循环体执行之后的增广矩阵。\n",
    "\n",
    "要求：\n",
    "1. 做分数运算\n",
    "2. 使用`\\frac{n}{m}`来渲染分数，如下：\n",
    " - $\\frac{n}{m}$\n",
    " - $-\\frac{a}{b}$\n",
    "\n",
    "增广矩阵\n",
    "$ Ab = \\begin{bmatrix}\n",
    "    8 & 3 & -5 & -9 & 1\\\\\n",
    "    -1 & 7 & 7 & 8 & 1\\\\\n",
    "    -5 & 7 & 6 & -7 & 1\\\\\n",
    "    1 & -2 & -5 & 2 & 1\\end{bmatrix}$\n",
    "\n",
    "$ --> \\begin{bmatrix}\n",
    "    1 & \\frac{3}{8} & \\frac{-5}{8} & \\frac{-9}{8} & \\frac{1}{8}\\\\\n",
    "   0 & \\frac{59}{8} & \\frac{51}{8} & \\frac{55}{8} & \\frac{9}{8}\\\\\n",
    "   0 & \\frac{71}{8} & \\frac{23}{8} & \\frac{-101}{8} & \\frac{13}{8}\\\\\n",
    "   0 & \\frac{-19}{8} & \\frac{-35}{8} & \\frac{25}{8} & \\frac{7}{8}\\end{bmatrix}$\n",
    "    \n",
    "$ --> \\begin{bmatrix}\n",
    "    1 & 0 & \\frac{-53}{71} & \\frac{-42}{71} & \\frac{4}{71}\\\\\n",
    "   0 & 1 & \\frac{23}{71} & \\frac{-101}{71} & \\frac{13}{71}\\\\\n",
    "   0 & 0 & \\frac{283}{71} & \\frac{1233}{71} & \\frac{-16}{71}\\\\\n",
    "   0 & 0 & \\frac{-256}{71} & \\frac{-18}{71} & \\frac{93}{71}\\end{bmatrix}$\n",
    "    \n",
    "$ --> \\begin{bmatrix}\n",
    "    1 & 0 & 0 & \\frac{753}{283} & \\frac{4}{283}\\\\\n",
    "   0 & 1 & 0 & \\frac{-802}{283} & \\frac{57}{283}\\\\\n",
    "   0 & 0 & 1 & \\frac{1233}{283} & \\frac{-16}{283}\\\\\n",
    "   0 & 0 & 0 & \\frac{4374}{283} & \\frac{313}{283}\\end{bmatrix}$\n",
    "   \n",
    "$ --> \\begin{bmatrix}\n",
    "   1 & 0 & 0 & 0 & \\frac{-257}{1458}\\\\\n",
    "   0 & 1 & 0 & 0 & \\frac{884}{2187}\\\\\n",
    "   0 & 0 & 1 & 0 & \\frac{-179}{486}\\\\\n",
    "   0 & 0 & 0 & 1 & \\frac{313}{4374}\\end{bmatrix}$   \n",
    "$...$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 71,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "   0,  -7,   2,   2 ||  1  \n",
      "  -7,  -9,   0,  -9 ||  1  \n",
      "  -7,  -4,   4,  -5 ||  1  \n",
      "  -7,  -9,  -1, -10 ||  1  \n"
     ]
    }
   ],
   "source": [
    "# 不要修改这里！\n",
    "A = generateMatrix(4,seed,singular=True)\n",
    "b = np.ones(shape=(4,1)) # it doesn't matter\n",
    "Ab = augmentMatrix(A.tolist(),b.tolist()) # please make sure you already correct implement augmentMatrix\n",
    "printInMatrixFormat(Ab,padding=4,truncating=0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "请按照算法的步骤3，逐步推演***奇异矩阵***的变换。\n",
    "\n",
    "在下面列出每一次循环体执行之后的增广矩阵。\n",
    "\n",
    "要求：\n",
    "1. 做分数运算\n",
    "2. 使用`\\frac{n}{m}`来渲染分数，如下：\n",
    " - $\\frac{n}{m}$\n",
    " - $-\\frac{a}{b}$\n",
    "\n",
    "增广矩阵\n",
    "$ Ab = \\begin{bmatrix}\n",
    "    0 & -7 & 2 & 2 & 1\\\\\n",
    "    -7 & -9 & 0 & -9 & 1\\\\\n",
    "    -7 & -4 & 4 & -5 & 1\\\\\n",
    "    -7 & -9 & -1 & -10 & 1\\end{bmatrix}$\n",
    "\n",
    "$ --> \\begin{bmatrix}\n",
    "    1 & \\frac{9}{7} & 0 & \\frac{9}{7} & \\frac{-1}{7}\\\\\n",
    "   0 & -7 & 2 & 2 & 1\\\\\n",
    "   0 & 5 & 4 & 4 & 0\\\\\n",
    "   0 & 0 & -1 & -1 & 0\\end{bmatrix}$\n",
    "    \n",
    "$ --> \\begin{bmatrix}\n",
    "    1 & 0 & \\frac{18}{49} & \\frac{81}{49} & \\frac{2}{49}\\\\\n",
    "   0 & 1 & \\frac{-2}{7} & \\frac{-2}{7} & \\frac{-1}{7}\\\\\n",
    "   0 & 0 & \\frac{38}{7} & \\frac{38}{7} & \\frac{5}{7}\\\\\n",
    "   0 & 0 & -1 & -1 & 0\\end{bmatrix}$\n",
    "    \n",
    "$ --> \\begin{bmatrix}\n",
    "    1 & 0 & 0 & \\frac{9}{7} & \\frac{-1}{133}\\\\\n",
    "   0 & 1 & 0 & 0 & \\frac{-2}{19}\\\\\n",
    "   0 & 0 & 1 & 1 & \\frac{5}{38}\\\\\n",
    "   0 & 0 & 0 & 0 & \\frac{5}{38}\\end{bmatrix}$    \n",
    "\n",
    "当c为4的时候，对角线及对角线以下的所有第四列元素的绝对值的最大值为0，所以是奇异矩阵。"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 2.3.3 实现 Gaussian Jordan 消元法"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 72,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 , 9/7 , 0 , 9/7 , -1/7\n",
      "0 , -7 , 2 , 2 , 1\n",
      "0 , 5 , 4 , 4 , 0\n",
      "0 , 0 , -1 , -1 , 0\n",
      "#########\n",
      "   1 & \\frac{9}{7} & 0 & \\frac{9}{7} & \\frac{-1}{7}\\\\\n",
      "   0 & -7 & 2 & 2 & 1\\\\\n",
      "   0 & 5 & 4 & 4 & 0\\\\\n",
      "   0 & 0 & -1 & -1 & 0\\\\\n",
      "------------------\n",
      "1 , 0 , 18/49 , 81/49 , 2/49\n",
      "0 , 1 , -2/7 , -2/7 , -1/7\n",
      "0 , 0 , 38/7 , 38/7 , 5/7\n",
      "0 , 0 , -1 , -1 , 0\n",
      "#########\n",
      "   1 & 0 & \\frac{18}{49} & \\frac{81}{49} & \\frac{2}{49}\\\\\n",
      "   0 & 1 & \\frac{-2}{7} & \\frac{-2}{7} & \\frac{-1}{7}\\\\\n",
      "   0 & 0 & \\frac{38}{7} & \\frac{38}{7} & \\frac{5}{7}\\\\\n",
      "   0 & 0 & -1 & -1 & 0\\\\\n",
      "------------------\n",
      "1 , 0 , 0 , 9/7 , -1/133\n",
      "0 , 1 , 0 , 0 , -2/19\n",
      "0 , 0 , 1 , 1 , 5/38\n",
      "0 , 0 , 0 , 0 , 5/38\n",
      "#########\n",
      "   1 & 0 & 0 & \\frac{9}{7} & \\frac{-1}{133}\\\\\n",
      "   0 & 1 & 0 & 0 & \\frac{-2}{19}\\\\\n",
      "   0 & 0 & 1 & 1 & \\frac{5}{38}\\\\\n",
      "   0 & 0 & 0 & 0 & \\frac{5}{38}\\\\\n",
      "------------------\n"
     ]
    }
   ],
   "source": [
    "# output of gj_Solve_in_fraction are only used to fill answer sheet for the two questions above\n",
    "\n",
    "from fractions import Fraction\n",
    "def grace_print(mf):\n",
    "    def fraction_markdown(tf):\n",
    "        if tf.denominator==1:\n",
    "            return str(tf.numerator)\n",
    "        else:\n",
    "            return \"\\\\frac{%d}{%d}\" % (tf.numerator, tf.denominator)\n",
    "    print '\\n'.join([' , '.join([one.__str__() for one in oneRow]) for oneRow in mf])\n",
    "    print '#'*9\n",
    "    print '\\n'.join(['   '+' & '.join([fraction_markdown(one) for one in oneRow])+'\\\\\\\\' for oneRow in mf])\n",
    "    \n",
    "\n",
    "def gj_Solve_in_fraction(A, b, decPts=4, epsilon = 1.0e-16):\n",
    "    rowNum, colNum = shape(A)\n",
    "    lineSys = augmentMatrix(A,b)\n",
    "    lineSys = [[Fraction(oneItem,1) for oneItem in oneRow] for oneRow in lineSys]\n",
    "    for c in range(colNum):\n",
    "        absCols = [abs(lineSys[r][c]) if r>=c else -1 for r in range(rowNum)] \n",
    "        if(max(absCols)<=epsilon):\n",
    "            return None\n",
    "        maxIndex = absCols.index(max(absCols))\n",
    "        swapRows(lineSys, c, maxIndex)\n",
    "        scaleRow(lineSys, c, 1/lineSys[c][c])\n",
    "        for inr in range(rowNum):\n",
    "            if inr==c: continue\n",
    "            addScaledRow(lineSys, inr, c, -1*lineSys[inr][c])\n",
    "        grace_print(lineSys)\n",
    "        print '-'*18\n",
    "    return [[r[-1]] for r in lineSys]\n",
    "# myA = [[8,3,-5,-9], [-1,7,7,8], [-5,7,6,-7], [1,-2,-5,2]]\n",
    "myA = [[0,  -7,   2,   2],[-7,  -9,   0,  -9],[-7,  -4,   4,  -5],[-7,  -9,  -1, -10]]\n",
    "myb = [[1],[1],[1],[1]]\n",
    "gj_Solve_in_fraction(myA, myb)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 73,
   "metadata": {},
   "outputs": [],
   "source": [
    "# TODO 实现 Gaussain Jordan 方法求解 Ax = b\n",
    "\n",
    "\"\"\" Gaussian Jordan 方法求解 Ax = b.\n",
    "    参数\n",
    "        A: 方阵 \n",
    "        b: 列向量\n",
    "        decPts: 四舍五入位数，默认为4\n",
    "        epsilon: 判读是否为0的阈值，默认 1.0e-16\n",
    "        \n",
    "    返回列向量 x 使得 Ax = b \n",
    "    返回None，如果 A，b 高度不同\n",
    "    返回None，如果 A 为奇异矩阵\n",
    "\"\"\"\n",
    "\n",
    "def gj_Solve(A, b, decPts=4, epsilon = 1.0e-16):\n",
    "    rowNum, colNum = shape(A)\n",
    "    lineSys = augmentMatrix(A,b)\n",
    "    for c in range(colNum):\n",
    "        #这里我觉得应该不会有审阅中提到的可能的bug\n",
    "        #因为在下面的列表表达式里，所有比c小的行，它们的c列的值我都取为了－1，\n",
    "        #所以下面的max(absCols)应该不会找到那些比c小的行，只会往下找，即便下面有多个相同的max值，我就取第一个被找到的行。\n",
    "        absCols = [abs(lineSys[r][c]) if r>=c else -1 for r in range(rowNum)] \n",
    "        if(max(absCols)<=epsilon):\n",
    "            return None\n",
    "        maxIndex = absCols.index(max(absCols))\n",
    "        swapRows(lineSys, c, maxIndex)\n",
    "        scaleRow(lineSys, c, 1.0/lineSys[c][c])\n",
    "        for inr in range(rowNum):\n",
    "            if inr==c: continue\n",
    "            addScaledRow(lineSys, inr, c, -1.0*lineSys[inr][c])\n",
    "            \n",
    "    matxRound(lineSys, decPts)\n",
    "    return [[r[-1]] for r in lineSys]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 74,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      ".\n",
      "----------------------------------------------------------------------\n",
      "Ran 1 test in 3.003s\n",
      "\n",
      "OK\n"
     ]
    }
   ],
   "source": [
    "# 运行以下代码测试你的 gj_Solve 函数\n",
    "%run -i -e test.py LinearRegressionTestCase.test_gj_Solve"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## (选做) 2.4 算法正确判断了奇异矩阵：\n",
    "\n",
    "在算法的步骤3 中，如果发现某一列对角线和对角线以下所有元素都为0，那么则断定这个矩阵为奇异矩阵。\n",
    "\n",
    "我们用正式的语言描述这个命题，并证明为真。\n",
    "\n",
    "证明下面的命题：\n",
    "\n",
    "**如果方阵 A 可以被分为4个部分: ** \n",
    "\n",
    "$ A = \\begin{bmatrix}\n",
    "    I    & X \\\\\n",
    "    Z    & Y \\\\\n",
    "\\end{bmatrix} , \\text{其中 I 为单位矩阵，Z 为全0矩阵，Y 的第一列全0}$，\n",
    "\n",
    "**那么A为奇异矩阵。**\n",
    "\n",
    "提示：从多种角度都可以完成证明\n",
    "- 考虑矩阵 Y 和 矩阵 A 的秩\n",
    "- 考虑矩阵 Y 和 矩阵 A 的行列式\n",
    "- 考虑矩阵 A 的某一列是其他列的线性组合"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "TODO 证明："
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 3  线性回归"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 3.1 随机生成样本点"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 75,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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Xb+AE1e4hUKDT1tY/7N5d33LOrA7+4MH05HiTNQd9+wZOIV/3ECjQG3XWK5Tt+JusOejb\nN3AK+bqHQIFGtT1lUleFddmOv8ltQPr2DZxCvu4hUKAxIadMQlRTjwYpqZ4g2bdv4Jyn3UFFqvJi\nv1GZHadxttcepyI4hmrqOtsQw9+DfhKV2Qit6vbaS0vSOedIjz22eu3QQ6Urrij2rTOGauq628DR\npGgCldmoVZWpnKpTJuefvz5ISMnv559fpKVxzOnX3QaOJkVIBApsqGquoWrS8pFHyl0fFcOcfgxt\nAOoSLFCY2RVm9pCZ3bnm2iVm9m0z+4aZfdbMnh2qfVhVdXlmqKRlDKtqYmgDUJeQI4qrJJ0ycu1G\nSce7+yslfVfS+9puFJ6qyjTKcKrqrLOS3z/1qeJTJlu2lLs+KoZVNTG0AahLsEDh7jdLenTk2hd9\n9VjVWyUd03rD8BRlp1HGXRZ76aXS5s3rr23enFwvau2c/kUXJaOftre/IK+Avog5R3GOpC+EbgTK\nT6OMW0m8sCBdeeX6b+NXXplcL5tU79v2F0AIQZfHmtmcpOvd/fiR64uS5iW93TMaaGY7Je2UpJmZ\nme0raWsRUZsyyzOrLost0oadO9cHoenp/CmdGJbKArEqujw2ukBhZmdL+n1JJ7l7yv6dT0UdRVya\n6pyrvG5TQQvog07WUZjZKZLeI+mtRYME4tPUip8qSXWWqQLjC7k89mpJt0g6zszuM7NzJX1c0jMl\n3Whmd5jZJ0K1D9U1teKnSqfPMlVgfCFXPZ3h7ke5+2Z3P8bdL3f3F7n7se7+6sHtHaHah/FUXfGT\nl6yu0umzTBUYX1RTT5hsG61QotMHwmBTQESjiSR4lZVSwKToZDIb4+n6ucpNbObXt9PhgBAIFD1R\ntLAs5mDSxAqlGHaSBbqOQNETRb45x16l3MQKJZbHAuMjUPREkW/ObU3DVB21NJGsZnksMD6S2T1R\nJBHcRpVyjMljTocD0pHMnjBFvjm3MQ0TY/KYXVyB8RAoeuSww1Z/3rLlqd/i25iGydqbsY7kccyJ\neKDPNoVuAMaXNt3zk5889XHDoLG4mHToU1Prv+2P+017aSmZxkqb3hp31DL6Nw4T8RIjBKBpjCh6\noMx0z8LC6sjiySeTa3WtflpczM6BjDtqaXJKi5EKkI9A0QNlawWa6nSz3s99/G/9TdVDhFwyTIBC\nVxAoOiatcymbpG6q0816v9nZ8V4377XHndIKlXyPvaYFWItA0SFZncupp5ZLUjfV6TaZLI/hjIs6\nRwAxrg4DMrl752/bt2/3Ptizx3121t0s+XfPnvX3z866JyFi/W342Lznjr7P9PT615iezn/OuH9D\nmfaVfe1x5H2mo+9d52eW9p7DG9AWScteoI8N3snXcetDoCjSEZmldyxm1d6v7k43772aCkxtta1o\nQClqair99aamxv2LgOIIFJEZZ7RQ5jExir3dRYJmnUHaPf21GFGgbUUDBTmKFhRJXGbNla+srM6H\nV5mnj2FlTew7uBap3K47r5OV4K8j8Q/UjUDRgiKJy7wOZ21xWZlN88ZZWVNngOnDDq51J9PZrBCd\nUmTYEfst9qmnrGmLtYnotLnycadqqk751J1TiDlHUUbdeZ0280RAGhWcemL32BZk7ew6NNxdVVrd\nXiNN2V1eq+4W29SRpOzgCsSl6O6xBIoWpO3FNGptJ1xXR131ddrYjhxAeGwzHpG1B/JkWZvYrWv+\nuurr9CGnAKA+BIqWDFfWZAWLtZ1w2klvO3YkUzdlkstVT4xrKtEawwosABUUSWTEfos9mb1WlcRu\nG8ng0cTqeefVn7jtQ0Ib6BPFXnAn6QpJD0m6c821IyXdKOmewb9HFHmtLgUK9/KrXZouWGujE4+9\n6A6YREUDRcipp6sknTJy7UJJX3b3F0v68uD33il7NGfTBWtVNqgrO40Ue9EdgGzBAoW73yzp0ZHL\np0vaPfh5t6S3tdqoSOUll+uY98/rxNNev0ohHwlyoMOKDDuaukma0/qpp/9a87Ot/T3luTslLUta\nnpmZqXc8FpmsqaHzzis+ZZQ33ZU1LbRlS/rrb9lSfhqJHAUQH8Weo/ANAsXg9/8s8jpdy1FUkdbR\n17VFdtb9WQEh67bRBnlUIgNxKRooghbcmdmcpOvd/fjB79+RdKK7P2hmR0n6irsft9HrxF5w15Si\nhXFFCu/SKqfPOiv99bOMU7kNoH1dLbj7vKQdg593SLouYFuiV3Tev0giOS3BnvX6T396EozWYkM7\noL+CBQozu1rSLZKOM7P7zOxcSRdLeouZ3SPp5MHvyFC0MK5qIjnt9Tdvlh5/fP1IwywpCGTvJqCf\nQq56OsPdj3L3ze5+jLtf7u6PuPtJ7v5idz/Z3UdXRXVOk9XIRSuvq1Zap73+4YdLjz22/nHu0g03\njP/3AIgTmwI2KG0zwOFOsW1/+65r91Y2DAT6o6s5il6pUshWVtERS9kivyzUQwCTh0DRkKWl7HMl\n6qpGHucEu6o4mQ2YPASKBgw78Cx1fftuY8QyquqOtAC6ixxFA/JOtKszR0G+AMA4yFEElDe1VOe3\nb/IFANpAoGhAVkc9O1vvFA35AgBtmPhA0USdQ9UOvGxbyBcAaEWRDaFiv1XdFHCcHU032uCu7AZ4\nse2uygZ+QP+pC5sC1qVqMrvIZnlpmiikq9qWJsRUKAigOUWT2RMdKKquGmqiU49pBVNMQQtAc1j1\nVEDVVUNZq5pWVp6aVyiad4hpBRPHlgJYa6IDRdmk87DTzxuEra2MLlM5HdMKppiCFoAIFElkxH4b\n54S7oknbtGRz1m14wlzRE+jKtqVpsSXWATRDJLPrlVdtPWqYV4gp71BWXbvNAohX0RzFpjYa0wdl\n5ueHUzQzM+nBpQtTOAsLBAYAiYnOUZSR1bnnHQkaU94BAKoiUBSU1em/4x3ZldFUTgPoA6aeChp2\n7mXn7ZnCAdB1BIoS6PQBTCKmngAAuQgUAIBcBAoAQC4CBQAgV5SBwszeZWZ3mdmdZna1mT0tdJsA\nYFJFFyjM7GhJfyRp3t2PlzQl6bfDtgoAJld0gWJgk6TDzGyTpGlJDwRuDwBMrOgChbvfL+nDkvZJ\nelDSj9z9i2FbBQCTK7pAYWZHSDpd0vMlPU/S083szJTH7TSzZTNb3r9/f9vNBICJEV2gkHSypO+7\n+353f1zStZJeP/ogd9/l7vPuPr9t27bWGwkAkyLGQLFP0glmNm1mJukkSXcHbhMATKzoAoW73ybp\n7yXdLumbStq4K1R7ip55DQB9FeWmgO7+AUkfCN2O4ZnXBw4kvw/PvJbYHBDA5IhuRBGTxcXVIDF0\n4EByHQAmBYEiR9bxp2WORQWAriNQ5Mg6/rQLZ14DQF0IFDk48xoACBS5OPMaACJd9RQTjj8FMOkY\nUQAAchEoAAC5CBQAgFwECgBALgIFACCXuXvoNozNzPZLWgncjK2SHg7chljwWazis1jFZ7Eqls9i\n1t03PKehF4EiBma27O7zodsRAz6LVXwWq/gsVnXts2DqCQCQi0ABAMhFoKhPsMOVIsRnsYrPYhWf\nxapOfRbkKAAAuRhRAAByESgaYGbvNjM3s62h2xKKmV1iZt82s2+Y2WfN7Nmh29Q2MzvFzL5jZt8z\nswtDtycUMzvWzG4ys2+Z2V1mdn7oNoVmZlNm9m9mdn3othRBoKiZmR0r6ZclTfo5eDdKOt7dXynp\nu5LeF7g9rTKzKUl/JelXJb1M0hlm9rKwrQrmCUnvdveXSTpB0h9M8GcxdL6ku0M3oigCRf0+Kuk9\nkiY6+ePuX3T3Jwa/3irpmJDtCeC1kr7n7v/u7o9J+rSk0wO3KQh3f9Ddbx/8/N9KOsijw7YqHDM7\nRtKvSbosdFuKIlDUyMxOl3S/u389dFsic46kL4RuRMuOlvSDNb/fpwnuHIfMbE7SayTdFrYlQX1M\nyZfJg6EbUhQHF5VkZl+S9NyUuxYlvV/JtNNEyPss3P26wWMWlUw9LLXZNsTHzJ4h6RpJF7j7j0O3\nJwQzO03SQ+6+18xODN2eoggUJbn7yWnXzewVkp4v6etmJiVTLbeb2Wvd/YctNrE1WZ/FkJmdLek0\nSSf55K3Dvl/SsWt+P2ZwbSKZ2WYlQWLJ3a8N3Z6A3iDprWZ2qqSnSTrczPa4+5mB25WLOoqGmNm9\nkubdPYaNv1pnZqdI+oikN7r7/tDtaZuZbVKSxD9JSYD4V0m/4+53BW1YAJZ8c9ot6VF3vyB0e2Ix\nGFH8sbufFrotGyFHgaZ8XNIzJd1oZneY2SdCN6hNg0T+OyX9g5Lk7WcmMUgMvEHSWZLePPi/cMfg\nGzU6ghEFACAXIwoAQC4CBQAgF4ECAJCLQAEAyEWgAADkIlAAAHIRKAAAuQgUAIBcBAoAQC4CBVAj\nM9tkZv9sZv9rZi8duW/n4OTDPwvVPqAKtvAAamZms5LukLQi6efd/Wdm9nIlGwPulXSiuz8Zso1A\nGYwogJq5+4qkcyW9StJfmtlhkv5W0k8lLRAk0DWMKICGmNlfSzpP0r9Ier2k35jwsxjQUQQKoCFm\n9jRJd0p6oaRPuvvOwE0CKmHqCWjOqyTNDH4+fnCYEdA5BAqgAWZ2uKSrJT2s5Dz110n6YNBGARXx\nDQdoxi5Js5Le4u7/aGavkXShmX3J3W8K3DagFHIUQM3M7FxJl0n6c3dfHFx7tpIls5slvdLdHwnY\nRKAUAgVQo0GR3V4lQeGNg7Ozh/e9TtLNkr7g7m8N1ESgNAIFACAXyWwAQC4CBQAgF4ECAJCLQAEA\nyEWgAADkIlAAAHIRKAAAuQgUAIBcBAoAQC4CBQAg1/8B+Q5HZQDCpREAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10fac9810>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# 不要修改这里！\n",
    "# 运行一次就够了！\n",
    "from helper import *\n",
    "from matplotlib import pyplot as plt\n",
    "%matplotlib inline\n",
    "\n",
    "X,Y = generatePoints(seed,num=100)\n",
    "\n",
    "## 可视化\n",
    "plt.xlim((-5,5))\n",
    "plt.xlabel('x',fontsize=18)\n",
    "plt.ylabel('y',fontsize=18)\n",
    "plt.scatter(X,Y,c='b')\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 3.2 拟合一条直线\n",
    "\n",
    "### 3.2.1 猜测一条直线"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 76,
   "metadata": {},
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "/usr/local/lib/python2.7/site-packages/matplotlib/cbook/deprecation.py:106: MatplotlibDeprecationWarning: Adding an axes using the same arguments as a previous axes currently reuses the earlier instance.  In a future version, a new instance will always be created and returned.  Meanwhile, this warning can be suppressed, and the future behavior ensured, by passing a unique label to each axes instance.\n",
      "  warnings.warn(message, mplDeprecation, stacklevel=1)\n"
     ]
    },
    {
     "data": {
      "image/png": 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cZ87M6NjZFuLz9V1EoSK7xC1qolDTUwlJ1vyy116FXVgomWTNSw1NXQBMnAhV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\nQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10fb66a90>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#TODO 请选择最适合的直线 y = mx + b\n",
    "m = 1.2\n",
    "b = 13\n",
    "\n",
    "# 不要修改这里！\n",
    "plt.xlim((-5,5))\n",
    "x_vals = plt.axes().get_xlim()\n",
    "y_vals = [m*x+b for x in x_vals]\n",
    "plt.plot(x_vals, y_vals, '-', color='r')\n",
    "\n",
    "plt.xlabel('x',fontsize=18)\n",
    "plt.ylabel('y',fontsize=18)\n",
    "plt.scatter(X,Y,c='b')\n",
    "\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 3.2.2 计算平均平方误差 (MSE)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "我们要编程计算所选直线的平均平方误差(MSE), 即数据集中每个点到直线的Y方向距离的平方的平均数，表达式如下：\n",
    "$$\n",
    "MSE = \\frac{1}{n}\\sum_{i=1}^{n}{(y_i - mx_i - b)^2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 77,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.36008547953\n"
     ]
    }
   ],
   "source": [
    "# TODO 实现以下函数并输出所选直线的MSE\n",
    "\n",
    "def calculateMSE(X,Y,m,b):\n",
    "    powList = [pow((y-m*x-b), 2) for x,y in zip(X,Y)]\n",
    "    return sum(powList)*1.0/len(powList)\n",
    "\n",
    "print(calculateMSE(X,Y,m,b))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 3.2.3 调整参数 $m, b$ 来获得最小的平方平均误差\n",
    "\n",
    "你可以调整3.2.1中的参数 $m,b$ 让蓝点均匀覆盖在红线周围，然后微调 $m, b$ 让MSE最小。"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 3.3 (选做) 找到参数 $m, b$ 使得平方平均误差最小\n",
    "\n",
    "**这一部分需要简单的微积分知识(  $ (x^2)' = 2x $ )。因为这是一个线性代数项目，所以设为选做。**\n",
    "\n",
    "刚刚我们手动调节参数，尝试找到最小的平方平均误差。下面我们要精确得求解 $m, b$ 使得平方平均误差最小。\n",
    "\n",
    "定义目标函数 $E$ 为\n",
    "$$\n",
    "E = \\frac{1}{2}\\sum_{i=1}^{n}{(y_i - mx_i - b)^2}\n",
    "$$\n",
    "\n",
    "因为 $E = \\frac{n}{2}MSE$, 所以 $E$ 取到最小值时，$MSE$ 也取到最小值。要找到 $E$ 的最小值，即要找到 $m, b$ 使得 $E$ 相对于 $m$, $E$ 相对于 $b$ 的偏导数等于0. \n",
    "\n",
    "因此我们要解下面的方程组。\n",
    "\n",
    "$$\n",
    "\\begin{cases}\n",
    "\\displaystyle\n",
    "\\frac{\\partial E}{\\partial m} =0 \\\\\n",
    "\\\\\n",
    "\\displaystyle\n",
    "\\frac{\\partial E}{\\partial b} =0 \\\\\n",
    "\\end{cases}\n",
    "$$\n",
    "\n",
    "### 3.3.1 计算目标函数相对于参数的导数\n",
    "首先我们计算两个式子左边的值\n",
    "\n",
    "证明/计算：\n",
    "$$\n",
    "\\frac{\\partial E}{\\partial m} = \\sum_{i=1}^{n}{-x_i(y_i - mx_i - b)}\n",
    "$$\n",
    "\n",
    "$$\n",
    "\\frac{\\partial E}{\\partial b} = \\sum_{i=1}^{n}{-(y_i - mx_i - b)}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "TODO 证明:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 3.3.2 实例推演\n",
    "\n",
    "现在我们有了一个二元二次方程组\n",
    "\n",
    "$$\n",
    "\\begin{cases}\n",
    "\\displaystyle\n",
    "\\sum_{i=1}^{n}{-x_i(y_i - mx_i - b)} =0 \\\\\n",
    "\\\\\n",
    "\\displaystyle\n",
    "\\sum_{i=1}^{n}{-(y_i - mx_i - b)} =0 \\\\\n",
    "\\end{cases}\n",
    "$$\n",
    "\n",
    "为了加强理解，我们用一个实际例子演练。\n",
    "\n",
    "我们要用三个点 $(1,1), (2,2), (3,2)$ 来拟合一条直线 y = m*x + b, 请写出\n",
    "\n",
    "- 目标函数 $E$, \n",
    "- 二元二次方程组，\n",
    "- 并求解最优参数 $m, b$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "TODO 写出目标函数，方程组和最优参数"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 3.3.3 将方程组写成矩阵形式\n",
    "\n",
    "我们的二元二次方程组可以用更简洁的矩阵形式表达，将方程组写成矩阵形式更有利于我们使用 Gaussian Jordan 消元法求解。\n",
    "\n",
    "请证明 \n",
    "$$\n",
    "\\begin{bmatrix}\n",
    "    \\frac{\\partial E}{\\partial m} \\\\\n",
    "    \\frac{\\partial E}{\\partial b} \n",
    "\\end{bmatrix} = X^TXh - X^TY\n",
    "$$\n",
    "\n",
    "其中向量 $Y$, 矩阵 $X$ 和 向量 $h$ 分别为 :\n",
    "$$\n",
    "Y =  \\begin{bmatrix}\n",
    "    y_1 \\\\\n",
    "    y_2 \\\\\n",
    "    ... \\\\\n",
    "    y_n\n",
    "\\end{bmatrix}\n",
    ",\n",
    "X =  \\begin{bmatrix}\n",
    "    x_1 & 1 \\\\\n",
    "    x_2 & 1\\\\\n",
    "    ... & ...\\\\\n",
    "    x_n & 1 \\\\\n",
    "\\end{bmatrix},\n",
    "h =  \\begin{bmatrix}\n",
    "    m \\\\\n",
    "    b \\\\\n",
    "\\end{bmatrix}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "TODO 证明:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "至此我们知道，通过求解方程 $X^TXh = X^TY$ 来找到最优参数。这个方程十分重要，他有一个名字叫做 **Normal Equation**，也有直观的几何意义。你可以在 [子空间投影](http://open.163.com/movie/2010/11/J/U/M6V0BQC4M_M6V2AJLJU.html) 和 [投影矩阵与最小二乘](http://open.163.com/movie/2010/11/P/U/M6V0BQC4M_M6V2AOJPU.html) 看到更多关于这个方程的内容。"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 3.4 求解 $X^TXh = X^TY$ \n",
    "\n",
    "在3.3 中，我们知道线性回归问题等价于求解 $X^TXh = X^TY$ (如果你选择不做3.3，就勇敢的相信吧，哈哈)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 78,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(0.9677, 13.0001)\n"
     ]
    }
   ],
   "source": [
    "# TODO 实现线性回归\n",
    "'''\n",
    "参数：X, Y\n",
    "返回：m，b\n",
    "'''\n",
    "def linearRegression(X,Y):\n",
    "    NX = [[one, 1] for one in X]\n",
    "    NY = [[one] for one in Y]\n",
    "    NXT = transpose(NX)\n",
    "    A = matxMultiply(NXT, NX)\n",
    "    b = matxMultiply(NXT, NY)\n",
    "    result = gj_Solve(A, b)\n",
    "#     print result\n",
    "    return [one[0] for one in result]\n",
    "\n",
    "m,b = linearRegression(X,Y)\n",
    "print(m,b)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "你求得的回归结果是什么？\n",
    "请使用运行以下代码将它画出来。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 79,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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RMtQ0SK6hYfXPCjKl9fffw803w5ZbwsCBLFq2Lid1eYC2M96kdlgd9ffmNv9osdeKKKaS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\nECkQM1sTmAZ0B25z90ERF0k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      "text/plain": [
       "<matplotlib.figure.Figure at 0x10faebfd0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# 请不要修改下面的代码\n",
    "x1,x2 = -5,5\n",
    "y1,y2 = x1*m+b, x2*m+b\n",
    "\n",
    "plt.xlim((-5,5))\n",
    "plt.xlabel('x',fontsize=18)\n",
    "plt.ylabel('y',fontsize=18)\n",
    "plt.scatter(X,Y,c='b')\n",
    "plt.plot((x1,x2),(y1,y2),'r')\n",
    "plt.text(1,2,'y = {m}x + {b}'.format(m=m,b=b))\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "你求得的回归结果对当前数据集的MSE是多少？"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 80,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.968703691926\n"
     ]
    }
   ],
   "source": [
    "mse = calculateMSE(X, Y, m, b)\n",
    "print mse"
   ]
  }
 ],
 "metadata": {
  "anaconda-cloud": {},
  "celltoolbar": "Raw Cell Format",
  "kernelspec": {
   "display_name": "Python 2",
   "language": "python",
   "name": "python2"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 2
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython2",
   "version": "2.7.12"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}