Skip to content

C
course-info
Commit da553227 by TeacherZhu
Options

Upload New File

parent 54a10210
Showing with 263 additions and 0 deletions
课件/0627Mini-homework1.ipynb.ipynb 0 → 100644
{
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Problem 1. Fibonacci Sequence\n",
"在课程里，讨论过如果去找到第N个Fibonacci number。在这里，我们来试着求一下它的Closed-form解。 \n",
"\n",
"Fibonacci数列为 1,1,2,3,5,8,13,21,.... 也就第一个数为1，第二个数为1，以此类推...\n",
"我们用f(n)来数列里的第n个数，比如n=3时 f(3)=2。\n",
"\n",
"下面，来证明一下fibonacci数列的closed-form, 如下：\n",
"\n",
"$f(n)=\\frac{1}{\\sqrt{5}}(\\frac{1+\\sqrt{5}}{2})^n-\\frac{1}{\\sqrt{5}}(\\frac{1-\\sqrt{5}}{2})^n$\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"// your proof is here ....\n",
"\n",
"可参考\n",
"http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibformproof.html\n",
"\n",
"\n",
"### 具体证明如下：\n",
"首先根据fibonacci数列的定义，我们可以得到如下的公式：\n",
"\n",
"$\\begin{pmatrix} f_{n+1} \\\\ f_n \\end{pmatrix}=\\begin{pmatrix}\n",
"1&1\\\\1&0 \\end{pmatrix}^n \\begin{pmatrix} f_1 \\\\ f_0 \\end{pmatrix}$\n",
"\n",
"令$A=\\begin{pmatrix} 1&1\\\\ 1&0 \\end{pmatrix}$,根据 $|\\lambda E-A|=0 $,我们可以求出$A$的两个特征值为\n",
"\n",
"$\\lambda_1=\\frac{1-\\sqrt{5}}{2},\\lambda_2=\\frac{1+\\sqrt{5}}{2}$\n",
"\n",
"进一步得到特征向量为\n",
"\n",
"$\\alpha_1=\\begin{pmatrix} 1\\\\ \\frac{1+\\sqrt{5}}{2} \\end{pmatrix},\n",
" \\alpha_2=\\begin{pmatrix} 1\\\\ \\frac{-1+\\sqrt{5}}{2} \\end{pmatrix}$\n",
" \n",
" 从而得到\n",
" \n",
"$P=\\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix},\n",
"P^{-1}=\\begin{pmatrix} \\frac{1-\\sqrt{5}}{2} & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & -1 \\end{pmatrix} $\n",
"\n",
"因为 $(PAP^{-1})^n=PA(P^{-1}P)A...(P^{-1}P)AP^{-1}=PA^nP^{-1}$,所以\n",
"$A^n=P^{-1}(PAP^{-1})^nP=P^{-1}\\begin{pmatrix} (\\frac{1-\\sqrt{5}}{2})^n & 0 \\\\ 0& (\\frac{1+\\sqrt{5}}{2})^n \\end{pmatrix} P $\n",
"\n",
"从而得出\n",
"$f(n)=\\frac{1}{\\sqrt{5}}(\\frac{1+\\sqrt{5}}{2})^n-\\frac{1}{\\sqrt{5}}(\\frac{1-\\sqrt{5}}{2})^n$\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### 证明如下:\n",
"$\\begin{pmatrix} f_{n+1} \\\\ f_n \\end{pmatrix}=\\begin{pmatrix}\n",
"1&1\\\\1&0 \\end{pmatrix}^n \\begin{pmatrix} f_1 \\\\ f_0 \\end{pmatrix}$\n",
"\n",
"令$A=\\begin{pmatrix} 1&1\\\\ 1&0 \\end{pmatrix}$,根据 $|\\lambda E-A|=0 $,我们可以求出$A$的两个特征值为\n",
"\n",
"$\\lambda_1=\\frac{1-\\sqrt{5}}{2},\\lambda_2=\\frac{1+\\sqrt{5}}{2}$\n",
"\n",
"进一步得到特征向量为\n",
"\n",
"$\\alpha_1=\\begin{pmatrix} 1\\\\ \\frac{1+\\sqrt{5}}{2} \\end{pmatrix},\n",
" \\alpha_2=\\begin{pmatrix} 1\\\\ \\frac{-1+\\sqrt{5}}{2} \\end{pmatrix}$\n",
" \n",
" 从而得到\n",
" \n",
"$P=\\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix},\n",
"P^{-1}=\\begin{pmatrix} \\frac{1-\\sqrt{5}}{2} & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & -1 \\end{pmatrix} $\n",
"\n",
"因为 $(PAP^{-1})^n=PA(P^{-1}P)A...(P^{-1}P)AP^{-1}=PA^nP^{-1}$,所以\n",
"$A^n=P^{-1}(PAP^{-1})^nP=P^{-1}\\begin{pmatrix} (\\frac{1-\\sqrt{5}}{2})^n & 0 \\\\ 0& (\\frac{1+\\sqrt{5}}{2})^n \\end{pmatrix} P $\n",
"\n",
"从而得出\n",
"$f(n)=\\frac{1}{\\sqrt{5}}(\\frac{1+\\sqrt{5}}{2})^n-\\frac{1}{\\sqrt{5}}(\\frac{1-\\sqrt{5}}{2})^n$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Problem2. Algorithmic Complexity\n",
"对于下面的复杂度，从小大排一下顺序：\n",
"\n",
"$O(N), O(N^2), O(2^N), O(N\\log N), O(N!), O(1), O(\\log N), O(3^N), O(N^2\\log N), O(N^{2.1})$\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"// your answer....\n",
"\n",
"\n",
"$O(1) --> O(\\log N)-->O(N)-->O(N\\log N)-->O(N^2)-->O(N^2logN)-->O(N^{2.1})-->O(2^N)-->O(3^N)-->O(N!)$\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Problem 3 Dynamic Programming Problem"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Edit Distance (编辑距离）\n",
"编辑距离用来计算两个字符串之间的最短距离，这里涉及到三个不通过的操作，add, delete和replace. 每一个操作我们假定需要1各单位的cost. \n",
"\n",
"例子： \"apple\", \"appl\" 之间的编辑距离为1 （需要1个删除的操作）\n",
"\"machine\", \"macaide\" dist = 2\n",
"\"mach\", \"aaach\" dist=2"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [],
"source": [
"# 基于动态规划的解法\n",
"def levenshtein_dp(s, t):\n",
" m, n = len(s), len(t)\n",
" table = [[0] * (n) for _ in range(m)]\n",
"\n",
" for i in range(n):\n",
" if s[0] == t[i]:\n",
" table[0][i] = i - 0\n",
" elif i != 0:\n",
" table[0][i] = table[0][i - 1] + 1\n",
" else:\n",
" table[0][i] = 1\n",
"\n",
" for i in range(m):\n",
" if s[i] == t[0]:\n",
" table[i][0] = i - 0\n",
" elif i != 0:\n",
" table[i][0] = table[i - 1][0] + 1\n",
" else:\n",
" table[i][0] = 1\n",
"\n",
" print(table)\n",
" for i in range(1, m):\n",
" for j in range(1, n):\n",
" table[i][j] = min(1 + table[i - 1][j], 1 + table[i][j - 1], int(s[i] != t[j]) + table[i - 1][j - 1])\n",
"\n",
" print(table)\n",
" return table[-1][-1]"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"输入字符串1：appl\n",
"输入字符串2：apple\n",
"[[0, 1, 2, 3, 4], [1, 0, 0, 0, 0], [2, 0, 0, 0, 0], [3, 0, 0, 0, 0]]\n",
"[[0, 1, 2, 3, 4], [1, 0, 1, 2, 3], [2, 1, 0, 1, 2], [3, 2, 1, 0, 1]]\n",
"1\n"
]
}
],
"source": [
"str1 = input(\"输入字符串1：\")\n",
"str2 = input(\"输入字符串2：\")\n",
"print(levenshtein_dp(str1, str2))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Problem 4 非技术问题\n",
"本题目的目的是想再深入了解背景，之后课程的内容也会根据感兴趣的点来做适当会调整。 \n",
"\n",
"\n",
"Q1: 之前或者现在，做过哪些AI项目/NLP项目？可以适当说一下采用的解决方案，如果目前还没有想出合适的解决方案，也可以说明一下大致的想法。 请列举几个点。\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"Q2: 未来想往哪个行业发展？ 或者想做哪方面的项目？ 请列举几个点。\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"Q3: 参加训练营，最想获得的是什么？可以列举几个点。\n",
"\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"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": "3.6.8"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment