From da5532276fdc83685a81483433143801e462f19b Mon Sep 17 00:00:00 2001
From: TeacherZhu <813664462@qq.com>
Date: Sat, 27 Jun 2020 18:57:27 +0800
Subject: [PATCH] Upload New File
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+{
+ "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
+}
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