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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",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
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"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"令$F_{n}$表示第N个Fibonacci number，令$F_{0}=0$\n",
"设存在$M=\\left(\\begin{array}{ll}A & C \\\\ B & D\\end{array}\\right)$使得$\\left(\\begin{array}{c}F_{n} \\\\ F_{n+1}\\end{array}\\right)=M\\left(\\begin{array}{c}F_{n-1} \\\\ F_{n}\\end{array}\\right)$有$\\left(\\begin{array}{c}F_{n} \\\\ F_{n+1}\\end{array}\\right)=\\left(\\begin{array}{c}A F_{n-1}+C F_{n} \\\\ B F_{n-1}+D F_{n}\\end{array}\\right)$令$M=\\left(\\begin{array}{ll}0 & 1 \\\\ 1 & 1\\end{array}\\right)$可求出通项公式$\\left(\\begin{array}{c}F_{n} \\\\ F_{n+1}\\end{array}\\right)=M\\left(\\begin{array}{c}F_{n-1} \\\\ F_{n}\\end{array}\\right) \\Rightarrow\\left(\\begin{array}{c}F_{n} \\\\ F_{n+1}\\end{array}\\right)=M^{n}\\left(\\begin{array}{c}F_{0} \\\\ F_{1}\\end{array}\\right) \\Rightarrow\\left(\\begin{array}{c}F_{n} \\\\ F_{n+1}\\end{array}\\right)=P D^{n} P^{-1}\\left(\\begin{array}{c}0 \\\\ 1\\end{array}\\right)$解其特征方程$\\operatorname{det}(M-\\lambda I)=0$得$\\lambda=\\frac{1 \\pm \\sqrt{5}}{2}$特征向量$\\left(\\begin{array}{c}1 \\\\ \\frac{1 \\pm \\sqrt{5}}{2}\\end{array}\\right)$\n",
"$P^{-1}=\\left(\\begin{array}{cc}\\frac{\\sqrt{5}-1}{2 \\sqrt{5}} & \\frac{1}{\\sqrt{5}} \\\\ \\frac{\\sqrt{5}+1}{2 \\sqrt{5}} & -\\frac{1}{\\sqrt{5}}\\end{array}\\right)$\n",
"$F_{n}=(1 \\quad 0)\\left(\\begin{array}{ccc}1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2}\\end{array}\\right)\\left(\\begin{array}{cc}\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{n} & 0 \\\\ 0 & \\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}\\end{array}\\right)\\left(\\begin{array}{cc}\\frac{\\sqrt{5}-1}{2 \\sqrt{5}} & \\frac{1}{\\sqrt{5}} \\\\ \\frac{\\sqrt{5}+1}{2 \\sqrt{5}} & -\\frac{1}{\\sqrt{5}}\\end{array}\\right)\\left(\\begin{array}{c}0 \\\\ 1\\end{array}\\right)$\n",
"$=(1 \\quad 0)\\left(\\begin{array}{ccc}1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2}\\end{array}\\right)\\left(\\begin{array}{cc}\\left.\\frac{1+\\sqrt{5}}{2}\\right)^{n} & 0 \\\\ 0 & \\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}\\end{array}\\right)\\left(\\begin{array}{c}\\frac{1}{\\sqrt{5}} \\\\ -\\frac{1}{\\sqrt{5}}\\end{array}\\right)$\n",
"$=\\frac{2^{-n}}{\\sqrt{5}}(1 \\quad 0)\\left(\\begin{array}{cc}1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2}\\end{array}\\right)\\left(\\begin{array}{c}(1+\\sqrt{5})^{n} \\\\ -(1-\\sqrt{5})^{n}\\end{array}\\right)=\\frac{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{n}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}}{\\sqrt{5}}$故\n",
"$F_{n}=\\frac{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{n}-\\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}}{\\sqrt{5}}$"
]
},
{
"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",
"$O(1),O(\\log N), O(N), O(N\\log N),O(N^2), O(N^{2.1},O(N^2\\log N), O(2^N),O(3^N), O(N!))$\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": 121,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Edit Distance of 'sunday' and 'saturday' is 3.\n"
]
}
],
"source": [
"# s1, s2 are two strings\n",
"def editDistDP(s1, s2):\n",
"\n",
" m, n = len(s1), len(s2)\n",
" dp = [[0 for _ in range(n+1)] for _ in range(m+1)]\n",
"\n",
" for i in range(m + 1):\n",
" for j in range(n + 1):\n",
" # i 为0 最少操作是Insert j字符\n",
" if i == 0:\n",
" dp[i][j] = j\n",
" # j 为0 最少操作是Insert i字符\n",
" elif j == 0:\n",
" dp[i][j] = i\n",
" # 当前字符相等时不需要编辑\n",
" elif s1[i-1] == s2[j-1]:\n",
" dp[i][j] = dp[i-1][j-1]\n",
" #不相等时需要编辑，从三种操作中选取最小的加一\n",
" else:\n",
" dp[i][j] = 1 + min(dp[i][j-1], # Insert\n",
" dp[i-1][j], # Remove\n",
" dp[i-1][j-1]) # Replace \n",
" return dp[m][n]\n",
"\n",
"s1 = \"sunday\"\n",
"s2 = \"saturday\"\n",
"edit_distance = editDistDP(s1, s2)\n",
"print(\"The Edit Distance of '%s' and '%s' is %d.\"%(s1, s2, edit_distance))"
]
},
{
"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",
"前期跟着咱们训练营做过一些广告点击率预测、chatbot等项目，目前尝试使用LSTM+crf做一些词性标注项目\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",
"医疗、金融、推荐系统等方面，想做信息检索、信息抽取、文本生成、机器翻译、问答系统的项目\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",
"主要想转行，想要获得对AI/NLP领域的深入了解，能实际解决一下NLP任务，对于用到的算法能够有一些自己的思考和理解，达到NLP工程师入门级水平\n",
"\n"
]
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"source": []
}
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"file_extension": ".py",
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课件/0606 Ensemble 模型实战--资料/代码/.gitkeep 0 → 100644
++ "b/\350\257\276\344\273\266/0606 Ensemble \346\250\241\345\236\213\345\256\236\346\210\230--\350\265\204\346\226\231/\344\273\243\347\240\201/.gitkeep"
课件/0607生活中的优化问题/.gitkeep 0 → 100644
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课件/0613Simplex Method与LP实战/.gitkeep 0 → 100644
++ "b/\350\257\276\344\273\266/0613Simplex Method\344\270\216LP\345\256\236\346\210\230/.gitkeep"
课件/0613Simplex Method与LP实战/lpsolve.py 0 → 100644
'''
'''
原题目：
有2000元经费，需要采购单价为50元的若干桌子和单价为20元的若干椅子，你希望桌椅的总数尽可能的多，但要求椅子数量不少于桌子数量，且不多于桌子数量的1.5倍，那你需要怎样的一个采购方案呢？
解：要采购x1张桌子，x2把椅子
max z= x1 + x2
s.t. x1 - x2 <= 0
1.5x1 >= x2
50x1 + 20x2 <= 2000
x1, x2 >=0
在python中此类线性规划问题可用lp solver解决
scipy.optimize._linprog def linprog(c: int,
A_ub: Optional[int] = None,
b_ub: Optional[int] = None,
A_eq: Optional[int] = None,
b_eq: Optional[int] = None,
bounds: Optional[Iterable] = None,
method: Optional[str] = 'simplex',
callback: Optional[Callable] = None,
options: Optional[dict] = None) -> OptimizeResult
矩阵A：就是约束条件的系数（等号左边的系数）
矩阵B：就是约束条件的值（等号右边）
矩阵C：目标函数的系数值
'''
from scipy import optimize as opt
import numpy as np
#参数
#c是目标函数里变量的系数
c=np.array([1,1])
#a是不等式条件的变量系数
a=np.array([[1,-1],[-1.5,1],[50,20]])
#b是是不等式条件的常数项
b=np.array([0,0,2000])
#a1，b1是等式条件的变量系数和常数项，这个例子里无等式条件,不要这两项
#a1=np.array([[1,1,1]])
#b1=np.array([7])
#限制
lim1=(0,None) #(0,None)->(0,+无穷)
lim2=(0,None)
#调用函数
ans=opt.linprog(-c,a,b,bounds=(lim1,lim2))
#输出结果
print(ans)
#注意：我们这里的应用问题，椅子不能是0.5把，所以最后应该采购37把椅子
课件/0613Simplex Method与LP实战/simplex.py 0 → 100644
import numpy as np
class Simplex(object):
def __init__(self, obj, max_mode=False):
self.max_mode = max_mode # 默认是求min，如果是max目标函数要乘-1
self.mat = np.array([[0] + obj]) * (-1 if max_mode else 1) #矩阵先加入目标函数
def add_constraint(self, a, b):
self.mat = np.vstack([self.mat, [b] + a]) #矩阵加入约束
def solve(self):
m, n = self.mat.shape # 矩阵里有1行目标函数，m - 1行约束，应加入m-1个松弛变量
temp, B = np.vstack([np.zeros((1, m - 1)), np.eye(m - 1)]), list(range(n - 1, n + m - 1)) # temp是一个对角矩阵，B是个递增序列
mat = self.mat = np.hstack([self.mat, temp]) # 横向拼接
while mat[0, 1:].min() < 0: #判断目标函数里是否还有负系数项
col = np.where(mat[0, 1:] < 0)[0][0] + 1 # 在目标函数里找到第一个负系数的变量，找到替入变量
row = np.array([mat[i][0] / mat[i][col] if mat[i][col] > 0 else 0x7fffffff for i in
range(1, mat.shape[0])]).argmin() + 1 # 找到最严格约束的行，也就找到替出变量
if mat[row][col] <= 0: return None # 若无替出变量，原问题无界
mat[row] /= mat[row][col] #替入变量和替出变量互换
ids = np.arange(mat.shape[0]) != row
mat[ids] -= mat[row] * mat[ids, col:col + 1] # 对i!= row的每一行约束条件，做替入和替出变量的替换
B[row] = col #用B数组记录替入的替入变量
return mat[0][0] * (1 if self.max_mode else -1), {B[i]: mat[i, 0] for i in range(1, m) if B[i] < n} #返回目标值，对应x的解就是基本变量为对应的bi，非基本变量为0
"""
minimize -x1 - 14x2 - 6x3
st
x1 + x2 + x3 <=4
x1 <= 2
x3 <= 3
3x2 + x3 <= 6
x1 ,x2 ,x3 >= 0
answer :-32
"""
t = Simplex([-1, -14, -6])
t.add_constraint([1, 1, 1], 4)
t.add_constraint([1, 0, 0], 2)
t.add_constraint([0, 0, 1], 3)
t.add_constraint([0, 3, 1], 6)
print(t.solve())
print(t.mat)
\ No newline at end of file
课件/project1/.gitkeep 0 → 100644
++ "b/\350\257\276\344\273\266/project1/.gitkeep"
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