Be insensitive, and keep a lifelong growth mindset.

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纽约州已经有十五万多的病例了,纽约市就过了八万,增长势头依然如同牛市中的纳斯达克指数。对比之下,加州接近两万,具体到湾区差不多有五千左右,更早发布隔离令,现在每天的增长也趋于平稳,一切似乎逐渐开始变得可控。

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日子一天天一周周过,转眼一年的四分之一就快过去,但进入2020以来整个世界发生了如此多惊天动地的变化,以至于三个月前还在满世界溜达记忆似乎已经如隔三秋。

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这周关注最多的依然是新冠病毒。美国的疫情越来越严重了,目前为止已经突破了3万,并且还在快速地增长。政策法规也变得越来越严格了,这周整个加州已经开始实行了Shelter in Place的就地避难令,号召所有人停止非必要的一切外出活动。不过往好的一方面想,希望更加严格的管控能够有效控制住疫情。

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CAP Theorem is a concept about distributed database system that we cannot build a general data store that is continually available, sequentially consistent, and tolerant to any partition failures.

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很久没有更新博客了,期间其实偶尔会想起有时间再继续写点什么,不过潜意识里还是没把这个太当回事儿,于是就总是因为这样那样的原因最后就都不了了之了。

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最近因为开始学Andrew Ng的Machine Learning在线课程,于是打算开始继续发布间断了很久的课程笔记,最大的目的其实还是能够激励自己能坚持学下去。结果在发表上一篇笔记的时候发现了两个问题:
第一,每次在clean, generate 和deploy的时候都会出大段的错误提示,而且其实这个问题已经出现了很长一段时间了。尽管最终貌似并不影响正常运行,但看起来还是很不舒服。
第二,也是更为重要的一个问题,就是Latex的数学公式在hexo deploy之后并没能按照预想地那样显示,而是仅仅显示了源码。

于是打算那就干脆今天彻底把问题都一起解决了。

刚开始还是打算在原来的基础上看能不能修复,结果很不幸地确实解决了一个问题又出现另一个问题,折腾了半天反而越来越混乱。所幸一不做二不休,干脆就一切都推倒了重新来过。

  1. 更新了Git到最新版本;
  2. 卸载并重新安装Node及npm到最新版本;
  3. 重新安装最新版本Hexo;
  4. 创建新的博客文件夹;
  5. 重新安装最新的NexT主题;
  6. 手动批量迁移以前的博客文章;
  7. 迁移更新博客程序和主题的配置文件。

最终更新完毕,虽然博客看起来还是老样子,但各个组件都已经是可以升到的最新,同时之前的两个问题也得以顺利解决。

P.S., 近一年整体工作都还是挺忙的,也是很久没有更新的一个原因。但最近读到《奇特的一生》中柳比歇夫的几个时间管理法则,其中特别有感悟的一点就是不要为了完成工作而放弃自己的兴趣爱好,长期而系统地坚持下去,终会有不小的收获,柳比歇夫自己就是一个活生生的例子。坚持写写东西,不仅可以倒逼这自己去输入,也是一个梳理自己思绪的过程,轻言放弃多可惜呀。

1. Introduction

1.1. What is Machine Learning?

Machine Learning Definition

  • Arthur Samuel (1959). Machine Learning: Field of study that gives computers the ability to learn without being explicitly programmed.
  • Tom Mitchell (998) Well-posed Learning Problem: A computer program is said to learn from experience E with respect to some task T and some performance measure P, if its performance on T, as measured by P, improves with experience E.

Machine Learning Algorithms

  • Supervised Learning
  • Unsupervised Learning
  • Other: Reinforcement learning, recommender system

1.2. Supervised Learning

In supervised learning, we are given a data set and already know what our correct output should look like, having the idea that there is a relationship between the input and the output.

Supervised learning problems are categorized into “regression” and “classification” problems. In a regression problem, we are trying to predict results within a continuous output, meaning that we are trying to map input variables to some continuous function. In a classification problem, we are instead trying to predict results in a discrete output. In other words, we are trying to map input variables into discrete categories.

Example 1:

Given data about the size of houses on the real estate market, try to predict their price. Price as a function of size is a continuous output, so this is a regression problem.

We could turn this example into a classification problem by instead making our output about whether the house “sells for more or less than the asking price.” Here we are classifying the houses based on price into two discrete categories.

Example 2:

(a) Regression - Given a picture of a person, we have to predict their age on the basis of the given picture

(b) Classification - Given a patient with a tumor, we have to predict whether the tumor is malignant or benign.

1.3. Unsupervised Learning

Unsupervised learning allows us to approach problems with little or no idea what our results should look like. We can derive structure from data where we don’t necessarily know the effect of the variables.

We can derive this structure by clustering the data based on relationships among the variables in the data.

With unsupervised learning there is no feedback based on the prediction results.

Example:

Clustering: Take a collection of 1,000,000 different genes, and find a way to automatically group these genes into groups that are somehow similar or related by different variables, such as lifespan, location, roles, and so on.

Non-clustering: The “Cocktail Party Algorithm”, allows you to find structure in a chaotic environment. (i.e. identifying individual voices and music from a mesh of sounds at a cocktail party).

2. Linear Regression with One Variable

2.1. Model and Cost Function

Model Representation

To establish notation for future use, we’ll use x^(i) to denote the “input” variables (living area in this example), also called input features, and y^(i) to denote the “output” or target variable that we are trying to predict (price). A pair (x^(i),y^(i)) is called a training example, and the dataset that we’ll be using to learn—a list of m training examples (x^(i),y^(i));i=1,…,m—is called a training set. Note that the superscript “(i)” in the notation is simply an index into the training set, and has nothing to do with exponentiation. We will also use X to denote the space of input values, and Y to denote the space of output values. In this example, X = Y = ℝ.

To describe the supervised learning problem slightly more formally, our goal is, given a training set, to learn a function h : X → Y so that h(x) is a “good” predictor for the corresponding value of y. For historical reasons, this function h is called a hypothesis. Seen pictorially, the process is therefore like this:

When the target variable that we’re trying to predict is continuous, such as in our housing example, we call the learning problem a regression problem. When y can take on only a small number of discrete values (such as if, given the living area, we wanted to predict if a dwelling is a house or an apartment, say), we call it a classification problem.

Cost Function

We can measure the accuracy of our hypothesis function by using a cost function. This takes an average difference (actually a fancier version of an average) of all the results of the hypothesis with inputs from x’s and the actual output y’s.

$$J(\theta_{0}, \theta_{1}) = \frac{1}{2m}\sum_{i=1}^{m} (\hat{y_{i}} - y_{i})^2 = \frac{1}{2m}\sum_{i=1}^{m} (h_{\theta}(x_{i}) - y_{i})^2$$

To break it apart, it is $\frac{1}{2}\bar{x}$ where $\bar{x}$ is the mean of the squares of $h_{\theta}(x_{i}) - y_{i}$, or the difference between the predicted value and the actual value.

This function is otherwise called the “Squared error function”, or “Mean squared error”. The mean is halved $(\frac{1}{2})$ as a convenience for the computation of the gradient descent, as the derivative term of the square function will cancel out the $(\frac{1}{2})$ term. The following image summarizes what the cost function does:

Cost - Intuition I

If we try to think of it in visual terms, our training data set is scattered on the x-y plane. We are trying to make a straight line (defined by $h_{\theta}(x)$ which passes through these scattered data points.

Our objective is to get the best possible line. The best possible line will be such so that the average squared vertical distances of the scattered points from the line will be the least. Ideally, the line should pass through all the points of our training data set. In such a case, the value of $J(\theta_{0}, \theta_{1})$ will be 0. The following example shows the ideal situation where we have a cost function of 0.

When $\theta_{1}=1$, we get a slope of 1 which goes through every single data point in our model. Conversely, when $\theta_{1} = 0.5$, we see the vertical distance from our fit to the data points increase.

This increases our cost function to 0.58. Plotting several other points yields to the following graph:

Thus as a goal, we should try to minimize the cost function. In this case, $\theta_{1} = 1$ is our global minimum.

Cost - Intuition II

A contour plot is a graph that contains many contour lines. A contour line of a two variable function has a constant value at all points of the same line. An example of such a graph is the one to the right below.

Taking any color and going along the ‘circle’, one would expect to get the same value of the cost function. For example, the three green points found on the green line above have the same value for $J(\theta_{0}, \theta_{1})$ and as a result, they are found along the same line. The circled x displays the value of the cost function for the graph on the left when $\theta_{0} = 800$ and $\theta_{1}=-0.15$. Taking another h(x) and plotting its contour plot, one gets the following graphs:

When $\theta_{0} = 360$ and $\theta_{1} = 0$, the value of $J(\theta_{0} + \theta_{1})$ in the contour plot gets closer to the center thus reducing the cost function error. Now giving our hypothesis function a slightly positive slope results in a better fit of the data.

The graph above minimizes the cost function as much as possible and consequently, the result of $\theta_{1}$ and $\theta_{0}$ tend to be around 0.12 and 250 respectively. Plotting those values on our graph to the right seems to put our point in the center of the inner most ‘circle’.

2.2. Parameter Learning

Gradient Descent

So we have our hypothesis function and we have a way of measuring how well it fits into the data. Now we need to estimate the parameters in the hypothesis function. That’s where gradient descent comes in.

Imagine that we graph our hypothesis function based on its fields $\theta_{0}$ and $\theta_{1}$ (actually we are graphing the cost function as a function of the parameter estimates). We are not graphing x and y itself, but the parameter range of our hypothesis function and the cost resulting from selecting a particular set of parameters.

We put $\theta_{0}$ on the x axis and $\theta_{1}$ on the y axis, with the cost function on the vertical z axis. The points on our graph will be the result of the cost function using our hypothesis with those specific theta parameters. The graph below depicts such a setup.

We will know that we have succeeded when our cost function is at the very bottom of the pits in our graph, i.e. when its value is the minimum. The red arrows show the minimum points in the graph.

The way we do this is by taking the derivative (the tangential line to a function) of our cost function. The slope of the tangent is the derivative at that point and it will give us a direction to move towards. We make steps down the cost function in the direction with the steepest descent. The size of each step is determined by the parameter α, which is called the learning rate.

For example, the distance between each ‘star’ in the graph above represents a step determined by our parameter α. A smaller α would result in a smaller step and a larger α results in a larger step. The direction in which the step is taken is determined by the partial derivative of $J(\theta_{0}, \theta_{1})$. Depending on where one starts on the graph, one could end up at different points. The image above shows us two different starting points that end up in two different places.

The gradient descent algorithm is:

repeat until convergence:

$$\theta_{j}:=\theta_{j}-\alpha\frac{\partial}{\partial\theta_{j}}J(\theta_{0}, \theta_{1})$$

where

j=0,1 represents the feature index number.

At each iteration j, one should simultaneously update the parameters $\theta_{1},\theta_{2},…,\theta_{n}$. Updating a specific parameter prior to calculating another one on the $j^{(th)}$ iteration would yield to a wrong implementation.

Gradient Descent Intuition

In this video we explored the scenario where we used one parameter $\theta_{1}$ and plotted its cost function to implement a gradient descent. Our formula for a single parameter was :

Repeat until convergence:

$$\theta_{1}:=\theta_{1}-\alpha\frac{\partial}{\partial\theta_{1}}J(\theta_{1})$$

Regardless of the slope’s sign for $\frac{\partial}{\partial\theta_{1}}J(\theta_{1})$, $\theta_{1}$ eventually converges to its minimum value. The following graph shows that when the slope is negative, the value of $\theta_{1}$ increases and when it is positive, the value of $\theta_{1}$ decreases.

On a side note, we should adjust our parameter $\alpha$ to ensure that the gradient descent algorithm converges in a reasonable time. Failure to converge or too much time to obtain the minimum value imply that our step size is wrong.

How does gradient descent converge with a fixed step size $\alpha$?

The intuition behind the convergence is that $\frac{\partial}{\partial\theta_{1}}J(\theta_{1})$ approaches 0 as we approach the bottom of our convex function. At the minimum, the derivative will always be 0 and thus we get:

$$\theta_{1}:=\theta_{1}-\alpha*0$$

Gradient Descent For Linear Regression

When specifically applied to the case of linear regression, a new form of the gradient descent equation can be derived. We can substitute our actual cost function and our actual hypothesis function and modify the equation to:

where m is the size of the training set, $\theta_{0}$ a constant that will be changing simultaneously with $\theta_{0}$ and $x_{i}, y_{i}$ are values of the given training set (data).

Note that we have separated out the two cases for $\theta_{j}$ into separate equations for $\theta_{0}$ and $\theta_{1}$; and that for $\theta_{1}$ we are multiplying $x_{i}$ at the end due to the derivative. The following is a derivation of $\frac{\partial}{\partial\theta}J(\theta_{j})$ for a single example:

The point of all this is that if we start with a guess for our hypothesis and then repeatedly apply these gradient descent equations, our hypothesis will become more and more accurate.

So, this is simply gradient descent on the original cost function J. This method looks at every example in the entire training set on every step, and is called batch gradient descent. Note that, while gradient descent can be susceptible to local minima in general, the optimization problem we have posed here for linear regression has only one global, and no other local, optima; thus gradient descent always converges (assuming the learning rate α is not too large) to the global minimum. Indeed, J is a convex quadratic function. Here is an example of gradient descent as it is run to minimize a quadratic function.

The ellipses shown above are the contours of a quadratic function. Also shown is the trajectory taken by gradient descent, which was initialized at (48,30). The x’s in the figure (joined by straight lines) mark the successive values of θ that gradient descent went through as it converged to its minimum.

3. Linear Algebra Review

3.1. Matrices and Vectors

Matrices are 2-dimensional arrays:

$\begin{bmatrix}
a & b & c\
d & e & f \
g & h & i \
j & k & l
\end{bmatrix}$

The above matrix has four rows and three columns, so it is a 4 x 3 matrix.

A vector is a matrix with one column and many rows:

$\begin{bmatrix}
w\
x \
y \
z
\end{bmatrix}$

So vectors are a subset of matrices. The above vector is a 4 x 1 matrix.

Notation and terms:

  • $A_{ij}$ refers to the element in the ith row and jth column of matrix A.
  • A vector with ‘n’ rows is referred to as an ‘n’ -dimensional vector.
  • $v_{i}$refers to the element in the ith row of the vector.
  • In general, all our vectors and matrices will be 1-indexed. Note that for some programming languages, the arrays are 0-indexed.
  • Matrices are usually denoted by uppercase names while vectors are lowercase.
  • “Scalar” means that an object is a single value, not a vector or matrix.
  • $\mathbb{R}$ refers to the set of scalar real numbers.
  • $\mathbb{R}^n$ refers to the set of n-dimensional vectors of real numbers.

Run the cell below to get familiar with the commands in Octave/Matlab. Feel free to create matrices and vectors and try out different things.

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% The ; denotes we are going back to a new row.
A = [1, 2, 3; 4, 5, 6; 7, 8, 9; 10, 11, 12]

% Initialize a vector
v = [1;2;3]

% Get the dimension of the matrix A where m = rows and n = columns
[m,n] = size(A)

% You could also store it this way
dim_A = size(A)

% Get the dimension of the vector v
dim_v = size(v)

% Now let's index into the 2nd row 3rd column of matrix A
A_23 = A(2,3)

3.2. Addition and Scalar Multiplication

Addition and subtraction are element-wise, so you simply add or subtract each corresponding element:

$$\begin{bmatrix}a&b\c&d\end{bmatrix} + \begin{bmatrix}w&x\y&z\end{bmatrix} = \begin{bmatrix}a+w&b+x\c+y&d+z\end{bmatrix}$$

Subtracting Matrices:
$$\begin{bmatrix}a&b\c&d\end{bmatrix} - \begin{bmatrix}w&x\y&z\end{bmatrix} = \begin{bmatrix}a-w&b-x\c-y&d-z\end{bmatrix}$$

To add or subtract two matrices, their dimensions must be the same.

In scalar multiplication, we simply multiply every element by the scalar value:
$$\begin{bmatrix}a&b\c&d\end{bmatrix} * x = \begin{bmatrix}ax&bx\cx&dx\end{bmatrix}$$

In scalar division, we simply divide every element by the scalar value:
$$\begin{bmatrix}a&b\c&d\end{bmatrix} / x = \begin{bmatrix}a/x&b/x\c/x&d/x\end{bmatrix}$$

Experiment below with the Octave/Matlab commands for matrix addition and scalar multiplication. Feel free to try out different commands. Try to write out your answers for each command before running the cell below.

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% Initialize matrix A and B
A = [1, 2, 4; 5, 3, 2]
B = [1, 3, 4; 1, 1, 1]

% Initialize constant s
s = 2

% See how element-wise addition works
add_AB = A + B

% See how element-wise subtraction works
sub_AB = A - B

% See how scalar multiplication works
mult_As = A * s

% Divide A by s
div_As = A / s

% What happens if we have a Matrix + scalar?
add_As = A + s

3.3. Matrix-Vector Multiplication

We map the column of the vector onto each row of the matrix, multiplying each element and summing the result.

$$\begin{bmatrix}a&b\c&d\e&f\end{bmatrix} * \begin{bmatrix}x\y\end{bmatrix} = \begin{bmatrix}ax+by\cx+dy\ex+fy\end{bmatrix}$$

The result is a vector. The number of columns of the matrix must equal the number of rows of the vector.

An m x n matrix multiplied by an n x 1 vector results in an m x 1 vector.

Below is an example of a matrix-vector multiplication. Make sure you understand how the multiplication works. Feel free to try different matrix-vector multiplications.

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% Initialize matrix A
A = [1, 2, 3; 4, 5, 6;7, 8, 9]

% Initialize vector v
v = [1; 1; 1]

% Multiply A * v
Av = A * v

3.4. Matrix-Matrix Multiplication

We multiply two matrices by breaking it into several vector multiplications and concatenating the result.

$$\begin{bmatrix}a&b\c&d\e&f\end{bmatrix} * \begin{bmatrix}w&x\y&z\end{bmatrix} = \begin{bmatrix}aw+by&ax+bz\cw+dy&cx+dz\ew+fy&ex+fz\end{bmatrix}$$

An m x n matrix multiplied by an n x o matrix results in an m x o matrix. In the above example, a 3 x 2 matrix times a 2 x 2 matrix resulted in a 3 x 2 matrix.

To multiply two matrices, the number of columns of the first matrix must equal the number of rows of the second matrix.

For example:

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% Initialize a 3 by 2 matrix
A = [1, 2; 3, 4;5, 6]

% Initialize a 2 by 1 matrix
B = [1; 2]

% We expect a resulting matrix of (3 by 2)*(2 by 1) = (3 by 1)
mult_AB = A*B

% Make sure you understand why we got that result

3.5. Matrix Multiplication Properties

  • Matrices are not commutative: A∗B≠B∗A
  • Matrices are associative: (A∗B)∗C=A∗(B∗C)

The identity matrix, when multiplied by any matrix of the same dimensions, results in the original matrix. It’s just like multiplying numbers by 1. The identity matrix simply has 1’s on the diagonal (upper left to lower right diagonal) and 0’s elsewhere.

$$\begin{bmatrix}1&0&0\0&1&0\0&0&1\end{bmatrix}$$

When multiplying the identity matrix after some matrix (A∗I), the square identity matrix’s dimension should match the other matrix’s columns. When multiplying the identity matrix before some other matrix (I∗A), the square identity matrix’s dimension should match the other matrix’s rows.

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% Initialize random matrices A and B
A = [1,2;4,5]
B = [1,1;0,2]

% Initialize a 2 by 2 identity matrix
I = eye(2)

% The above notation is the same as I = [1,0;0,1]

% What happens when we multiply I*A ?
IA = I*A

% How about A*I ?
AI = A*I

% Compute A*B
AB = A*B

% Is it equal to B*A?
BA = B*A

% Note that IA = AI but AB != BA

3.6. Inverse and Transpose

The inverse of a matrix A is denoted $A^{-1}$. Multiplying by the inverse results in the identity matrix.

A non square matrix does not have an inverse matrix. We can compute inverses of matrices in octave with the $pinv(A)$ function and in Matlab with the $inv(A)$ function. Matrices that don’t have an inverse are singular or degenerate.

The transposition of a matrix is like rotating the matrix 90° in clockwise direction and then reversing it. We can compute transposition of matrices in matlab with the transpose(A) function or A’:

The transposition of a matrix is like rotating the matrix 90° in clockwise direction and then reversing it. We can compute transposition of matrices in matlab with the transpose(A) function or A’:

$$A = \begin{bmatrix}a&b\c&d\e&f\end{bmatrix}$$

$$A^T = \begin{bmatrix}a&c&e\b&d&f\end{bmatrix}$$

In other words:

$$A_{ij}=A_{ji}^T$$

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% Initialize matrix A
A = [1,2,0;0,5,6;7,0,9]

% Transpose A
A_trans = A'

% Take the inverse of A
A_inv = inv(A)

% What is A^(-1)*A?
A_invA = inv(A)*A

“你曾说真理与光明,是你背井离乡的理由,虽然人间烟火多烦忧,虽然光阴总难留”


入职Facebook已经两个月了,完成了Bootcamp,进到了具体的组。今天就借此时机先来谈谈近期对于新公司的一些感受。

首先在工作环境上,Facebook相对来说还算是一个更年轻的公司。开放式的工作环境, 充满设计感的装饰元素,风格各异的壁画涂鸦,免费三餐,全天无限制饮料小吃供应,早晚班车接送,甚至还提供了理发,洗衣等服务,如此一般的环境的确让员工免除了很多后顾之忧,更能够专心,高效及富有创造力地投入到工作之中。同时,为了营造平等交流的氛围,公司刻意隐去了在头衔上的等级区分,无论是高级工程师还是初级工程师,清一色的都叫XXX Engineer,还有就是无论你是最刚毕业的职场新人,还是某个大部门的VP,在办公区间上也毫无差别,同样的区域,同型号的桌椅。

再单独说说Bootcamp, 无论你最终要从事的是哪一个方面的具体工作,都需要参加这个为期六到八周的培训,这已经成为了这里工程师文化的很重要的一个方面。期间的课程会涵盖了公司各个技术栈及业务部门,让人对于整个Facebook的工程体系有一个全局的了解,每个人也会被分配到一些小的技术任务,通过解决这些问题来进一步熟悉和融入实际工作。同时,每个人都会配备专门mentor,配合每天各种主题的office hour,包治各种疑难困惑,就为了让新人更快更好地融入公司。

不过Bootcamp仅仅是一个开始, 公司内部常年开放各类技术及非技术的课程,干货满满,对所有员工开放,为了方便不同office的人免于奔波,课程都提供线上直播。公司内部换组也是非常方便,甚至公司层面还鼓励员工这么做,鼓励员工去挖掘和投身于兴趣所在。此外,各类活动诸如Hackathon, offsite, 讲座等等也是层出不穷。因此对于职业发展而言,Facebook的确是一个提供了丰富资源的平台。

整体工作节奏上来说也要快不少,这也是Facebook本身所倡导的文化。”Move Fast and Break Things.” 压力一定会有,不过这也自然会让人grow faster, and make bigger impact.

其实近两个月都没有更新Blog,新入职之后的忙碌的确是原因之一。但同时,这并不是一个足够有说服力的理由。前两天在听李一诺在她的《诺言》栏目里聊到她的时间管理论,其中有一条给我留下了挺深的印象。

Your activities will expand to fill the time you have.

也就是说不管你做什么,时间总是会被填满的。想要更多的时间当然不现实,于是时间管理就变成了一个选择问题,每天就这么些时间,我到底拿来干嘛?在想都没想就开始抱怨“我太忙,没有时间”的时候,倒不如认真想想这件事儿,是不是真的想做。如果是,那就别想那么多,先把时间占上,然后再去据此安排其他次重要的事情,剩下的各种琐碎的杂事要么就见缝插针,要么就干脆放弃。道理其实正如很多人都听过的那个小试验一样,想要把大石块,小石子,沙子,还有水都装进同一个瓶子里,那么一定是把大石块先填上,其次是小石子,再次是沙子,最后再倒水。反之,可能就永远失去了放进大石块的机会。

P.S,开篇和结尾的歌词都是取自于以ACSSY最新发布的会歌《我们》,词曲都很棒,学弟学妹们好赞!


“告别青春我的朋友 没人能在这里停留 我们走过冬夏与春秋 向前奔 不回头”