I wrote this post for lecture 17 in Andrew Ng's lecture collections on Machine Learning. In my previous post, we discussed Markov Decision Process (MDP) in its simplest form, where the set of states and the set of actions are both finite. But in real world application, states and actions can be infinite and even continuous. For example, if we want to model states of a self-driving car in a 2D plane, we must at least have the position \((x, y)\), the direction \(\theta\) of the car pointing to, its velocity \((v_x, v_y)\) and the rate \(r\) of change in \(\theta\). So the states of a car is at least a 6 dimensional space. For actions of a car, we can control how fast it goes in direction \(\theta\) and we can also control \(r\). Thus the actions have dimension 2.

In this post, we consider only continuous states with finite actions. Indeed, actions space usually has much lower dimension than states space, so in case of continuous actions, we might just discretize the actions spaces to get a finite set of representatives of actions. One may argue that we can also discretize the states space. Yes, we can do it, but only when the dimension \(n\) of state space is small enough: if we discretize each dimension into \(k\) parts, then there would be \(k^n\) many states. If \(n\) is large, then \(k^n\) is not feasible. This is so called the curse of dimensionality. Moreover, discretizing states space usually results in lack of smoothness.

This article is my notes for 16th lecture in Machine Learning by Andrew Ng on Markov Decision Process (MDP). MDP is a typical way in machine learning to formulate reinforcement learning, whose tasks roughly speaking are to train agents to take actions in order to get maximal rewards in some settings. One example of reinforcement learning would be developing a game bot to play Super Mario on its own.

Another simple example is used in the lecture, and I will use it throughout the post as well. Since the example is really simple, so MDP shown below is not of the most general form, but only good enough to solve the example and give the idea of what MDP and reinforcement learning are. The example begins with a 3 by 4 grid as below.

周末晚（14号）我终于去看了久闻的歌剧魅影，果然十分震撼。

This article is my notes on Principal Component Analysis (PCA) for Lecture 14 and 15 of Machine Learning by Andrew Ng. Given a set of high dimensional data \(\{x^{(1)}, \dots, x^{(m)}\}\), where each \(x^{(i)} \in \R^{n}\), with the assumption that these data actually roughly lie in a much smaller \(k\) dimensional subspace, PCA tries to find a basis for this \(k\) dimensional subspace. Let's look at a simple example:

来到路村的第一件事是长胖。某人的厨艺实在了得，此次记录的是她的代表作之一——烤猪排。

这次比赛不难，但是我在最简单的两题各错了一次。

第一题Flood Fill

我记得高中的时候学到Flood Fill这个词的时候有种莫名开心，可能是因为这个名字很形象地描述了DFS的过程吧。

class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        """
        :type image: List[List[int]]
        :type sr: int
        :type sc: int
        :type newColor: int
        :rtype: List[List[int]]
        """
        visited = set()
        n = len(image)
        m = len(image[0])
        color = image[sr][sc]

        def dfs(r, c):
            image[r][c] = newColor
            visited.add((r, c))
            for dr, dc in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
                x, y = r+dr, c+dc
                if x < 0 or y < 0 or x >= n or y >= m or image[x][y] != color:
                    continue
                if (x, y) in visited:
                    continue
                dfs(x, y)

        dfs(sr, sc)
        return image

最近学会了手工鸡蛋面，然后就开始各种黑暗尝试。我第一次的黑暗尝试是在面中加葱，所以揉出来的面是绿色的。

This article is my notes on the topic of factor analysis. These notes come out of lecture 13 and 14 of Andrew Ng's online course. Roughly speaking, factor analysis models some \(n\) dimensional observed data with the assumption that these data are actually from some \(d\) dimensional plane in \(\R\), up to some Gaussian distributed errors. Let's make it more precise.

Suppose we have a set of observed data \(\{x^{(1)}, \dots, x^{(m)}\}\) implicitly labeled by some latent random variable \(z \in \R^d\) where

Factor analysis model tries to model \(P(x)\) using the assumption that

for some \(\mu \in \R^n, \Lambda \in \R^{n \times d}\) and diagonal matrix \(\Psi \in \R^{n \times n}\). These \(\mu, \Lambda\) and \(\Psi\) are parameters of the model.

In this article, I will collect my notes on Expectation-Maximization algorithm (EM) based on lecture 12 and 13 of Andrew Ng's online course. Given a set of unlabeled data points EM tries iteratively to determine the distribution of data, assuming that all data points are implicitly labeled (unobserved latent variables). For simplicity, we shall limit ourselves to the case where there are only finitely many implicit labels.

Description of the problem

Given a set of unlabeled data \(\{x^{(1)}, \dots, x^{(m)}\}\), our goal is to determine \(P(x)\), the distribution of \(x\), with the following assumptions.

Assumptions 1 and 2 will gives us a set of parameters \(\theta = (\phi_1, \dots, \phi_j, a_1,\dots, a_j)\) and

We want to find this set of parameters so that the likelihood function

is maximized. Or equivalently, the log likelihood function below is maximized:

where

这次比赛考细节／模拟。

第一题Longest Word in Dictionary

我是用trie做的。首先将单词按照长度排序，然后将单词一一放进一个trie中。用trie是因为将单词放入trie中的同时也可以判断单词的前面部分是否在trie里（这也是按长度排序的原因）。

class Solution(object):
    def longestWord(self, words):
        """
        :type words: List[str]
        :rtype: str
        """
        words.sort(key=lambda x: len(x))
        tree = {'#': {}}
        ans = ''
        for word in words:
            p = tree
            ok = True
            for c in word:
                if c not in p:
                    p[c] = {}
                ok = ok and '#' in p
                p = p[c]
            p['#'] = {}
            if ok and (len(word) > len(ans) or word < ans):
                    ans = word
        return ans