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

在这篇博客中，我将使用softmax模型来识别手写数字。文章的第一部分是关于softmax模型的理论推导，而第二部分则是模型的实现。softmax的本质是一个线性模型，所以推导所需要的理论在我之前的一篇博客Generalized Linear Model已经详细介绍过了。softmax是逻辑回归(logistic regression)的推广：逻辑回归使用Bernoulli分布（二项分布），而softmax使用多项分布。

这次比赛之后，下次再遇到O(n²)的题目，而且n < 1000的时候，一定要用C++！这是泪的教训啊😭

第一题Degree of an Array

在统计每个数字频率的同时，记录这个数字出现的最左和最右位置(l, r)。那么包含这个数字的最短数组的长度为r-l+1。

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        cnt = {}
        f = 0
        for i, n in enumerate(nums):
            if n not in cnt:
                cnt[n] = [1, i, i]
            else:
                c, a, _ = cnt[n]
                cnt[n] = [c+1, a, i]
            f = max(f, cnt[n][0])
        return min(v[2]-v[1]+1 for v in cnt.values() if v[0]==f)

这次比赛又是对Python不友好。最后一题同样的实现Python会TLE，但是C++就AC。不过也有可能我的算法不是太好。值得一提的是，我发现比赛时我第三题的算法是错误的，但是还是过了。

第一题Binary Number with Alternating Bits

比较直接，代码如下。

class Solution(object):
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        b = bin(n)[2:]
        return all(b[i]!=b[i+1] for i in range(len(b)-1))

This article is my notes on support vector machine for Lecture 7 and 8 of Machine Learning by Andrew Ng.

Intuition

In a binary classification problem, we can use logistic regression

where \(g\) is the sigmoid function with a figure of it below.

Then given input \(x\), the model predicts \(1\) if and only if \(\theta^x \ge 0\), in which case \(h_\theta(x) = g(\theta^T x) \ge 0.5\); and it predicts \(0\) if and only if \(\theta^T x < 0\). Moreover, based on the shape of sigmoid function, if \(\theta^T x >> 0\), we are very confident that \(y=1\). Likewise, if \(\theta^T x << 0\), we are very confident that \(y=0\). Therefore, we hope that for the training set \(\{(x^{(i)}, y^{(i)})\}_{i=1}^m\), we can find such a \(\theta\) that \(\theta^T x^{(i)} >> 0\) if \(y^{(i)}=1\) and \(\theta^T x^{(i)} << 0\) if \(y^{(i)}=0\).

This article is my notes on generative model for Lecture 5 and 6 of Machine Learning by Andrew Ng. What we do in logistic regression using generalized linear model is that, we approximate \(P(y|x)\) using given data. This kind of learning algorithms is discriminative, in which we predict \(y\) based on the input features \(x\). On the contrary, generative model is to model \(P(x|y)\), the probability of the features \(x\) given class \(y\). In other words, we want to study how the features structure looks like given a class \(y\). If we also learn what \(P(y)\) is, we can easily recover \(P(y|x)\), for example, in the binary classification problem,

where \(P(x) = P(x|y=0)P(y=0) + P(x|y=1)P(y=1)\).

In this article, we are going to see a simple example of generative model on Gaussian discriminant analysis and Naive Bayes.

这次比赛跪了。。

第一题Repeated String Match

二分查找。首先算出最少需要多少次才有可能使得B是子字符串：必须要保证重复后的A的长度不小于B的长度。我们可以简单算出最少需要重复：

这题竟然WA了2次(因为typo)，TLE一次(一开始用j＝10i)。。。

class Solution(object):
    def repeatedStringMatch(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        i = (len(A) + len(B) - 1) / len(A)
        j = 2 * i
        ans = -1
        while i <= j:
            k = (i+j) / 2
            tmp = A * k
            if tmp.find(B) != -1:
                ans = k
                j = k - 1
            else:
                i = k + 1
        return ans

This article is a companion article to my another post Generalized Linear Model. In this article, I will implement some of the learning algorithms in Generalized Linear Model. To be more specific, I will do some examples on linear regression and logistic regression. With some effort, google search gives me some very good example data sets to work with. The datasets collected by Larry Winner is one of the excellent sets, which will be used in the article.

The implementations here use Python. Required 3rd party libraries are:

• Requests: use to get online data
• Matplotlib: use for plotting
• NumPy: use for matrix calculation

这次比赛做得还算可以，只是最后一题看错题目WA了3次😭

第一题Baseball Game

class Solution(object):
    def calPoints(self, ops):
        """
        :type ops: List[str]
        :rtype: int
        """
        points = []
        for op in ops:
            if op == 'C':
                points.pop()
            elif op == 'D':
                points.append(points[-1]*2)
            elif op == '+':
                points.append(points[-1]+points[-2])
            else:
                points.append(int(op))
        return sum(points)