Retrofitting Analysis

To figure out the process of retrofitting[1] objective updating, we do the following math.

Forward Derivation

\[
\psi(Q) = \sum_{i=1}^{n}\left[ \alpha_i||q_i-\hat{q_i}||^2 + \sum\beta||q_i-q_j||^2 \right] \\
\frac{\partial \psi(Q)}{\partial q_i} = \alpha_i(q_i-\hat{q_i}) + \sum\beta(q_i-q_j) = 0 \\
(\alpha_i+\sum\beta_{ij})q_i -\alpha_i\hat{q_i} -\sum\beta_{ij}q_j = 0 \\
q_i = \frac{\sum\beta_{ij}q_j+\alpha_i\hat{q_i}}{\sum\beta_{ij}+\alpha_i}
\]

Backward Derivation

This is how I understood this updating equation.

In the paper[1], it has mentioned "We take the first derivative of \(\psi\) with respect to one qi vector, and by equating it to zero", hence we get follow idea:
\[
\frac{\partial\psi(Q)}{\partial q_i} = 0
\]

And,

\[
q_i = \frac{\sum\beta_{ij}q_j+\alpha_i\hat{q_i}}{\sum\beta_{ij}+\alpha_i} \\
\alpha_iq_i - \alpha_i\hat{q_j} + \sum\beta_{ij}q_i - \sum\beta q_j = 0 \\
\alpha_i(q_i-\hat{q_j})+ \sum\beta_{ij}(q_i-q_j) = 0
\]

Apparently,
\[
\frac{\partial\psi(Q)}{\partial q_i} = \alpha_i(q_i-\hat{q_j})+ \sum\beta_{ij}(q_i-q_j) = 0
\]

Reference

Faruqui M, Dodge J, Jauhar S K, et al. Retrofitting Word Vectors to Semantic Lexicons[J]. ACL, 2015.

Retrofitting Analysis的更多相关文章

  1. IJCAI 2019 Analysis

    IJCAI 2019 Analysis 检索不到论文的关键词:retrofitting word embedding Getting in Shape: Word Embedding SubSpace ...

  2. Why many EEG researchers choose only midline electrodes for data analysis EEG分析为何多用中轴线电极

    Source: Research gate Stafford Michahial EEG is a very low frequency.. and literature will give us t ...

  3. Automated Memory Analysis

    catalogue . 静态分析.动态分析.内存镜像分析对比 . Memory Analysis Approach . volatility: An advanced memory forensics ...

  4. Sentiment Analysis resources

    Wikipedia: Sentiment analysis (also known as opinion mining) refers to the use of natural language p ...

  5. Call for Papers IEEE/ACM International Conference on Advances in Social Network Analysis and Mining (ASONAM)

    IEEE/ACM International Conference on Advances in Social Network Analysis and Mining (ASONAM) 2014 In ...

  6. 主成分分析(principal components analysis, PCA)

    原理 计算方法 主要性质 有关统计量 主成分个数的选取 ------------------------------------------------------------------------ ...

  7. 《利用Python进行数据分析: Python for Data Analysis 》学习随笔

    NoteBook of <Data Analysis with Python> 3.IPython基础 Tab自动补齐 变量名 变量方法 路径 解释 ?解释, ??显示函数源码 ?搜索命名 ...

  8. Python for Data Analysis

    Data Analysis with Python ch02 一些有趣的数据分析结果 Male描述的是美国新生儿男孩纸的名字的最后一个字母的分布 Female描述的是美国新生儿女孩纸的名字的最后一个字 ...

  9. 使用SQL Server Analysis Services数据挖掘的关联规则实现商品推荐功能(七)

    假如你有一个购物类的网站,那么你如何给你的客户来推荐产品呢?这个功能在很多电商类网站都有,那么,通过SQL Server Analysis Services的数据挖掘功能,你也可以轻松的来构建类似的功 ...

随机推荐

  1. 实体类与数据库字段不匹配问题,java.sql.SQLSyntaxErrorException: Unknown column 'xxx' in 'field list'

    控制台报错 ### Error querying database. Cause: java.sql.SQLSyntaxErrorException: Unknown column 'user_nam ...

  2. 忘记oracle的sys用户密码如何修改以及Oracle 11g 默认用户名和密码

    忘记除SYS.SYSTEM用户之外的用户的登录密码 CONN SYS/PASS_WORD AS SYSDBA; --用SYS (或SYSTEM)用户登录 ALTER USER user_name ID ...

  3. c# 6.0、c#7.0、c#8.0新特性

    官方: https://docs.microsoft.com/zh-cn/dotnet/articles/csharp/whats-new/csharp-6 https://docs.microsof ...

  4. Codeforces Round #581 (Div. 2) B. Mislove Has Lost an Array (贪心)

    B. Mislove Has Lost an Array time limit per test1 second memory limit per test256 megabytes inputsta ...

  5. Summer training round2 #1

    A:水 B:求两个三角形之间的位置关系:相交 相离 内含 ①用三个点是否在三角形内外判断    计算MA*MB.MB*MC.MC*MA的大小 若这三个值同号,那么在三角形的内部,异号在外部 #incl ...

  6. centos7安装kong和kong-dashboard

    1.安装Kong yum install -y https://kong.bintray.com/kong-community-edition-rpm/centos/7/kong-community- ...

  7. 树上倍增求LCA详解

    LCA(least common ancestors)最近公共祖先 指的就是对于一棵有根树,若结点z既是x的祖先,也是y的祖先(不要告诉我你不知道什么是祖先),那么z就是结点x和y的最近公共祖先. 定 ...

  8. 微信小程序 点击事件 传递参数

    wxml: data-参数名="值" bindtap="函数名" <view class="buy-button {{cap_select == ...

  9. 在JavaScript中,++在前和++在后有什么区别

    一.++可以与输出语句写在一起,++写在变量前和写在变量后不是一个意思++ i 和 i ++ 区别在于运算顺序和结合方向. 在JavaScript中有两种自加运算,其运算符均为 ++,功能为将运算符自 ...

  10. HDU - 6582 Path (最短路+最小割)

    题意:给定一个n个点m条边的有向图,每条边有个长度,可以花费等同于其长度的代价将其破坏掉,求最小的花费使得从1到n的最短路变长. 解法:先用dijkstra求出以1为源点的最短路,并建立最短路图(只保 ...