毛线岛民的博客

原创 CS152 Review--Complex Pipelining

1.This problem deals with an out-of-order single-issue processor that is based on the basic RISC-V pipeline and has a floating-point unit. The FPU has one adder, one multiplier, and one load/store unit. The adder has a two-cycle latency and is fully pipeli

原创 CS162 Lecture2 --Synchronization

Process Management APIpthread: posix(portable operating system interface)join(pid) - suspends execution of the calling thread until the target pid terminatesexit – terminate a processfork – copy the current process>0>0>0: Returned to p

原创 verilog基础

Verilog Modules and Instantiationport declarations (input,output, or inout)wirewire a;wire [7:0] d; // 8-bit wire declarationbuilt in gates(and, or, xor, not, nand, nor, xnor)Modules can be instantiated inside other modules. The syntaxused is&l

原创 计算机组成3 --RISC V Single-Cycle Control and Pipelining

BasicsInstruction TableInst[31:0]BrEqBrLTPCSelImmSelBrUnASelBSelALUSelMemRWRegWEnWBSeladd**+4**RegRegAddRead1 (Y)ALUsub**+4**RegRegSubRead1ALU(R-ROp)**+4**RegReg(Op)Read1ALUaddi**+4I*R

原创 计算机组成2 --RISCV Assembly

RISC-V ArchitectureRegister Size: 1 word = 32 bits = 4 bytes32 registers:x0(holds the value zero)-x31Byte OrderingLittle Endian(x86, ARM)RISC-V InstructionsAssembly Syntax[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-f5d93vCZ-1648721107169)(http

原创 计算机组成1 --Number Representation

Logical OperationsShift operationsLeft Shiftthrow away extra bits on left, fill with 0’s on the rightRight Shiftthrow away extra bits on rightlogical shift: fill with 0’s on the leftarithmetic shift: replicate most significant bit(x >> k

原创 Digital Systems Recall

Combinational LogicCircuit BasicsThree terminals: Source, Gate, and DrainSwitch action: if voltage on gate terminal is (some amount)higher/lower than source terminal then conducting pathestablished between drain and source terminalsNot GateNAND a

原创 离散数学复习--Modular Arithmetic

DivisibilityDefinitionLet x,y∈Zx,y\in Zx,y∈Z with x≠0x\neq0x​=0We say that a divides ba|b, if there exists some k∈Zk\in Zk∈Z such that b=akb = akb=ak.TheoremProof EasyCorollaryLet a,b,c∈Za, b, c \in Za,b,c∈Z. If a|b and a|c, then a|mb+nc for any

原创 采样定理与奈奎斯特极限

采样定理对信号X(f)在频域进行理想采样，时域体现为周期性搬移X(w)∑n=−∞+∞δ(w−nws)X(w)\sum_{n = -\infty}^{+\infty} \delta(w-nw_s)X(w)∑n=−∞+∞​δ(w−nws​)⟺T∑m=−∞+∞X(t−mT)\Longleftrightarrow T\sum_{m= -\infty}^{+\infty} X(t- mT)⟺T∑m=−∞+∞​X(t−mT)当ws≥2wmw_s\geq2w_mws​≥2wm​时，信号通过理想低通滤波器可以无失

原创 Sex education season 3, Silicon Valley

New wordsspectrum ['spɛktrəm] n. 光谱alfresco --fresh 露天的

原创 Javascript基础复习

BasicPrimitive typesBoolean (true, false)Number (12, 1.618, -46.7, 0, etc.)String (“hello”, “world!”, “12”, etc.)Null (no value)Undefined (declared but not yet assigned a value)SymbolBigIntConst,let and varhoistscopevarOnly declarat

原创 Leetcode 10 Regular Expression Matching

ProblemGiven an input string s and a pattern p, implement regular expression matching with support for ‘.’ and ‘*’ where:‘.’ Matches any single character.​​​​‘*’ Matches zero or more of the preceding element.Example 1:Input: s = “aa”, p = “a”Output:

原创 Leetcode hot 100

1 Top K frequent ElementsGiven an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.Example 1:Input: nums = [1,1,1,2,2,3], k = 2Output: [1,2]Example 2:Input: nums = [1], k = 1Output: [1]

原创 美国生活Tips

Stock看租房Work薪资等级

原创 CS186 Lecture3 --Index & Hashing

B+ TreeStructureNode[…, (Pi,KiP_i,K_iPi​,Ki​), …]P1P_1P1​K1K_1K1​P2P_2P2​…Pn−1P_{n-1}Pn−1​Kn−1K_{n-1}Kn−1​KnK_nKn​Leaf NodeFor i = 1, 2, …, n–1, pointer PiP_iPi​ points to a file record with search-key value KiK_iKi​PnP_nPn​ points

原创 cs186 lecture2 --Data Models

Data ModelsLogical data independence: Maintain views when logical structure changesPhisical data independence: Maintain logical structure when phisical structure changesER ModelKey constraintsarrow: 可以divided(从尾到头at most one)#mermaid-svg-SUoQXSri8CA

原创 JHU An Introduction to the Theory and Practice of Poker Lecture1 --Rules of Poker

原创 TOEFL Speaking

Listening1.One of our students, Stewart Telford, has a nineteenth-century banjo—a real antique.[æn’tik] 古董2.meteorology [,mitɪə’rɑlədʒi]n. 气象学Listen to part of the lecture in a meteorology class.3.mediatea primary way that heat energy is transferred

原创 TOEFL Writing

ProblemNowadays high school students are generally lack of Interest in classroom learning.Which of the following methods do you think would be the most effective to help teachers improve high school students’ interest in classroom learning?using more t

原创 Ceceil Hotel

New wordsdose [dəʊs] n. 剂量；一剂，一服

原创 cs186 Lecture1 --Relational Model & SQL

SQL (sequel/structured query language)Basic ConceptCreateDatabaseset of named RelationsRelationSchema(description/metadata):unique attribute names,atomic typesInstance:can change very oftenAttribute(Column,Field)Tuple(Record,Row)Primary key

原创 算法设计与分析--Space-for-time tradeoffs

Two AlgorithmsInput EnhancementPrepocess the input(or its part) to store some info to be used later in solving the problemCounting SortsString Searching AlgorithmsPreconstructuringPrepocess the input to make accessing its elements easierhashi

原创 算法设计与分析--Transform and conquer

Notion of Heap1.shape(查找方便2n2n2n and 2n+12n+12n+1)The binary tree is complete (except rightmost leaves)2.The parent dominance*(方便排序)

原创 Java Application Design--Exceptions & IO

Read a fileopen the file;(共享文件同时正在写)determine its size;(文件=键盘)allocate that much memory;(JVM64MB)read the file into memory;(杭州下雨软盘失效)close the file;

原创 Java Application Design--Containers

IntroductionTwo Containers:Collection and Map1.Collection:list:hold the element in a particular sequenceset2.Map:key-value object pairsCollectionFunctionListDeque[dek]–double ended queue

原创 算法设计与分析--Decrease and Conquer

下标排列问题Josephus Problem

原创 Java Application Design--Class

Primitive Data TypesTypeStorage(bits)Min ValueMax Valuebyte8-128127short16−215-2^{15}−215215−12^{15}-1215−1int32−231-2^{31}−231231−12^{31}-1231−1long64−263-2^{63}−263263−12^{63}-1263−1float32Approximately-3.4e+38 with

原创 Database网络开发

1.Linux三大查询命令man,info,–helpman lsman passwdinfo pwdwho --helpman page:manual page

原创 Numrical Analysis Summary

Evaluating a polynomialDirect form:O(n2n^2n2)Horner’s method:O(n)

原创 Linux网络分析工具ping、traceroute、nslookup

Pingping, ping6 - send ICMP ECHO_REQUEST to network hostsping [-LRUbdfnqrvVaAB] [-c count] [-i interval] [-s packetsize] [-t ttl] [hop ...] destination参数解释参数解释[ -c count]Stop after sending count ECHO_REQUEST packets.[ -i interval]Wa

原创 TOEFL Reading

两个“不是”绝对不是简单的对每个段落的段首句的改写不是所有段落在选项中都有对应，有时正确选项可能并不包含文章的所有要点（可以去掉列举的）两个“是”选项是按文章顺序的一个正确选项可能是对文章 2+个自然段的整体概括常见错误无中生有，选项中出现了文章中没有出现的关键词错乱搭配，两个相距较远的段落（第 2 段和第 6 段）的细节被拼合在同一个选项中(必须按顺序)偷换概念，选项中关键词都对，但是关键词之间的关系不对。

原创 3 Basic Strategy

二分查找在闭区间查找[begin,end],注意对找不到的处理int binary_find(vector<int>& nums, int target){ int l=0,r=nums.size()-1;//Notice while(l<r){ int mid=l+(r-l)>>1; if(nums[mid]==target)return mid; else if(nums[mid]>targe

原创 The 100 season 4

New wordsbail [beɪl] v.保释grunt [grʌnt] n. 咕哝；呼噜声[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-q1AWBIlr-1593920407552)(https://s3-us-west-2.amazonaws.com/secure.notion-static.com/e0de753d-4255-4048-9dda-fcb9e3c941fe/Untitled.png)][外链图片转存失败,源站可能有防盗链机制,建议将图片保存

原创 离散数学复习--图论基础

基础知识图的概念1.设图G=(V,E)，u,v∈Vu,v\in Vu,v∈V，(u,v)∈E(u,v)\in E(u,v)∈E若u=v,称e为环若(u,v)不止一条，称这些边为平行边。平行边的条数称为重数无环和平行边的图称为简单图。

原创 算法设计与分析--时间复杂度分析

递归结构分析1

原创 离散数学复习--代数系统综合

1 证明左与左对应，右与右对应2 求子群要罗列完整

原创 逆透视变换车道线检测

摘要随着社会的进步和发展，安全驾驶和汽车的简便操作越来越受到人们的重视。苹果、谷歌和特斯拉等公司相继推出了无人驾驶系统，国内的厂商也对此领域产生了高度兴趣。基于人工智能和大数据的无人驾驶技术，被认为是消除交通安全问题的最终方法。从美国DARPA 挑战赛到中国未来智能车大赛，无人驾驶技术在全世界兴起了一阵热潮。对于无人驾驶系统来说，车道线检测是非常重要的技术。到现在为止，已经有一些基于机器视觉的无人驾驶技术得到了应用。对于车道线的检测方法大致可以归结为三种思想：基于车道线模型的方法和基于车道线特征的方法。

原创 基于CNN-LSTM命名实体识别和关系抽取联合学习

Word2vecGensim训练Word2vec步骤1 将语料库预处理：一行一个文档或句子，将文档或句子分词（以空格分割，英文可以不用分词，英文单词之间已经由空格分割，中文预料需要使用分词工具进行分词，常见的分词工具有StandNLP、ICTCLAS、Ansj、FudanNLP、HanLP、结巴分词等）；2 将原始的训练语料转化成一个sentence的迭代器，每一次迭代返回的sentence是一个word（utf-8格式）的列表。可以使用Gensim中word2vec.py中的LineSen

原创 离散数学复习--代数系统基础

重排定理

原创 王道计算机网络(3)TCP可靠传输、流量控制、拥塞控制

可靠传输流量控制拥塞控制慢开始：起始慢slow start threshold