This section aims to explain how the clone() syscall is executed when the systme is under the NUMA architechture. System Call Handler for clone() If the call is made by the glibc wrapper clone(), the clone syscall handler is used If syscall(SYS_clone3, ...) is used (no wrapper provided by glibc), the clone3() syscall handler is used. Both path ends in a kernel_clone() call. #ifdef __ARCH_WANT_SYS_CLONE #ifdef CONFIG_CLONE_BACKWARDS

The note of the lecture, Approximation and Online Algorithms. Expect typo, grammar errors, and rarely false information. Introduction This course aims to relate the theories to non-theories: Theories Non-Theories Complexity <br> Data Structure <br> Discrete Math Databases <br> Machine Learing <br> Operating Systems

Efficient Distributed Secure Memory with Migratable Merkle Tree IEEE International Symposium on High-Performance Computer Architecture (HPCA), 2023 Introduction Many hardware assisted enclaves exists, such as: Intel SGX AMD SEV ARM CCA RISC-V Penglai

Q1. Whats the difference between "Hypermedia" and "Multimedia" ? Multimedia An integration that is construct by media needing a device (multimedia deliver systems) to represent. E.g., Images, Audio, Video... Can be stored Linear or Non-linearly Hypermedia Hypermedia is a system that links multimedias, providing each user ability to access other resources. E.g., WWW Can only be stored non-linearly Q2. What extra information is multimedia good at conveying

$$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)} \newcommand{\funcsub}[3]{\mathrm{#1}{#2}(#3)} \newcommand{\excp}[1]{\mathrm{E}(#1)} $$ $$ \require{boldsymbol}

$$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)} \newcommand{\funcsub}[3]{\mathrm{#1}{#2}(#3)} \newcommand{\excp}[1]{\mathrm{E}(#1)} $$ $$ \require{boldsymbol}

東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０１７年　第三問</span> $$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)} \newcommand{\funcsub}[3]{\mathrm{#1}{#2}(#3)} \newcommand{\excp}[1]{\mathrm{E}(#1)} \newcommand{\cdf}[1]{\mathrm{F}(#1)} $$

東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０１７年　第一問</span> $$ \require{cancel} $$ $$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)}

東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０２２年　第三問</span> $$ \require{cancel} $$ $$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)}

$$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)} \newcommand{\funcsub}[3]{\mathrm{#1}{#2}(#3)} \newcommand{\excp}[1]{\mathrm{E}(#1)} $$ $$ \require{boldsymbol}

東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０２２年　第一問</span> $$ \require{cancel} $$ $$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)}

東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０１９年　第三問</span> $$ \require{cancel} $$ $$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)}

東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０１９年　第一問</span> $$ \require{cancel} $$ $$ \newcommand{\prob}[1]{\mathrm{P}(#1)} \newcommand{\probsub}[2]{\mathrm{P}{#1}(#2)} \newcommand{\func}[2]{\mathrm{#1}(#2)}

NCKU PD1 OJ 會考成績計算 The Comprehensive Assessment Program for Junior High School Students(CAP), is one of the current Senior high-school entrance program in Taiwan. The exam regulations and grading rules are following: The CAP consists of 5 subjects: Chinese Language, English, Mathematics, Natural Science, and Social Studies. Base on the result of each subject, each are devided into "Excellent" (Grade A), "Fair" (Grade B), and "Improvement Needed" (Grade C) three ranks.Within grade A and B, each will be further divided into three smaller levels "A++/B++", "A+/B+", and "A/B". Each rank corresponds to different amount of points, and each level (within grade A and B) corresponds to a different amount of credits. Both are summed up to get the total points/total credits. The point/credit chart is as following: The subjects Chinese Language, Social Study and Natural Science are graded by the number of corrent answers.

２０１７年　第一問 ２０１７年　第三問 ２０１８年　第一問 ２０１９年　第一問 ２０１９年　第三問 ２０１８年　第三問 ２０２０年　第三問 ２０２２年　第三問 ２０２２年　第一問

MIT OCW 18.06SC Linear Algebra - Unit 2 Least Squares, Determinants and Eigenvalues $$ \require{enclose} \require{textmarcos} \newcommand\hlight[1]{\enclose{roundedbox}[mathcolor=gold]{#1}} \newcommand\bmat[1]{\begin{bmatrix}#1\end{bmatrix}} \newcommand\vmat[1]{\begin{vmatrix}#1\end{vmatrix}} \newcommand\inv[1]{#1^{\text{-}1}} \newcommand\transpose[1]{#1^\textrm{T}}

MIT OCW 18.06SC Linear Algebra - Unit 1 Ax=b and the 4 foundamental subspaces $$ \require{enclose} \require{textmarcos} \newcommand\hlight[1]{\enclose{roundedbox}[mathcolor=gold]{#1}} \newcommand\bmat[1]{\begin{bmatrix}#1\end{bmatrix}} \newcommand\inv[1]{#1^{\text{-}1}} $$

Operating System Chapter 01Introduction to Operating System Component of a Operating System Kernel — The program that is always running on the computer. Middleware — Frameworks that ease application development and provide feature. System Software — Aids the kernel. Component of a General-Purpose Computer

Claims Chapter 1.1, Mathematical Preliminaries and Notation Chapter 1.2, Three Basic Concepts Chapter 1.3, Some Applications Chapter 2.1, Deterministic Finite Accepters Chapter 2.2, Nondeterministic Finite Accepters []

BOTTOM-UP-HEAPSORT, a new variant of HEAPSORT beating, on an average, QUICKSORT (if n is not very small) Number of Comparison of Sorting Algorithms General Comparision Sort Worst Case Analysis For a input of $n$ different entries, there are at most $n!$ permutations, which only 1 of them is the sorted list. If a algorithm always fisinhs after $f(n)$ steps (comparisons), it can at most distinguish $2^{f(n)}$ different permutaions, due to the fact there are only 2 result of each comparison, eliminating half of the possibilities. Therefore, the algorithm must at least yield $2^{f(n)}\geq n!$, or $f(n)\geq\log_2{n!}$ Since the number of comparisons is definetely a whole number, we can assume the lower bound of it at worst case as $\big\lceil\log{n!\ }\big\rceil$, then use the Strling's Formula to simplify: $$\begin{align}