# Quick Calculation 3.8

A First Course in String Theory

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Show that this condition fixes uniquely $\displaystyle{\alpha = \gamma = 1/2}$, and $\displaystyle{\beta = - 3/2}$, thus reproducing the result in (3.90).

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Eq. (3.93):

$\displaystyle{l_P = (G)^\alpha (c)^\beta (\hbar)^\gamma}$

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$\displaystyle{l_P = \left( \frac{l_p^3}{m_p t_P^2} \right)^\alpha \left( \frac{l_P}{t_P} \right)^\beta \left( \frac{m_P l_P^2}{t_P} \right)^\gamma}$

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\displaystyle{\begin{aligned} 3 \alpha + \beta + 2\gamma &= 1 \\ -\alpha + \gamma &= 0 \\ - 2 \alpha - \beta - \gamma &= 0 \\ \end{aligned}}

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var('a b c')

solve([3*a+b+2*c==1, -a+c==0, -2*a-b-c==0], a, b, c)


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\displaystyle{\begin{aligned} \alpha &= \frac{1}{2} \\ \\ \beta &= \frac{-3}{2} \\ \\ \gamma &= \frac{1}{2} \\ \\ \end{aligned}}

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Eq. (3.90):

$\displaystyle{l_P = \sqrt{\frac{G \hbar}{c^3}}}$

— Me@2022-06-23 10:46:22 AM

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# Schrodinger cat’s misunderstanding

Schrodinger’s cat, 3.2

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In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The main point of the Schrödinger’s cat thought experiment is NOT to prove that there should also be superposition for macroscopic objects. Instead, the main point of the thought experiment is exactly the opposite—to prove that regarding a superposition state as a physical state leads to logical contradiction.

— Me@2022-06-15 07:19:36 PM

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1 photo
HK\$10

— ~2011

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# 數學教育 7.5.3

A Fraction of Algebra, 2.3

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— Me@2022-06-21 11:51:20 AM

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# C-h ?

 C-h w command-name

 C-h k key-sequence 

C-h f function-name 

— Me@2022-06-18 12:52:10 PM

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# Ex 1.27 Identifying total time derivatives

Structure and Interpretation of Classical Mechanics

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From equation (1.112), we see that $\displaystyle{G}$ must be linear in the generalized velocities

$\displaystyle{ G(t, q, v) = G_0(t, q, v) + G_1(t, q, v) v }$

where neither $\displaystyle{G_1}$ nor $\displaystyle{G_0}$ depend on the generalized velocities: $\displaystyle{\partial_2 G_1 = \partial_2 G_0 = 0}$.

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So if $\displaystyle{G}$ is the total time derivative of $\displaystyle{F}$ then

$\displaystyle{ \partial_0 G_1 = \partial_1 G_0 }$

For each of the following functions, either show that it is not a total time derivative or produce a function from which it can be derived.

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[guess]

a. $\displaystyle{G(t, x, v_x) = m v_x}$

\displaystyle{ \begin{aligned} G_0 &= 0 \\ G_1 &= m \\ \\ \partial_0 G_1 &= 0 \\ \partial_1 G_0 &= 0 \\ \\ \end{aligned} }

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\displaystyle{ \begin{aligned} \partial_0 F &= 0 \\ F &= k_0(x, v_x) \\ \\ \partial_1 F &= m \\ F &= m x + k_1(t, v_x) \\ \end{aligned} }

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Let

\displaystyle{ \begin{aligned} k_0(x, v_x) &= mx \\ k_1(t, v_x) &= 0 \\ \end{aligned} }

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Then

\displaystyle{ \begin{aligned} F &= mx \\ \end{aligned} }

[guess]

— Me@2022-06-17 05:10:38 PM

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# Schrodinger’s cat, 3.1

It is natural to ask why ordinary everyday objects and events do not seem to display quantum mechanical features such as superposition. Indeed, this is sometimes regarded as “mysterious”, for instance by Richard Feynman. In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The modern view is that this mystery is explained by quantum decoherence. A macroscopic system (such as a cat) may evolve over time into a superposition of classically distinct quantum states (such as “alive” and “dead”). However, the state of the cat is entangled with the state of its environment (for instance, the molecules in the atmosphere surrounding it). If one averages over the quantum states of the environment—a physically reasonable procedure unless the quantum state of all the particles making up the environment can be controlled or measured precisely—the resulting mixed quantum state for the cat is very close to a classical probabilistic state where the cat has some definite probability to be dead or alive, just as a classical observer would expect in this situation.

Quantum superposition is exhibited in fact in many directly observable phenomena, such as interference peaks from an electron wave in a double-slit experiment. Superposition persists at all scales, provided that coherence is shielded from disruption by intermittent external factors. The Heisenberg uncertainty principle states that for any given instant of time, the position and velocity of an electron or other subatomic particle cannot both be exactly determined. A state where one of them has a definite value corresponds to a superposition of many states for the other.

— Wikipedia on Quantum superposition

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It is natural to ask why ordinary everyday objects and events do not seem to display quantum mechanical features such as superposition. Indeed, this is sometimes regarded as “mysterious”, for instance by Richard Feynman.

Superposition is not “mysterious”. It is “mysterious” only if you regard “a superposition state” as a physical state.

Only observable states are physical states. Any observable, microscopic or macroscopic, is NOT a superposition.

A superposition is NOT observable, even in principle; because the component states of a superposition are physically-indistinguishable mathematical states, aka macroscopically-indistinguishable microscopic states.

(Those component states, aka eigenstates, are observable and distinguishable once the corresponding measuring device is allowed.)

They are indistinguishable because the distinction is not defined in terms of the difference between different potential experimental or observational results.

Actually, the distinction is not even definable, because the corresponding measuring device is not allowed in the experimental design yet.

— Me@2022-06-15 11:51:22 AM

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# 超整理法

— 「超級」整理．超高效率

— 天下雜誌173期

— 孫曉萍

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— Me@2022-06-14 03:54:57 PM

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2022.06.14 Tuesday ACHK

# 烹飪事實

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「善意謊言」的最大問題是，「善意」不代表「善行」，更不代表「善果」。到病人最終知道真相時，只加倍傷心。

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「我明日下午見，一位異性老朋友。你一同去嗎？」

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— Me@2022-06-12 01:08:49 PM

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# Quick Calculation 3.7

A First Course in String Theory

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The force $\displaystyle{\vec F}$ on a test charge $\displaystyle{q}$ in an electric field $\displaystyle{\vec E}$ is $\displaystyle{\vec F = q \vec E}$. What are the units of charge in various dimensions?

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Eq. (3.74):

$\displaystyle{E(r) = \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{Q}{r^{d-1}}}$

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\displaystyle{ \begin{aligned} \vec F &= q \vec E \\ &= \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{qQ}{r^{d-1}} \\ \\ \\ [\vec F] &= \left[ \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{qQ}{r^{d-1}} \right] \\ &= \left[ \frac{1}{1} \frac{q^2}{r^{d-1}} \right] \\ &= \left[ \frac{q^2}{r^{d-1}} \right] \\ \end{aligned}}

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\displaystyle{ \begin{aligned} [r^{d-1} \vec F] &= \left[ q^2 \right] \\ \\ \left[ q \right] &= [\sqrt{r^{d-1} \vec F}] \\ \\ &= \sqrt{m^{d-1} N} \\ \\ \end{aligned}}

— Me@2022-06-08 11:09:27 AM

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Lorentz–Heaviside units (or Heaviside–Lorentz units) constitute a system of units (particularly electromagnetic units) within CGS, named for Hendrik Antoon Lorentz and Oliver Heaviside. They share with CGS-Gaussian units the property that the electric constant $\displaystyle{\epsilon_0}$ and magnetic constant $\displaystyle{\mu_0}$ do not appear, having been incorporated implicitly into the electromagnetic quantities by the way they are defined. Heaviside-Lorentz units may be regarded as normalizing $\displaystyle{\epsilon_0 = 1}$ and $\displaystyle{\mu_0 = 1}$, while at the same time revising Maxwell’s equations to use the speed of light $\displaystyle{c}$ instead.

Heaviside–Lorentz units, like SI units but unlike Gaussian units, are rationalized, meaning that there are no factors of $\displaystyle{4 \pi}$ appearing explicitly in Maxwell’s equations. That these units are rationalized partly explains their appeal in quantum field theory: the Lagrangian underlying the theory does not have any factors of $\displaystyle{4 \pi}$ in these units. Consequently, Heaviside-Lorentz units differ by factors of $\displaystyle{\sqrt{4\pi}}$ in the definitions of the electric and magnetic fields and of electric charge. They are often used in relativistic calculations, and are used in particle physics. They are particularly convenient when performing calculations in spatial dimensions greater than three such as in string theory.

— Wikipedia on Lorentz–Heaviside units

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# Visualize

visualize ~ feel all at once

— Me@2016-09-28 08:20:18 PM

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visual ~ same-time representation ~ spatial representation

— Me@2016-06-30 07:38:28 AM

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# Emacs, 3

jarcane on Oct 10, 2014 | next [–]

“They used a manual someone had written which showed how to extend Emacs, but didn’t say it was a programming. So the secretaries, who believed they couldn’t do programming, weren’t scared off.”

easytiger on Oct 10, 2014 | parent | next [–]

I was reading a book on a long train journey from Paris to Nice by train that I read usually once a year. My girlfriend couldn’t understand why I was reading a kids book as it had an elephant on the front. I told her it wasn’t, but was in fact one of the most profound books about teaching you to think in a new way.

So I let her read it and she got about half way through and she totally got it and loved it. No harder than doing a crossword or a sudoku for the first time.

The book is: little-schemer

— My Lisp Experiences and the Development of GNU Emacs (2002)

— Hacker News

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2022.06.07 Tuesday ACHK

# 飯餸湯

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（但是，如果你不是俊男，那就只是兩場誤會，通常。）

— Me@2022-06-06 05:12:48 PM

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# 3Blue1Brown

(defun 3b1b ()

(interactive)

(setq is-python-mode (string= major-mode "python-mode"))

(if (not is-python-mode)
(print "This is not a python file.")

(print buffer-file-name)

(setq the-command (format "%s %s %s"
"manim -p"
buffer-file-name
"JustAShape"))

(print the-command)

(shell-command the-command)
)
)

(global-set-key (kbd "C-p") '3b1b)

(global-set-key (kbd "C-/") 'comment-region)

(global-set-key (kbd "C-.") 'uncomment-region)


— Me@2022-06-05 04:00:37 PM

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# Ex 1.26 Lagrange equations for total time derivatives

Structure and Interpretation of Classical Mechanics

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Let $\displaystyle{F(t, q)}$ be a function of $\displaystyle{t}$ and $\displaystyle{q}$ only, with total time derivative

$\displaystyle{D_t F = \partial_0 F + \partial_1 F \dot Q}$

Show explicitly that the Lagrange equations for $\displaystyle{D_t F}$ are identically zero, …

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[guess]

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial}{\partial \dot q} L (t, q(t), \dot q(t)) \right) - \frac{\partial}{\partial q} L (t, q(t), \dot q(t)) &= 0 \end{aligned}}

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Eq. (1.114):

\displaystyle{ \begin{aligned} D_t F (t, q, v, a, ...) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \\ \end{aligned} }

\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] (t) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }

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Put $\displaystyle{ D_t F (t, q, Dq, ...) }$ into the Lagrange equation:

\displaystyle{ \begin{aligned} &\frac{d}{dt} \left( \frac{\partial}{\partial \dot q} D_t F (t, q, Dq, ...) \right) - \frac{\partial}{\partial q} D_t F (t, q, Dq, ...) \\ \\ &= \frac{d}{dt} \left[ \frac{\partial}{\partial \dot q} \left( \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \right) \right] \\ &- \frac{\partial}{\partial q} \left( \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \right) \\ \\ &=\frac{d}{dt} \left[ \partial_2 \partial_0 F + \partial_2 (v \partial_1 F) + \partial_2 (a \partial_2 F) + ... \right] \\ &- \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] \\ \\ &= \partial_0 \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] \\ &+ \partial_1 \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] v \\ &+ \partial_2 \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] a + ... \\ &- \left[ \partial_1 \partial_0 F + \partial_1 (v \partial_1 F) + \partial_1 (a \partial_2 F) + ... \right] \\ \\ &= ... \\ \end{aligned} }

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Assume that $\displaystyle{\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}}$:

\displaystyle{ \begin{aligned} &\frac{d}{dt} \left( \frac{\partial}{\partial \dot q} D_t F (t, q, Dq, ...) \right) - \frac{\partial}{\partial q} D_t F (t, q, Dq, ...) \\ \\ &= \frac{d}{dt} \left( \frac{\partial \dot F }{\partial \dot q} \right) - \frac{\partial}{\partial q} \frac{d F}{dt} \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{\partial}{\partial q} \frac{dF}{dt} \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{\partial}{\partial q} \left( \frac{\partial F}{\partial t} + \frac{\partial F}{\partial q} \dot q + \frac{\partial F}{\partial \dot q} \ddot q + ... \right) \\ \\ \end{aligned}}

\displaystyle{\begin{aligned} &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t} + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q + \frac{\partial F}{\partial q} \frac{\partial}{\partial q} \dot q + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + \frac{\partial F}{\partial \dot q} \frac{\partial}{\partial q} \ddot q + ... \right) \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t} + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q + \frac{\partial F}{\partial q} (0) + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + \frac{\partial F}{\partial \dot q} (0) + ... \right) \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t} + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + ... \right) \\ \\ &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) \\ \\ &= 0 \\ \\ \end{aligned} }

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Prove that $\displaystyle{\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}}$.

\displaystyle{\begin{aligned} &\frac{\partial \dot F}{\partial \dot q} \\ \\ &= \frac{\partial }{\partial \dot q} \left( \frac{\partial F}{\partial t} + \frac{\partial F}{\partial q} \dot q + \frac{\partial F}{\partial \dot q} \ddot q + ... \right) \\ \\ &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right) \dot q + \frac{\partial F}{\partial q} \frac{\partial }{\partial \dot q}\dot q + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q} \right) \ddot q + \frac{\partial F}{\partial \dot q} \frac{\partial }{\partial \dot q}\ddot q + ... \\ \\ &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right) \dot q + \frac{\partial F}{\partial q} (1) + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q} \right) \ddot q + \frac{\partial F}{\partial \dot q} (0) + ... \\ \\ &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right)\dot q + \frac{\partial F}{\partial q} + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q}\right) \ddot q + ... \\ \\ &= \left[ \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial t} + \left( \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial q} \right)\dot q + \left( \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial \dot q}\right) \ddot q + ... \right] + \frac{\partial F}{\partial q} \\ \\ &= \frac{d}{dt} \frac{\partial F }{\partial \dot q} + \frac{\partial F}{\partial q} \\ \\ \end{aligned}}

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If $\displaystyle{ L' = L + D_t F }$ depends on $\displaystyle{ \{t, q, Dq\} }$ only, then $\displaystyle{ F }$ will depend on $\displaystyle{ \{t, q\} }$ only.

\displaystyle{\begin{aligned} \frac{\partial \dot F}{\partial \dot q} &= \frac{d}{dt} \frac{\partial F }{\partial \dot q} + \frac{\partial F}{\partial q} \\ \\ &= \frac{d}{dt} (0) + \frac{\partial F}{\partial q} \\ \\ &= \frac{\partial F}{\partial q} \\ \\ \end{aligned}}

[guess]

— Me@2022-06-03 09:04:37 PM

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# PS2

CPU: Core 2 Duo 3.2 GHz / Core-I series
GPU: GeForce 9600 GT
2GB RAM

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2022.06.03 Friday ACHK

# Programming Lab

Time: 9:30am-12:30pm ([T]hursday)

Venue: SHB123

— ~ 2011

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2022.06.03 Friday ACHK

# 電腦病毒

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— Me@2022-06-02 11:52:12 AM

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