# Laplace’s Determinism

If everything is determined (by its causes), there is no free will.

If everything is random (aka not determined), there is also no free will.

If there is free will, it is neither cases.

— Me@2011.08.20

— Me@2018-02-27

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# 軟硬智力 2.2

Hardware intelligence is part software intelligence due to the basic education got at very young age, for it is the time when the basic intellectual hardware structure of the brain formed.

— Me@2011.08.19

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# 機遇再生論 1.7

$P(A_m)= 1 - (1 - \frac{1}{N})^m$

$P(A_m)$
$= 1 - (1 - \frac{1}{N})^m$
$= 1 - (1 - \frac{1}{52!})^{10,000,000}$

$1.239799930857148592 \times 10^{-61}$

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「（在一次牌都未洗的時候問）洗牌 二千萬 次，起碼一次洗到原本排列 A 的機會率」

「（在一次牌都未洗的時候問）洗牌 一千萬 次，起碼一次洗到原本排列 A 的機會率」。

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$P(A_m)$
$= 1 - (1 - \frac{1}{N})^m$
$= 1 - (1 - \frac{1}{52!})^{10,000,000 \times 2}$
$\approx 2.479599861714297185 \times 10^{-61}$,

$P(A_m)$
$= 1 - (1 - \frac{1}{52!})^{10,000,000 \times 10,000,000}$
$\approx 1.2397999308571485923950342 \times 10^{-54}$

$m = 10,000,000^3, P(A_m) = 1 - (1 - \frac{1}{52!})^{10,000,000^3} \approx 1.2398 \times 10^{-47}$

$m = 10,000,000^4, P(A_m) \approx 1.2398 \times 10^{-40}$

$m = 10,000,000^8, P(A_m) \approx 1.2398 \times 10^{-12}$

$m = 10,000,000^9, P(A_m) \approx 0.000012398$

$m = 10,000,000^{10}, P(A_m) = 0.9999999...$

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$P(A_{10,000,000^{10}}) = 0.9999999...$

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（問：你的意思是，即使我洗了（例如）一千萬牌，仍然得不回原本的排列 A，只要我洗多一千萬次，得回 A 的機會，就會大一點？）

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$P(A) = \frac{1}{N}$

$(N = 52! \approx 8.07 \times 10^{67})$

$P(A_{10,000,000})$
$\approx 1.2398 \times 10^{-61}$

$P(A_{10,000,000})$
$\approx 1.2398 \times 10^{-61}$

$P(A_{20,000,000})$
$\approx 1.2398 \times 10^{-54}$

— Me@2018-02-23 08:21:52 PM

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# On Lisp

Lisp is an especially good language for writing extensible programs because it is itself an extensible program.

Because Lisp gives you the freedom to define your own operators, you can mold it into just the language you need. If you’re writing a text-editor, you can turn Lisp into a language for writing text-editors. If you’re writing a CAD program, you can turn Lisp into a language for writing CAD programs. And if you’re not sure yet what kind of program you’re writing, it’s a safe bet to write it in Lisp. Whatever kind of program yours turns out to be, Lisp will, during the writing of it, have evolved into a language for writing that kind of program.

— On Lisp: Advanced Techniques for Common Lisp

— Paul Graham

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2018.02.21 Wednesday ACHK

# Problem 14.3b6

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring ($\alpha' M^2 = 1$ and $\alpha' M^2 = 2$). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

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When $\alpha' M^2 = 2$, by Equation (14.54), the possible states are

$\{ \alpha_{-2}^I, \alpha_{-1}^I \alpha_{-1}^J, d_{-1}^I d_{-1}^J \} | R_a \rangle, || ...$

$\{ \alpha_{-1}^I d_{-1}^J, d_{-2}^I \} | R_{\bar a} \rangle, || ...$

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For $\alpha_{-2}^I | R_a \rangle$, the number of states is 8.

For $\alpha_{-1}^I \alpha_{-1}^J | R_a \rangle$, the number of states is $\frac{8 \times 7}{2} + 8 = 36$.

For $d_{-1}^I d_{-1}^J | R_a \rangle$, the number of states is $\frac{8 \times 7}{2} = 28$.

For $\alpha_{-1}^I d_{-1}^J | R_{\bar a} \rangle$, the number of states is $8 \times 8 = 64$.

For $d_{-2}^I | R_{\bar a} \rangle$, the number of states is 8.

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However, since each of $a$ and ${\bar a}$ has 8 possible values, there is an additional multiple of 8.

The total number of states is $8 \left[ 8 + 36 + 28 + 64 + 8 \right]$.

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You can check this answer against Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

— Me@2018.02.20 10:57 AM

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# The language of Change 1.2

Energy conservation, 6.2 | Energy 5.2

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time ~ change

energy ~ the ability of _keeping_ changing

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constant velocity ~ the amount of an object’s change of position, measured with respect to its observer’s unit of change, is constant

$s = \Delta x$

$v = \frac{s}{\Delta t} = \frac{\Delta x}{\Delta t}$

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kinetic energy ~ the amount of the ability of keeping changing an object’s position

$\frac{1}{2} m v^2$ ~ the square of (the amount of change of position, relative to the observer’s unit of change)

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Energy difference is _not_ exactly a measurement of the amount of change, time interval is.

— Me@2018-02-20 09:39:30 AM

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# Cat

— Me@2018-02-19 02:33:42 PM

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# 潛行凶間 16

Inception 16

1. 清醒夢

1.1 有些人在某些時候，在夢知道自己在發夢，卻又可以保持住，發夢的狀態。

1.2 那些人之中的部分人，在那些清醒夢時候的部分時候，甚至可以控制著，那些夢境的劇情演變。

（CPK：未。不過，我的姐姐試過。）

《潛行凶間》中的意念，你可以假想，有七成是真的。

2. 多層夢

3. 多重自我

3.1 每一個人，其實有超過一個自我。

3.2 而每一個自我，其實有超過一個層次的意識。

— Me@2018.02.18

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# Creative constraints

Imagine you were asked to invent something new. It could be whatever you want, made from anything you choose, in any shape or size. That kind of creative freedom sounds so liberating, doesn’t it? Or … does it?

If you’re like most people you’d probably be paralyzed by this task. Why?

Brandon Rodriguez explains how creative constraints actually help drive discovery and innovation.

With each invention, the engineers demonstrated an essential habit of scientific thinking – that solutions must recognize the limitations of current technology in order to advance it.

Understanding constraints guides scientific progress, and what’s true in science is also true in many other fields.

Constraints aren’t the boundaries of creativity, but the foundation of it.

— The power of creative constraints

— Lesson by Brandon Rodriguez

— animation by CUB Animation

— TED-Ed

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We cannot change anything until we accept it. Condemnation does not liberate, it oppresses.

— Carl Jung

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# Twelve-step program

A twelve-step program is a set of guiding principles outlining a course of action for recovery from addiction, compulsion, or other behavioral problems. Originally proposed by Alcoholics Anonymous (AA) as a method of recovery from alcoholism, the Twelve Steps were first published in the 1939 book Alcoholics Anonymous: The Story of How More Than One Hundred Men Have Recovered from Alcoholism. The method was adapted and became the foundation of other twelve-step programs.

As summarized by the American Psychological Association, the process involves the following:

– recognizing a higher power that can give strength;

– examining past errors with the help of a sponsor (experienced member);

– making amends for these errors;

– learning to live a new life with a new code of behavior;

– helping others who suffer from the same alcoholism, addictions or compulsions.

— Wikipedia on Twelve-step program

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We cannot change anything until we accept it. Condemnation does not liberate, it oppresses.

— Carl Jung

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# 深淵 2

— 尼采

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As soon as men decide that all means are permitted to fight an evil, then their good becomes indistinguishable from the evil that they set out to destroy.

— Christopher Dawson

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2018.02.16 Friday ACHK

# Problem 14.3b5

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring ($\alpha' M^2 = 1$ and $\alpha' M^2 = 2$). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

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When $\alpha' M^2 = 2$, by Equation (14.37),

$M^2 = \frac{1}{\alpha'} \left( - \frac{1}{2} + N^\perp \right)$

$N^\perp = \frac{5}{2}$

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$\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M,$

$b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M,$

$b_{-5/2}^I,$

$\alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M,$

$\alpha_{-1}^I b_{-3/2}^J,$

$\alpha_{-2}^I b_{-1/2}^J,$

$\alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

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There are $\frac{8 \times 7 \times 6 \times 5 \times 4}{5!} = {8 \choose 5} = 56$ states for

$\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times \frac{8 \times 7}{2} = 224$ states for

$\{ b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8$ states for

$\{ b_{-5/2}^I \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times {8 \choose 3} = 448$ states for

$\{ \alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8 = 64$ states for

$\{ \alpha_{-1}^I b_{-3/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8 = 64$ states for

$\{ \alpha_{-2}^I b_{-1/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $\left( \frac{8 \times 7}{2!} + 8 \right) \times 8 = 288$ states for

$\{ \alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

So the total number of states is 56 + 224 + 8 + 448 + 64 + 64 + 288 = 1152.

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You can check this answer against Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

— Me@2018-02-16 03:22:13 PM

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# The language of Change

Energy conservation, 6 | Energy 5

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time ~ change

energy ~ the ability of causing change

Assuming

1. a system of one single particle

2. has only kinetic energy

3. and that kinetic energy is conserved.

conservation of energy ~ an object’s potential amount of change of position, measured with respect to its observer’s unit of change, is constant

$s = \Delta x$

$v = \frac{s}{\Delta t} = \frac{\Delta x}{\Delta t}$

— Me@2018-02-15 02:21:20 PM

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Note:

The above argument has a bug:

If the mass m is constant, the kinetic energy $E_K$ should be proportional to velocity squared $v^2$, instead of velocity $v$.

$E_K = \frac{1}{2} m v^2$

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However, the above argument is still technically correct:

When $E_K$ is constant, $v^2$ is constant. In turn, the magnitude of $v$ also remains unchanged.

— Me@2018-02-19 09:37:24 PM

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# Definition of Time, Prime

Energy 4

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time ~ change

energy ~ the ability of causing change

— Me@2018-02-15 10:20:25 AM

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# Importance, 2

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When you give a lot of importance to someone in your life,

you tend to lose your importance in their life.

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2018.02.14 Wednesday ACHK

# 機遇再生論 1.6

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（而這個意思，亦在「機遇再生論」的原文中，用作其理據。）

$P(A) = \frac{1}{N}$

$P(\text{not} A) = 1 - \frac{1}{N}$

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$P(A) = \frac{1}{N}$

$P(\text{not} A) = 1 - \frac{1}{N}$

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(問：那樣，為什麼要問多一次呢？）

「如果洗牌兩次，起碼一次洗到原本排列 A 的機會率是多少？」

$A_2$ = 兩次洗牌的結果，起碼一次洗到原本排列 A

$A_2$ 的互補事件為「不是 $A_2$」：

= 兩次洗牌的結果，不是起碼一次洗到原本排列 A

= 兩次洗牌的結果，都不是排列 A

$P(\text{not} A_2) = (1 - \frac{1}{N})^2$

$P(A_2)$
$= 1 - P(\text{not} A_2)$
$= 1 - (1 - \frac{1}{N})^2$

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$P(A_m)= 1 - (1 - \frac{1}{N})^m$

$P(A_m)$
$= 1 - (1 - \frac{1}{N})^m$
$= 1 - (1 - \frac{1}{52!})^{10,000,000}$

$1.239799930857148592 \times 10^{-61}$

— Me@2018-01-25 12:38:39 PM

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