迷宮直昇機 5.2

Ken Chan 時光機 1.4.2

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但是,他的學校不正常——在中四和中五的校內考試測驗中,不斷地考核高考課程。所以,他在中四時代開始,已經要鑽研,高考程度和大學程度的物理。

結果,到真正會考時,由於「只會」考核中五程度的東西,他會突然覺得十分容易。

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其實,那種快感不難體會。

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假設,你現在是中學四年級。

你想像,如果突然之間,那份中四的數學卷,換成中一程度的數學卷。你會有什麼感受?

你不單會覺得如釋重負,而且會有信心,有機會取滿分;即使,那份中一測驗卷的課題,在中一以後,從來未刻意溫習過。

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(問:我又不敢說,我一定會滿分。)

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我沒有說「一定」。我只是說「覺得有機會」。

試想想,現為中四的你,面對一份中四的數學卷,你夠不夠膽說「有機會取滿分」?

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(問:不太敢,因為,那幾乎沒有可能。但是,我明白你的意思。現在要我做回中一的測驗卷,又真的會覺得,一定會高分,可能會滿分。)

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其實,那種情形,對現為中四的你而言,並不只是一個「假設」,因為,你現正修讀, G.Maths(核心數學)和 A.Maths(附加數學)。

「附加數」比「基礎數」而言,艱深非常,大部分人也覺得十分辛苦。但是,亦正正是因為「附加數艱深非常」,你才會覺得「核心數容易萬分」。

那又為什麼,會有這個現象呢?

— Me@2018-12-09 09:03:02 PM

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2018.12.09 Sunday (c) All rights reserved by ACHK

EQL5-Android

Common Lisp for Android App Development 2018

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The first step to set up EQL5-Android is to install Qt.

d_2018_12_08__21_55_22_PM_

In Ubuntu, if you do not need the most updated Qt, you can just install the Qt-Creator using apt-get.

— Me@2018-12-08 09:18:19 PM

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2018.12.08 Saturday (c) All rights reserved by ACHK

Problem 14.5c3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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Open String:

\displaystyle{   \begin{aligned}  N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\   L_n^{\perp} &= \frac{1}{2} \sum_{-\infty}^{\infty} \alpha_{n-p}^I \alpha_{p}^I~~~(n \ne 0) \\   L_0^{\perp} &= \alpha' p^I p^I + N^\perp   \end{aligned}  }

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Closed String:

\displaystyle{   \begin{aligned}  N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\   \bar N^\perp &= \sum_{p=1}^{\infty} \bar \alpha_{-p}^I \bar \alpha_p^I \\   L_0^\perp &= \frac{\alpha'}{4} p^I p^I + N^\perp \\   \bar L_0^\perp &= \frac{\alpha'}{4} p^I p^I + \bar N^\perp   \end{aligned}  }

— A First Course in String Theory

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We have the following well-known bosonic string mass formulae \displaystyle{\alpha' = \frac{1}{2}}:

open string:

\displaystyle{\frac{1}{2} M^2 = N - a}

closed string:

\displaystyle{\frac{1}{8} M^2 = N_L - a}
\displaystyle{\frac{1}{8} M^2 = N_R - a}

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

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What is the meaning of \displaystyle{\alpha'}?

How come \displaystyle{\alpha' = \frac{1}{2}}?

— Me@2018-12-07 10:43:10 PM

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2018.12.07 Friday (c) All rights reserved by ACHK

Double slit experiment, 8

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Although the screen itself is a photon position detector, it gets no which-way information. So it can get an interference pattern.

— Me@2012-04-09 7:24:23 PM

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2018.11.27 Tuesday (c) All rights reserved by ACHK

PhD, 2.1

故事連線 1.1.3 | 碩士 3.1

這段改編自 2010 年 4 月 18 日的對話。

.

(安:你的意思時,不鼓勵年青人,攻讀研究院?)

不是。

研究工作,有其獨特的至尊好處。我反而覺得,每個人都應花兩年時間,經歷一次。換句話說,你可以考慮,先讀一個研究式的碩士,然後才盤算,自己適不適合再攻讀博士。

不過,你要留意,無論是攻讀碩士還是博士,你在之前選擇指導教授時,都要格外小心。你要保證,你的指導教授,有品德和有才能,幫你把與論文課題沒有直接關係的工作,全部推開。當然,那樣可靠的教授,萬中無一。

— Me@2012.08.20

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在從來未與一位教授共事過的情況下,就選他為自己的碩士或博士論文導師,是十分危險的事。

那足以令你,自此不能再於學術界發展,抱憾終生。

(問:在那位教授成為你的論文導師前,又怎會共事過呢?)

可以在之前,有過其他形式的工作關係,直接或間接地,知道他的才德如何。

(問:你意思時,在本科時,就選修他的課?)

可以那樣說。聽他的講課和做他所予之功課,就可以知道他,工作態度認不認真、思考清不清晰 和 顧不顧及他人感受。但是,那也只是第一重的「測試」,先決而未充分;因為,那只是他的「公眾形象」而已。當他作為你的論文導師,你作為他的研究生時,你再也不是,和他的「公眾形象」相處了。

有很多人的「個人形象」和「公眾形象」,都是差天共地的。

(問:即是表裡不一?)

有時,甚至是「表裡相反」。

(問:那應該怎麼辦?)

— Me@2018-11-24 08:17:17 PM

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2018.11.25 Sunday (c) All rights reserved by ACHK

Common Lisp for Android App Development 2018

An REPL called “CL REPL” is available in the Google Play Store. But itself is not for developing standalone Android apps, unless those apps are Common Lisp source code files only.

However, “CL REPL” itself is an open source GUI app using Common Lisp and Qt. So by learning and using its source, in principle, we can create other Android apps using Common Lisp with Qt.

The library that “CL REPL” uses is EQL5-Android.

— Me@2018-11-23 04:07:54 PM

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2018.11.23 Friday (c) All rights reserved by ACHK

Problem 14.5c2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = 4k} with the right-moving R- states with \displaystyle{\alpha' M_R^2 = k}.

c) Are there tachyonic states in heterotic string theory?

~~~

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— This answer is my guess. —

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The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}

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The left R’+ sector:

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}

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The right-moving NS+ states:

NS+ equations of (14.38):

\displaystyle{\begin{aligned}  \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\  \alpha'M^2=1, ~~~&N^\perp = \frac{3}{2}: &\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\\  \alpha'M^2=2, ~~~&N^\perp = \frac{5}{2}: &\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}~& \\  &&\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle \end{aligned}}

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The R- states (that used as right-moving states):

Mass levels of R- and R+ (Equations 14.54):

\displaystyle{\begin{aligned}  \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle~~&||~~|R_{\bar a} \rangle \\  \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\  \alpha'M^2=2,~~~&N^\perp = 2:~~~~&\{ \alpha_{-2}^I,~\alpha_{-1}^I \alpha_{-1}^J,~d^I_{-1} d^J_{-1} \} |R_{a} \rangle,~&|| \\  &&\{\alpha_{-1}^I d_{-1}^J,~d_{-2}^I \} |R_{\bar a} \rangle~~&||~~ ... \\ \end{aligned}}

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There are no tachyonic states in heterotic string theory, since neither of the right-moving parts (NS+ and R-) has states with \displaystyle{\begin{aligned} \alpha' M^2 < 0\end{aligned}}.

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— This answer is my guess. —

— Me@2018-11-22 12:00:30 PM

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2018.11.22 Thursday (c) All rights reserved by ACHK

The problem of induction 3

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In a sense (of the word “pattern”), there is always a pattern.

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Where if there are no patterns, everything is random?

Then we have a meta-pattern; we can use probability laws:

In that case, every (microscopic) case is equally probable. Then by counting the possible number of microstates of each macrostate, we can deduce that which macrostate is the most probable.

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Where if not all microstates are equally probable?

Then it has patterns directly.

For example, we can deduce that which microstate is the most probable.

— Me@2012.11.05

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2018.11.19 Monday (c) All rights reserved by ACHK

Insomnia, 5

mingyeow 47 minutes ago [-]

I had signs of insomnia in the last couple of years. Tried all the usual techniques, and I realized those just do not work for one simple reason – that the moment I do not fall asleep, I become frustrated. And that frustration in turn makes sleeping much harder.

So, I tried doing the exact opposite, and my sleep has been tremendously better since.

Essentially, I force myself to keep my eyes open. When I catch myself falling asleep, I actually force myself to keep them often again. Before I know it, it is morning.

It sounds counter-intuitive, but all my insomnia problems are over, including on a plane. Would love to know if it works for others, or if it is just a silly thing that works for only me.

— Insomnia: To Pursue Sleep So Hard You Become Invigorated by the Chase

— Hacker News

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2018.11.20 Tuesday ACHK

Ken Chan 時光機 1.4.1

迷宮直昇機 5.1

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至於第三個技巧,則實在是最宏觀,亦可能是最重要。

Ken Chan 說:「你們現在會盤算,在會考中,各科太概會有什麼目標,奪取什麼等級的成績。但是,我當年全部科目,只有一個目標,就是要『攞 full』」。

那就即是要,全部科目中的每一科,不只要甲等,而是要滿分;因為在當年,如果可以在全部科目奪得滿分,考生會獲頒一張特別證書。他那時就為了,那張證書而努力。

但是,那是如何辦到的呢?

考生在中五時參加「會考」,中七時參加「高考」。所以,正常的學校,會在中四和中五的校內考試測驗中,考核會考課程。

但是,他的學校不正常——在中四和中五的校內考試測驗中,不斷地考核高考課程。所以,他在中四時代開始,已經要鑽研,高考程度和大學程度的物理。

結果,到真正會考時,由於「只會」考核中五程度的東西,他會突然覺得十分容易。

— Me@2018-11-18 10:06:43 PM

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2018.11.18 Sunday (c) All rights reserved by ACHK

defmacro, 2

Defining the defmacro function using only LISP primitives?

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McCarthy’s Elementary S-functions and predicates were

atom, eq, car, cdr, cons

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He then went on to add to his basic notation, to enable writing what he called S-functions:

quote, cond, lambda, label

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On that basis, we’ll call these “the LISP primitives”…

How would you define the defmacro function using only these primitives in the LISP of your choice?

edited Aug 21 ’10 at 2:47
Isaac

asked Aug 21 ’10 at 2:02
hawkeye

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Every macro in Lisp is just a symbol bound to a lambda with a little flag set somewhere, somehow, that eval checks and that, if set, causes eval to call the lambda at macro expansion time and substitute the form with its return value. If you look at the defmacro macro itself, you can see that all it’s doing is rearranging things so you get a def of a var to have a fn as its value, and then a call to .setMacro on that var, just like core.clj is doing on defmacro itself, manually, since it doesn’t have defmacro to use to define defmacro yet.

– dreish Aug 22 ’10 at 1:40

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2018.11.17 Saturday (c) All rights reserved by ACHK

Problem 14.5c

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with \displaystyle{\alpha' M_L^2 = k} with the right-moving NS+ states with \displaystyle{\alpha' M_R^2 = k}.

~~~

In heterotic (closed) string theory, there are left-moving part and right-moving part. Then, what is the meaning of “at any mass level \displaystyle{\alpha' M^2 = 4k} of the heterotic string”?

— Me@2018-11-11 03:44:18 PM

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Type IIA/B closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322

\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}

\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}

\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}

These are the reasons that any mass level of the heterotic string is always in the form \displaystyle{\alpha' M^2 = 4k}.

— Me@2018-11-12 03:09:11 PM

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Equation (14.77):

p.322

closed string sectors: (NS, NS), (NS, R), (R, NS), (R, R)

\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)

\text{type IIB}:~~~(NS+, NS+), ~(NS+, R-),~ (R-, NS+), ~ (R-, R-)

What is the difference between Type IIA/B closed superstrings and heterotic SO(32) strings?

— Me@2018-11-12 03:15:46 PM

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The five consistent superstring theories are:

d_2018_11_15__09_36_16_am_

  • The type I string has one supersymmetry in the ten-dimensional sense (16 supercharges). This theory is special in the sense that it is based on unoriented open and closed strings, while the rest are based on oriented closed strings.
  • The type II string theories have two supersymmetries in the ten-dimensional sense (32 supercharges). There are actually two kinds of type II strings called type IIA and type IIB. They differ mainly in the fact that the IIA theory is non-chiral (parity conserving) while the IIB theory is chiral (parity violating).
  • The heterotic string theories are based on a peculiar hybrid of a type I superstring and a bosonic string. There are two kinds of heterotic strings differing in their ten-dimensional gauge groups: the heterotic E8×E8 string and the heterotic SO(32) string. (The name heterotic SO(32) is slightly inaccurate since among the SO(32) Lie groups, string theory singles out a quotient Spin(32)/Z2 that is not equivalent to SO(32).)

— Wikipedia on Superstring theory

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2018.11.15 Thursday (c) All rights reserved by ACHK

Detecting a photon

In the double-slit experiments, how to detect a photon without destroying it?

— Me@2018-11-10 08:07:29 PM

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Artlav: I’ve been thinking about the double slit experiment – the one with single photons going thru two slits forming an interference pattern never the less. Now, one thing i was unable to find clarification for is the claim that placing a detector even in just one of the slits to find out thru which slit a photon passed will result in the disappearance of the interference pattern. The question is – how does such detector work? How can one detect a photon without destroying it?

Cthugha (Science Advisor): Well, in the kind of experiment you describe, the photon will usually be destroyed by detecting it. However, in some cases it is possible to detect photons without destroying them. Usually one uses some resonator, for example some cavity, in which photons go back and forth and prepare some atom in a very well defined spin state. Now the atom falls through the cavity perpendicular to the photons moving back and forth and the spin state of the atom after leaving the cavity will depend on the number of photons because the spin precession will be a bit faster in presence of photons. If you do this several times, you will get a nondestructive photon number measurement. However, these are so called weak measurements, so this means you do not change the photon states if you are in a photon number eigenstate already. The first measurement however might change the photon state from some undefined state to a photon number eigenstate.

Reference: physicsforums double-slit-experiment-counter.274914

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2018.11.10 Saturday ACHK

凌晨舊戲 2.2

這段改編自 2010 年 4 月 18 日的對話。

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繼而,甲版本的瀕死經驗,可信度亦會高一些;因為,如果甲所講述的「境界極高道理」,其實是他自己的創作的話,一般而言,他並沒有動機,去把功勞賦予一個,虛構的瀕死經驗。

(問:但是,道理又怎樣為之「境界極高」呢?)

這個問題,跟前一個問題性質一樣,所以,答案跟前一個答案相若,都是用「好樹結好果」這個大原則。

能結出好果的,就為之「好樹」。能治療(或舒緩)病患的,就為之「好醫生」。

同理,如果一個建議,或者個觀點,能大大改善,讀者的心靈世界,甚至實際生活的話,那就為之「境界極高道理」。

(以下的「瀕死經驗者」,是「自稱經歷瀕死經驗人仕」的簡稱。同理,「瀕死經驗後」,是「所宣稱的瀕死經驗後」的縮寫。)

有部分瀕死經驗者,在瀕死經驗後,性格太幅度地改善,道德超常地提升。那樣,他口中的道理,可信度自然甚高。

(問:那樣,你又如何呢?

雖然,你未曾有過瀕死經歷,但是,你有閱讀過,所謂由瀕死經歷中,領悟到的道理。你有沒有嘗試,應用過它們?結果是「好果」?還是「壞果」?)

有部分是好果,受用無窮。但是,不幸地,亦有部分是壞果,受害無窮。那就是為什麼,我剛才再三強調,

如果你閱讀那些文章的話,要小心一點,因為那類文章良莠不齊——當中有些文章發人深省,有些則謊話連篇。

(問:那樣,你又可否舉例,哪些是帶來「好果」的真道理,而哪些卻是帶在「壞果」的假道理?)

不太可以,因為,沒有足夠的上文下理,個別道理說出來,很容易造成誤解。但是,要有足夠的上文下理的話,除了此刻需要,長篇大論外,聽者的人生閱歷,不能太少。

(問:即是話,要足夠老?)

可以這樣說。但是,我可以給你一個大方向。

年輕人,很容易以為,動聽的(所謂)道理,就是真確的道理。

只要你能夠提防這種錯覺,你就已經可以,避免大量的錯誤。

— Me@2018-11-08 11:24:23 AM

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2018.11.08 Thursday (c) All rights reserved by ACHK

defmacro

SLIME, 2

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Alt + Up/Down

Switch between the editor and the REPL

— Me@2018-11-07 05:57:54 AM

~~~

defmacro

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(defmacro our-expander (name) `(get ,name 'expander))

(defmacro our-defmacro (name parms &body body)
  (let ((g (gensym)))
    `(progn
       (setf (our-expander ',name)
	     #'(lambda (,g)
		 (block ,name
		   (destructuring-bind ,parms (cdr ,g)
		     ,@body))))
       ',name)))

(defun our-macroexpand-1 (expr)
  (if (and (consp expr) (our-expander (car expr)))
      (funcall (our-expander (car expr)) expr)
      expr))

.

A formal description of what macros do would be long and confusing. Experienced programmers do not carry such a description in their heads anyway. It’s more convenient to remember what defmacro does by imagining how it would be defined.

The definition in Figure 7.6 gives a fairly accurate impression of what macros do, but like any sketch it is incomplete. It wouldn’t handle the &whole keyword properly. And what defmacro really stores as the macro-function of its first argument is a function of two arguments: the macro call, and the lexical environment in which it occurs.

— p.95

— A MODEL OF MACROS

— On Lisp

— Paul Graham

.


(our-defmacro sq (x)
  `(* ,x ,x))

After using our-defmacro to define the macro sq, if we use it directly,


(sq 2)

we will get an error.

The function COMMON-LISP-USER::SQ is undefined.
[Condition of type UNDEFINED-FUNCTION]

Instead, we should use (eval (our-macroexpand-1 ':


(eval (our-macroexpand-1 '(sq 2)))

— Me@2018-11-07 02:12:47 PM

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2018.11.07 Wednesday (c) All rights reserved by ACHK

Problem 14.5b2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) … Keep only states with \displaystyle{(-1)^{F_L}=+1}; this defines the left R’+ sector.

Write explicitly and count the states we keep for the two lowest mass levels, indicating the corresponding values of \displaystyle{\alpha' M_L^2}. [This is a shorter list.]

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— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

If we define N^\perp in a way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

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\displaystyle{\begin{aligned}  (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\  (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

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\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\  \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\  \end{aligned}}

— This answer is my guess. —

— Me@2018-11-06 03:39:15 PM

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2018.11.06 Tuesday (c) All rights reserved by ACHK

Monty Hall problem 1.6

Sasha Volokh (2015) wrote that “any explanation that says something like ‘the probability of door 1 was 1/3, and nothing can change that…’ is automatically fishy: probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance.”

— Wikipedia on Monty Hall problem

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2018.11.02 Friday ACHK