眾害取其輕 10.1

The least of all evils, 10 | 地獄之路 3.2 | Good intentions 3.2

這段改編自 2021 年 12 月 15 日的對話。

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在那愚善的年代,我受到誤導,以為:

1. 有問題一定要解決。

錯。

2. 「難受」一定要避免。

錯。

3. 任何情況下,都要避免自己或他人的難受。

錯。

這三句也十分荒謬。

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在 2010 年我發現,正確的講法應該是,眾害取其輕;意思是,有時那些「害」可以零,有時不可以;未必受你主觀的控制。

可以稱得為「害」的,一定會令你十分慘。所以,「眾害取其輕」亦稱「眾慘取其輕」。凡事選「害處最小」、「慘情最細」的那一條路。

留意,「最小」不代表「很小」;「最小」仍然可以「很大」,視乎情境而定。例如,富人甲、乙、丙,分別有財產一千億元、一千億減一元 和 一千億減二元。三人之中,丙的財產是「最少」,只有一千億減二元;但仍然是「很多」。

.

為什麼「害」或「慘」,有時不可以零呢?

因為雖然眾多選項中,你選哪一個,是你主觀的決定,但是,那「眾多選項」中,有沒有「零害之選」,有時是外在客觀的現實,不受你左右。

如果在當時的情境,某個選擇已是「害處最小」,那就唯有推持,因為,如果你移除那「最小的害」,你就必然換上「更大之惡」。例如,通常而言,「上班」是苦,但「不上班」更慘。兩害相權取其輕,選「上班」。

然後,從「上班」這選擇中,內部再作「眾害取其輕」;即是在你眾多可能的工作中,選慘情最小的那一份。那可能是你當時的職位,亦可能不是。

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「眾害之最輕」有時可以零,有時不可以零。你只能客觀面對,不能主觀判斷。換句話說,「眾害之最輕」不一定「可以零」,亦不一定「不可以零」。

記住,主觀意願不成理據。任何人在,沒有足夠理據的情況下,認為某事選擇的「眾害之最輕」必為零,或者認為「眾害之最輕」必不為零,都是狂妄;要麼是蠢,要麼是壞,通常又蠢又壞。

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不可為零的眾害之最輕,簡稱「必要之惡」。

愚善的人往往認為,任何情況下,任何人的任何痛苦,都可以驅除,或應該驅除。他們移除「必要之惡」時,引發「不必要之惡」,帶來更大的痛苦。例如,移除工作之痛,卻換來捱餓之苦。

又例如,任何學問,必有必須背誦的地方。但是,竟然有人提倡「背誦是苦,所以應該只需要理解,而毋須背誦」。我年輕時錯信這點,誤了學業。正確的取向是:

1. 理解後背誦

2. 有用的東西就背

3. 無用的東西就不背

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愚賤之人的邪惡,則是另一個極端。他們認為,任何情況下,痛苦都無可避免。

以「眾害取其輕」為名,把「不必要之惡」標籤為「必要之惡」,讓自己及他人承受著大量,不必要的痛苦。

— Me@2022.09.24 10:56:23 AM

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2022.09.25 Sunday (c) All rights reserved by ACHK

Haskell mode, 2

Euler problem 3.3

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The goal of this blog post is to install an advanced Haskell mode, called LSP mode, for Emacs.

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1. Open the bash terminal, use the following commands to install the three packages:

sudo apt-get install elpa-haskell-mode

sudo apt-get install elpa-yasnippet

sudo apt-get install elpa-which-key

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2. Read and follow the exact steps of my post titled “Haskell mode“.

.

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— Me@2022.09.20 12:49:29 PM

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2022.09.21 Wednesday (c) All rights reserved by ACHK

Functional Differential Geometry

Chapter 1: Introduction

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(define ((Gamma w) t)
  (up t (w t) ((D w) t)))

(define q-rect
  (up (literal-function 'x_0)
      (literal-function 'y_0)
      (literal-function 'x_1)
      (literal-function 'y_1)
      (literal-function 'F)))

(show-expression (q-rect 't))

(show-expression ((Gamma q-rect) 't))





(define ((Lfree mass) state)
  (* 1/2 mass (square (velocity state))))

(define ((sphere->R3 R) state)
  (let ((q (coordinate state)))
    (let ((theta (ref q 0)) (phi (ref q 1)))
      (up (* R (sin theta) (cos phi))
          (* R (sin theta) (sin phi))
          (* R (cos theta))))))

(define ((F->C F) local)
  (up (time local)
      (F local)
      (+ (((partial 0) F) local)
         (* (((partial 1) F) local)
            (velocity local)))))

(define (Lsphere m R)
  (compose (Lfree m) (F->C (sphere->R3 R))))

((Lsphere 'm 'R)
 (up 't
     (up 'theta 'phi)
     (up 'thetadot 'phidot)))

(show-expression
 ((Lsphere 'm 'R)
  (up 't
      (up 'theta 'phi)
      (up 'thetadot 'phidot))))



(define ((L2 mass metric) place velocity)
  (* 1/2
     mass
     ((metric velocity velocity) place)))

(define ((Lc mass metric coordsys) state)
  (let ((x (coordinates state))
        (v (velocities state))
        (e (coordinate-system->vector-basis
            coordsys)))
    ((L2 mass metric)
     ((point coordsys) x) (* e v))))

(define the-metric
  (literal-metric 'g R2-rect))

(define L
  (Lc 'm the-metric R2-rect))

(L (up 't (up 'x 'y) (up 'vx 'vy)))

(show-expression
 (L (up 't (up 'x 'y) (up 'v_x 'v_y))))



(define gamma
  (literal-manifold-map 'q R1-rect R2-rect))

((chart R2-rect)
 (gamma ((point R1-rect) 't)))

(show-expression
 ((chart R2-rect)
  (gamma ((point R1-rect) 't))))



(define coordinate-path
  (compose
   (chart R2-rect) gamma (point R1-rect)))

(coordinate-path 't)

(define Lagrange-residuals
  (((Lagrange-equations L)
    coordinate-path) 't))

(show-expression
 Lagrange-residuals)





(define-coordinates t R1-rect)

(define Cartan
  (Christoffel->Cartan
   (metric->Christoffel-2
    the-metric
    (coordinate-system->basis R2-rect))))

(define geodesic-equation-residuals
  (((((covariant-derivative Cartan gamma) d/dt)
     ((differential gamma) d/dt))
    (chart R2-rect))
   ((point R1-rect) 't)))

(define metric-components
  (metric->components
   the-metric
   (coordinate-system->basis R2-rect)))

(- Lagrange-residuals
   (* (* 'm (metric-components
             (gamma ((point R1-rect) 't))))
      geodesic-equation-residuals))

(show-expression
 (- Lagrange-residuals
    (* (* 'm (metric-components
              (gamma ((point R1-rect) 't))))
       geodesic-equation-residuals)))

— Me@2022.09.18 04:25:12 PM

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2022.09.18 Sunday ACHK

Posted in FDG

The Ship of Theseus

_jgvg on July 23, 2018 | next [–]

This reminds me of someone who once told me that the vast majority of philosophical problems are self-inflicted and caused by the intentional use of vague and ambiguous definitions, and therefore could be solved trivially simply by agreeing on a definition for something.

One example given was the Theseu’s Paradox (or The Ship of Theseus).

The entire problem only exists because you’re intentionally using a vague term in the problem, therefore causing the problem itself.

At least that was the commentary, I’m not an expert in the subject, and I’m sure many will disagree. But then, even the disagreeing is just pointlessly creating a problem for the sake of the discussion.

zukzuk on July 23, 2018 | parent | next [–]

I think you’re missing something really profound here. I wrote my masters thesis on identity problems in version control systems (Git, Mercurial, Subversion, and the like). These systems are forced to take a stance on the Ship of Theseus problem — they’re tracking the identity of files/documents whose entire contents, name, and format may change over time.

Different systems have chosen to take very different stances on this. As it turns out, taking a stance on what makes a file a file (or a ship a ship) — being explicit rather than vague — does not solve the problem. Not even close. In fact, one of the many little things that make Git a brilliant piece of software is that it refuses to take a stance on identity. Git is intentionally vague about what makes a file a file. It doesn’t actually track it at all — file identity is determined retroactively. It’s a question for which Git intentionally doesn’t have a straight answer.

What I think is really fascinating about all of this is that computers are forcing us to deal with age-old philosophical problems head on. These are not just interesting riddles anymore. They’re real problems that need concrete solutions. Event more interestingly, there is something in this that establishes computing as categorically different from just pure mathematics. Computing forces us to bridge abstraction with the real world — to answer questions like what is the relationship between an abstract model and the thing that is being modelled. This to me is really exciting, and what follows from it can’t be easily dismissed with a comment like “most philosophical problems only exist because language is vague”.

— Was Wittgenstein Right? (2013)

— Hacker News

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2022.09.15 Thursday ACHK

地獄之路 3

Good intentions 3

這段改編自 2021 年 12 月 15 日的對話。

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贓官可恨,人人知之,清官尤可恨人多不知,蓋贓官自知有病,不敢公然為非;清官則自以為不要錢,何所不可,剛愎自用,小則殺人,大則誤國,吾人親目所見,不知凡幾矣。

唉!天下大事壞於奸臣手上的十之三四,壞於不通世故的君子手上的倒有十分之六七!

— 老殘遊記

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(JC: 不記得何時,看過一個紀錄片,主持人到非洲做義工。

他說在那兒,如果有人問你拿食物,千萬不要即場給予,因為那樣的話,求助人身邊的人會,向你一湧而上地求助,令你即時有生命危險。)

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所以,要成功做到好事,單單善意是不夠的,還要智力和策略。

記住,地獄不乏善意,而天堂,則需要善行。

心地好而沒有智力,比壞人更加危險,因為他可以在雙方不知不覺間,置你於死地。

地獄是善意之地,而天堂,則是善行之境。

— Me@2022.09.13 12:01:08 PM

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2022.09.13 Tuesday (c) All rights reserved by ACHK

Photons in expanding space

If a photon (wave package) redshifts (stretches) travelling in our expanding universe, is its energy reduced?

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Since you say you’re talking about what happens locally (in a small volume), I’ll answer from that point of view. The usual formulation of energy conservation in such a volume is that energy is conserved in an inertial reference frame. In general relativity, there are no truly inertial frames, but in a sufficiently small volume, there are reference frames that are approximately inertial to any desired level of precision. If you restrict your attention to such a frame, there is no cosmological redshift. The photon’s energy when it enters one side of the frame is the same as the energy when it exits the other side. So there’s no problem with energy conservation.

The (apparent) failure of energy conservation arises only when you consider volumes that are too large to be encompassed by a single inertial reference frame.

To be slightly more precise, in some small volume V=L^3 of a generic expanding Universe, imagine constructing the best possible approximation to an inertial reference frame. In that frame, observers near one edge will be moving with respect to observers near the other edge, at a speed given by Hubble’s Law (to leading order in L). That is, in such a frame, the observed redshift is an ordinary Doppler shift, which causes no problems with energy conservation.

If you want more detail, David Hogg and I wrote about this at considerable (perhaps even excessive!) length in an AJP paper.

— Photons in expanding space: how is energy conserved?

— answered Aug 16, 2011 at 15:31

— Ted Bunn

— Physics Stack Exchange

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2022.09.11 Sunday ACHK

Combination Lock 3

jancsika on July 26, 2018 | next [–]

I think the article makes “epiphany” sound rather extraordinary. But in the normal course of learning music there are plenty of them which we tend to ignore for some reason.

For example: if you learned to ride a bike as a kid, you’ll remember a moment where your balance just “locked in” and there you were riding for an arbitrary amount of time.

— Some suddenly become accomplished artists or musicians with no previous training

— Hacker News

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2022.09.10 Saturday ACHK

排列組合 1.2

nCr, 0.2

這段改編自 2010 年 7 月 27 日的對話。

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1.4.2 另外一種設計定義的想法是,由 \displaystyle{n!}算式取靈感。

既然 \displaystyle{n} 是多少,\displaystyle{n!} 就有多少項相乘,那樣,零的階乘,\displaystyle{0!},理應只有零項相乘,\displaystyle{~a^0~}

\displaystyle{0! = a^0}

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但是,「零項相乘」的數值,又應該是什麼呢?

「零項相乘」即是「乘了等如沒有乘」,所以是一,因為任何數乘了一,效果都等於沒有乘;所以,

\displaystyle{a^0 = 1}

\displaystyle{0! = 1}

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「零項相乘」的正式學名是,「空積」或「零項積」。

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1.4.3 如果要詳細一點,去理解「空積」的話,可以先嘗試了解「次方」的意思。

首先,\displaystyle{a} 五次方的意思,就是有五個 \displaystyle{a} 相乘。

\displaystyle{a^5 = aaaaa}

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如果在其之後,再乘 \displaystyle{a^3} 的話,就即是再乘多三個 \displaystyle{a}

\displaystyle{a^5 a^3 = (aaaaa)(aaa)}

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所以,

\displaystyle{a^5 a^3 = a^{5+3}}

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既然,在其之後乘以 \displaystyle{a}三次方,就是乘多三個 \displaystyle{a}

那樣,在其之後乘以 \displaystyle{a}三次方,就可以理解成乘少三個 \displaystyle{a},即是:

.

你想想,自出生以來,你學過什麼操作,會有刪除因子的效果呢?

就是分母:

\displaystyle{a^5 a^{-3} = \frac{aaaaa}{aaa}}

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所以,所謂「負三次方」的運算的方法,就是把那三次方,放於分母之中。

\displaystyle{a^{-3} = \frac{1}{a^3}}

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然後,我們可以再研究,一個抽象一點的問題:

\displaystyle{a}次方,又會是什麼呢?

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我們可以這樣想,如果 \displaystyle{a^{1/2}} 存在的話,它必須達到什麼效果呢?

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如果暫時想不到的話,可以改為思考:「\displaystyle{a} 的五次方,乘了半次 \displaystyle{a},再乘半次 \displaystyle{a}」 的話,即是總共乘多了,多少個 \displaystyle{a} 呢?

\displaystyle{a^5 a^{1/2} a^{1/2}}=?

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那很明顯是,總共乘多了一次 \displaystyle{a}

\displaystyle{a^5 a^{1/2} a^{1/2}}=a^5 a^1

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所以,

\displaystyle{a^5 a^{1/2} a^{1/2}}=a^6

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亦即是話,

\displaystyle{a^{1/2} a^{1/2}}=a^1

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究竟這個所謂的「半次 \displaystyle{a}」存不存在呢?

可以這個想,你自出生以來,有沒有學過什麼數字符號,自乘後會等如一次 \displaystyle{a} 呢?

有,那就是 \displaystyle{a} 的平方根。

\displaystyle{\sqrt{a} \sqrt{a}}=a^1

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所以,所謂的「半次 \displaystyle{a}」,其實就是,平方根 \displaystyle{a}

\displaystyle{a^{\frac{1}{2}} = \sqrt{a}}

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言歸正傳,我們再來研究,所謂的「\displaystyle{a} 零次方」,其實是什麼?

顧名思義,即是乘 \displaystyle{a} 的次數為零,不要乘也。

再想想:

\displaystyle{a} 五次方乘以 \displaystyle{a} 的正三次,就是乘多三個 \displaystyle{a};所以,結果就是 \displaystyle{a} 的八次方。

\displaystyle{a^5 a^3 = (aaaaa)(aaa) = a^8}

同理,\displaystyle{a} 五次方乘以 \displaystyle{a}三次,就是乘三個 \displaystyle{a};所以,結果就是 \displaystyle{a} 的二次方。

\displaystyle{a^5 a^{-3} = a^2}

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那樣,\displaystyle{a} 五次方乘以 \displaystyle{a}次,理應就是「不乘多亦不乘少」,維持原來的 \displaystyle{a} 五次。

\displaystyle{a^5 a^{0} = aaaaa = a^5}

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換句話說,「乘以 \displaystyle{a} 零次」的效果,就是「乘了等如沒乘」。

你想想,自出生以來,有沒有學過什麼數字符號,會有「乘了等如沒乘」的效果呢?

— Me@2022.09.07 08:09:14 PM

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2022.09.10 Saturday (c) All rights reserved by ACHK

Way

There is always a WAY.

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(defun sum-of-squares(x y &key (a 1) (b 1))
  (+ (* a x x) (* b y y)))

(deftype candidate()
  '(simple-array boolean (*)))

(defmacro get-sieve-function(a b mod-results)
  `(lambda(x y)
     (let* ((n (sum-of-squares x y :a ,a :b ,b))
        (r (mod n 12)))
       (when (and (<= n limit)
          (not (null
            (member r ,mod-results))))
     (setf (aref candidates n)
           (not (aref candidates n)))))))

(defun get-atkin-prime-candidates-map(limit)
  (let* ((lmt (isqrt limit))
         (candidates (make-array
              (1+ limit)
              :initial-element nil))
         (stage1 (get-sieve-function 4 1 '(1 5)))
         (stage2 (get-sieve-function 3 1 '(7)))
         (stage3 (get-sieve-function 3 -1 '(11))))
    (declare (type candidate candidates))
    (declare (optimize (speed 3)))
    (loop for x from 1 to lmt do
      (loop for y from 1 to lmt do
    (progn
      (funcall stage1 x y)
      (funcall stage2 x y)
      (when (> x y)
        (funcall stage3 x y)))))
    candidates))

(defun atkin-sieve-map(candidates)
  (let ((len (length candidates)))
    (loop for i from 1 to (1- len)
      when (aref candidates i)
        do (loop for j from 1
             for n = (* j i i)
             while (< n len)
             do (setf (aref candidates n) nil))))
  (setf (aref candidates 2) T)
  (setf (aref candidates 3) T)
  candidates)

(defun eratosthene-sieve-map(limit)
  (let ((candidates
      (make-array
       (1+ limit) :initial-element T)))
    (declare (type candidate candidates))
    (declare (optimize (speed 3)))
    (progn
      (setf (aref candidates 0) nil)
      (setf (aref candidates 1) nil))
    (loop for i from 2 to limit
      when (aref candidates i)
        do (loop for j from (+ i i) to limit by i
             when (aref candidates j)
               do (setf (aref candidates j) nil)))
    candidates))

;; This is free software released
;; into the public domain.
;;
;; ykm

(defun get-primes(limit
          &key
            (generator :eratosthene))
  (let ((candidates
      (case generator
        (:atkin (atkin-sieve-map
             (get-atkin-prime-candidates-map
              limit)))
        (:eratosthene (eratosthene-sieve-map
               limit))
        (otherwise
         (error "not valid type
(:atkin :eratosthene)")))))
    (declare (type candidate candidates))
    (loop for i from 0 to (1- (length candidates))
      when (aref candidates i)
        collect i)))

(format t "soe: ~d
" (time (length (get-primes 12345678))))

(format t "soa: ~d
" (time (length (get-primes 12345678
                :generator :atkin))))

(defmacro log10 (x)
  `(/ (log ,x) (log 10)))

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— Me@2022.09.06 03:24:36 PM

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2022.09.06 Tuesday (c) All rights reserved by ACHK

Homogenous function

A function f is homogenous of degree n if and only if f(ax) = a^n f(x).

— 1.8 Conserved Quantities

— Structure and Interpretation of Classical Mechanics

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\displaystyle{\begin{aligned}     D_a f(ax)     &= \frac{d}{da} f(ax) \\     &= \frac{d}{da} a^n f(x) \\     &= f(x) \frac{d}{da} a^n  \\     &= n a^{n-1} f(x) \\     \end{aligned}}

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\displaystyle{\begin{aligned}     D_a f(ax)     &= \frac{d}{da} f(ax) \\     &= \frac{d(ax)}{da} \frac{d}{d(ax)} f(ax) \\     &= x \frac{d}{d(ax)} f(ax) \\     &= x D_{ax} f(ax) \\     \end{aligned}}

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\displaystyle{\begin{aligned}     x D_{ax} f(ax) &= n a^{n-1} f(x) \\    x D_{x} f(x) &= n f(x) \\    \end{aligned}}

— Me@2022.09.05 06:28:00 PM

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2022.09.06 Tuesday (c) All rights reserved by ACHK

11X

~~~

... 中化口試
巴士:11X

~~~

10k trimmer pot: 7.31k

~~~

3150, 4260: Samuel

3170: Danny, 達
3260: Dennis

~~~

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2022.08.29 Monday (c) All rights reserved by ACHK

Presentation 基本原理 1.2.2.3

反不相關推薦 3.3

這段改編自 2010 年 4 月 24 日的對話。

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每一節課應該,只有一個重點。在該課中,你要用不同的句子、字眼、例子,來重複釐清和闡述,那個重點。講課時,一切的細節,都要圍繞著,那個重點來運行。

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(安:但是,有時正正就是需要,解釋兩個要點的關係。那又如何做到,一節課只講一要點呢?)

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「一課一要點」只是大概而言的企圖,毋須百分百執行。

另外,(甲、乙)兩個要點的關係,本身就已成一個要點(丙)。

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(安:你的意思是,第一、二、三節課,分別講甲、乙、丙?)

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漫畫化來說,正確。但是,你千萬不要,鑽牛角尖地問:「何謂『一個』要點?多大的要點,才為之『一個』呢?」

那是相對於,當刻的聽眾而言。一位聽者的「一個」要點,對於另外一位來說,可以是「兩個」。所以,你有責任,為當刻的聽眾,提供適合他們的「一個」要點。

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另外,在那堆要點之中,你不單是選「一個」,而且還要選,對他們而言,最有用的那一個。那往往是在,他們知識邊緣的東西,即是介乎知與不知之間。

換句話說,究竟是眾多要點中的哪一個呢?

就是聽眾當刻,剛剛有能力理解的數個要點之中,最重要、最根本的那一個。

— Me@2022-08-28 07:56:28 PM

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2022.08.29 Monday (c) All rights reserved by ACHK

It doesn’t matter

Euler problem 3.1

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The bad news is: You cannot make people like, love, understand, validate, accept, or be nice to you.

The good news is: It doesn’t matter.

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

(defmacro list-head (lst)
  `(car ,lst))

(defmacro list-tail (lst)
  `(cdr ,lst))

(defmacro last-item (lst)
  `(car (last ,lst)))

(defun good-reverse (lst)
  (labels ((rev (lst acc)
         (if (null lst)
         acc
         (rev
          (cdr lst)
          (cons (car lst) acc)))))
    (rev lst nil)))

(defun prime-sieve-a-list (input-lst)
  (labels ((sieve-iter (go-lst acc-list)
         (if (not go-lst) 
         acc-list        
         (if (> (sq (list-head go-lst))
            (last-item go-lst))

             (append (good-reverse acc-list)
                 go-lst)
             
             (sieve-iter
              (remove-if #'(lambda (x)
                     (=
                      (mod x (list-head go-lst))
                      0))
                 (list-tail go-lst))
              (cons (list-head go-lst)
                acc-list))))))

    (sieve-iter input-lst '())))

(defun range (max &key (min 0) (step 1))
  (loop :for n :from min :below max :by step
    collect n))

(defmacro prime-sieve (n)
  `(prime-sieve-a-list (cons 2 (range (1+ ,n)
                      :min 3
                      :step 2))))

(time (length (prime-sieve 1234567)))

;; 0.764 seconds of real time
;; 95360

(time (length (prime-sieve 12345678)))

;; 20.128 seconds of real time
;; 809227

— Me@2022-08-27 07:59:30 PM

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2022.08.28 Sunday (c) All rights reserved by ACHK

3.1 Lorentz covariance for motion in electromagnetic fields, 2

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Is \displaystyle{\frac{d p_\mu}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds}} gauge invariant?

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~~~

… the defining property of a Lorentz transformation, \Lambda^\mu_{\;\;\nu}:

\eta_{\mu\nu} \Lambda^{\mu}_{\;\;\alpha}  \Lambda^\nu_{\;\;\beta} = \eta_{\alpha\beta}

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… 4-vectors and (Lorentz)-tensors are transformed like this:

U'^\mu = \Lambda^\mu_{\;\;\nu}U^\nu

and

F'_{\mu\nu} = \Lambda_\mu^{\;\;\alpha} \Lambda_\nu^{\;\;\beta}F_{\alpha\beta}= \Lambda_\mu^{\;\;\alpha} F_{\alpha\beta} (\Lambda^{-1})^{\beta}_{\;\;\nu}

where we have used the conventional notation

\displaystyle{\Lambda_\nu^{\;\;\mu} = (\Lambda^{-1})^\mu_{\;\;\nu}}

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Let us then take your equation and apply \displaystyle{\Lambda_\sigma^{\;\;\mu}} on both sides (recall this Lorentz transformation does not depend on \displaystyle{\tau}), and try rewriting everything in terms of prime quantities:

\displaystyle{  \begin{aligned}  m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau}     &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\     &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\    m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha} ((\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta}) U_\alpha\\    &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big)\\    &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\    &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu\Big) U'_\beta\\    &= e F'_{\sigma\nu}\eta^{\nu\beta} U'_\beta \\    \\    m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e F'_{\sigma\nu} U'^\nu \\  \end{aligned}}

This “game” can always be done with contracted indices, …

— answered Jul 7, 2020 at 15:06

— ohneVal

— Lorentz invariance of the Lorentz force law

— Physics StackExchange

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How come \displaystyle{ \begin{aligned}     \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}?

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Note that

\displaystyle{  \begin{aligned}    \sum ... \sum \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu    &= \sum ... \sum \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu  \\    \end{aligned}}

does not mean that

\displaystyle{ \begin{aligned}     \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}

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Also

\displaystyle{ \begin{aligned}     \sum \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \sum \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}

does not mean that

\displaystyle{ \begin{aligned}     \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu     &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\     \end{aligned}}

.

\displaystyle{  \begin{aligned}    A  = \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu U_\beta'    &=     \sum_{\alpha, \beta} F_{\mu\alpha}\eta^{\alpha\alpha}\Lambda^{\;\;\beta}_{\nu=\alpha} U_\beta' \\    &=     \sum_{\beta}     \left(     - F_{\mu 0} \Lambda^{\;\;\beta}_{0}    + F_{\mu 1} \Lambda^{\;\;\beta}_{1}    + F_{\mu 2} \Lambda^{\;\;\beta}_{2}    + F_{\mu 3} \Lambda^{\;\;\beta}_{3}      \right)     U_\beta' \\        \\    \end{aligned}}

\displaystyle{ \begin{aligned}     B = \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu U_\beta'    &=     \sum_{\alpha, \beta}     F_{\mu\alpha}     \eta^{\beta \beta}\Lambda^{\;\;\alpha}_{\nu=\beta} U_\beta      \\     &=     \sum_{\alpha}     F_{\mu\alpha}     \left(    - \Lambda^{\;\;\alpha}_{0} U_0'    + \Lambda^{\;\;\alpha}_{1} U_1'     + \Lambda^{\;\;\alpha}_{2} U_2'     + \Lambda^{\;\;\alpha}_{3} U_3'     \right)     \\       \end{aligned}}

At the first glance, it seems to be unlikely that

\displaystyle{  \begin{aligned}    \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu U_\beta'    &= \sum_{\alpha, \beta, \nu}  F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu U_\beta'    \\    \end{aligned}},

because while in A, for any \beta, U_\beta'‘s have visible negative terms; in B, only U_0'‘s do.

Without additional mathematical properties among those physical quantities, the identity is impossible to prove.

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Just for future reference:

From the invariance of the spacetime interval it follows

\displaystyle{\eta =\Lambda ^{\mathrm {T} }\eta \Lambda}

— Wikipedia on Lorentz transformation

.

\displaystyle{  \begin{aligned}    \eta_{\mu \nu} = \eta'_{\mu \nu} &= \Lambda^{\alpha}{}_{\mu} \Lambda^{\beta}{}_{\nu} \eta_{\alpha \beta} \\     &= (\Lambda^T)_{\mu}{}^{\alpha} \eta_{\alpha \beta} \Lambda^{\beta}{}_{\nu} \\ \\     \eta_{\alpha\beta} &= \Lambda^{\mu}_{\;\;\alpha} \Lambda^{\nu}_{\;\;\beta}\eta_{\mu\nu} \\    \eta^{\alpha\beta} &= \Lambda^{\alpha}_{\;\;\mu} \Lambda^{\beta}_{\;\;\nu}\eta^{\mu\nu} \\    \end{aligned}}

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\displaystyle{  \begin{aligned}    m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\     & = e \Lambda_\nu^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\    \end{aligned}}

Inserting the identity \sum_\beta (\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta} into the expression:

\displaystyle{  \begin{aligned}    m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big) \\      &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\    \end{aligned}}

This path does not work. Also, the formula \displaystyle{ \Lambda^\alpha_{\;\;\beta} U_\alpha = U'_\beta } is plain wrong!

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Inserting the identity \sum_\beta (\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta} into the expression before (actually without) lowering the index:

\displaystyle{  \begin{aligned}    m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau}     &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\    &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha} ((\Lambda^{-1})^{\alpha}_{\;\;\beta} \Lambda^\beta_{\;\;\nu})  U^\nu \\    m \frac{{\rm d}U_\sigma'}{{\rm d}\tau}      &= e (\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha} (\Lambda^{-1})^{\alpha}_{\;\;\beta}) (\Lambda^\beta_{\;\;\nu}  U^\nu) \\    &= e {F'}_{\sigma \beta} {U'}^\beta \\    \end{aligned}}

— Me@2022-08-23 12:03:54 PM

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2022.08.26 Friday (c) All rights reserved by ACHK

High level machine language

susam 16 days ago

The first article in this issue of BYTE has a very interesting characterization of Lisp that I have not come across before. I mean, famous quotes like “Lisp is a programmable programming language” by John Foderaro and “The greatest single programming language ever designed” by Alan Kay are often mentioned in articles about Lisp. But in this issue of BYTE, the article “An Overview of LISP” by John Allen at page 10 has something very interesting to say. Excerpt from the article:

“The best description of the LISP programming language is that it is a high level machine language. That is, it shares many of the facets of contemporary machine language –the necessity for attention to detail and the freedom to manipulate the machine’s data and programs without restriction– yet LISP is high level in that the language contains the expressive power and convenience of traditional high level languages. The contradiction is resolvable: a LISP machine is just a higher level machine whose data items are organized differently from the binary bit patterns of most machines, and the LISP programming language is the assembly language for this machine.”

Consider the Emacs Lisp (Elisp) interpreter for example. Elisp interpreter is the Lisp machine. It understands Elisp symbolic expressions, the language of this machine. With enough code written in this machine’s language, we get this fine editing and productivity software known as Emacs!

aap_ 16 days ago

This exactly matches my thoughts. It seems that machine language and LISP are the only two languages (that I know anyway) where code and data are fundamentally the same kind of thing.

— Byte Magazine: LISP (1979)

— Hacker News

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2022.08.22 Monday ACHK

Daily life is boring

地獄篇 3.2 | 開山祖師牛 6.4

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YOU are the missing MAGIC in your daily life.

— Me@2016-08-24 12:32 AM

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2022.08.22 Monday (c) All rights reserved by ACHK