The favorite place to find work

fecak 4 months ago

The problem with job boards is that unless they are rather unknown, everyone is using them. Applying through boards is essentially like getting into the back of a line and hoping you get noticed.

When I coach job seekers on finding new work, I typically encourage them to be careful not to spend too much time on the boards, and instead rely on their networks/meetups or personal research.

Using LinkedIn to search for open jobs is similar to using any other site, but it’s greatest value is as a research tool.

Say you’re a Python programmer in a suburb somewhere and you’re looking for a new gig. If you use LinkedIn to search “Python” and set a geographic preference, your results will likely be other Python programmers in the area. Where do they work? Where did they used to work? Sometimes LinkedIn will offer other profiles in the sidebar (“people also viewed”) – click those and see where they worked. Now you’ve got a list of companies that have employed Python devs, so you can do a bit more research to see if they are the type of place you might want to work – and pay no attention to whether or not they have any jobs listed on their site.

Once you found some companies that interest you, use LinkedIn to figure out the best person to reach out to. Might be their CTO if it’s a small shop, could be an internal recruiter or hiring manager for a larger firm. Make the approach, tell them why you’re interested in the company, and make a soft close to try and get them to agree to a conversation.


iridium 4 months ago

Something I read a while ago that stuck with me – ‘When you are looking for a new opportunity, you are really just looking for a person.

This reframing totally changed how I look for new jobs, and what suprised me more was how willing people were to refer me, even if they had never met me.


fecak 4 months ago

That is great advice, and to take it a step further I’d say not to focus on getting an interview but rather a conversation.

If people you haven’t met are referring you, you are already doing something right.

— Ask HN: What is your favorite place to find work? | Hacker News



2018.03.21 Wednesday ACHK

機遇再生論 1.8

如果你在洗完一千萬次牌後,發現原本排列 A 還未重新出現,然後問:

現在開始,再洗多一千萬次牌的話,至少一次洗到原本排列 A 的機會率是多少?


\approx 1.2398 \times 10^{-61}

但是,如果你在洗完一千萬次牌後,發現原本排列 A 還未重新出現時,問_另一個_問題的話,答案就會截然不同:

剛才,我洗了一千萬之牌,仍然回不到 A。

我決定,現在開始再洗牌,多不只一千萬次,而是二千萬次牌的話,至少一次洗到原本排列 A 的機會率是多少?


\approx 1.2398 \times 10^{-54}


(問:那我不需要在「洗完一千萬次牌後,發現原本排列 A 還未重新出現時」,才問_另一個_問題,因為,事先透過運算,就已經知道,那機會十分之微。


我決定,現在開始洗牌二千萬次,至少一次洗到原本排列 A 的機會率是多少?



如果你在現在,一次牌都未洗時,打算將會洗牌的次數越多,相對於現在的你而言,至少一次洗到原本排列 A 的機會率,就會越高。


洗牌 10,000,000^{10} 次,起碼一次洗到原本排列 A 的機會率是多少?


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





在洗了一次牌後,如果已知結果不是排列 A,餘下的洗牌次數中,起碼一次洗到原本排列 A 的機會率,再不是

P(A_{10,000,000^{10}}) 了,


P(A_{10,000,000^{10} - 1})


(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?)




只要不斷洗牌,回到原本排列 A 的機會,就會越來越高。




如果你在現在,一次牌都未洗時,打算將會洗牌的次數越多,相對於現在的你而言,至少有一次洗到原本排列 A 的機會率,就會越高。



任何一個組合排列 A,假設有極長的時間,去作極多次的變動,只要那「極多次」足夠多,相對於現在的你而言,那「極多次」之中,「至少有一次回到排列 A」 的機會率,會極度高。






— Me@2018-03-20 02:26:35 PM



2018.03.20 Tuesday (c) All rights reserved by ACHK

Dog 3


“But there are homeless dogs everywhere,” the old man replied. “So your efforts don’t really make a difference.”

The little boy looked at the dog and stroked him. “But for this little dog, it makes all the difference in the world.”



2018.03.16 Friday ACHK



Quark and Sakonna are sharing the same cell.

I hope you’re happy.

I am a Vulcan. My emotional state is irrelevant.

Well I’m a Ferengi. And my emotional state is very relevant. And right now, I’m miserable. And it’s all your fault.

You were well paid for your assistance.

Not well enough. Look, why don’t you just tell them what they want to know?

Sakonna just stares at him.

DEEP SPACE: “The Maquis, II” – 02/17/94 – ACT FOUR 42.


(continuing, talking confidentially) I know the Cardassians can’t be trusted. I know that the Central Command would like nothing better than to destroy the Federation colonies in the Demilitarized Zone.

Then you agree with our position?

Not for a second.

Why not?

Because your position is… (searching for the right word) Illogical.

This takes Sakonna off guard.

Do you propose to lecture me on logic?

I don’t want to, but you leave me no choice. It all comes down to the third Rule of Acquisition.

(off her blank reaction)

You don’t know that one, do you?

I am not well versed in Ferengi philosophy.

Remind me to get you a copy of the Rules. You never know when they’ll come in handy. Now, the third rule clearly states,

Never pay more for an acquisition than you have to.

Logical. But I fail to see how that applies to my situation.

DEEP SPACE: “The Maquis, II” – 02/17/94 – ACT FOUR 43.


You want to acquire peace. Fine.

Peace is good. But how much are you
willing to pay for it?

Whatever it costs.

That’s the kind of irresponsible spending that causes so many business ventures to fail.

You’re forgetting the third rule. Right now peace could be bought at a bargain price and you don’t even realize it.

I find this very confusing.

Then I’ll make it so simple that even a Vulcan can understand. The Central Command has been caught red-handed smuggling weapons to their settlers. So from now on, every ship approaching the Demilitarized Zone will be searched. Without the support of the Central Command, the Cardassian settlers won’t be so eager to fight.

You forget the weapons they already have.

They have weapons… you have weapons… everyone has weapons. But right now, no one has a clear advantage. So the price of peace is at an all-time low. This is the perfect time to sit down and hammer out an agreement. Don’t you get it… attacking the Cardassians now will only escalate the conflict and make peace more expensive in the long run. Now I ask you, is that logical?

Quark sits down, pleased with his performance.

— Star Trek: Deep Space Nine



2018.03.16 Friday ACHK

Lisp in Lisp

; The Lisp defined in McCarthy's 1960 paper, translated into CL.
; Assumes only quote, atom, eq, cons, car, cdr, cond.
; Bug reports to

(defun null. (x)
  (eq x '()))

(defun and. (x y)
  (cond (x (cond (y 't) ('t '())))
        ('t '())))

(defun not. (x)
  (cond (x '())
        ('t 't)))

(defun append. (x y)
  (cond ((null. x) y)
        ('t (cons (car x) (append. (cdr x) y)))))

(defun list. (x y)
  (cons x (cons y '())))

(defun pair. (x y)
  (cond ((and. (null. x) (null. y)) '())
        ((and. (not. (atom x)) (not. (atom y)))
         (cons (list. (car x) (car y))
               (pair. (cdr x) (cdr y))))))

(defun assoc. (x y)
  (cond ((eq (caar y) x) (cadar y))
        ('t (assoc. x (cdr y)))))

(defun eval. (e a)
    ((atom e) (assoc. e a))
    ((atom (car e))
       ((eq (car e) 'quote) (cadr e))
       ((eq (car e) 'atom)  (atom   (eval. (cadr e) a)))
       ((eq (car e) 'eq)    (eq     (eval. (cadr e) a)
                                    (eval. (caddr e) a)))
       ((eq (car e) 'car)   (car    (eval. (cadr e) a)))
       ((eq (car e) 'cdr)   (cdr    (eval. (cadr e) a)))
       ((eq (car e) 'cons)  (cons   (eval. (cadr e) a)
                                    (eval. (caddr e) a)))
       ((eq (car e) 'cond)  (evcon. (cdr e) a))
       ('t (eval. (cons (assoc. (car e) a)
                        (cdr e))
    ((eq (caar e) 'label)
     (eval. (cons (caddar e) (cdr e))
            (cons (list. (cadar e) (car e)) a)))
    ((eq (caar e) 'lambda)
     (eval. (caddar e)
            (append. (pair. (cadar e) (evlis. (cdr e) a))

(defun evcon. (c a)
  (cond ((eval. (caar c) a)
         (eval. (cadar c) a))
        ('t (evcon. (cdr c) a))))

(defun evlis. (m a)
  (cond ((null. m) '())
        ('t (cons (eval.  (car m) a)
                  (evlis. (cdr m) a)))))

— Paul Graham



2018.03.15 Thursday ACHK

Density matrix, 4

Consider a system that is in a mixed state. The system has 0.3 of probability in a pure state |\psi_1 \rangle and 0.7 of probability in another pure state |\psi_2 \rangle. Then the density matrix \rho is

0.3 | \psi_1 \rangle \langle \psi_1 | + 0.7 | \psi_2 \rangle \langle \psi_2 |

In the most general cases, neither |\psi_1 \rangle nor |\psi_2 \rangle is an eigenstate. So we cannot expect that \rho is diagonal.

For example, if each of the pure state |\psi_1 \rangle and |\psi_2 \rangle is a superposition of two eigenstates (|\phi_1\rangle, |\phi_2\rangle), then

| \psi_1 \rangle = \frac{1}{\sqrt 2} |\phi_1 \rangle + \frac{1}{\sqrt 2} |\phi_2 \rangle

| \psi_2 \rangle = \frac{1}{\sqrt 3} |\phi_1 \rangle + \sqrt{\frac{2}{3}} |\phi_2 \rangle



= 0.3 | \psi_1 \rangle \langle \psi_1 | + 0.7 | \psi_2 \rangle \langle \psi_2 |

= 0.3 \left( \frac{1}{\sqrt 2} |\phi_1 \rangle + \frac{1}{\sqrt 2} |\phi_2 \rangle \right) \left( \frac{1}{\sqrt 2} \langle \phi_1 | + \frac{1}{\sqrt 2} \langle \phi_2 | \right)

+ 0.7 \left( \frac{1}{\sqrt 3} |\phi_1 \rangle + \sqrt{\frac{2}{3}} |\phi_2 \rangle \right) \left( \frac{1}{\sqrt 3} \langle \phi_1 | + \sqrt{\frac{2}{3}} \langle \phi_2 | \right)


For simplicity, assume that the eigenstates \{ |\phi_1\rangle, |\phi_2\rangle \} form a complete orthonormal set.

If we use \{ | \phi_1 \rangle, |\phi_2 \rangle \} as basis,


= 0.3 \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{bmatrix} + 0.7 \begin{bmatrix} \frac{1}{\sqrt 3} \\ \sqrt{\frac{2}{3}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 3} & \sqrt{\frac{2}{3}} \end{bmatrix}

= \frac{0.3}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + \frac{0.7}{3} \begin{bmatrix} 1 & \sqrt{2} \\ \sqrt{2} & 2 \end{bmatrix}


— Me@2018.03.12 11:51 AM



2018.03.14 Wednesday (c) All rights reserved by ACHK

Mixed states

To me the claim that mixed states are states of knowledge while pure states are not is a little puzzling because of the fact that it is not possible to uniquely recover what aspects of the mixed state are subjective and what aspects are objective.

The simple case is this:

Let’s work with a spin-1/2 particle, so there are states:

|0 \rangle
|1 \rangle
|+ \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle + |1 \rangle \right)
|- \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle - |1 \rangle \right)

The mixed state corresponding to 50% |0> + 50% |1> is the SAME as the mixed state corresponding to 50% |+> + 50% |->.

— Daryl McCullough

— Comment #13 November 19th, 2011 at 2:00 pm

— The quantum state cannot be interpreted as something other than a quantum state


\frac{1}{2}_c | + \rangle \langle + | + \frac{1}{2}_c | - \rangle \langle - |

=\frac{1}{2}_c \left( \frac{1}{\sqrt{2}}_q | 0 \rangle + \frac{1}{\sqrt{2}}_q | 1 \rangle \right) \left( \frac{1}{\sqrt{2}}_q \langle 0 | + \frac{1}{\sqrt{2}}_q \langle 1 | \right)+ \frac{1}{2}_c \left( \frac{1}{\sqrt{2}}_q | 0 \rangle - \frac{1}{\sqrt{2}}_q | 1 \rangle \right) \left( \frac{1}{\sqrt{2}}_q \langle 0 | - \frac{1}{\sqrt{2}}_q \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{\sqrt{2}}_q \frac{1}{\sqrt{2}}_q \left( | 0 \rangle + | 1 \rangle \right) \left( \langle 0 | + \langle 1 | \right)+ \frac{1}{2}_c \frac{1}{\sqrt{2}}_q \frac{1}{\sqrt{2}}_q \left( | 0 \rangle - | 1 \rangle \right) \left( \langle 0 | - \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{2}_q \left( | 0 \rangle + | 1 \rangle \right) \left( \langle 0 | + \langle 1 | \right) + \frac{1}{2}_c \frac{1}{2}_q \left( | 0 \rangle - | 1 \rangle \right) \left( \langle 0 | - \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{2}_q \left( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | + | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{2}_q \left( 2_c | 0 \rangle \langle 0 | + 2_c | 1 \rangle \langle 1 | \right)

= \frac{1}{2}_q | 0 \rangle \langle 0 | + \frac{1}{2}_q | 1 \rangle \langle 1 |

— Me@2018-03-11 03:14:57 PM


How come the classical probabilities \frac{1}{2}_c of a density matrix in one representation can become quantum probabilities \frac{1}{2}_q in another?

\frac{1}{2}_c | + \rangle \langle + | + \frac{1}{2}_c | - \rangle \langle - | = \frac{1}{2}_q | 0 \rangle \langle 0 | + \frac{1}{2}_q | 1 \rangle \langle 1 |

1. Physically, whether we label the coefficients as “classical probabilities” or “quantum probabilities” gives no real consequences. The conflict lies only in the interpretations.

2. The interpretation conflict might be resolved by realizing that probabilities, especially classical probabilities, is meaningful only when being with respect to an observer.

For example,

\frac{1}{2}_c | + \rangle \langle + | + \frac{1}{2}_c | - \rangle \langle - | = \frac{1}{2}_q | 0 \rangle \langle 0 | + \frac{1}{2}_q | 1 \rangle \langle 1 |

represents the fact that the observer knows that the system is either in state |+\rangle \langle+| or |-\rangle \langle-|, but not |0 \rangle \langle 0| nor |1 \rangle \langle 1|.


\frac{1}{2}_c | 0 \rangle \langle 0 | + \frac{1}{2}_c | 1 \rangle \langle 1 |

represents the fact that the observer knows that the system is either in state |0 \rangle \langle 0| or |1 \rangle \langle 1|, but not |+\rangle \langle+| nor |-\rangle \langle-|.

— Me@2018-03-13 08:10:46 PM



2018.03.14 Wednesday (c) All rights reserved by ACHK

潛行凶間 16.2

Inception 16.2

這段改編自 2010 年 8 月 13 日的對話。



2. 多層夢









— Me@2018-03-08 08:32:52 PM



2018.03.08 Thursday (c) All rights reserved by ACHK


keenerd 9 hours ago

> Three practices set romance writers up for success: they welcome newcomers, they share competitive information and they ask advice from newbies.

That last one is interesting. People entering a field do so for a reason. It might just be to do a job and get paid, but it could also be because no one is making what they want. Newbies unconsciously represent gaps in the market where someone with better execution could make a killing.

Brb, got to dive beginner programming forums and look at the types of projects that are so pressing that someone is willing to learn how to program.

— What gig workers can learn from romance writers

— Hacker News



2018.03.08 Thursday by ACHK

Charlie Brown


Charlie Brown: I thought being in love was supposed to make you happy…

Peppermint Patty: Where’d you get that idea?



2018.03.02 Friday ACHK

Shape of a program

(defun bad-reverse (lst)
  (let* ((len (length lst))
	 (ilimit (truncate (/ len 2))))
    (do ((i 0 (1+ i))
	 (j (1- len) (1- j)))
	((>= i ilimit))
      (rotatef (nth i lst) (nth j lst)))))

It used to be thought that you could judge someone’s character by looking at the shape of his head. Whether or not this is true of people, it is generally true of Lisp programs. Functional programs have a different shape from imperative ones. The structure in a functional program comes entirely from the composition of arguments within expressions, and since arguments are indented, functional code will show more variation in indentation. Functional code looks fluid on the page; imperative code looks solid and blockish, like Basic.

Even from a distance, the shapes of bad- and good-reverse suggest which is the better program. And despite being shorter, good-reverse is also more efficient: O(n) instead of O(n^2).

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

— p.30

— On Lisp

— Paul Graham



2018.03.02 Friday ACHK

Problem 14.4a

Closed string degeneracies.

For closed string states the left-moving and right-moving excitations are each described like states of open strings with identical values of \alpha' M^2. The value of \alpha' M^2 for the closed string state is four times that value.

How come?


Equation (14.78):

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

Mass M is two times that value (M = 2 M_L or M = 2 M_R). So M^2 is four times.

— Me@2015.07.14 12:47 PM


Not necessarily so.

— Me@2015.07.16 08:14 AM


Instead, it is just due to this definition.

— Me@2015.07.30 09:19 AM


p.322 “In the closed string theory the value of the mass-squared is given by…”

Equation (14.78) is actually a definition of the mass-squared of a closed string state.

Also, consider

p.322 “As befits closed strings there is also the level-matching condition \alpha_0^- = \bar \alpha_0^- on the states. This condition guarantees that the left and right sectors give identical contributions to the mass-squared: \alpha' M_L^2 = \alpha' M_R^2.”


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

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

— Me@2018.03.02 11:05 AM



2018.03.02 Friday (c) All rights reserved by ACHK