jackmaney

Please tell me where you work. I don't ever want to be on the same team as you.

Pandoc is fantastic! Just a few days ago, I used it in a makefile to keep markdown, HTML, and PDF versions of my resume in sync.

Nice site, but it hijacked my back button. There's no excuse for that.

What contract?? You can't find a single example of any such contract that I've ever signed.

Then let them fear. Loyalty between employer and employee (in either direction) died out many, many years ago.

Can anyone name a company that, in 2016, is known for keeping a significant number of their employees for their entire career?

> If the work is good and the company is good, why leave?

Possible reasons, off the top of my head:

1. Being approached by a recruiter or hiring manager and end up being given an offer you can't refuse (eg much more money, dream job, etc).

2. Your spouse or significant other has to move to another city for their job, you don't want to leave them, and your current employer won't let you work remotely.

2a. Ditto with needing to move to care for an older and/or ill family member.

3. A shift of passion into a new career.

4. Health problems that cause such an extended leave of absence that you may as well leave.

5. I've encountered some managers who are leery of people who have spent "too long" at one place, wondering, eg "Why did s/he stay there for ten years? Did this person just find a spot to coast and collect a paycheck? Why hasn't s/he moved on to bigger and better things?" I think this line of thinking is just as knee-jerk as being leery of "job hoppers".

Dementia is the single thing I fear the most. Without my intelligence, I have literally no marketable skills (my body became unfit for heavy manual labor years ago).

I also watched my grandmother go through dementia in the last months before she passed away. It was hell.

> You'd Be Happier If You Ate Less

Yes, and the poor would be wealthier if they had more money.

What is garbage like this article doing on HN?

Yeah, I'm actually pretty damn disappointed that NPR would print such garbage. They used to be more respectable than that. Now, they'll apparently publish any idiotic claim, whether or not it has any evidence whatsoever to back it up.

It's a shame that this site still has the annoying pop-overs asking for my email address, along with at least one bouncing and annoying "Click on me!" button.

> If someone isn't in front of you banging on their keyboard you have no assurance that they're actually doing work.

If someone is in front of you banging on their keyboard, you need not have any assurance that they're doing work, either.

PC growth also slowed because, in general, hardware upgrades aren't needed as frequently as they used to.

Post-device era? I'm still waiting for that alleged "Post-PC era" to arrive. I'm not holding my breath.

Agreed. I spent several seconds wondering if there was either a subdivision of Microsoft or yet another overvalued start-up with the name of Street.

I guess in the latter case, the name would be Streetly or Streetr or something equally dumb.

Denoting {a + b sqrt(-5) for a, b integers} by Z[sqrt(-5)], it's worth pointing out that you can actually use the norm to get a bit of the FTA--namely that every nonzero nonunit factors into irreducibles. By x being "irreducible", I mean that x is a nonzero nonunit element of Z[sqrt(-5)] and whenever x = yz for y,z in Z[sqrt(-5)] then either y or z is a unit.

So, denoting x = a + b sqrt(-5) in Z[sqrt(-5)], we define the norm[1] of x as follows:

    N(x) = a^2 + 5b^2

Note that, in this case, N is a function from Z[sqrt(-5)] to the natural numbers.

For x, y in Z[sqrt(-5)], it turns out that the following properties of the norm hold (proving them is fairly straightforward, as it's essentially "plug-and-chug" combined with a bit of reasoning about how things work in the natural numbers):

1. N(xy) = N(x)N(y)

2. N(x) = 0 if and only if x = 0

3. N(x) = 1 if and only if x = 1 or -1 (ie x is a unit).

So, if x is a nonzero nonunit that isn't irreducible, then we can, by definition, write x = y*z where y and z are also nonzero nonunits. Applying strong induction via the norm, we can show that x can be written as a product of irreducibles. Of course this product is, in general, not uniquely determined.

[1]: https://en.wikipedia.org/wiki/Field_norm

There's a difference between equal and equitable.

That's why we need fines that are based upon a percentage of revenue or profit within, say, the past year. A fine of, eg, 15% yearly revenue over FY2015 would make any company sit up and pay attention.

Why not? There's plenty of turnover in tech jobs, and talented developers mostly find themselves in a seller's job market.

I had the same impression. I then imagined some marketing douchebag trying to sell the C-level suits on the idea of a "Three Inch Punch of Productivity".

An interesting idea, but it would be difficult at best to prevent false positives on sites with dynamic elements.

No business has a fundamental right to exist. So, if one of your workers leaves, starts his/her own company, and puts you out of business, then tough.

Your business has no right of any kind to exist. Period.

Isn't that how the "invisible hand" of the "free market" (or some such garbage) is supposed to work?

I closed the tab as soon as the audio started playing automatically. If I want to watch a video or listen to an audio clip, I will indicate that by pressing a play button. Until then, please stop shoving your unwanted sounds into my ear holes.

There's a pretty good handle on it, in the form of the ideal class group[1].

[1]: https://en.wikipedia.org/wiki/Ideal_class_group

Z[sqrt(-3)] does not have unique factorization. However, Z[w] does, where w = (-1 + sqrt(-3))/2 denotes a primitive third root of unity.

In particular,

    2 * 2 = (-1 + sqrt(-3)) * (-1 - sqrt(-3))

gives two irreducible factorizations of 4 in Z[sqrt(-3)] (you can use norm arguments in Z[w] to show irreducibility). Note that the right hand side can also be written as

    (2w) * (2w^2)

however, since w is not in Z[sqrt(-3)], the two factorizations above are distinct in Z[sqrt(-3)]. They become the same factorization in Z[w].

Edit: Another way to think about why Z[sqrt(-3)] doesn't have unique factorization is because it isn't integrally closed[1]--that is, there is a monic polynomial (ie a polynomial with a leading coefficient of 1) with coefficients in Z[sqrt(-3)] that doesn't have roots in Z[sqrt(-3)]. In particular, since w^2 + w + 1 == 0, w is a root of the polynomial x^2 + x + 1, which is monic over Z[sqrt(-3)]. It turns out that any unique factorization domain[2] is integrally closed. Since Z[sqrt(-3)] is not integrally closed, it is not a UFD.

[1]: https://en.wikipedia.org/wiki/Integrally_closed_domain

[2]: https://en.wikipedia.org/wiki/Unique_factorization_domain

> InMobi was subject to a $4 million civil penalty, which was lowered to $950,000 "based on the company's financial condition," the FTC said in a statement issued Wednesday, without elaborating.

Why should the penalty be reduced based upon InMobi's finances? If the fine bankrupts them, all the better!

Edit: Honestly, I don't know why companies can't be fined based upon a percentage of revenue taken over, say, the past year. A fine of 10% yearly revenue would actually make these bastards sit up and pay attention.

A nice discussion, but a bit of a nitpick: in Z[sqrt(-5)], 2, 3, 1+sqrt(-5), and 1-sqrt(-5) are actually irreducible, not prime.

In an integral domain D, a nonzero element x is called irreducible if x is not a unit and whenever x = ab (for a, b in D), then one of a or b is a unit.

On the other hand, a nonunit element x in D is called prime if for all a, b in D, if x divides ab then x divides a or x divides b (by "x divides ab", I mean that there's some element--call it y--in D such that xy = ab).

In Z (or any unique factorization domain[1]), these concepts coincide. In Z[sqrt(-5)], however, there are irreducible elements that are not prime. In particular, 2 is irreducible in Z[sqrt(-5)], but it isn't prime, since 2 divides (1+sqrt(-5))*(1-sqrt(-5)), but 2 divides neither 1+sqrt(-5) nor 1-sqrt(-5).

[1]: https://en.wikipedia.org/wiki/Unique_factorization_domain

I've worked on a few projects where I've needed to know pairwise distances among a set of addresses.

About three years ago, I was at a place that used Qlikview, and it was the worst software I've ever used in my life. It's technically able to handle larger datasets than Excel, but it's much slower than Excel at any kind of analysis or aggregation, it doesn't make use of SQL (even though the data it handles is tabular...instead, it uses a home-brewed, mangled, and inferior query language), and its visualizations and UI are sub-par compared to Tableau.

Good riddance.

Just once, I'd love to see one of these tech companies call these ridiculous bluffs.

"Fine. You want to try and force us to carry local French TV, and fully pay for the production of said local TV programs? We will be cutting off all Netflix access to France, effective immediately."

It would take less than ten minutes for these cowardly bureaucrats to cave.