A problem that only affects newbies huh?

Let’s say that you are writing code intended to be deployed headless in the field, and it should not be allowed to exit in an uncontrolled fashion because there are communications that need to happen with hardware to safely shut them down. You’re making a autonomous robot or something.

Using python for this task isn’t too out of left field, because one of the major languages of ROS is python, and it’s the most common one.

Which of the following python standard library functions can throw, and what do they throw?

bytes, hasattr, len, super, zip

Oh, I’ll try to describe Euler’s formula in a way that is intuitive, and maybe you could have come up with it too.

So one way to think about complex numbers, and perhaps an intuitive one, is as a generalization of “positiveness” and “negativeness” from a binary to a continuous thing. Notice that if we multiply -1 with -1 we get 1, so we might think that maybe we don’t have a straight line of positiveness and negativeness, but perhaps it is periodic in some manner.

We can envision that perhaps the imaginary unit, i, is “halfway between” positive and negative, because if we think about what √(-1) could possibly be, the only thing that makes sense is it’s some form of 1 where you have to use it twice to make something negative instead of just once. Then it stands to reason that √i is “halfway between” i and 1 in this scale of positive and negative.

If we figure out what number √i we get √2/2 + √2/2 i

(We can find this by saying (a + bi)^(2) = i, which gives us (a^(2) - b^(2) = 0 and 2ab = 1) we get a = b from the first, and a^(2) = 1/2)

The keen eyed observer might notice that this value is also equal to sin(45°) and we start to get some ideas about how all of the complex numbers with radius 1 might be somewhat special and carry their own amount of “positiveness” or “negativeness” that is somehow unique to it.

So let’s represent these values with R ∠ θ where the θ represents the amount of positiveness or negativeness in some way.

Since we’ve observed that √i is located at the point 45° from the positive real axis, and i is on the imaginary axis, 90° from the positive real axis, and -1 is 180° from the positive real axis, and if we examine each of these we find that if we use cos to represent the real axis and sin to represent the imaginary axis. That’s really neat. It means we can represent any complex number as R ∠ θ = cos θ + i sin θ.

What happens if we multiply two complex numbers in this form? Well, it turns out if you remember your trigonometry, you exactly get the angle addition formulas for sin and cos. So R ∠ θ * S ∠ φ = RS ∠ θ + φ. But wait a second. That’s turning multiplication into an addition? Where have we seen something like this before? Exponent rules.

We have a^(n) * a^(m) = a^(n+m) what if, somehow, this angle formula is also an exponent in disguise?

Then you’re learning calculus and you come across Taylor Series and you learn a funny thing, the Taylor series of e^x looks a lot like the Taylor series of sine and cosine.

And actually, if we look at the Taylor series for e^(ix) is exactly matches the Taylor series for cos x + i sin x. So our supposition was correct, it was an exponent in disguise. How wild. Finally we get:

R ∠ θ = Re^(iθ) = cos θ + i sin θ

What god formula?

No, I just understand math. So yes.

Well, 13 microarcseconds is the resolution they claim to be shooting for. The nearest star is 4.2 light-years away. 13 microarcseconds at 4.2 light-years is 2500km, the earth is about 12742 km in diameter. So we can theoretically take an approximately 5x5 pixel image of Proxima Centauri b.

If you are taking requests, I am curious how ridiculous The Longest Journey would be.

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No one tell OP that the ml in lemmy.ml is for Marxist Leninists.

Too bad you’ll never receive that option from any manufacturer.

The scam is that they are actually doing the work, getting paid well

Listen. I know that there are some really shitty stuff going on in North Korea, and very real threats that their government is capable of, and it sucks for the people living there who have to do this work under threat of death.

But if you say that “the scam” is they’re doing work and receiving full pay for work done, I’m going to make fun of you. Oh no, someone outside of the West did work and was slightly less exploited by capital than usual in the process. Horror upon horror.

Most recently, other than Trump, George HW Bush lost the election while incumbent. Prior to that it was Jimmy Carter.

The next most recent person to win the election but lose the popular vote was George W Bush, prior to that is was Harrison back in 1888.

Please don’t tell me you, unironically, actually use the Carmack rsqrt function in the year of our Linux Desktop 2024.

Also if you like, you can write unsafe Rust in safe Rust instead.

std::men::transmute

(please attend to primaries next time…)

So… should I have voted for Marianne Williamson or Dean Phillips, keeping in mind Dean Phillips formally withdrew from the race before my state’s primary, and Marianne Williamson couldn’t have won if she had sweeped every state after and including mine?

I think the problem is mostly that the US system of elections is turbo mega fucked.

Not quite. Their “malicious” extension only got a few hundred installs. Using the data gathered by that extension and via other means they were able to locate other actually malicious extensions. Those total in the millions of installations.

Through this process, they have found the following:

1,283 with known malicious code (229 million installs).
8,161 communicating with hardcoded IP addresses.
1,452 running unknown executables.
2,304 that are using another publisher's Github repo, indicating they are a copycat.

To add, let’s do some math!

Let s be the total annual salary of every employee using Adobe. Our goal is to find the productivity ratio r such that changing to Gimp and open source more generally is a net positive from the standpoint of productivity and labor.

s/r will be the total annual salary after changing over, because (for instance) if r = 0.8 then LTT will need to either hire or work his existing hires 1/0.8 times longer, giving (at best, ignoring overtime and so on) s/r as the new labor cost.

We then subtract the current labor cost to get the switching cost s/r - s, and if this is greater than $10,000 then the switch is not worth it.

For instance, let’s say LTT employs 1 person at $50k/year. He’s a bit of a skinflint. We solve for r and arrive at a ratio of 5/6 or 83.33%.

If we have a different world where LTT hires 10 people and pays each of them $100k, we solve for r and get about 99%.

In other words, the switch is worth it only if the labor cost is small, so the extra labor is not very expensive, or the difference between the two software is negligible.

For those curious: Gothic 1.

I’ve never heard of it before and it doesn’t look like my type of game. Anyone played it?

Cassette Beasts not Beats ;)

What about Elisa? I was under the (potentially mistaken) assumption that Elisa was the successor of Amarok.