5,809 karma · joined February 25, 2010
This does seem to reduce the likelihood that Navalny wasn't poisoned, or that he was poisoned by someone unconnected to the government. MattGaiser's argument is that the implied threat is the goal, which is helped by confirmation of the above.
https://aws.amazon.com/about-aws/whats-new/2019/12/introduci...
Random historical context.
Being legally compelled to provide data isn't the same as being 'willing'.
Completely ignoring the educational content itself, well made educational videos bring community good will. This helps in both purchasing decisions and recruitment. Honestly, this basically can come from a PR / recruiting budget. Lecturer's aren't that expensive.
Now, for ML specifically, expanding the set of ML engineers has a dual benefit:
1) Creates more awareness and therefore demand for ML cloud services. Opensource? No problem, amazon will cheerfully sell you gpu enabled ec2 instances.
2) Expanded pool of candidates for hiring, and increased likelihood that internally, someone will propose a viable ML solution/product.
The ML sector is still growing rapidly, anything that fuels its growth is going to benefit Amazon simply by virtue that they're always going to be in the running for any business purchasing decision, and probabilistically, they'll get a percentage cut of that growth. Education helps grow it, and educational videos are pretty dang cheap.
No imtringued does not.
imtringued wrote:
> This is basically a reverse audit but the software engineer has no authority conduct such an audit and even if he was allowed to, the business has no obligation to give him the necessary information to determine whether the MCAS will be used unethically.
imtringued is saying that it would be impossible for a software engineer to determine whether what they were asked to do was ethical or not.
I seriously doubt that it was the intent of the author to accuse Matt Cutts as someone who spun blame onto others and avoided responsibility.
They appear to be funding Haskell development.
The value of their work doesn't end with their death. This means that they can sell the rights prior to dying and receive some of that value.
https://en.wikipedia.org/wiki/Postal_Accountability_and_Enha...
https://ips-dc.org/how-congress-manufactured-a-postal-crisis...
log_k(c) = e*log_k(m)
log_k(c)*e^-1 = log_k(m)
Raising k by both sides yields k^(log_k(c)*e^-1) => mYou are correct that the main TEXT section used to typically not be position independent.
https://tauri.studio/docs/about/security
Unfortunately, they don't explicitly say what their threat model is, but you may be able to infer it from their feature list.
I am surprised they list "Decompilation is Difficult" as a feature. I'd be hesitant to rely on this for long.
What about process explorer?
https://docs.microsoft.com/en-us/sysinternals/downloads/proc...
You're not about to guess what 'commutative group' is from it's name any more than you're going to guess what an 'abelian group' is, unless you have already learned what commutative or abelian means in the specific context of algebra.
I'm always baffled that programmers, who freely adopt frameworks and languages with mostly meaningless names (which incidentally makes them easier to google for), continue to insist that names are the major stumbling block in learning math. It's utter nonsense. You pick up the names and terminology quickly enough (or look them up if you forget); the hard part is applying them.
Which the parent says is explained via decoherence. Why do you feel that decoherence does not account for the observed phenomenon?
Let's take a few steps back and talk about forces, and work. Gravity isn't really a key part of your question.
Suppose you had a uniform downward force field like so:
| | | | | | | |
v v v v v v v v
| | | | | | | |
v v v v v v v v
| | | | | | | |
v v v v v v v v
Those are vectors pointing downwards. Now suppose you built a miniature roller coaster. Your first one is super boring | | | | | | | |
v v v v v v v v
car
|---\
<-------------------->
| | | | | | | |
v v v v v v v v
It just goes back and forth, perpendicular to the field. Now, if you really grease up the tracks, and have really good bumpers on the end (noiseless, don't heat up, a perfect elastic collision), the car will just bounce back and forth for a very long time, and not speed up or slow down much at all (and ideally, none at all).There is a concept of 'work' in physics. When a force points in the direction of motion, we say that a force does work. The total work done along a path is simply the sum (integral) of all the parts of the path, multiplied by their length, multiplied by the amount of force in the direction of the path. This definition chosen so that the amount of kinetic energy gained, or lost, is equal to the work done.
Since the force is perpendicular for the above path, the 'work' done is said to be zero.
Now let's have a more interesting path. Let's give the car a kick.
car
<- /---|
/--------<-------\
| |
v ^
| |
\------->--------/
Here we have two parts, top and bottom, where the force does no work. However, we have a part going down, and a part going up. The car will accelerate going down, and decelerate going up. If we grease the tracks up,
then the car will always be traveling at the same speed on the top track. It will also always be moving faster on the bottom track.We say that the force does 'work' when the car is on the tracks going down and going up.
Now, the total amount of work that the force does, for any loop, ends up being zero. This is clear here, since the distance the force is applied downwards, is the same as the distance of the force applied upwards.
What is slightly less obvious is that any looped track we could build here would have the same property. If we had something like
/---\
| |
\ |
\ |
\|
(and ignore the mechanics of those super sharp turns) we would find that the work done was still zero. The downward diagonal bits would end up still adding up the same amount of work as the upward straight bits (bit of vector math shows this).The fact that the work done on any loop is zero makes this particular force a 'conservative force'.
Not all force fields need to be conservative! Suppose our field was only present on the left side.
/---|
/--------------------\
| | | |
v | v |
| |
| | | |
v | v |
| |
| | | |
v | v |
\--------------------/
This would only do work going down. This would mean that the track car would keep getting accelerated faster and faster.Potential energy only makes sense as a concept in the presence of conservative fields. You define a reference point, sometimes chosen to be at infinity, and define the potential energy at each point to be the amount of work done to bring the object to that test point.
However, you don't need to just be a constant field to be conservative. Gravity, as expressed by Newton, is a conservative field. Regardless of how you arrange a bunch of objects or planets, looped paths will always have zero net work done. This is because the mathematical form of Gravity can be epxressed as the gradient of a scalar (takes a position, gives a number) function.
You seem happy with the idea that locally on earth, energy is conserved because you get the energy back when you travel in a loop. This doesn't change when you travel in a more complicated path from one planet to another. The force field does work on the object to accelerate it when it moves in the direction of the force, and decelerates it when you pull it back.
The big takeaway point here is that energy, as a concept, is built on top of (at least classically) the more primitive concept of 'force'.
"Snowpack treats bundling as a final, production-only build optimization. By bundling as the final step, you avoid mixing build logic and bundle logic in the same huge configuration file. Instead, your bundler gets already-built files and can focus solely on what it does best: bundling."