How do I make an espresso shot in a moka pot?

Question: Ive tried brewing espresso in a moka pot but I end up making coffee that tastes like drip coffee. Not sure why, but is it because the water level in the moka pot (I fill it up just below the opening hole) is too much? I'd like to get the consistency of an espresso shot made by an espresso machine (thick and creamy). I've been following tutorials online but nothing changes!
Comment: The short answer is: you can't :(
Answer: You can create more pressure if you will use finer grind (but be careful). I even tamp a bit (but be careful again). You can also buy Bialetti Brikka, which has a pressure valve - it helps to build higher pressure and create crema. However, with moka pot you'll never get real espresso, you can only try to brew something similar.
Comment: thanks for the comment. yes i was wondering as well if a moka pot can actually be used to make espresso.
Answer: It is a common misconception to think that a Moka Pot can make espresso like coffee. Espresso machines use up to 9 bars of pressure to push the water through the grounds. The Moka Pot only creates around 1,5 bar of pressure (for comparison, the pressure of your tap water lies probably between 3 and 6 bar). It is simply impossible to get anything close to an espresso (although it is often considered to be the closest espresso alternative without an expensive machine supposedly due to the residue coffee in your cup that gives it a slightly similar mouthfeel and body).
You can try to use very finely ground coffee, but that's about the only thing and it still won't make you an espresso. For that you need to invest in a machine. The good ones start at 500$/€. You can of course buy one for 100$/€, but they are not worth the money in my opinion. Tamping and other tweaks are not recommended because you might block the water from passing through the grounds, building up pressure until it is explosively released. If you are lucky through the pressure valve, but I've seen a Bialetti literally "erupt", spraying coffee through the entire kitchen and several meters high to the ceiling (slightly more than 3m to be precise, because that's the height of the ceiling where to coffee eruption came to a forceful halt).
Comment: great advice. yes im looking to buy myself an espresso machine!
Comment: @abunesh Yes if you want real espresso at home there won't be an alternative. However I would consider making non espresso coffee at home. In order to offset the cost of the machine, the grinder (if you are serious about espresso it will cost as much as the espresso machine), occasional maintenance, you would need to drink lots of it. It might just be cheaper or at least more convenient to get your espresso fix at your local coffee shop, which will likely also taste better depending on the quality of the coffee shop of course. Depends on what you want, but I think it's worth thinking about.
Comment: Also if you think a response answers your question you can up vote and/or accept them.


Unintentionally remapped keys on a MacBook Air (2012)

Question: One day my MacBook Air mysteriously remapped the 'U', 'I' and 'O' keys to all be the alt/option key. I've no idea what triggered this, I took it to the apple shop and it seems to stay that way when they did the hardware test. Does anyone know why this could be, and hopefully recommend a way of fixing it?
Comment: Does an external keyboard work OK?
Answer: Try resetting the NVRAM and the SMC on your Mac then test those keys and see if they are still remapped.
NVRAM aka PRAM
https://support.apple.com/en-us/HT204063

How to reset NVRAM
Shut down your Mac, then turn it on and immediately press and hold
  these four keys together: Option, Command, P, and R. You can release
  the keys after about 20 seconds, during which your Mac might appear to
  restart.
On Mac computers that play a startup sound, you can release the keys
  after the second startup sound. On iMac Pro, you can release the keys
  after the Apple logo appears and disappears for the second time.

SMC (System Management Controller)
https://support.apple.com/en-us/HT201295

How to reset the SMC on Mac notebooks
First, determine whether the battery is removable. Most older Mac
  notebooks have removable batteries. Mac notebooks that have
  nonremovable batteries include MacBook Pro (Early 2009 and later), all
  models of MacBook Air, MacBook (Late 2009), and MacBook (Retina,
  12-inch, Early 2015 and later). Learn more about Mac notebook
  batteries.
If the battery is nonremovable:

Choose Apple menu > Shut Down.
After your Mac shuts down, press Shift-Control-Option on the left side of the built-in keyboard, then press the power button at the same
  time. Hold these keys and the power button for 10 seconds. If you have
  a MacBook Pro with Touch ID, the Touch ID button is also the power
  button.
Release all keys.
Press the power button again to turn on your Mac.

If the battery is removable:

Shut down your Mac.
Remove the battery. If you need help removing the battery, contact
  an Apple Authorized Service Provider or Apple Retail Store.
Press and hold the power button for 5 seconds.
Reinstall the battery.
Press the power button again to turn on your Mac.


Can an independent subset of a theory be completed to axiomatize the theory while preserving independence?

Question: Let $L$ be a first-order language, and let $T$ be a theory in $L$, that is, a consistent and deductively closed set of sentences. Let $I$ be a subset of $T$ which is independent. An independent set of sentences is one where no sentence is redundant. We are not assuming that the deductive closure of $I$ is $T$, merely that $I$ is a subset of $T$. My question is, is there always a set $I'$, where $I \subseteq I' \subseteq T$, such that $I'$ is also independent, and the deductive closure of $I'$ is $T$? Or, is there a counterexample for some language $L$ and theory $T$ and set of sentences $I$?
Comment: If $T$ is countable, the usual strategy for giving an independent axiomatization also gives an affirmative answer to this question. If $T$ is uncountable, then even proving the existence of an independent axiomatization in the first place is hard. See https://math.stackexchange.com/questions/3289207/nonexistence-of-independent-axiomatization-of-a-theory-in-a-finite-language/3289290#3289290.
Comment: @NoahSchweber: I don't think the usual strategy for giving an independent axiomatization delivers the goods if it is required to give a result that contains some given subset of the theory. Primo Petrii's answer looks good to me.
Comment: @RobArthan You/he might be right, I'll take a look later today.
Comment: @RobArthan Belatedly looks like you're right.
Answer: If I am not mistaken, this should be a counterexample:
The language contains the constants $\omega$ and a unary predicate $r(x)$.
Let $S=\{r(i):i\in\omega\}$.
Let $T={\rm cl}(S)$, the closure of $S$ under logical consequences.
Let $I=\{r(0)\vee r(i+1):i\in\omega\}⊆T$.
Clearly $I$ is independent
Let $I'$ be such that $I ⊆ I'⊆ T$ and $T={\rm cl}(I')$.
By compactness, there is some $I''⊆I'$ finite such that $I''\vdash r(0)$.
Then $I''\vdash I$, hence $I'$ is not independent.


Pay a very high price

Question: What are English idioms for "you have to pay a very high price to get the thing that you want"?
Comment: Sorry, user626528, that *is* an idiom. Did you mean something like "synonymous phrases" or roughly what, please?

Remembering that "you have to pay a very high price to get the thing that you want" is *not* an English idiom, but a lengthy phrase which happens to include such an idiom, perhaps as a translations from another language… where specifically did your original phrase come from, please?
Comment: @RobbieGoodwin does it really matter where it comes from? I just need some idioms that have this meaning.
Comment: Then why did you ask here, rather than using a search engine, please?
Comment: @RobbieGoodwin if I can't ask questions here, what is this site for?
Comment: @RobbieGoodwin I know no search engine capable to recognize idioms. And by the way, I think that you should be spending more time trying to answer questions and less time trying to push people around.
Answer: You can use the metaphor "to pay a high price for something".
Alternatively, use the proverb There is no such thing as a free lunch, which means it is impossible to get something for nothing, there is always a price to pay.
If you mean that you have to pay a very high price to get the things you want and you have doubts about it, or you are not sure if it's worth it, you can say The game is not worth the candle. This idiom is a bit old-fashioned though.
Answer: This one looks very close to this meaning: Give an arm and a leg for (something)

Pay a very high price


Is it ever explained where Elsa got her powers?

Question: I recently saw the movie Frozen, and got to wondering about Elsa's amazing Ice Powers.
Where exactly did they come from?  It's apparent that there's some type of curse from the dialogue between her parents and the Trolls, and possibly that it's an inherited curse from her father's knowledge of how to control it by wearing gloves, but does anything beyond that indicate how Elsa (and possibly the whole family) came to be cursed this way?
Comment: The X-Gene, obviously.
Comment: @MichałPolitowski Disney doesn't have the rights for that. It was from exposure to the Terrigen Mist.
Comment: It's inborn, not cursed.
Comment: @MichałPolitowski - http://www.youtube.com/watch?v=Dach1nPbsY8
Comment: Elsa's powers are a way of nature compensating for her innate hotness.
Comment: Well, Jack Frosts had a sister. As for she grew up and the fill the emptyness in her heart, the sisters first daughter received powers to resemble Jack's. Or how Pixar drops eggs all the time ...
Comment: I was going to say Terrigenesis but @MichałPolitowski got there first :)
Comment: https://www.youtube.com/watch?v=dooAjI6yOhg
Answer: In-universe answer:
It seems quite plausible that Elsa is descended from someone who was cursed by ice magic. That power may have then transferred down the family line.
As you can see from the image below, the King's library has a book detailing the risks of being hit by ice powers. In this book we see a picture of a man being cured by trolls. The runes next to the picture state that his condition was caused by being...

"...to stone brought forth (by) powers of the moon sky"

presumably in much the same way that Rapunzel's powers relate to a drop of sunlight falling from the sky.

Given the fine clothes he's wearing and the royal crest on his tunic, it's quite possible that he is of noble birth and hence, an ancestor of Elsa's. Assuming some of the residual 'ice magic' was left in this man's system after he was cured it would explain why one of his descendants was born with those same ice powers.
Out-of-Universe answer:
Screenwriter Jennifer Lee specifically spoke to this point in an interview with Scriptnotes. The origins of Elsa's powers (planetary alignment, predestination, etc) were intentionally obscured in order to simplify the story.
In a nutshell, she has powers because the plotline requires her to have powers.

Jennifer: And that there's a connection between them, but yet it's not in your face, but it's just something that… — Because what I
  always loved about, particularly Pixar films for me, was that
  everything just added up. And everything had a special little, "Oh my
  god, oh my god, wait, and that, and that!" And it was my favorite
  thing and we wanted to make kind of every time we had a scene trying
  to say what is that that's maximum, why is it here. If there's
  anything extraneous we got to get rid of it.
But yet adding all that flavor, so that's why. But to move onto Elsa,
  it was an exhausting process coming to the simplicity of her powers. At times we had a narration by a troll, who used to have a Brooklyn
  accent for no reason other than I miss Brooklyn. You know, no reason.
  But, we had this whole explanation like when Saturn is in this
  alignment with such-and-such on the thousandth year a child will be
  born and blah, blah, blah. And then –
John: Ultimately you almost throw it away with one line. So, the
  line is just like, "Was she born with the powers or was she cursed?.
  And it's born with it and that's the last piece of it.
Aline: It's so great.
Jennifer: And that's it. But I think part of what it was is if anything about us felt like it was like, "Oh, god, like okay, we have
  to say this," then we didn't want to say it. And then also we found
  the more you explained the more questions you had about magic and the rules. It was like, argh. You know?
Comment: That the King immediately knows what to do suggests personal or family experience with someone getting zapped by frosty magic—I'm putting my money on it being a family trait that's skipped a few generations :-)
Comment: It just occurred to me that, out of universe, the answer all along was to *let it go*.
Comment: The out of universe detail is nice. Your in-universe answer suggesting remnants of a curse leading to a descendant being born with it is a stretch and doesn't add anything (In fact seems to be debunked by the interview excerpts.) I don't think Tangled serves as precedence, as Rapunzel was actually in-utero when Queen Primrose was treated with the flower and absorbed it's powers directly.
Comment: _"It was like, argh. You know?"_ Nobody can bloody speak any more.
Comment: When do we see the powers transferred from one person to another? Other than by birth, which is a supposition by evidence that I allow for.  Also, while this lends credence to it being an inherited trait, we can't realy determine that they 'cured' the fellow in the picture by this image alone.  For all we know, that's an evil Troll who *created* the curse, and the other trolls are helping his decendents out of a need to atone for their wicked 'cousin''s behavior.
Comment: @LightnessRacesinOrbit - "*such-and-such on the thousandth year a child will be born and blah, blah, blah*"
Comment: @Richard I'm glad you were able to decipher that interview. I don't understand it at all.
Comment: In-Universe states in the movie she was born with her powers.  Past that is pure speculation.  Towards other "lore"  there is one other with the exact same power, but does not take the warm inside and fear thing the same as Elsa.  Again, pure speculation since there is no connection past Repunzel and Nose guy joining the Coronation.
Comment: Of course, the REAL reason they didn't explain it is because now they have a plot for Frozen 2.
Comment: The patient in the picture does appear to have Arendelle's crest on his tunic, but is wearing an unadorned helmet not a crown. Seems as likely to just be a soldier or guard as a descendant. Just as plausible explanation to the quick and ready access to the book... King: 'Hey, my daughter has these strange ice powers, maybe I should research them...'
Comment: @Mr.Mindor - Ah, but we do see the powers transferred, suggesting it's at least a possibility.
Comment: @zibbobz - the out of universe answer clearly contradicts the in universe one. I'm merely pointing out that the powers of magic seem to be transferable. It's a possible answer, not a definitive one.
Comment: @Richard Fair enough.  It IS a good answer.
Comment: "It was like argh"
Comment: If the picture shows someone being cured of the curse, why didn't the trolls cure Elsa? It could be that it shows someone injured by the powers being healed, like Anna. Also, idk about you but being born with ice powers would be a blessing, not a curse
Answer: When they visit the trolls, the Grand Pabbie asks "Born or cursed?" Elsa's dad answers "Born." Aside from this, there is no other reference to how she got her powers. Here's a link to the screenplay. On page 6 is where they visit the trolls.
EDIT AFTER FROZEN 2 RELEASE:

 Elsa was given her powers by the spirits in the North so that she could right the wrongs committed by her grandfather, and calm the angry nature spirits.
Comment: I was going to chime in as an authority with two young daughters as well, but then for the first time in forever, I decided to just let it go.
Comment: It's the final. I've got an 8 year old daughter. I've seen this movie half a thousand times. He says she's born with it.
Comment: @corsiKa: Bad puns are bad. Put 2 quarters in the bad pun jar, please.
Comment: Pretty solid credentials. I accept this as fact...unless something else emerges to debunk it. (like, say, a sequel)
Comment: Odd, I remember it being a curse from the movie...are you sure that's the final screenplay draft?   I'll check it personally when I get home but...my memory must either be foggy, or it must've been changed at some point.
Comment: Yeah, if they come out with a sequel that explains it, I'd edit my answer as needed.
Comment: @Zibbobz - The debunker has arrived!
Comment: @Richard what has been debunked?
Comment: @Mr.Mindor - "there is no other reference to how she got her powers" - Aside from the picture of her ancestor getting zapped with ice powers, obviously :-)
Comment: @Richard: except it's not. It's a "possible" relative. There's no proof of it.
Comment: @Richard, I tend to agree that it is pretty good evidence, but not solid proof beyond what we already suspected - it does lend some crednece to an inherited trait, and that is good to know, but the solid answer is "she was born that way".  I'd recommend keeping that new information though - it's very releveant to this question.
Comment: Hopefully they will clarify this someday. It still doesn't really make sense.

Spirits never give powers to natives, but see Iduna do something selfless, and then either wait a decade and somehow track her down after she leaves, or otherwise "bless" her so that not her, but her future firstborn, Elsa, has ice powers, hoping that Elsa eventually hears a siren to return and somehow fix stuff by turning to ice and stuff for reasons.

Would have connected better if Ahtohallan had been the river (or source thereof) that had been dammed.
Answer: Frozen is cited as being inspired by Hans Christian Andersen's The Snow Queen.
However, the link seems pretty loose, and the titular Snow Queen's powers don't seem to have a specified origin beyond simply being her nature.
Furthermore, Elsa's character doesn't really bear any resemblance to the character of the Snow Queen, and in general there is very little similarity between the plots.
As MyCodeSucks points out, the only reference to the power's origin in Frozen is that Elsa was born with it.
Comment: Originally, it was much closer, but it changed a lot in development.
Answer: Here's a quick summary of the history we learn from Frozen 2:

 There was a group of people called the Northuldra living in an enchanted forest to the north of Arendelle. These people had a great connection to magic, living and working with the four elementals (water, fire, earth, and air) up until the elementals were upset by conflict between the Northuldra and the Arendellians, which was started by Elsa and Anna's paternal grandfather, King Runeard of Arendelle. (Then) Prince Agnarr is saved by one of the Northuldra, Iduna, who becomes his wife (and of course they are Elsa and Anna's parents).

As for what this makes Elsa:

 Elsa is the fifth elemental, born with her powers as a gift from nature due to her mother selflessly saving her father. She is destined to be a bridge between the mundane and magical worlds.

Which means that her powers aren't from a curse after all.
Comment: Indeed. And while this entirely retcons where her powers come from, it's still an answer worthy of an upvote.
Answer: The Grand Pabbie asks

Born with the powers or cursed?

and the King immediately says

Born. And they're getting stronger.

However, when Anna tells Elsa that her ice powers have set off an eternal winter "everywhere", Elsa loses hope and sings,

Oh, I should have known, I can't be free/No escape from the storm
  inside of me/I can't control the curse/Anna, please, you'll only make
  it worse!

Basically, Elsa calls it a curse because she sees it as one, but it isn't; she was born with it.
Comment: Isn't it posible that she was born with a curse? I don't see why they have to be mutually exclusive.
Answer: I think elsa inherited her powers from someone such as a grandfather, who was cursed with ice powers( notice the book). Maybe they're going into the guy after all. And also notice that when Pabbie asks if she was born with them or cursed. The king says,"Cursed, born." He hesitates then says "born". So maybe she was somehow cursed around her birth. What do you think?
Comment: I interpreted the hesitation as him waiting for an answer after asking if she had been cursed. "Cursed?" No answer. "Born."
Answer: Elsa I think got cursed by a evil witch or she got cursed from the trolls in the book. Based upon in Frozen how it shows you in the picture from a troll about a man that looks like he either got hit by this curse that he got from the troll or he was being healed.
Comment: @cde - The runes (next to the picture) specifically state that he's being cured, not cursed by troll : http://lingaspect.wordpress.com/2014/03/13/deciphering-the-runes-book-in-frozen/
Comment: @cde - I've updated my answer to reflect the content of the runes.
Comment: The picture shows a man being cured of the ice magic. Note that the magic is leaving him, in the same way that it left Else.
Comment: @Richard a still image leaves no context.  I can see that image easily being interpreted as cursed rather than uncursed.
Comment: @Richard see, that's the type of information that is desperately needed in an answer. Some animator's attention for detail is amazing. I normally expect things like random runes in a 3 second screenshot to be gibberish ipsum lorem, not relevant. And it's added connection to Tangled is interesting too. You should definitely answer this question with that info.
Comment: @richard the thing is the troll in that picture seems ominous, dark, without any positive/happy expression. A shadowed clawed creature with (imho) aggressive eyes typically denotes a negative situation. Without your explanation, I honestly think that man is being cursed, not cured.
Answer: Maybe she was young and was out playing, maybe the girl from the ice queen wanted to get rid of her curse so she cursed Elsa and ran away.
Or one of her ancestors had ice powers; notice the book and the way the Elsa's dad knew exactly where to go.
The boring answer she was born with it.
Comment: Welcome to SciFi.SE Keely.  Your answer is largely opinion-based.  On SciFi.SE, we're looking for answers backed up by facts from canon (unless otherwise stated, and then only in special circumstances).  These are the rules of the site.  To find out more rules about answering questions, try http://scifi.stackexchange.com/help/answering


How can I force netcat to send my input immediately, not just on newlines?

Question: How can I force netcat to send my input immediately, not just on newlines? I want to test an HTTP parser manually and check how it behaves when header lines are spread across multiple packets.
Comment: try `socat` as listed in this [answer](http://superuser.com/a/331595/122284)
Answer: Use CtrlD, which is set by default as the tty eof key. When pressed in the middle of a line, it will give to netcat everything that has been input at that point.
The buffering is actually done by the tty layer and not handled by nc at all. stty -icanon && nc ... would disable the buffering and allow nc to see the data as it is entered into the terminal, at which point it will be sent right away. (Note that the stty and nc commands must be run together, otherwise the shell itself would likely reenable it when displaying its prompt.)
Comment: @FinalContest: The buffering is done by the tty layer. `stty -icanon && nc ...` or `stty raw && nc ...` would disable it. [Note that it must be run together, otherwise the shell itself would reenable it when displaying the prompt.]
Comment: @FinalContest: Yeah, `raw` changes a whole bunch of tty options, one of them being the handling of special "control" keys at tty level. (It is really a "raw" mode, in that it passes _everything_ to the program.) You can use `stty -a -F /dev/pts/XX` from another tty to see the current parameters.
Comment: I am using `stty -icanon -echo && nc ...` otherwise every char I typed in nc get echoed twice
Comment: +1, but is there no option for opening netcat up the way that it does not buffer anything?
Comment: @grawity: awesome, the former works fine, but the latter gets stuck... ctrl-c does not work anymore to quit the session, and I cannot get ctrl-z to work on it either. Is that expected? Also, please update your answer so that we can clean the comments up.


Attach and retrive multiple pdf files to post or page

Question: I need to create two sections that will respectively contains only some pdf to downloads. I've never used pdf inside a wordpress websites, my question is how I can display them or in any case had a list of them? I was thinking about attaching the files to the relative post/page, this to avoid confusion between the different files that I will display on the two different sections. Can anyone suggest me a smart and fast way to achieve this (files attaching)?
Answer: You should definitely use Advanced Custom Fields (ACF) plugin. Go for the PRO version, since you might need it's functionality (it costs less than 20 USD) and will be one of the best WordPress related investments.
You should look at the Repeater field (two of them, one for each section) and put a File field as repeater's row. They you need to display these fields on the front-end.
The documentation is amazing and easy to understand even for a complete beginner.
Comment: Ok,I know about that plugin, but unfotunately my client will not pay for an extra plugin to have this possibility so I was looking for a reference or a code snippet that will let me achieve this with php and wp core functions
Comment: @sialfa You can definitely code everything that ACF offers on your own. Make some custom meta fields, code the upload forms in the back end and and attempt to display it on the front end, but it will take you much, much more time and effort. There is no point in reinventing the wheel. 20 bucks is really nothing. But if you somehow can't afford it, there are still ways to get the Pro version free, right? But we do not support that over here.
Comment: It's not a question of the needed time, the point is that if a client have a budget, it's supposed that the developer respect it. Moreover a professional developer needs to know how to create custom components and not use a plugin of a third party to do something in wordpress. Plugins are helpful but sometimes they are created from developers for user with no programming skills in general. This is why I don't like to work with wordpress. Too many developers are using plugins but don't know how to make the things on their own.
Comment: @sialfa Are you not charging your client? Are you working for free? If you charge 10 cents an hour it is reasonable to code it yourself. If not, ACF is 100 times cheaper. So yes - if you want to respect client's budget, buy ACF for 20 bucks. Also, I am a web developer, I work with wordpress 8 hours a day, and again - there is absolutely no point in reinventing the wheel. I like to use my own code instead of bloated plugins, but it is not the case of ACF.


Did the Starcraft Dominion Terrans evolve directly from the U.S.S.R?

Question: The Terran backstory (animation scenes) in Starcraft has multiple Russian Generals. Why? Did that arm of Terrans evolve directly from the U.S.S.R?
Comment: What is the point of the image?
Answer: There are 3 main terran factions that appear in the original game. The Terran Confederacy, The Sons of Korhal, and The UED (United Earth Directorate). Only the UED have the Russian Generals that you mentioned. And they make a point in the game to say that UED are the only Terran fleet directly from earth. It might stand to reason that Earth is now a USSR modeled form of government (or the UED are based out of Russia), but not all Terran forces follow this model.
Answer: The Terran came from Earth but got lost in space.  With their existing knowledge, they formed governments and a military.  Then there was a revolution turning the government into a dictatorship.
First of all the Terran's didn't evolve from anything from earth.  They are from Earth, in a matter of speaking.  Their "race" is from Earth.  As you see from different characters, like Duran, Raynor, Kerrigan, Mengsk,.
Blizzard probably made the Generals look Russian to make a reference that the Soviets have a dictatorship, just as Mensk is the Dictator of the Terran Alliance.
Comment: ahem, _had_ a dictatorship...
Comment: Funny you should mention Duran...
Answer: Short answer: no.
In the original game manual, pages 26-27, there's a description of Earth history leading up to the launch of the sleeper ships (emphasis mine):

By the end of the twenty-first century, mankind had seen bold and unprecedented changes within the world. Radical new technologies were surfacing at incredible rates.... In the wake of the eradication of Communism from the Eastern Nations, nuclear weapons quickly became available in abundance. The international power structure... was blasted apart as third-world nations rose to challenge [the superpowers].
Popular sentiment held that the world was plummeting towards an inevitable catastrophe as overpopulation and genetic alteration swept across the globe.... Horrific acts of terrorism and violence erupted between the corporate sector and the humanist factions, resulting in forced police actions across the globe.... Ultimately, the precarious balance of world power exploded into international pandemonium.
On November 22, 2229, the United Powers League was founded. The UPL was to become the ultimate incarnation of the vision of a unified humanity held by the now-defunct United Nations.... The UPL was founded on the basis of "enlightened socialism", but often resorted to harsh, fascist police actions to maintain public order.

From this, it seems unambiguous that the old social order, nations and all, had essentially fallen apart by the time the UPL came around, with special emphasis on the collapse of Communism. The UPL itself seems to have foundational ethics from all over the place; certainly it has some Stalinist elements, some Communist ideas, but it isn't a direct descendant of any one single Earth ideology or nation.
When they realized that the Terrans in the Koprulu sector had discovered intelligent alien life, the UPL was reorganized into the UED with the express aim of defending Earth by any means necessary (read: purge all the xenos).
Answer: The only characters in StarCraft with a Russian accent is Alexei Stukov, one of the leaders of the United Earth Directorate invasion from Earth to the Koprulu sector. Stukov, due to his accent, can be assumed to have had his origins from the old Russia on Earth.
That does not mean that Earth was governed by Russia.
Actually, in the StarCraft lore, it is stated that the story departs from a point where the USSR never broke. It's vaguely explained that The world politics made some twists based on some new religion, related with pure-human (humans with no implants and so), supremacy.
Criminals and outcasts were sent in 4 big ships as a colonial experiment, and some of them survived, founding 3 colonies in different systems of the Koprulu sector.
The game lore starts a few hundred years later.


What can I do about cracks in a stone foundation?

Question: Considering purchasing a brick farmhouse in mid-Michigan built in 1885. There are large visible cracks in the foundation. Anyone have thoughts on the cause and possible repairs?
Comment: Freeze thaw cycles over the past 130 years, causing minor unrepaired cracks to expand. Erosion of the ground around the foundation.  Those are probably two of the possible causes. Repair is going to depend on the extent of the damage. Cost is off topic.
Comment: There isn't a 19th-century stone foundation in existence that doesn't have large, visible cracks. They probably didn't occur recently. Evaluate the interior of the home for the resulting symptoms, and assess from there. It's not a question for a internet forum.
Answer: You need to get a professional home inspector onsite. He should be capable of telling you whether these are cracks of structural significance or not, and discuss remediation measures (not cost thereof).
Answer: It's amazing to me how residential construction is so over designed. Your pictures clearly show how the corner of the foundation has cracked and fallen away from the structure. Obviously, it's not "needed" to hold up the walls of the house, or it would have fallen down by now. (Are there large cracks in the wall on the other side of this area?  ...No, probably not.)
These stone foundation systems are constructed by mortaring stones together. However, no reinforcing steel (rebar) is used. So, the foundation can withstand vertical loads placed on it, but it can't withstand "soft spots" in the soil. When the soil settles away over time, it tends to crack due to the weight of the foundation.
While that portion of the foundation is not structurally required, it's unsightly and will continue moving (sinking) away from the house. It needs to be repaired. There is equipment that can "raise" portions of a foundation. They inject special foam into the ground that pushes it up and into place, then it is bolted to the rest of the foundation.
Also, good building practices would require a new pressure treated plate with anchor bolts that bolt the foundation to the stud wall construction.
The house has been there so long without any major problems, it's hard to say this is all mandatory. However, if not fixed, it will get worse. Soon, the crack will open up and allow small animals and horses to crawl in...and you don't want that.


What's the difference between Aetna payflex and a regular health savings account

Question: I'm currently enrolled in payflex has which doesn't roll over and constantly asks me for eobs to prove that my expenses are tax deductible. My mom told me that you can open up a HSA with your bank. What's the difference?
Comment: Payflex encompases a bunch of different accounts. Are you sure you have an HSA? When you say that it "doesn't roll over," it sounds like an FSA, not an HSA. FSAs need to be used up by the end of the year, or you lose the funds in the account. Is that what you have?
Answer: Payflex is an FSA (flexible spending account) administrator. These types of accounts have been around for quite a while and are characterized by letting you spend pre-tax dollars on qualified health care expenses. The downside as you see is that it is a user it or lose it design. With an FSA you provide documentation to the administrator and they then determine the eligibility of the expense, the benefit of that is that you don't need to retain the documentation after you have been paid.
HSA (health savings accounts) are newer and act more like a bank account as they roll from year to year. Like an FSA they enjoy the same pre-tax benefit. However here you administer your own payments so you need to maintain records showing that the reimbursements are qualified (I.e. those EOB statements). The one downside to an HSA is that they can only be used in conjunction with a high deductable health plan (essentially you are self-insuring for the base expenses and the insurance covers big ticket items, hence why the money rolls over as it is allowing years of low expenses to offset years of higher expenses).
The bottom line is since you have an FSA at work you are not eligible for an HSA. If you want an HSA you will need to change to a high deductable health care plan at the next open enrollment period (assuming that your employer offers one).
Comment: Note that Payflex appears to be an Aetna brand, not necessarily a FSA (though it seems to be in this case). I have an Aetna HSA with a Payflex branded debit card for accessing the funds.
Comment: @RickGoldstein Payflex is also used by Anthem. Though it is true that they may be more than just FSA, but in this case it is the FSA aspect that is the focus of the question.
Comment: @RickGoldstein is it possible to open an HSA not through work. I missed my enrollment period.


RGB or CMYK for resume/portfolio

Question: Is it best to use RGB color-space when you don't know if the end-user of the document is going to be printing the document on an inkjet or laser printer? For example, even though my home printer has 4 cartridges (CMYK) it still uses RGB 'drivers'.
Comment: Both inkjet and colour laser use CMYK ink/toner.
(apart from some specialist ones)
Answer: You do not need to work in CMYK... Unless you need to.
The only reason somebody needs to work in CMYK is when your project needs to be printed on a commercial print, (offset print).
In some few cases, you can send the file in CMYK if, again the provider really needs it. Some inkjet plotters do use the data from a CMYK file.
But most of the time, home, office, small copy shops, laser printers, inkjet printers use the RGB values and re interpret them to whatever values they need. Sometimes it is CMYK, other times is CcMmYK, some other CcMmYKk when you have Light Cyan, Light magenta, light black.

But normally you need to simulate colors, so you do not end with fluorescent colors if you will not use special inks.

Sayed that I agree with Emilie.

They most likely will not print your work.

A CMYK file can potentially look bad on a visualizer that do not support CMYK profiles.

RGB files compress better than CMYK ones.
Answer: I would not assume the person receiving your resume will print it, unless they've told you so.
Personally I would stick to RGB for a few reasons:
Your resume will likely be viewed on screen. I don't know any studios who print out all the resumes they receive, it's too much. If they DO print your resume out, at this point I would think the interview will make the difference.
Second, you will benefit from a larger gamut which will make your work look better than CMYK.

RGB or CMYK for resume/portfolio


How to rename an executable from its details

Question: I have loads of .exe files that got renamed to file1.exe, file2.exe, etc. When you look at a file's Details under Properties (in Windows Explorer), it has the original file name under "Product Name".
I am trying to find a way to batch rename them all back to the product name.
There are tools that can do this with images, music, etc., but I can't find anything out there that will do it for applications. This would be (fairly) straightforward if there were some way to access the Properties → Details in a command line (batch) context.
Answer: You can use PowerShell to do this.
$newname=(get-item .\file1.exe).VersionInfo.ProductName
Rename-Item .\file1.exe -NewName "$newname.exe"

EDIT:
Here's a one-liner that should accomplish what you need:
(gci).name | Where-Object {$_ -like 'file*.exe'} | foreach { $name=(get-item $_).versioninfo.productname;rename-item $_ -NewName "$name.exe"}

Answer: to make it like ProductName_ProductVersion.exe :
(gci -Path "E:\test" -Filter *.exe).FullName | %{$NewName = (Get-Item $_).VersionInfo.ProductName +"_" + (Get-Item $_).VersionInfo.ProductVersion;Rename-Item $_ -NewName "$($NewName).exe"}

and if you want to make it like ProductName_FileVersion.exe just replace ProductVersion with FileVersion in code
DO NOT forget to replace E:\test with true path


Why is (dot) used in arguments for parameters?

Question: Currently trying to learn swift and I am a bit confused to why some arguments are written with a dot before it and some don't. For example
To add a button to a UIAlertView object
alert's name = simpleAlert
buttons title = "YES"
button's style = Default
handler = nil
why is the appropriate way to write this is
simpleAlert.addAction(UIAlertAction(title:"YES",style: .Default, handler: nil))

instead of
simpleAlert.addAction(UIAlertAction(title:"YES",style: Default, handler: nil))

Answer: It's a short way of writing UIAlertActionStyle.Default (UIAlertActionStyle being an enum and Default being a value of it). The UIAlertView already knows you are gonna use a UIAlertActionStyle for that parameter so you can just write that enum's value.
Answer: simpleAlert.addAction(UIAlertAction(title:"YES",style: .Default, handler: nil))

can also be written  as a
 simpleAlert.addAction(UIAlertAction(title:"YES",style: UIAlertActionStyle.Default
, handler: nil))

UIAlertActionStyle is the Enum
enum UIAlertActionStyle : Int {

    case Default
    case Cancel
    case Destructive
}

so .Default is just a simplification which means Enum type is Already given we just have to pass the proper case which we want to use.
Answer: As others have already said, that is a shorthand for UIAlertActionStyle.Default.
Skipping the name of the enum and simply listing the value (.Default instead of UIAlertActionStyle.Default) is an example of Swift "type inference".
Type inference means that if the compiler can infer (figure out) the type of some value/parameter, you don't have to state it explicitly. In the example above, the value you provide for style must be a value of type UIAlertActionStyle, so you can skip that part.


How to convert an HDR Image created / rendered on Blender to a format acceptable on social media and websites?

Question: I rendered a HDRi on Blender from my ambient - to look like real state presentation - and I want to convert this HDRi file into a format acceptable on Facebook and / or other media as they doesn't open HDRi's.
I don't even know what is the format of 360 (equirectangular, spherical actually) photos generated from cameras and phones, otherwise I could do my research on it :(
Any help is appreciated! Thanks!!
Comment: Surely this is better asked on Facebook or other social media sites.
Comment: You can save your HDRI as .png and/or .jpg if you want to. They should work fine on facebook. No need for .exr or anything fancy like that.
Comment: Hey :). So you want to render a 360° photo/video and upload it to FB?
Answer: Sorry for the links only answer, but this is too long for comments.

You are mixing two different concepts:
1. HDRi
stands for High Dynamic Range image, that has larger colour values than those that can be displayed by any screen. They are used mostly as environment textures to light scenes.
2. 360 images
concept of environment projection https://wiki.panotools.org/Projections
Watch https://www.coursera.org/lecture/360-vr-video-production/equirectangular-projection-cubemaps-uTv8r
Most HDRi is mapped used in 3D and VR is using equirectangular projection, but is not mandatory.
As to how to do this in blender, it has been asked before:
Blender panoramic render
Is it possible to render a panorama in perspectives other than full 360°x180°?
equirectangular rendering for Blender Internal?
how do I bake an animated sky box using cycles?


Add "parent" class to parent menu items

Question: Is there a good way to add a class of "parent" to list items that have children using wp_nav_menu?
I've found a couple of solutions online but non seem to have the desired effect.
Answer: See Wordpress codex http://codex.wordpress.org/Function_Reference/wp_nav_menu#How_to_add_a_parent_class_for_menu_item
Use this function:
add_filter('wp_nav_menu_objects', function ($items) {
    $hasSub = function ($menu_item_id, &$items) {
        foreach ($items as $item) {
            if ($item->menu_item_parent && $item->menu_item_parent==$menu_item_id) {
                return true;
            }
        }
        return false;
    };

    foreach ($items as &$item) {
        if ($hasSub($item->ID, &$items)) {
            $item->classes[] = 'menu-parent-item'; // all elements of field "classes" of a menu item get join together and render to class attribute of  element in HTML
        }
    }
    return $items;
});

Answer: You can do it with a bit of regex and a custom menu walker although I've not tested this thoroughly:
class My_Walker_Nav_Menu extends Walker_Nav_Menu {
    function start_lvl(&$output, $depth) {
        $indent = str_repeat("\t", $depth);
        $output = preg_replace( "/(.*)(\<li.*?class\=\")([^\"]*)(\".*?)$/", "$1$2$3 has-submenu$4", $output );
        $output .= "\n$indent
\n";
    }
}

What this does is every time we start a new level within the menu we look at the last opening
 in the string and add 'has-submenu' to the class attribute. Unfortunately at the time of generating the list item itself we don't know if it has children or not.
To use the above just pass in the walker argument where you declare your menu eg:
wp_nav_menu( array( 'theme_location' => 'header', 'walker' => new My_Walker_Nav_Menu() ) );

Answer: This may not be the ideal solution, depending on what you need the class for, but you could add a "parent" class via JavaScript. The line below uses the jQuery :has() selector:
$('#nav-id').find('li:has(ul)').addClass('parent');

Add "parent" class to parent menu items


How can I use clustering algorithms to bin highly skewed data process?

Question: I have a large set of multi dimensional data.The data points are highly skewed and  not smoothly distributed.I want to divide the data set to some finite number of bins.I have approached this problem with some clustering algorithms(like KMeans and Mean shift).But as there are many clustering algorithms I am not sure which one would be most efficient. Again I can apply clustering to multi dimensional data only.But how should I approach when the data set consists of single dimensional data(like a large 1-D array) only?
Comment: It would be helpful to know what your goal is. Does your data actually have clustered structure and you want to identify it? Or do you want to do something more like vector quantization (i.e. just bin the data in a multidimensional way)? What was unsatisfying about the approaches you tried, and what does 'efficiency' mean in your application?
Answer: You should try other distance measures.
For example Canberra and Clark distances often work much better with skewed distributions. So they are worth a try.
Try loading your data into ELKI, and run OPTICSXi with Canberra distance. That has worked very well for me as a first attempt to cluster the data when exploring a new data set.


Can you mine smaller cryptocurrencies - e.g. coins found in an ICO etc?

Question: E.g. any of the coins found here:
https://coinmarketcap.com/
Comment: Many cryptocurrencies are mineable, some are not. On Coinmarketcap, any coin supplies marked with an asterisk are not mineable.
Answer: "Can you mine smaller cryptocurrencies?"
if they are mineable: obvious yes. (otherwise they can not be used to execute transactions so they would not have any benefit.)
but there are also some not mineable cryptocurrencies, for example ripple or cryptocurrencies which have 100% proof of stake.
"... e.g. coins found in an ICO etc?"
nearly all ICOs use tokens. a token is not a mineable coin. a token exist due to a smartcontract of another cryptocurrency. so the creator of the smartcontract can decide who gets how many tokens.
Answer: You can mine those: https://coinmarketcap.com/coins/views/filter-non-mineable/
But those coins are not minable cause they are Ethereum tokens or the people behind it don't want you to mine them (Ripple for example)


How can I tell that a certificate is a root certificate?

Question:
Are these certificates root certificates or not and how can I disable them if they are?
Comment: Welcome to Ask Different! :)  I hope you come to find this site has a lot to offer! In case you haven't already, it's worth taking the time to read the [tour](https://apple.stackexchange.com/tour). All the best!
Answer: You may want to refer to Lists of available trusted root certificates in iOS on Apple's website. This resource also provides links to lists of root certificates for specific versions of iOS, macOS, watchOS and tvOS.
Typically, to remove a root certificate on an iOS device one would follow this process:

Open Settings
Tap on General
Tap on Profiles (NOTE: if the Profiles screen is empty then there's nothing to delete)
Select the Profile you want to delete
Tap on Delete Profile
Enter your passcode
Tap on Delete again to confirm the deletion

Based on your screenshots and tags it's clear you're referring to a device that is controlled by a Mobile Device Management (MDM) process. If you can't remove the certificate(s) you want to with the above process you will need to contact your employer/school/etc to seek assistance.
Answer: The list of installed profiles in your screenshot does not contain any root certificates. There are two certificates at the bottom, but those are signing certificates, and also not root certificates.


How can I remove orphaned start menu entries?

Question: For some reason I have a bunch of left-over start menu entries from applications (e.g, Gnome Do, Synapse, Leafpad.) and was wondering if  there's a way to remove all orphaned entries for applications that are no longer installed?
Comment: Try https://github.com/hatkidchan/fbrokendesktop-fix-quotes
Answer: You can use the following command to list application icons that link to nonexistent programs:
for i in {/usr,~/.local}/share/applications/*.desktop; do which $(grep -Poh '(?<=Exec=).*?( |$)' $i) > /dev/null || echo $i; done

I suspect you'll find that most of yours are customized icons in your home folder, since these are not automatically cleaned up by the package manager. If this is the case and you'd like to trash  them all at once, you can use a modification of the previous command:
for i in ~/.local/share/applications/*.desktop; do which $(grep -Poh '(?<=Exec=).*?( |$)' $i) > /dev/null || trash $i; done

Or, of course, browse to ~/.local/share/applications/ in Nautilus and trash them via the graphical interface.
Comment: That did the trick! Is there anyway I can turn that into a shortcut that I can run when such a thing happens again?
Comment: the answer by intika ( https://askubuntu.com/a/1185127/148272 )  expands on this and scans subfolders too :)
Answer: If by start menu you mean the 'Applications' menu, you can edit it by going to:
System -> Preferences -> Main Menu.
You can add, rearrange, edit and delete entries at will.
Comment: Hi Eugenio. That's correct, but I have to go through each one, confirm if it really is orphaned or not, and then delete it. I'm looking for a way that would do all of this for me, as I have about 40+ entries that I want to remove.
Comment: Probably the most noob-friendly answer. +1
Answer: When an application is uninstalled it should remove menu items it installed. And it work this way on my machine.
If it works differently, it is a bug of a particular package or a problem on your system. I cannot believe that you have more than 40 entry orphaned. It would be the symptom of a problem.
Comment: It's very strange, I've never had this problem until my upgrade to Natty Narwhal. I may downgrade to 10.04LTS, and wait a short while until 11.04 is a little more bug-free.
Comment: There are programs that actively install themselves but which don't have an actual installer per se. They get orphaned if the binary is moved.
Answer: Alternative:
Following @ændrük answer, here is a similar solution that support spaces in path and subfolders, this can be used on a specific folder or the the entire disk to look for broken .desktop files (shortcut)
find {/usr,~/.local,/usr/local,}/share/applications/ -name '*.desktop' | while read -r line; do which $(grep -Poh '(?<=Exec=).*?( |$)' $line) > /dev/null || echo $line;  done

For the entire disk
find / -name '*.desktop' | while read -r line; do which $(grep -Poh '(?<=Exec=).*?( |$)' $line) > /dev/null || echo $line;  done

Note: that any method can show false positive check twice found files before deleting them
Answer: Well, this one is hard but i want, at least, to tell you what's the problem.
Gnome stores the installed applications in /usr/share/applications with files ended in .desktop (so vlc program is launchad by vlc.desktop)
Inside each one, there's a "Category" section in which you can set in what part of the Applications menu that program you want it to be. The thing is: the only way i think it can be done what you are asking for is using a bash script that ask if the package related to that .desktop is installed or not. if it's not, delete that .desktop file. When you delete the file, it will be out of the menu, and when you app is in a category, that one isn't shown.
That one is a bit out of my scope because every command (dpkg, apt-get, etc..) always return a good exit when they say the package isn't installed, so we have to read the exit from that app searching for a String (f.i. "not installed"), check it, and if the String exists, it means the application isn't installed and the .desktop can be deleted.
So, if you want, you could just enter the folder (/usr/share/applications) and click on files that correspond to the programs. They'll dissappear from the Applications menu. To do it automatically, i guess we need a good bash script on the track i said.
I hope this helps in some way.
Comment: Hi David, thanks for your help! That tip about going into /usr/share/applications is certainly a time saver compared to deleting the broken links one-by-one in the menu editor.
Answer: I found on the internet a very smart person who knew of a program called Alacarte (easy GNOME menu editing) program. It installed using the software manager. It's so easy to use. Just click on to highlight the uninstalled program entry you want to get rid of and click delete on the right and it's done.


Multiple views in illustrator

Question: I work on some detailed graphics.
I would like to have two windows of the same file open at once, one zoomed in and the other at 100%. This is easy.
But what I can't figure out is how to remove ALL "stuff" from the 100%
I don't want any guides, rulers, Bounding Boxes handles or such only my pixles.
This works in PS using hide extras on the 100% view. But in Illustrator hiding the bounding box for example hides it in both views.
Am I missing something here?
Comment: Just curious but are you using Illustrator's Navigator?
Comment: @Matt_2.0  i'm not sure what you mean "using"? Yes I have it open. But I usualy move around using combination of holding space and zooming using the wheel. Don't see what this has to do with anything?
Answer: You should be able to control the display of guides, edges, artboard, print tiling, and rulers per-view. Simply switch to the window you want to change and hide/show options. At least I can here with AI 16.0.4 (CS6)
The Bounding Box is special for some reason. The bounding box visibility is not window independent and will either show or hide for all windows. Truth be told, I hate the bounding box in general. It's not needed and was only added to make Corel users more comfortable when switching apps. The only real use for only the bounding box is to resize area text boxes easier.
Comment: Thanks for the fast response. I seem to have got rid of most of the overlays, except the 9-slice lines. Also don't I need the bounding box to scale symbols, or did I miss something?
Comment: You really don't *need* the bounding box for anything. If you need to scale via dragging just tap the e key to select the Free Transform Tool.

Multiple views in illustrator


Merging header content of all pages to the body in a word file

Question: I have generated some reports in pdf format using a sweave code in R. Those reports require slight editing. For that purpose, I converted those reports into Word files using Adobe Acrobat Reader. But after conversion, some portion of the text on the top gets included in header section. That poses some problems. So my question is, is there some way in Word through which I can remove headers from all the pages while merging the content in the header with the body or to not create headers while conversion of pdf to Word file?
Answer:
Some pdf readers give option to suppress repeated headers on each page while converting to doc. PDF nitro reader had such option in earlier versions but that feature has now been made paid in latest version. or check whether some online pdf to word convertor site is providing that option.
You can use ocr also, some ocr-s also give option to repeated headers,like Tessaract which is free. or find if any online ocr site is giving that option.

These will not merge headers to body text, these will rather put the header/ footer only once. If you want to merge header/ footer to text, it is as good as select text extraction tool, copy entire text from pdf and put that to a word document.


Doubt about open sets in the Product topology

Question: So we know that in the product topology a basis element is gonna be of the form $\prod U_\alpha$ where  $ U_\alpha \neq X_\alpha $ for a finite number of $\alpha$. So my thing is that we know that a product of closed sets in the product topology is gonna be closed so why is not true that if i do $ \prod X_\alpha - \prod F_\alpha = \prod X_\alpha-F_\alpha$ im gonna get a contradiction. I know this isnt true i just cant seem to grasp it, Thanks in advance.
Comment: We don't have $(A\setminus B)\times (C\setminus D) =(A\times C) \setminus (B\times D)$.
Comment: Yeah thats what i tought , do you know a counterexample ?
Comment: Any proper subsets yield a counterexample.
Answer: Here's a graphical demonstration:

Here on the left you see $X_1$ (black), $F_1$ (red) and $X_1\setminus F_1$ (green). Note that I moved the latter two horizontally, so you can see all three, but actually they are all at the same line. Similarly, you see $X_2$, $F_2$ and $X_2\setminus F_2$ on the bottom.
In the area you see the products. The full rectangle is $X_1\times X_2$. The red rectangle in the middle id $F_1\times F_2$ Therefore $(X_1\times X_2)\setminus(F_1\times F_2)$ is everything of the big rectangle around the red rectangle.
On the other hand, $(X_1\setminus F_1)\times(X_2\setminus F_2)$ consists only of the four green rectangles.
Answer: Because $\prod_\alpha X_\alpha\setminus\prod_\alpha F_\alpha\neq\prod_\alpha(X_\alpha\setminus F_\alpha)$. In fact, $\prod_\alpha X_\alpha\setminus\prod_\alpha F_\alpha$ is the union of all the products $\prod_\alpha Y_\alpha$ where $Y_\alpha=X_\alpha\setminus F_\alpha$ for one specific $\alpha$, whereas for all other $\alpha$'s you have $Y_\alpha=X_\alpha$.
Comment: Oh yeah thats true , Thanks.
Answer: If $x \notin \prod_{\alpha \in J} F_\alpha$ this means that $x(\beta) \notin F_\beta$ for some $\beta \in J$, and then the basic open set $\prod_{\alpha \in J} O_\alpha$ with $O_\beta=X-F_\beta$ and all other $O_\alpha = X_\alpha$ is a neighbourhood of $x$ that misses $\prod_{\alpha \in J} F_\alpha$ entirely.


How is clock gating physically achieved inside an FPGA or ASIC?

Question: It is bad idea to add logic gates in clock signal path. How is clock gating achieve in FPGA and ASIC designs and how does it prevent glitch in the output signal i.e the gated clock as it is enabled or disabled?
Comment: This is a few lines that asks for mountains of effort in reply. The site is not for free personal tutoring. This information is readily available to be researched on the Internet. Please edit your question and detail your situation and what you already know, show all that you discovered for yourself on the subject.
Comment: Look, all you have to do is read the user manuals then. Put the effort in yourself instead of demanding personal reports from the community here. Unsurprisingly, this is being voted to closed.
Comment: In FPGA it usually isn't, you usually design with an enable input to a continuously clocked block. ASIC is a different matter.
Comment: How the clock enable is implemented in a particular FPGA or ASIC is probably considered company IP (intellectual property), and so details are not readily available.  Though I bet if you just do some web searching you can find design ideas.
Comment: Look all I want to know, for specific FPGA e.g Intel Cyclone 10, or Xilinx Spartan 6, how is clock gating achieved physically using the hard IP that exists inside it.
Answer: Is it a bad idea to gate clocks? It depends.
In the ASIC there's well-understood timing for clock paths, so it's reasonable to instance a standard cell on the clock tree to gate a sub-region's clock. On ASIC then, not only is clock gating not 'a bad idea', it's widely used as a means to save power.
Not so much with the FPGA. In fact, it's never a good idea to create gated clocks directly out of FPGA fabric logic; the synthesis tools will warn you about it if not outright forbid it. Why? The resulting inserted skew becomes impossible to manage at higher frequencies, even if the gated clock doesn't glitch (which it will without careful design.)
This brings up a common issue: modeling ASIC clock gating on FPGAs. It isn't really feasible to just define the clock gate in HDL and hope for the best. It needs special handing.
You can model ASIC-like gated-clock behavior in your FPGA using clock-enable flops for your synchronous blocks. This can be dealt with as a synthesis option in your flow, which will identify the gated clock domain and convert its flops to FDCEs. Vivado example: https://support.xilinx.com/s/article/982650?language=en_US
Some FPGAs do support clock gating, using dedicated clock gate resources with predictable timing and glitch-free behavior. More here from Xilinx (look for BUFGCE): https://www.xilinx.com/support/documentation/user_guides/ug572-ultrascale-clocking.pdf ; other FPGAs will be similar.
Note that with BUFGCE, you're still obliged to meet clock setup/hold from clock enable to clock rising edge. Still, that's an easier constraint to meet than making an asynchronous clock gate out of fabric logic. If you're modeling an ASIC you have to account for the difference between BUFGCE and whatever standard cell you're using in the ASIC.
Finally, you asked how clock gating is actually done. Tl, dr: the enable is latched to prevent disturbing the clock pulse with a glitch. More here: https://anysilicon.com/the-ultimate-guide-to-clock-gating/
Answer:
It is bad idea to add logic gates in clock signal path.

Do not add logic gates to your signal path. Strategically placed clock-gating and clock-divider are sometimes exceptions, but should be used minimally if ever. The exceptions are more likely to be acceptable for large ASICs where power is a concern. FPGAs and small ASICs generally discourage any logic added to the clock.

How is clock gating achieve in FPGA and ASIC designs and how does it prevent glitch in the output signal i.e the gated clock as it is enabled or disabled?

On ASICs clock gating is achieved with a negedge flip-flop and an AND gate (low when disabled) or posedge flip-flop and an OR gate (high when disabled). Glitches are avoided because the toggle to the enable/disable happens when there is no impact on the output. If you find an FPGA that supports clock gating, then it likely does it the same way.
If you are to use clock gating, it needs to be done to large modules; not individual flip-flops. Adding any logic to a clock creates additional timing  offset between gated-clock circuits those using the clock directly. If your synthesis tool does not have great clock gating support, then there is a chance it will mess up your clock tree and will require async FIFOs between modules of different gated-clocks even if the are derived from the same clock.
Comment: *Do not add logic gates to your signal path* You mean 'Do not add logic gates to your clock path'.
Comment: FPGAs that support clock gating are far from rare these days, they do it using clocking buffers and the synthesis tools recognize the same HDL constructs that you describe for ASIC gating. Clocking buffers are basically the same gates you describe optimized for glitchless transitions between having the clock enabled and disabled.
Comment: Last time I worked on this it was a half-latch rather than a whole flipflop, but the principle of latching the enable during the active half of the clock is the same.
Comment: Agree with pjc50, using TSMC standard cells (28nm, 16nm, 7nm) I've only seen clock gates implemented as latch + AND (I think) gate.
Answer: It is bad idea to add logic gates in clock signal path.
In an FPGA, not only is it a horrible idea, the design software should not let you do it.
how does it prevent glitch in the output signal i.e the gated clock as it is enabled or disabled?
It doesn't. What you do instead of gating the clock is to enable the logic which the clock drives. Don't worry about gating the input logic to a flip-flop, either. Just use a flip-flop with an enable input and drive that.
Comment: Will using flip flop with clock enable, or an enable input actually result in power saving?


Safari crash and refusal/failure to reload

Question: Recently, Safari unexpected crashed while I was using my desktop.  Now, when I try to reload, it keeps saying it "unexpectedly quit" and will not reload (even when I push the reload button).  I have restarted my computer multiple times but it will not open Safari and I keep getting the same error messages over and over.  Any help would be appreciated.  Thanks
Answer: 1st thing to try is go to ~/Library/Safari/ & move the contents of the Extensions folder somewhere else.
Try again.
If successful, quit, add back one at a time, restarting Safari between each to eliminate which one was responsible.
Alternatively, go to ~/Library/Saved Application State/ & delete the entire folder called com.apple.Safari.savedState


SubSonic "Save" improvement suggestion

Question: [We are using ActiveRecord.]
While running Sql Server Profiler, we noticed that a simple "Save" was preceded by a good bit of database activity. We found that the SubSonic core runs through all the properties and saves their values in a dictionary before it actually does the save.
We typically "extend" our data objects in partial classes that include some properties that require database activity.  This is where the extra activity is happening, and, as far as we can tell, unnecessarily.
We changed where the script creates a property from each database column so that it now decorates these properties with [LocalData].  We then changed the core (Extensions.Objects.ToDictionary) to skip the GetValue if this attribute is missing.
In one test with a lot of save-type database activity, we saw the elapsed time drop from 21 seconds to 2, and the database activity drop by about two thirds.
There are two questions in all this is:  1) was there a feature we missed that would make our change unnecessary, and 2) if not, do you know of some other ToDictionary-related processes that will now break?
Thanks!
Comment: Properties requiring database activity - this may be better implemented as a method (and a read-only property not requiring database activity). Usually it's not considered good practice to have "expensive" properties.
Answer: Would the [SubSonicIgnore] attribute have achieved the same thing?

SubSonic "Save" improvement suggestion


what does % of parent mean in a site collection storage metric report

Question: So I'm looking at a storage metric report for a site collection and I see one document library that is using up 99.98% of the parent...what does that mean and would this have any affect on saving documents if the % of site quota only lists this library as taking up 3.11%?
Answer: I would suggest checking out SharePoint Designer, in association to your site. Under the "Site Information" panel, it will actually tell you specifically, the "Total Storage Used" and the "% of Available Storage Used". See, below.

Similarly, you could use powershell, to get the actual numbers. In terms of the % of Parent, it's the % of the parent's directory used. However, all results I tend to see show "100%' of their parent, as items have a parent-child relationship. As mentioned, your best result would be utilizing the actual storage metrics.
Answer: The "% of Parent" column indicates the amount of storage that an item takes up relative to the rest of the items in that scope (i.e. that item's parent). In your case, 99.98% of the used space in that view is being taken up by that document library. Drilling into the document folder will show the breakdown of usage scoped to that folder. The "% of Parent" column will show which files or folders are using the most space proportionally to each other only relative to that folder.
You can continue to save documents until the total usage of the site reaches the site collection's quota. Looking at which folders and items are using the most space (via the "% of Site Quota" column) can help you find and remove large documents and folders if you ever hit the site storage quota. The "% of parent" column can then help you choose which specific files to remove.
Comment: Thank you that was the explanation I needed to see, it confirms my assumptions.


Desktop sharing from Ubuntu 18.04 to Mac

Question: I've been searching the web sporadically for a month or two trying to figure out how to remote share my Ubuntu desktop to my Mac laptop. I've been able to connect over VNC and over RDP but neither one works enough to be even serviceable. I either get a blank window, or maybe just one window. I've run through a dozen articles on various services and whether to use SSH or not etc.
So bottom line, what's a dead simple way to share my Ubuntu 18.04 desktop to Mac? The simplest way I found so far is to use "Desktop Sharing" but it doesn't work on 18.04 as attested by the half dozen reviews of Desktop Sharing in the Ubuntu Software App.
Answer: What about TeamViewer? It's completely free for personal use, and you can remote share via Linux, Windows, MacOS, Chrome OS, mobile, etc.


Is there a way to scale all polygons up or down?

Question: I am looking for a way to take a layer and effectively scale and translate all the vector polygons.  Specifically, I am trying to make a miniature map of North American States and Provinces that will geographically overlay a small region, such as a town. For example, I'd like to size the polygons to fit over New York City.  Then I could have an app that let's me know what "State/Province" I am in as I drive around NYC using GPS.  To do the scaling/translating, it would be helpful to use a control, such as the mouse wheel, to scale the layer and drag it until it visually fits the region I'm interested in, but any method would be helpful.  I can probably do it blindly in Java by operating on all the vertices and changing the coordinates, but it would involve a lot of trial and error.  I am wondering if there is a tool in QGIS3 that would help.
Comment: "Operating on all the vertices and changing the coordinates" pretty much describes the needed process, exactly. What have you tried? Where are you stuck?
Answer: In edit mode, the "Move Feature" and "Scale Feature" seem to be exactly what I'm looking for.


Is OpenGL coordinate system left-handed or right-handed?

Question: I am trying to understand the OpenGL coordinate system. However, some tutorials say the default coordinate system is left handed (see http://www.c-sharpcorner.com/UploadFile/jeradus/OpenGLBasics11172005014307AM/OpenGLBasics.aspx) and others say it is right handed (see http://www.falloutsoftware.com/tutorials/gl/gl0.htm). Which is correct? I understand that we can transform one to the other by mirroring but I would like to know the default coordinates.
Comment: Doesn't this depend entirely on how you write your transforms in the shaders and is therefore entirely up to you?
Comment: Related http://programmers.stackexchange.com/a/88776/12693
Comment: Just my two cents http://www.evl.uic.edu/ralph/508S98/coordinates.html, it has  a few self explanatory images.
Comment: I don't suppose you'd consider updating your accepted answer?
Answer: There is some confusion here.

OpenGL is right handed in object space and world space.
But in window space (aka screen space) we are suddenly left handed.
How did this happen?
The way we get from right-handed to left-handed is a negative z scaling entry in the glOrtho or glFrustum projection matrices.  Scaling z by -1 (while leaving x and y as they were) has the effect of changing the handedness of the coordinate system.
For glFrustum,

far and near are supposed to be positive, with far > near.  Say far=1000 and near=1. Then C= -( 1001 ) / ( 999 ) = -1.002.
See here for more details and diagrams.
From an orthographic perspective, glOrtho generates a matrix like this:

Here, left, right, bottom and top are just the coordinates for left vertical, right vertical, bottom horizontal, top horizontal clipping planes (resp).
The near and far planes, however, are specified differently.  The near parameter is defined as

Near: The distance to the nearer depth clipping plane.  This distance is negative if the plane is to be behind the viewer.

and far:

zFar  The distance to the farther depth clipping plane. This distance is negative if the plane is to be behind the viewer.

Here we have a typical canonical view volume

Because the z multiplier is (-2/(far-near)), the minus sign effectively scales z by -1.  This means that "z" is turned left handed during the viewing transformation, unbeknownst to most people as they simply work in OpenGL as a "right handed" coordinate system.
So, if you call
glOrthof(-1, 1, -1, 1, 10, -10) ; // near=10, FAR=-10,

Then the NEAR PLANE is 10 units ahead of you.  Where are you?  Why, at the origin, with the x-axis to your right, the y-axis on top of your head, and your nose pointing down the negative z-axis (that's the default "By default, the camera is situated at the origin, points down the negative z-axis, and has an up-vector of (0, 1, 0).").  So the near plane is at z=-10.  The far plane is 10 units behind you, at z=+10.
Comment: +1 for the whole answer, -1 (wait, rather -2) for *"The way we get from right-handed to left-handed is because the forward vector is negated in the standard implementation of gluLookAt(), which is the function everybody uses to create the view matrix. When you negate the forward vector, this has the effect of changing the handedness of the coordinate system."*, which is mere rubbish...
Comment: ...Negating the forward vector in the `gluLookAt` computation has nothing to do with changing the handedness of the coordinate space, that's just how this matrix is computed (and in fact +z is indeed *"backward"* in right-handed space). `gluLookAt` does nothing else than compute an ordinary rigid body transformation (rotation+translation) in right-handed space. It is the projection matrices used by the fixed function pipeline (and usually by shaders, too) that perform the actual handedness change in their negation of the z-component, as you already noted in your answer.
Comment: _We're suddenly left-handed_ from clip space onwards not just window space.
Comment: @bobobobo Your saying if you compute your view matrix using rotation and translation functions only (which I guess you wouldn't accuse of changing handedness), instead of `gluLookAt`, there wouldn't be a change of basis while `gluLookAt` does one? Definitely not (since you probably know that not *"everybody"* uses `gluLookAt` for view matrix computation), since as you should know `gluLookAt` does nothing else than a bunch of rotation and translation calls (be they even disguised as the *"high level"* term *"change of basis"*) with absolutely no reflections involved.
Comment: "the forward vector is negated in gluLookAt" - can you confirm that this is that the case? Not gluPerspective or glFrustum or glOrtho?
Comment: *"Your discussion in the previous 2 comments appears high level and I don't think you really understand what you're talking about."* - Sorry to counter that one with a classic *"no, you might not be aware of what you're talking about"* (sorry, but *"high level"*, what the...?). Of course the 3x3 part is a change of basis, as every ordinary rotation is, this doesn't neccessarily imply a change of handedness, and in the case of the standard rigid body transformation that `gluLookAt` creates it certainly doesn't.
Comment: Yes, in the last few lines, `matrix2[2] = -forward[0]; .. ` flips the direction of the forward vector.
Comment: @ChristianRau The upper 3x3 of the lookAt matrix is a change of basis, or just 3 vectors. It consists of a new forward, new right, and new up vector. In the code linked, the forward components are computed as  `forward[0] = center3D[0] - eyePosition3D[0];`, but in the final assignment into the matrix they are suddenly inverted `matrix2[2] = -forward[0];`. Your discussion in the previous 2 comments appears high level and I don't think you really understand what you're talking about. That said, [still haven't resolved w/perspective projection](http://programmers.stackexchange.com/a/88776/13572).
Comment: Hmm, it appears that the reason the forward vector is negated is because OpenGL's default forward vector is (0,0,-1). For _that_ reason it may be true that the lookAt matrix _does not_ change the handedness of the coordinate system.
Comment: What is the convention on using your fingers to represent axis? I see different illustrations assigning different axis to different fingers. I use the Ogl/left hand analogy to align my hand with the NDC (x to the right, y to the top, z into the scene/screen). Index is X, thumb is Y, middle is Z?
Answer: By default the Normalized Device Coordinate is left-handed.
The glDepthRange is by default [0, 1] (near, far) making the +z axis point into the screen and with +x to the right and +y up it is a left-handed system.
Changing the depth range to [1, 0] will make the system right-handed.
Quoting a previous answer from Nicol: (the strike-through is my work, explained below)

I'm surprised nobody mentioned something: OpenGL works in a left-handed coordinate system too. At least, it does when you're working with shaders and use the default depth range.
Once you throw out the fixed-function pipeline, you deal directly with "clip-space". The OpenGL Specification defines clip-space as a 4D homogeneous coordinate system. When you follow the transforms through normalized device coordinates, and down to window space, you find this.
Window space is in the space of a window's pixels. The origin is in the lower-left corner, with +Y going up and +X going right. That sounds very much like a right-handed coordinate system. But what about Z?
The default depth range (glDepthRange) sets the near Z value to 0 and the far Z value to one. So the +Z is going away from the viewer.
That's a left-handed coordinate system. Yes, you can change the depth test from GL_LESS to GL_GREATER and change the glDepthRange from [0, 1] to [1, 0]. But the default state of OpenGL is to work in a left-handed coordinate system. And none of the transforms necessary to get to window space from clip-space negate the Z. So clip-space, the output of the vertex (or geometry) shader is a left-handed space (kinda. It's a 4D homogeneous space, so it's hard to pin down the handedness).
In the fixed-function pipeline, the standard projection matrices (produced by glOrtho, glFrustum and the like) all transform from a right-handed space to a left-handed one. They flip the meaning of Z; just check the matrices they generate. In eye space, +Z moves towards the viewer; in post-projection space, it moves away.
I suspect Microsoft (and GLide) simply didn't bother to perform the negation in their projection matrices.

I did strike one part since it diverged from my findings.
Either changing the DepthRange or the DepthFunc and using the ClearDepth(0) works but when using both they cancelled out each other back to a left-handed system.
Comment: @SigTerm why do you think correcting a false statement is not worth it?

I have spent several hours troubleshooting vertex coordinates and matrix rotation generation just to find out that the default coordinate system in fact was not right-handed.
Comment: The question was whether it was a left- or right-handed system.

This answer has more details than that, explaining two other ways to get it right-handed, adding a -1 in a matrix transformation is yet another way of doing it.

But none of that matters if you believe the system is right-handed by default, you can only change something if you know what you are changing from.

Still
Comment: I found the answer very enlightening, and it came up in searches so it was worth it for me.
Comment: The depth range can't be [1, -1]. I think you meant [1, 0]. Also, [this has been pointed out before](http://programmers.stackexchange.com/questions/17519/why-does-directx-use-a-left-handed-coordinate-system/88776#88776).
Comment: Your answer is great, too bad it is to a DirectX question, you should repost it here!
Comment: I don't think this was worth resurrecting 2 years old question.
Comment: @phq: "why do you think" Because it is a minor implementation detail that can be easily fixed by adding a "-" sign.
Comment: As I understand that, NDC is always left-handed, while window coordinates can be left-handed (default) or right-handed depending on depth range.
Comment: @Kos not only the DepthRange but it and the DepthFunc together, note the last paragraph about them cancel each other out.
Answer: ONLY NDC
You should only notice that OpenGL only knows NDC! and that is a left-handed coordinate system.
No matter what coordinate system you use -- left-handed or right-handed axis-coordinate system -- all need to be mirrored to NDC. If you like, you can totally handle world-space as left-handed coordinate system.
Why do we usually use right-handed coordinate system in world-space?
I think it`s conventional. It just does. Maybe it just  want to distinguish from DirectX.
Comment: OpenGL predates DirectX by a few years, so OpenGL were certainly not trying to distinguish themselves from DirectX. The lead developer of DirectX, Alex St. John, has said that Microsoft's decision to go for a left-handed co-ordinate system was "in part out of personal preference" and "an arbitrary choice."
Comment: It might be nicer if repeating it to include the words "normalised device coordinates" though.
Comment: I think this doesn't get repeated enough. All these diagrams showing "modeling transform", "view transform", "perspective transform" fail to show they're not an intrinsic part of OpenGL. You could even do all your transforms manually, with or without shaders, with or without matrices.
Comment: This is not entirely correct. OpenGL also knows the clipping coordinate system, since clipping happens in that. The clip coordinate system has the same handedness as NDC coordinate system tho.
Answer: The book "WebGl Programming Guide" by Kouichi Matsuda spends almost ten pages on "WebGl/OpenGl: Left or Right Handed?"
According to the book:

In practice, most people use a right-handed system
OpenGl actually is a left-handed system internally
Internally, more deeply it's actually neither. At the very bottom OpenGl doesn't care about the z-value. The order in which you draw things determines what is drawn on top (draw a triangle first, then a quad, the quad overrides the triangle).

I don't fully agree with the "it's neither" but that's probably a philosophical question anyway.
Comment: `The order in which you draw things determines what is drawn on top (draw a triangle first, then a quad, the quad overrides the triangle).`

This is true, in the absence of depth testing. In the presence of depth testing, the z value of a fragment is compared with the value in the depth buffer, and the fragment discarded if it fails the depth test, so the quad may or may not overwrite the triangle depending on their depths and the depth function used.
Answer: Opengl is definitely left-handed. You see a lot of tutorials stating the opposite because they are negating the z-value in projection matrix. When the final vertices are computed inside vertex shader, it's converting the vertices that you pass from client-side (right-hand coord) to left-handed, and the vertices will then be passed to geometry shader and fragment shader. If you use right-hand coordinate system in client-side, Opengl doesn't care. It only knows normalized coordinate system, which is left handed.
Edit: If you don't trust me, just experiment in your vertex shader by adding a translation matrix, and you can easily see if Opengl is left-handed or not.
Comment: I think you're right about people thinking OpenGL is right-handed because of tutorials that negate the z in the projection matrix, however I don't think it's inherently left-handed either.
There is nothing in the OpenGL spec that indicates either way. The only justification I can find for saying it's left-handed is that the default depth testing function is `GL_LESS` which would typically mean that objects with a lower z value are seen to be closer - but depth testing itself is disabled by default anyway so it really comes down to how your application uses the GL.
Answer: By using OpenGL's built in projection and transformation functions, observing the movements on screen follow the rules of the right-handed coordinate system. For example, if an object in front of your view is translated in the positive z direction, then the object will move towards you.
The depth buffer is quite the opposite, and this is where the NDC (Normalized Device Coordinates) come into play. If passing GL_LESS into the glDepthFunc means that pixels will draw when they are nearer to you than what's already in the depth buffer, then pixels are considered to live in a left-handed coordinate system.
There's one more coordinate system, and that's the viewport! The viewport's coordinate system is such that +x points to the right, and +y points down. I think by this point the handedness is moot since we're only dealing with x, y at this point.
Lastly, gluLookAt under the hood has to negate the look-at vector. Since the math assumes a vector is pointing in a positive direction towards the object it's looking at, and a camera looks down -z, the look-at vector must be negated so that it aligns with the camera.
Something to chew on. It doesn't make much sense to call the z direction of a right handed coordinate system a forward vector :). I think Microsoft realized this when they designed Direct3D.


Who is winning is this game of chess?

Question:
What do you think are the chances for the black to win?
Answer: It's game over for Black. Down by material, mating attack, zero counterplay etc. The Black's rook is under attacked.

1...Rd6 2.e5 Qxe5 3.Qg4+ Qf5 4.Qxh5 winning easily

1-0 unless White makes a horrible mistake.
Answer: Black is down in material by a couple of pawns and the black King is very weak.
Unless there are any immediate tactics (doesn't look like there are, but haven't checked with an engine), white should win this.

Who is winning is this game of chess?


Mechanical waves with slower speed than sound

Question: Sound is mechanical waves of high and low air pressure transmitting with 300m/s. Are there high and low air pressure waves transmitting with lower speeds? How are they produced?
Comment: See here http://physics.stackexchange.com/questions/193246/how-can-shock-waves-travel-faster-than-sound#_=_
Answer: The speed of mechanical sound waves through the air at 0 degrees C is 331 m/sec.  But sound can travel at many different speeds, depending on the medium it propagates through and the temperature and pressure, among other variables.
What we call sound is any mechanical wave within the range of human perception that is transmitted to our eardrums via the air.  Here is a table of speeds of sound waves through several different materials: http://hyperphysics.phy-astr.gsu.edu/hbase/tables/soundv.html.  You'll notice that in liquids and solids, the speed is faster than in air.  Generally, a lattice of molecules transmits mechanical waves faster than gas because they impact each other more quickly than a loose configuration.
Generally, the speed of sound in any medium may be modeled as:
Velocity = √ (elastic property / inertial property)
Elasticity is the tendency of a material to snap back to its original shape after deformation.  Inertia is the difficulty of moving a material.  As temperature rises, velocity of sound waves will also rise, as the air molecules become more energetic.  If the air is humid, sound travels faster because water vapor has a lower molecular mass than dry air, and therefore the average molecule has less inertia to overcome.  Here is a calculator which allows you to compute the speed of sound in air under various temperatures: http://hyperphysics.phy-astr.gsu.edu/hbase/sound/souspe3.html#c1
As you can see from the calculator, you can significantly reduce the speed of sound by lowering the temperature of the air.


What is exactly Max Born rule?

Question: I have thought Max Born rule as one of the axiom of quantum mechanics that says norm square of wavefunction gives the probability density. But I also found written somewhere that the rule says that probability of eigenstate is norm square of the amplitude of the eigenvalue corresponding to the eigenstate. First I thought maybe they are same thing but I am unable to find connection between them (former one takes the square norm of wavefunction whereas latter one do the same for amplitude of eigenvalue). Which one is actually Max Born rule? And also I want to know how we know the other one is true.
Comment: isn't $\psi(x)$ the amplitude for the eigenstate of the $\hat x$ operator?, which means each statement is equivalent.
Comment: "But I also found written somewhere" - where? It's difficult to answer this question without context.
Comment: The second sentence in the wikipedia article that you link to tells it like it is:

In its simplest form, it the Born rule] states that the probability density of finding a particle at a given point is proportional to the square of the magnitude of the particle's wavefunction at that point.
Comment: @probably_someone here is the link http://www.preposterousuniverse.com/blog/2014/07/24/why-probability-in-quantum-mechanics-is-given-by-the-wave-function-squared/
Answer: They are the same thing. Be careful, I think the other book you're reading by 'amplitude of eigenvalue' means probability amplitude of measuring that eigenvalue, not the amplitude of the eigenvalue itself (otherwise really large values of an observable would have really high probabilities which makes no sense).
Comment: actually I read on this site. See the fourth point. http://www.preposterousuniverse.com/blog/2014/07/24/why-probability-in-quantum-mechanics-is-given-by-the-wave-function-squared/
Answer: It is the same thing. The way I approach it is to start with a Hilbert space spanned by position states (actually I start a bit further back, by justifying Hilbert space, but the present argument starts with Hilbert space).
For any ket $|f\rangle \in \mathbb H$ we can define the magnitudes of the coefficients $\langle|f\rangle$ in position space according to the Born rule,
$$ {|\langle x|f\rangle|^2 \over \langle f|f\rangle}=P(x|f)$$
If $|f\rangle$ is normalised this reduces to
$$ |\langle x|f\rangle|^2 =P(x|f)$$
and $ \langle x|f\rangle$ is the probability amplitude.
When we do a measurement, $K$, we get a definite result, a terminating decimal
or $n$-tuple of terminating decimals read off the measurement apparatus. Let
the possible results be $k_i$ in $\mathbb Q^n$ for $i=1, \dots, m$. The $k_i$ are taken to be distinct; if $i \neq j $ then $k_i \neq k_j $.
We assume that the dimension $\mathbb H$ of is greater than $m$.
Each physical
state is associated with a ket, labelled by the measurement result, so that if the
measured result is $k_i$ then the state is $|k_i\rangle$. The empirical determination of
$|k_i\rangle$ requires that we draw from experimental data the value of
the inner product $\langle k_i|f\rangle$ for arbitrary $|f\rangle$.
Without loss of generality, $|k_i\rangle$ and $|f\rangle$ are normalised. By assumption,
measurement of $K$ is reducible to a set of measurements of position, so that
each $k_i$ is in one to one correspondence with the positions $y_i$ of one or more
particles used for the measurement (e.g. $y_i$  may be the positions of one or
more pointers). Then
$$|\langle k_i|f\rangle |^2= |\langle y_i|f\rangle |^2=P(y_i|f) = P(k_i|f) $$
is the probability that a measurement of $K$ has result $k_i$, given the initial ket
$|f\rangle$. It follows from $\langle x|y\rangle = \delta_{xy}$ that
$$\langle k_i|k_j\rangle = \langle y_i|y_j\rangle = \delta_{ij}$$
meaning that if the result is $k_i$ it is definitely $k_i$ and cannot at the same time be $k_j$with $i \neq j$.
To relate this to the eigenstates of a Hermitian observable we note that measurement with result, $k_i$, implies a physical action on a system and is represented
by the action of an operator, $K_i$, on Hilbert space. If a quantity is
measurable we require that there is an element of physical reality associated with
its measurement, meaning that the configuration of matter necessarily becomes
such that the quantity has a well-defined value. In practice this means that, in the
limit in which the time between two measurements goes to zero, a second measurement
of the quantity necessarily gives the same result as the first. It follows
that $K_i$ is a projection operator
$$ K_i=|k_i\rangle\langle k_i| $$
This is the projection postulate. The expectation of the result from a measurement of $K$, given the initial normalised ket, $|f\rangle$, is
$$ \langle K \rangle = \sum\limits_i k_i P(k_i|f)=\sum\limits_i \langle f|k_i\rangle k_i \langle k_i|f\rangle = \langle f |K|f\rangle$$
where Hermitian operator,
$$K= \sum\limits_i |k_i\rangle k_i \langle k_i|$$
is an observable with eigenvalues $k_i$.
(this has been extracted from my book The Mathematics of Gravity and Quanta and my published paper The Hilbert space of conditional clauses)

What is exactly Max Born rule?


Creating an Application Launcher for GNOME 3 in Ubuntu

Question: I have Ubuntu 12.04 LTS installed on my machine. I also have installed GNOME 3. I then installed the IDE IntelliJ without issue. However I installed it to a custom location and now the only way to run it is through the terminal via "./idea.sh". I would like to create an application launcher so I can launch IntelliJ via the applications list in the GNOME 3 windows manager. How would I do such a thing via the terminal, not through a 3rd part utility. I would also prefer to have the correct icon for the launcher, not a generic one. Thank you for the help.
Answer: Run Intellij, then go to Tools > Create Desktop Entry.
Comment: It's perhaps the right answer for IntelliJ, however, and considering the title of the question, the chosen answer *is* the best as it is generic and it works.
Answer: create a file called intellij.desktop in the directory /usr/share/applications/
my file looks like this
[Desktop Entry]
Name=IntelliJ IDEA Community Edition
Comment=Free Java, Groovy, Scala and Android applications development
Exec=/path/to/your/bin/idea.sh
Path=/path/to/your/bin
Terminal=false
Icon=intellij-idea-ce
Type=Application
Categories=Development;IDE

for more details check this website, http://standards.freedesktop.org/desktop-entry-spec/desktop-entry-spec-latest.html
edit: mlissner (Thank you!) mentioned that intelliJ in particular implemented a feature to add that Icon/Launcher automatically. Just open IntelliJ IDE, go to Tools and click Create Desktop Entry...
Answer: On Ubuntu 20 (Focal Fossa), create a launcher with alacarte.
The generated .desktop file is automatically placed in ~/.local/share/applications.
Move it from there to /usr/share/applications/ if it is to be accessible to all the computer's users.
Answer: You can create launcher using following steps
step 1) Install gnome-panel
    sudo apt-get install --no-install-recommends gnome-panel
step 2) run gnome-desktop-item-edit
    gnome-desktop-item-edit ~/.local/share/applications --create-new
step 3) fill the required values in widget