What is this web on the surface of the Sun?

Question: I was going through my Social Media Feed and found the attached post too frequent. The caption reads this is the best image of our Sun. Just as an example, the Universe Today's This is the Highest Resolution Image Ever Taken of the Surface of the Sun
Why? What exactly are the black lines that appear to be like kind of a web, and will such patterns be observed if in case the star was not the Sun but some other star? Are they thought to be common?
Comment: It's the SWW (Sun-Wide-Web) ;-)
Comment: With a little luck you can observe similar convection columns (of which here the upper side is visible) in a cup of coffee with milk. Pour the milk carefully, don't stir.  I sometimes could get them going by *very gently* blowing over the surface, to cool it.
Comment: @Michael no, the Tholians have attacked, and encased the sun in their web.
Comment: I've made some edits to make your post fit the site's style a bit better, feel free to edit further or roll back. Also read [Can we  to have emoji in posts?](https://astronomy.meta.stackexchange.com/q/515/7982)
Comment: Now I'm curious what the occasional bright patches between cells are. E.g. lower part of the image, just to the right of centre.
Comment: related: [What do the colors in this "web on the surface of the Sun" image mean?](https://astronomy.stackexchange.com/q/34915/7982)
Comment: That's a still photo.  Search for the video version.  E.g., https://www.youtube.com/watch?v=znBesUwVOok (about 20 seconds from the start.)
Comment: @CJDennis Exactly what you'd expect - even warmer plasma, coming up in a much tighter convection column ("tighter" being relative, of course - it's still the area of a reasonably small country; the smallest bright "dots" are still the size of a city).
Comment: "Spins a web, any size", indeed.
Answer: The dark lines are colder areas at the edge of the convection cells, where the cooled down plasma sinks towards the inside of the Sun. Now "colder" for the surface of the Sun, is still pretty hot, as explained here.
The yellow parts are where the plasma rises to the surface.
Each yellow spot (which is actually the size of a country) is called a granule, and this web-like appearance is called granulation.
In the outer part of the Sun (the convection zone in the image below), there is convection, that is hotter plasma floats towards the top, cools down at the surface, and sinks back down, like in a lavalamp.

The existence of a convective zone in the outer part of the star is determined by the mass of the star, and all stars with a convective zone in their upper layer are thought to have such granulation patterns. So stars like our sun, or smaller have these patterns.
For larger stars, though, the convective zone is in the inner part of the star, and the outer part of the star is the radiative zone, so there might not be the same patterns on the surface.
Comment: I've just asked [What do the colors in this "web on the surface of the Sun" image mean?](https://astronomy.stackexchange.com/q/34915/7982)
Comment: @Luaan Ah - so it's not like "Hey boys, found our landing site!" cool, it's "melts your face off a few microseconds later" cool.
Comment: Phil Plait has written an excellent article on this image here https://www.syfy.com/syfywire/the-sun-up-close-and-very-personal-first-light-for-the-daniel-k-inouye-solar-telescope
Comment: Exactly how cold is this "cooled down" plasma?
Comment: @corsiKa Luminosity depends on the fourth power of temperature, so the anwer is probably "not really all that much cooler". It does produce a rather dramatically visible effect. Note that the sunspots, which appear almost black, are only about 1000-2000 K colder than the ~5700 K of the bulk of the photosphere - still brighter than pretty much anything you will see on Earth, with a few exceptions like lightning. I couldn't find any decent numbers, though - perhaps the temperature is too variable, or too hard to measure accurately.
Answer: Usernumber's explanation of the light and dark regions is correct, but there is more detail to be added about granulation on other stars.
Granulation is expected on other stars with surface convection zones, but the properties and timescales of the granulation can be quite different.
On the Sun, the granules appear and disappear in timescales of 10-30 minutes and the granules have a characteristic diameter of around 1500 km. There are thus about 4 million of these visible on the solar photosphere.
The size of the granules is expected to vary as the gravitational scale height in the photosphere, which is proportional to $T_{\rm eff}/g$. Thus stars with lower temperatures (K- and M-stars) are expected to have smaller granules, but stars with lower surface gravities (subgiants and giants) are expected to have much bigger granulation patterns (Cranmer et al. 2014).
In fact, given that gravity scales as $R^{-2}$, the ratio of the radius of the star to the size of a granule gets smaller as gravity decreases. Thus giants are expected have far fewer, but bigger granules.
The timescales are also different. The frequency of granulation appears to scale with the peak frequency of p-mode oscillations, which in turn scales as $g/\sqrt{T_{\rm eff}}$, and so cooler stars have higher frequency granulation, but giants, with 1-2 orders of magnitude lower surface gravity have much more slowly changing granulation patterns (Kallinger et al. 2014).
The truth of the above has been basically confirmed using disk-integrated variability seen in stars monitored by the Kepler satellite.
Of course, the granulation pattern cannot be imaged in distant stars, except in those stars with the largest radii and largest granulation patterns. There have been claims that surface brightness variations on Betelgeuse are due to granulation, but the first really believable images are of the close hypergiant $\pi^1$ Gruis (Paladini et al. 2017). This star is half the temperature of the Sun and it's gravity is about $10^5$ times lower. According to the ideas above, the granules should be 50,000 times bigger than on the Sun, i.e. a diameter of 75 million km.
The radius of $\pi^1$ Gru is about 250 million km, so its surface will be covered by only around 100 granules, roughly in agreement with what is observed (see below).

VLT near infrared image of $\pi^1$ Gru (ESO).
Comment: Ah! The source for the image in [this answer](https://astronomy.stackexchange.com/a/34791/7982) does indeed claim the bright spot imaged by ALMA on Betelgeuse to be "a convective hot spot on the photosphere". This image of π¹ Gru is much more convincing, but still I wonder [How do they know the non-uniformities seen on the disk of π¹ Gru are real and not artifacts?](https://astronomy.stackexchange.com/q/34922/7982)
Comment: @uhoh There is a subsection in the paper addressing this point.
Answer: I'll add to @usernumber's answer some graphics. Unfortunately we can't yet "has YouTubes" for some reason so I'll just add the links.
There are two videos of the Sun linked in Phil Plait's Bad Astronomy article

DKIST first light high-resolution video of solar granules
DKIST First light video of solar granulation (wide angle).

Here are the same kind of convection cells shown in more familiar settings:

Bénard-Marangoni convection in a cacao cup
Convection Currents in Miso Soup
Convection of coffee and milk


Why is it said that speed always remains constant in circular motion?

Question: One can cover circular motion with different speeds at different positions, right? The only aim is to complete his circular motion, right?
Comment: Said where? Your question could use a reference and a quote from the source to clarify the context/origin.
Comment: No, it's not. Tangential speed is a vector quantity and vectors do have _direction_ and magnitude. In every circle position speed direction changes, thus speed vector **is not a constant.**
Comment: Most highschool textbooks only deal with scenario which op has described above so
Comment: You can, but you tend to need calculus to do so. Once the system has reached steady state, at sufficiently high RPMs, it's a good approximation since inertia helps keep things constant.
Answer: Circular motion at constant speed is just the simplest type of circular motion, so the first examples of circular motion that you come across often have constant speed to keep them simple. In more complicated examples of circular motion the speed can change with time and with position. And when you progress to orbital mechanics then you have examples of objects in elliptic or hyperbolic orbits where both the speed and the radius of the orbit change with time.
By analogy, the simplest type of linear motion has constant speed - but you can have linear motion with varying speed and constant acceleration or even with varying acceleration - an example of the latter is simple harmonic motion.
Comment: "*Circular motion with constant magnitude acceleration is circular motion at constant speed*". This is not true. One could have acceleration with a constant magnitude where the vector always has a tangential component as well.
Comment: You're assuming the tangential component stays constant. One could change the tangential and centripetal accelerations accordingly so that the acceleration maintains a constant magnitude while also maintaining circular motion. Holding the derivative of $a$ to be $0$ for circular motion would mean that $\dddot\theta=-2\dot\theta^3$.
Comment: 'Constant acceleration' If we consider centripetal acceleration ( magnitude is same, direction changes) as the direction is different  the acceleration is always varying right ? can u give an example of constant acceleration?
Comment: @Angelinevarghese An example of linear motion with constant acceleration would be an object falling under gravity, or a ball rolling down an inclined plane.
Comment: Actually i asked about circular motion
Comment: @Angelinevarghese Then you have answered your own question. Circular motion with constant *magnitude* acceleration is circular motion at constant speed because the acceleration vector is always perpendicular to the velocity vector. Motion with acceleration that is constant in magnitude and direction is parabolic motion, not circular motion.
Comment: @BioPhysicist A tangential component to the acceleration will change the speed. Then either the radius changes and the motion is not circular or the centripetal acceleration changes and the acceleration does not have a constant magnitude.
Answer: In uniform circular motion, the velocity has a constant magnitude. In this case the acceleration always points towards the center of the circle (perpendicular to the velocity) and has a magnitude exactly equal to $v^2/r$.
In the more general sense, we can have acceleration that has a tangential component that points along the circle (parallel/anti-parallel to the velocity). In this case we will have a velocity that also changes in magnitude in addition to changing direction.
For the more mathematically inclined reader, the acceleration vector in polar coordinates is
$$\mathbf a=(\ddot r-r\dot\theta^2)\,\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$
However, for circular motion it is the case that $\dot r=0$ and $\ddot r=0$, so we have
$$\mathbf a=-r\dot\theta^2\,\hat r+r\ddot\theta\,\hat\theta$$
In uniform circular motion (constant speed), $\ddot\theta=0$ also, and so we get what was stated above, that the acceleration only has a centripetal component equal to $r\dot\theta^2=v^2/r$.
In general $\ddot\theta\neq 0$, and so we have an additional component to the acceleration that is tangent to the circle. In this case the speed is not constant because $\dot\theta=v/r$ is changing. The magnitude of the acceleration vector is
$$a=\sqrt{r^2\dot\theta^4+r^2\ddot\theta^2}$$
Answer: I want to address what you said about the aim being to "complete his circular motion". I assume you mean that the particle is moving in a circle with a constant radius. Let $v$ be the speed of the particle, $a_{||}$ be the component of the acceleration in the direction of the velocity of the particle, and $a_{\perp}$ be the component of the acceleration perpendicular to the velocity.
The condition to not change the radius of an orbit is:
$$ a_{\perp} = \frac{v^2}{ r}.  $$
If $a_{\perp} < \frac{v^2}{r}$ then there isn't enough force to keep the object in orbit and the radius of the orbit will increase. If $a_{\perp} > \frac{v^2}{r}$ then there is so much force pulling the object in and the radius will decrease.
If we demand the radius $r$ to be constant, then $a_{\perp}$ only depends on the speed of the particle $v$, and the easiest situation is having both of them to be constant.  Which is why "they" always say a circular motion keeps the speed constant. However we also see that we can keep the particle with constant radius if we increased $a_{\perp}$ together with $v$. We need to increase them together in this very particular way though such that when $v$ is doubled, $a_{\perp}$ is quadrupled. Note that increasing $a_{\perp}$ means increasing the pull (the force) on the particle in the direction perpendicular to the velocity, so this might not always be possible. For example, increasing the force might be possible if the particle is held in orbit by a string, but will not be possible if the particle is held in orbit by gravity.
Answer: The "aim" of circular motion is indeed to complete to circulate at a constant radius to the source of the force. The speed doesn't have to be constant.
Take a piece of string with a mass attached in your hand. Now give the mass on the string (while its length stays constant, which is a rather idealized piece of string) a circular motion. You can also attach the string to a rotating motor. You can make the circular speed vary by adjusting it with your hand, or even better by varying the speed of the rotation of the motor. But it's easier to use your hand.
When the speed of the mass is getting higher you feel the centrifugal force pulling at your hand getting bigger and vice-versa.
Another example. A mass attached to string (again with constant length). Attach the string to a nail on a table. Give the mass a push and due to the friction force (which is a non-conservative force, though this is not of real importance here), the mass will make a decelerating circular motion.
In both cases, the force on the mass is not directed towards the center of the circle on which the mass moves.
So, it is not always said that the speed in a circular motion is constant. Mostly though this is said in the context of astronomy and the math simplifies for describing a circular motion, but not in the case I mentioned, which is a non-uniform circular motion.
Note that exact circular motion exists only in theory. For the math, see the answer given above by @BioPhysicist.
Answer: The equation modelling the most general kind of spiralling motion in a plane is:
$$ r(t) = A(t) e^{i \theta(t)}$$
Where $ A(t)$ is the radius of circle as a function of time, $ \theta(t)$ is angle of circle as a function of time. Now,
$$ \frac{dr}{dt} = v(t) =  \frac{dA}{dt} e^{ i \theta(t) }  + iA \frac{d \theta}{dt} e^{ i \theta(t) }$$
Now, the first term in the expression denotes radial velocity $ v_{rdl} (t)$ and the second denotes $ v_{tngt} (t)$.
$$v_{rdl} = \frac{dA}{dt} e^{ i \theta(t) }$$
$$ v_{tngt} = i A \frac{ d \theta}{dt} e^{ i \theta(t) } = i \frac{ d \theta}{dt} r(t)$$
Now, suppose we thought of complex numbers as vectors, we could think of a multiplication of 'i' to analogous to rotating the vector ninety degrees. So, if $ r(t)$ denotes the position vector, multiplying it with 'i' gives a vector perpendicular to it. In other words this denotes tangential component of the velocity.
So, the angle the rate at which the particle traverses the angle can be any function of time. Maybe it can start spinning faster and faster as times goes and could be given as:
$$ \frac{ d \theta}{dt}  = e^{t}$$
So, like an exponential increase of angle speed, and hence lead to large tangential velocities as you've noted.
Hope this helps

In the regular circular motion, we say that $ A(t) = A$ and $ \theta(t) = \omega t$ where $ \omega $ is some constant

Reference:
Page-10,20  of Tristam Needham's visual complex analysis book
This website
Comment: Although your solution might be technically right I don't see how introducing complex numbers is appropriate to answer this question.
Comment: *radius of circle ... angle of circle* Spirals are not circles.
Comment: It models the physical situation very well and each part catches all elements of the motion. After op sees this, I can not imagine him having any more cases on the different kind of spiral motion.

As for why I just introduced complex numbers, I thought this answer was an opportunity to showcase the use of complex numbers in modelling physics situations..
Comment: Not a precise explanation but what I meant is a spiral can be thought as a circle with increasing varies as the angle changes


How do Bitcoin Wallets Maintain Address Balances?

Question: Does a Bitcoin software wallet (not the Bitcoin core wallet) read the entire blockchain to gather balance information? Or does it read the UTXO set? Or some other data source? Does it need to read it every time it wants to retrieve balance information, or does it simply read the last transactions posted to the blockchain?
Answer: A full node wallet collects from peers the whole blockchain and uses that data to construct its copy of the UTXO set.
From that it can calculate balances for addresses it has generated.
As it receives new transactions it updates its UTXO set and balances.

A lightweight (SPV) wallet might only collect block header information from configured servers. It will probably only fetch full information for specific transactions that are relevant to it.
The header information can be validated by the lightweight wallet to check that it is not being deceived by a malicious server. This is necessarily less rigorous than the validation/checking performed by a full node.
From the received subset of full transaction data the lightweight node can calculate balances for addresses of interest to it.
Comment: Thank you!  But would you mind explaining how a non-full node wallet calculates and maintains balances?
Comment: @Jamie, I know less about lightweight nodes, but I have added to the answer what I know about them.
Comment: RedGrittyBrick Thank you!  Just to make sure I am using the correct terminology, what is the difference between: light wallet; lightweight wallet; software wallet and SVP wallet?
Comment: I recently revised my answer to [What kinds of wallet are there](https://bitcoin.stackexchange.com/q/81240/13866) maybe that covers it?
Comment: Thank you Sir !


Microsoft OneNote hidden on installing Windows 10

Question: After upgrading my machine to Windows 10, I am not happy with what it has done to "Microsoft OneNote". I used to have a lot of notes on this. Now they are nowhere to be seen. It has forced me to waste time setting up a Microsoft "account", but after doing that and starting OneNote within it, the old notes are still lost? Archived? Hidden? Deleted? Does anyone know how I can get them back?
Answer: This is a common issue for Windows 10 upgrades.
Be sure to launch OneNote 2016 (NOT the app called just OneNote) - this version is more likely to contain your old notes. Should this be your solution, you can follow these steps to change your default OneNote version to the 2016 version, as mentioned in this Microsoft support article:

Press the Windows button in the lower-left corner of your screen, and then choose Settings.
In Settings, choose System > Default apps, scroll to the bottom of the list, and choose Set defaults by app.
In the list under Set your default programs, find the version of OneNote that you'd like Windows to use as the default, and then choose
  Set this program as default. For example, choose OneNote (desktop) if
  you want to always open notebooks in OneNote 2016.
Click OK to save your changes.
Comment: Just include the steps to set the default version of OneNote.  That is all you have to do to make this an acceptable non-link only answer or don't that is entirely your choice.  *Answers like this appear in the review queue, actually all your answers would appear there, because you have less then 100 reputation.*  When this answer contains the instructions I also will reverse the vote.
Comment: @Jonas The reason we want people to include the relevant information is because of Link Rot: Through a combination of companies going under, products being EoL'd and websites being redesigned, a link that works fine today might not work years, months or even weeks in the future. If someone sees this post in 3 years after Microsoft releases Office 2019 and changed their link scheme, it should still be equally helpful as it is today, and including the relevant information is an important aspect of that.
Comment: @NateKerkhofs Very reasonable in retrospect. I'll stick to doing it this way in the future.

Again, thanks.
Comment: This answer should include the relevant information not just link to an external website.
Comment: It isn't just a link to an external website. I provided an answer directly on here. If this truly solves the problem, I provide a link, mind you with context, to a Microsoft support page detailing how to change the default OneNote version to the working one.
Comment: I downvoted this answer for the reason Ramhound provided. Once you type the steps, I have no problem removing the downvote.
Comment: All right, I've added the actual steps needed to change the default app. Didn't knew this was the preferred way of doing things, so thanks for letting me know.
Answer: There are 2 possible causes for this.
OneNote usually synchronizes your notes online, using your Microsoft account. Either you have not synchronized your notes before, or you used a different Microsoft account.
In the second case, you should simply use the old Microsoft account to gain access to your notes.
In the first case, if you haven't ever synched the notes they will still be on your disk. The most likely issue is that you are now using a different application, the Windows Store App called "OneNote" instead of the Office Application called "OneNote 2010" (or whatever version you're using). To launch the correct program, click in the search bar on the left side of the task bar on the bottom of your screen, and type "OneNote 20". The search bar should then suggest to launch the correct application - if it doesn't do so instantly, wait for a few seconds.


Why is it that my car smells like it's overheating but my temperature gauge says that it's not?

Question: I have a 2006 Nissan Maxima with a hundred and sixty-eight thousand miles on it. The water pump works fine but for some reason everytime I get out of my car it smells like it's overheating and yet my temperature gauge says it's not. When I start my car in the morning the temperature gauge was smoothie rise but then stops at the Middle Point and never goes any higher what could causes or what could cause the smell out of it overheating. The only reason I know what it smells like when a car overheats is because I actually blew an engine once and that's the exact same smell. Thanks one more thing the radiator is also fine.
Comment: Welcome to Motor Vehicle Maintenance & Repair! I'd bet you have an oil leak which is getting burned off on one of the exhaust headers. If you open your hood and take a look around, especially right after you've driven it, where it's coming from might be self evident. The engine itself can run just fine with an oil leak, as long as you ensure the proper amount is in there (ie: replace what is lost). Take a look around and see if you can tell anything.
Answer: What do you mean by smell of overheating?
To a random person not knowledgeable in cars, it could be a burning smell. That's burning oil so you have an oil leak to find, then.
However, for car people, the smell of overheating usually means the smell of coolant. If you can smell that, you are losing coolant and need to find the cause. Start from the simplest: check that your radiator cap is tight. If it is, pressure test the coolant system to see if it's leaking and if it is, use UV dye to find the source of the leak.
Answer: It could be that your catalytic converter is overheating, from misfire or rich condition. Engine misfire can be caused by weak spark which might not set a code. Or you could have a sluggish oxygen sensor which might set a code.
Use a vacuum cleaner hose as a sniffer. Stop the car out of the wind, lie down beside the car and extend the hose toward the cat while sniffing the other end.
Answer: You may have water leaking from the head gasket onto the exhaust manifold (header).  Will immediately vaporise and cause the smell.
It's the coolant in the system that smells (antifreeze)
Comment: A leak of coolant from any source will produce a "hot engine" smell.


Opening data connection

Question: In c#.net when you open a connection to a database, in my case it's oracle, is there a lot of overhead involved? I'm guessing there isn't because of connection pooling. So am I correct in saying that everytime I open a connection it actually grabs an open connection out of the pool and if there are no available connections in the pool it will then open a new connection?
Thanks!
Answer: You are correct. There are a lot of good articles explaining ADO connection pooling. For exmaple, MSDN - SQL Server Connection Pooling (ADO.NET), which says:

Connection pooling reduces the number
  of times that new connections must be
  opened. The pooler maintains ownership
  of the physical connection. It manages
  connections by keeping alive a set of
  active connections for each given
  connection configuration. Whenever a
  user calls Open on a connection, the
  pooler looks for an available
  connection in the pool. If a pooled
  connection is available, it returns it
  to the caller instead of opening a new
  connection. When the application calls
  Close on the connection, the pooler
  returns it to the pooled set of active
  connections instead of closing it.
  Once the connection is returned to the
  pool, it is ready to be reused on the
  next Open call.
Comment: Thanks for the response! I just wanted to make sure there wasn't anything else going on.
Answer: The first time you open the connection there will be a lot of overhead.  If you are using connection pooling and you use the exact same connection string, the next time you open the connection it should use a connection from the connection pool and be a lot quicker.

Opening data connection


Does implied consent protect an organization's ability to discipline former members?

Question: Somewhat inspired by this question, I found this website. To summarize, they imply that they would be legally protected from civil suits from former members who are being disciplined. Would this actually work? My first thought is that it doesn't make sense that a person who leaves cannot revoke their consent, especially since this wasn't a binding contract of any sort.
Comment: The church would discipline it's *former* member by... withholding Communion?  Shunning?
Comment: It should be possible to understand the question without following the links.  As it is now, I have little idea what you're asking about.
Comment: @RonJohn Exactly. Most forms of religious punishment are non-justiciable.
Comment: @ohwilleke it occurs to me that the former church member might *work for* a church member in good standing (either in an owner-employee relationship or a manager-employee relationship.  In such a case, shunning might entail firing the worker or creating a hostile work environment.  Church members could probably be sued for that under labor laws.
Comment: @RonJohn The effect would be minimal even then, due to employment at will. Maybe the employee would have a valid claim for unemployment insurance. Even then, a close call.
Comment: @ohwilleke but isn't religious affiliation a [protected employment class](https://www.eeoc.gov/religious-discrimination)?
Comment: @RonJohn I have doubts that a person being shunned by an employer out of an employer's religious conviction would be framed that way.
Comment: @RonJohn the text of the article seemed to imply some form of harassment or other potentially damaging action such as public shaming. Specifically, it mentions churches being sued for defamation, invasion of privacy, and intentional infliction of emotional distress.
Answer: In Guinn v. Church of Christ, plaintiff withdrew from the church after an internal investigation of her conduct. The church apparently held as a matter of religious doctrine that she must repent of her sins, also that withdrawing from the church is doctrinally impossible. The transgressions were widely publicized within the church; a lawsuit over outrage and invasion of privacy ensued. The upshot of the appeal is that the church can be held liable for post-withdrawal actions, but before that, the church has a privilege to communicate such transgressions (the church is not subject to secular judicature of its actions w.r.t. its members).
Contract law is not relevant here, what matters is that a person can knowingly and intelligently waive their right to litigate against a party, and while one is a member of the church which has such a waiver as part of their disciplinary doctrine, one cannot sue the church for its doctrinary actions as long as the actions do not constitute a threat to public safety which would justify state interference. Although the church argued that church membership is irrevocable, the court found that "Just as freedom to worship is protected by the First Amendment, so also is the liberty to recede from one's religious allegiance". Given that plaintiff had withdrawn consent yet the church subsequently announced the transgressions without her consent, the church was thus found to be liable.
In Stepek v. Doe, the court similarly affirmed that a church enjoys a privilege against charges of defamation, when the plaintiff continues to operate within the church, not having left the church.
So it can "work", to some extent. The person can always withdraw consent; the person has no legal recourse in case consent has not been withdrawn. The article is correct as far as it goes, which is not far enough: it errs in not stating what the legal consequences of of the transgressor leaving the church are.


React Native Could not open URL No Activity found to handle Intent

Question: I'm working with React Native and Square Point of Sale. I'm using Web API for Android.
Also I have installed Square Point of Sale.
This is my code:
componentDidMount() {
    let url ="intent:#Intent;action=com.squareup.pos.action.CHARGE;package=com.squareup;S.browser_fallback_url=https://my.website.com/index.html;S.com.squareup.pos.WEB_CALLBACK_URI=https://my.website.com/index.html;S.com.squareup.pos.CLIENT_ID=sq0ids-yOurCLieNtId;S.com.squareup.pos.API_VERSION=v2.0;i.com.squareup.pos.TOTAL_AMOUNT=100;S.com.squareup.pos.CURRENCY_CODE=USD;S.com.squareup.pos.TENDER_TYPES=com.squareup.pos.TENDER_CARD,com.squareup.pos.TENDER_CARD_ON_FILE,com.squareup.pos.TENDER_CASH,com.squareup.pos.TENDER_OTHER;end";

    Linking.openURL(url).catch(err => console.error("An error occurred", err));
}

This is my error message:
Could not open URL 'intent:#Intent; ... end': No Activity found to handle Intent { act=android.intent.action.VIEW dat=intent: flg=0x10000000 }

What I'm doing wrong?
Answer: I tried building using the web API but found that the flow wasn't great. The user doesn't always make it back to the app; sometimes they get stuck on the website.
I built react-native-square-pos to make things easier for the next developer.  It uses Native Modules to communicate with the Android / iOS Square PoS APIs.. so you don't have to write Java or Objective-C.
Answer: The intent:#Intent that you're using there isn't really going to work from within React Native, since that is used for going from a web browser on an Android device into the Square POS app.
The Android Web API intent is intended to do a hand-off from your browser to the Square POS application, not from application to application.
It sounds like what you'd be wanting to do is use the Square Point of Sale SDK for Android with your React Native application. You'll want to take a look at https://facebook.github.io/react-native/docs/native-modules-android.html about how to link a native module with your React Native application.


Imagemagick: Convert to fixed width, proportional height

Question: Using the imagemagick convert command, how can I resize an image to a fixed width and proportional height e.g. using the -resize or the -thumbnail option?
Answer: Use -resize 100x to resize images to 100 pixels in width while maintaining the height's aspect ratio.
Read the fine ImageMagick manual for details.
Comment: imagemagick comes with `mogrify` which will alter original image. I found it easiser to copy the images into a new folder then run `mogrify -resize 512x *.jpg`
Comment: mkdir will complain if the directory already exists. To avoid this, add the -p flag "no error if existing" to mkdir. `mkdir -p thumbnails` explained in this question: http://stackoverflow.com/questions/4906579/how-to-use-bash-to-create-a-folder-if-it-doesnt-already-exist
Comment: How do you resize to specified width, but original height?
Answer: Imagemagick geometry: 'width'x'height'
If you leave one part empty, this means resize proportional.
Examples:
100x200       # width = 100, height = 200
300x          # width = 300, height = proportional
x300          # width = proportional, height = 300

Comment: `100x200` will create an image with theses dimensions. The original picture will keep its ratio W/H. `100x200\!` will also create an image with theses dimensions, but image inside will be deformed to stick the new dimensions.
Comment: @MTranchant & chrise : AFAICS, `-geometry 100x200` behaves as `min(100x, x200)` (ImageMagick 184.108.40.206, Fedora 23).
Answer: For nubie's like me...
Windows: Powershell: for Facebook Photos for example:
Right click windows 'Start', select 'Terminal'
C: cd to directory that you require to do conversion from:
ie: cd Users/OneDrive/Pictures
C: mkdir FBName ;creates sub-directory 'makedir ./FBName' does the same.
C: magick name.jpg -resize 2048 ./FBName/newname.jpg
; Will resize 'name.jpg' with a max size of 2048 (long edge, portrait or landscape) into the subdirectory FBName that you created previously. Linux would replace 'magick' with 'convert'


Spring PDF generation + i18n

Question: I have a spring application that overloads an AbstractView in order to generate a PDF file. That file should be internationalized too, so I am adding appropriate strings from resources. There are a lot of code in such implementation.
So mabe there is another way to implement it? I know that Flying Saucer can do that conversion but does it provide an i18n?
Ideally I want some template engine with i18n support and PDF generation tools.
Does anybody familiar with such engine?
Thank you
Answer: Perhaps IText or JasperReports is what you need.
Comment: I am using iText already. And doesjasperReports support i18n?
Answer: It's not trivial, since for unicode characters to be presented in PDF, they must either come from one of provided fonts, or from embedded font. You must prepare yourself the set of fonts, that will support all languages that may be used.
Comment: This behaviour is caused by PDF specification itself.

Spring PDF generation + i18n


Can a dynamic circuit extract all energy from an ultra capacitor?

Question: I understand that the energy extraction from an ultra capacitor drops the voltage to the point where the load can't function. But, is it possible to create a converter that can continuously present a consistent voltage to the load using a dynamic dc-to-dc circuit to allow all the energy in the ultra capacitor to be utilized, even if the current drops? What are the issues?
Comment: I'm mostly sure I know what you mean, but what do you mean by "even if the current drops"?
Answer:
[...] to allow all the energy in the Ultra Capacitor to be utilised [...]

No. The closest that you can get, to that, is by using a synchronous boost converter (if your cap voltage is always lower than your target output voltage), or a synchronous buck/boost converter (if your cap voltage may be above and below the target output voltage).
A synchronous converter uses only controlled switches, instead of controlled switches plus diodes. The benefit is reduced losses, because the main current does not cause a (relatively big) voltage drop across any diode. Only across Rds_on of the controlled switches, which may be in the milliohm range, and will cause much lower voltage drops (and therefore power losses).
As Vin approaches zero, Iin will have to grow substantially, so as to keep Vin·Iin above Pout (which usually you will want to keep constant). Those high input currents will eventually cause that your Vin voltage will just drop across the first devices of your converter. So, below a certain threshold for Vin, you will not be able to deliver any more power to the output. The converter will still drain your cap, but that energy will serve no purpose any more, at the output.
Comment: Excellent thanks, but is such a circuit easy or expensive to build?
Comment: @Anthony It is not expensive, but not easy, either. You should not do it on a breadboard, and the PCB layout, being  it not difficult, is not trivial for someone new to electronics. I don't know what's your experience doing PCBs. // If you can buy a "step-up" or "step-up/down" dc-dc converter already done, it will be best. I can't recommend any specific one, because I don't know the specs.
Answer:
But, is it possible to create a converter that can continuously present a consistent voltage to the load using a dynamic dc to dc circuit to allow all the energy in the Ultra Capacitor to be utilised, even if the current drops?

All the energy? No. But you should be able to extract most of it with the approach you described.
Let's say you have a 5V output and the capacitor is a 0.5F 5V abs max so you start with 4.5V on the capacitor, and can use a boost converter that will operate from 0.8V - 6V, so the capacitor is the constraint for the upper voltage, and the boost converter is the constraint for the lower voltage.
Energy extracted from the capacitor is 1/2 0.5F * (4.5V^2 - 0.8V^2) = 4.9J; energy remaining in the capacitor is 1/2 * 0.5F * 0.8V^2 = 0.16J.
The amount of energy presented to your circuit is 4.9J *  the efficiency of the converter; you can get DC/DC converters that operate in the 90%+ range fairly easily these days.
Answer: You can "asymptote towards zero" but you can never get there.
 In practice getting 95% + of available energy is 'easy enough'.
Energy in cap is proportional to V^2.
 Say Vstart = Vs, Vfinish = Vf.
 So

End energy as a fraction of start energy is (Vf/Vs)^2.
eg
5V -> 1V you have (1/5)^2 = 1/25 = 4% left
5V -> 0.5V = (0.5/5)^2 = 0.25/25 = 1% left

If you really really really MUST get all possible energy out you can use "energy harvesting" circuits that will accept perhaps 10 mV in and boost it to say 5V BUT the efficincy is unlikely to be more than sau 50% and the energy available is very small.
If you start at eg 5V then stopping at 0.5V = 1% is probably very ample in most cases, and 1V / 4% is probably often acceptable.
Comment: Thanks!, The purpose is to capture the energy quickly, like an ultra capacitor can, and over time transfer to Lithium Ion batteries slowly so that we treat the battery nicely.
Answer: My company makes such a thing for holding up the DC bus of a variable-frequency drive during an outage. Exactly what you're describing: use a boost converter to present the load with a fixed voltage, even as the cap voltage drops. I understand that this is done in many other lower-power applications, as well.
As others point out, the problem is that using all the energy isn't practical. The first problem is efficiency; there will always be some loss, so you can't get all the energy into the load. But beyond that, the boost ratios involved become impractical beyond a certain low input voltage.
Some of the components in a boost converter are sized based on the input current to the converter, including the choke and the boost switch. If you assume a fixed load current, that means that a boost converter doing a 10:1 boost has to be much beefier than a boost converter doing a 2:1 boost. Doing a 10:1 boost means your input current is 10x your output current!
And with capacitors, the further down you go in voltage, the less energy you're actually getting out. For example, our boost converters are rated for ~2:1 boost, so we typically only drain the caps to half voltage, meaning we've gotten 3/4 of the available energy (1/2 C V^2) out. We could get more energy, say by draining the caps to 1/4 voltage. But we'd have to rate the converter for a 4:1 boost, which would mean all our components would have to be up-sized for higher peak currents. Control of a boost converter at a high boost ratio also becomes problematic. And all we'd get for our additional expense would be another 12.5% of the available energy. Diminishing returns.
Now, that was just for going from 2:1 to 4:1. Imagine doing that at a 10:1 boost. You get even less energy, but you've had to scale your components up by a factor of ten now. And no matter how low you go, you've still got some energy left in the caps, because there comes a point where your boost converter simply won't pull any more because of the non-ideality of the switch.
So can you get all the energy out? No.
Should you even try to get out as much energy as theoretically possible? No.
Is it still useful to use a boost converter to get a fixed voltage output off an ultracap? Absolutely.


The training error of best hypothesis

Question: Let $\mathcal{X}$ and $\mathcal{Y}$ denote the domain set and label set respectively. Also let $\mathcal{D}$ be a distribution over $\mathcal{X}$ and $f:\mathcal{X} \to \mathcal{Y}$ be the true labeling function. So for generating the data, we first sample from $\mathcal{D}$ and then apply function $f$. We define the error of a prediction rule $h:\mathcal{X} \to \mathcal{Y}$ to be, $$L_{\mathcal{D} , f}(h) = \mathcal{D}(\{x:h(x) \not = f(x) \}).$$Suppose that the realizability assumption for hypothesis class $\mathcal{H}$ holds which means that there exists $h^* \in \mathcal{H}$ such that $L_{\mathcal{D} , f}(h^*) = 0$. We collect $m$ samples $S=\{(x_1,y_1),(x_2,y_2),\dots , (x_m,y_m) \}$ and compute the error of $h^*$ on $S$: $$L_S(h^*) = \frac{|i\in [m]: h^*(x_i) \not = y_i|}{m}$$ Now we have:$$\mathbb{P}[L_S(h^*) = 0] = \mathcal{D}^m(\{(x_1,x_2,\dots, x_m):h^*(x_1) = f(x_1) \ \text{and} \ h^*(x_2) = f(x_2) \ \text{and} \dots  h^*(x_m) = f(x_m)\})$$ If the samples are independent then, $$\mathbb{P}[L_S(h^*) = 0] = \mathcal{D}(\{x_1: h^*(x_1) = f(x_1)\})\mathcal{D}(\{x_2: h^*(x_2) = f(x_2)\}) \dots \mathcal{D}(\{x_m: h^*(x_m) = f(x_m)\}) = 1.$$ What about the case in which samples are dependent? How can it be proved that $\mathbb{P}[L_S(h^*) = 0] = 1$?
Answer: tl;dr : The Assumptions Avoid the Proof
What is assumed avoids having to prove either the sample dependent or independent case. It merely follows by what is assumed.
The distribution $\mathcal{D}$ is irrelevant in this case because the actual distribution involved in your $\mathbb{P}[L_S(h^*) = 0] = 1$ is a Dirac distribution with all the probability on $L_S(h^*) = 0$ as follows from assumptions of $h^*$ in "realizable" learning. Absolute certainty lies in determinism.
Note that how you wrote the equality between the expressions using $\mathcal{D}$ to define $\mathbb{P}[L_S(h^*) = 0]$ is incorrect because $\mathcal{D}$ is only about the probability of witnessing a sample $x \in \mathcal{X}$, not about the probability of error. As written $x $ ~ $ \mathcal{D}$ does not necessarily equal probability 1, but comparing if $h^*(x) = f(x)$ is independent of the probability of $x$ occurring as they are defined, and so that is 1 by def of $h^*$.
Below, I explain in detail my point without a proof because that quickly turns into an open question in research of statistical learning if you keep relaxing distribution constraints from i.i.d. to beyond ergodic and stationary process into a non-stationary stochastic processes.
Note if you were to practically find $h^*$, $\mathcal{D}$ and the dependency between its samples absolutely matters.
An Explanation Without Formal Proof
I believe the issue lies in where the probability distributions actually affect anything, in particular $L_S$ and $L_{\mathcal{D}, f}$. This is often an issue when people use $P(\cdot)$ without specifying what equals what, or which probability distribution things are relative to. To help be clear what source of uncertainty I am referring to, I will continue (as from  my prior answer here) to refer to $\mathcal{D} = (X, Y)$ as a joint distribution of input $X$ and output $Y$ as random variables, and $f(X) = Y$.
As per your question and the assumption of "realizable learning", the only distribution that could matter is $X$. I explain that now.
If $h^*(x) = f(x) = y$ for all $x \in \mathcal{X}$, which we are given, then the error in prediction from $h^*(x)$ wrt $f(x)$ is zero, always, regardless of any random or purposeful assortment, permutation, or poking and prodding of these functions' definitions. If given the same $x$ as input, they produce the same $y$, as that's what their equality means.
The formal theorem for this is the Data Processing Inequality, as defined in Ref. [2], Section 2.8

Theorem 2.8.1 (Data-processing inequality) If $X \rightarrow Y \rightarrow Z$ [is a markov chain of three random variables, $X,Y,Z$], then $I(X;Y) \ge I(X;Z)$.

This means that as long as there is only one source of uncertainty (as in entropy), in our case $X$ because assuming "realizable learning" means $f(X) = Y$ is a deterministic function and there is no other source of error, thus allowing zero error in the loss, then any deterministic transformation of that random variable can only have equal expected uncertainty or less. Given $h^*(X) = f(X) = Y$ by definition and "realizable learning" assumption, then all the uncertainty (probability) in $Y$ is fully determined by $X$ as transformed by $f$.
So that's the reason the loss is zero in addition to the fact that this magical $h^*$ is plucked from the imagination and not defined on how it is obtained. The curse of being too general to be useful, but yet given these properties assumed, we can reason about such an object.
This is why I stated in my last answer that if $h^*$ yields zero error on all of $\mathcal{X}$, then applying that hypothesis to any subset of $\mathcal{X}$ also yields zero error. Doesn't matter about the distribution $X$ over the space $\mathcal{X}$, because we have this magically correct $h^*$ and the $f(\cdot)$ and the losses are independent of $\mathcal{D}$ after $h^*$ is assumed.
If you want to know how to actually find $h^*$, that is an entirely different problem ripe with model assumptions including that the hypothesis space you considered even includes a correct hypothesis that could yield zero error (this is identifiability, and related is the concept of parameter or hypothesis space misspecification).
In this context, to address dependency of samples across time you need to use stochastic processes, where a process is a function over time, and a stochastic process is essentially how the random variable changes over time. There the dependency on prior states can involve either observed or latent (hidden) states.
This is now a time series problem, and here be monsters. ;)
Further Reading
For more reading on finding $h^*$, I provide a snippet from this answer that is relevant after updating the reference numbers here. Recall I refer to the hypothesis space as the parameter space.

identifiable [Ref. 3], meaning that the model is able to represent the observed process and is typically meant as a single parameter is point-identifiable from the information available in the observations, and
"consistent" or is a "well behaved" parameter space with a "well
behaved" update function that can converge to the true parameter.

This is discussed at length in [Ref. 4, Ch.6--9], specifically with regard to Bayesian modelling. Unfortunately, this is not open access.
What you are seeking in such a case is a mathematical and algorithmic formalization of inductive reasoning and inference, which has been of great interest for centuries. The Bayesian modeling approach tends to be involved in the probabilistic mathematical induction. Some relevant books of interest include Ref. 5 Ch. 5 & Ref. 6 Ch. 2.

References
 = Closed Access.  = Open Access.

Shalev-Shwartz, Shai and Ben-David, Shai. "Understanding Machine Learning: From Theory to Algorithms". May 2014. Cambridge University Press. ISBN 978-1-139-95274-3

 https://www.cs.huji.ac.il/~shais/UnderstandingMachineLearning/

Cover, Thomas M. and Thomas, Joy A. "Elements of Information Theory". John Wiley & Sons, Inc. 2006. 2nd Ed. ISBN-13 978-0-471-24195-9

 https://doi.wiley.com/10.1002/047174882X

Lewbel, Arthur. 2019. "The Identification Zoo: Meanings of Identification in Econometrics." Journal of Economic Literature, 57 (4): 835-903.

 preprint:
https://www.bc.edu/content/dam/bc1/schools/mcas/economics/pdf/working-papers/wp957.pdf

Ghosal, Subhashis, and Aad Van der Vaart. Fundamentals of nonparametric Bayesian inference. Vol. 44. Cambridge University Press, 2017.

 https://scholar.google.com/citations?view_op=view_citation&hl=en&user=u2tifuYAAAAJ&citation_for_view=u2tifuYAAAAJ:HDshCWvjkbEC

Li, Ming, and Paul Vitányi. An introduction to Kolmogorov complexity and its applications. Vol. 3. New York: Springer, 2008.

 https://link.springer.com/book/10.1007/978-3-030-11298-1

Hutter, Marcus. Universal artificial intelligence: Sequential decisions based on algorithmic probability. Springer Science & Business Media, 2004.

 https://link.springer.com/book/10.1007/b138233
Public related slides: http://www.hutter1.net/ai/suaibook.pdf
Comment: I admit i haven't yet read through those lecture notes, and this is from mostly recall besides looking up the equations in Ref 1. ATM, I am not convinced. I'd have to think on this more and reread both Ref 1 and the notes. Far too often do I find things like this in the literature on a mismatch on what distribution we're focusing on. If there are proofs for this in the notes, then as long as they're correct it does not matter whether I am convinced or not. lol ;)
Comment: Note that the realizability assumption says that $$L_{\mathcal{D} , f}(h^*) = \mathcal{D}(\{x:h^*(x) \not = f(x) \}) = 0.$$ This proposition is different from $\forall x\in \mathcal{X}: h^*(x) = f(x)$. In fact the latter implies the former but **not** vice versa. Also the expression is correct since $L_S(h^*)=0$ occurs if and only if $$h^*(x_1) = f(x_1) \ \text{and} \ h^*(x_2) = f(x_2) \ \text{and} \dots  h^*(x_m) = f(x_m)$$ occurs. Thus the events are equivalent and their probabilities the same.
Comment: Actually the [lecture notes](https://stat.ethz.ch/~fadouab/MTML.pdf) which I referred to, have used the same argument.
Answer: In general you just can write
\begin{eqnarray*}
\mathbb{P}[L_S(h^*) = 0] &=& \mathcal{D}(\{x_1: h^*(x_1) = f(x_1)\})\\
&& \mathcal{D}(\{x_2: h^*(x_2) = f(x_2)\}|h^*(x_1) = f(x_1))\\
&& \mathcal{D}(\{x_3: h^*(x_3) = f(x_3)\}|h^*(x_1) = f(x_1),h^*(x_2) = f(x_2))
 \dots,
\end{eqnarray*}
and all of the terms have to be 1.
This is because $\mathcal{D}(\{x: h^*(x) = f(x)\})=1$ by assumption, and therefore $\mathcal{D}(\{x: h^*(x) = f(x)\}|C)=1$ must hold for any $C$ with $P(C)>0$ as $P(A)=P(A|B)P(B)+P(A|B^c)P(B^c)$, which can only be 1 if $P(A|B)$ is 1 as long as $P(B)\neq 0$. (Note that there is a certain mix-up in notation between your use of ${\mathcal{D}}$ and my use of $P$ here, and I'm too lazy to fix that. The argument shouldn't be affected.)
Comment: @S.H.W I have added that to my answer.
Comment: @S.H.W Without reading all the detail over there, I don't think the comment is incorrect. It should basically remind you that a set with probability/measure zero is still nonempty and you can find something there. But this doesn't affect the probability statement.
Comment: @S.H.W Yes. Correct.
Comment: Thanks. Could you explain why the conditional probabilities are equal to $1$, please?
Comment: Great answer, thanks! Also I think [this question](https://stats.stackexchange.com/questions/186619/does-an-unconditional-probability-of-1-or-0-imply-a-conditional-probability-of-1) is related.
Comment: Sorry about asking this question, but why [CechMS's comment](https://stats.stackexchange.com/questions/304991/confused-about-the-realizability-assumption-and-equations-of-upper-bound#:~:text=I%20don%27t%20understand%20your%20assertion) is incorrect? According to his/her comment we can select training set $S$ such that $h^*$ and $f$ be completely different on $S$. Is it because of we don't know the training set $S$ in advance?
Comment: Thanks. So the correct implication is$$\mathcal{D}(\{x:h^*(x) = f(x) \}) = 1 \implies \mathcal{D^m}(\{(x_1,x_2,\dots,x_m):h^*(x_1) = f(x_1), h^*(x_2) = f(x_2),\dots ,h^*(x_m) = f(x_m)\}) = 1$$ which holds for every measure $\mathcal{D}$. Is my understanding correct?

The training error of best hypothesis


What prevents a coder from brute forcing a wifi?

Question: I know that a coder is able to connect to a wifi through code so my question is: Is there anything that will make it difficult to brute force a wifi password?
Comment: Depends on the auth method. Home routers use pre-shared key (PSK) which has no protection. An attacker can even capture the login packets and brute force the password offline. This is why you should have a long, complex password.
Answer: What makes WiFi encryption harder to brute force?
First, a strong password, preferably not set to the router default. The best password is set using a cryptographically secure random number generator (CSRNG), to the full 63 ASCII or 64 Hex chars. You can also use WPA-802.1X mode, however, this requires a Authentication server, and therefore is generally used for large networks.
However, a password built from five and more words would be far easier to remember, without significantly compromising the security of the key.
Another issue is WiFi Protected Setup, specifically, the PIN method significantly reduces the entropy for an attacker. The PIN is a 8 digit number checked in two chunks. This reduces the entropy from around 50 bits (5 word password) to less than 15 bits (2 times a 4 digit PIN), a factor of more than 34 billion. Turn that off, and you'll be relatively safe.
Answer: Only the size of the key/passphrase, really. Given long enough, it's always possible. An access point could theoretically have some rate limiting built into it, which would help against authentication schemes that can't be attacked offline, but in practice the only solution is to use a long key/passphrase. You also need strong encryption, of course. (Do not use WEP; it's broken. Similarly, do not use any variant of MS-CHAP if using WPA-Enterprise.)
When discussing brute force attacks, it's important to consider the size of the search space (the collection of possible values). This is often discussed in terms of entropy, usually discussed in the base-2 logarithm of the number of possible values. For example, if there's about a billion (10^9) options, that's 30 bits of entropy (2^30 is roughly a billion). Compare this value with rate that your attacker can try guesses at; in general, it's probably worth assuming at least a thousand guesses per second for an online attack (though in practice home WiFi access points can't support that many).
Completely random English letters and numbers, case-sensitive, are about six bits of entropy per character, but the letters in English words are only a couple bits per character; there are usually only a handful of possible options for the next character in a word. This means, in practice, that human-memorable passwords are usually not very good, but passphrases, especially ones composed of randomly-chosen words, can be quite good.
XKCD has a pretty good illustration of this concept.
Answer: You can consider allowing connections only to whitelisted mac addresses and then even brute forcing the password won't succeed, but this is only a defence against casual attacks as mac spoofing is relatively simple to achieve.
Comment: To someone who's willing to put the effort into brute-forcing a wifi password, MAC filtering isn't even a speed bump.
Comment: I disagree. Brute forcing is easy, just write a simple script to guess passwords. Mac address spoofing requires the attacker to snoop and then force a deauth on top of the password issue. But, in the general scheme of things I agree it's only a mild protection against trivial attackers.
Comment: Nevertheless, it is easy to spoof a MAC address. Considering the fact that 99,99% people use ready-to-use tools, it is as easy as running bruteforce atack. Therefore: can't be consider as a good security control in this case.


How to print Facebook friends list

Question: I need to print out my friends list on Facebook. However their instructions just bring me to a blank page. Something is wrong on my account since this happens on whatever PC, laptop I use. Is there a way I can print it off without using Facebook's instructions of entering settings and download a copy?  Also perhaps is there a way to print off my contacts from Messenger?  This would take the place of printing my friends list on Facebook. Facebook has ignored my support help request since Dec 7.
Answer: Easy way to print friend list is to download a copy of your data and look for Friends into data and print the list.
Other way is using Facebook developer tool. See the graph API overview of friend list. A user access token with read_custom_friendlists permission is required. Learn here => how to use graph API.
Another way is to create a new Yahoo! email account and import your Facebook contacts with Yahoo! contacts. Then print from Yahoo! contacts, it will include person name and email id.
Answer: I think what you are looking for is :
Friends Manager CRM
Try that, if it works well for you.
Good Luck.
Comment: First line of that link is '**Friends Manager CRM doesn't synchronize with your Facebook friends anymore**'.

How to print Facebook friends list


What types of magic has been displayed in A song of Ice and Fire / Game of Thrones Universe?

Question: What types of magic has been displayed in the the A song of Ice and Fire / Game of Thrones Universe?
Answer: The ASoIaF Wiki has categorized them pretty well:

Elemental control: Like the priests of  R'hllor, they can control elements like fire(like Melissandre and Thoros).
Divination: Some people can see the future (like Maegi, see also: greensight)
Skinchanging: People can 'enter' the mind of animals (and humans) (like Bran, Varamyr)
Raising the dead: The priests of R'hllor have brought people back from the dead (like Thoros resurrected Beric when the latter was killed by the Hound)*
Necromancy: Some can raise the dead and make them do their willing (like Qyburn planned to do with Gregor Clegane and those skeleton thingys Bran found North of the Wall)
Glamouring: Some can change their appearence (like Melissandre)

Also, it would be good to point out that the Citadel claims that there is no magic left in the world. They claim it is just a mythical power.
There was supposed to be magic in Valyria as well, but it is said that it perished along with Valyria.

It is said that they could set dragonglass candles to burning with strange, unpleasantly-bright light. With the obsidian candles, they could see across vast distances, look into a man's mind, and speak with one another though they were half the world apart
--A Feast for Crows, Chapter 45, Samwell.

The most powerful magic is said to be originated from the Shadow Lands.

There are firemages who can conjure ladders from the air that are made of fire and can stand forty feet high. The mage can climb up the ladder, each rung disappearing behind him, leaving nothing but silvery smoke. When he reaches the top, the ladder and he are both gone.
--A Clash of Kings, Chapter 40, Daenerys

Moreover, people believe that the children of the forest
used to practice magic.
*This kind of magic might have blood as a requirement; Mirri Maz Duur sacrificed a horse to bring Khal Drogo back from the dead.
Comment: @Dreamwalker I don't remember that specific example - I was thinking more of the faceless assassins that Arya meets at Harrenhall.
Comment: What about in game of thrones tv?
Comment: All of the above categories have been mentioned on the TV Show. I'm not 100% sure about the glamouring. I'll add examples from the show.
Comment: Hasnt the show shown new powers? Teleportation and bilocation? [Warlocks and little girl warlock}
Comment: I don't remember glamoring in the TV show so far.
Answer: There are many kinds of magic in the books and TV series. Here's my list of them. I left out anything I could explain through natural means such as wildfire, which is probably just simple chemistry. I deliberately omitted the act of burning people alive to appease the Lord of Light because I am not convinced that is magic, but could be a mere power play by the Red Woman. Nor am I convinced that what the Faceless Men do is real magic.
Reanimating the dead. (The dead body does the bidding of whomever controls it.)

The Children

Resurrection. (The dead body becomes their former self.)

Kissed by Fire
Home

Converting the dead into wights. (I consider this different than merely reanimating the dead because the wights can be reduced to skeletons and still function, while reanimated dead like Gregor Clegane probably could not function after being skeletonized.)

Winter is Coming - a dead wildling girl is converted to a wight.
The Pointy End
Hardhome

Transforming babies into White Walkers

Oathkeeper

Teleportation.

Valar Morghulis
Valar Dohaeris

Bilocation.

The Ghost of Harrenhal
A Man Without Honor
Valar Morghulis

Hatching dragon eggs.

Fire and Blood - hatching dragon eggs requires a blood sacrifice

Drinking poison and living

The North Remembers

Visions (both greensight and seeing events in fire)

The North Remembers
The Night Lands
What is Dead May Never Die
The Ghost of Harrenhal
Valar Morghulis
Dark Wings, Dark Words
The Climb
Second Sons
The Lion and the Rose
The Red Woman
Home
Oathbreaker
The Door
The Winds of Winter

Foretelling the future.

The North Remembers
Valar Morghulis
The Climb
The Wars to Come

Unharmed by fire.

A Golden Crown (I'm referring to Daenarys, and obviously not her brother who died in this episode.)
The Book of the Stranger

Warging.

Dark Wings, Dark Words
The Rains of Castamere
The Lion and the Rose
First of His Name
The Watchers on the Wall
Mother's Mercy - When Jon Snow warged into his direwolf, Ghost.
The Red Woman - When Jon Snow warged into his direwolf, Ghost.
The Door

Birthing shadow demon assassins.

Garden of Bones

Shadow demon assassination.

The Ghost of Harrenhal

Protective cave against Wights

The Children

Glamouring

The Red Woman

Being saved from becoming a wight

Blood of My Blood

This list goes only through the sixth season. (More to come!)


How to create rough/printed looking textures

Question: I want to achieve the speckled black dots around the image and text alongside the overall rough texture on the shapes. Tried using spatter and texture overlays however I can not achieve the desired texture of the below image. Fully aware that this will have been printed and rescanned afterwards
Comment: how about just randomly adding dots with your mouse / pen? It's how it was done before everything was a filter or a button. Sorry for the snarkiness, but I'm actually serious. Sometimes it's faster to just do it than go on the internet to try to find a faster way.
Answer: The basic paper texture can be made using a layer above with any paper-like texture as a grey (rgb 128 128 128), and then set the layer blend to "difference" with a low opacity.
The fuzzy edges from ink bleed, you might try using "color range" selection with a high fuzziness on the ink, dupe that to its own layer directly above the drawing layer and then play with blur and/or texture tools. Then leave that to overlay the original drawing with low opacity.
The sample looks like an actual drawing where the artists' hand and arm was touching the paper and has smeared the ink. You can try using smear tools to imitate this on a local basis on a duplicate drawing layer under the original drawing layer, with a color adjustment. Maybe start with a basic blur tool first.
Comment: Note the one section of the type which is sideways is just using an ink stipple technique. This is not an artifact of the paper/scanning process.


How can I prove that without further assumptions Chebyshev's Inequality can not be improved?

Question: I have found some examples on the web for specific random variables such $X$ (a discrete type)  with probabilities $1/8$, $3/4$ and $1/8$ at the points $x=-1,0,1$ with $\mu=0$ and $\sigma=1/2$. Then, when $k=2, 1/k^2=1/4$ and $\Pr(|X-\mu|\ge k\sigma)=\Pr(X\ge1)=1/4$, so it attains the upper bound $1/4$ and it can´t be improved without further assumptions. I would like to know if there is a way or if you have any ideas to prove that  the inequality cannot be improved without further assumptions about the distribution of any random variable
Answer: Try a discrete distribution with probabilities $\dfrac1{2k^2},1-\dfrac1{k^2},\dfrac1{2k^2}$ at points $-k\sigma,0,k\sigma$.


How to calculate Chern number of a band numerically?

Question: I have a tight-binding Hamiltonian with an aperiodic potential which shows non-trivial topological properties. I wanted to calculate the Chern number of it's bands. How can I do it numerically?
Answer: I hope this is not too late to do some good.  You have many options to compute a Chern number numerically.  There are several real-space formulas and a formula based on scattering theory.  Let me discuss some of the real-space formulas.
The first is called, in physics anyway, the Bott index.  This will work if the following hold:  you are on a regular lattice; the Hamiltonian involves hopping terms only to nearest and next-nearest neighbors; the ratio of the norm of the Hamiltoninan to the gap size is about 10 or smaller.
To compute the Bott index, take a lattice model of about 15-by-15 and impose periodic boundary conditions.  Now take the (diagonal) position observables, scale them so the width and length is $2\pi$, and form exponentials $U_0=\mathrm{exp}(iX)$  and $V_0=\mathrm{exp}(iY)$.  Do a full eigensolve of the Hamiltoninan and find the projector $P$ corresponding to all energy below a certain gap.  Now form invertible matrices $U= PU_0P + (I-P)$ and $V= PU_0P + (I-P)$.
Compute all eigenvalues of $VU{V^\dagger}{U^\dagger}$, take the log of all of these, now sum them up.  The real part of this complex number will be the Chern number (or minus the Chern number).
Be sure to repeat this with a larger system size, to get some evidence you are avoiding the effects of small system size.  As to why, and when, the Bott index equals the Chern number, you can look here:  "On the equivalence of the Bott index and the Chern number on a torus, and the quantization of the Hall conductivity with a real space Kubo formula" by
Toniolo, Arxiv:1708.05912.
A recent decription of the Bott invariant is here:  "Topological Photonic Quasicrystals: Fractal Topological Spectrum and Protected Transport" by Bandres, Rechtsman, and Segev, Phys. Rev. X, 6(2), 011016, 2016.  Sample code implementing this in Matlab is here:   http://digitalrepository.unm.edu/math-statsdata/5/
This invariant will fail you if you have a small gap between bands.  If the gap is about 1/100 of the norm of the Hamiltonian, you will need system size like 100-by-100, and so the dense matrix methods above are too slow and memory intensive.  If you are going beyod the classical band insulator, say working on a quasicyrstal, then periodic boundary conditions will be difficult to define so as to preserve the gaps.  The mathematical version Bott index is twenty years old and not suited to large matrix calculations.
A more modern approach is to use a local index.  For example, there is an index due to Kitaev that was computed as a local index here:  "Amorphous topological insulators constructed from random point sets" by Mitchell, Nash, Hexner, Turner & Irvine, Nature Physics (2018)
doi:10.1038/s41567-017-0024-5.
My take on a local index, which works with sparse matrix calculations and so on rather large systems, is detailed here:  "K-theory and pseudospectra for topological insulators", Annals of Physics Volume 356, 383-416, 2015.  Sample code is also here:  http://digitalrepository.unm.edu/math-statsdata/5/
There are issues in how to rescale the position observables with combining these into a single matrix.  These are just now getting resolved in joint work with Scultz-Baldes available on the Arxiv.
I suggest you try the Bott index unless you have an exotic situation.  Best of luck.
Comment: Thanks a lot for the answer, Professor. I will browse through the resources you attached, specially the Bott Index. I have obtained some strange numerical results with quasi-crystals and that's why I needed to calculate the Chern number/winding number to get a clearer perspective. It's true what you said about the gaps disappearing when I suitably want to impose the periodic boundary conditions.
Comment: Fulga, Pikulin and I computed an index for a quasi-crystal using open boundaries and the newer, local index formula.  I find this preferable than dealing with states where gaps should be.  This is the strong topological insulator mentioned in our paper on weak topological phases: "Aperiodic Weak Topological Superconductors" in PRL, 116, 257002 (2016).  Email me if you need more information.
Comment: @TerryLoring Is there something like a Bott index for a 1D system? I have a discrete finite 1D system that lacks translational invariance. It appears to present edge states, but I would like to calculate an appropriate topological index. Any suggestions? Thanks :)
Comment: I just noticed you posted a question on this.  Let me gather my thoughts and give a proper answer.  There are a few indices you might use.


Linear mixed effects with covariates

Question: I have a group of participants (Subj) all of whom have a covariate measure (Pars). Each participant saw a series of images (Image) that belonged to one of three categories (Stim). To each image, participants made a response (Resp) that was classified as a behavior (Beh). I have combined Stim and Beh (StimBeh), and for each of these Stimulus-Behaviors I have a parameter estimate (Est). I would like to see what the effect of the covariate is in the parameter estimate for the different types of behaviors associated with the different stimuli types, when treating subject and image as random effects. I believe StimBeh should be a fixed effect. I have been unable to make the model converge.
Comment: Does the model with `(1 | Subj)` rather than `(Pars | Subj)` converge ? Also please don't post a picture of output, Post the output itself. How many subjects and images do you have ? And why do you appear to be subsetting the data based on values of the outcome ?
Comment: You might do better on https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models if no replies here.
Comment: I do not understand your nesting (or non-nesting) structure. You say that for "each of these Stimulus-Behaviors I have a parameter estimate (Est)." That implies that StimBehav represents the lowest level of your hierarchy, yet it is repeated, just as subjects are repeated. Can you be more specific about your data, perhaps posting a sample of it for us to look at?
Answer: Combining Stim and Beh is a strange approach. I may be missing something, but I would have incorporated Stim and Beh (or possibly Resp) as covariates. Combining them imposes an interaction without allowing each in the model as a main effect. Finally, if your parameter lies in $[0,1]$, you might want to transform the response.
Comment: Ah of course. I have updated that.
Without covariates the model runs fine, but it still will not converge with covariates. I tried increasing the number of iterations to no avail. Should I conclude that I don't have enough data for the model complexity?

Linear mixed effects with covariates


Is it possible to create a Post and change the author with Wordpress?

Question: is it possible to create a post, logged in as admin, and changed the name from the author? after publishing? or is it possible without login to other users , to create a post, and change the name from the author? You know what i mean?
Comment: You could certainly do it by editing the database. I'm not sure about in the admin UI.
Comment: Are you trying to do this, or are you trying to work out if someone has done this to you?
Answer: Are you trying to do this programmatically? It's certainly possible to be logged in and change the author of any post.  It's even part of the "Quick Edit" menu, or you can go to the post, select "Edit". The post author select menu is at the bottom of every post (or page) in the Classic Editor or in the right column in the Gutenberg editor.


Why is an s-box input longer than its output?

Question: I don't understand where the extra bits are coming from in this article about s-boxes.  Why doesn't the s-box take in the same number of bits for input as output?
Answer: It is the way s-boxes work. They can be m * n  ==>  m bit input , n bit output.
For example, in the AES S-box the number of bits in input is equal to the number of bits in output.
In DES, m=6 and n=4.
The input is expanded from 32 to 48 bits in the first stages of DES. So it is be reduced to 32 bits again by applying one round of S-box substitution. Thus no information is lost here.
The Wikipedia article on itself can be a bit confusing. It will make people think that information is lost. You should read the article in conjuncture with implementation details of some encryption algorithm using s-boxes.
Answer: What extra bits?  They are going from 6 to 4.
EDIT: Whoops! I'm an idiot.  This is kinda like a 2nd grade multiplication table.  They strip the outer bits off of the 6-bit block to be encypted, and leave the middle 4.  Just like a table for an arithmatic operation, they go down one side, and find the outer bit sequence, then across the top and find the middle ones.  To answer your question, it could input and output the same number of bits, but this s-box is just set up to do it the way it does.  Its arbitrary.


Proper use of Rhubarb

Question: What is the proper way to prepare rhubarb as to not poison anyone? What parts of the plant is poisonous? and what general precautions should be done when making food involving rhubarb?
Answer: Use only the stalks; the leaves are poisonous (or at least toxic if eating in high quantity)
Cut up the stalks in chunks from the bottom up to the where it becomes green (to keep the result as pink as possible) and cook down with sugar and enjoy in a crumble or a pie (with strawberries).
I've ate raw stalk dipped in sugar when I was a kid.
from http://www.rhubarbinfo.com/poison:
Precautions for rhubarb gardening
Trim leaves from stalk immediately.
Don't use stalks from frost bitten plants.
Wash the stalks well.
Children should be taught to eat only the rhubarb stalks, preferably under supervision
Answer: My favorite way to eat rhubarb:
is to chop off the top and bottom (poisonous/unappetizing parts)
wash it
chomp it down
There's no real need for sugar as it's slightly sweet if your body needs it, and if it doesn't it still tastes great.
Comment: It is also rather tart - at least in a baking situation, sugar is advisable to balance the tartness....
Comment: depends on your taste buds, and what your body wants. study a little herbalism

Proper use of Rhubarb


Where can I find the log of all mounted volumes?

Question: To be more specific, I mean the file that keeps a history of every disk image/volume ever mounted. I know it exists because I found it once, but it was a long time ago and I have no idea how to find it again. It is similar to the InstallHistory.plist file, which logs every program ever installed.
Answer: You can find the information in /var/log/system.log (or using the Console application). But the information is not kept forever.
For example:
Aug  4 07:34:34 ******** kernel[0]: hfs: mounted External on device disk3
Aug  4 07:34:44 ******** kernel[0]: hfs: mounted Time Machine on device disk5

and and a DMG
Aug  5 07:53:25 ******** kernel[0]: hfs: mounted Xcode on device disk6s2

After a while system.log is compressed and archived. To search in all the stored logs
cd /var/log/
(sudo cat system.log; sudo gzcat system.log.*) | grep mounted

Comment: I feel like it was specifically a list of DMGs that had been mounted, not external volumes.
Comment: In my system log I also find external volumes
Comment: Me too, that's why I don't think that's the list I remember. I remember it showing just the DMG files that had been opened and mounted. Maybe some sort of Disk Image Mounter log or something.


understanding tilda,slash command in bash

Question: I have confusion in the meaning of command ~/
if x is a directory, then what is the meaning of ~/x in bash.
Please guide me
Answer: Unix shells commonly, and bash does, expand ~ to the path to the current user's home directory. This path is often also available as the HOME environment variable, so ~ and $HOME are normally equivalent.
So when you give a path ~/x, you are saying "the path to my home directory, followed by /x". You could also write $HOME/x and get the same result.
If your home directory is at /home/umar, then this expands to /home/umar/x, which is a valid path. If x exists in your home directory, the pointed-at file or directory also exists and can be used (assuming sufficient permissions, which you will normally have to anything in your home directory).
Answer: ~ is a path relative to the user, and represents the logged on users home directory root (or more specically, the path indicated by the logged in user's environment configuration). if I run a standard debian system as the user Bob, then my profile/home exists by default at /home/bob. In this scenario, ~ would evaluate as /home/Bob. ~ is both a convenient shortcut, and also a means to abstract away the root location of the users profile, if it is somewhere unpredictable on the filesystem.
One benefit to using ~, is that you can write scripts that execute the same way for any user regardless of what their username is or whether their profile is stored in the default location. By abstracting away the actual file system paths to the user profile, scripts can be portable across systems that handle profiles differently, or allow a user to move the profile to a non-standard location.
Comment: "the logged on user's home directory" is not necessarily completely accurate. Consider for example the case of logging in, then using `sudo -i`. One could argue that the logged on user is still the original user (whoever is at the keyboard) whereas the *current effective* user is that of (almost certainly in this simple case) root.
Comment: `sudo -i` is a special hack designed for SU operations specifically. if you run `sudo -s`, you will observe the default behavior, where `~` switches to /root (at least on debian systems). Sudo tries to preserve the real users environment info, which is not really normal, but necessary in that narrow circumstance. added commend about environment to address your concern.


"Such as me" or "such as myself" — which one is correct?

Question: Which is correct, "such as myself" or "such as me"?
Comment: It entirely depends on what you are saying. Many contexts would allow for either, but if you can give us an example of a sentence we can comment more precisely.
Answer: Either. Such as myself is perhaps more emphatic.
There are people who insist on such as I (on the specious justification that the phrase represents such as I am). Hardly anybody says this, but it is always acceptable in formal contexts.
Answer: "Myself" is a Reflexive pronoun, so it should be used as a Direct or Indirect Object. The wording of "Such as myself" specifies a non-reflexive pronoun, so we can narrow it down to "Such as I" or "Such as me."
"Such as I/me" is an abbreviation for 'Such a person as I am." So, in order to choose between Such as I and Such as me, we should change it to the long version to check.
"Such a person as me am." That doesn't work at all.
"Such a person as I am." That does work. Therefore, you shouldn't use either of those, but instead use "Such as I."
Comment: Nonsense. This argument was always false: there is no reason whatever to regard _such as me_ as an abbreviation for anything, since _as_ is a perfectly good preposition. The most you can say is that there are some people who will insist on _such as I_.
Comment: I'm sorry, I should have worded that as, "if you regard Such as I as an abbreviation." This is because the origin of Such as I is Such a person as I am. Even if it weren't (but I believe it is) Such a person as I am is still a valid replacement for Such as I, just a bit longer. This is why I made that argument.
Comment: Yes, but the question was about _such as me_ and _such as myself_, which are a different construction. Most of your answer is irrelevant to the question, and ends up with a bald statement which is wrong.
Comment: I see. I concede. I'll do some more research on this and possibly get back to you.


Importing wallets to cardano-node

Question: when I'm upgrading cardano-node and cardano-wallet, I saved the wallets (.sqlite, sqlite-shm, sqlite,wal files) created in the node (cardano-node/data folder). After upgrading to the new node 1.35.3 I paste the wallet files to the same directory, but the cardano-wallet wallets list command isn't able to recognize the files. What should I do?
Answer: Make sure your cardano-wallet is pointing to the right database path by passing --database /path/to/your/db
Generally speaking you should manage cardano-wallet in a 'daemonized' manner (e.g. using systemd), this way you update your binary and restart the daemon, leaving the configuration untouched.
Answer: If you still have the mnemonic and the passphrase of the wallet, you can recover it with cardano-wallet.

Importing wallets to cardano-node


Latin Modern vs cm-super?

Question: I've read various answers to questions on here that suggest the original Computer Modern fonts have been superseded. Sometimes it is suggested we use the cm-super package, sometimes that we instead use latin modern fonts. Which is better? Which is easier to get working? (getting new fonts to work in LaTeX seems a daunting task...) Which look better?
Answer: Latin Modern is derived from cm, cm-super also.

cm-super is a vectorization of cm bitmap fonts but mainly automatically done.
The vectorization of Latin Modern is handmade.
Latin Modern has revised metrics.
Latin Modern provides more glyphs, especially diacritical characters. There are 72 text fonts, each of them containts more than 700 glyphs. In addition there are 20 math fonts. cm-super comes with more fonts but fewer glyphs, check its README file.
Latin Modern development goes on.

Update: Now there's also a Latin Modern Math font, available since July 2011.
Comment: I read that the vectorization was through MetaType1, which involves a METAPOST macro package, some \*nixy scripts, and changing the sources to *use* the macro package. Now, that does involve some manual labour, but surely "the vectorization is handmade" is a bit of a stretch? I mean, it's not like they hand-vectorized each glyph of each shape in each optical size -- they adapted the font code to (essentially) vectorize itself, and perhaps did some manual touch-ups afterwards, as I understand it.comments may only be edited for 5 minutes(click on this box to dismiss)
Comment: Will the LM Math fonts be included in the `lmodern` package or a separate package?
Comment: @You — if you look inside `lmodern.sty` you can see it includes code to set up the maths alphabets. Because they're a drop-in replacement for the default CM fonts, it doesn't take many lines of code.
Comment: Is there such thing as `Latin Modern` font to set? So far I found that this works `\usepackage{fontspec}\setmainfont{Latin Modern Math}` but doesn't sound right to use a Math font for main font. Does it?
Answer: In addition to what Stefan wrote:

Both are easy to use in any modern TeX distribution. The distributor has taken care of this for you
cm-super takes up more space in your distribution, but in my testing lmodern generally takes up more space in your generated PDF. This is because lmodern uses font subroutines, which all must be put in the PDF, even if you only use a subset of glyphs
lmodern is available in the same optical sizes as computer modern, cm-super is available in more optical sizes (this 'feature' of cm-super switched is off by using the fix-cm package (highly recommended))
With cm-super it's recommended to use the fix-cm package to fix a lot of broken design decisions in cm-super (and in addition this makes the final PDF a bit smaller)
Neither cm-super or lmodern are as well-hinted (for onscreen reading) as the bluesky computer modern fonts. Hinting is simply too hard to get right and the lmodern developers don't focus on it. Both cm-super and lmodern each have their hinting flaws, and I'm not sure which I prefer
As I understand it, neither cm-super or lmodern will give exactly identical page breaks as the traditional computer modern, but lmodern will be closer
Comment: The bluesky computer modern is the standard PostScript type 1 conversion of computer modern, that all(?) modern TeX distributions should use when producing PostScript or PDF output. They were professionally hinted by Y&Y under contract to Blue Sky, and originally sold commercially (free TeX distros used the inferior bakoma fonts) until the AMS bought the rights and released them to the TeX community. See http://www.math.utah.edu/~beebe/fonts/bluesky.html
Comment: what is this bluesky computer modern you mention?
Comment: The [type 1 spec](https://www-cdf.fnal.gov/offline/PostScript/T1_SPEC.PDF) tells about subroutines. This [latin modern paper](http://www.tug.org/TUGboat/Articles/tb27-0/jackowski.pdf) describes how latin modern uses them.
Comment: I'm no expert, but I did a little digging - pdftex in TexLive 2010 will indeed use the "blue sky" version of `cmr` by default (located in `texmf-dist/fonts/type1/public/amsfonts/cm/`). However, upon choosing `[T1]{inputenc}`, the implementation defaults to `cm-super`.
Comment: @MattB the `cm` fonts don't exist in t1 encoding (see above), so latex is set up to switch to `cm-super` if you ask for t1 `cm`.  i use lm for any actual document i want to look knuthian, and largely ignore `cm`/`cm-super`, nowadays
Comment: What are these 'font subroutines'? Certainly LM subsets in PDF. To be honest, I don't understand what these even *could* be, if we're talking about type1 fonts. Really mystified. (I assume you're right, since nobody has said otherwise in 7 years, but that doesn't make me any less mystified.)
Answer: If you're saying "Meh, all these differences don't really seem to matter to me", here's something that might be relevant for you, especially if you're writing in German:
cm-super's sharp s ⟨ß⟩ is ugly imho. Latin Modern replaced it with the original Computer Modern one, which is prettier. This is actually an easy way to tell the two apart – if there is a ⟨ß⟩ in the text you're looking at.
Edit: Following Philippe's comment about accents, I added some to the picture. Note that this now is not a street name anymore that you're likely to find in Germany ...

(both at 12pt, zoomed in)
Comment: The ß of cm-super is more a ligature of long and short/round s: ſs → ß, while the ß of lmodern is more a ligature of long s and _curved_ z: ſʒ → ß.
Comment: Another easy way to tell apart cm-super and lmodern is the accents. In cm-super, accents on `é` or `è` are badly positioned (they seem to be falling off the letter) and less slanted.
Comment: Indeed; I think I remember reading that the cm-super ß glyph was controversial since it didn't follow the original CM design. It does seem like an odd choice because an innocent `\usepackage[T1]{fontenc}` line would give the obviously different output.
Comment: Do you have any idea why they changed it?
Comment: @doncherry — no idea, it was before my time, so to speak.
Answer: One more addition: cm-super contains Cyrillic fonts and lmodern doesn't.
Comment: As the word "Latin" in "Latin Modern" hints ;-)
Answer: As Lev Bishop explained, both cm-super and lmodern have problems with hinting; the PDF files may look bad in some PDF viewers.
Therefore I'd suggest that you consider the third option as well: \usepackage{ae}. Looks good on screen and works everywhere without installing anything new.
Comment: ae is not a good choice. If you use non-ascii characters copy & paste will be broken in the pdf. If you don't use such chars than you don't need T1 encoding and the ae fonts.
Comment: you don't get eth and thorn (icelandic), or the various guillemets, but if `ae` covers your needs, it is a good choice.

another possibility is the 'european computer modern' commercial fontset from micropress, inc. Price is about $150. Don't know what the quality of the hinting is like.
Comment: @Ulrike: Yes, copy & paste will be broken, but at least the PDF file *looks good*, both on screen and on paper.
Answer: The Latin Modern fonts are terribly hinted (or maybe not hinted at all)? This is LMRoman12-regular on Windows 7:

That ain't right.
I have managed to get a decent Computer Modern style font with hinting in a modern format! Here's how:

Download Latin Modern.
Open the Latin Modern .otf fonts in FontForge (you must run FontForge as admin for some reason; obviously high quality code... /s).
Go to Element -> Font Info -> General and set the Em size to 1024.
File -> Generate Fonts
Format: TrueType
Uncheck "Validate Before Saving" (Latin Modern has some errors apparently).
Save the font.
Download Freetype's TTF autohinter: https://freetype.org/ttfautohint/doc/ttfautohint.html.
Autohint the TTF with it. I used all the default settings.
Behold!

Before:

After:

I am really impressed with the autohinting. I have not converted all the fonts yet, or checked out how to make bold/italic/slant/whatever work.
I put all the fonts converted by this procedure at http://concentriclivers.appspot.com/LatinModernTTFHinted.7z [broken link].
This is the best way to use latin modern in Microsoft Word.
Comment: So can you get this into the cm-super package? Or, is this not a good idea to always do?
Comment: Any chance to have the auto-hinting in LaTeX's lmodern package? Or is this less of a problem with LaTeX?
Comment: Yeah I never noticed this as a problem with LaTeX. No idea why.

Latin Modern vs cm-super?


Website using different fonts across different browsers

Question: The navigation board items on my website at www.amsa.mn uses the wrong font when displayed with Firefox. The correct font is loaded when I access the website with Safari or Chrome.
Is there any way to force Firefox to load the correct font?
Comment: I checked the site using Chrome and FF, and the nav font looks pretty much the same. Are you sure its working for Chrome as well?
Comment: The same for me - I see the same font in both Chrome and FF. However, I suspect I'm seeing the wrong font in both browsers. Can you supply a screen shot of the preferred output? Glancing over your CSS I can't see where you are loading the font (I might have missed it)? There is a reference to "Arimo" - but I don't have this on my machine, so I assume it's defaulting to sans-serif for me?
Answer: You need to convert the "Arimo" font to a web-font with a converter like http://www.fontsquirrel.com/fontface/generator. Your CSS references the Arimo font. Since the font is not available via CSS, it's attempting to load it locally. Most people won't have Arimo installed, so will see a default font instead.
Comment: Whilst I agree with your answer, how does this explain why Chrome displays the font OK, when Firefox does not? (I'm assuming the OP is testing these browsers on the same machine.)
Comment: I'm not sure. It may have something to do with the fact that he's using the font tag. For some reason, FF is omitting the font tag. Chrome isn't. Now, that doesn't make a ton of sense, but it's the only thing I could find that's different between the two.
Answer: Looks fine when I check on firefox, I guess you have already found a solution? If not...
It is probably a declaration issue, or not having the correct file type of font that Firefox uses.
As NickWebman answered, go to somewhere like fontsquirrel.com, into their font-face section and either find the font you want and download it or convert one of your existing fonts.
Make sure to create/download all the different file types so your font will display in all browsers as the core browsers tend to prefer different file types - .woff, .ttf, .svg, .eot etc
TrueType Fonts for Firefox 3.5+ , Opera 10+, Safari 3.1+, Chrome 188.8.131.52+
EOT fonts for Internet Explorer 4+
WOFF fonts for Firefox 3.6+, Internet Explorer 9+, Chrome 5+
SVG fonts for iPad and iPhone

One solution for Internet Explorer is to seperate it from the normal declaration e.g.

@font-face {
    font-family: 'ProximaNova';
    src: url('../fonts/ProximaNova.eot?') format('embedded-opentype');
    font-weight: normal;
    font-style: normal;
}


How do I wipe my computer completely clean and remove ubuntu?

Question: I have decided to switch to Windows. I would like to completely remove ubuntu, and all the other files on my computer. How do I do this? I basically want to wipe my computer clean.
Answer: formatting does not delete the data, you need to write data to the drive. Boot a live CD and either write zeros to the hard drive with dd or use a utility such as shread
dd if=/dev/zero of=/dev/sda iflag=nocache oflag=direct bs=4096

or use a utility such as scrub
scrub /dev/sda

http://linux.die.net/man/1/scrub
See Securely erase hard drive using the Disk Utility
https://wiki.archlinux.org/index.php/Securely_wipe_disk
You then format the hard drive and install your os of choice
Answer: That would be reformatting the drive.  You can probably do this automatically when installing windows.  Otherwise, you can also use a utility program do do it ahead of time.  You could boot an OS from a live DVD, to get a running OS without having your harddisk mounted, and then use a disk utility from that live DVD's OS to nuke the the disk.  Better make sure you don't have any data.
Comment: formatting does not delete data
Answer: You could try to run a live CD of any OS, like the Ubuntu one and use the Disk Manager to "format" the partition again.
Comment: formatting does not delete data
Answer: If all you're after is clearing the existing data so it won't interfere with a clean install of Windows, creating a new partition table and partitions along with formatting the partitions is sufficient; Windows installers at least up through 7 would do this automatically if you answer the prompts that way.
If you're concerned about ensuring data on the existing partitions can't be retrieved, you need to use a "data shredder" or "mil-spec erase" utility; in my experience, these are often bootable Live CDs with a limited Linux OS and (often auto-running) utilities tailored specifically to erasing data.  I have one called "Boot and Nuke" that boots from CD and erases all hard disks attached to the system with a secure erase.  Due to the nature of secure erase (overwriting all of the data surface of the device, many times, with patterns and random data) this process takes a long time (overnight, at least, with modern terabyte class drives).
Unless you're trying to remove evidence of a crime (doing so is a crime in itself!) or national security related data, this is probably an unnecessary length.  For most purposes, it's quite sufficient to boot from the Windows media and tell the Windows installer that you want to erase the hard disk and partition it from scratch.
Answer: Simply boot from the Windows installation media and follow the prompts to use the entire drive for your new installation.
Answer: I will help you to wipe your hard drive such that it is completely indistuishable from a new hard drive.  Asking us how to reinstall Windows is "off topic" for this site.  Please rephrase your question to omit that aspect of your request.
The super simple way to wipe your hard drive to like new status (no partitions at all), is to boot your computer to literally any ubuntu Live Boot boot (media) CD/usb-flash.  It doesn't have to be remotely up to date, so whatever you installed ubuntu with should be more than adequate.
Once your computer is booted to a live boot CD you merely open one program already on the live boot media.  It is called Disk Utility, and it will do exactly what you want to do with absolutely no thinking or grief.  You don't even have to open terminal!  Detailed step-by-step instructions below for both opening the utility and what to do once you have it open, but my guess is that you could adequately wipe your hard drive with that utility even without the following instructions.
Dumbed down usage instructions:

Open Disk Utility (press the windows key on your keyboard and immediately type DISK and then press enter)
First select your hard drive from the left side menu within that window.
Then maximize the window so you can see all its options.
Choose the lower right option called FORMAT VOLUME.
In the window which appears, change the selection in the drop-down menu next to "TYPE" to EMPTY.  (please don't try to give it a name.  You cannot since there is no partition there to name)
Press format, and the drive will be completely wiped, devoid of data, partitions or even partition tables.

This type of erasure is not quite the same as SHRED nor is it as time consuming.  If  you turned your drive over to the CIA they could probably recover all your data before sending you to a secret prison... or perhaps afterwards, I won't speculate.  If you want that type of file destruction and have the time, I recommend shred instead.  The method I've described is much faster and will ensure that windows will install correctly.
One last item I'd like to mention.  When you first open Disk Utility, it will offer you a peek into the "SMART" status of your drive, and can tell you if your hard drive is healthy.  All operating systems which install to hard drives rely upon the hard drive working correctly.  If you have had an unhappy experience with Ubuntu, there's a good chance the cause was actually that the drive is sick.  In which case, Windows won't work right either.