Could a habitable gas ring be possible around an Earth-like world?

Question: I have been reading through Larry Niven's The Integral Trees, which concerns a gas ring around a neutron star with a habitable center that humans became stranded on., The atmosphere of which was stripped from a nearby gas giant. I was wondering if such a gas ring could exist around a world that is just like earth aside from there being no moon.
How could this come about? I was thinking a series of ice comet impacts could have blown part of the atmosphere away from the planet where it became locked into orbit, meanwhile the planet's atmosphere was replenished due to the water from the comets.
This does seem unlikely but thought I would ask.
Comment: Ring systems, as I understand it, are generally believed to form by the breakup of one or more small satellites, which may initially be captured. Look up *Roche limit*. (We have plenty of questions and answers involving that even right here on the site.)
Comment: Not a duplicate - drastically different science due to planet size.
Answer: Mr. Niven's book allowed us to suspend our disbelief because of the gravitational pull of the neutron star.  An horrendous amount of matter could be held in orbit around the star — enough to permit the idea of a habitable jelly-filled donut with the star at the center.
Regrettably, you have two really big cons for a planet the size of earth to have such a thing.

We're probably not be big enough.  A simple gas within the ring would simply bubble off into space no matter how much there was.  Give the ring enough mass to hold the atmosphere together (you know, like a jelly donut...) and you might have enough competing gravity to simply yank the ring to the ground.  That's why the neutron star was such a brilliant idea.  There was so much gravity in play the ring was believable.
The bigger problem, however, was one Mr. Niven didn't need to worry about: the Van Allen Radiation Belts (why?  Because (a) a neutron star's magnetosphere is massive and (b) a neutron star isn't spitting out solar wind like our boring G-type star).  That protective aspect of our magnetosphere permits life on Earth at all.  Those belts are a whole lot closer to the Earth than the moon, and a whole lot closer than would permit a habitable ring system.  This is where I just ache to say it, but without those rings, you can have all the atmosphere you want and the ring still won't be habitable.

HOWEVER Please do not give up on the idea!  I believe you can overcome the problems with some adjustments to your planet.

Make your planet larger than earth.  Indeed, make it as large as astonomers believe a habitable world can be.  The larger the world, the more substantial the ring system, and the larger the radiation belts.
Modify the core such that a stronger magnetic field is generated.  Make it more dense, or a higher percentage of iron, or inject the core with an unusually high oxygen component to induce greater ferrousnessness... ness...  Or have the core spin faster (which would even be believable with a larger world).  Anything to justify a larger magnetic field.
Finally, you might consider really dense "rocks" as part of your ring.  Something to exert a bit o' gravity to hold the ring together and keep the atmosphere from bubbling off into space.  Make the rocks magnetically repellant so they create a shield around the jelly donut with the atmosphere at the center.

And if none of that works, consider a habitable moon around a gas giant sporting a habitable ring between the planet and the moon.
Comment: I had not considered the idea of moving the setting to a habitable moon. Really love the idea. The sky of the world would look cool, having both a ring and the giant when facing it.
Answer: Nope. Gas torii are real things (there's one around Jupiter created by Io, and Enceladus produces one around Saturn), but they are wispy things. They are the result of gas being ejected or stripped from an orbiting body, so right away we can say that no such thing could exist around an Earth with no moon, because you need the moon to create the ring! Stripping gas from a moon produces such a ring because the individual gas particles remain in orbit around the primary, pretty close to the orbit they started in--that being the orbit of the moon itself!--with the relatively minor perturbations caused by whatever stripped them from the moon causing them to spread out into a torus.
There is, however, no restoring force holding those particles together as a coherent ring against internal pressure. They persist only due to a combination of being so thin that any individual gas particle is highly likely to complete a whole orbit without actually bumping into any other gas particle, and being continually replenished by the source moon. If a ring were thick enough to have noticeable pressure in the middle, it would simply blow itself apart.
Note that Niven's Smoke Ring is fed by the continual loss of gas from Goldblatt's World, which is stripped off by Voy's (the neutron star's) tides. It takes a huge continuous input of new gas to maintain it. Even so, the Smoke Ring is not particularly plausible itself. Niven gets away with it because the story is good, the Rule of Cool is in operation, and the environment is exotic enough that the average reader can't immediately recognize it as implausible. After all, who knows what can happen in the extreme environment around a neutron star stripping gas from a close gas giant? In reality, what should happen is that you get a hot, violent accretion disk, which spirals inwards due to loss of energy to internal friction from tidal shear--not a cozy, habitable Smoke Ring at all! Try to put it around an Earthlike planet instead, and familiarity kills your suspension of disbelief.
You could, however, go with an artificial gas torus, held in by a thin, solid, artificial membrane, with active stationkeeping. That's the approach taken by the Orion's Arm worldbuilding project, for example.
Answer: Small gas cloud of near atmospheric pressure in vacuum dissipates almost instantly. Why Niven's torus does not? Even though it is huge, it's self-gravitation is too weak to hold it together - gas has low density. It is held together by tidal forces from neutron star.
A planet would be several orders of magnitude lighter than star, and it's tidal forces would be accordingly smaller. So it would be able to hold torus with 1/1000 pressure of Niven's torus at best.
Answer: A gas ring is possible, but would be most likely in a highly reactive environment. Perhaps caused by high levels of meteoritic bombardment, extreme volcanism or similar. A major problem would be reactive elements such as oxygen reacting with other elements in the vicinity especially hydrogen. So yes to the gas ring but no to an oxygen rich gas ring, unless you can invent some form of special pleading.


Why no margin in FireFox?

Question: http://dl.dropbox.com/u/18795563/BLOGDESIGNNEW/index.html
In FireFox, Why is there no margin on the left of the hgroup in the header? It works fine in Chrome and Safari. How do i make it work in FireFox?
Comment: Chrome and Safari use the same rendering engine, so it's not as if it'd be two versus one.
Answer: This looks like a margin to me.
Answer: after inspecting the html code of your page, I figured out the following issue in #topheader hgroup css:
overflow: visible:; Remove the column from the end of visible. i.e. overflow: visible;
Answer: Which version of Firefox? Firefox4 DOES show the margin but FF3.x may not because it does not have built-in styles for the new HTML5 elements. You must add 'display:block' for proper rendering.
Answer: I believe the problem is that the hgroup's default display setting is inline. Setting it to block will fix your problem with the margin.
#topheader hgroup {
  display:block;
}

Answer: Margin does not work in most case. and that is becaus it should calculate distance with elements outside. and this is dificult.
you should try padding instead of margin:
#topheader hgroup {
padding: 1em;
overflow: visible;
}

also change : to ; ;)
this will work in every browsers.

Why no margin in FireFox?


Quasars to a galaxy

Question: Do quasars collapse and make a galaxy or do they eventually spew out enough matter to make one? Because I heard they are the center of galaxies, and I wonder if they can collapse from their own gravity or from becoming unstable, and if so, what would happen?
Comment: This video from PBS Space Time might help - [Why Quasars are so Awesome](https://www.youtube.com/watch?v=3TZEp_n3eIc)
Answer: A quasar is a name given to a particular state of activity of an active galactic nucleus - characterised by high levels of accretion onto a supermassive black hole.
i.e. Quasars are not "objects" in their own right and if the accretion stops then so does the quasar-like activity and you end up with a more dormant galactic nucleus, like the one in our own Milky Way galaxy.
Thus quasars do not "make a galaxy", although there are interesting but poorly understood correlations between the properties of galaxies, their supermassive back holes, their history of mass accretion and the properties of the galaxies they reside in. They thus probably play an important role in galaxy formation and evolution, principally through the injection of momentum and energy from their winds, jets and radiation.
There is also no sense in which they can collapse. A supermassive black hole is as compact as it can be. It can of course continue to accrete matter and get bigger, but most supermassive blackholes have a mass that is only a small fraction of the mass of the galaxy they are at the center of.
Comment: Thank you for clearing that up, quasars are a new subject to me.

Quasars to a galaxy


Does temperature affect nuclear decay?

Question: I'm wondering if the temperature could affect the half-life in an element. For example, Carbon 14 has a half-life of 5,730 years. Is this always true or only true for standard conditions of temperature and pressure? Also, if that does affect, how does it affect?
Comment: Temperature is not very meaningful for atomic nuclei. This is because a nucleon *can* be in an excited state (just like an electron orbiting an atom can be excited to a higher orbital), but the energy of this excitation corresponds to at least 10^7K, normally even much higher (google nucleon resonances for more info). Therefore, for most practical purposes a high temperature will not affect the nucleon too much. But if the temperature *is* >10^7K, all sorts of nuclear reactions begin to happen, as they do inside stars.
Comment: As I said, since the minimum excitation energy for a nucleon is in the ballpark of millions of kelvin, there is virtually no observable difference for nucleons at 1, 10, 100, 1000 or even 10000 K. Only an exponentially small fraction of them (e.g $e^{-1000}$) will be in an excited state. $e^{1000}$ is a huge number, orders of magnitude more than the number of atoms in the universe. So far all intents and purposes - zero temperature. (cont'd)
Comment: @LLlAMnYP And what about very low temperatures, close to zero kelvin?
Comment: I can imagine that *maybe*, excited electronic orbitals with their altered EM field interact ever so slightly with the nucleus, but the general behavior of a nucleus is governed by strong and weak nuclear forces, whereas this interaction is electromagnetic, so also not much of an avenue to change things.
Answer: Temperature affects the half-life via time dilation. If the half-life of a nucleus as observed in its rest frame is $\tau(0)$, then in some other frame in which it moves at a speed of $v$, it will be:
$$\tau(v) =\frac{\tau(0)}{\sqrt{1-\dfrac{v^2}{c^2}}}$$
Because the typical thermal speeds of atoms at room temperature are much smaller than the speed of light, we may expand the above expression to leading order. This yields:
$$\tau(v) =\tau(0)\left[1 + \frac{v^2}{2c^2}+\cdots \right]$$
The average of the square of the speed over the velocity distribution of atoms at temperature $T$ follows from the equipartition theorem. The average energy in each degree of freedom of a gas is $\frac{1}{2} k T$, this means that the average of the kinetic energy is $\frac{3}{2} k T$, it then follows that:
$$\left\langle v^2\right\rangle = \frac{3 k T}{m}$$
Therefore, the leading temperature dependence of the half-life is given by:
$$\tau =\tau(0)\left[1 + \frac{3 k T}{2mc^2}+\cdots \right]$$
For carbon-14 at $25 ^{\circ}$ C the term $ \dfrac{3 k T}{2mc^2}$ is $2.95\times 10^{-12}$. Since the experimental error in the half life is about 0.7% this effect is too small to be observed.
Comment: Very interesting. So, if I understand correctly, any stationary object can suffer time dilation via (extremely high) temperatures?
Comment: Nevermind: https://physics.stackexchange.com/questions/258278/does-heat-cause-time-dilation
Answer: Pierre Curie checked this out in 1913 by cooling radium in liquid hydrogen and found that the half life was the same as room temperature...radioactive decay is independent of temperature. I am seeing other answers that provide more detail, but this I would say is the simple answer.

Does temperature affect nuclear decay?


Switch a condenser dryer from drawer to pipe

Question: How does one switch a condenser (ventless) dryer such as the Electrolux edc78550w from collecting the water in the drawer to pushing it through the back pipe?
Comment: Did you ever get this resolved? If so, please give a check-mark to the answer or write up your own answer explaining what you did to get it fixed and give yourself a check mark. That will help others with this kind of problem know that this has a resolution and is a good place to look for their answer.
Answer: I've done this before.  On the back of my machine there was a drain plug and I simply attached a hose to this and ran the other down the drain in the floor.
On my machine the drawer was located at the top of the machine so there was actually a little pump that pumped the water from below up to the drawer. If you don't have a drain in the floor you might have to use a hose going from this pump.
Comment: Thanks. I see a red pipe coming out of the back of the machine close to the top. How does the dryer know when to pump through it, and when to use the drawer? In other words, why does it not always drip through it? (I don't have enough reputation to upvote).
Comment: On my machine it was very easy to take off the various body panels to see what was going on. The main point is water doesn't go uphill, so depending on where your drain is you need to tap in to the system so you can get rid of the water. I think all dryers have a drain plug low down somewhere in on the machine, so if you have a drain in the floor or can raise the machine a little you could also use that. I had a drain in the floor so I used that one because it was the easiest to access and it meant that there was never any water for the pump to send to the tray.
Comment: Thanks. But I do have a drain. The question is - the dryer doesn't know whether or not I attached a pipe to the short pipe protruding from the top of the machine. So how does it know whether to push water through that pipe or not?
Comment: The condensate pumps usually have a check valve so once they pump the water is out the tube, a sensor or a float activate the pump once the water level drops the pump shuts off. The check valve holds the water in the pipe but worse case it drains back into the pan and continues to fill until the switch turns the pump back on.


is it possible to use copper coil instead of pads in air cooler

Question: The pads are used in air coolers to cool the air but make it more humid.
Is it possible to use copper / aluminum coils instead of pads to get cold and dry air?
Comment: Can you provide photos/links/generally more information? It's difficult to know exactly what you're talking about...
Comment: am talking about Evaporative cooler
like this:
https://www.breezair.com.au/wp-content/uploads/2016/01/Breezair_FEAC-Diagram-web-800x349.jpg
Comment: vote to close. insufficient prior research, unclear what OP is asking.
Answer: Copper or aluminum coils are used not in evaporative coolers (which add humidity); they are used in heat pump-style coolers that have compressors, pumps and freon refrigerant (and subtract humidity). These two types of coolers are completely different in how they work and their cooling components cannot be interchanged.
Comment: i know copper coils are used in air conditioner, but why is not possible(or possible) to use them in  evaporative coolers instead of pads(wood wool or honeycomb), in order to prevent adding humidity. when air hit the circulated water in the copper coils, the air exist become cooler with no humidity change
Comment: OK- I understand you now. Here's an explanation: when you cool down air with cold refrigerant running through coils, the moisture in the air condenses onto the coils if they are below the dew point temperature, and it runs off them for disposal. The same thing would happen if you ran cold water through the coils. the problem then becomes this: how do you make that water cold in the first place? One way to do it is by spraying water into the airflow (this is more or less what an evaporative cooler does)  then the water cools down by evaporation . on to next comment-
Comment: ... and extracts heat from the air both because of the water's heat capacity and because of its heat of evaporation. But unless the water feed is ice-cold, the cooling process will be dominated by the heat of evaporation and not by the coldness of the feed water. Therefore: since heat transfer is better if the water flow simply runs through the incoming air (as it does in a wetted pad) than it would be if it were inside a metal pipe, evaporative coolers use recirculated water in direct contact with the air.
Comment: thermodynamically, the evaporative cooler can be analyzed for effectiveness in comparison to a refrigeration-type cooler; this exercise can be found in any HVAC textbook.
Comment: How do you make this cold water in the first place?
I do not want to make the water cold, just circulate the water in copper coils using a pump (to lift it from the tank)
When the air hit the copper coils, the water absorbe the heat of air (evaporation), making the air more cooler and the rest of the water in copper coils is become more cooler
Answer: Reverse cycle air conditioners, heat pumps by another name, are closed system coolers whereby heat is transferred form one side of a divide to another, ie from inside a room to outside the building, when cooling.
Evaporative air coolers operate on a totally different principle. Water is evaporated and dispersed into the air to cool the air by absorbing heat. In doing so they increase the humidity of the air inside the room. Because of this, such coolers are not suited to environments with humid air, such as tropical regions and coastal regions. They are best suited to areas with dry air, such as deserts.
If the pads of an evaporative air cooler were to be replaced with a coil, as you ask, there is nothing to evaporate. The only way for such a system to work is to turn the evaporative air cooler into a refrigerative air conditioner (also known as a reverse cycle air condition or a heat pump). This defeats the purpose of having an evaporative air cooler. You may as well just acquire a reverse cycle air conditioner.
Comment: "If the pads of an evaporative air cooler were to be replaced with a coil, as you ask, there is nothing to evaporate." why!? why the water inside coils can't be evaporating?
Comment: Sure it can, theoretically, but then the whole cycle changes, as described above... in practical use, however, it wouldn't all evaporate when inside a pipe, which is why other fluids are used as refrigerant most commonly


Can you marry other players?

Question: My friends and I are doing a server roleplay and we have the 1.12.2 version of MCA. Me and one friend want to marry each other but we don't know how to. Can you marry other players? If so, then how?
Answer: According to the MCA github to get married you use the command <code>/mca propose <PlayerName></code>.
Comment: @Joachim the tags on the question states that this question refers to the mod Minecraft comes alive, I looked at the GitHub as I could find a reference to a proposal command but couldn't find what that command was
Comment: You're right, didn't catch that immediately :)

Can you marry other players?


Disadvantages of using same key for encryption and decryption in symmetric cipher?

Question: In a symmetric cipher, the same key is used for encryption and decryption. What are some disadvantages of using the same key?
Comment: As you say, all symmetric encryption algorithms use the same key for encryption and decryption. Therefore, are you asking for disadvantages of symmetric encryption in general?
Comment: "symmetric" means same key for encryption and decryption
Comment: Can you clarify what you are asking? Sometimes unclear questions get closed on stackexchange. Don't fuss about it. Just update the question and it will get reopened
Answer: The disadvantage of symmetric encryption
Symmetric encryption always use the same key for encryption and decryption - that is the very definition of it. That has one major downside.
If the person doing the encryption and the decryption are not the same, they have to somehow securely share the key. If A generates a random key and encrypts a message for B with it, how does he get the key to B? To do this securely, he has to transmit the key out of bound, or encrypt it with B's public key using asymmetric encryption.
Obviously asymmetric encryption does not suffer from this disadvantage, since B can freely share his public key with anybody without loosing any confidentiality.
The disadvantage of asymmetric encryption
One word: Performance. It is slower than symmetric encryption. Therefore it is in general just used to encrypt a symmetric key that is used to encrypt the rest of the message.
Answer: The main disadvantage of using a shared key in encryption is that you cannot use it to ensure non-repudiation. If you got a message and you are able to decipher it, there is no proof that the sender did encrypt it, because one can still argue you encrypted it yourself.
Answer: This is fairly easy to understand. Simply symmetric encryption use single key for encryption and decryption while asymmetric use different keys (4 different keys).
Main disadvantage of symmetric key is the damage, if someone get your key. He can read your messages(sent/receive). Because you use the same "compromised key" for encryption and decryption.(Read both sent and receive)
But in asymmetric key sender(X) and receiver(Y) has two different private keys for decryption. Let's assume someone take the 'X' private key. So he can read the messages sent by 'Y'. But still key can't read messages send by 'X'. Because for that he needs to know the 'Y' private key.(Read only sent or receive)
Summary: damage is massive when you lost your symmetric key. In Asymmetric, you need to know the both private keys to do the same damage.
Other point is there is no way to safely share the secret key with both parties
Comment: I disagree. This disadvantage is not inherent in the use of symmetric keys and can be mitigated, with the following key usage protocol:
Alice always encrypts with symmetric key Key1, and Bob always encrypts with a distinct symmetric key Key2.
Then in the event of compromise of Key1, only messages sent from Alice are revealed. To reveal all messages would entail both Key1 and Key2 to be compromised.
Comment: You misunderstood the concept here.I don't say all messages. Compromise of Key 1, attacker can read messages sent from Alice as well as messages received by Alice. Because Alice use Key 1 for both encryption and decryption. If they user shared secret key(Key3), then reveal all the messages.
Comment: I do understand very well actually. The compromise argument applies (I agree with that) because the same key is used for both encryption and decryption. My point is that this disadvantage is not inherent to the use of symmetric keys - which your answer implies - because it can be avoided easily.


Associate Professor to Assistant Professor

Question: Is it possible that an associate professor in one university be designated as assistant professor after a year in the same university? What could be the consequences of it?
Answer: Within one university this would be a demotion so it is possible but rare, and could happen if the appointment and ensuing titles were given by courtesy, i.e. the appointment is not a regular full-time appointment.  Rank associated with this courtesy appointment might then be tied to some specific commitments.
I've never seen this happen for a regular position: I would guess that such demotion of a regular faculty member would lead to dismissal or resignation.  For courtesy appointments my experience is that these are either renewed or not, but I know of one place where such a "demotion" is technically possible.
Answer: Caveat: This is a view from the US. In other countries with different traditions things may differ.
I think that would be rare. It might be possible (unlikely) if the person changed fields and departments and needed to start over. I've never heard any instance of it. An associate professor in the US has tenure. An assistant does not.
In order for such a thing to work the person would have to give up tenure, which would be very unlikely. There can hardly be any incentive to do so, other than a change of field/department.
If it were forced, the consequences would likely be a lawsuit as it is a civil, not criminal matter. If it were by mutual consent then there are no consequences.
But switching universities is a different matter. It is possible to give up tenure and the rank of associate professor at one place and become an untenured assistant professor at another. But the incentives would need to be pretty strong for a person to want to do this. More commonly a switch would retain the rank but with a probationary period built in, normally a small number of years, before tenure is granted.

Additional caveat. There are a few places in the US that do not have a system of tenure at all. People normally serve on multi year contracts that don't automatically renew. In such a place pretty much anything is possible and there are few consequences other than higher turnover. See, for example: https://eric.ed.gov/?id=ED424814
Comment: Tenure and promotion are distinct processes at several US universities, including my own, at which one is eligible for promotion to associate professor the year before one is eligible to apply for tenure.  (US academia even contains instances, albeit fewer, of tenured assistant professors.)  Nevertheless I agree with "I think that would be rare."
Comment: To be clear, the incentives for giving up tenure when moving to another university are usually that the new university is substantially better than the old one. This might convince someone that going back from Associate to Assistant Prof is a worthwhile move.
Comment: Since pretty much every school's faculty negotiates their own system, the caveats never stop. I think there are more than a few schools that have a non tenure-track system of ranks alongside the tenure track, I'd presume to allow them to pay better salaries to untenured faculty.


Would it be possible to publish a journal paper about data-visualization website?

Question: I am graduated with a bachelor's degree in computer science. I want to know whether is that possible to publish a paper about my final-year project? The final-year project was about a website which contains data visualization of one company's data. This works had done by help of my supervisor during my bachelor time. Does anyone know about it?
Comment: How did your supervisor answer this question?  (You _did_ ask your supervisor this question, didn't you?)
Comment: Does it solve a problem or shed new light? This is with respect to existing body of knowledge.  If it's just a paper that says "i've done this", then it's no different to a brochure.
Comment: @moh89 You can publish anything as long it is written down.  The only issue is 'where' it is published. For reputable researchers, they only publish when they have something new to say.  Otherwise, it's a waste of time.
Comment: @JeffE he said it could be a kind of revision of work paper,no problem to publish it
Comment: @Prof.SantaClaus your perspective is logical to me  and i respect it but is it all the publication in the world adding knowledge all the time? in my case lets say i m new to web analytics and i want to review an existing web analytic system to learn, could not be possible for me to study review papers?
Answer: This would entirely depend on the contribution your website and techniques involved add to the state of the art. Is any element of it new and novel? I.e. something done nowhere else. It may be worthy of a conference paper for example, if he contribution is weaker.
What is your supervisors opinion? Have they hinted at this possibility?
Edit: The company's data may also feature as an issue during publication. Would they be happy for it to be used and if not can you describe your work well enough without it? i.e. if you use a figure in your paper that shows their data, such as a screenshot.
Comment: my supervisor was satisfied to participate in the publication, and he said : there should be no problem for publishing such a paper, but i wondered for continuing my masters degree,could be a case to mark down my resume for such a review work paper?
Comment: I don't think it would harm your resume. The explanation of what you was doing is clear, so there would be no gap in your CV. If you to go into academia publications will help.


Want to dissolve adjacent polygons but no common edges

Question: I have a bunch of parcel data, I want to join the polygons basing on the same owner name.
Even they are not touched, I want merge them together, such a 100 feet tolerance. The purpose is to show the somebody's properties without too much separated lines.
Like this
I have some programming experience, but just start to try to get the tool by python script, and tried arcgis add_in wizard.
If there is some idea about the script work flow, it will be really helpful!
Answer: I would try this as a test, if it gives you a good result, put it into a SearchCursor (arcpy or arcpy.da) that runs through all owners:

Select out the parcels for a single owner into a new feature class
Buffer those polygons by 100 ft (51 ft, or 50.1 ft, may be better)
Dissolve by owner to merge overlapping buffers (or do the dissolve as part of Buffer)
Buffer negatively i.e. by -100 ft to bring the buffers back close to their original size and boundaries

Alternatively, if you have an Advanced level license then the Aggregate Polygons tool (from the Cartography toolset) with its ORTHOGONAL option may be what you need.
Comment: Great idea!
But I don't want the around end point. Is there any solution for it? Thank you very much!
Comment: I think you must mean the rounded corners for which I do not currently have a suggestion.  Note that I have just included an alternative in my Answer which should not suffer from rounded corners but requires and Advanced level license.
Comment: Thanks very much, that is what I am just doing now, trying to using the "Aggregate Polygons" tool with my owner name attribute.
Answer: If you have access to the dissolve tool in the Datamangement Toolbox/Generalization, try using that. Select the file you want to dissolve and click on 'Environments/XY Resolution and Tolerance'. Set the XY Tolerance to for example 100 feet. Be aware this whole process leads to generalization of your data and you might have to play around with the tolerance value to achieve a result you are happy with.
Comment: I cannot find the 'Environments/XY Resolution and Tolerance' options in the tool dialogue
Comment: If you are in 10.1 and you have found and started the dissolve tool you'll see an Environments... button just next to the Cancel button. This next window will allow you to do various settings, one of which will be XY Resolution and Tolerance.
Answer: ArcMap: Dissolve
Or you can manually select the parcels and use Editor: Merge. This could be scripted for large files. Dissolve is built for this task, and I believe it comes with every license of ArcMAP.
Comment: If you try to merge parcels/polygons across space, aren't you creating false data if they become one large block? Are there symbolism tricks you can do to get at what you want to do? such as hiding all non-relevant parcels or not using borders?
Comment: But in geometry, they are still separated polygons. I don't want any lines or gaps in it.
Comment: Yes, we ignore the minor false of the data. Cause not all the data I want dissolve, only the specific, so, no borders does not work, I think.


What is the distribution of the following limit?

Question: Assume $x \in \mathbb{R}$. We already know that
$$\lim_{\epsilon \to 0+} \frac{1}{x-i\epsilon} - \frac{1}{x+i\epsilon} = 2\pi i \delta_x.$$
Here $\delta_x$ denotes the Dirac distribution. If we consider a slightly more irregular limit
$$\lim_{\epsilon \to 0+}\frac{\log^m(x-i\epsilon)}{x-i\epsilon} - \frac{\log^m(x+i\epsilon)}{x+i\epsilon},$$
what will be the result distribution? Here $m \ge 1$ is an integer.
Comment: WolframAlpha suggests it will be the same: https://www.wolframalpha.com/input/?i=Limit+of+Log%5Bx+-+I+a%5D%5Em%2F%28x+-+I+a%29+-+Log%5Bx+%2B+I+a%5D%5Em%2F%28x+%2B+I+a%29+as+a+-%3E+0%2B+
Answer: This question already has an impeccable answer but I would like to give a second one in the hope that it might be of interest to the OP.  Before so doing, I will list the properties of distributions (on the line) that I require:

Any "reasonable" function can be regarded as a distribution (in our context, this means locally integrable);

Any "reasonable" notion of convergence for functions is stronger than distributional convergence (in our context, local $L^1$-convergence);

Any distribution is differentiable and we can always interchange the operations of taking limits and distributional convergence in a sequence of distributions.

I will start with the (familiar) case $m=0$ of the required equation since this displays the essence of the problem.  We choose the branch of the logarithm function (on the plane minus the negative real axis)  $\ln z= \ln |z| + i \arg z$ with the argument function ranging from $]-\pi,\pi[$.
Thus for positive $\epsilon$, $$\ln (x+ i \epsilon)= \ln |x+i\epsilon|+i\arg (x +i\epsilon).$$  If we substract this from the corresponding equation with $-\epsilon$, differentiate with respect to $x$ and take the limit we get the required expression on the LHS.  On the right hand side, we first take the limit, which is $-2\pi i H (-x)$ (with $H$ the Heaviside function), and THEN differentiate. This gives the required equation.
The more general case can be proved similarly—starting from the equation
$$\ln^m(x+i\epsilon)=(\ln(|x+i\epsilon|)+i \arg(x+i\epsilon)^m,$$ subtracting this from the version for
$-\epsilon$, differentiating and taking the limit on the LHS, resp., taking the limit and then differentiating on the RHS.
Added as an edit: I should mention that I learned this proof during a lecture course on the elementary theory of distributions given by J. Sebastião e Silva in Lisbon over 50 years ago.
Answer: $\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}\newcommand{\R}{\mathbb{R}}$Let
\begin{equation*}
    K_\ep(x):=\frac{\ln^m(x-i\ep)}{x-i\ep}-\frac{\ln^m(x+i\ep)}{x+i\ep}.
\end{equation*}
Let us show that, in an appropriate weak sense,
\begin{equation*}
    \frac{K_\ep}{\ln^m\ep}\to2\pi i\de_0 \tag{1}
\end{equation*}
(as $\ep\to0$).
Take any bounded continuous function $g\colon\R\to\R$. Then
\begin{align*}
    &\int_\R dx\,K_\ep(x)g(x)=\int_\R du\,[L(u-i)-L(u+i)]g(\ep u),
\end{align*}
where
\begin{equation*}
    L(z):=\frac{(\ln\ep+\ln z)^m}z=\sum_{k=0}^m\binom mk \ln^k\ep\,\frac{\ln^{m-k}}z.
\end{equation*}
So,
\begin{align*}
    &\int_\R dx\,K_\ep(x)g(x)=\sum_{k=0}^m\binom mk \ln^k\ep\; I_{m-k}(\ep), \tag{2}
\end{align*}
where
\begin{equation*}
I_j(\ep):=\int_\R du\,\Big(\frac{\ln^j(u-i)}{u-i}-\frac{\ln^j(u+i)}{u+i}\Big)g(\ep u).
\end{equation*}
For each $j=0,1,\dots,m$ and all real $u$ with $|u|\ge2$,
\begin{equation*}
    \ln^j(u\pm i)-\ln^j u=\int_0^1 i\,dt\,j\frac{\ln^{j-1}(u\pm ti)}{u\pm ti}\ll
    \frac{\ln^{j-1}|u|}{|u|},
\end{equation*}
where $a\ll b$ means $|a|=O(b)$; here and in what follows, the constant in $O(\cdot)$ may depend only on $g$ and $m$.
So,
\begin{align*}
&\int_{|u|\ge2} du\,\Big(\frac{\ln^j(u-i)}{u-i}-\frac{\ln^j(u+i)}{u+i}\Big)g(\ep u) \\
&=\int_{|u|\ge2} du\,\ln^j u\,\Big(\frac{1}{u-i}-\frac{1}{u+i}\Big)g(\ep u)+O(1) \\
&=\int_{|u|\ge2} du\,\frac{2i\ln^j u}{u^2+1}\,g(\ep u)+O(1)
=O(1). \end{align*}
Also, clearly,
\begin{equation*}
    \int_{|u|<2} du\,\Big(\frac{\ln^j(u-i)}{u-i}-\frac{\ln^j(u+i)}{u+i}\Big)g(\ep u)=O(1).
\end{equation*}
So,
\begin{equation*}
    I_j(\ep)=O(1) \tag{3}
\end{equation*}
for each $j=0,1,\dots,m$.
Moreover,
\begin{equation*}
I_0(\ep)=2i\int_\R \frac{du}{u^2+1}g(\ep u),
\end{equation*}
\begin{equation*}
\int_{|u|\le1/\sqrt\ep} \frac{du}{u^2+1}g(\ep u)
=\int_{|u|\le1/\sqrt\ep} \frac{du}{u^2+1}(g(0)+o(1))\to\pi g(0),
\end{equation*}
\begin{equation*}
\int_{|u|>1/\sqrt\ep} \frac{du}{u^2+1}g(\ep u)
\ll\int_{|u|>1/\sqrt\ep} \frac{du}{u^2+1}\to0.
\end{equation*}
So,
\begin{equation*}
I_0(\ep)\to 2i\pi g(0). \tag{4}
\end{equation*}
Collecting (2), (3), and (4), we get
\begin{align*}
    &\frac1{\ln^m\ep}\,\int_\R dx\,K_\ep(x)g(x)\to2i\pi g(0).
\end{align*}
Thus, (1) is proved.


IDE best supportive for sparse many line code

Question: I use Allman brace placement. Most of the time the code I write is sparse, involves a lot of scrolling up and down. Clearly an IDE that has multiple windows viewing the same file is a must. Quick code folding helps. Breadcrumbs. A minimal minimap on the side helps with navigation. I'm looking for an IDE that makes it easier to look at different parts of your program, especially when programs are more than several hundred lines. Any plugins that make the process easier. What IDE is best suited for the above needs?
Comment: I know this is *software* recommendations, but if it weren't I'd suggest getting a 4K monitor with as large a screen as you can reasonably handle. My main Terminal window is 90 lines × 110 characters, and the monitor can support 3 windows like that side by side.  If my eyes weren't so old, I could use a much smaller font. A few hundred dollars will pay for itself in productivity almost at once.
Answer: CudaText editor is not an IDE, but it supports plugins, and LSP Client plugin exists. With LSP plugin and LSP Server for C++ language, you will get partly what is needed.
https://wiki.freepascal.org/CudaText_plugins#LSP_Client
Then CudaText can split single editor window to 2 parts (for the same file).
Menu "View / Split tab / Split vertically".
Answer: Visual Studio Code or it's FOSS sibling VSCodium is one of the best gratis option available today.
It can handle large code files, built in it has the following features you've mentioned:

Folding
minimap
breadcrumbs
side-by-side editing which also allows to have multiple live views of the same file.

Together with all the bells and whistles needed in an IDE, the feature list is huge, and the release notes always introduce something new.
And probably one of the most interesting implementation of this excellent editor, is the self-hosted web version of it - code-server, which makes it so accessible that all you'll need is a browser to access the editor.


What are the mathematical topics that I need for Machine Learning & Data Science?

Question: I would like to start studying ML & DS, but I feel I am a bit lost, so I don't really know what to study, what the prerequisites are, I mean I know I should study linear algebra, calculus, and statistics, but do not know the exact amount I need. I would be thankful if someone could provide me with a suitable roadmap to start, and an exact description of what I should do (What book to start with, whether to watch videos or read books .. etc)
Answer: The field(s) of ML and DS are very broad. Some topics are related to "classical" statistics (which require good knowledge in standard algebra and matrix calculation), while other fields expand into computer science and programming. Things like image classification (or "computer vision") for instance have been developed mostly in the computer science community, while other predictive models (linear regression etc) are widely used in "classical" statistics. As the fields/topics in ML/DS are very diverse, you will eventually need to focus on some fields according to your needs and interests.
However, a good starting point surely is "Introduction to Statistical Learning" (avaliable online) which also comes with R "labs" (also Python labs are available).
In case you want to dive deeper into some topics, you may refer to "Elements of Statistical Learning", which is more formal and allows you to go into more details.
For Reinforcement Learning you may have a look at the introduction by Sutton and Barto.
For more applied work related to neural nets, you may also follow the Keras Blog. Often links to various papers and other sources are provided, so that you can go into the details by topic if you are interested.
Answer: Start learning about the datasets, Data Cleaning, EDA (Exploratory Data Analysis), Visualization, some basics of linear algebra and statistics. Also learn libraries like pandas, numpy. These are some of the pre requisites before starting ML and DL.
For ML start with Regression and classification. Once you get grasp on them. Apply the learning in kaggle Competitions. You can also learn courses for free in kaggle. You will eventually learn the next steps as you move forward.
However, Hands-on Machine Learning with scikit learn, Keras and tensorflow by O'Reilly would be a great book for beginners as it covers almost all the topics in a practical approach.


Command with infinite coordnates?

Question: I am making a command that is /execute @e[type=Snowball] -603 4 170 /summon PrimedTnt So whenever I throw a snowball at the specific coordnates, it will blow up. Is there a way to make it so there are infinite coords so no matter where the snowball lands it blows up no matter what?
Thanks
Answer: You're misunderstanding the purpose of the coordinates a little bit. The command as you have it will summon a primed TNT at -603 4 170 for every snowball. No matter where those snowballs are in the world, that is where the TNT will go.
What you'll want to do is use tilde notation to specify relative coordinates:
/execute @e[type=Snowball] ~ ~ ~ summon PrimedTnt

What this will do is summon primed TNT at the coordinates of every snowball entity. Be careful about putting this in a clock without some sort of /kill @e[type=Snowball] command, unless you want to have TNT summoned along the entire trajectory of the snowball.
Comment: Ok Thanks! Yes I know what you mean I already put that command into a command block next to a repeater the '/kill @e[type=Snowball]'
Comment: how would you set it up then? i mean the full redstone and stuff
Comment: I did /testfor @e[type=Snowball] comparator 5 ticks then I did the execute command and then I connected a repeater to that and did /kill @e[type=Snowball]

Command with infinite coordnates?


Visualizing XYZ data in ArcGIS Pro?

Question: I am pretty new to ArcGis Pro and would like to import (near surface geophysical data) which are available as x/y/z/p(arameter/value). The x/y locations are in UTM so no local coordinate system is needed to work around.
I am sure this is possible but couldn't find anything on the ArcGis ressources page.
Comment: Firstly you need create file with delimiter (csv, or excel file). Then add to arcmap. There is button for Add XY Data. Click on it and it will visualize z-value. Here is a link: https://blogs.esri.com/esri/arcgis/2010/04/28/xyz-file-into-arcmap/
Answer:
Save the file as a CSV or excel sheet with the four columns (x,y,z,p)
Go to ArcScene use add x,y data from file menu, add data option.
Use the appropriate columns for x,y, z values
After adding the data to the map , right click on the corresponding layer and select properties.
In the properties there is a tab for base heights. In that use elevation from values in the layer's features. This will render the layer in 3 dimension.
Use the symbology section and choose appropriate colours based on the 'p' attribute for rendering the colours based on parameters.

Visualizing XYZ data in ArcGIS Pro?


Set Associative Cache - duplicate Tag

Question: In a set associative cache (in this case 4-way) what happens when you try to read an entry with e.g. the tag 0x3B and this tag appears two times within the same set. Given the first is invalid, would you search on for the second (which is valid) or simply get a miss? And why do would there be a duplicate line at all?
Answer: The CPU tests all caches ways and selects the one that is both valid and have a matching tag address. It may occur concurrently on all ways or be sequential (with a preferred way, depending on the address)
When invalidating cache lines, only the valid bit needs to be cleared, the tag address value do not need to be modified, so it may contain stray values.
(Cache lines may be invalidated by the operating system for coherency with some DMA peripherals, because of self-modifying code, by hardware cache coherency algorithms like MEI/MESI, etc. Depends on the architecture.)
You cannot have several lines that are simultaneously valid and with the same tag address [and different data].
It may occur at power-up, if registers are not automatically reinitialized, and the boot sequence may have to manually clear the cache before enabling it using special accesses and/or instructions.
Comment: The CPU tests the cache?
Comment: Well, test is not a good term. The cache controller _compares_ the address of the current address with the address field of the tags and, if it matches and if the corresponding valid bit is set, then it is a hit and the read or write operation can be completed.
Answer: The cache usually performs the tag comparison in parallel, and the same tag is not supposed to be in two different locations. However, since the valid bit is off at one place, it may be the case that the same tag appears more than one, where only one of the places is valid.
In fact, if the valid bit is off there is no reason to even compare that tag—even if it matches, the result will be a miss, similar to the case the tag was not found).
So the most rational behavior here, is that the "valid" bit enables the selection of that cell, and cells with valid=0 are excluded form the tag comparison circuit.

Set Associative Cache - duplicate Tag


What's so good about the Bazaar Bargain?

Question: When I see it on TF2 wiki, I could only find info about its disadvantages from missing people. What makes the Bazaar Bargain so good that I would rather pick that than the stock sniper?
Answer: Bazaar Bargain is one of those items that just depend on your playstyle and skill.If you keep running around getting one-two headshots and then die it is pretty much useless to you,but if you keep getting in good positions with nice headshots and killstreaks you will have a nice bonus without even noticing the disadvantage.
The -40% starting charge rate sounds like a big disadvantage,but really it is not.It just means that after you set up in a nice position you will have to wait 4.6 seconds before the first shot instead the 3.3s , which is negligible considering the benefit.
Now if you ignore the starting charge,you can see on this graph the benefits.
You can also ignore Huntsman since it is more of a mid-range weapon due the high projectile drop.
Also note that the graph ignores the de-buff of the bazaar for the first shot completely,meaning that the charge time for 0 heads is actually 4.6s but rest of the values stay the same.

And before July 10th the starting charge rate was -20% which was like 0.8s,really negligible.
Comment: To add to this already fantastic answer, headshots from either rifle with no charge kills 5 of the 9 classes in one shot. Though the bazaar bargain is not a particularly popular weapon, it's certainly by no means underpowered in the hands of someone who can consistently headshot.
Comment: I'm trying to reconcile the chart here with your words... the first point on the chart for the BB is at 3.3s, you state that you have to "wait 4.7 seconds" for the first shot (i.e. zero heads). The chart seems to indicate it behaves exactly like the normal sniper rifle with no heads. Which is it?
Comment: The chart uses pure rifle stats(without de-buffs) in order to pre-calculate all the exponential values.In this case(without de-buff) the starting charge is same as it is with ordinary rifle 3.3s.If you add -40% de-buff on the first shot you get ~4.7s(which is the actual time for 0 heads)...Edited the answer since it really does make the graph look wrong without explanation.


Right way to get paid from US company to India Company

Question: I am a US Citizen and have moved recently to India and started my own company. My Client is in US has asked me to Fill W8-BEN form.
I want to check will W8-BEN be applicable to my Indian company and how can I avoid double taxation .
What is right and legal way to get payed from US customers.
Answer: If you're a US citizen - you do not fill W8-BEN, you fill W9.
If you're a US citizen - you pay your taxes in the US on your worldwide income. Check with an Indian tax accountant as to how to claim credit on these taxes from India.
If you're a US citizen - Indo-US tax treaty is most likely not applicable for you.
From the US tax perspective, you may be better off as a sole proprietor than working under your own company. However, you should check it with a properly licensed (US-licensed EA/CPA) tax adviser, and also consider Indian tax consequences.
Reporting ownership in foreign corporations is quite a burden for US citizens.


Prove that if $v$ is an eigenvector for A, than $v$ is also and eigenvector for $adj(A)$.

Question: Let $A$ be a complex square matrix and $\operatorname{adj}(A)$ its adjugate (so the entries of $\operatorname{adj}A$ are minors of $A$, up to sign).
Prove that if $v$ is an eigenvector for $A$, than $v$ is also an eigenvector for $\operatorname{adj}(A)$.
Comment: I can't do it but this should be put on hold.
Comment: Can you provide your defintion of the adjugate matrix?
Comment: $A^{-1} =\frac{1}{detA} adj(A) $
Answer: An argument that works over any commutative ring:
It is well-known that $\operatorname{adj} A$ can be written as a polynomial in $A$. In other words, there exists a polynomial $p$ (in one variable) over the ground ring such that $\operatorname{adj} A = p\left(A\right)$. (This $p$ can be described explicitly: We have
\begin{align}
p\left(t\right) = \left(-1\right)^{n-1} \sum_{i=0}^{n-1} c_{n-1-i} t^i ,
\end{align}
where $c_j$ denotes the coefficient of $t^{n-j}$ in the characteristic polynomial $\det\left(tI_n-A\right)$ of $A$. See Theorem 5.14 in my The trace Cayley-Hamilton theorem for a proof. Alternatively, you can find proofs at https://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a .)
Now, let $v$ be an eigenvector of $A$, and let $\lambda$ be the corresponding eigenvalue. Thus, $A v = \lambda v$. Consider the polynomial $p$ defined above, which satisfies $\operatorname{adj} A = p\left(A\right)$. From $A v = \lambda v$, we can easily obtain (by induction) that $A^i v = \lambda^i v$ for each nonnegative integer $i$. Hence, $p\left(A\right) v = p\left(\lambda\right) v$ (since the polynomial $p\left(t\right)$ is a linear combination of powers $t^i$). Thus, $v$ is an eigenvector of the matrix $p\left(A\right)$. In other words, $v$ is an eigenvector of the matrix $\operatorname{adj} A$ (since $\operatorname{adj} A = p\left(A\right)$).
Answer: The easiest (and the most natural) solution is to note that $\operatorname{adj}(A)$ is a polynomial in $A$ (as indicated in darij grinberg's answer). For a more elementary approach, you may consider the rank of $A$. Suppose $A$ is $n\times n$. There are three possibilities:

$\operatorname{rank}(A)=n$. This case is easy, as $\operatorname{adj}(A)=\det(A)A^{-1}$.
$\operatorname{rank}(A)<n-1$. This is case is also easy, as $\operatorname{adj}(A)=0$.
$\operatorname{rank}(A)=n-1$. Then the left or right null spaces of $A$ are one-dimensional and $\operatorname{adj}(A)=uv^T$ for some eigenvector $u$ in the right null space of $A$ and some eigenvector $v$ in the left null space of $A$. Now, suppose $x$ is an eigenvector of $A$.

If $Ax=0$, since the null space of $A$ is one-dimensional, we must have $x=cu$ for some scalar $c$. Thus $\operatorname{adj}(A)x=uv^T(cu)=(v^Tu)x$.
If $x$ is an eigenvector of $A$ corresponding to some nonzero eigenvalue $\lambda$, then $0=(v^TA)x=v^T(Ax)=\lambda v^Tx$. Hence $v^Tx=0$ and $\operatorname{adj}(A)x=uv^Tx=0$.
Comment: @Widawensen I don't think so. E.g. when $A=0\oplus X$ and $B=Y\oplus 0$ are two block matrices with conformable sizes and $X,Y$ are square sub-blocks, $e_1$ is not necessarily an eigenvector of $B$ and $e_n$ is not necessarily an eigenvector of $A$.
Comment: I wonder whether using your method could be proved more general statement i.e. if $AB=BA=0$ and $rank(A)+rank(B)=n$ ( I'm not sure whether this condition is really  needed) then A,B have the same eigenvectors ( ... if you regard   necessary I can ask a new question).
Comment: Yes, it seems to be a good counterexample.
Answer: This is an interesting question, in fact at first I thought it wasn't true when the eigenvalue is $0$, but it turns out that it is !
We have the well-known formula $adj(A)A = \det(A)I_n$. From this it follows at once that if $v$ is an eigenvector of $A$ with nonzero eigenvalue $\lambda$, $v$ is an eigenvector of $adj(A)$ with eigenvalue $\det(A)/\lambda$.
If $Av =0$ however, this does not work so we have to use a trick : perturb $A$ by $\epsilon>0$ to get $A_\epsilon = A +\epsilon I_n$. Then $v$ is an eigenvector of $A_\epsilon$ with eigenvalue $\epsilon$ which is nonzero. It follows that $adj(A_\epsilon)v = \frac{\chi_A(-\epsilon)}{\epsilon}$.
Note that since $A$ has $0$ as an eigenvalue, its determinant is $0$ so $\chi_A(0) =0$. Therefore, letting $\epsilon \to 0$ in the previous equality gives the derivative of $\chi_A$ : $adj$ is continuous so that it yields $adj(A)v = -\chi_A'(0)v$
(We even have an explicit formula for the eigenvalue !)
One might wonder whether this generalizes to other fields, since I used the topology of $\mathbb C$ (or $\mathbb R$) to let $\epsilon$ tend to $0$;  the only problem is that you have to be careful in the last step : you want to first work in $K(X)$ if $K$ is your base field, then notice that the fraction $\det(A+XI_n)/X$ is actually a polynomial (because $\det(A)=0$), so we may work in $K[X]$, and then let $X=0$.
(Morally you can work in the ring if dual numbers $K[\epsilon]$ but the technicalities seem more annoying to write)
Answer: It is because if A, B are similar square matrices, adj(A) and adj(B) are also similar (§).
If v is an eigenvector of A, then you pick a basis e_1...,e_n with v=e_1, the result will be obvious in that basis.
In order to prove (§), define K:=the field of fractions of Z[X_1,...,X_{2n^2}] you may consider for every nxn invertible matrices F,G with coefficients in K, the identity (GFG^{-1})^{-1} = 1/det(F) adj(GFG^{-1}) = 1/det(F)G*adj(F)G^{-1}=GF^{-1}G^{61} which yields to
adj(GF*adj(G)) = det(G)^{n-2}Gadj(F)*adj(G). This holds for every invertibles F,G and so it holds for any F,G with coefficients in Z[X_1,...,X_{2n^2}] and so this polynomial identity holds in any commutative ring. The result can be deduced from this (by dividing again by the determinant of the base change matrix)
Comment: Looks like this post needs a LaTeX hero :)
Comment: Welcome to MSE. For some basic information about writing mathematics at this site see, *e.g.*, [basic help on mathjax notation](/help/notation), [mathjax tutorial and quick reference](//math.meta.stackexchange.com/q/5020), [main meta site math tutorial](//meta.stackexchange.com/a/70559) and [equation editing how-to](//math.meta.stackexchange.com/q/1773).


Re-enable editable file path in Nautilus

Question: I'm using Gnome Shell with Ubuntu 11.04. I'm liking it a lot so far, but the only issue is the fact I can't type a file path into Nautilus; I only get the silly breadcrumb buttons. Can someone point me to a solution on how to revert back to having a nice, normal, useful address bar please?
Answer: I have not figured out a permanent solution myself, but once you have nautilus open, you can do
ctrl + l

you will get access to the address bar, and you can type the full file path in there.
Another option would be to type the following when opening nautilus itself, either through the run command (alt+f2) or from console.
nautilus path/to/folder

Answer: This page explains how to bring back the address bar. In short:
$ gconftool-2 --type=Boolean --set /apps/nautilus/preferences/always_use_location_entry true

Answer: I just found this page, the removal of the location bar setting makes me nuts...
http://www.webupd8.org/2010/05/use-text-mode-location-bar-in-nautilus.html
Answer: One more method, in addition to those already given, is to start typing / in nautilus and it will bring up the editable file path.
This seems to me by far the easiest method that makes me not miss the button at all.


where does the photon go after scattering?

Question: My question is about photo electric but it could be applied to other daily routine phenomenon. As we know rest mass of photon is zero. When a photon strikes the metal surface it transfers its energy to the electrons. Whether electron will be emitted or not, it depends upon work function. But my question is about that photon which was hit on metal surface. Does that photon vanished? Does that photon turned into nothing? where does it go after scattering? When I study this I only find the story about the emitted electrons but not about photons after collision. Am i missing some basic concept?
Comment: Where does the flame go after you blow out a candle? Where does the sound of someone's voice go after it has reached your ears? When something disappears in one location, you have to think carefully about whether or not it's reasonable to infer that it must have gone somewhere else (as opposed to being gone altogether). The intuition that "all things must be somewhere, so if it's not here then it must be somewhere else" is justified by the various physical conservation laws (mass, energy, momentum, charge, etc.) and spatiotemporal continuity. (1/2)
Comment: The law of conservation of mass tells us that the amount of mass of a closed system is constant; mass can neither be created nor destroyed. Suppose you weigh a pot of water before and after boiling it for some time. You'll notice that it weighs less afterwards. From the law of conservation of mass, you can infer that since the mass of the missing water couldn't have simply been destroyed, and that the missing mass will be found if we look in the right place (collect and weight the water vapor). (2/3)
Comment: Another process where mass is conversed is the [electrolysis of water](http://en.wikipedia.org/wiki/Electrolysis_of_water), i.e. the conversion of water into oxygen and hydrogen: $2\text{H}_2\text{O}\rightarrow2\text{H}_2+\text{O}_2$. The mass before and after the reaction is conserved, but the number of molecules is not. It makes no sense to ask "where did the H2O molecules go?" because after the reaction, the H2O molecules aren't anywhere. They're simply gone. In much the same way, photon number isn't conserved either. I apologize for the lengthy comments. Hopefully they'll still be helpful.
Answer: In particle interactions the total number of particles is not conserved. For example in a collision in the LHC two photons collide and many hundreds of particles are created in the collision.
There are still some conserved quantities, for example lepton number is still conserved so you cannot just create an electron. You need to create an electron and positron together so the total lepton number doesn't change (the electron has number +1 and the positron -1, so they add to zero).
However the number of photons is not conserved. Photons are bosons and they are their own antiparticle so no particle number conservation law is violated when you create a photon. Specifically an accelerating electron can emit any number of photons, and the corollary of this is that an electron can absorb photons and be accelerated. This is what happens in the photoelectric effect. An electron in the metal absorbs the photon and its energy is increased by the photon energy (so energy is conserved). The electron will in turn collide with other electrons in its vicinity, and in a small percentage of cases enough energy is transferred to another electron that it can escape from the surface.
Note that not every photon falling on the metal emits a photon. Far from it in fact as the quantum yield is usually around $10^{-5}$ i.e. only one photon in 100,000 manages to eject an electron. In the other 99,999 cases the energy of the photon just ends up heating the metal.
Answer: Have you read the wiki link

electrons are only dislodged by the photoelectric effect if light reaches or exceeds a threshold frequency, below which no electrons can be emitted from the metal regardless of the amplitude and temporal length of exposure of light. To make sense of the fact that light can eject electrons even if its intensity is low, Albert Einstein proposed that a beam of light is not a wave propagating through space, but rather a collection of discrete wave packets (photons), each with energy hf. This shed light on Max Planck's previous discovery of the Planck relation (E = hf) linking energy (E) and frequency (f) as arising from quantization of energy. The factor h is known as the Planck constant.1

In this formulation, the electron absorbs the total energy and momentum  of the photon.
In non metals electrons are in bound states around the nuclei forming the stable atoms. A photon will interact with an electron and transfer its total energy and momentum if its energy matches a quantum energy difference in higher levels of the atoms potential well, or has enough energy to kick it completely out.
In metals, there exists the Fermi level and the fermi function which characterize the collective but still quantum mechanical behavior of the electrons which create the conductivity of metals. In this case it is easy for a photon to kick an electron out because the energy levels have a very small magnitude.
In both cases the photon disappears. There is no conservation of photons quantum number. It is a boson and many of them can exist in the same energy state and they also may give all their energy up to electronic and other transitions and disappear.
Answer: First: momentum is conserved, energy is conserved, electric charge is conserved, and a few other things.
Particles and particle number can, in some instances, flit into and out of existence.
The photon is absorbed by the electron. I.e. the photon ceases to exist, its energy and momentum have to go somewhere (the electron and the bulk material). In the photoelectric effect if the electron cannot absorb the energy of the photon (i.e. it cannot exist at that energy level because of atomic orbitals) then the photon just passes through without absorption. This is the case with glass, visible light does not have enough energy to excite an electron so it just passes through.

Does that photon vanished?

Basically yes, it ceases to exist, but some of its properties must be conserved.

Does that photon turned into nothing?

It's not nothing. It turns into the extra energy in an electron.


Sitecore 8.1 Multivariate testing optimization shows no test status

Question: I have set up a Multivariate test in Sitecore 8.1. In Experience Editor i have a notification,it says item has a active test and Optimization tab shows red button but status shows no tests. I did check in the analytic database no test record found. I did deploy the Test from Workflow and its running. I am getting all options to stop, cancel and suspend. Sitecore is not testing my variations can any one help me to resolve this?
Comment: From Sitecore's Launch pad, chose the Experience Optimisation tool. In there what is currently displayed for your test?
Comment: Experience Optimisation Tool Shows nothing, Not even active test which is running. In Experience Editor under Optimisation view shows red symbol shows for running test.
Answer: From what you are describing it sounds like the Sitecore_testing_index is erroneous. Perhaps the index is invalid, not built or not present. This index is required for Sitecore's Experience Optimisation to find the Tests in their various states.
I recommend getting a default copy of Sitecore.ContentTesting.Solr.IndexConfiguration.config from a fresh install and copy it to the path App_Config/Include/ContentTesting.
Once that is done, rebuild the Index via the Index Manager. You should then be able to find tests in the Experience Optimisation.
If you're still having trouble I recommend replacing the entire contents of App_Config/Include/ContentTesting with the files from a fresh install. And re-indexing.
I also recommend having a look against the Item which has the Multivariate test it should be within the default field Page Level Test Set. It should have values something along the lines of this

If it is populated but you are still unable to the Test, follow this guide how to create Multivariate Tests in Sitecore to identify where something is going wrong.
Comment: I am having Sitecore.ContentTesting.Lucene.IndexConfiguration and Sitecore_testing_index working fine. I am able to generate fresh index file through Index Manager. As i am running Multivariate test both page level Test set and content test fields are empty.
Comment: Let me know why these field values are not populated? where i am going worng?
Comment: I am able to do Page tests successfully but Multivariate test on Component with different data source throwing below error Exception:         System.ArgumentNullException
Message: Value cannot be null.
Parameter name: contentItem
Source: Sitecore.ContentTesting
   at Sitecore.ContentTesting.TestConfiguration..ctor(Item contentItem, ID deviceId, TestDefinitionItem test)
   at Sitecore.ContentTesting.Data.SitecoreContentTestStore.LoadTest(TestDefinitionItem testDefinitionItem, ID deviceId)
   at Sitecore.ContentTesting.Pipelines.GetChromeData.GetRenderingTest.


Failed to resolve: com.android.support:appcompat-v7:27.+ in IntelliJ IDEA

Question: I am getting this error while i try to build an adroid project in IntelliJ IDEA (Error:(26, 13) Failed to resolve: com.android.support:appcompat-v7:27.+
Install Repository and sync projectShow in File)how can i fix it :
Comment: Can't see the error; post the image, not a link.  I'm guessing that you are using Maven but IntelliJ can't find the JAR in the repositories you've asked it to look in.
Comment: Just click the 'Install repository...' link in the error message.
Comment: just click on  Install Repository and sync project.
Answer: Click on the "Install Repository and sync project" link.
Or you can try to fix your build.gradle manually.
Add this in build.gradle in the project's root directory:
allprojects {
    repositories {
        // add below
        maven { url "https://maven.google.com" }
    }
}

Comment: I tried it ,  in android studio it worked but in IntelliJ nothing seems to be happening
Comment: can you add both of your build.gradle files to the question?
Comment: is this a good idea to solve the problem? shouldn't we remove the '+' after '27.' and replace it with a full version number?
Answer: For Gradle 4.1+ you can use the following (used by Android Studio 3.0.1):
allprojects {
    repositories {
        // add below line
        google()
        jcenter()
    }
}

I'm not sure why Intellij IDEA doesn't prepopulate this dependency as it does with Android Studio (will raise a bug and reference it here shortly), but Google's Maven repository is required as it contains the Android libraries, see link.  Please upvote IDEA-187121 (bug I raised on YouTrack), if you want to see this fixed.
@Gavriel's suggestion of clicking "Install Repository and sync project" doesn't work at least with IDEA 2017.3.4.  The suggestion to include "maven { url "https://maven.google.com" }" does work for projects using all gradle versions, but is quite verbose.


Surgical strikes by India

Question: Surgical strikes against Pakistani terrorists have been happening for a long time by India. What used to be clandestine operations are now publicised by the present government. What brought about this change?
Answer: This article lays out the reason it's making headlines:

While exchanges of artillery and light weapons over the line are very common, intentional incursions by aircraft have not been publicly acknowledged since the two countries fought a war in 1971.

Surgical strikes by India


How to install new ATmega firmware via the ISP pins?

Question: While uploading .hex firmware directly onto boards that allow this via CURA, this is not always an option for all machines (for lack of a functioning USB-connectable port for example).
On both my TronXY X1 board (Melzi 2.0 V5 with ATMEGA1284P) and the Creality Ender 3 (Creality 3D v.1.1.2 with Δ ATMEL  ATMEGA1284P  AU 1726) I spotted these 6 pins labeled ISP that are supposed to allow to connect an Arduino or similar to flash firmware. According to the RepRap Wiki, Melzi is a derivative of the Sanguinololu and both are Sanguino Compatible.
After creating the new firmware in Arduino IDE and making sure it compiles for the correct chip in the board, how can I use an Arduino Uno or other microcontroller to flash the new firmware  via the ISP to my printer?
This is asking specifically about ATmega based controller boards, using the Arduino toolchain.
Comment: There isn't a one-size-fits-all answer for this. It would be much better to have separate questions for separate types of board, because a single answer can't cover all possible scenario's you're imagining. Keep in mind that [you should only ask practical, answerable questions based on *actual problems that you face*](https://3dprinting.stackexchange.com/help/dont-ask). You aren't actually facing the problem of having a board you can't upload to using Cura.
Comment: Not all printers are created equal. The process would be very different for a board based on, say, an AtMega2560 than a board based on a Cortex M3. It also depends on what (if any) bootloader is present on the board. I think the question is too broad in its current state.
Comment: @TomvanderZanden true, this is meant as a generic "I have some printer and managed to get the firmware to compile - hw do I get it onto the printer? I have this other microcontroller around". If you can add to the question or reitterate it in a better form, feel free to do so - we pretty much need to answer these basic questions to be able to answer more difficult answers more quickly by interlinking.
Comment: @TomvanderZanden I know it is supposed to work, but I fail to understand why it is supposed to be better than for example directly loading it via CURA or a different method to send the compiled .hex to the board... After all you have to *open your machine*... But it is a rather universal solution and would work for pretty much any machine - even ones that have no possibility to connect it to the PC, for exampe if the USB port was broken.
Comment: @TomvanderZanden - the ATmega on the controller of the Ender-3 does not have a bootloader, apparently, as per [TH3D Unified Firmware Package](https://www.th3dstudio.com/knowledge-base/th3d-unified-firmware/): *CR-10, Ender 2, and Ender 3 Printers need the bootloader. You will need to follow the CR-10 bootloader guide to flash it before you can load the firmware. You can use our [Uno kit](https://www.th3dstudio.com/product/arduino-uno-bootloader-flashing-kit/) to flash the bootloader on your board.* The Uno kit is the ISP.
Answer: Considering AVR based controller boards...
You might not need an additional Arduino, to serve as an ISP. It really depends upon the board. There are basically three scenarios:

Controller board which is Arduino based - Arduino Mega 2560 with a RAMPS 1.4 - with a bootloader
Controller board which is Arduino based - Arduino Mega 2560 with a RAMPS 1.4 - without a bootloader
Controller board which is AVR based board (no separate Arduino board) - as is the case with an Ender-3 and the TronXY board1 - without a bootloader

Option 1 is the easiest, and you simply upload with the USB cable, directly from the Arduino IDE.
Options 2 and 3 is where you would require an In-circuit Serial Programmer (ISP). Such as a USBasp: USBasp - USB programmer for Atmel AVR controllers

or you can use another Arduino, see Arduino as ISP and Arduino Bootloaders.
An ISP device basically converts the USB to TTL signals (along with a serial protocol (JTAG or derivative)) which then connects to the ICSP2 interface on the controller board:

Scenario 1
As stated above, you simply connect the board to the PC/Mac using the USB cable and, within the Arduino IDE, you hit Upload - ensuring that you have the correct Board (i.e. Mega2560) and Port selected in the Tools menu.
Scenario 2
For the Mega2560 the pins the MOSI, MISO and SCK are broken out on to are:
Arduino / Genuino Board    MOSI            MISO             SCK       Level
Mega1280 or Mega2560    51 or ICSP-4    50 or ICSP-1    52 or ICSP-3    5V

To program the Mega2560 using an Uno, connect as follows:

or using the ISP connector (in this diagram the electrolytic capacitor3 is not shown):

Take note of the electrolytic capacitor3 on the programming device:

The 10 µF electrolytic capacitor connected to RESET and
  GND of the programming board is needed only for the boards that have
  an interface between the microcontroller and the computer's USB, like
  Mega, Uno, Mini, Nano. Boards like Leonardo, Esplora and Micro, with
  the USB directly managed by the microcontroller, don't need the
  capacitor.

or using the Mega's ICSP connector, from Atmega bootloader programmer (see original image):

Note D10 is used, not RESET4
Scenario 3
So, to use an Arduino Uno on your controller board, which is scenario 3 (an AVR based board, but not an Arduino board per se), you would need to connect the Uno to the ICSP of the controller board like this:
ICSP     Uno

MOSI     D11
MISO     D12
SCK      D13
GND      GND
5V       5V
RESET    D10

or, like this, from Chaos Drucker - CREALITY ENDER3 MIT MARLIN AUSSTATTEN  (see original image):
There is an error in the image and the red line should go to D10 and not D9 as shown

Note, again, that the RESET on the Arduino Uno's ISP connector is not used (and Pin 10 is used to connect to the controller's ICSP's RESET pin instead)4.
Configure the Arduino IDE
Then in the Arduino IDE you will need to select the Programmer to be used, in this case Uno as an ISP, like so:

Once you have connected correctly, and selected the programmer, then you should be able to hit compile and upload as usual.

Footnotes
1 Which both use the same board - see this answer to What voltage does the Creality Ender-3 run at?
2 In case you are wondering, see What's the difference between ICSP vs ISP?:

The difference between ISP and ICSP is a hyphen.
ISP stands for In-circuit Serial Programming, and ICSP stands for In
  Circuit Serial Programming.
Any arrangement that allows you to program a microcontroller while it
  is in a circuit using a serial protocol can be called ISP or ICSP.

See also Wikipedia - In-system programming.
3 See When using an Arduino as an ISP, is the capacitor required or not?
4 Why? From Can the Arduino Uno become an AVR programmer and use the ICSP header to program the target board:

First of all, as configured, the RESET pin can not be used as an output pin. In order to use it as an I/O pin, you would have to program the RSTDISBL fuse. But once you do that, as jippie said, it becomes near-impossible to reprogram your Arduino (if you have the socketed through hole variety of the MCU, you could pull it out and put it into a high voltage programmer; in all other cases, it's effectively impossible).
So in theory, you can use the ICSP header the way you suggest. In practice, it's under almost any conceivable circumstance a terrible idea.
Comment: thanks a lot - also I suggest to put up a note reminding to CHECK YOUR CONNECTION WIRES in case some odd error pops up.


What does leveling up weapon mods/attachments do?

Question: What is the difference in Far Cry 3? Say you have the F1 and have the Marksman's Mod Level 3 for it, what happens if you go up a level to Marksman's Mod Level 4 or 5? Does it increase the stats boosted and decreased?
Answer: After searching for some time I found a forum post from 2012/2013 which discusses the differences between weapon mods.
As you know, weapon mods affect your weapons in different ways:

Hunter's Mod - Decreases gun sway and recoil. Smaller magazine.
Warrior's Mod - Increases damage and penetration. Decreases accuracy.
Marksman's Mod - Longer range and improved accuracy. Slower handling.
Ranger's Mod - Faster reloading and aiming. Less ammo carried.
Stalker's Mod - Larger magazine. Slower movement and weapon
switching.

And all of these mods have five levels Mk1 through Mk5, for a total of 25 mods available through decoding.
On the Ubisoft forums user Detrooit asks (grammar fixed):

I know the difference from the Weapon Mods, and every Mod has an Mk Level from 1 till 5.
  After my decoding I get, for example,  the Ranger Mod Mk5, but I still have the Ranger Mod Mk2. The level is changing, but what does it exactly effect?

And users KasDaDon21 responds (grammar fixed):

I have studied an I have a found an answer. As you know each type of mod does something different. The mod does what the description says, but the higher the Mk the stronger the effect of the mods description.

This would mean a gun modified by a Mk1 Hunter's Mod, which decreases weapon sway, would still have more sway than the same gun modified by a Mk5 Hunter's mod.
Earlier in the forum post users also confirmed that upgrading the Mk of the mod does not decrease the negative effects of the mod. For instance, the Hunter's mod, which also decreases magazine size, would maintain the same small magazine size regardless of whether the mod is Mk1 or a Mk5.
In summary, a Mk5 mod increases the benefits of a weapon mod with no effect on the downsides. A Mk5 mod will be objectively better than a Mk1 weapon mod of the same category.


Save only one type of attachment

Question: I have the following code which saves all attachments to a given folder which is what I want most of the time, but I have a need to save ONLY one type of attachment in a particular instance, say only the PDF or only the XLS.
What do I need to add to my code to do this.
My code:
Public Sub SavePayRoll()
' Save data from Payroll Service
Dim objOL As Outlook.Application
Dim objMsg As Outlook.MailItem 'Object
Dim objAttachments As Outlook.Attachments
Dim objSelection As Outlook.Selection
Dim i As Long
Dim lngCount As Long
Dim strFile As String
Dim strFolderpath As String
Dim strDeletedFiles As String

    ' Get the path to folder
    strFolderpath = CreateObject("WScript.Shell").SpecialFolders(16)
    On Error Resume Next

    ' Instantiate an Outlook Application object.
    Set objOL = CreateObject("Outlook.Application")

    ' Get the collection of selected objects.
    Set objSelection = objOL.ActiveExplorer.Selection

' The attachment folder needs to exist
' You can change this to another folder name of your choice

    ' Set the Attachment folder.
  strFolderpath = "C:\DrBox\"

    ' Check each selected item for attachments.
    For Each objMsg In objSelection

    Set objAttachments = objMsg.Attachments
    lngCount = objAttachments.Count

    If lngCount > 0 Then

    ' Use a count down loop for removing items
    ' from a collection. Otherwise, the loop counter gets
    ' confused and only every other item is removed.

    For i = lngCount To 1 Step -1

    ' Get the file name.
    strFile = objAttachments.Item(i).FileName

    ' Combine with the path to the Temp folder.
    strFile = strFolderpath & strFile

    ' Save the attachment as a file.
    objAttachments.Item(i).SaveAsFile strFile

    Next i
    End If

    Next

ExitSub:

Dim shel As String
shel = strFolderpath

Set objAttachments = Nothing
Set objMsg = Nothing
Set objSelection = Nothing
Set objOL = Nothing

Dim retVal As Long
    retVal = Shell("explorer.exe " & shel, vbNormalFocus)

End Sub

Thanks in advance
Answer: You can try to check if the strFile has the extension that you want, e.g
' Get the file name.
strFile = objAttachments.Item(i).FileName

' Only save pdf files
If strFile.EndsWith("pdf") Then
    ' Combine with the path to the Temp folder.
    strFile = strFolderpath & strFile

    ' Save the attachment as a file.
    objAttachments.Item(i).SaveAsFile strFile
End If

Answer: Thanks @Dat your suggestion gave an error but you put me on the right track, My code now reads
dim sFileType as string

 ' Get the file name.
strFile = objAttachments.Item(i).FileName
sFileType = LCase$(Right$(strFile, 4))
MsgBox (sFileType)
' Only save pdf files
     If sFileType = ".pdf" Then
' Combine with the path to the Temp folder.
strFile = strFolderpath & strFile

' Save the attachment as a file.
objAttachments.Item(i).SaveAsFile strFile
    End If

    Works fine!

Save only one type of attachment


What options do you recommend for language translation on content driven Web sites?

Question: Please read the whole question. I'm not looking for an approach to managing multi-lingual content, but I'm looking for a way to actually get that multi-lingual content. This usually falls within technical recommendations on most projects I work on, and I hope someone can offer some help. We are working with a client now who has the personnel to physically translate content, and each language has a separate domain, hosting, CMS, etc. For those clients that want to do business internationally though, and have no personnel, what are some approaches or services to use? Is it cheaper to farm this out to people, or to do it dynamically, and what are the technical drawbacks? Any advice you can offer is helpful. I know this isn't strictly a programming question, but I think it falls within the technical realm.
Answer: I have yet to see a dynamic translation service that would be suitable for the content of a professional website.  Language translation is not (yet) a mechanical activity - it requires thought and analysis.  Your clients would best be served by outsourcing translation (or hiring a translator).
Answer: Normally you would need two teams: a forward translation and a backward translation and you do this in parallel (3-4 four teams would be good) - you then check where the backward translation are in agreement (hopefully this is at 75-85%), you then get an expert to give your recommendations on the remaining 15-25%. You could theoretically set this up in Mechanical Turk.


How can I make Firefox open magnet-links in Transmission?

Question: I can't find out how to make Firefox open magnet-links in Transmission.
There is no option to make it the standard program for magnet-links in its properties, and as I am not experienced with the filesystem so I can't add it as the standard program in Firefox either.
Comment: https://trac.transmissionbt.com/wiki/MagnetLinks These instructions should still be good as gconf still has the url handler settings.
Answer: Open up Firefox and its preferences window.  Go to the "Applications" tab and you should see "magnet" on the list.  Click on "Always ask" and if Transmission does not appear as an option then click on  "Use other" and navigate to /usr/bin on your file system.  Look for "transmission" or "transmission-gtk" and click that.
Update: Firefox has changed so the above no longer works.  Instead, when you click on a magnet link, a "Launch Application" window will appear, from which you can choose to always associate an application (make sure to check the checkbox at the bottom to always use the chosen app).  To find the torrent app, choose "file system" on the left side, and then navigate to /usr/bin as described above.
Comment: In my case, "magnet" didn't show up in the applications tab, so I had to do this via about:config as explained [here](http://ubuntuforums.org/showthread.php?t=1416277)
Comment: Thanks! If the magnet does not appear in the list of applications click on any magnet link, choose an application, type "/usr/bin/transmission-gtk". Works on Xubuntu 14!
Comment: In ubuntu 21.10, try setting `firefox > preferences > applications > magnet` to `"Use System Handler (default)"` and see what that does (it obviously depends on the system :-) ). When I tried setting `/usr/bin/transmission-gtk` as the handler, clicking on magnet links did absolutely nothing. Setting firefox to "use system handler (default)" however did work.
Comment: @michael 's comment worked for me!  In Ubuntu 22.04 it is Settings > General > Scroll down to Files and Applications > Search for magnet > select "Use System Handler (default)"
Comment: Note in transmission the torrent comes up with a question mark, but it still continues to download. Just discovered this as I download Germany demolish Brazil.
Comment: Unfortunately in recent Firefox there is no "magnet" entry in "Applications". Also "Launch Application" window doesn't appear when the link is clicked.
Comment: This change may have been reverted, "magnet" is in the applications setting (in FF) and was the only way I could find to change the setting (which I had initially set via the popup).
Comment: Same as @michael and DanglingPointer here, setting default app to transmission-gtk does nothing but letting the system handle it works just fine
Answer: How to set default apps in Chromium / Chrome / Firefox / Ubuntu Linux
Situation: You hate the default app that opens when you double click a file or when you open it with chromium browser and you want to change it.

In Unity (and probably Gnome 3 Shell), some default apps can be changed through gnome-control-center: System settings > System info > Default applications
You can also right-click on a file > select Properties > Open with. See what applications/programs are listed there. You can also add them and set them as default.
If none of the above work, you can use xdg-utils (xdg-mime) to set a default application for a specific Mime Type

First, you have to figure out what text string to use as a "mime type". An easy way to find it is to look into the default application's desktop file.
Example: Chromium opens magnet links and bittorrent files with transmission-gtk by default. You want to use deluge. The .desktop files are usually in folder /usr/share/applications/. We use the cat command to look into the file /usr/share/applications/transmission-gtk (a line that starts with "MimeType"):
$ grep 'MimeType' /usr/share/applications/transmission-gtk.desktop
MimeType=application/x-bittorrent;x-scheme-handler/magnet;

Bingo! x-scheme-handler/magnet and application/x-bittorrent are the mime types we're after.
How to use the xdg-mime command? Simple: xdg-mime default myapp.desktop mymimetype
..where we substitute myapp.desktop and mymimetype with the values from above.
Following the example above, we want magnet links and bittorrent files to open with Deluge (deluge-gtk) application. First you have to install the deluge-gtk package (otherwise the .desktop file won't exist). Then you have to execute:
xdg-mime default deluge.desktop x-scheme-handler/magnet
xdg-mime default deluge.desktop application/x-bittorrent
grep -i deluge ~/.local/share/applications/*

The last command (grep) should return "application/x-bittorrent=deluge.desktop" and the other mime type for magnet. Otherwise, something went wrong (maybe the desktop file didn't exist?).
And that's about it! Close and open chromium-browser again (or logout/login).
For the sake of documentation, the files that have to do with mime types and default apps are:
~/.local/share/applications/defaults.list
~/.local/share/applications/mimeapps.list
~/.local/share/applications/mimeinfo.cache
/usr/share/applications/mimeinfo.cache
/usr/share/applications/mimeapps.list
/usr/share/applications/defaults.list

Answer: I tried a couple of approaches and the only on that worked for me is described here: https://support.mozilla.org/en-US/questions/1012864#answer-609391
All you have to do is to open about:config in Firefox, and add boolean value network.protocol-handler.expose.magnet with value false. The next time you click on a magnet link, you'll be prompted to choose your favorite application to open the link.
I'm using Ubuntu 17.10 and Firefox 56
Comment: has this been deprecated?  Useful answer but might need updating.
Answer: For me, with transmission-daemon the solution was the following (according to this documentation).

Create an executable file /usr/local/bin/magnet with these commands:
sudo tee /usr/local/bin/magnet >/dev/null <<EOF
#!/bin/sh
exec transmission-remote -n yourTransmissionUserName:YourTransmissionPassword --add "$1"
EOF
sudo chmod +x /usr/local/bin/magnet

Then go to Firefox preferences (Alt + E + N) and select it after finding the magnet file type on the Applications tab:
Answer: For a solution that worked for me:
https://support.mozilla.org/en-US/questions/965921
"Click on the help menu and select "Restart with Add-ons Disabled" and then select the reset option in the dialog box."
Follow the on-screen prompts. Then restart Firefox.
When you click on a magnet link the usual "Choose a program" type of dialog box appears. Browse for /usr/bin/transmission-gtk
Works in Ubuntu 13.04
Answer: If you use Transmission for torrents the only thing you must do is when the window opens and ask you with which program you want that Firefox open magnet links with, you drag & drop the transmission icon from your side bar in the Firefox dialog box and that's all. I have Ubuntu 14.04 (Trusty Tahr), and it works fine.


No power to media display

Question: I have a 2016 Ford Focus we bought at auction, so I have no history. The radio/media display seems to have no power. We had it tested by the dealership and they said it needed a new screen. We replaced that and still it seems that power is not getting to the unit. We have power to the heat and AC and we have power to the hazard button in the same area. But nothing to the screen or stereo in general. I've checked the fuse and it is fine. Does anyone have any suggestions on how to troubleshoot this?  Do I need to take it back to the dealership?
Comment: Have you checked for power on the power connectors on the back of the media unit?
Comment: It would be useful if you added some photos of the media unit both front and rear.
Answer: It is very common for the stereo/cd unit behind the bezel to be the fail point in this vehicle. It is an internal electrical failure which affects radio/cd/media/sync/info functions. New replacement from the dealership could cost you upwards of $600. I personally recommend getting a used one from an wrecking yard if available, generally costing less than $50. Also they are very easy to swap out with little skill and simple tools which would save you even more $. There several decent YouTube videos on this procedure. Hope this helps.

No power to media display


Is there a way to duplicate objects betwenn two objects?

Question: Is there an add-on or other way to  duplicate specified number of objects between other two objects? Modo has this function and for 3dsmax I have a script: https://youtu.be/A1rFqZxpxjc
Comment: I have not seen an option to do it like the video but the Duplicate (DupliVerts) and Array Modifier should get you the result you need. But you will have to either create a path for the Duplicates to follow or provide the offsets to create the Array.
Answer: If you only want to do a single object between two existing objects, you can use 3D cursor snapping. In object mode, that is respective to the object origins.
So you'd pick the two objects you'd want to place the new object in between. Then snap cursor to selected. (This places the 3D cursor at the average position of both object origins.) Then pick and duplicate an object. Then snap it to the 3D cursor location.
By the way, the default menu option for 3D cursor snapping (2.7x) is Shift+S.
Alternately you may want to look into the Oscuart Tools add-on. It allows distribution of objects. That may function more closely to what you're wanting. I'm not exactly familiar with the workflow involved there, but it'd be worth looking into.


many rational points on an algebraic curve

Question: Given polynomial $f(x,y)$ with integer coefficients, may be reducible, but without linear factors. For positive integer $n$ denote by $a_n$ the number of points $(x,y)\in \frac1n \mathbb{Z}^2$ on a curve $f(x,y)=0$. May it appear that $a_n$ tends to infinity (when $n$ increases taking all positive integer values), but is always finite? Similar question: if we consider only bounded part of our curve, say $\{ (x,y):f(x,y)=0,x^2+y^2 < R^2 \}$, and define $b_n$ as the cardinality of the intersection of this set with lattice $\frac1n \mathbb{Z}^2$, may $b_n$ tend to infinity?
Comment: Regarding the question about bounded points, if a curve has infinitely many rational points then these points are dense (at least in one connected component of the curve). So the only problem I can think of is the case of a cubic curve with a bounded real component.
Answer: I'll assume the curve is irreducible, otherwise look the components.
If your $a_n$'s are unbounded then your curve has infinitely many rational points, so has genus zero or one. In the case of genus one (or genus zero with more than two points at infinity), the $a_n$ will always be finite and, by finiteness of $S$-integral points, the $a_n$ are bounded if e.g. $n$ ranges through integers with a fixed set of prime factors. So $a_n$ does not go to infinity with $n$. But, even in this case, the sequence $a_n$ can be unbounded. E.g. take an elliptic curve with rank one, such as $y^2=x^3+3$.
Maybe you should look at the case of the unit circle for both of your problems. That's not covered by what I wrote above and probably you can figure out precisely what is happening in this case.
Comment: For the unit circle $a_p$ equals 4 for prime $p=4k+3$. I think, a priori the problem can not be reduced to components, because different components may have many points with different denominators.


Refresh behavior when signing into acc from admin

Question: In my product, there is a feature where an admin can sign into the customer's account and manage their settings. (ex: Admin can sign into user's account and change their name) I currently have the design where the dashboard is exactly the same, with a header on top letting the user know what acc they are signed into and an "X" to leave that account. I am curious about the expected behavior when the user refreshes on this page. Should the admin stay signed into the customer acc, or since it looks like an overlay with the "X" should it close and be back to the admin acc? Here is a rough wireframe of my design:
Comment: @Devin It looks like an overlay to the user, on the code however, it is not an overlay. The code replaces the entire header/body and the "back to admin" reverts those changes. There is not a second window on top of the other. It does, however, look like an overlay because the "back to admin" button takes you to the page you were on before you signed into the customer acc, simulating an overlay.
Comment: *"looks like an overlay"* or *"it's an overlay"*? Two different things and the answers are different as well
Answer: You should keep the user with the account they were using. This is very common in server panels when using an admin account and acting as another user; if you reload, you'll still be in that user account. Just imagine the chaos if you were taken to your admin account without noticing it!
Another common example is Google Analytics and Google Ads. You may have a master account and then different subaccounts and even properties. When using either a subaccount or a property, you'll always remain in that subaccount until you log out or end the session.