Why do so few players play razz?

Question: I don't understand why razz is not more popular. Is it an especially difficult game to learn or play? Or are there other causes that keep the game from becoming popular?
Comment: Not many players play razz, because it is unpopular. They just don't like it as much as other games. Or the game does not get a lot of exposure.
Comment: I agree that this question is pretty opinion based, but a big factor into why NLHE is the big game, that I feel is not opinion based, is because ESPN came along and NLHE is way more exciting than a limit game from a TV perspective. Essentially ESPN can capture the highlights, make things seem way more exciting that a normal game of poker is, and a lot of people got their first introduction to poker through TV poker. Before ESPN started televising poker the way it is today, the 'other' games, Razz, 7-card, etc were way more popular. A casino isn't going to spread a game that won't make money.
Answer: Razz has never been a popular game.
Razz is a form of seven card stud. The best hand is a "wheel" which is a 5 high straight (a-2-3-4-5). They also play a 2-7 version. These days in brick and mortar poker rooms, the only time you see razz is when it is included at a mixed games table.
In Nevada, from the 1960's through the mid-late 80's the only regular Razz game was at the Stardust in Las Vegas (AKA the Algiers in the movie Casino). Just like was depicted in the movie, the poker room there had a number of lowlife hustlers working in teams. The razz game was famous for being full of cheating and colluding players. The game was a limit game, typically 15/30  with a dollar ante. Razz may also have been played on a hit and miss basis at tournaments and other rooms. But for the most part the game was considered dirty so most card rooms just didn't promote or spread it.
In the mid-late eighties a few events happened. Organized crime was about done in Las Vegas, so the cheating game of Razz wasn't so protected anymore at the Stardust. The Las Vegas Hilton opened a large poker room of 26 tables in their new super book, and solicited razz players to bring the game over there. By this time the Stardust was way beyond it's heyday (see Swingers), and wasn't attracting very many players of any kind. California opened their poker rooms through a court ruling to Stud and Holdem, and Indian gaming began, and poker started to spread. Atlantic city opened poker rooms about this time. So there was some pressure for Razz players that were serious poker players trying to make a living to move on to greener pastures. (Poker at Wiki)

At the Hilton they reduced the ante in the game by half to fifty cents. This had the effect of cooling the collusion some. But by this time the player pool consisted of mostly aging poker players. Literally as the poker players slowly died off the game died off. Other markets and other games like Holdem and Omaha were also slowly draining some players away, and also capturing any of the new players that came along. For the decade or more before the new markets came along poker was low growth enough in Las Vegas to be virtually stagnant. The Hilton poker room home of the only regular Razz game in the world in the late 80's was slowly loosing business to the new emerging markets. Then in 1989 the boom in Las Vegas started with the opening of Steve Wynn's Mirage. This killed the room at the Hilton and consequently the Razz game.
Razz was occasionally played at the Mirage. But it never became a regularly spread anywhere as near as I could tell. I never overheard players talking about some razz game here or there. When I was running my websites between 98 and 2008, one of which listed about six hundred poker rooms, all of which we surveyed by phone for the types of games they played we never found a regular razz game.
For all practical purposes Razz is a dead game. No regular cash games, only to be found at an occasional tournament, WSOP has a $1500 Razz event still, and for cash  almost exclusively just one of the games in a mixed game stack. I have played it online, but that was a decade ago and really don't know if it is offered anywhere in that venue.
So that short history of razz, should answer part of your question, it is not popular now, because it has never been popular.
So lets discuss why it has never been popular, and in my opinion will never be popular and is forever slated to be a niche game.
Of all the variations of poker, Razz is the most simple minded boring game of them all. Its a game of simple ranks of hands, one does not need to consider suites, or straights. Since it is a seven card game, pairs that are bad, are not so bad. One can still make a clean low with two pair, three of a kind, flushes, and straights in their hand. Drawing to a hand is not a bad thing, everyone is drawing. Everyone misses draws and bluffing, mostly desperate bluffing is the norm. These things raise the general frustration level of the game. This simplicity also narrow the EV between the skilled player and the unskilled player.
Like any poker game, Razz is fun to learn, fun to play for awhile, but when you get down into the grind of the game, Razz sucks. So many negative factors all come together to make it suck. With low EV for even the best of players things get desperate and the players gravitate toward cheating. The players get nasty because the ones that keep coming back to the game are the worst sort of degenerate, lowlife addicted gamblers, or successful cheats. Not even the worlds best Razz player stands a chance against the cheats. And it does not take a great player to get as good as you can get at Razz. It also does not take a great mind to cheat as well as one can cheat in the game.
Razz will always remain a niche game, because when it is spread on a regular basis the game degenerates into a really dirty game to play more so then any other game. That's the bottom line as to why it has never been popular and never will be.
Comment: Ok, thank you for extended answer. Regards, @misdeed!

Why do so few players play razz?


Power of a Multiple Linear Regression

Question: I have a set of models that are the result of a multiple linear regression. I would like to calculate the power for each of these models. I found this tutorial on calculating the power using G*Power. Since the data has already been collected, I am using the Post Hoc type of Power Analysis, and I was hoping to use the "Determine" button to automatically calculate the effect size. However, I see that one need to provide a partial R2, but I would think that for a multiple linear regression (which I think would have multiple partial R2 values), it would make more sense to provide the R2 for the resulting model. Am I misunderstanding the difference between R2 and partial R2? If so, how do I obtain the partial R2 to use in G*Power?
Comment: It's questionable whether one should do post-hoc power analysis at all. See e.g [this article](http://effectsizefaq.com/2010/05/31/what%E2%80%99s-wrong-with-post-hoc-power-analyses/). This refers to post-hoc as in after the *analysis* has been done, not just after the sample has been collected.
Comment: Are they worried about power or over-fitting? If your result was significant, you had enough power to detect the effect size that you found. But if your N was less than about 10 times the number of variables, that can be problematic.
Comment: Interesting. Someone else has already done analysis on this data, where we did achieve significant results. However, someone else was questioning whether we had enough samples to validly perform a multiple linear regression. Does the presence of significance indicate that we do have enough samples, or is there another way to determine this?
Comment: Our N is greater than 10 times the number of variables for some models, and less than 10 times the number of variables for other models. What does having an insufficient N indicate, even if you do get significance?
Comment: That would indicate possible overfitting of the data. To give an extreme example, if you have N-1 independent variables, you will get a perfect fit even if all the data is random.
Answer: In GPower, you do a power for an R2 in multiple regression by doing the partial R2 with no predictors in the baseline model.
To do this, set the total number of predictors to 1, and the number of tested predictors to 1. You're then testing the model against an intercept only model, with an R2 of zero.
(You can always think of regression models in this way - you're testing against no predictors, and looking at the change in R2.)
Comment: Thanks for the explanation. I had never heard of regression as being a change in R2.

Power of a Multiple Linear Regression


'"How to change the song in one time?" Audio Spectrum Visualizer (After Effect)

Question: I made an audio spectrum visualizer similar (After Effects) style like this one with multiple solid layers. But I have a problem: how to change the song in all layers at once? Every layer has different colours, glow and all other effects but the song is the same, so now if I want to change the song I have to go into every solid layer and change FX Audio Spectrum > Audio Layer > Song Name.
Comment: I'm really not sure what you are asking!
Comment: I ask how to make preset for change song in multiple layers at the same time
Comment: What do you mean by "Change song" though?
Comment: I think OP wants to replace the audio source in all layers in one go
Comment: Exactly Luciano i want replace the audio source in all layers in one go
Answer: Using the Audio Spectrum plugin in After Effects, you can reference a layer (using the "Audio Layer" parameter) on your timeline as an audio source. To update all referenced audio sources in your project at once, you can precomp the audio track and use the new nested precomp as your audio source instead.

1) On your timeline, highlight your audio track. Hit CTRL + Shift + C (CMD + Shift + C on OSX) to Pre-compose the audio.
2) In the Pre-compose menu, name it "Audio Reference" or something similar, then select "Move all attributes into the new composition" and hit "OK"
3) This new composition will be added in your Project panel. Use this composition instead of your audio file when working in After Effects.

To update the audio file project-wide with a couple of clicks, just open the reference composition you created and swap out the audio file inside. This will change every instance of that composition in your project.
Hope this helps! Let me know if you need clarification.
Answer: The easiest way to do this is:

you move only audio file to another folder or disk (any audio, either music / recorded files.).
Re-open the project file, it will show you (after effects warning: 1 file is missing and blah blah blah.)
just click "OK"
double click on the specific audio which you moved/ needed replacement!.
It will show up your PC folders, and all you gotta do is select
the audio you modified/ New audio and just add it.
it will automatically refresh and load the effects you put in to your project. [It will take time accordingly with the audio file size]

This is just for the Audio effects though!


phd admission- both professor from same university agreed to be mentor

Question: 2 professors from the same university same department have agreed to be my mentor. But none have funds, prof.1 may give me TA, while prof.2 can give me RA. Which one should I choose? Can I put 2 applications with 2 different SOPs and then decide later which prof. to go for?
Comment: How can a professor with no funds give you an RA?
Answer: I would counsel against two different SoP statements. If they are compared then it makes you look like you don't really know what you want. The university might be fine with two separate applications (or not).
But more important than the source of funding and the activities to earn the funds is the question of what would the research be with each of these people and can one of them better support your career. Make your decision on that basis.
Bot the TA and the RA experience can be valuable for an future academic, but in different ways. But that is less important than the research opportunities.
I'll guess that the TA position is probably easier to manage your time than the RA position, but that isn't necessarily true, depending on the nature of the university and of the field and department.
You might consider trying to set up a three way conversation to see if the two profs have some advice for you about which would be more suitable and advantageous. But that assumes that you know something about them first and are sure that it won't generate bad feelings.


Acceleration-distance graphs

Question: This question, I solved quite theoretically differentiating the equation of given graph $$(v=mx+v_0)$$
taking $m$ as slope of given line. Note that $m$ is a non-zero negative number. Thus $$a=m^2x+mv_0$$ which means that acceleration as a function of distance when graphed should have a positive non-zero slope and a negative y intercept (because $m$ is negative)
rendering option A as correct. But what I don't get is the intuition behind it. I've always solved graphs intuitively without equations and stuff and my first guess was C because the acceleration must be like constant? for the velocity distance graph to turn out to be a straight line? It's kinda my guess that if acceleration itself varied with distance then velocity graph would be curved (basically intuition, no hard proof). But then C has slope $0$ which disagrees with my intuition. If I'm moving with a constant retardation, my velocity will keep on decreasing linearly innit? Please help me understand without involving maths.
($m=\frac{-v_0}{x_0}$ to be precise)
Comment: There must be more to this question than you present.  Can you give us the text of the complete problem?  Also, your expression for $a$ can't be correct if $m$ is the slope of some line.  Advice:  please don't post links to off-site locations. The targeted location might disappear.  Especially in this case, as the image is stored in a Google drive.  Are you prepared to guarantee that you will *never* delete or move the image?  Broken links on this site do happen, and they render the question or answer useless.
Comment: Rew, it would be a good idea for you to develop more trust in following mathematical solutions rather than depending on "intuition".  In my experience, there are many physics problems where the math takes unusual and non-intuitive twists and turns before turning into an answer that can be associated with a physical phenomenon.
Answer: The acceleration is not constant (it depends on x which changes with time).  For a constant a, you would need v = vo + mt, (not mx).

Acceleration-distance graphs


Building an OCX with VS.NET?

Question: What happened to OCX's? Are they no longer possible to create with the latest tools?
I need to create an ActiveX control that I can place in Word, PowerPoint and Excel documents. If I use VS.NET, I can only create a UserControl DLL with COM Interop, but I don't think I can add that using the "More Controls" toolbox of PowerPoint's Developer Tab.
What would you do?
Answer: Yes, you can still create them. But you can't create an OCX with a .NET language. Need to be unmanaged C++ (or VB). The DLLs with COM interop you can create in C# or VB.NET are just .NET objects that are invoked via CCW.
Comment: I'd say that's very likely. With the move to .NET, COM is really a legacy technology now in any form. There's little reason these days to use an OCX. A CCW-invoked .NET assembly is almost always a better solution now.
Comment: Are they on their way out (OCXs)?
Comment: What is CCW?  I assume you're referring to something other than concealed carry.
Answer: You can still use .NET to create ActiveX components. However, this is experimental and not recommended. If you still want to try it you can find information in the blog of Andrew Whitechapel:

Using Managed Controls as ActiveX Controls

But as Andrew says:

This seems to work OK for Excel (with
  the very limited testing I've done),
  partly works with PowerPoint, but
  fails miserably with Word.

So if you don't want to spend a lot of time debugging strange errors I would recommend you to use the old VB 6.0 and Visual Studio 6.0 to build you ActiveX components.
Answer: Use a C++ or VB compiler which can target native code + ActiveX.
Comment: yup. they were so easy to create in VB6.
Answer: Depending on what you need to do, an Add-in may be more appropriate than an OCX, and you can certainly create those with .NET. You should also look at the Visual Studio Tools for Office, not as a direct replacement for an OCX, but only to understand the new capabilities. It may show you new approaches to the problems you're now solving with an OCX.

Building an OCX with VS.NET?


Is mathematical physics theoretical physics?

Question: I was reading about mathematical physics on a university website and they said "an education in theoretical physics is..." appearing to imply mathematical physics is theoretical physics. Is this the case? I want an education in theoretical physics or the closest thing to it. The problem is I live in New Zealand and the universities here don't teach it other than this one potentially.
Comment: Does this answer your question? [Difference between theoretical physics and mathematical physics?](https://physics.stackexchange.com/q/56293/)
Comment: The phrase "mathematical physics" can mean a lot of different things depending on who is using the term for which purposes. It could mean what is known as "theoretical physics" to laypeople (which is more precisely referred to as "high energy theory" or "high energy particle theory"). However, "mathematical physics" can often be used a specific subfield of mathematics/physics, which is not really concerned with the same things as a high energy theory. I'd have to see the webpage to judge how they are using the term.
Comment: You can study theoretical physics at any university which teaches physics.
Answer: Both deal with theory rather than with experiments. The most striking difference is the level of rigour being put in. For example: In theoretical physics you solve differential equations, in mathematical physics you solve them as well, but you also prove that the solution exists and is unique or, if it is not unique, you consider all possible solutions and try to give them physical interpretation.
Theoretical physicists sometimes use mathematical notions which are ill defined, for example, the Gauss integral with purely imaginary exponent. They don't care that much about mathematical rigour (but, obviously, still much more than an experimental physicist would do), use the mathematical tools, if they seem useful intuitively, and deal with new physics. Mathematical physicist would rather deal with the problem of making the Gauß integral with imaginary exponent well defined by considering some suiting limit.
But it's not like mathematical physics are just cleaning up after theoretical physicists, which left some mathematical mess while pushing forward the understanding of physics. While making things rigorous, often new questions emerge, dealing with them might give a better understanding of physics.
Comment: @Vadim Why would writing equations and ignoring divergences be about understanding nature, while writing equations and understanding how to properly deal with the divergences wouldn't? It's not as simple as that.
Comment: I think it is not only about rigor with equations: theoretical physics is *physics*, whereas mathematical physics is a branch of math - it is not about understanding the nature, i.e., formulating the equations, but about solving them.
Answer: Mathematical physics

In the most general sense mathematical physics means mathematical methods used in physics (i.e., those parts of mathematics that are actually useful in physics applications). E.g., one may be interested in developing new methods of solving sine-Gordon equation and analyzing its solutions - without actually changing the physical situations where equation arises and its applicability to such situations.
As @user1379857 mentioned in the comments, sometimes mathematical physics simply means theoretical physics, as opposed to the experimental physics
In some places mathematical physics has more specific meaning - e.g., in Russia the title is common to the books and courses dealing specifically with the second order partial differential equations.

Theoretical physics

In the most general sense it is doing physics with equations, rather than in a lab, i.e., as opposed to experimental phsyics. Sometimes computational physics is treated as a separate branch as well.
The term may more specifically address to the high energy physics, particle physcis or cosmology, thus excluding the theoretical fields that are considered "more mundane" although not necessarily less complex -such as condensed matter theory, quantum optics, etc.


sp.contract can get only explicit string as an entrypoint

Question: I'm trying to call another contract from mine.
It seems like I can't feed sp.contract with an entrypoint string that comes from an argument (it works perfectly fine with an explicit string like "XXX" instead of "entryPoint") because it might be missing when my contract is getting called.
How can I do it anyway?
import smartpy as sp

@sp.module
def main():
    class caller(sp.Contract):
        def __init__(self):
            pass

        @sp.entrypoint
        def call(self, contractAddress, entryPoint, str1, str2):
            c = sp.contract(sp.list[sp.string], contractAddress, entryPoint).unwrap_some()
            sp.transfer([str1, str2], sp.mutez(0), contractAddress)

Answer: Do not specify the entrypoint.
An address can hold the entrypoint information. In this case it has the form KT1XXX%<entrypoint_name>.
@sp.entrypoint
def call(self, contractAddress, str1, str2):
    c = sp.contract(sp.list[sp.string], contractAddress).unwrap_some()
    sp.transfer([str1, str2], sp.mutez(0), contractAddress)

For example, one can call this entrypoint with contractAddress being sp.to_address(sp.self_entrypoint("my_entrypoint"))


Filling column with consecutive and static numbers in QGIS

Question: I've created a new field and used @rownumber to have a numbered index starting from 1. However, when I delete a feature of the layer, the numbered index is assigned once again to all the features so that I cannot preserve the initial numbers for each feature. Is there a way to preserve the numbered index even if some fetaures are deleted later?
Comment: Have you created a virtual field? because if you have then this is the reason the numbering changes since it is a calculated field and the value is not actually stored in the layer. you need to create a new regular field to store the `@rownumber`
Comment: it doesn't make a difference as long as you have not created a virtual field. If it is a "normal" field then there is no reason for the assigned values to change. (the calculated value is stored, not the expression that will be calculated on the fly which is what virtual fields do)
Comment: I've created a new field and then used the expression. Maybe I have to create the new field and use the expression at the same time in the field calculator
Answer: Try maximum("ID-field")+1, provided your ID-field is of type integer.


Why is it required for Test Administrators and Proctors to constantly walk around the room during testing?

Question: I ask this because from my experience, a TA and/or a Proctor moving around the testing room looking over my shoulder is distracting and breaks my concentration. I have been in both positions, TA and Proctor and I try to only move when a student needs something,  such as a pencil or paper. My administrator seemed to get angry today because I was standing in one place too long. I also noticed that I have a better view of testers when I can see them all from a fixed point not turning my back on part of them throughout the test. Does anyone else see moving around  the room as distracting to testers?
Comment: Mostly speculation. If you stay in one place, you're only getting one angle of the students. Presumably, there are cheating methods that rely on limited line-of-sight from the proctor. Staying still for extended periods of time makes it easier for students to determine and exploit your blind spots.
Comment: @Roger: I almost feel that anyone who gets all *Ocean's Eleven* about cheating on an exam may deserve to get some rewards for their good planning and hard work.  Only almost.  But the "cheater savant" is a rare bird indeed in my part of the world.  If I have ever had one in front of me, then I was fooled by the thoroughness of their camouflage: they make sure not to do well enough on the exam for me to notice.
Comment: Cheaters NEVER "deserve" to get rewarded for their work. Ever.
Comment: Wait, that's required?  Oops. // On a different note, I once had a professor who walked around the room, looked at my work and just said "Hmmmm...."  *That* was quite unnerving.
Comment: I also find it incredibly distracting, especially in a small class where rotations around the room are done in quick succession. Your point about visibility is also accurate, in my opinion. That said, it really doesn't make sense to me why they're walking around so much. Then again, I've only been in the position if one being *tested*, so I may be missing something.
Comment: Are you sure that your administrator did get angry? Is it possible that they didn't care about you moving or not? I would say that if you have been specifically told to move around, you can ask the person who told you the reason why. If not, don't assume that it is a 'requirement'.
Answer: It is not globally required: e.g. I do not move around the room unprompted when proctoring an exam.  So it is probably a matter of local academic culture and/or preference of the (head) course instructor.
In these kinds of fine points, academic culture differs so strongly from place to place as to make it clear that there are not always deep, well-thought out reasons for doing things one way or another: many institutions do things in a certain way because they "always" have done so.  The academic culture shock that I experienced in moving from the US to Montreal was considerable.
For instance, I believe that I first learned the word "invigilator" when someone introduced himself to me as an invigilator for my course.  If you had told me before I arrived in Montreal that final exams there have professional invigilators, I might not have believed you.  In the US, if you are teaching a course without a "common final" (I imagine that common finals occur only in a small minority of all courses taught, but I don't know for sure) then you, the instructor, are very likely the sole inviligator/proctor/administrator/TA in the room when an exam is taking place.  (Not that having professional invigilators is a bad idea: if you have the infrastructure in place, why not?  But I assume that they get paid at least a little bit, and this would never fly in the state university at which I currently reside.)
So understand that when I say that I do not move around the room when giving a final exam, I mean no one does.  It is not really clear to me what cheating I might be missing out on by only looking at the students from the front of the room.  I should say that I am used to a classroom environment with stadium seating and small flip-up desks so I can see a lot from the front of the room.  I also teach a subject -- math -- such that coming in with a cheat-sheet or something like that would not make things that much easier.  Even consulting the internet on your cell phone while in the bathroom would not help that much -- one does need to show one's work, after all -- unless the student was unusually insightful and well-prepared about how to cheat in this way.  In my local academic culture, cheaters are not insightful or well-prepared...
Comment: invigilators are typically grad students and drawn from the TA ranks. If they are paid it is at the same sort of rates they are paid to TA
Comment: @Kate Gregory: At McGill, invigilators are not only paid but also unionized: http://www.mcgill.ca/students/exams/invigilation.  Apparently about 90% of them are students (which I didn't know until just now; the small number of invigilators I worked with were much older than I was).
Answer: Moving around is probably not required (to do so would be micromanaging on the institution's part), but as others have pointed out, it depends on the "local academic culture" and most importantly on the instructor or whoever is in charge of proctoring the exam.
As a TA of various roles, when I've given or proctored an exam, I tend to move a lot. Here's why:

Depending on the test and how much time is left for the students to finish, there can be a lot of questions, which makes me move a lot.
I've found that moving around encourages students to ask a question. (Most of my questions are from people as I pass by them.) I think they'd rather not raise their hand if I'm in the front of the room or not paying attention because it would slow them down. If I'm near them, they can ask their question quickly and move on. This is my primary reason for moving around when not asking questions.
Moving around tells my students that I'm eager to help them with their questions. It shows I'm interested and mindful of them. It's an accurate impression, but I want them to see it.
I've caught a ton of wandering eyes by moving around. It's easier for me because I can't see the eyes of the students in the back rows well enough from the front of the classroom (15-25 m away). Surprisingly, watching students from behind can be useful because they are sometimes more careless, or don't see me watching their heads turn towards their neighbor's work.
Moving around gives my students the impression that I'm watching them more closely, and I hope that this (if nothing else!) discourages them from cheating.
Comment: 6. If the exam has been going on for a long time or your even have to proctor two or more sequential exams, moving around keeps your attention high.
Comment: I second #2.  I also like to go around periodically to check on how far the students have gotten part-way through.  This helps me gauge my test-writing acumen.
Answer: When I have proctored exams, I tended to move around the room for two reasons:

Answering questions: in a good-sized class, there's a lot of motion just from answering questions.
When not answering questions, I would sometimes move around the class just because I was feeling rather bored and restless.

Looking for cheaters didn't actually enter into it.
Comment: @ArtGalaxy If the students aren't allowed to ask anything, then the only thing I can think of is that they really want you to look for cheaters... or at least to make the students *think* you are looking.  And I suppose you'll at least be uniform in your impact if you're moving around.
Comment: This testing situation is a state test where teachers and students are not allowed to discuss the test at all. Sure during the beginning of test when everyone is getting settled, then as students complete the test, we go to them and retrieve the tests and other materials. My issue is why am I required to meander around the room during the test to look over students' shoulders when I can see everyone from one fixed point.
Comment: @Roger-I could understand the point of moving according to angle or vantage point; however, in this room, the desks and students were in individual desks with large tops and the students hands, legs, bodies, and feet are visible from a single vantage point. Also, no bags, purses, cell phones, etc.are allowed in the testing room. Constant or near constant walking just seems an unnecessary distraction for the students testing.
Comment: @jakebeal- I think you're probably correct in your reasoning. I have a bit of a stubborn streak that won't allow my mind to rationalize disturbing students who are testing in order to fulfill a requirement that doesn't make a whole lot of sense.
Comment: I took an entrance exam once where the proctor kept rattling the coins in his pocket.  After half an hour of that, I was ready to strangle him.  - - -  Could you ask the administrator your question, without alienating him/her?
Comment: I am sure that the administrator was angry because of the way he handled a situation in the testing room where I was a proctor. He angrily unlocked the door of the room where I was proctoring a test and rudely gestured to me with his hands to move around.


is it possible to write two threads in arduino & switch between data & AT command mode dynamically?

Question: I need to recieve data from a android phone and also estimate distance of phone from the HC 05 module , So i want to write two threads or functions. In one part i want recieve data from android phone store it in a variable and in other enter AT command mode to see RSSI of the phone.
Answer: You can switch between command and "online" modes, but you cannot use both at once. While in command mode you can't be receiving data, and while in online mode you can't be sending commands.
You would have to periodically switch between the two in your program at a time when you know you won't be receiving data (or you don't mind losing that bit of data).
And no, the Arduino doesn't have the concept of threads. There are a number of libraries that do emulate threads to a certain extent, but they are (usually) cooperative multi-tasking (one "thread" has to manually "yield" to another) and as such are pretty pointless, since you can get the same effect with a decent finite state machine.
Answer: Receiving serial characters is partially done in hardware and partially in the Arduino firmware.
The only thing you need to do here is:
- check if you've received a full message.
- handle the message (if a full message has been received)
For the ultrasonic sensor, you'll have to:
- Start a measurement.
- wait for a response.
- convert the response.
You can't do these exactly at the same time. But none of these will/should take up 100% of your "processor time".
So we'll have to "share the time" of your microcontroller.
Though, timing may be critical in some of these parts. So for this, we will have to make sure that there is no delay in, for example, handling the ultrasonic sensor.
(I'm not sure, and can't check since I'm on my phone, if Arduino uses interrupts for receiving characters from serial, which may mess up my logic below if these interrupts aren't prioritized well or the time an arduino RX interrupt takes so long it will influence your ultrasonic sensor readings significantly.)
But, you'll want to handle the ultrasonic responses as fast as possible, since this timing indicates the distance between an object. And you don't want to use processor speed while waiting for the result.We will probably want to use an interrupt here (if polling doesn't give a good enough result). See the Arduino page in interrupts, or use a HC-sr04 library that is non-blocking or interrupt/event-based
Now you can, if the ultrasonic sensor doesn't require any attention. Check and handle your serial buffer. (I believe there even is an option in the Arduino library to set a callback for a line end).
By doing the above. You can be sure that your ultrasonic readings are stable. And as long as the ultrasonic routines don't consume all processing time (which they shouldn't), you'll have time to handle your serial messages.
Also have a look at state machines and non-blocking code, to avoid making your "subroutines" block eachothers execution while waiting for a result.


What would happen if you reduced the coupling of $SU(2)$ in the standard model to zero?

Question: Ultimately, my goal is to find a free parameter that you could change in order to significantly reduce the strength of, or eliminate, the weak interaction. Would such a modification leave other parts of the Standard Model unchanged?
Comment: This question (v2) seems quite broad.
Answer: It is straightforward to see, even though your ultimate vision should be in trouble. I assume you mean decrease the coupling g of just SU(2), and leave the EM coupling e and the Higgs v.e.v. v alone, which cannot be done.
You then just look at the formulas:
$$\cos \theta_\text{W} = \frac{g}{\,\sqrt{g^2+g'^2\,}\,}, \qquad     \sin\theta_\text{W} = \frac{g'}{\,\sqrt{g^2+g'^2\,}\,}  \\
 e=g\sin\theta_W= g'\cos \theta_W\\
 m_\text{Z} = \frac{m_\text{W}}{\,\cos\theta_\text{W}\,}, \qquad  m_W= {ev\over 2\sin\theta_W}, \qquad G_F= 1/(v^2\sqrt{2}). $$

As $g\to 0$, the Weinberg angle increases to π/2, its cosine vanishes, and its sine goes to 1.

But note, the EM charge e  cannot stay invariant, since $e\to g$.

The Fermi constant stays put; the mass of the W  goes to zero (it stops coupling!), and the mass of the Z goes  to $g'v/2$.

NB aside: If you are seeking decoupling, the opposite limit, $g\to \infty$, paradoxically is better behaved, and often taught in class: in that case, e can stay unchanged, $e=g'$, since $\theta_W=0$, the cos is 1, and $m_Z=m_W=\infty$, while the Fermi constant is what it always was: the old Fermi theory! So you may think of EW unification as a descent from infinite to a finite g...
Comment: Ahhh, okay i see
Comment: Thank for your answer!
Comment: Wait - I have a question. In your postscript, are you saying that you can eliminate the weak interaction by increasing $g$ to $\infty$? What does it mean to have bosons with infinite mass? Perhaps it would be also interesting to explore a more intermediate case - what if $g$ was significantly larger?
Comment: Not eliminate! Hide the gauge bosons thereof! The Fermi constant stays the same, so the four-fermion interaction is still what it was for 40 years (1933-1973), when they didn't quite know how heavy the vector bosons were! Plug *g* s twice, 10 times, 100 times the real value!
Comment: Wouldn't the weak force still exist where you have virtual particles mediating the interactions? Although a temporary disturbance in the quantum field with infinite mass is crazy to think about...
Comment: As I said, the 4-fermi interaction, and beta decay, etc... would *still* be operative at low energies; only the virtual vector bosons would he absent. You are doing yourself a disservice if you try to visualize these phenomena as disturbances in quantum fields.  I gathered you were familiar with the mainstream working of the standard model....


Should low frequency categories be eliminated or amalgamated in multiple correspondence analysis?

Question: This is just a theory question so no code or example is required I think. I want to know if there is a rule of thumb for what constitutes low frequency in MCA? Should the variable be eliminated? Could adjacent categories be amalgamated instead? And what consequences occur if the analysis goes ahead without taking action? Thanks
Comment: Not so much expertise on MCA around here it seems... I don't think there's a rule of thumb. Generally the impact of low frequency categories will depend strongly on the behaviour of few observations, and may be much larger than you'd want it to be. I however expect that "how small is too small" depends on the overall dimension of the problem. If there are few other variables and categories, fewer observations will be needed. I'd use amalgamation if the resulting category makes sense in terms of its meaning. "Adjacent" implies you have ordinal variables, which may make it easier.
Comment: Just another comment: the webpage at STDHA (http://www.sthda.com/english/articles/31-principal-component-methods-in-r-practical-guide/114-mca-multiple-correspondence-analysis-in-r-essentials/) which describes the use OF MCA through R (quite usefully) is the source for the suggestion that low frequency category variables should be dropped. But Costa et al (2013) (https://www.hindawi.com/journals/jar/2013/302163/) says that MCA does not pre-require the kind of assumptions that Pearson and Fisher tests require.
Comment: You could also do sensitivity analysis, i.e., compare solutions with and without the category in question (or with and without variable or amalgamation), and then try to understand from looking at the data whether the impact of the low frequency category is harmful and inappropriate.
Comment: Also, I have seen https://stats.stackexchange.com/questions/575818/how-many-observations-should-there-be-for-the-less-frequent-level-of-a-binary-va. I'm hoping for  more detail on this.
Comment: Thanks Christian, all good suggestions. Yes MCA doesn't seem to be well-known compared to other techniques.
Answer: Le Roux and Rouanet (2005, p.216) suggest a 5% rule of thumb to identify low-frequency categories. What to do once you identified these low-frequency categories depends on the circumstances of your research. Le Roux and Rouanet mention three options: grouping, deletion, and a third method called "specific multiple correspondence analysis".
Low-frequency categories are influential in multiple correspondence analysis, and will often appear remotely from the center of the graph of variables (an exception to that may be when you have a variable with many different categories, hence with many low-frequency categories). In turn, other categories will be pulled toward the center of the graph. This may, or may not, be a desirable property for the purpose of your analysis.
For instance, if you're interested in analyzing these specific categories, it would make little sense to merge or delete them (except, maybe, for the purpose of some sensitivity analysis as suggested in the comment section). However, if these categories are of little interest in themselves compared to others, and can be meaningfully merged with another category, then it may be a good option.
Merging or deleting may have unexpected effects on other categories, so that's something to consider carefully relative to the research question you may have.
Besides merging categories or discarding observations, you can also use a method called "specific multiple correspondence analysis". Contrary to simple deletion, specific MCA won't remove the observations belonging to these low-frequency categories, but it will take away the influence these categories have on the MCA.
Specific MCA might be appropriate if you have several low-frequency categories, otherwise its effect might be almost indistinguishable from merging categories or removing observations. (NB: this remark comes from simulations I ran. I'm confident in the results of those simulations, but as it's an empirical result and not a theoretical one, it may not apply in all circumstances.)
For a theoretical discussion of "specific MCA", see Le Roux and Rouanet (pp. 203-214). For software implementations, there are at least three R packages offering this method:  soc.ca, GDAtools, and FactoMineR. I don't know if other languages or statistical software offer this method, but it may be a question for softwarerecs if you're interested.

Reference
Le Roux, B., & Rouanet, H. (2005). Geometric Data Analysis : From Correspondence Analysis to Structured Data Analysis. Springer Netherlands. https://doi.org/10.1007/1-4020-2236-0


Raspberry Pi Zero As A Mountable USB Stick?

Question: Is it possible to format a Raspberry Pi Zero to automatically mount as a hard drive when plugged into a computer via USB? If so, are there any available tutorials or resources to help me achieve this?
Comment: Stephan, welcome to the RPi flavored corner of Stack Exchange. As is, your question is way too broad to be reasonably answered in this format. I suggest you take the [tour](https://raspberrypi.stackexchange.com/tour) and take a look at [what kind of questions](https://raspberrypi.stackexchange.com/help/on-topic) work well here.
Comment: Your project has a ton of moving parts. If you need help from us, you'll need to break it down into much, much smaller and more manageable pieces. Presenting as a mountable device, being powered by a battery, and interfacing with Dropbox are all completely separate projects and can be rather involved in their own right.
Comment: Thanks for the advice! I'll try moving forward with just one part at a time.
Answer: I used this to create a 1.5 MB Pi USB memory stick:
g_mass_storage - To have your Pi Zero appear as a mass storage device (flash drive), first create a mini filesystem in a file on your Pi with:
    dd if=/dev/zero of=./piusb.bin bs=512 count=2880

and set it up as a fat32 filesystem with
    mkdosfs ./piusb.bin

Then load the drive use:
    sudo modprobe g_mass_storage file=./piusb.bin stall=0

Of course you can make the piusb.bin file as large as you want your USB file system to be.  512 bytes * 2880 = 1.5 MB
Thanks to this Raspberry Pi forum posting I found that if you use g_multi instead of g_mass_storage (or in fact g_serial and g_ether) you can use all of the USB gadgets!!!
Therefore you want the following in your /etc/rc.local:
    /bin/sleep 5
    /sbin/modprobe g_multi file=/home/pi/piusb.bin

As an added plus you can now also use screen to get to the console of your Pi using the command:
    screen /dev/ttyACM0 115200

You may need to enable serial console using the command:
    sudo systemctl enable email@example.com


Compare and contrast Google Drive and Dropbox

Question: What's the practical difference between Google Drive and Dropbox?  What features does one offer that the other doesn't?  Specifically, how well does sharing and permissions work in Google Drive (Dropbox is fairly straightfoward)?
Answer: One could sit and write all the features here, but you could also just google around as there are hundreds of post on this right now. The one from The Verge has a more elaborated answer.
Answer: I will list a few here but as the user Bibhas has stated, they can be easily found on Google here.

Google Drive could be considered more reliable. Google hosts way more then just users files and have been hosting it for ages. They have handled the terabytes of YouTube videos, and are one of the biggest internet companies out there. I would say Google is more reliable for having that safety feeling.
On the other hand, Google has been known to push there privacy policy as far as they can, and they could possibly be snooping in your files, specifically your text files. Of course this is possible for Dropbox as well, but they have less reason to want to do this.
Dropbox only offers 2GB for a free user, where as Google Drive gives you 5GB to start with. Yes with Dropbox you can earn a lot more space for free by refering, but lets be honest not to many people have the time to get that many people to sign up.
If you need more storage and want to buy more, Google Drive is only $2.49 for 25GB, or $4.99 for 100GB. Dropbox costs $9.99 for 50GB and $19.99 for 100GB
More people use Dropbox. ~45 Million
Google Drive implements Google Docs, so you also have all those tools along with Google Drive
You can use OCR to search documents to find a specific bit of text in it.

Of course this list is not complete. Others feel free to add on to this!
Answer: The key thing about Google Drive, for me, and the real relevance of this topic to this stack is they are the first link up the file types with the web applications that open them. Just as an OS UI generally knows what to do with an open file request, so should your web drive.
And does it really make sense for every web application to have a separate UI for file access? I think not, it's not that you use one web application with each storage provider, you use one storage provider with each web application. Each of the storage providers should be offering a standard open/save in JavaScript (Google do this for open), so if you pick one provider, you get a standard interface everywhere for them.
What I'd love to see is the storage providers standardise on a web application manifest and to all provide their file access widget in JavaScript. This would make it far easier for web application providers to integrate with the storage providers.
Answer: I want to point out here that I wish Google Drive gave users the option to open files in Docs or via a native app like Pages. The lack of this option is why I use Insync. Insync syncs all Google Docs as .DOC files so I can open them in Pages if necessary. It's rarely necessary, but it's nice to have the option.


Reading files from android application

Question: I have 9000 files in the memory card, I have created an application that read each file and parse it in to a String and put that String in a HashMap in order to avoid repetition,and save the names of only unrepeated files, but that operation is taking about 7 to 10 seconds.. is there any way to make that faster.. I tried to find better method but I didn't find anything...
Object : get the names and Strings of all the unrepeated files..
Is there any way to do that faster using java, new IO, or NDK, is it worth to install NDK and try that using C language and pointers..
I appreciate any help..
Answer: Its obviously that even just loading of 9k files takes some time. Imagine there a lot of android devices and there are many devices which are not as fast as device which you are using to test. So the operation gonna take even more time on such devices. The problem is a numerous of small files. Maybe you should pack it in one, lets say a JSON or XML? Reading of 1 file is more faster and reliable.

Reading files from android application


Regression for non linear data

Question: I have a non-linear dataset. How can I do regression in orange? Orange does not have polynomial regression widget.
Answer: You have to add higher order polynomial terms to your dataset yourself. Accordingly, you should add new columns to your dataset which consists of higher order polynomials with the desired combination of the existing features. It means that if you have already a CSV file which contains two columns for different samples, you have to add your desired extra feature columns which are the polynomial combination of the existing features. Then use simple regression.
Answer: Orange has a Polynomial Regression widget in the Educational add-on.
Answer: In R: Try regression with a higher-order polynomial by adding I(variable^power) to the regression formula. You don't need to add another column if you use polynomial terms in the lm function. e.g, if you want to add the quadratic of a certain variable called feature:
lm(response_variable~feature + I(feature^2), data=Orange)

Regression for non linear data


Failed to restore snapshot - IndexShardRestoreFailedException file not found in elasticsearch

Question: I am using 3 node cluster setup with the elasticsearch 1.3.1, i have 17 indices each one is having min 0.5 M (1Gi)  documents and 1.4 M (3 Gi) max. now i would like to try the snapshot and restore process in my cluster. i used the following REST calls to do the same...
To create a repository:
curl -XPUT 'http://host.name:9200/_snapshot/es_snapshot_repo' -d '{
    "type": "fs",
    "settings": {
        "location": "/data/es_snapshot_bkup_repo/es_snapshot_repo"
    }
}'

Verified the repository:
curl -XGET 'http://host.name:9200/_snapshot/es_snapshot_repo?pretty' the response is
{
  "es_snapshot_repo" : {
    "type" : "fs",
    "settings" : {
      "location" : "/data/es_snapshot_bkup_repo/es_snapshot_repo"
    }
  }
}

done the SNAPSHOT using
curl -XPUT "http://host.name:9200/_snapshot/es_snapshot_repo/snap_001" -d '{
  "indices": "index_01",
  "ignore_unavailable": "true",
  "include_global_state": false,
  "wait_for_completion": true
}'

the response is
{
  "accepted": true
}

then I am trying to restore the snapshot by the request
curl -XPOST "http://host.name:9200/_snapshot/es_snapshot_repo/snap_001/_restore" -d '{
  "indices": "index_01",
  "ignore_unavailable": "true",
  "include_global_state": false,
  "rename_pattern": "index_01",
  "rename_replacement": "index_01_bk",
  "include_aliases": false
}'

ISSUE:
As I informed I have 3 nodes. the index which I am trying to take snapshot & restore is has 6 shards and 2 replicas.
Most of the shards and its replicas are restored properly, but sometimes 1, sometimes 2 primary shards and its replicas restoring is not happen. those primary shards are in the INITIALIZING state. I allow the cluster to relocate them for more than an hour but the shards are not relocating to the correct node... I got the following exception in my node.
the restore process trying to place the shard in the other 2 nodes... but it can't possible...
[2014-08-27 07:10:35,492][DEBUG][cluster.service          ] [node_01] processing [
  shard-failed (
    [snap_001][4],
    node[r4UoA7vJREmQfh6lz634NA],
    [P],
    restoring[es_snapshot_repo:snap_001],
    s[INITIALIZING]),
    reason [Failed to start shard,
    message [IndexShardGatewayRecoveryException[[snap_001][4] failed recovery];
  nested: IndexShardRestoreFailedException[[snap_001][4] restore failed];
  nested: IndexShardRestoreFailedException[[snap_001][4] failed to restore snapshot [snap_001]];
  nested: IndexShardRestoreFailedException[[snap_001][4] failed to read shard snapshot file];
  nested: FileNotFoundException[/data/es_snapshot_bkup_repo/es_snapshot_repo/indices/index_01/4/snapshot-snap_001 (No such file or directory)]; ]]]:
done applying updated cluster_state (version: 56391)

Could anyone help me to overcome this issue and please correct me if I done any mistake in these process...
FYI I am using master node to pass the curl request
Comment: This issue is fixed, i made a mistake in the fileSystem location. actually we need to point a shared file system. i gave a local location of each node. Now i changed the location to an shared mount folder [which is accessible by all nodes] and this issue is fixed, Thanks
Comment: The snapshot will be stored in a common location of all nodes. the restore operation also done :)
Answer: We need to provide a shared file system location which should be access by all the elasticsearch nodes with read & write permission.


Positive inuequality of Laplacian with a variable coefficient

Question: Let $0 < a_0 \leq a(x)$ be a smooth function on $\mathbb{T}=[0,2\pi]$,  $a(0)=a(2\pi)$, whether it holds that
$$
\int_\mathbb{T} a(x)|\partial_x \varphi|^2 \geq \int_\mathbb{T}\frac{\partial^2_x a }{2}\cdot |\varphi|^2dx
$$
for all $\varphi \in \dot{H}^1_{per}(\mathbb{T})$? More precisely,
$$
\varphi(0)=\varphi(2\pi), \quad \int_\mathbb{T} \varphi dx =0.
$$
Prove it or give a counterexample please.
Comment: counterexample: $\phi(x)=\cos(x)$ $a(x)= 2+\sin(Nx)$ gives $$ 2\pi > N^2\frac{\pi}{2}.$$
Comment: This reads like homework.  Voted to close.
Comment: Thanks for your answer. For your example, the right hand side is 0, which does not give a contradiction.
Comment: It's a question of notations. If $\partial_x^2 a =(\partial_x a)^2$ my counter example works. If $\partial_x^2 a = \partial_{xx} a$ the answer below gives the counter example.
Answer: The inequality does not hold.
Take $a,\varphi$ smooth, periodic, then integrations by parts yield
$$
0
  = \int_\mathbb{T}(a\varphi^2)_{xx}
  = \int_\mathbb{T}(a_{xx}\varphi^2+a(\varphi^2)_{xx}+2a_x(\varphi^2)_x)
  = 2\int_\mathbb{T}a_{xx}\varphi - 2\int_\mathbb{T}a(\varphi^2)_{xx}
  = 2\Bigl(\int_\mathbb{T}a_{xx}\varphi^2-\int_\mathbb{T}a\varphi_x^2\Bigr)
    - 2\int_\mathbb{T}a(\varphi_x^2+2\varphi\varphi_{xx})
$$
that is
$$
  \int_\mathbb{T}a\varphi_x^2 - \int_\mathbb{T}a_{xx}\varphi^2
    = -\int_\mathbb{T}a(\varphi_x^2+2\varphi\varphi_{xx})
$$
and there is no reason the right-hand side should be positive. Take for instance $\varphi=\cos x$ and $a=1+b$ with $b$ non-zero (and big) only where $\varphi_x^2+2\varphi\varphi_{xx}=\sin^2x-2\cos^2x$ is positive.
[update] The inequality does not hold even in its general version. Say you want to prove that $\int a (\varphi_x)^2\geq\lambda\int a_{xx}\varphi^2$, then by integration by parts $\int a_{xx}\varphi^2=\int a(\varphi^2)_{xx}$ and the supposed inequality reads $\int a((\varphi_x)^2-\lambda(\varphi^2)_{xx})\geq0$. By localization (with a smooth cut-off function), it is sufficient to find a $\varphi$ (not necessarily periodic, by the same localization argument) such that $(\varphi_x)^2-\lambda(\varphi^2)_{xx}<0$ on an open set. This is easy, take $\varphi(x)=e^{a e^{bx}}$ and choose $a,b$ suitably.
Comment: Thanks for your nice answer. For your post, I make some correctons on previous problem. See above. Your example is not sufficient now. The question arised in the study whether the operator $-a\partial_x^2$ is positive (under the above assumptions). But now it seems little is known.


Just sitting around

Question: What are the alternate ways of telling a person that I am doing nothing in the moment? I have tried using 'just sitting around'.
Comment: Idiomatically, *[I'm] **just hangin' / hanging** [out, about, around]* is increasingly common among younger speakers. Or more extreme, ***just vegging*** (mindlessly *vegetating*).
Answer: There isn't a definitive way of saying this that I can think of. To my knowledge most ways of expressing that you're not doing anything in particular right now are idiomatic.
With that said here are a few idioms that you could use if you want to vary your vocabulary a little:

Just chillin'
Nothing much
Hanging out
Twiddling my thumbs
Nothing in particular

There are loads of ways to express this, these are one's I'd use.

Just sitting around


Where to close the cursor , when using an adapter and an asynctask?

Question: I am using AsyncTask to retrieve some information from a database.
Hence inside the doInBackground method, I write the following code:
dbhandler = new DbHandler(TestActivity.this);
db = dbhandler.getReadableDatabase();
Cursor cursor = db.rawQuery("SELECT sys_id as _id,statuses,user,created_at FROM status ORDER BY sys_id desc", null);
Log.d(TAG, "cursor got count: " + cursor.getCount());
return cursor;

I use this cursor in the postExecute method which is:
adapter = new TimelineAdapter(TestActivity.this, cursor);
listTimeline.setAdapter(adapter);
dbhandler.close();
Log.d(TAG, "Getting here ...");

Now the problem is: if I keep this as it is, I get an error in logcat, saying that the cursor has to be closed. But when I write cursor.close() in postExcecute() method, I am not able to retrieve a single record though my cursor has a finite count.
Answer: Let the Activity handle it, by calling startManagingCursor().


GNUHealth installation fails

Question: Today trying to install GNUHealth on my Fedora 27 (32bit). after creating user I tried to verify PostgreSQL.
here's a file pg_hba.conf in /var/lib/pgsql/data. This shows following lines

In this file i found no line containing local all all trust
gpg --recv-key 0xC015E1AE00989199 returns this
gpg: directory `/root/.gnupg' created
gpg: new configuration file `/root/.gnupg/gpg.conf' created
gpg: WARNING: options in `/root/.gnupg/gpg.conf' are not yet active during this run
gpg: keyring `/root/.gnupg/secring.gpg' created
gpg: keyring `/root/.gnupg/pubring.gpg' created
gpg: no keyserver known (use option --keyserver)
gpg: keyserver receive failed: bad URI

gpg --with-fingerprint --list-keys 0xC015E1AE00989199gives
gpg: /root/.gnupg/trustdb.gpg: trustdb created
gpg: error reading key: public key not found

After downloading the detached signatures I tried to verify the package by gpg --verify gnuhealth-3.2.9.tar.gz.sig gnuhealth-latest.tar.gz
this too shows
    gpg: error reading key: public key not found

Then I wrote this of installation ./gnuhealth-setup install

 this returns 2018-05-20 07:08:03 [INFO] Starting GNU Health 3.2.9 installation...
2018-05-20 07:08:03 [INFO] Checking requirements
 -> Looking for wget... wget is /usr/bin/wget
 -> Looking for 2to3... 2018-05-20 07:08:03 [ERROR] 2to3 command not found. Please install it or check your PATH variable
2018-05-20 07:08:03 [INFO] Bailing out !
2018-05-20 07:08:03 [INFO] Cleaning up temp directories at
2018-05-20 07:08:03 [INFO] removing base dir at

So I failed in installation. what should be done in this?
Answer: This is because 2to3 is not installed by default and can be installed by using
yum install python-tools

Now running ./gnuhealth-setup install should allow it to complete

GNUHealth installation fails


Book, possibly YA, about a young girl with fey powers, sought by fey royalty to destroy the human world

Question: A young girl finds out she has fey powers, and the fey queen wants to use her to destroy the human world. The dark fey king turns out to be her uncle and trains her. Meanwhile she falls for the guy who is supposed to turn her in to the fey queen to get access back to fey lands. She is helped by two male pixies I believe.
Comment: I recommend going to https://scifi.meta.stackexchange.com/questions/9335/how-to-ask-a-good-story-id-question for help in improving your question. I'm making some modifications, but there are things only you can answer like what kind of book it was and where you read it.
Answer: Darkness of Light (2013) by Stacey Marie Brown.
It's the accepted answer to this Goodreads identification request, posted the same day as this very SFF question and worded the same.
The Goodreads blurb itself isn't that much of a match, but still:

Ember Brycin has always known she's different. After finding her mother brutally murdered, her mind begins to see things that shouldn't exist. The more she feels her sanity slipping, the more bizarre and inexplicable things constantly happen around her.
When the latest school explosion lands her in a facility for troubled youth, she meets Eli Dragen—a hot as hell, darkly mysterious, bad boy from a notorious biker gang. Their connection is full of passion, danger, and secrets. Secrets that will destroy her life and lead her down a path she could never imagine.
A pawn in a war between the Fae.
In the middle of Light and Dark, Ember finds another world where truth and knowledge are power and no one can be trusted.
No one.

Found with the Google query fantasy book "fey queen" pixies uncle which returned the above ID request.


Is a sentence's deep structure representative of i-language?

Question: That is to say, is the deep structure supposed to be what's happening in our head when we speak a language? Or is this just to make our model of a grammar consistent?
Comment: No. Not even Chomsky believes that all that stuff actually happens, and certainly nobody else would. "Deep" (or any other non-obvious abstract) "structure" just means a common anchoring point for multiple synonymous utterances with different regular variations ("transformations", "rules", "alternations", "derivations", etc.) in their syntax. Like finding the fundamental group of a topological space.
Comment: *Not even Chomsky believes that all that stuff actually happens, and certainly nobody else would* -- unfortunately, I don't think this is true. I recently asked this very question of a fairly well-regarded linguist, who has published in Language etc. (don't want to name names here), and he replied very forcefully that transformational grammar was intended as a model of what actually happens in the brain when we produce and parse language, and that there'd be no point in it otherwise. I don't know how widespread that position is among syntacticians, though.
Comment: As long as they're useful, use'em. When they're not, don't. Think about it. Transformations ([sample here](http://www.umich.edu/~jlawler/haj/Preliminarybufattenedlistoftransformations.pdf)) are pretty regular, and don't change meaning. If you could analyze any sentence into one of a (small, in a mathematical sense) set of basic sentences through a series of relations, wouldn't it be worthwhile? That's the goal, and it keeps receding, like all goals, as we find out more stuff and our earlier ideas are refined. But there is progress.
Comment: Fewf! So I guess my next question would we be: why even posit transformations in the first place? It doesn't seem very parsimonious. Sure there's factors like valency, theta role assignment etc. but I don't see these concepts as being 'wrong' if they're not linear.
Comment: There is a certain vogue in Minimalism -- the latest brand -- to call it "biolinguistics" and talk about "mutations" in the brain, but this is clearly only metaphor, not biology. They're just covering their assumptions, the most important of which is "Universal Grammar".
Answer: Deep structure appears to be very real. The modern research on the topic of syntax supports this notion to such a degree that the syntactic movement of constituents is basically a truism. Some fairly revealing, though obvious interpretable, data on the topic of the essentiality of deep structure exists in many fields. One of those fields of research is that of syntactic complexity.
Many experiments have demonstrated that until a syntactic constituent has found its 'original' location, the mind uses up resources 'holding on' to that constituent. Experiments will, for example, produce paradigms were an individual is presented with an apparently moved constituent early on in the sentence, and then later down the sentence place ambiguous anchor, that immediately after processing a certain word, there arises a possibility for there to be an appropriate location for the moved constituent to be located. As the sentence continues, following words disambiguate the potential 'deep location' into an incompatible location. This causes a fair amount of processing strain, and this strain is incremental as more and more 'false origins' are presented.
While this doesn't exactly prove that there is a neurological 'structure', it certainly supports the hypothesis. Almost all investigation into the topic reveal very sophisticated phenomena, going from people 'perceiving' prosodic qualities which are not there in anchoring locations, to people conceptualizing the utterance in its basic-most structure.
Deep structure seems to be so representative of our I-language that we will even commit ungrammatical utterances during communication to grammatical and deep structure-like sentences. Chomskyans, and many other linguists, really believe the deep structure to be the intensional (the 'actual' structure) of I-grammar, and for movement to be executed by the parser, which is a component that operates with the grammar, but isn't the grammar. I'm not sure, but I think that evidence for deep structure is even found in aphasiac studies.
The way that you say to make our model of grammar consistent, is I think to mean 'to make our external analysis of grammar' consistent. That is what we call 'extensional' analysis. Most modern linguists are postulating theories concerning the 'intensional' (structural) properties of language.  They're saying "that's what's actually happening in the brain".
Sorry for going on so long!!
Thank for reading!
Answer: The distinction to keep in mind is competence and performance. Chomsky is exclusively concerned with competence, ie. the knowledge of the language system. Deep structure is what defines the competence but it is not active in the process of performance (except in some very constrained contexts such as garden path sentences. Nobody assumes that we're actually generating sentences from some deep structures during the process of utterance. However, the deep structure places constraints on what can be said and what is said.
Of course, Chomsky (and most generative linguists) don't actually use the concept of deep structure any more. The earlier concepts of deep structure have moved to the notion of Universal Grammar. There are some theories like the Prague Functional Generative Description framework that postulate something akin to deep structure (they the FGD people call it tectogrammatic level) but they still would not claim this to be a part of performance.
However, your summary and the actual question are contradictory. I-Language is NOT "what's happening in our head when we speak" that's a description of performance while i-language (for Chomsky, if I understand him correctly) is still a part of competence. It's just a different way of looking at what competence is. It is contrasted with e-language to cover all the externalities like communicative and social knowledge.
Finally, it's precisely this sort of question that has led many linguists (myself included) to abandon the performance/competence distinction and postulate the usage-based model of language.
Comment: Even if one thinks "Universal Grammar" is hogwash (as most linguists do), deep structures (whether called that or not) are still useful, because they link derived structures by rules, and the rules are clear, even if UG isn't. (BTW, "i-language" is another concept that most linguists don't believe in, either, so your question is not going to get a general answer)


Does the sum over reciprocals of all sums of squares converge

Question: There is a useful result when studying lattices that the sum $\sideset{}{'}\sum_{m,n}(m^2 + n^2)^{-r/2}$ converges for $r > 2$, where the sum is taken over all pairs of integers $(m,n)\neq(0,0)$. I'm curious whether the condition $r > 2$ is really necessary or not. Does the sum $$\sideset{}{'}\sum_{m,n}\frac{1}{m^2 + n^2}$$ converge? Can we say what happens for fractional powers, or more generally, for which $x > 0$ the sum $$\sideset{}{'}\sum_{m,n}\frac{1}{(m^2 + n^2)^x}$$ converges?
Comment: One relevant fact is that every prime of the form $4k+1$ can be written as the sum of two squares. So a lower bound for the first sum is $\sum_{p\equiv1\pmod4} 1/p$, which diverges.
Comment: just curious, what does the primed sum notation $\sideset{}{'}\sum_{m,n}$ mean?  (finding it hard to search for this... searching for 'primed sum' gets me info on sums of primes)
Comment: @postylem The prime means to take the sum over all lattice points except for $(0,0)$. I've seen this notation used before a few times although I can't come up with a reference.
Comment: Ah okay! I think I've seen similar notation used for other cases where the sum is "not over the whole space" for some meaning relevant to the specific use case.  But I also can't recall where I saw that.
Answer: We can use Euler products to rewrite the sum as the product
\begin{align}
\sum_{\mathbb{Z}^2 \ni (m, n) \neq (0, 0)} \frac{1}{m^2+n^2} &= 4 \sum_{k \geq 1} \frac{r_2(k)}{4k} \\
&= 4 (1 + 2^{-1})^{-1} \prod_{p = 4j+1} (1 + p^{-1})^{-2},
\end{align}
where $r_2(n)$ is the number of representations of $n$ as a sum of two squares and the last product is taken over primes. The second product diverges by Dirichlet's theorem on arithmetic progressions:
$$
\prod_{p = 4j+1} (1 + p^{-1})^{-2} = \biggl[\,\prod_{p = 4j+1} (1 + p^{-1}) \biggr]^{-2}
$$
diverges, because
$$
\sum_{p = 4j+1} \frac{1}{p}
$$
diverges.
So the set of all real numbers for which the sum
$$
\sum_{\mathbb{Z}^2 \ni (m, n) \neq (0, 0)} (m^2+n^2)^{-r/2}
$$
converges, really is the interval $r \in (2, \infty)$.
Answer: For rough estimates you only need to know that the map
$$(x,y) \mapsto \sqrt{x^2 + y^2}$$
is a norm so comparable with any other norm on $\mathbb{R}^2$.
Hence you can ask for a ( any ) norm $\|\cdot \|$ on $\mathbb{R}^2$ for which $s$ the series
$$\sideset{}{'}\sum_{m,n} \frac{1}{\|(m,n)\|^s}$$
is convergent. Now choose a convenient norm, $\|\cdot \|_{\infty}$.  The norm of every integral point is a natural number, and moreover each sphere of radius $N$ has an easy to see number of integral elements which is of order $N$. So now we have to see for which $s$ the series
$$\sum_{N\ge 1}\frac{N}{N^s}$$
which gets us $s > 2$.
Answer: There are a variety of ways
to show that
$\sideset{}{'}\sum_{\!m,n} \frac{1}{m^2 + n^2}$
diverges
depending on how
the limit is taken.
Here is one
that shows that the sum over
the first $n^2$ integer
pairs from
$1$ to $n$
grows like
$\ln(n)$
and so diverges.
In particular,
this shows that the sum over
$1\le m\le N,
1\le n \le N
$
grows like
$\frac{\pi}{4}\ln(N)+O(1)$.
I have checked the estimate
for $s_1(n)$, below,
numerically.
\begin{align}
s(n)
&=\sum_{k=1}^n \sum_{j=1}^n \dfrac{1}{k^2 + j^2}\\
&=\sum_{k=1}^n \left(\sum_{j=1}^k \dfrac{1}{k^2 + j^2}+\sum_{j=k+1}^n \dfrac{1}{k^2 + j^2}\right)\\
&=\sum_{k=1}^n \sum_{j=1}^k \dfrac{1}{k^2 + j^2}+\sum_{k=1}^n \sum_{j=k+1}^n \dfrac{1}{k^2 + j^2}\\
&=s_1(n)+s_2(n)\\[2ex]
s_1(n)
&=\sum_{k=1}^n \sum_{j=1}^k \dfrac{1}{k^2 + j^2}\\
&=\sum_{k=1}^n \dfrac1{k^2}\sum_{j=1}^k \dfrac{1}{1 + (j/k)^2}\\
&=\sum_{k=1}^n \dfrac1{k}\dfrac1{k}\sum_{j=1}^k \dfrac{1}{1 + (j/k)^2}\\
&\approx\sum_{k=1}^n \dfrac1{k}\int_0^1\dfrac{dx}{1+x^2}
\qquad\text{(Riemann sum)}\\
&=\sum_{k=1}^n \dfrac1{k}\dfrac{\pi}{4}\\
&=\dfrac{\pi}{4}\sum_{k=1}^n \dfrac1{k}\\
&\gt\dfrac{\pi}{4}\ln(n)
\qquad\text{(Harmonic sum)}\\[2ex]
s_2(n)
&=\sum_{k=1}^n \sum_{j=k+1}^n \dfrac{1}{k^2 + j^2}\\
&\ge\sum_{k=1}^n \sum_{j=k+1}^n \dfrac{1}{k^2 + n^2}\\
&=\sum_{k=1}^n \dfrac{n-k}{k^2 + n^2}\\
&=\dfrac1{n}\sum_{k=1}^n \dfrac{1-k/n}{1+(k/n)^2}\\
&\approx \int_0^1 \dfrac{1-x}{1+x^2}dx\\
&=\dfrac14(\pi - \log(4))\\
&≈0.43882\\
\end{align}
Answer: It is certainly true that there are some amazing identities (due to Euler and others) that reduce this question to simpler ones, there are also more direct, "geometric" ideas, perhaps roughly attributable to Minkowski in our times, though certainly people understood them earlier.
So, first, one might imagine some sort of integral test to apply here. In fact, with some trouble this can be justified, and it gives the "obvious" answer.
Somewhat more subtly-and-precisely, and which would amount to a proof of some version of an integral test in this case, on geometric grounds we can argue that any annulus of inner radius $r$ and outer radius $R$ contains at most (a constant times!?!) $\pi(R^2-r^2)$ (the area) lattice points. Then the conclusion is easy. Making this (or something equivalent) precise is not so hard.
The more difficult cases, due to Minkowski (at latest) involve error terms...
This approach applies as well to the $n$-dimensional case, to see that $\sum_{v\in \mathbb Z^n} 1/(v_1^2+\ldots+v_n^2)^s$ is absolutely convergent exactly for $\Re(s)>n/2$...
Answer: Take the sum for n < m and for n >= m which is the same as the sum for n <= m because of symmetry.
We have $m^2 \le n^2 + m^2 \le 2m^2$, and $(m^2)^{-r/2} \ge (n^2 + m^2) ^{-r/2} \ge (2m^2 )^{-r/2} $. For every m there are 2m -1 values n < m or n <= m, so for every fixed m we have a sum between $2m (m^2) ^{-r/2}$ and $2m (2m^2) ^{-r/2}$ or between $2 m^{1-r}$ and $2^{1-r/2} m^{1-r}$.
The sum over all m converges iff the exponent 1-r is less than -1 or r> 2.


A French registered car in the UK caught up in travel restrictions with owner now out of the UK

Question: A person had  traveled from France by car to visit relatives in the UK and then had to fly out to Ireland on a family medical emergency ( an expected short duration. ) leaving their French registered car parked up on the road in the UK. COVID travel restrictions then came in both in Ireland and UK forcing the person to remain outside of the UK.
Is there anyway that the French registered car could be moved to a safe location legally when the owner who is not UK national is outside of the UK?
Comment: I'm not sure that the fact that car is registered in France or that the owner isn't a UK citizen make much difference.  Apart from the fact a UK citizen would probably be able to return to the UK more easily.
Answer: The owner (now in Ireland) could send the keys by courier (FedEx, UPS, DHL, et al) to a friend, or their aforementioned relatives, in the UK and have them drive the car to a safe location.
Comment: Insurance would also be a concern.  The person doing the driving would need to be insured to drive this car, most likely via some sort of open drive policy or with more difficulty by having the car owner add them to their insurance policy.
Comment: Of course one needs to make sure this is an allowed activity during lockdown.
Comment: There's also a lively market in short-term motor vehicle insurance in the UK.  A quick web search will turn up a number of providers who will do you a one-day policy for not very much money, which the friend could ask the owner to reimburse.


First Job as a self-taught graphic designer

Question: What are some good projects to show in your portfolio for a self-taught designer? And what is a good way to find a first job in a company/ graphic design studio as a self-taught graphic designer?
Answer: First of all you shouldn't different a self-taught designer portfolio to a professional designer portfolio.
The quality and the complexity of the projects may be different but the concept it is the same.
So take ispiration from professional designers portfolio.
The projects you present in your portfolio should be realistic, even better if real case studies.
Start with really small businesses in your local area, or work for charity, this will make you works on real projects and enlarge your portfolio.
If possible work on project on the same graphic bench the company/ graphic design studio you want to work with are involved.
For example if you want to work for apple you should deflect your project on hi-tech product design and interface design, instead if you want to work with Nike you should work on t-shirt design and sport marketing graphic design.
Those are only examples but it's better if your portfolio is specialized in the same as your aspiring job.


Are there pre-warp civilisations within Federation space

Question: Are there planets that have pre-warp civilisations in Federation space? If so, what happens when they reach warp capability? Does their planet and surrounding space no longer belong to the Federation?
Comment: According to Memory Alpha - Malcor III a pre-warp civ (seen in TNG S4E15) is in / near Federation space and were made aware of the Federation (albeit accidentally) and Picard proposes peace (if not explicit membership just yet)
Comment: What level of canon do you want? Only worlds clearly indicated to be in Federation Space in dialogue on Trek shows/movies, or would you count pre-warp civilizations placed in Federation space by [Star Trek: Star Charts](http://memory-alpha.wikia.com/wiki/Star_Trek:_Star_Charts)?
Answer: From TNG we know of at least one such planet, Mintaka 3, with its proto-Vulcan civilization.  This planet was featured in the third season episode "Who Watches the Watchers?"
The Federation has a policy of strict non-interference with pre-warp cultures, which precludes them owning non-aligned planets with such cultures on them.  As we saw in the fourth season episode "First Contact" the Federation positions observers within a culture years in advance if it looks like that civilization is about to venture out into space at warp speeds.  Again, this is not to claim or renounce any sort of ownership, but rather to prepare to make first contact and welcome the civilization into a larger galactic civilization.
Comment: According to the aforementioned Star Trek Star Charts - Mintaka III is in the Beta Quadrant so without some futher clarification that might or might not count as Federation territory I think. http://memory-alpha.wikia.com/wiki/Mintaka_III
Comment: What happens if the civilisation does not want to be part of the federation
Comment: @Darren: nothing. They do their thing, the Federation does theirs.
Comment: It looks like this answer is incorrect--according to the index [here](http://www.startrekcrossindex.com/SCepisodesw.html) Mintaka III from "Who Watches the Watchers?" has "II" for its page number, which according to the site's system means it comes from the second fold-out map at the back of *Star Trek Star Charts*, I checked that map and it shows Mintaka (which apparently is a planet orbiting Delta Orionis) lying just beyond the last of a series of concentric circles representing "known space", well outside the blue region that represents Federation territory on this map.
Answer: The planet Malcor III was shown to be on the verge of developing warp in the episode "First Contact" (early in the episode a scientist from the planet asks her government for funding for a program to physically build a prototype design), so prior to that episode it would have been pre-warp, and it seems to have remained a pre-warp civilization afterwards since the government leader said at the end "The warp program will have to be delayed. We will divert more resources to education and social development to prepare for the day when we are ready".
I found Malcor listed here in an online index to the book Star Trek Star Charts, which says it's on p. 47 of the book, and when I checked that page in my copy I found that it did lie in the blue region that represents Federation space. Below is a scan I made--areas marked in blue are Federation territory, pink is the Cardassian Union, yellow is the Tzenkethi Coalition. Malcor appears at the bottom, just to the left of the Betreka Nebula, it appears to be in a blue region, and lies approximately between two starbases.

As for the question about what happens when they reach warp-capability, that's shown in the episode--information is gathered about their culture by disguised Federation officers (Riker in the episode), and then first contact is made (whether they would be offered Federation membership would probably depend on how peaceful and unified their planet is). From the transcript here, Picard tells the lead scientist on the warp project, "We've been monitoring your progress toward warp-drive capability. When a society reaches your level of technology and is clearly about to initiate warp travel, we feel the time is right for first contact. We prefer meeting like this, rather than a random confrontation in deep space." As for what happens if the planet wants nothing to do with the Federation, that's answered in this later exchange between Picard and the planet's leader:

DURKEN: And if I should tell you to leave and never return to my world?
PICARD: We will leave and never return. Chancellor, we are here only to help guide you into a new era. I can assure you we will not interfere in the natural development of your planet. That is, in fact, our Prime Directive.

Meanwhile, as I said in a comment, Kyle Jones' answer is incorrect--Mintaka III from "Who Watches The Watchers?" appears in the second fold-out map at the back of Star Trek Star Charts, which has an inset showing all of "known space", Mintaka lies out past the last of a set of concentric circles centered on Earth, and is well outside the blue area that represents Federation Space. Scan below--it's in the little bubble at the bottom right (click image to expand).
Comment: Needs more hand-drawn circles
Answer: Just because a planet is within the territorial boundaries of Federation space doesn't mean that the planet "belongs to" to the Federation, regardless of its level of technological development. As a pre-warp society, it would presumably be under Federation protection from outside influence so that it can develop as it will. "Post-warp", it would either be a Federation member with all attendant benefits and obligations, or a non-member where there would still need to be some form of treaty that both respects the world's sovereign status and obligates that world to not interfere with Federation activities occurring beyond some territorial limit.
Answer: Given the structure and laws of the Federation  (their high standards for entry), their space has to enclose many inhabited worlds that do not belong to it because they either don't qualify for membership or aren't even inclined to apply for it.  I doubt they claim the stellar system of an inhabited non-member world even when it's a pre-space-travel society but they may very well claim the space between the systems, meaning that one of the selling points for joining the Federation is that it gives the new member less restricted access to interstellar travel.  First Contact did in fact show us how the Federation handles a society's transition from pre-warp to post-warp when a planet is deemed to be otherwise qualify to be offered Federation membership.  There seems to be an assumption in TNG that planets like Earth who achieve warp drive before political unity are the exception, not the rule due to Star Trek's Planet of Hats tendencies.
In cases where planet achieves warp drive without political unity or under a dictatorship then I suppose the Federation would probably make diplomatic contact with the parties who have warp drive and subtlely steer them toward democracy and unification.


How will you measure speed in km/hr in space?

Question: How is  the speed of an object in space measured? Also more importantly how do you measure your own speed in space? On the road we use a speedometer which tells us the speed easily. How is it done in space?
Comment: speed in km/h is a very impractical unit of speed in space. Much more often you'll see km/s or AU/day
Answer: The main question is "relative to what?"
For space probes and the like, the speeds that matter are be either with respect to the Earth, the target object(s) (Mars, some asteroid, Space station, etc.), and/or the Sun (or Solar system barycenter). These speeds are measured mostly by Doppler shifts in

radio waves emitted by a radar the probe carries, reflected by the surface of some target
the communication signal between probe and Earth (see for instance, the deep space network).

Other methods have been used (image analysis between consecutive images taken by the space probe, the temperature of the heat shield on atmospheric entry, etc.) but these are all much less precise than Doppler measurements.
Space telescopes will measure redshift to some object (star, galaxy, etc.) (which is very similar to Doppler), which is more an indication of how fast that object is moving with respect to the entire solar system, rather than just the space telescope. Parallax methods are also used (see @IgnacioVazquez-Abrams's answer), but such methods can only be used for objects relatively close by (the parallax for most galaxies is too small to measure).
Other methods include Cepheid variables, and of course the famous Type 1a supernovae, which were used to conclude that the expansion of the universe is accelerating. But these are primarily measures of distance, and only crude measures of speed -- for objects at large distances, redshift is the only accurate way to measure the speed with respect to those objects.
Answer: Galileo Galilei had a similar question like yours. Only, back then he was asking how to measure the speed of the ship on which you are moving.
He came to the following conclusion - you cannot measure the speed of the ship if you don't look outside the ship - to the stars, passing islands and so on. This is what we now call Galileo's principle of relativity.
Back to your qestion - if there is nothing around you and you also don't see any stars from your spaceship (because of dust around) and your gadgets don't register any fields around the ship (magnetic and electric) there is no way to measure your speed :)
Relativity means that speed as a concept always needs somebody else, somebody with relation to whom this speed is calculated - in practical life it's always the ground.
If I have a ship and I'm moving there is no speed yet, speed appears when my ship is moving through something - planetary system, system of stars, galaxy and so on.
To sum up the speed (as well as many other things in life) is always relative and not absolute.
Comment: In general please avoid CAPITALS on the web - it's considered SHOUTING RUDELY.  Use the **bold** and _italic_ formatting instead for emphasis.
Comment: @StephenG I don't think that is a constructive comment, especially on a 4.5 year old post...
Answer: Parallax with various objects is measured, and known distances are used to triangulate your position in 3 dimensions at two moments in time. Since you have both the space displacement and the time displacement, you can just divide to calculate the speed.
Answer: One way to measure the speed of an object in space is to see how fast it moves against a lit background. For example, Mercury moving in front of the sun. We could see a black blob (Mercury) move across the bright sun, and record how long it took. Then we could say that it takes $x$ amount of seconds for Mercury to cross over the sun. If we also knew more information on their sizes then we could get that speed of Mercury into km/hr. This site has more analytical ways to calculate that if you're interested in how they get certain numbers.
Furthermore, if we are on a rocket ship and flying through space, we could look out the window and do the same thing in a sense by looking at planets and suns and seeing how fast we pass them. Or, we could still get km/hr even if the rocket ship had no windows. If we know how much fuel we're exhausting over time, we could calculate that as km/hr. It's like the car, it has a built in system to determine how fast you are going ... it doesn't shoot a laser at the passing scenery to determine speed.
Answer: One possibility is to measure your speed relative to the ions around you. Aeroplanes in the early days had the same problem. They used ventura pipes where the drop in air pressure was a function of speed. This was air speed not ground speed. A tail wind meant their air speed was in accurate. They did spot checks on land marks to Guage the air speed. Space is full of ions. Don't know how but maybe if these ions passed through a copper coil it would induce a current. Another possibility is to put multiple GPS salellites around the sun at different distances and sending the signal in 360 degrees.


Create Read Only Admin

Question: Is there an easy way to create an admin user that has read only permissions? That is, they can login and see everything an admin can, but they cannot change anything at all.
Answer:
Is there an easy way...

AFAIK, no, not at all.
If using a single site installation, you'd need to carefully disable (not hide, disable) all possible user actions that modify things. And this has to be done screen by screen (Posts, Pages, Comments, Appearance, Plugins, etc).
For example, in the All Posts screen (/wp-admin/edit.php):
/**
 * Remove all row actions Edit|Quick Edit|Trash|View
 */
add_filter( 'post_row_actions', 'wpse_56560_remove_row_actions', 10, 2 );
function wpse_56560_remove_row_actions( $actions, $post )
{
    global $current_user;
    if( 'pseudo-admin' == $current_user->user_login )
        return array();

    return $actions;
}

/**
 * Remove the Trash link from All|Published|Trash
 */
add_filter( 'views_edit-post', 'wpse_74488_remove_trash_link' );
function wpse_74488_remove_trash_link( $views )
{
    global $current_user;
    if( 'pseudo-admin' == $current_user->user_login )
        unset( $views['trash'] );

    return $views;
}

In the Post Edit screen (/wp-admin/post.php?post=ID&action=edit):
/**
 * Remove the Publish meta box
 */
add_action( 'do_meta_boxes', 'wpse33063_remove_meta_box' );
function wpse33063_remove_meta_box()
{
    global $current_user;
    if( 'pseudo-admin' == $current_user->user_login )
        remove_meta_box( 'submitdiv', 'post', 'side' );
}

In the previous examples, all function names contain the ID of the WPSE Question where they were taken from.
Also, refer to this Gist to see how to redirect users if they try to access a page using a direct URL.

If using a Multisite installation, @BrianFegter Answer to this Question: How can I open up my administrative panel to everyone?, has this nice suggestion:

I would use WP Multisite and allow users to register a test site. Woo Themes does this with their playground feature.
Comment: The problem with this is they can still edit the posts if they type in the correct link. Is it not possible to remove caps but keep the post type visible? If I make a custom post type and only give a user the read cap, it doesn't show up... Only when they have edit/delete caps as well.
Comment: I don't quite understand what you want to do with the caps, but to block users from direct access we have to redirect them in `admin_head` or another suitable hook (see Gist that I mentioned)

Create Read Only Admin


Will I get punished for having gay thoughts?

Question: I know that being gay in Islam is a very major sin and I would never let my self fall into doing this and engaging into this act no matter what my feeling and desire tell me. What I want to know if I have gay thoughts purposely or unintentionally will I get punished for it. Also I would like to mention I don't have these thoughts about everyone who's the same gender as me but just one particular person. And that's how I know I'm not really gay.
Answer: Thinking about bad thoughts is sinning, but this ummah is forgiven from it's bad thoughts as long as it does not speak of its thoughts, or act upon it.
(Sahih bukhari 6664)
Answer: everyone thinks things you wouldn't get sent to prison for thinking about killing someone though you would be watched more carefully i like to think the same will apply to your situation as long as you dont commit any sins your golden most of the time

mr.H


Synchronize LDAP data between two separate AD domain controllers into a single ADAM instance

Question: Would like to ask for some guidance.  I have 2 Active Directory domain controllers in separate domains with a trust established.  I also have an application that supports LDAP authentication, but not multiple LDAP servers. I did some research on ADAM (Active Directory Application Mode) and would like to clarify a few things.  1. When creating a partition within ADAM, must I create a partition matching the naming context of AD.  In other words, if your AD is DC=abc,DC=org your ADAM naming context should be DC=abc,DC=org. Is this true?  If so, how do I get around synchronizing to the second AD DC.  Will I need another partition? if so, then i think i'm back to square one as far as application LDAP auth is concerned.  Is there a better way to combine users from multiple Active Directory instances?
Comment: You don't provide details of yor OS. I assume Windows 2003 is in use here. You don't have to create partitions with same FQDN as AD. I assume you are asking for ADAMSync purposes? Please provide more details of what you are attempting
Comment: The 2 AD DC's are Windows 2003 and Windows 2008.  I was thinking to create a totally separate partition with a different naming context other than the ones on either DCs.  I would like to synchronize data from both Active directories into a single ADAM partition to be able to authenticate against that adam partition. I remember reading taht in order to synchronize between AD and ADAM, i would need to have identical partition name as the AD context name.  The limitation is that the application is only capable of authenticating against only one LDAP instance.
Answer: ADAMSync supports different DNs for source and destination DN. So you can have DC=Domain,DC=Com for source and O=Domain,C=US for ADAM partition as an example. No problems here.
What you will run into issues with is syncing both domains to the same partition. You wont be able to have DC=Domain1,DC=COM and DC=Domain2,DC=Org both sync to O=Domain,C=US. You can only have one of those sources sync to the the destination partition.
The reason why you will run into issues is because adamsync needs to store the configuration on the destination ADAM partition.  You cannot store two configurations simultaneously on the NC. In this case, if you chose DC=Domain1,DC=COM , it will write the relevant configuration to do that sync. If you try to store DC=Domain2,DC=COM , the previous configuration for DC=Domain1,DC=COM  is lost.
You do have the option of doing a /install for the DC=Domain1,DC=COM related XML, then doing a /sync. Next /install the DC=Domain2,DC=COM XML and do a /sync for that. So you keep swapping between configurations and doing /syncs.
This works but IMHO is not elegant. it may suit the purposes of some.
Comment: I have gotten this to work without an issue.  Syncing both domains - One out in each to a single partition like O=Domain,C=US.  ADAM allows you to do this without an issue.  I cant go into much detail at this point due to time constraints, but can answer any questions you may have
Comment: Added clarification to my reply.
Comment: Thank YOu for your reply. This is my ultimate goal here.  If you have a few mins, plese take a peek at the document, it describes a potential solution to what i'm planning on doing, but does not coincide with your answer :)
Comment: http://community.igniterealtime.org/docs/DOC-1534