look into Delzant Polytope

Question: A Delzant polytope in R^n by definition is a simple, rational, and smooth convex polytope in R^n (Ana Cannas da Silva's book for notions.) Do you guys have any insight of the definition, for example, anything we can say about the shape? They satisfy some rigidity conditions? All related comments are welcome!
Answer: The standard model of a vertex which satisfies the Delzant condition is the positive "quadrant" $x_i \geq 0$ of $\mathbb{R}^n$ near the origin. In general a polytope is Delzant if and only if every vertex can be taken to this standard model by some element of $\mathrm{GL}(n, \mathbb Z)$.
The motivation for this definition, coming from symplectic geometry, is the following fact, due to Archimedes. Consider the standard radius-one sphere in 3-space enclosed in a cylinder of radius one. Project the sphere outwards onto the cylinder, orthogonal to the cylinder's axis. Archimedes' theorem says that this map preserves areas. In terms of Delzant polytopes, you should think of the sphere as 2-dimensional toric manifold. The corresponding "polytope" is the interval $[-1,1]$. Archimedes' theorem says that if you take the cylinder $S^1\times [-1,1]$ and collapse the circles $S^1\times\{\pm 1\}$ you obtain the sphere as a symplectic manifold. To see how to generalise this to higher dimensions, take the n-fold product of the cylinder $S^1\times [0,\infty)$ and consider the map to $\mathbb{R}^n$ given by forgetting the $S^1$-factors. The image is the positive "quadrant" mentioned in the first paragraph above. Archimedes' theorem tells us that if we collapse the preimage of the boundary of this quadrant in a controlled way we obtain something which admits a smooth symplectic structure. I.e., over the orgin we collapse the whole $T^n$-fibre, more generally, over each coordinate $r$-plane we contract the corresponding orthogonal copy of $T^{n-r}$ inside $T^n$. (To do this recall each $S^1$-factor corresponds to a direction in $\mathbb{R}^n$).
Now the meaning of Delzant's condition should be clear: given a Delzant polytope P we built a symplectic manifold by taking $P \times T^n$ and then collapsing parts of the boundary. To decide how, we map each vertex to the standard model described above and collapse as in that case. Archimedes theorem tells us the resulting object is a smooth symplectic manifold with a torus action.
Comment: Thank you! Your first paragraph is brief and sharp! In your second paragraph you said "more generally, over each coordinate r-plane we contract the corresponding orthogonal copy of T^(n−r) inside T^n"; that is the same as to contract the corresponding copy of T^r inside T^n, right? For example, take a 3-fold product (labeled x,y,z in order) of S^1 \cross [0, \infinity), then over the xy-coordinate plane, we contract the last copy, i.e., the z copy of S^1 inside T^3 = S^1 \cross S^1 \cross S^1, right?
Comment: Firstly, credit where credit's due; this cute modern interpretation of Archimedes' theorem was explained to me by Richard Thomas.

Secondly, your 3-dim example is correct, but the preceding statement  - that contracting the orthogonal T^{n-r} in T^n it is the same as contracting T^r - is not. The first gives an r-torus, the second an (n-r)-torus. This is evident in your example: we must contract the third S^1 - leaving a 2-torus - and not the product of the first and second S^1 which would leave a single S^1.

Hope that helps!
Comment: Thank you! It definitely helped me. It is clear to me now and sorry I misunderstood the language at first. Also sorry for the delay as I was offline over the weekend.
Answer: Well, they are fairly rigid in that the normals to their faces have to be lattice vectors. So the only deformations are to slide the faces perpendicular to themselves. (On the other hand, you may always do this, so they are not completely rigid.) And "simple" is a restriction on their shape. The image of a Delzant polytope under $GL\_n(\mathbb{Z})$ is another Delzant polytope, so it only makes sense to ask about properties which are $GL\_n(\mathbb{Z})$ invariant. That tends to rule out obvious geometric properties (e.g. curvature) so I'm not sure what else to say.
UPDATE: In the theme of "all related comments are welcome", here is a speculation that I haven't thought about seriously. I do know one geometric property which is preserved by $GL_n(\mathbb{R})$. For $K$ a compact convex set, let $A$ be the volume of the smallest ellipsoid containing $K$, and $a$ the volume of the largest ellipsoid contained in $K$. Let Define $s:=a/A$ to be the sphericity of $K$. (Not sure if this is the standard name.) So $s=1$ for ellipsoids, and smaller for everything else. Delzant polytopes tend to be kind of fat, so I wonder if one could prove a lower bound for their sphericity which was better than for ordinary polytopes.
Comment: It is not true that any GL_n(Z) element takes a Delzant polytope to another, e.g. the triangle in the plane with vertexes (0,0) (0,1) (2,0) is not but the triangle with vertexes (0,0) (0,1) (1,0) is. But for those with determinate 1 or -1 it is true. I think the smoothness condition is subtle. That the normals to the faces are lattice vectors follows basically from the rational condition. I do not quite get why the ONLY deformations are to slide the faces perpendicular to themselves but I agree with you we can always do that.
Comment: The definition of GL_n(Z) that I am familiar with is that a matrix M is in GL_n(Z) if M and M^{-1} have integer entries. This forces determinant \pm 1.
Comment: How would you smoothly change a lattice vector while keeping it a lattice vector? If you say that you can't, then the normals are fixed and the only deformations are to slide the facets while keeping the normals the same. By the way, this isn't a purely rhetorical question. I would be interested in knowing of some other notion of deformation here.
Comment: David:  There are a zillion different ways to gauge the roundness of a convex body; no one of them will be called "the" sphericity.  I'm not sure if yours has a name, but are you partial to that choice?  For instance you could look at the Banach-Mazur distance, which is the ratio of the circumradius to the inradius (concentrically), minimized over the action of GL(n,R).
Comment: Another more similar roundness statistic is, in your notation, (Vol K)/a.  This one is called "the volume ratio".  But again, you can be infinitely creative in defining roundness statistics; usually you would pick one for a purpose.
Comment: As I said, I haven't thought about this seriously at all.
Comment: I feel the Banach-Mazur distance is a formidable notion. :)
Answer: As "all related comments are welcome" I should perhaps say that if you add a "reflexive" condition (the dual polytope is also lattice polytope) we get smooth Fano ones of which there are finitely many in each dimension, but the number grows pretty quickly. The attempts to classify these have been moderately successful (see www.imf.au.dk/publications/phd/2008/imf-phd-2008-moe.pdf for example), but they are still somewhat wild.
Comment: Great! Thank you! Lattice polytopes are not necessarily Delzant, but reflexive lattice polytopes must be!

look into Delzant Polytope


Two-fold coverings $S^n \rightarrow X$

Question: Let $S^n \rightarrow X$ be a two-fold covering. Is it true that $X$ is homeomorphic to $\mathbb{R}\mathrm{P}^n$? If so, how does one prove it?
Comment: You should check the book "Involutions on Manifolds" by Lopez de Medrano. I remember that he has a detailed discussion of free involutions on spheres of dimension $\ge 5$, but I do not remember in which category: Smooth or topological. If you work in the smooth category, there will be examples which are not diffeomorphic to the standard $RP^n$ (at least in dimension 4 - Fintushel-Stern and Capell-Shaneson, and, most likely, in all higher dimensions).
Answer: There are different answers in the categories of homotopy type, topological, PL, and smooth manifolds. Let's start from the bottom and go up.
Suppose the finite $n$-dimensional CW complex $X$ has 2-fold cover homeomorphic to $S^n$. The map $X \to \Bbb{RP}^\infty$ classifying the fundamental group factors through $\Bbb{RP}^n$ by cellular approximation. Pulling back the universal bundle restricted to $\Bbb{RP}^n$ we still get the same bundle on $X$, but get an equivariant map $S^n \to S^n$ whose induced map on the quotient is the given map; note that the action on the sphere in both cases is free. If this map were of degree 1, we'd be finished: the map $f: X \to \Bbb{RP}^n$ clearly induces an isomorphism on all homotopy groups up to degree $(n-1)$, the degree statement implies it on degree $n$, and you can use a (particularly nasty) form of the Hurewicz theorem to conclude the map is a homotopy equivalence. Unfortunately I see no particular reason to believe it is degree 1; however, you can modify an "inverse map" from $\Bbb{RP}^{n-1}$ on the top degree cell so that it is a homotopy equivalence; the details are in Wall's "Surgery on compact manifolds". Conversely a manifold homotopy equivalent to $\Bbb{RP}^n$ is the quotient of a homotopy $S^n$ by some $\Bbb Z/2$ action.
In fact the classification of $n$-manifolds homotopy equivalent to $\Bbb{RP}^n$ but not (homeomorphic, PL homeomorphic, diffeomorphic) is the same as the classification of free $\Bbb Z/2$ actions on (possibly exotic) $S^n$ of the appropriate type up to the appropriate kind of equivalence (equivariant homeomorphism, diffeomorphism, etc).
TOP/PL/Smooth. These all agree in dimensions up to 3. It is classical then that there is only $\Bbb{RP}^2$ and a corollary of Perelman's proof of the elliptization conjecture says that there is only $\Bbb{RP}^3$ in dimension 3.
In dimension 4, PL = Smooth, but there are non-smoothable 4-manifolds, and most topological 4-manifolds admit many distinct smooth structures. In dimension 4 there are exactly two manifolds homotopy equivalent to $\Bbb{RP}^4$ up to homeomorphism; one is not smoothable. See Hambleton-Kreck-Teichner Up to diffeomorphism (even when the universal cover is diffeomorphic to $S^4$) this is still open, but there are a few inequivalent smooth structures on $\Bbb{RP}^4$ with universal cover known to be standard. I don't know if there are infinitely many yet. Probably not.
In dimensions at least $5$ this is answered by surgery theory. Wall has a complete description in his book for PL $\Bbb{RP}^n$s; I suspect there is only minor change from the topological setting. (I do not really want to decode his notation and calculate the number of $\Bbb{RP}^n$s in each degree. It's finite except in dimensions equivalent to $3 \mod 4$.) His book will also provide a good reference to what's known smoothly in high dimensions and where to look.
Comment: @RobinStoll No, it's just not true then. If $X \simeq \Bbb{RP}^n$ and $X$ is a manifold, then its universal cover is seen to be homotopy equivalent to $S^n$ and is a manifold, but thanks to the now-solved Poincare conjecture in all dimensions, the universal cover must be homeomorphic to $S^n$. All of these fake $\Bbb{RP}^n$s can be given CW structures.
Comment: Actually in the second paragraph $X$ just has to be a finite CW complex with a double cover homotopy equivalent to $S^n$.
Comment: Thanks for the detailed answer! I have two more question:
Firstly, I did not quite understand how exactly the question if there are spaces homotopy equivalent but not isomorphic to $\mathbb{R}\mathrm{P}^n$ (in some category) relates to the original question of coverings by $S^n$.
Secondly, do you see some way to get a homeomorphism (not just a homotopy equivalence) between $X$ and $\mathbb{R}\mathrm{P}^n$ in the CW-setting?

Two-fold coverings $S^n \rightarrow X$


Neo4j Using a custom evaluator to get the depth a node was retrieved at

Question: I'm using the latest version of Neo4j to build up a graph of nodes and relationships with the Java API.
My problem is that I need to traverse the nodes to a certain depth. There may exist a relationship between two nodes at the same depth in the database, but I don't want to return that relationship in the traversal.
I tried to make a custom Evaluator by implementing the Evaluator interface but the only method it overrides is Evaluation evaluate(Path path) . It doesn't appear to have the notion of depth associated with it.
I would really appreciate some advice on how to either associate a node with its depth (when traversing from a particular node) or prune a relationship where two nodes are in the same level.
Answer: The path has a length(), which is the depth. The length is equal to the number of relationships in the path, i.e. number of nodes - 1.
Answer: You can use Evaluators.atDepth() to get a predefined Evaluator that only includes paths with a certain depth.
In your custom evaluator you can simply check the length of the passed path parameter to decide if you want to include this path or not e.g. with:
Evaluation evaluate(Path path) {
    return path.length() == 4 ? Evaluation.INCLUDE_AND_PRUNE : Evaluation.EXCLUDE_AND_CONTINUE);
}

Answer: Have you tried Cypher for that, something like
start n = node(1) match p=n-[*4]->(x) return x, length(p)

?


Drive cycle on 2001 Saab 95 with codes P1312 & P1110

Question: I replaced DIC, did a basic drive cycle to remove codes, but didn't remove codes. All my monitors are in ready state. The check engine lite was on, however, now it's off, I still show codes on scanner. O'Reilly's Auto Parts will not clear codes at this point, says do drive cycle. What am I doing wrong? Thank you
Comment: Please edit your question and stop YELLING at everyone!
Comment: Is your car a Saab 95 or a Saab 9-5 ... I'm pretty sure you mean a 9-5, so put that in the tags.
Answer: This sounds like proper OBD-II software behavior.  The CARB OBD-II rules state that for most trouble codes the check engine light should go off after two passed tests and then the codes should erase from memory after 40 passed tests.
The fact that the light has gone off on its own is a good sign that the repairs were correct. No need to clear the codes, just let the on board software resolve the issue.
Note that OBD software logic has changed over time so it is important to know what year the vehicle was made.  In the first few years, 96-98, it was not uncommon for the software to not follow the rules.
Comment: THANKS FRED, I TRIED ERASING MY CODES AND IT DID ERASE CODES, HOWEVER IT ALSO ERASED MY CATALYST MONITOR, EVAP. MONITOR,  AND O2 SENSORS, AND I RECEIVED YOUR SUGGESTION AFTER MY ERASE, TOMORROW I'LL CLOSE THE REST OF THE MONITORS AND O2S'. MY SCANNER SOUNDED AS IF THE CODES WOULD BE THE ONLY ITEMS ERASED. THANKS AGAIN
Comment: I suggest you not use the clear command. It clears many things. Just drive the car and let the onboard software runs it tests. If the CEL does not come back on in the next few trips you are done.   And turn your caps lock off.
Comment: FRED, IT STILL HAS NOT RESET CODE P1110, AND CEL IS ON, DO I CHANGE VACUUM VALVE?


What significance does an index_id < 256000 have?

Question: In a certain tutorial I read the author is filtering sys.indexes based on the predicate index_id < 256000. What does this accomplish?
Comment: Maybe they copied the code from [`sys.sysindexkeys`](http://www.g-productions.nl/index.php?name=sysindexkeys)
Comment: @Martin Oh, yuck.
Comment: @AaronBertrand - and also in `sys.selective_xml_index_paths`, `sys.xml_indexes`, `sys.sysindexes` but I suppose these will just get updated if the magic number is no longer valid.
Comment: @Martin I wouldn't bet on it. Especially for the backward compatibility views. What a horrible way to demonstrate how to retrieve metadata...
Answer: This is based on the misconception that XML indexes are currently the only type that could ever have an id scheme that is >= 256000 (at least based on their observation; this scheme is not documented AFAIK, so not even sure if it's intentional). Probably fine in current versions, but who knows what type of index will be added next and where its id scheme will start? If you want to exclude XML indexes, you are now also excluding something else. Spatial indexes, for example, seem to start at id = 384000. If the query above is intending to include spatial indexes but not XML indexes, they're going to be in for a surprise.
A much better filter would be:
WHERE type <> 3;

...or even better, since it is self-documenting...
WHERE type_desc <> N'XML';

And now when you want to also exclude, say, spatial indexes, your query changes to...
WHERE type_desc NOT IN (N'XML', N'SPATIAL');

...instead of having to figure out what numeric range the id values for spatial indexes might occupy (or not). Good luck with that.
These are pretty clearly documented in sys.indexes (Transact-SQL). I see no reference to this magic number and I highly recommend you point your tutorial author here so they can see that this magic number is not something they should be relying on (never mind teaching others to rely on).
Comment: +1 that's an awful bad habit.  Forget about `index_id`.  Especially since more accurate data for determining type is sitting right next to it... literally.
Comment: It's probably a design error of SQL Server to give out index_id's with this regularity. They should be randomized so that nobody can mistakenly rely on them.
Answer: According to book "Microsoft SQL Server 2012 Internals"By Kalen Delaney, Craig Freeman, The XML index's index_id begins numbering with 256000. So to get all type indexes information (querying sys.indexes) but skipping XML indexes you can place filter like that.
SELECT * FROM sys.indexes WHERE index_id <256000

Same result set can be achived by placing filter on type column of sys.indexes. For XML type of indexes type = 3.
SELECT * FROM sys.indexes WHERE type <> 3

or
type_desc column can also be used.
SELECT * FROM sys.indexes WHERE type_desc <> 'XML'

Comment: @swasheck question is that, why figure 256000 is used, and not what is safe. For best practice definitely i would like to go with Aaron
Comment: do you have official documentation for this claim?
Comment: "What does this accomplish?" Technically the answer would be "nothing." Circumstantially, it's a way that people have correlated filtering out XML indexes.
Comment: i have it right here. which page? also - i respect the heck out of those authors but i'm not sure that counts as "official documentation."
Comment: That is a casual observation, at best, by Kalen at a specific point in time, with no knowledge that this was actually intentional, never mind no future-telling ability to determine whether some new index type will be > 256000 in the future. It is not something Microsoft intended you to rely on, which is why you won't find any reference to it in the official documentation. And agree with @swasheck, while that book is definitely a valuable resource, it is not official documentation.
Comment: @swasheck Its Chapter 9, Page 461


Windows 8 Pixel density large screen

Question: I have Windows 8 Media Center Edition running on an HTPC and would like to make the Metro icons 'larger' on my 62" 1080i TV.  I've found a blog article that discusses how Windows determines pixel density based on resolution, however I do not see a way to change the default assigned values for my resolution via the UI.  I know this can be done as I saw a 2012 Build video where they demo'd a large screen touch display and mentioned that you can over-ride the pixel density via the registry, I just don't know what key.  Help anyone?
Answer: Check this article:
Adjusting Windows 8 display size/scaling for high DPI/PPI monitors

right-click over the desktop and select Screen resolution
Select Make text and other items larger or smaller from the bottom
of the display settings.
here, you can change the scaling for your display.

(image source: nvidia.custhelp.com)
Comment: This does not affect the tile size on the standard metro desktop.  I would like to have the tiles be larger, as they are not readable on a 62" screen at 1080i
Comment: ok. I haven't tried to connect my win8 to my tv.

Windows 8 Pixel density large screen


Install Brother MFC295CN drivers to Ubuntu

Question: I'm trying to get the drivers for my printer going, and not doing well. I tried the following: http://ubuntuforums.org/showthread.php?t=1377270 and the sudo dpkg -i PACKAGE.deb format. It installs the printer, but then doesn't see the printer via USB. Ideas?
Answer: You have to follow the instructions on this page. I installed a Brother HL2250DN just one hour ago and it works perfectly.
Because the instructions are a bit hard to follow, here the summary for Ubuntu (with updated instructions for cups):
(0. If you are a 64-Bit Ubuntu user:)
sudo apt-get install ia32-libs lib32stdc++
1. Turn on the printer
2. Download the LPR and cupswrapper driver from this page (you need the .deb packages).
3. Open a terminal by pressing CTRL+ALT+T.
4. Install the LPR driver
In the terminal type dpkg  -i  --force-all  downloaded-lpr-driver where "downloaded-lpr-driver" is the filename of the downloaded deb for the LPR driver (something like mfc295cnlpr-1.1.3-1.i386.deb)
5. Install the cupswrapper driver
In the terminal type dpkg  -i  --force-all  downloaded-cups-driver where "downloaded-cups-driver" is the filename of the downloaded deb for the cupswrapper driver (something like mfc295cncupswrapper-1.1.3-1.i386.deb)
6. Test your installation
dpkg  -l  |  grep  Brother
should show both drivers to be installed.
7. Setup cups
Point your Browser to http://localhost:631/printers and click on your printer. In the Administration dropdown choose Modify Printer. The next step wants you to choose the connection type.
If you're using USB to connect the printer, the connection should allready be set to something like usb://Brother/(your printer's model name). If your Printer is not listed, go to http://localhost:631/admin and add your printer.
If you're using a network connection, choose from Discovered Network Printers or if your printer was not discovered, select "LPD/LPR Host or Printer" or "AppSocket/HP JetDirect" and insert lpd://(Your printer's IP address)/binary_p1 with the correct IP of your printer.
See these examples for working configurations.
Comment: I went through the steps provided.  I did it both with the USB cable and the network cable.  The printers show in both the http://localhost:631/printers page and the system settings / printer page.  All indicates the printer is available and idle.  However, when I send something to the printer, it flashes in the job queue as processing, these disappears as if it printed.  However, there isn't any action from the physical printer.
Answer: You can install the printer drivers easily with the free utility that Brother provides. Retreive and run the file by opening up the terminal and running these commands (make sure you acknowledge and accept the license agreement, which can be found at the page I linked):
wget http://download.brother.com/welcome/dlf006893/linux-brprinter-installer-2.0.0-1.gz
tar xf linux-brprinter-installer-2.0.0-1.gz
rm linux-brprinter-installer-2.0.0-1.gz
chmod a+x ./linux-brprinter-installer-2.0.0-1
sudo ./linux-brprinter-installer-2.0.0-1

You will be walked through the installation. Once complete, you will have an item in your home folder named linux-brprinter-installer-2.0.0-1, which you can safely delete.


What is the difference between a plenary and a keynote talk?

Question: I am organizing a workshop and have invited a featured speaker to give a synthetic talk. What is the appropriate term for this role?
Comment: "Synthetic talk"?
Answer: They are both used (often interchangeably in my field). A plenary is a talk which does not have anything scheduled against it. A keynote is an invited talk in a conference or session.
Answer: I came up with this answer while asking the question:
Both terms indicate a sort of synthesis, but plenary seems to refer to being fully attended whereas keynote appears to indicate setting a common theme.
Answer: Keynote; A theme by an authority person with credentials on topic and knoledge of participants and their reason for being here. Speaker offers challenges.
Plenary; Speaker/ Facilitator with authority remarks to stimulate input and discussion by all to come to conclusion for action.


How reliable / respected are George Friedman / Stratfor's interpretations of geopolitics?

Question: Recently, a talk by Stratfor's chairman George Friedman at the Chicago Council on Global Affairs attracted some attention in my homeland Germany, among other things because it appears he frankly supports suspicions that within German media and politics would immediately be dubbed conspiracy theories. To get some background for assessing his interpretations of geopolitics, my question is: How respected is George Friedman in US-american geopolitical circles? Are his claims and interpretations usually taken seriously? Since he runs an intelligence company, one could expect his earnings depend crucially on how respected he is. On the other hand, maybe it's also lucrative to sometimes overstep the bounds of reputable political analysis, just to gain some publicity.
Comment: @cpast the name George Friedman wasn't known to me either but Stratfor is a pretty known company that often gets quoted in respectable newspapers (Which I suspect is also the best possible answer)
Comment: Thanks @user45891, yes I was surprised by cpast's comment because I also had the impression that Stratfor is moderately well known. Though the first time I heard about them it was because Anonymous had hacked into their servers and threatened to publish their subscribers list. So personally you'd say that since Friedman is the founder, CEO, and chairman of this moderately well-known company, you'd take his interpretations seriously?
Comment: Can you give an example of something that would be considered a "conspiracy theory"?
Comment: @user45891 - that's a VERY valid statement. By itself, neither Germany nor core Russia is a geopolitical threat to the USA. Combined, they can easily control most of Eurasia and thus scale up to become such a threat. One might quibble over some areas of this but as a concept it's preeminently not-insane. Whether conscious US posture had always derived from that concept is a different story.
Comment: How respectable is who now? (the fact that I, someone who pays a fair bit of attention to US politics, have never heard of this guy should tell you something)
Comment: @ThePompitousofLove the statment A. Donda most likely is referring to is `The primordial interest of the United States [...] has been the relationship between Germany and Russia, because united they are the only force that could threaten us, and to make sure that doesn't happen.` from somewhere in this video: https://www.youtube.com/watch?v=QeLu_yyz3tc
Comment: The question should be how effective is Stratfor in predicting future poltical developments. How well known, or popular they are, or may be, is irrelevant since popularity very often lacks substance. They should be judged by the results they produce. In that respect I think they are fairly successful, though at times I think they have ignored analyses that they have gotten wrong. What is also important to note is that unlike any other outfits that I've seen, there is a methodology that they use, a paradigm. One can debate the effectiveness of their methods, but to be sure what they produce is
Comment: let's not have extended discussions in the comments
Answer: After all this discussion, it is worth answering the question decisively.
George Friedman is an academic, who runs a side business providing popular commentary on International Relations. He has a PhD and is very much a part of the U.S. national security commentariat, and relatively well respected although, obviously, not everyone agrees with him. Because he does dedicate a certain amount of time to popular writing and lives outside of Washington, D.C. he is not nearly as influential as many people might think, but he is absolutely not a crock.
That said, his assertion that a core interest of the U.S. is to prevent the emergence of a Eurasian hegemon, through the unification of Russia with Germany is uncontroversial in the U.S. Furthermore, preventing such a unification is also a core interest of Britain, France and essentially every European or Asian country as well.
In fact, whether one is a realist with geopolitical leanings--which Friedman is--or a liberal, there is a strong case to ensure that neither Germany nor Russia dominate the other. From a realist perspective, were one country be able to marshal the forces of all or most of continental Europe, it is the only extant political situation that could legitimately threaten the U.S.'s national security (without using nuclear weapons). From a liberal perspective, the only way that either Germany of Russia could dominate the other would be through a collapse of domestic liberal democratic mechanisms which serve as safeguards against threats to other democratic countries. Either way, it is in the U.S.' national interest that neither Germany nor Russia come to dominate the other.
Perhaps most importantly, this is not at all controversial among German scholars and the German National Security establishment either, as it is the direct basis for the high degree of confidence in U.S. security guarantees to Germany. Because Russia has the largest nuclear arsenal in the world, and Germany has been a good global citizen and a valuable partner, the U.S. need not fear that Germany will ever dominate Russia--and I am certain that Friedman had no intention of implying so. Therefore, Germany can rest completely assured that the U.S. will defend it against Russia because the only possible threat to U.S. national security comes from Russian domination of Germany, which neither the U.S. nor Germany wants.
Comment: *collapse of domestic liberal democratic mechanisms* Sounds pretty much what happened the last 5 years in Russia.
Comment: Thanks for the answer. – A comment: In my understanding, "united" doesn't necessarily mean "forming a political union" (I can't see that happening at all), but also "standing together on matters of common interest". With respect to that, publicized understanding in Germany would be that of course we are a souvereign country acting based on our own interests, and if those interests happen to align with those of russia there can be united action – it just so happens that our interests align much more with the US than with russia.
Comment: Compared to that, what Friedman says sounds like no, German souvereignty is bogus, US foreign policy is directed at never to let such an alignment happen. A concrete example are current economic sanctions against russia, which strongly hurt parts of the German economy, but not in the same way the economy of the US. The "conspiracy theory" would be that Germany is not so much a valued partner, but a good little soldier that needs to be kept in line. It has now been a while since I watched that talk, but to German ears it sounds like that's what Friedman is saying.
Comment: While I wouldn't be surprised if George came off as thinking that, I'm sure he doesn't. Sovereignty, especially of a major power like Germany, us a fundamental assumption of the realist paradigm, which George is a major proponent of.
Comment: Nice answer, but come on... "whether one is a realist with geopolitical leanings, or a liberal"?  Nice jab there.


Image URL saving

Question: Here is how my project works, I have a UIScrollView below it is a button addButton, when clicked redirects you to the AGImagePickerController (For those who doesn't know AGImagePickerController,it is a multiple image picker). Then you clicked the images(single or multiple images). When you press DONE, it will save the images in NSCachesDirectory. It will show the image chosen in UIScrollView (Which can be deleted). When you pressed the addButton again, it will show you the image picked a while ago with a checkMark.
PROBLEM : When I delete the image in the UIScrollView the image that was deleted in the AGimagePickerController is still checked. Needed is when deleted in the UIScrollVIew will also be remove in the AGimagePickerController.
What I wanted to do is, save its image by its URL then put it in a folder inside my NSCachesDirectory so I can load it easily, but I dont know where to start, since Im arranging my images in the UIScrollView by names with integers. Hope someone could suggest what to do. Thankyou Very Much.
NOTE : To those who have read this, please comment on what part of the code you want me to post here, or the part on which your having a problem with. Thankyou again.

CODE:
Here is my DONE part:
for (int i = 0; i < info.count; i++) {

    //NSLog(@"%@", [info objectAtIndex:i]);
    NSArray *paths = NSSearchPathForDirectoriesInDomains(NSCachesDirectory, NSUserDomainMask ,YES );
    NSString *documentsDir = [paths objectAtIndex:0];
    NSString *savedImagePath = [documentsDir stringByAppendingPathComponent:[NSString stringWithFormat:@"oneSlotImages%u.png", i]];

    ALAssetRepresentation *rep = [[info objectAtIndex: i] defaultRepresentation];
    UIImage *image = [UIImage imageWithCGImage:[rep fullResolutionImage]];

    //----resize the images
    image = [self imageByScalingAndCroppingForSize:image toSize:CGSizeMake(256,256*image.size.height/image.size.width)];

    //----fix image orientation
    image  = [image fixSlotOrientation];

    NSData *imageData = UIImagePNGRepresentation(image);
    [imageData writeToFile:savedImagePath atomically:YES];

}

Then in my AGIPCAssetsController.m(Where the retaining of the image checkmark part)
- (void)loadAssets
{
    count = 0;
  [self.assets removeAllObjects];

    AGIPCAssetsController *blockSelf = self;

    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_LOW, 0), ^{
        // Where I make an array to put the `ALAsset Url`
        selectedPhotos = [[NSMutableArray alloc] initWithArray:blockSelf.imagePickerController.selection];

        @autoreleasepool {
            [blockSelf.assetsGroup enumerateAssetsUsingBlock:^(ALAsset *result, NSUInteger index, BOOL *stop) {
                // ALAssetRepresentation* representation = [result defaultRepresentation];
                // NSUInteger assetindex= [blockSelf.assetsGroup numberOfAssets];

                if (result == nil)
                {
                    return;
                }

                AGIPCGridItem *gridItem = [[AGIPCGridItem alloc] initWithAsset:result andDelegate:blockSelf];
                if(count < blockSelf.imagePickerController.selection.count)
                {
                    if ( blockSelf.imagePickerController.selection != nil &&
                        [result isEqual:[selectedPhotos objectAtIndex:count]])
                    {
                        gridItem.selected = YES;
                        count++;
                        NSLog(@" %@",result);
                    }
                }
                [blockSelf.assets addObject:gridItem];

            }];
        }

        dispatch_async(dispatch_get_main_queue(), ^{

            [blockSelf reloadData];

        });

    });
}

Answer: i think cause both of them have the same variable reference ,
check your variable reference and compare between both and let me know what u found .
Answer: Just an idea and not an exact solution. On Done button action, is it possible to save the image with name as some unique property of rep. Probably [rep filename] or [rep url] can be used as image name while saving in place of @"oneSlotImages%u.png". And in loadAssets read all the images stored in cache and compare its file name against the file names of images in selectedPhotos and remove it from the selectedPhotos array if it is not present in cache.
Comment: I have other images in NSCaches. So I renamed them as `@"oneSlotImages%u.png"`
Comment: What exactly is info? Is it the array of all selectedPhotos?

Image URL saving


How do i model this plastic handle?

Question: Another question from a sad rookie:
I managed to properly model this 5 liter bottle (Thanks for the amazing guidance from a user in my previous question) and now i need a cap and a handle for the bottle. The cap is pretty easy to model but this plastic handle is a different story. I don't even know where to start — it is such a complicated object. I was thinking about modeling it with proportional editing — no luck so far... Any tips and tricks for this kind of work?
Comment: I am newbie in Blender, but if I am not mistaken you may create straight handle first then bend it with curve as well using curve modifier
Comment: Great thinking man! I still have to develop this kind of understanding of modeling. Thank you!
Answer: This is how I would do it:

Create a bezier curve for the in the shape of the handle's profile. Life is easier if you put the object's origin at one end.

Create another curve in the shape of the handle.  To simplify things, I left the origin at the World Center and used a mirror modifier.

With the handle shape selected, go into the Curve properties, Geometry section and create a bevel, using the profile curve as the bevel object.

That should give you the basic geometry.  You could convert to mesh, add a solidify modifier and make any tweaks you wanted from there.
Answer: You probably should use curves to do this, as @R Palo has shown, because it makes it easier to go back and alter moves you made earlier; ( it's less destructive ). But this is the sort of way I would hack a one-off together for myself, if I knew there were going to be no changes.

X across, Y up. All the modeling is done flipping in and out of 2 levels of subdivision (Ctrl0 and 2 on the keyboard)

Starting with a 16-sided nGon circle, I Inset, verts J joined as shown
V rip the longitude, delete the inner face and right half, assign a Mirror modifier in X with 'Merge' and 'Clip' checked.
(From now on working on the left half,) select the horizontal radial
edge, CtrlShiftR spread a couple of
loops on either side to the right width to extrude a handle.
EX Extrude the
handle,AltM merge the inner-rim  vertices, and
X > Dissolve the edge to make the kite-shape shown..
CtrlR cut a loop across the handle, ready to
extrude downwards.
Extrude downwards, and put the 3D cursor roughly at the center of the arc you intend to make at the bottom. AltR Spin the bottom edge through a negative angle to roughly form the arc. Go into the tool's settings to adjust the angle, center and number of segments. (Center Z should be 0)
Same as 4, spin from the downward facing edge. SX0 Scale the vertices near the center line, so they weld together
CtrlR put in a couple of edge loops as shown.

E Extrude the whole shape in Z to form the base thickness of the handle... then  select the shown region of faces to EZ the finger-grip.
On the other side, select all the faces of the original ring and I inset, so you can select a region of faces to extrude down Z to make the neck.

Now you could just put a Bevel Modifier by 'Angle' to bevel all the sharp edges, but it would probably be more accurate to go round CtrlR putting in holding loops, GG sliding them close to soft edges to make them sharper.
Once you have some habits, and get to know how to avoid various 'gotchas', this stuff is a lot quicker to do than to describe.)

One option, when finished, is to bend the whole thing using a Lattice Deform modifier.
Comment: @IlezSnakeEyes don't tick me until you have! ( and.. psst... _whispers_ anyway, I think you should give it back to Palo... I really wouldn't mind)
Comment: Very detailed answer.

How do i model this plastic handle?


How do I make enough money to live and keep a roof over my head?

Question: I've been playing Papers Please, doing my best to go along with the more idealistic elements of the regime, without bowing to the more horrific totalitarian number-pushing and individuality quashing aspects, only to find myself destitute and at the mercy of the totalitarian forces I just mentioned.
However, despite my sympathy for the cause and all that, I keep getting letters saying they're taking the house and everything I own... And even when I'm not getting the government turning up at my doorstep with their best kept prole-quashing boots on, I find I can barely afford to feed myself, let alone pay for unexpected bills like doctors and dentists or look after family (I think there was a dentist bill last time I played...) -- despite occasionally picking up a little bit of off-the-books work or a cypher-puzzle here and there.
So: if anyone has any good, basic tips as to how to make enough money to survive, preferably while avoiding getting carted off to prison for my highly subversive activities checking over documents and asking people questions, it would be appreciated.
I've been looking for guidance online, but I'm still very much struggling to provide for my basic needs.
Comment: Related: [How can you make more money and upgrade flats](http://gaming.stackexchange.com/q/127237/4797)
Answer: First, if you accept the 1000 dollar EZIC gift, you will be found out 100% of the time... The only tactic to support your family is to get really good at examining the people quickly. For every successful entrant/reject, you get $5.
Just have to get really good at identifying the seals, required paperwork, identifying fake city names, or city names that have the wrong province, etc. It takes practice. There is a cheat sheet of the visual stuff somewhere, but I haven't played in a while, but google may turn something cool up for you.
Edit: This is the kind of thing I'm talking about. There are lots out there, so find one that works for you.
Comment: True, but doing that too many times and you'll get fined. You get $10 for detaining when the option arises, but you lose the potential $5 from a successful interrogation. Personally I just detained only the ones that should be, seems to maximize my profits.
Comment: also, remember you can detain people for extra money
Answer: You can also alternate between not feeding your family and not paying for heat. Your family will always be either hungry or cold this way, but it seems they they don't really bother, as long as either status lasts at most a day. (Maybe pay for both if someone is sick)
Also note that you won't lose any money for the first two tickets you receive. So there is no need to be too pedantic (especially with hard to check stuff like issuing city if you don't know it by heart) until you got your second warning.
There are some more regularities that can help: there is always at most one thing wrong with the paper given to you, so if you have to ask for one of the documents and then receive it, everything else will be fine; the first character (sometimes the first two) usually introduces the new rules of the day, so you can usually skip all checks that you already had to do the day before; if you get a card for a brothel, all paperwork seems to check out (as far as I have played); and probably some more patterns I have not yet noticed.
Comment: I've played the game for a while. I might go back to it. It seems quite similar to Don't Starve, though.
Comment: *"Also note that you won't lose any money for the first two tickets you receive."* While this is *technically* correct, it is misleading as this does *not* mean there is no difference at all for the first two mistakes. The two warning citations have no penalty in that you don't *lose* salary, but you still miss out on the salary that you otherwise would have earned if you did not make that mistake. Once the citations begin charging penalties, not only do you miss out on earning salaries, but you are also *losing* previous salaries.


Advantage of clock enable over clock division

Question: I have an FPGA design which uses different clocks. There is a 100 MHz reference clock provided by an oscillator. The reference clock is used in a DCM (Xilinx FPGA) to generate 3 related clocks, 100 MHz, 50 MHz and 10 MHz (no phase offset).
In situations like this the advice of tool vendors and collegues is to use only one clock, the 100 MHz one, and instead of the other clocks create clock enables which are only active every 5th / 10th clock cycle.
What's the advantage of doing this? Yes, there will only be one clock and timing analysis will become easier, but that's a tool issue and should not dictate fundamental design decisions.
Especially for a design with a high device utilization I can see advantages for divided clocks, since e.g. the 10 MHz clock domain will have a much easier time to achieve timing closure and will "leave more room" for the placement and routing of the logic in the 100 MHz clock domain. With clock enables I'd carelessly give away this advantage.
Comment: Your last paragraph isn't actually true. You would tell timing analysis that it is a multi-cycle clock (i.e. on clocks ones every n cycles) and as long as you don't use both rising and falling edges (bad idea anyway) in the design, then you lose nothing by running at 100MHz+CE - it takes into account the clock enable when calculating the allowed delays.
Comment: @Tom: If I only have one 100 MHz clock how do I easily define multicycle paths for e.g. all the logic running at 10 MHz? Are you thinking of constraining all module instances which should be running at the lower clock frequency? If there is a CDC within one or more modules the constraints will quickly grow in number and become messy (and thus be a burden when migrating to a different plattform or changing the design hierarchy).
Comment: It depends on the timing tool. TimeQuest allows you to specify multi cycle paths based on anything driven by a specified clock enable register (not every register individually). You would have to find the documentation of your tool to see what it supports.
Answer: There are several advantages to this methodology that I can think of:

Clock Network - Firstly you only have one clock rather than three. This means that there is less competition for global and local clock routing resources. There are usually only a small number of low-skew clock trees, so minimising usage requirements can help routing.
ALM Restrictions - Depending on the structure of your ALMs/Logic Blocks, you may only be able to clock each block from a single clock - i.e. you can't have two registers on two different clock domains in the same block. If this is the case on your FPGA, then using only a single clock-enabled clock domain could allow logic to be more tightly packed in a design that has a high % of blocks used. Tighter packing can then lead to a higher chance of meeting timing as everything is closer together.
Clock Domain Transfer - Another important consideration is going between clocks. If you need to transfer data between the clock domains, then there are two options.
If you have no reference between the clocks as would be the case if you divide the entire clock (you don't know which edge of the fast clock corresponds to the slow clock), then your transfer must be asynchronous which implies the additional headache of FIFOs for data, and synchroniser chains for control signals.
On the other hand, if you use a clock enable to slow things down, you know exactly when the slow clock and fast clock are going to intersect - you know this by monitoring the clock enable signal. For example if you want to transfer data from your /10 domain to your /1 domain, then you don't need FIFOs to do it - you just say that the clock enable signal is also a valid signal. Because they are the same clock, no clock domain transfer is needed.
It is possible to do domain transfers from PLL clocks if you can keep track of the clock edges - for example going from a /1 to a /2 and vice versa clock is easy because you can potentially compare the fast and slow clocks directly to synchronise. However this depends on the structure of the FPGA - some won't easily allow the clocks to be fed in to look up tables as data inputs.
Valuable Resources - In smaller FPGAs, PLLs are few and far between. For example Spartan-6 LX9 FPGAs if I recall correctly only have two PLL sites! Ideally you want to save these for things like external interfaces (LVDS, Memory, etc.) and not be using them for general clock division. Why use a valuable PLL for nothing but integer clock division when it is possible to do it in general logic.
Furthermore, using clock enables, your division is simply a counter which can be implemented anywhere on the FPGA. PLLs on the other hand are located in specific areas. Say you need a local clock on one side of the chip, but the only free PLL is on the other side. You have to use not only a remote PLL, but also valuable global clock routing to get the clock all the way to the other side of the chip where it is used locally. If instead you build a counter for clock enables, it can be placed right next to the logic, thus reducing usage of valuable resources.

Those are just the reasons off the top of my head. I will try to thing of some more, and also I'll add in some "why is PLL favourable to CE" arguments as well.

In answer to your final point about timing. In practice, using a slow clock or a clock enabled fast clock doesn't affect routing too much. Really your designs should have all registers use the rising edge of a clock (or all use the falling edge) but not mix and match (except maybe in DDR I/O buffers). As a result, both clocks still have their sensitive edges at the same time. The fast clock has n times more, but the clock is not enabled, so these are ignored.
This means you can tell your timing analysis tool that either the clock is multi-cycle - i.e. only one of n cycles is valid - or you tell it that it is a slower clock with a narrow duty cycle. Both of these ways let the tool know that it can allow multiple fast clock periods of setup and hold time in its calculations (because the register values aren't going to change when the clock is not enabled).
Comment: OK, so as I see it, it boils down to a trade-off b/w either using more global clocking and routing resources in case a reference clock is actually divided, or using more general logic and routing resources to generate and distribute the clock enables. Depending on the architecture and which resources are scarce/abundant both techniques can be viable solutions. Thanks for shading some light on a design aspect of FPGA clocking which many people repeat like a mantra without actually giving any objective arguments.
Comment: One thing I would think would be useful, but I've not often seen done, would be to have everything which is run off the primary clock switch on e.g. the rising primary clock edge, and have all divided clocks switch at the falling primary clock edge.  Unless one is pushing the limits of device speed, that would allow signals to cross clock domains without delay (signals passing between divided domains could be cleaned up without delay by using a single register triggered by the rising master clock edge).  Are such techniques used much?


Confused about direction of current in bjt question

Question: In solution of this question they said that current through 3k and 9k resistors are same...
Here my confusion is
Why Ib of Q3 is zero? Because currents for both resistors are same only when Ib
is zero.
Also base voltage of Q3 is 3V given in solution, but i think that it should be voltage across 9k resistor (which is 9V as per voltage divider)
Comment: You shouldn't accept a answer so fast.  You don't know what others might say that haven't even had a chance to see the question yet.  Give it a day or so.
Answer: For reference and to protect against edits to the question, here is the schematic being discussed:

they said that current through 3k and 9k resistors are same
That is a legitimate first approximation.  It is basically saying that the gain of Q3 is infinite.  Since the base current is the collector current divided by the gain, the base current is 0 in this approximation.
Assuming that the transistor gain is infinite is often useful just to get a basic idea what the circuit is doing.  Once you solve for the voltages and currents with this simplifying assumption, you get some idea what parts are sensitive to transistor gain.  Then you can go back and figure things more accurately in that area.
Another point is that good BJT circuits work with the transistor gains anywhere from the guaranteed minimum to infinite.  Therefore analyzing at infinite gain is verifying one of these criteria.  Often there is little difference between a gain of, say 100, and infinite.  Whether the collector current is 99% of the emitter current or 100% of the emitter current shouldn't make much difference in a good design.
Let's see what we get with infinite gain, then compare that to the fixed gain of 100.  With infinite gain, the base current is 0, so there will be 9 V across the 9K resistor, and 3 V across 3K.  This leaves about 8.3 V across 2.77K, for a collector and emitter current of 3 mA.
With a gain of 100, the resistance connected to the emitter is reflected to the base times (gain + 1).  The base therefore looks like a 280 kΩ load to -11.3 V.  The base voltage then comes out to -3.07 V.  The voltage across the emitter resistor is 8.23 V, and the emitter current 2.97 mA.  Again using a gain of 100, the collector current is therefore 2.94 mA.
This is a good example showing how the infinite gain approximation can lead to useful first pass results.  With Q3 having infinite gain, I0 is 3.00 mA.  With it having the finite gain of 100, I0 is 2.94 mA.  If this transistor has a minimum guaranteed gain of 100 at this operating point, then the rest of the circuit has to be able to tolerate the 2.94 to 3.00 mA variation.  That's life with real BJTs.  Fortunately, that level of uncertainty should not be a hardship.
Comment: Can you give me some insight about voltage divider confusion?
Comment: @user: I don't know what you mean.
Comment: In solution they are saying that Base Voltage of Q3 is 3V, but i think that base voltage is voltage across 9K resistor as it connected across base, am i wrong?
Comment: @user: The base will be at about -3 V *relative to ground*.  Generally when a absolute voltage is given, it is implied to be relative to ground.
Answer: The solution should have said that the currents are nearly the same. The base current is much less than the 1 milliamp flowing through the 3k and 9k resistors.


How can I run ALL my Apps "as administrator" by default in Windows 7?

Question: I know that I can change the property in advanced setting of the shortcut to run one app as administrator.
But is possible to do that for all my application? There is a general setting or something somewhere that can tweak that?
Comment: I know it's many years later, but I just found this other question and answers to be helpful: http://superuser.com/questions/1002262/run-applications-as-administrator-by-default-in-windows-10?rq=1
Answer: You can disable User Account Control (UAC) entirely from the control panel; this will mean all programs (installed, downloaded, background, everything) will effectively run with the privileges you currently have to grant manually (via UAC pop-up or compatibility settings).
I think this is what you are asking for?
However, I highly recommend against doing this; it lowers the security of your entire system by allowing any program to modify your system.
In Windows 7 you can adjust the level of UAC to suit your needs. In descending order of strength, the settings for Win7 are:

Always notify
UAC pop-up for changing certain Windows settings and for programs that require elevated rights.
Notify me only when programs try to make changes to my computer
Doesn't request permissions when changing Windows settings, they are granted automatically. Programs still require using a UAC pop-up.
Notify me only when programs try to make changes to my computer (do not dim my desktop)
As above, but screen is not dimmed. Potentially leaving the UAC pop-up windows open to manipulation by other software.
Never notify (I suspect this is what you are after, but please reconsider!)
If logged on as an admin-level user everything is automatically given elevated permissions.
If you are logged on as a standard-level user, anything that requires elevated permissions will automatically be denied (with no opportunity to enter a password in a UAC pop-up).
Comment: Vista allows notifying without prompting for a password, just asking for permission. Does one of these settings do that in Windows 7?
Comment: @datatoo - all of them (well, except "never notify", for obvious reasons), if you're an admin-level user you just have a allow/deny pop-up to respond to (no need to enter a password). If you're a standard-level user then the UAC pop-up requires an admin-level user to provide a password.
Comment: thanks, I didn't realize that was the cause. It is the safe passwords that make it a hassle when you are a poor typist
Answer: 1 - Add User To Administrators Group And Remove From Users Group:

Log In As Administrator
Go To Run ( WinKey + R )
Type "control userpasswords2"
Select Your Account And Click On "Properties"
Select "Group Membership" Tab
Select "Administrators"
Click OK And OK

2 - Disable Approval Mode

Log In As Administrator
Go To Run ( WinKey + R )
Type "secpol.msc"
Go To Local Polices > Security Options
Find "User Account Control: Switch to the secure desktop when prompting for elevation"
Open It Click Disable It And Click Ok.
Restart Computer( Very Important Log Off Doesn't Work )

Now Every App Will Be Run As Administrator In All Administrator Users
Comment: the info on how to access secpol.msc was very useful
Comment: Thank you this is a godsend. Confirmed that this also works in Win 10
Comment: I recently reinstalled Windows 7 Home. When I try to run `secpol.msc`, the file is not found. I searched for it and found 2 copies, in `c:\Windows\winsxs\amd64_microsoft-windows-s..ytools-ex.resources_31bf3856ad364e35_6.1.7600.16385_en-us_61a93666a0ac0769\ ` and `c:\Windows\winsxs\amd64_microsoft-windows-s..ment-policytools-ex_31bf3856ad364e35_6.1.7600.16385_none_b55447455ac6a57a\ ` however running either of them tells me `MMC could not create the snap-in.` ([image](https://i.stack.imgur.com/YUmOF.png)) **Thoughts?**
Answer: You can always just logon as the "Administrator" account.
Or, just set UAC to never prompt you, and then all the apps that need admin rights will get it, without you needing to do anything. Just search for "User Account Control" on the Start Menu.
Just so you know though, this is very dangerous. Any malware will have full admin access!
Answer: If you are on windows 10 home you can use this to enable the local security policy
https://www.majorgeeks.com/content/page/how_to_enable_local_security_policy_in_windows_10_home.html
Comment: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).
Comment: While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference.  Link-only answers can become invalid if the linked page changes. - [From Review](/review/late-answers/1108077)

How can I run ALL my Apps “as administrator” by default in Windows 7?


"Always/never" Example of colloquialism

Question: When someone in conversation says something like "you always forget [x]" or "you never remember to [y]", is the use of the word always/never when it is understood that it is not literally always or never, considered a colloquialism?
Comment: It's just hyperbole or exaggeration. Nothing about it is particularly colloquial.
Answer: As Mitch commented, this is hyperbole.
From MW

extravagant exaggeration (such as "mile-high ice-cream cones")
Answer: I believe it is an example of a colloquialism because "never remember" or "always forget" are basically slang terms for having a poor memory. But I would not use it in a academic paper or any serious type of writing. Although, for fictional writing it is appropriate if you are trying to set a scene or trying to portray your subjects through their verbal etiquette.
Answer: In the sense that the phrase segments are an aspect of common speech - yes, they are absolutely colloquialisms.
I find it interesting that the difference between the statements is that "always" refers to an active behaviour where as "Never" refers to a lack of action.  So in the sense of the phrase meanings "never remember" would seem the best fit to the intent of the expression.
Answer: From Garner's Modern American Usage 3rd Ed:
colloquial
A good deal of confusion has surrounded this word. Traditionally, lexicographers used it to denote that a word so labeled is typical of cultivated conversation or informal speech, as opposed to the most formal style of written prose. E.g.: "The sensible man speaks colloquially most of the time. When he wants to be formal or unusually impressive he tries to speak as he thinks he writes. But on these occasions he often makes a pompous ass of himself." Bergen Evans, "Your Speech Is Changing" (1959).
    But many dictionary users mistook the label as indicating a departure from high grammatical standards—even though the dictionary writers who used the label had no such intention. Hence colloquial, and especially the corresponding noun colloquialism, gradually took on negative connotations. In the second half of the 20th century, most lexicographers dropped the labels. At the same time, mediocre writers strove for hyperformal stiffness, as another authority noted: "Most of us, when we write, have a fear of dropping into colloquialism, and so go to almost any lengths of stilted periphrasis to avoid it." G.H. Vallins, Good English: How to Write It 145 (1951).
    Things have gradually changed, partly as a result of the electronic age: modern communications are increasingly informal. We have come closer to developing a style of speakable writing—one that is natural, idiomatic, and comfortable. Perhaps, after a period of degenerate connotations, colloquial will become a term of praise.

"Always/never" Example of colloquialism


How does electromagnetic induction happen?

Question: Electric and magnetic fields are different from each other(i think i am correct).. How does changing magnetic field induce electric current???
Comment: Electric fields and magnetic fields are just different components of the *electromagnetic field*.  The (classical) relationship between the field's components is described by Maxwell's equations.
Answer: Alfred and Sam29 are quite correct, but the Maxwell equations and saying electric and magnetic fields are just different components of one thing, the electromagnetic field, is just a (nicely consistent) way of representing what is observed in nature.
It does not answer the question of why it is observed in nature.
I'm not sure there is a good answer (other than "it just is"), but I would propose a poor answer:
I assume you have no problem with the principle of the electromagnet.
If you run a current through a wire, a magnetic field is observed around the wire.
This can be exploited to make powerful magnets.
Turning on the current causes the field to appear, and turning it off makes it disappear.
Turning the current on half-way gives you half of the magnetic field.
Then, all you have to do is see there's no reason to suppose the relationship is one-way only.
Then if you supply the magnetic field from outside, if it's two-way, you should see a current, and you do.
There's a simple demonstration:
Take a flashlight battery, and take a short piece of wire connected to one end of the battery, and tap it on the other end.
You see a tiny spark.
Now do the same thing with a long piece of wire in a coil (making an electromagnet).
You get a fat spark.
Why? Because when you disconnect the wire from the battery, the magnetic field in the electromagnet doesn't just disappear, its change induces current in the coil.
Again, it may be easier to see that, because of conservation of energy and whatnot, the relationship between current and magnetic field has to be two-way.
Then, Maxwell's equations make a beautifully concise explanation of that two-way relationship.
Answer: As Alfred correctly pointed out, both the electric and magnetic fields are fundamentally the same field, the electromagnetic field - they are just different components. By this logic, naturally, if you were to somehow change one component (by moving a magnet or something of that sort), it would lead to a change in the other.
Keep in mind that this obviousness is now easier to see because of Maxwell's laws. Before them, many physicists tried to answer the same question using their own approaches, and they failed. Electromagnetic theory gives answers in perfect harmony with experiments. I suggest you look up the laws, and that would answer your question better than any of us here could.
Answer: Electric fields and magnetic fields are very related. This can be summarized as: a changing magnetic field creates a changing electric field (and vice versa). Moving the magnet closer changes the  magnetic field at the coil, creating an electric field in the wire.
Have you learnt about the force of a magnetic field on a charge yet? If you have, then here's another way to think of it. Imagine instead of moving the magnet into the coil, you moved the coil to the magnet. By the Lorentz force, the charges in the wire would  move in a spiral e.g along the wire.
Moving the wire through the magnetic field is physically the same as increasing the magnetic field at the coil. That's why changing the magnetic field creates a current.
Comment: Why it is like that ? I mean why a charged conductor produce magnetic field or why when we cut magnetic flux, it induces current  ? what is the reason ?


What to do when iTunes answer is PC-specific

Question: I came across an answer that is obviously directed at Windows users of iTunes. Is that okay? What can be done to improve the answer, even if my personal PC knowledge is lacking?
Answer: iTunes for Windows is very much on-topic for this site. iTunes is an Apple product, and many of its users use the Windows version.
If a question asks about iTunes for Mac and the answer comes in about iTunes for Windows, that is obviously a less-helpful answer. Similarly if the question is about iTunes for Windows but the answer is Mac-centric.
The real trick is when the question doesn't specify platform. Many users on this site (myself included) will give Mac-centric answers to a generic question (not on principle but because we forget iTunes for Windows exists!), but that isn't always what the original poster actually wants. So if you're more knowledgable about the Windows version of iTunes, please do contribute to answers that help Windows users.
Otherwise, you (specifically as someone who knows more about Windows than Mac) can do two things to improve the answer:

help improve the QUESTION by commenting to find out if the OP (original poster) is actually looking for a Mac or PC solution
use the expertise make the Windows-side answers as good as possible, and let people who are more familiar with the Mac side of things do the same.
Comment: Very helpful, thank you! Also, I embarassingly switched Mac and PC in the second sentence of the question. I have fixed this.
Comment: Edit is your friend @myhd - you can always link to an edit you have made in [chat] should you want some eyes to confirm you're on the right track.


Tokens in menu link and title

Question: menu_token module is working fine to achieve the purpose. Does the module token_filter has this functionality? I can use tokens in wysiwyg using token_filter but cannot make it working for menu title or url since there is no setting I could find. Since the comment in drupal issue says that it was solved using token_filter module, I want to know the way it was achieved.
Answer: No, not by itself. The Token Filter module's functionality adds a token input filter to text formats in Drupal. The only field types that can use text formats are the 3 Text (formatted) field types, not the Link field type.
One can only speculate on the comment's solution you refer to. Text Formats/Input filters are designed to allow HTML input that should be filtered for presentation. It would be an odd-use-scenario to apply that to a link field.
Comment: Any alternate to `menu_token` module?
Answer: Approach to solve the problem-

Loop through menu links (optionally title) in custom module
Process the tokens using \Drupal::token()->replace() method

Tokens in menu link and title


How is the authenticity and integrity of the various chips inside laptops and mobile phones ensured by their vendors?

Question: Modern laptops and mobile phone platforms are built around a main, beefy SoC, which generally supports Secure Boot for its firmware and also has a unique hardware identity that is used to attest to a remote management system that it is genuine and not tampered. Yet there are a plethora of chips, many coming from a 3rd party compared to the SoC vendor or platform vendor, that may be tampered in the supply chain, or in a repair shop, either by replacing their firmware with a malicious one, or by replacing the chip altogether. Such examples could be NVRAM chips, chipsets, 4G (5G) modems, DSPs, audio chips etc., and I won't even mention the PC-world mainstays of GPUs, SSDs, network cards etc.
There are a few industry initiatives that promote or require that such chips are authenticated and preferably their firmware integrity attested, such as OCP Cerberus, Google Titan, DMTF SPDM, NIST SP800-193 etc. but it seems they are more common for the server domain. Some vendors even interpose the loading of these 3rd party components' firmware by a platform Root of Trust (RoT) they control, to make sure no unverified firmware can even reach them.
Searching online for the approaches of top smartphone and laptop vendors, I noticed they largely focus only on the integrity of the main SoC (e.g. based on Google Titan mobile, Apple T2, HP ESC, Microsoft Pluton etc.), and don't verify much the other chips, especially if they're soldered to the motherboard.
So I would like to know if such chip authenticity solutions are deployed also in consumer devices, ideally with some real-world examples to understand their working principle. And if not, why not? Are the threats of replacement/tampering the 3rd party chips not that high?
Answer: Consumer devices mainly prioritize securing the main SOC, while efforts exist to ensure third-party chip firmware integrity via isolated environments like Intel SGX, Qualcomm SEE, and ARM PSA. Examples include Intel SGX for application security and Qualcomm SEE & ARM PSA for overall device security.
Comment: How does e.g. Intel SGX ensure third-party chip firmware integrity? Or any of the other mentioned ones, as they're all "main SoC" technologies.


Graphics card boot loop dependent on same-generation motherboard change

Question: I recently purchased a Geforce GT 730 to use on my Core i5 machine with an H61H2-M2 (H61 LGA 1155) motherboard, but installing it results in a startup loop where the computer fails to POST.
The screen comes alive with the BIOS logo and basic info, but it doesn't allow any input, and after a few seconds the power cuts off (no attempt to load the OS is visible). A few seconds later it attempts to start again, and the boot loop is engaged.
I transferred the card to my Core i7 machine with an MSI P67A-G45 (B3) (P67 LGA 1155) motherboard, and it worked fine. A noticeable difference was that this time the graphics card showed a brief identification screen before booting just fine, no loop.
The power supply in the i7 machine was considerably more powerful than I had in the i5, so to eliminate that variable I transferred the card and the power supply to the i5 computer, but the boot loop remained.
I suppose the most notable difference between the motherboards is the chipset, H61 vs P67. But is that really enough of a difference to decide support for a budget graphics card? What other variables decide compatibility.
Answer: The answer appeared to be that the PCI-E slot was bad. A new motherboard solved the issue.


Move photos to SD card on HTC Desire 310

Question: I have an HTC Desire 310 and my phone memory is full and can't update apps. How to move photos from my phone memory to an SD card?
Answer: Pretty much any file browser in the Play Store allows you to move files. Just download one. Find the folder. Long press on it, tell it to move. Then navigate to the folder on the SD card and off you go.


Free potable water law in UK and EU

Question:
In Britain, all licensed premises are required by law to provide
  "free potable water" to their customers upon request.

What legislation covers this?
Is this the case in other European countries? I am particularly interested in whether this is a legal requirement in Italy.
Comment: It is worth noting that the requirement is for free water *to their customers*.  If you just pop in to ask for a glass of water, you are not a customer; you have to buy something else to qualify.
Answer: In the UK, this is covered in the Licensing Act 2003 (Mandatory Licensing Conditions) (Amendment) Order 2014, which amends the Licensing Act 2003 (Mandatory Licensing Conditions) Order 2010.
Mandatory Licensing Condition 2 states:

The responsible person must ensure that free potable water is provided on request to customers where it is reasonably available.

Note that this applies only to premises which are licensed to sell alcohol.
Comment: This partly answer the question... Thank you.
Comment: @Leo: No problem. :-) I'm not familiar with the law in other European countries, so I'll leave that to others to answer.
Answer:
France is an often cited example of a country that mandates restaurants to serve tap water for free,
while Germany and Austria do not have any such legislation.

In Austria it is nevertheless more usual than in Germany for restaurants to not charge guests for tap water (according to the linked German newspaper articles).
In Germany, expect to be charged also for tap water - though my last experience in Germany is also: no charge for tap water alongside a big bowl of coffee.

I'm not entirely sure about the legal situation in Italy, but the very least it is in my experience customary for restaurants to provide tap water for free (sometimes automatically, sometimes after asking).

There have been some rumours that the EU may mandate free tap water in restaurants and the like, but the AFAIK current status of that directive proposal speaks about

encouraging [rather than mandating] the free provision of water intended for human consumption in public buildings, restaurants, [...]
Comment: This partly answer the question... Thank you.

Free potable water law in UK and EU


Why are some mistakes counted as "experimental error"?

Question: I have always thought (and been told) that experimental errors and uncertainties do not include mistakes which can be fixed by just conducting the experiment more carefully. Yet we still include parallax error and zero error when talking about experimental error. Can these two not simply be removed by conducting the experiment carefully? Or is it that these two might happen despite our care?
Answer: Zero error can be usually fixed by re-calibrating measurement device or changing it into a newer not damaged device. It's one-time fix. However story is far more complex about parallax error - you can't be sure that you are looking at device readings at a right angle. Besides usually parallax error will be lost in other more fundamental error source which is half-value of measurement device smallest scale interval. For example. Consider that you are measuring square's perimeter with a ruler which has smallest interval of $1 mm$. Your ruler readings are (10  + 10 + 10 + 10) mm. Then because measurement errors have tendency to add-up, your total measurement error of perimeter is
$$ \frac{1}{2}mm+\frac{1}{2}mm+\frac{1}{2}mm+\frac{1}{2}mm = 2mm $$
Thus your perimeters measurement result would be $$40mm\pm2mm$$
Comment: Probably because calibration can't guarantee **100%** accuracy. Every device will be _slightly differently_ calibrated
Comment: Apologies for the late reply , but if zero error can be corrected by taking care that our instrument is not damaged and is calibrated, why is that still talked about when talking about experimental errors? I understand your point about parallax.


Nginx reverse-proxy cache invalidation strategies

Question: I've been using Nginx 0.8.5 as a caching reverse-proxy to build out a custom cdn-like solution to speed up our site's load time. It works great so far except that cache invalidation is really cludgy, even with the cache purge module installed.
For example, sometimes we may need to invalidate 100 urls at once, and this requires 100 different requests to get sent out to each Caching reverse-proxy server.
This hardly works with just two cache servers set up, and our plans are to scale to 20 servers, so we need to find a better solution. Any ideas?
Simply reconfiguring Nginx won't work, and I'm sure that we will need to custom-code something. And we don't want to get involved in custom Nginx module development.
Answer: maybe this approach will help you: Purging the Cache(in the end)
basically instead of using the purge module they just show you how to delete the nginx files directly, so you can make some easy script that will get urls and will purge them directly

The naming convention of the cache is based on the variables we set for the "fastcgicachekey" directive... Passing this string through MD5 hashing... get the directory and delete the files


How to know if a JR train has unreserved seats?

Question: On Japan, most trains have at least one car for unreserved seats but not always. Is there a way to know if a train has such seats? Unless I missed it the JR machines to book tickets and check time schedules don't give you that information.
Answer: Japan Guide goes into which trains offer unreserved seating and which don't:

Most shinkansen trains offer both non-reserved seats (自由席, jiyūseki) and reserved seats (指定席, shiteiseki) in separate cars. Only the Hayabusa, Hayate and Komachi trains on the Tohoku Shinkansen and Hokkaido Shinkansen and the Kagayaki trains on the Hokuriku Shinkansen are fully reserved and do not carry non-reserved seating.

If you're not sure, you can check the train you'll be taking ahead of time on HyperDia.
If there is an "unreserved seat" option under "seat fee", then that train has unreserved seating. Otherwise, it doesn't. Note that you still have to pay a "seat fee" even for unreserved seating. This is because the "fare" is actually the same as a regular train, and then the "seat fee" is a combination of the "limited express fee" and seat reservation fee.
Note that if you're on holiday in Japan and travelling with the JR Pass, you can request a free seat reservation by visiting the ticket desk at the station.
Comment: +1 for Hyperdia, a must-have for train travel in Japan and great if you have a JR rail pass


How is 'bolt-on' the same or different from add-on, extension, or module?

Question: I hear the term 'bolt-on' used in many contexts, specifically in my organization where PeopleSoft is used extensively.  I don't know exactly how it differs from an 'extension' of an existing product, a new module within an application, or an 'add-on' like you would purchase for a product you own.  Is there a more formal definition of 'bolt-on' that I'm not familiar with?
Comment: "Bolt-on" is used to identify a component that is quickly and easily added to another system. This distinguishes it from components that require more sophisticated integration or customization.
Comment: It is a fancy word for module, extension, etc.
Answer: Most of those terms only have meaning in a specific context. In particular, I've only heard bolt on used with respect to PeopleSoft, where it means a separate application that integrates with the PeopleSoft ecosystem. Here's an article about the difference between bolt-ons and customizations (tl/dr: a customization is a change to some part of the PeopleSoft system, a bolt-on is a separate thing that uses the PeopleSoft system).
The other terms you used, module, extension, and add-on also need to be considered in a particular context. For example, in some contexts a module could be some source code component that needs to be compiled into a program, while in others it may just be a functional component of the program that can be added without modifying the original.
If you're unsure of what someone in your office means by a particular term, ask them. If you feel you can't do that without looking uneducated, check the documentation for the pieces of software that they're talking about.
Answer: As M. Dudley stated in a comment below the answer, "bolt-on" is used to identify a component that is quickly and easily added to another system. This distinguishes it from components that require more sophisticated integration or customization.

I would also add that this is mainly a slang term, and can also mean anything added to a system hastily, without regard for suitability or esthetics.


Most Efficient Commercial Aircraft Design?

Question: What was the most efficient aircraft design (based on factors such as lift, drag, and fuel consumption) but not developed or scrapped due to issues with long term viability and maintenance, operations and repair?
Comment: *but not developed* - We can't truly know that, because companies generally don't publish their designs until they are being developed. I'm sure someone will come up with some things that reached the prototype stage, but this is inherently an unknowable thing.
Comment: And *not developed* means we do not have data. From design to a certified aircraft there are many changes. Simulations cannot give you accurate figures. And you forget one important part of equation: engines (which are usually designed or modified for a specific aircraft, so to calculate drag and efficiency is impossible)
Answer: The most efficient designs, based on lift, drag and fuel consumption, for heavier than air craft, were the ones that could fly slowly enough and still have sufficient Reynolds numbers to generate lift efficiently.
Some, like the Handley Page HP 42 could lift 15 pounds per horsepower and fly at 100 mph.  The contemporary Douglas DC-3 could lift 10 pounds per horsepower but fly it twice as fast.
The C-130J Hercules can lift 8 pounds per horsepower but fly it at 400 mph.
Aircraft gain advantage in True Air Speed by flying higher, thereby faster, to maintain the same Indicated airspeed in thinner air.
However, at altitude, a (turbine) compressor is needed to provide enough air to maintain enough thrust for the same IAS.
So the most efficient design may be ... a jet powered aircraft with an excellent, glider-like lift to drag ratio, flown at altitude.
Modern airliners are designed in precisely this way$^1$.
$^1$ ton-miles per gallon fuel consumed is one way of looking at it.  The faster plane can then become more efficient.


Is there such a thing as wall lights that you can plug into a power outlet?

Question: There is a room in our house with a fluorescent bulbs from the dark ages and they've stopped working a while back. We've tried replacing the starter, it worked for a moment but stopped, maybe a loose connection somewhere. We can't get an electrician to come check it out for the time being, so what I'm wondering is, are there any wall lights that you can plug into a power outlet and hang on the wall? This might be a stupid idea, but is there? Googling only brought out accent lights, but I want something a little more powerful.
Comment: If you want something more powerful, you should just go for a floor lamp.  Wall sconces are going to be limited to one light, usually in the <40 watts.
Answer: Yes, absolutely: what you want are called "wall mount plug-in lamps". You have a number of choices, e.g.:
See that pipe going downward? It's a conduit that dresses up the extension cord and protects it from getting yanked on. At the bottom is a standard plug.
These come in all sorts of shapes and sizes. I have one in my house which takes a 3-way bulb (50/100/150W) so you can get a lot of light out of them, and they often have hinges to let you swing them in and out.
Generally you will need to drill 2 holes into your drywall, hammer in a set of drywall anchors, and then attach a plate to it. (All the parts are included in the light.)
If you're renting, check with your landlord before drilling holes in the wall--but often it's easy enough to use spackle or compound to fill the holes you made, and then paint over it.
Comment: Thank you so much, this is exactly what I was looking for.


Is Lepton Flavour Universality an accidental symmetry of the Standatd Model?

Question: Is Lepton Flavour Universality an accidental symmetry of the Standatd Model? If it is, why? How does it emerge from the Standard Model?
Answer: It is not accidental, at least in the sense that this term is used.
The accidental symmetry is the symmetry that is forced on you by the renormalizability - i.e. for a given field set you can't write the interaction that would violate such symmetry. On the other hand nonrenormalizable interactions or renormalizable interactions with extra fields (these two options are related) may violate symmetry. The example is the baryonic number conservation which is easily violated in many extensions of the standard model.
On the other hand the lepton universality is the property of the non-abelian gauge invariance. In qft the non-abelian gauge interaction is determined solely by the representation of the field and the coupling constant, common for all the fields. I.e. you can't have $SU(2)$ doublet interacting with $W$ boson 1% stronger than another doublet.
Of course this constraint is weaker for $U(1)$ interaction, such as hypercharge field. But if you change somewhat $Y$ for some lepton, this may ruin the gauge invariance of the mass terms and the triangle anomaly cancellation.
Comment: @AbelGutiérrez choosing lepton couplings to  Z W be the same for the three flavors is compatible with the data and thus no change is needed to the standard model,  except if the data will show differences between expected calculations from the SM and the data, which is what is being checked at the LHC.
Comment: Thanks @OON .One thing I don't understand is why lepton flavour universality is assumed in the SM. (probably I need more theoretical knowledge). Why it is assumed if there is evidence that is not "conserved"? Why the leptons coupling to W,Z must be the same? If evidence turns into discovery, why people talk about possible new particles instead of changing the couplings?
Comment: I understand that there are some observables that preserve lepton flavour universality, but others seem to do not. Is that right? That is my confusion.
Answer:
In the Standard Model of particle physics the three charged leptons are identical
copies of each other, apart from mass differences, and the electroweak coupling of the gauge
bosons to leptons is independent of the lepton flavour. This prediction is called lepton flavour
universality (LFU) and is well tested. In tree level decays, any violation of LFU would be a clear
sign of physics beyond the Standard Model

The standard model developed slowly over the last century by continually fitting the data, using axiomatic assumptions and testing them against data. One of these axiomatic assumptions was LFU .
LFU is not an accidental symmetry, it is the result of the assumed couplings within the standard model.
From the introduction in the link

In the Standard Model of particle physics (SM), the electroweak gauge bosons Z and W ± have
identical couplings to all three lepton flavours. This means that branching fractions of decays
involving different lepton families do not depend on lepton flavour but differ only by phase
space and helicity-suppressed contributions
Comment: Dear, Anna v. I will repeat my criticism for some other answers of yours. You again answer the theoretical question with "the standard model was developed to fit the data, if it would not fit the data you would change the model". The aim of the theoretical question is rather to understand how some particular properties of the model are related to each other. And what exactly should be changed if this property of the model does not agree with experiment.
Comment: @OON In this particular answer there is the theoretical part in the second quote, "it is the identical couplings to all three lepton flavors for the verti ces of Z and W exchange" that imposes the universality lfu to the SM.
Comment: @OON you say "The aim of the theoretical question is rather to understand how some particular properties of the model are related to each other"  and I accept it, but the aim of main stream physics is to model mathematically nature with validated models, which is something theoretically inclined people tend to ignore, prefering to play with the mathematics. if LFU was not evident in the data , a different mathematically model would be necessary.  That is what makes experiments look for violations of LFU, to find new physics.
Comment: I prefer @OON answer but agree with anna v last comment. I look at Physics as a model to describe nature based on mathematics... but maybe today mathematical speculations become useful in future (or in a parallel universe, idk). Anyway, anna v maybe you can contribute to my comments at OON answer.


Why should $\mathcal{A}$ not have elements of size $0$ to form $\text{Seq}(\mathcal{A})$?

Question: In Flajolet's symbolic method, given a combinatorial class $\mathcal{A}$, $\text{Seq}(\mathcal{A})$ is defined as $\mathcal{E}+\mathcal{A}\times \mathcal{A}+ \mathcal{A}\times\mathcal{A}\times \mathcal{A}+...  $ and assuming that the generating function of $\mathcal{A}$ is $A(z)$ then the generating function of $\text{Seq}(\mathcal{A})$ is $1+A(z)+A(z)^2+A(z)^3+...=\frac{1}{1-A(z)}$.In the bibliography it is requested that $\mathcal{A}$ be a combinatorial structure without objects of size $0$, but it is not clear to me why, could someone explain it to me (I think the same thing happens with the $\text{Cyc}( \mathcal{A})$ operation).
Answer: If there is an object $z\in \mathcal A$ with weight zero, then there would be infinitely many sequences in $\text{Seq}(\mathcal A)$ with weight zero. Namely,
$$
(z),(z,z),(z,z,z),\dots,(z,z,\dots,z),\dots
$$
all have weight zero. A combinatorial class is only allowed to have finitely many elements of each weight, so $\text{Seq}(\mathcal A)$ is not a well-defined combinatorial class in this case.
Comment: @user19872448 Same is true for $\text{Cyc}$ and $\text{MSet}$. There might be others.
Comment: Is it the only class in which this case occurs?
Comment: so, to define $\text{Seq}(\mathcal{A})$ $\mathcal{A}$ must not have a neutral element, but if I define $\mathcal{B}=\text{Seq}(\mathcal{A})$ then $\mathcal{B}$ does have a neutral element ?
Comment: @user19872448 That's right.


Webkit Update in JAVAFX

Question: Is it possible to update the webkit in the javafx *netbeans*?
Answer: Source Code for Open Source Software Components: JavaFX
Modified Webkit source code
Answer: Java snapshot releases generally utilize more recent versions of webkit than Java standard releases, so you could try a snapshot release and see if it exposes the WebKit functionality you seek.

There are currently no instructions provided by the JavaFX team on how you might yourself integrate an updated or modified WebKit version into JavaFX.  I think any attempts to do so will require the Java portions of the JavaFX web node to be open sourced as there is likely a tight JNI based integration between the Java code and native code for the web node.  Perhaps such instructions might be provided when the JavaFX web node is completely open sourced.  You could request details from the open-jfx mailing list on how to do this as well as an eta for complete open sourcing.

Webkit Update in JAVAFX


Are all Gromov hyperbolic spaces proper metric spaces

Question: Here a proper metric space is a metric space such that all closed balls are compact. My question is are all Gromov hyperbolic spaces proper metric spaces? I only know the rudimentary definitions of these things which is why I am asking, I didn't see it referenced anywhere and I was curious.
Answer: No.  For instance, a hedgehog space with infinitely many spines is Gromov hyperbolic but is not proper (for instance, any closed ball around the center point contains an infinite closed discrete subspace by picking one point in each spine, each the same distance from the center point).
Answer: Even simpler, any infinite set $X$ endowed with the discrete metric, ie.
$$(x,y) \mapsto \left\{ \begin{array}{l} 0 & \text{if $x=y$} \\ 1 & \text{otherwise} \end{array} \right.,$$
provides an example of a non proper Gromov hyperbolic space.


Are there any languages without Semivowels?

Question: Semivowels seem to be pretty common around the world's languages, I wonder if there are languages which don't have them.
Answer: There is a minor terminological distinction between "glide" and "semivowel", in that "glide" includes the laryngeals [h, ʔ, ɦ] which would not be called "semivowels". The classic non-laryngeal glides a.k.a. semivowels are [j w ɥ], which are the non-syllabic correspondents of the vowels [i u y]. A symbolic/terminological lacuna is the lack of a "semivowel" version of back unrounded [ɯ]: but the symbol <ɰ> is available and describes a "velar approximant". In IPA terminology, [j] is a palatal approximants and [w, ɥ] are labial-veolar and palatal-velar approximants (there's no such thing as a semivowel in IPA), but there are also "approximants" like ɹ that would not be called "semivowels". The main point here is that until you define what you mean by "semivowel", it's impossible to determine if all languages have them. Semivowel is not a well-defined technical term.
Sticking to [j w ɥ], the answer is, from what I can determine, that all languages have one of those sounds at some level. The problem is that one can often analyze any segment out of existence by positing some rule that e.g. derives [j] from /i/ in some context. If you change the question to "do any languages not have minimal contrasts involving at least one of [j/i w/u ɥ/y]", the answer is, yes, vast numbers. Very few languages contrast vowels and corresponding glides.
Comment: So they exist in all Languages?
Comment: do you have a source for laryngeals being glides, because I have never heard that?
Comment: @Tristan I can see [h, ɦ], the laryngeal fricatives, as glides (they're basically vowels anyway), but [ʔ] is a stop - the literal opposite of a glide!


DDOS attacks and multi-threading

Question: Does a host create a separate thread for each TCP SYN request? Or it depends on the socket implementation for the host?
Comment: you need to edit your question body to include the relation with DDoS. Do you want to see if you can solve DDoS by multi-threading?
Answer: Except for user space network stacks the operating system fully handles the complete TCP handshake in the kernel and notifies the application only once the connection is fully established. The kernel part is handled in a lightweight way, i.e. usually there are no kernel threads involved. What's happening in user space after the connection is established depends on the application. Some applications create a new thread or a new process for each new connection while others handle it  inside the same thread and process.
Comment: Thanks @Steffen, so when a DDOS attack occurs, the attackers tries to flood various ports of the host. At this stage there are no threads assigned then how does the host keep track of multiple tcp requests?
Comment: @Rads:  each TCP connection is defined by source and destination IP and port. The host just needs to maintain the current state for a connection based on this tuple which does not require any use of threads.
Comment: Nice answer.  Went into a bit of detail why it's not inherently security-relevant (or might be!) in my answer. But wanted to congratulate you on your pretty concise answer.
Answer: The way a network stack handles multiple connections is absolutely Operating System specific. In fact, it's even networking hardware-specific, in that most better network cards can do a lot of the TCP handling themselves.
You cannot, hence, generally assume that a network stack is inherently multithreaded or not – from a pure flow-logical point of view, a network card gets and processes one ethernet packet after the other, and thus, you can always process them sequentially in software.
However, there might be advantages (mainly: being able to split the workload across CPU cores) to making your operating system handle networking in a multi-threaded way. Note that this threading is likely to not have the shape of user-land threads like you probably know them, but of operating system tasks; but there's no guarantee that TCP handling for existing connections isn't done in user land within the context of the process that is using a socket. It's an OS design decision.
On the other hand, there's good reasons to not multithread networking, especially in low-load situations or when there's only one dominant receiver or sender of packets in your user space (mainly: synchronization is hard, and thus, often more expensive than just parsing a TCP header, and you'd actually lose more than you win when deciding which CPU core should handle a packet after one core has already touched it). This again, is often solved very adaptively by operating systems, or on high-end network cards, even in hardware that can be told to transfer packet contents to different memory regions and thus allow certain traffic to be "sticky" to one specific core.
Since this is the information security SE and you implied the topic of DDOS: everything is highly adaptive. But it's save to assume that the state-checking in modern operating systems can pretty much be done before the next network card interrupt comes in, and then, the handling/generation of SYN-ACK packets can done normally in e.g. a main event loop or in worker tasks.
SYN flooding used to be an effective method some 20 years ago. I just don't know whether that is still the case – modern machines just don't care when the table of connections they have to handle in a SYN-but-not-yet-ACK state hits several hundred kilobytes (gaaasp!). Sure, looking up the few SYN requests of legitimate clients might take hundreds of nanoseconds or even several microseconds longer in a halfway efficient data storage format for attack scenarios where the attacker has not gotten a 10Gbit/s lane to your server... My impression is that a mediocre laptop can these days do stateful firewalling (at least with Linux' default conntrack), NATing and TCP handling at full 1Gbit/s wire speed without breaking into too much of a sweat. Saturating these capabilities with entries to a boring table of half-open connections will be pretty hard.

DDOS attacks and multi-threading