How can you change the "inherit rotation, scale location" for a vertex or empty or an object, not just a bone?

Question: When you parent bones to bones, you have options of inheriting rotation, scale and location. But, I don't see those options when I do things like parent an object to an empty or an empty to a vertex. Where are these options?
Answer: There is no such option. Objects always inherit their parent's transformation. You can use constraints to selectively copy transformations from other objects to work around this, though.
Comment: Using constraints would be better, as it doesn't need any external add-on. I've updated my answer for this.
Comment: There is no direct option, but you could imitate the same behavior using animation nodes
Answer: Just to amplify @dr. 's answer, the 'Child Of' constraint gives Object parenting behaviour, with options for each of the basic transforms, in each of the axes.
The 'Copy XXX' constraints give more options: how to combine the target's transforms with the owner's existing ones, and in which space the copied transforms are evaluated.


Implementing a web server on a microcontroller

Question: I am trying to implement a web server on to a micro controller as a self learning project.I am considering Atmega16 and Atmega88 as primary options.Need to know which will be the best choice.
Answer: Like this one?
There is example using Atmega88.
Answer: Perhaps obviously the one with the greatest memory resources.  Neither part in your short-list supports Ethernet so neither is particularly suited to network applications; you would need an external Ethernet solution - the controller on that is likley to be more powerful and more expensive than the ATMega16/88 in any case.
You could of course use the UART and PPP or SLIP, but that is rather restrictive.


Surfaces contained in a ball

Question: In this Paper there is a proof that a closed plane curve of length
$L$ and curvature bounded by $K$ can be contained inside a circle of radius
$L/4 - (\pi - 2)/2K$. Are there similar results for smooth surfaces in $\mathbb{R}^3$?
Comment: @LoïcTeyssier A negative bound here is not *a priori* absurd.  If $K$ is too small compared to $L$, there may not be any closed curves of the sort in the question --- the ends of a segment of length $L$ can't be brought together without introducing a non-negligible amount of curvature.
Comment: The bound can become negative for small values of $K$ and fixed $L$?
Comment: From the Chakerian et al. paper: In 2D, the extreme is a *racetrack*, the convex hull of two circles of the same "radius $1/K$ whose centers are a distance $L/2 - \pi/K$ apart."
Comment: My bad, I didn't read the key word "closed". Now it makes sense.
Answer: A possible generalization is an upper bound for the diameter of a surface $S \subset \mathbb{R}^3$ in terms of its area $A(S)$ and its gaussian and mean curvatures $K$, $H$. Let me give an answer in the case of positively curved surfaces.
We define an ovaloid as a compact connected surface $S \subset \mathbb{R}^3$ with $K(p) >0$ for all $p \in S$. If $S$ is an ovaloid then it is orientable, and by using Hadamard-Stoker theorem it follows that its inner domain is a convex subset of $\mathbb{R}^3$ and that its Gauss map $N \colon S \to \mathbb{S}^2$ is a diffeomorphism. In particular, $S$ is a topological sphere.
The following classical estimate for the diameter of an ovaloid is due to Bonnet. Modern proofs are based on Hopf-Rinow theorem, see for instance
S. Montel, A. Ros, Curves and Surfaces, Theorem 7.45.

Theorem (Bonnet). Let $S \subset \mathbb{R}^3$ be a closed connected surface whose Gaussian curvature $K$ satisfies $\inf_{p \in S} K(p) >0$. Then $S$ is compact, that is $S$ is an ovaloid, and its diameter satisfies $$\textrm{diam} \, S \leq \frac{\pi}{\sqrt{\inf_{p \in S} K(p)} }.$$

Now, by using the isoperimetric inequality we can show that for any ovaloid the inequality $$\bigg(\int_S H(p) \bigg)^2 \geq 4 \pi A(S)$$ holds, see Exercise 10 p. 196 in the book quoted above. Substituting into the inequality provided by Bonnet's theorem, we finally obtain
$$\textrm{diam} \, S \leq \frac{\big(\int_S H(p) \big)^2}{4 \, A(S) \sqrt{\inf_{p \in S} K(p)} }.$$
Answer: Sufficient conditions, in terms of geometric invariants of the domains, such as volumes, surface areas and curvature integrals of the boundaries of domains, under which one  domain can be moved by an isometry inside the other domain in a 3-dimensional space of constant curvature, are given in http://www.jstor.org/stable/119206 (Sufficient Conditions for One Domain to Contain Another in a Space of Constant Curvature, by J. Zhou).

Surfaces contained in a ball


Progressive Web Applications

Question: Does Salesforce support the development of progressive web applications via VisualForce, Lightning or any other means? Progressive web applications are web applications that utilize service workers, caching, background sync, web push, application manifests, responsive design and APIs for things like media and geolocation to create web applications that look, feel and behave like just like mobile applications.
Answer: Just running down the checklist I think at least most of that should be possible in visualforce/lightning. It just depends on the needs of the company and the developers skill level. There are some items on the check list that don't seem to apply to websites but that could just be me being a bad web developer.
For instance:

Load various pages in the PWA with an airplane mode enabled. Ensure
  the app presents some content even when offline. Use Lighthouse to
  verify the start URL responds with a 200 when offline.

I'm not really sure that's feasible when the entirety of Salesforce exists in the cloud. Even when I try to think of ways to accomplish it, I can't help but feel like they result in a security risk (such as downloading meta-data or data to a local system which you probably don't want to do).
For the most part though, I don't see why you can't develop a Progressive Web Application in Salesforce. Lightning is probably better suited as SLDS is already set up for an enhanced mobile experience.
Update
See comments - The answer is generally yes, you can build a PWA on Salesforce.
Comment: Your example is accomplished via [Offline Caching Using the HTML5 manifest Attribute](https://developer.salesforce.com/docs/atlas.en-us.pages.meta/pages/pages_html_features_cache_manifest.htm). Many of the other features are also available.
Comment: And I'll come back and give you a +1 when the clock resets. I ran out of votes.
Comment: I see what you mean about the manifest. It'll definitely ensure that "some" content displays (which I guess is what the checklist item is looking for) but it would produce a relatively nonfunctional experience as they would still not have access to records or live data. And My point remains that you would not want to locally cache data for security reasons.
Comment: @gNerb data caching are actually few of the requirements as I have seen whenever discussing offline capabilities. Consider an example, where you want to mention a contact from the same company who's seated just besides you on a plane, but you can't just because you are in airplane mode. Or a field service agent who wants to view a part installation manual while fulfilling a service request in a basement of building where there's no network.
Comment: I completely understand the benefits, but that exposes data into a machine that has potential to be less secure. I'm not saying you can't do it, I just wouldn't want to.

Progressive Web Applications


What is the Relationship between the Rank and the Number of Eigenvalues?

Question:
for (a) it says it can't be rank 1 because there is only one non zero eigenvalue? I understand it can't be 0 or 3 but I don't understand how you can conclude we have 2 just by knowing how many eigenvalues we have.
Comment: Recall that the rank is the dimension of the image of $B$, where we view $B$ as a map from $V\rightarrow V$, $V$ being our underlying vector space. Eigenvectors of distinct eigenvalues are linearly independent. Can you show the result from here?
Comment: Yes, they are the same thing, just different names. (Similarly, kernel and null space mean the same thing.)
Comment: Eigenvectors corresponding to distinct eigenvalues are linearly independent. So if $x$ and $y$ are eigenvectors corresponding to eigenvalues $1$ and $2$, then $Bx = x$ and $By = 2y$ are also linearly independent, hence the dimension of the image (aka the rank) of $B$ is at least $2$.
Comment: @Bungo I must be missing something really obvious. I understand Bx = x By = 2y need to be linearly independent. But What does this say about B?
Comment: $Bx$ and $By$ are elements of the image of $B$, and they are linearly independent. In general, if a vector space contains $k$ linearly independent vectors, its dimension must be at least $k$. (Here $k = 2$.) This is because any linearly independent set of vectors can be extended to a basis if it is not already a basis.
Comment: Image is the samething as column space correct?
Answer: Since $B$ is $3\times 3$, and has $3$ distinct eigenvalues, it is diagonalizable. So $B$ is similar to
$$
\begin{pmatrix}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix}.
$$
Since rank is a similarity invariant, it is now clear $B$ has rank $2$.
Comment: @curiousBiggie If you take a basis of eigenvectors, the matrix acts by scaling these basis vectors. The image is then spanned by the image of the eigenvectors, and it's enough to take the span by the eigenvectors with nonzero eigenvalue, because those corresponding to eigenvalue $0$ get sent to $0$. So the eigenvectors with nonzero eigenvalue are a spanning set of the image. Since eigenvectors with distinct eigenvalues are linearly independent, in this case they also form a basis for the image.
Comment: This makes sense but I want to understand it more conceptually. Why does two disticnt nonzero eigenvalues = dimension two?
Answer: It's not how many, it's the fact that $B$ has an eigenvalue $0$. Each nonzero vector in the kernel is an eigenvector of $B$ associated to the eigenvalue $0$, so if there's an eigenvalue $0$, the kernel has dimension at least one, and hence the image (by the rank-nullity theorem) has dimension at most two.
Since there are two other distinct nonzero eigenvalues and $B$ is $3\times 3$, its image must have dimension exactly two (and hence its rank is two).
EDIT: Clarification.

Lemma: Let $A$ be an operator and for each $i=1,\dots,n$ let $v_i$ be an eigenvector of $A$ associated to an eigenvalue $\lambda_i$. If the $\lambda_i$ are all distinct,then $\{v_1,\dots,v_n\}$ is linearly independent.

Proof: We do this by induction on $n$.
For the base case, we take $n=2$.
Let
\begin{align}\tag{1}\label{base1}
c_1v_1+c_2v_2=0
\end{align}
be a linear combination of the $v_i$ that equals $0$.
Applying $A$ to $\eqref{base1}$ yields
\begin{align}\tag{2}\label{base2}
A(c_1v_1+c_2v_2)=\lambda_1c_1v_1+\lambda_2c_2v_2=0
\end{align}
Now, multiply $\eqref{base1}$ by $\lambda_1$ to obtain
\begin{align}\tag{3}\label{base3}
\lambda_1c_1v_1+\lambda_1c_2v_2=0
\end{align}
Now, the expression $\eqref{base2}-\eqref{base3}$ becomes $(\lambda_2-\lambda_1)c_2v_2=0$.
Since the $\lambda_i$ are all distinct, $(\lambda_2-\lambda_1)\neq0$ and it follows that $c_2=0$.
Substituing back into $\eqref{base1}$, we conclude that $c_1$ too is $0$, and hence $\{v_1,v_2\}$ is linearly independent.
Now, suppose the statement is valid for all $m$ with $2\leq m \leq n$.
We will show it holds for $n+1$.
Indeed, let
\begin{align}\tag{4}\label{induc1}
\sum_{i=1}^{n+1}c_iv_i=0
\end{align}
be a linear combination of the $v_i$ that equals $0$.
Then
\begin{align}\tag{5}\label{induc2}
A\left(\sum_{i=1}^{n+1}c_iv_i\right)=\sum_{i=1}^{n+1}\lambda_ic_iv_i=0
\end{align}
and multiplying $\eqref{induc1}$ by $\lambda_{n+1}$ yields
\begin{align}\tag{6}\label{induc3}
\sum_{i=1}^{n+1}\lambda_{n+1}c_iv_i=0
\end{align}
Calculating $\eqref{induc2}-\eqref{induc3}$, the term in $v_{n+1}$ vanishes and we're left with
$$\sum_{i=1}^n (\lambda_i-\lambda_{n+1})c_iv_i=0$$
Here we may apply the induction hypothesis: $\{v_1,\dots,v_n\}$ is linearly indepdent, so the coefficients of the linear combination above must be all $0$.
Since the $\lambda_i$ are all distinct, for each $i=1,\dots,n$ we have that $\lambda_i-\lambda_{n+1}\neq0$ and hence it follows that $c_i=0$.
Finally, substituting them back into $\eqref{induc1}$ yields $c_{n+1}v_{n+1}=0$, so $c_{n+1}$ too is $0$.
It follows that all coefficients of $\eqref{induc1}$ are $0$, and hence $\{v_1,\dots,v_{n+1}\}$ is linearly independent. $\square$
Comment: I don't see the leap from two distinct nonzero eigenvalues to dimension exactly two.
Comment: See the edit. It contains a proof that 'Eigenvectors associated to distinct eigenvalues are linearly independent'. When the operator $A$ is symmetric, they're even orthogonal! It's a nice exercise.
Comment: Notice that if $v_i$ is an eigenvector of $B$ associated to an eigenvalue $\lambda_i\neq0$, then $v_i$ is in the image of $B$. Indeed, $$v_i=B\left(\frac{1}{\lambda_i}v_i\right)$$
Hence, if $v_1,\dots,v_n$ are eigenvectors of $B$ associated to distinct and nonzero eigenvalues, then the image of $B$ contains them all and (by the lemma) has dimension at least $n$. Is the connection clearer now?
Comment: yes thank you!!
Answer: You know that the eigenvector associated with the $0$ eigenvalue is in the kernel of $B$.
And you know that the eigenvectors are independent.  The space spanded by the set of eigenvectors associated with the non-zero eigenvalues has dimension equal to the number of non-zero eigenvalues.
Comment: Yes I understand that but what does eigen space of non-zero eigevnalues have anything to do with kernel of B(I CLEARLY understand the 0 part).
Comment: They are not in the kernel!  Only vectors parallel to that 0 eigenvector are in the kernel.
Comment: Most of people who answered are saying two distinct nonzero eigenvalues imply dimension of two. How do you make this leap?
Comment: 2 non-zero eigenvalues indicates 2 independent vectors that are not in the kernel.
Answer: Suppose that a matrix has rank $1$. This means there is a vector $w$ such that $f(u)$ is a multiple of $w$ for all $u\in U$.
Therefore the only eigenvectors in $V$ are the multiples of $U$. So only one non-zero eigenvalue can exist (since if $f(u)=\lambda(u))$ then $f(ku)=\lambda(ku)$.
Since in this case we have more than $1$ non-zero eigenvalue we can conclude the rank is greater than $1$.


QGIS - split polyline

Question: I need to create a shapefile with crossed streets. I can't split the lines to number each segment with the house number ranges.
Comment: You have two different questions here. For splitting lines in QGIS, see http://gis.stackexchange.com/questions/48461/ Another term for this is planarize. For the tracing part, what are you tracing from? A scanned image? Another vector layer?
Comment: Welcome to gis.stackexchange Cristian! Please note that the format of this site limits each thread to exactly one question. I've therefore removed the second question. Please open a separate thread for it.
Comment: Thanks Chris W! I´ll check the link. Then, i´ll draw the lines over the street´s map handly.
Answer: Everybody thanks a lot. Finally, I did it. Was to easy (when you know how). In QGIS 2.8.1 the tool is SPLIT LINES WITH LINES (on the V. Overlay, on Geoalgorithms, from Advanced Interface). I came there thanks to Chris W advice.
Comment: Well Maybe next time express your issue with a picture! I'm still not sure what the problem is like =/ because it's easy to solve.

QGIS - split polyline


How to get Service Provider, not the Network Name

Question: I have an Application (Operator Widget) which Displays the Name of the Network you are currently Connected to.
The Problem is that some Providers either share their Network with others (eg. T-Mobile UK uses 3's Network in some areas) or don't have an own Network and use the one of another Provider (eg. "bob" uses the Network of "A1").
The Problem is that the getNetworkOperatorName() always returns the Name of the Wireless Network (so T-Mobile Users will see 3 as their network...) which confuses Users.
Android itself on the Other hand shows the Service Providers name, what is the Function to do this?
Answer: Did you try getSimOperatorName()


Make shipping optional ubercart

Question: How to provide option for users to select whether to ship a product or not at time of chekout using ubercart in drupal ?
Answer: Make a new flat rate shipping quote called 'No shipping' with zero cost. Then in the Delivery method pane at checkout the user can choose between your shipping option and your no shipping option.
You make shipping quotes at /admin/store/settings/quotes specifically the 'Add flat rate quote' link /admin/store/settings/quotes/methods/flatrate/add
Obviously you need both the ubercart shipping quotes module and ubercart fulfilment flat rate module enabled.

Make shipping optional ubercart


Do I have to crop images in Photoshop before placing them into InDesign

Question: I have to place about 200 images into a Layout in Indesign, most images need to be cropped. Is there any disadvantage if I don't crop the images first in Photoshop? The Layout is for a book design which will be printed offset.
Comment: Hi Sylvia, Welcome to GraphicDesign.StackExchange. We hope you enjoy sharing knowledge and experience.
Comment: The printing method does not affect your preparation of the artwork, per se.
Answer: Not really. In fact, there is an advantage by not doing so which is you can reframe in InDesign and have room for playing with them.
But if your images need a "mechanical" crop, for example, all were well taken and framed correctly, but you need a square proportion, you can benefit for some automatization, like a Ps action.

In the old days where you could send the base files (Indesign) and you were limited by space, yes, the "norm" was to crop and resize before so you sent smaller files.
But normally these days you send a PDF that automatically crops and resample the images (and the bandwidth is enough where you do not really need to save space)
Answer: No, you do not need to crop in Photoshop, just link them as they are in ID and use the crop in ID. This will not affect print quality at all.
Answer: Always size an image to the right number of pixels using Photoshop. While reducing or enlarging using InDesign is possible; doing this will increase the amount of time it takes to output the job. (Time is money).
Comment: Especially if you have to do a lot of them to process.
Comment: This seems completely backwards to me. Since illustration sizes and crops are often tweaked as you work on the layout in InDesign in order for everything to match up properly, making pixel-perfect versions of each illustration in Photoshop would be significantly **more** time-intensive than just making the changes directly in InDesign. I would actually say the best advice is the exact opposite of what you give: ***always*** size and crop images directly in the layout for decrease the amount of time spent working on the job. The output time increase pales in comparison.


ASP.NET multiline textbox issue with javascript and AJAX (Jquery)

Question: In my web application I have a textbox called "Address" which is multiline and have 3 rows and 250 characters.
I am calling a javascript function in which i am calling AJAX to send the data to a webservice for some processing but AJAX is unable to process variable which contains multiline text.
I am reading multiline textbox value like this in javascript function.
var puid = document.getElementById('<%=userid.ClientID%>').value;
var paddress = document.getElementById('<%=xaddress.ClientID%>').value;

and passing like this.
$.ajax({
    type: "POST",
    url : "DataService.asmx/UpdateProfile",
    data: "{'puserid': ' " + puid +   'padd': ' " + paddress +  " '}",
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    success: OnSuccess,
    error: OnError
});

If I comment paddress parameters in ajax "data:" it works fine other wise ajax goes on "OnError" ...Oh i just figured out that my address has 'B' Area (apostrophe) in it thats why it its giving this problem. So how to parse apostrophe as textbox value and read in javascript variable and write back in database similary.
Answer: I'd use the jQuery Post function instead. Your GET request URIs shouldn't have new line chars in them.
Answer: var puid = document.getElementById('<%=userid.ClientID%>').value;
var paddress = document.getElementById('<%=xaddress.ClientID%>').value;

puid  => string value
paddress => string value

$.ajax({
  type: "POST",
  url : "DataService.asmx/UpdateProfile",
  data: {'puserid': puid  , 'padd': paddress },  need = >","
  contentType: "application/json; charset=utf-8",
  dataType: "json",
  success: OnSuccess,
  error: OnError
});

Comment: Oh i solved my problem by writing var newadd = paddress.replace(/'/g, "\\'"); :)


My sites is related to User Profiles?

Question: I need to know whether my sites is related to user profiles or not. Will we face any difficulties without using my sites? I referred this link https://technet.microsoft.com/en-us/library/ff382643.aspx for overview of My sites, but I need more Clarification about my sites.
Answer: MySites are using the UserProfile Service. You can't use MySites without UPS, however you can use the UPS without configuring MySites.
For further information the answer to this question pretty much says everything:
What happens if I don't configure mysites?
Answer: You can setup the User profile without MySites, but still you need mysite host.
MySites are the User's personal Site collection where they can store documents, create list etc.
MySite Host is the Root Level Site collection with MySite Host Template. You need mysite host in order to User profile work properly.

MYsite Host required for User profile pictures, as photo saved on the Mysite host's user profile picture library.
People search also need the MySite host.

So, You can create a site collection with MySite host template, Use that as mysite host location in your UPA's settings. You should disable self service site creation so that no user can create their mysite.

My sites is related to User Profiles?


Why changing font in Inkscape does not work?

Question: I am trying to change the font of "x" in the figure below. I selected it with "Select" (the arrow icon) and then clicked on "Create and Edit Text" (the A icon) and I think the thin dashed line around x means that it is selected. Then I expected to be able to adjust its properties in the text toolbar on top, but to no avail. I changed the font to 10 size and nothing happened. I changed it "Perpetua" again nothing. Why? As far as I can deduce from the click menu the image is not grouped nor frozen.
Comment: Sorry, but we know nothing of the structure of your drawing. The x can be a symbol, a clone or a path or something which is got from elsewhere and its "read only" in Inkscape. It can be under a multilevel group or clipping path. It can be a linked object. Show what the Objects panel says of your x . Try to select it also with the node tool to see is it a path.
Comment: Thanks! Indeed, I checked Objects panel and that x is a path. Your comment answers your question, so if you make it into an answer I will approve it.
Comment: There's the same already as an answer and as a well written one. Accept it to mark "no problem any more, case closed."
Answer: Looks like the x might not editable text, for example text which has been converted to outlines will no longer be editable as text.
You can tell by clicking and dragging on the text with the text tool. Should look like this if it's editable text, it will be highlighted.

You can also tell by simply selecting the text box using the Select Tool (S), and check the status bar at the bottom of the edit window. If it's a text object it will say so.

If you can't do that, but can select the paths using the Edit Path by Nodes Tool (N) and it looks like this below, then you have outlines, and not editable text.

All may not be lost however, if you know what the font is and already have it installed, you can just type it again, and delete the old one.  Simply choose the Text tool, click and type the letters you want. Highlight the text, and change the font to the one you want, set the size, and click and drag to reposition it.


Help to translate SQL query to Relational Algebra

Question: I'm having some difficulties with translating some queries to Relational Algebra. I've a great book about Database Design and here is a chapter about Relational Algebra but I still seem to have some trouble creating the right one:
Thoes queries I've most difficuelt with is these:
SELECT COUNT( cs.student_id ) AS counter
FROM course c, course_student cs
WHERE c.id = cs.course_id
AND c.course_name = 'Introduction to Database Design'

SELECT COUNT( cs.student_id )
FROM Course c
INNER JOIN course_student cs ON c.id = cs.course_id
WHERE c.course_name = 'Introduction to Database Design'

and
SELECT COUNT( * )
FROM student
JOIN grade ON student.f_name = "Andreas"
AND student.l_name = "Pedersen"
AND student.id = grade.student_id

I know the notation can be a bit hard to paste into HTML forum, but maybe just use some common name or the Greek name.
Thanks in advance
Mestika
Answer: "and here is a chapter about Relational Algebra"
Where ???  This doesn't seem to point to anything.
At any rate, the examples you give are examples of what the literature-with-an-algebraic-perspective usually calls "aggregations", or "summaries", or some such.
In contrast with the "basic" operators such as JOIN, PROJECT, etc., the consensus on how to deal with such "aggregate operators" is relatively small.  Keep in mind that there is no such thing as "the" relational algebra, and that different implementations are completely free to choose which set of algebraic operators they make available to their users !!!
Comment: I think "and here is a" should really be "and there is a" (as "in the book"). I cannot help much otherwise, I fear :)


What is the Chomsky level of pattern matching?

Question: Pattern matching in programming languages as for example in Racket is some kind of parsing. Which level in the Chomsky hierarchy does this parsing technique have? Is it type-3 like regular expressions or type-2 or even lower?
Answer: It depends on what types of patterns are supported.  If it is only classical regular expressions, as in theoretical computer science, it is the regular languages.  If it is Perl-compatible regular expressions (PCRE), then it is the context-sensitive languages; see Which languages do Perl-compatible regular expressions recognize?.
See also When is a regexp not a Regular Expression? and Expressiveness of modern regular expressions and How much more powerful are regexes in modern programming languages compared to regular expressions from the theory of computation? and What can be proven regarding the differences in power between unary ECMAScript regex functions and primitive recursive functions?.


Are ICAO and FAA airport signs and markings the same?

Question: Are ICAO and FAA airport signs and markings the same or are there differences? If so what are the differences?
Comment: [Here are the FAA sign standards](https://www.faa.gov/airports/runway_safety/news/publications/media/QuickReferenceGuideProof8.pdf)
Comment: Reference [this Jeppesen document](http://ww1.jeppesen.com/documents/aviation/business/ifr-paper-services/airport-signs.pdf)
Comment: This question seems very broad. The full list of airport signs and markings is extensive; if the answer is *NO*—they are not the same—then an exhaustive comparison would likely be too broad for this exchange. I am not voting to close, as yet, but I think you would do well to narrow your question's scope, for example to certain signage, or to runway markings.
Comment: Thank you for your comment. I was hoping to find something similar to this but for ICAO ... https://www.faa.gov/airports/runway_safety/news/publications/media/QuickReferenceGuideProof8.pdf
Answer: Principle
ICAO Airport markings are defined by ICAO in Annex 14 of Chicago Convention, Aerodromes.
ICAO also requires at section 5.8 Notification of differences in the AIP of Doc 8126 Aeronautical Information Services Manual, that differences are explained in the country AIP, Section GEN 1.7 Differences from ICAO Standards, Recommended Practices and Procedures.
US regulations are not entirely compliant with ICAO SARP, and differences exist for airport marking.
Example of ICAO SARP
For the touchdown zone lights, ICAO states in Annex 14, at paragraph 126.96.36.199:

The lateral  spacing between the innermost lights of a pair of
  barrettes shall be equal to the lateral spacing selected for the
  touchdown zone  marking. The longitudinal spacing between pairs of
  barrettes shall be either 30 m or 60 m.

US example
As US regulations are not fully compliant, the differences are explained in GEN 1.7 of the US AIP, page 129:

U.S. does not require the lateral spacing of touchdown zone lights
  to be equal to that of touchdown zone marking when runways are less
  that 45 meters wide.

In general
All differences can be found at this location, sorted by ICAO reference and chapter. Differences related to Annex 14 start on page 124 (actually labeled GEN 1.7 81). Listing all of them (15 pages in the AIP) would be too broad for this answer.


How can I get After Effects to recognize newly saved changes in an imported Illustrator file?

Question: I've updated the imported Illustrator file that I am using in After Effects with some new layers and other stuff, but the changes are not seen in my compositions. How can I get my compositions to recognize the changes?
Answer: Right click the Illustrator file in the Project panel and select Reload Footage.

If you imported your Illustrator file as Footage, then that will work with no problem; you'll see your new layers and all changes etc.
Unfortunately if you imported your Illustrator file as a Composition (meaning you have a separate project resource for each layer as in the above screenshot) reloading those layers won't import your new layers. Also if you imported with Retain Layer Sizes, your layers within the composition won't update they're position, only the content.
The only real solution to that is to reimport as a new composition and copy over the new layers to the original composition.
Answer: I had similar problems updating an Illustrator file and it wasn't updating. After trying the methods above, I then deleted the source file in the Project window. The warning box told me that the element was being used in (x) compositions. I deleted anyway, then immediately performed 'Undo'. Like magic, the file finally updated correctly. Note that this was a layer in an Illustrator file that I had previously imported as a composition along with the other layers in the same file.
Comment: "Like magic, the file finally updated correctly. Note that this was a layer in an Illustrator file that I had previously imported as a composition along with the other layers in the same file."
- Dang; just tried it and didn't work for me. Guess I'm lacking the magical touch...
Answer: I found that if you just rename the files to the original name and put the files in the original location it will recognise the files and it will use the files in After effects
Comment: You should just use Reload Footage, check the other answer. If you rename files you have to replace them in the project, it's just more work.
Answer: Make sure that the illustrator file saved correctly. If it's in a new location or under a new name, After Effects will just keep reloading footage using the old file, not knowing a new version exists
Answer: I am still new to AE but I have found that you can drag+drop layers from multiple folders in the projects window into a new folder AND you can drag+drop layers  from multiple project folders into an open composition and just simple pick-whip new layers into existing null objects that contain old animations.  The composition not so much "link" with the layers folders themselves but directly with the layers WITHIN the folders.
However, AE also creates invisible links between an open project  and .ai files...so if you rename, move or delete .ai files, this breaks links and there seems to be no way to simply replace layers that are compositions and not "footage".  Quite a headache.
Comment: Welcome to GD.SE!


Custom user store manager class not showing

Question: I am new to doing wso2 development. We are running version 5.3 and I am trying to experiment with writing a custom secondary user store. Our primary one is an LDAP server and we would like to configure our system to also use a, database backed, a secondary custom user store.
to get this project off the ground I have been following the instructions on
https://docs.wso2.com/display/IS530/Writing+a+Custom+User+Store+Manager. When placing the jar in the dropins directory my custom store manager does not appearing on the list of user stores.
After a bit more digging, I found this article Custom user store manager class not showing in user store drop down. This appeared to address my exact issue. However after adding the CustomUserStoreManagerServiceComponent class, I am still unable to get the plugin to appear in the drop down.
The source for this project is here https://github.com/terellda/src-questions. I know the project is incomplete but I am trying to get this first issue addressed so that I can continue.
Answer: This is a Custom User Store manager. Use this as a Sample.
https://github.com/Manukam/wso2-custom-user-store


Custom field populated at form handler submission time even if field is not present on the form

Question: We have a custom field in Pardot that is synced with a field in Salesforce. The setting  Keep this field's type and possible values (for dropdowns, radio buttons, checkboxes) in sync with the CRM is enabled. In Salesforce there is a default value for this field.
When a prospect submits a certain form handler, I can see in the Audits section that this field is filled with the default value, even though this field is not present in the form handler (nor in the form on the website). Moreover, the time this field is populated is the same the form handler is submitted and the source is the same - Web Tracking and not CRM, as I would expect if it was filled because of the sync with the CRM.
Any clue on why this happens?
Answer: 2 things to consider:

Is the visitor's browser hitting the form handler, or are you POSTing to the form handler behind the scenes (server-side submit).   If you are doing a direct POST, the form handler won't have any reference to the visitor's pardot visitor ID and can't determine the source campaign. If this is the case, you might want to look at using the Pardot API to "assign" a visitor reference to a prospect to see if you get your desired results. https://developer.pardot.com/kb/api-version-3/visitors/
When the prospect is synced to Salesforce, and Salesforce doesn't get a value for a field with a default value, it will set the value and then possibly cause another sync back to Pardot, making it appear as if the value "came from nowhere".


Evaluate $\int_0 ^x \frac{\sin\left(\frac{1}{2}-n\right) t}{\sin\frac{t}{2}}\,dt$.

Question: Evaluate
$$\int_0 ^x \frac{\sin\left(\frac{1}{2}-n\right) t}{\sin\frac{t}{2}}\,dt$$
for $x\in(0,2\pi)$ and $n\in\mathbb Z$. From trigonometric formulas, we have:
$$\int_0 ^x\cos nt\, dt-\int_0 ^x \frac{\cos\frac{t}{2}}{\sin\frac{t}{2}}\, \sin nt\,dt$$
What can we say for the second integral? I was not able to solve it.
Answer: Only a partial answer for $n\in\mathbb{N}$, but maybe helpful nonetheless. Call $f_n(x)=\int_0^x dt \sin(nt)\cot(t/2)$ and consider the generating function
$$
F(z,x)=\sum_{n=0}^\infty f_n(x)z^n=\int_0^x dt\cot(t/2)\frac{\mathrm{i} \left(-1+e^{2 \mathrm{i} t}\right) z}{2 \left(-z+e^{\mathrm{i} t}\right) \left(-1+e^{\mathrm{i} t} z\right)}=
$$
$$
=-\frac{z}{2}\int_0^x dt\ \frac{\left(1+e^{\mathrm{i} t}\right)^2 }{\left(-z+e^{\mathrm{i} t}\right) \left(-1+e^{\mathrm{i} t} z\right)}\ ,
$$
using the Euler's representation of the cotangent. The real part of it can be computed with the help of Mathematica, yielding eventually (for $0<x<2\pi$ and $|z|<1$)
$$
F(z,x)=-\frac{(x+\pi ) z+2 (z+1) \arctan\left(\frac{(z-1) \cot \left(\frac{x}{2}\right)}{z+1}\right)-x+\pi }{2 (z-1)}\ .
$$
[This agrees with numerics]. The generating function can be represented as
$$
F(z,x)=\varphi\left(\frac{z-1}{z+1};x\right)+\frac{x}{z-1}\ ,
$$
where
$$
\varphi(\xi,x)=-\frac{1}{\xi}\arctan(\xi\cot(x/2))-\frac{x+\pi}{2\xi}\ .
$$
With any symbolic software, you can obtain a closed form solution for any $n$ by simply Taylor-expanding this result around $z=0$. For example:
$$
f_3(x)=x+2 \sin (x)+\sin (2 x)+\frac{1}{3} \sin (3 x)\ ,
$$
and based on some experiments, one might conjecture a general form as
$$
f_n(x)=x+\sum_{k=1}^n \frac{n_k}{m_k}\sin(k x)\ ,
$$
with $n_k$ and $m_k$ small integers. A closed-closed form solution might be obtained after some painstaking work on the generating function (using also the great simplification that $\arctan(\cot(x/2))=(\pi-x)/2$). But that's not for me tonight.
Answer: It is sufficient to  consider an integral of the form $$\int \frac{\sin \left(\left(2k+1\right)t\right)}{\sin t}dt.$$ One has $$\frac {\sin\left(\left(2k+1\right)t\right)}{\sin t}=\frac{e^{i\left(2k+1\right)t}-e^{-i\left(2k+1\right)t}}{e^{it}-e^{-it}}=2\left[\cos 2kt +\cos \overline{2(k-1)}t + \cos \overline{2(k-2)}t + \cdot\cdot\cdot+\cos 2t \right]+1.$$ Each of the terms are easily integrated, but can't think of a further simplification.


What are the advantages of going through the practice league?

Question: What are the advantages of going through the practice league?  Aside from the practice and that battle.net can find better matches?
Answer: The practice league is usefull if you never played SC2 multiplayer. It uses Novice maps where the entrances to both main bases are blocked by destructible rocks which prevent you from being rushed and shuts down many cheesy builds like cannon rush. Through this you have some more time to build up and get used to your buildings and units until you engage with your opponent.
Although when you get out of the practice league you maybe will have to readjust your play since there no longer is any early game protection.
IMO its better go do some custom games vs AI or people you know to get used to multiplayer and the maps.
Comment: Also practice-leage runs on "Normal" game-speed if I recall correctly. So the game runs a bit slower. Good for getting started, but it's a noticeable change when the game suddenly runs on "Faster" in ladder.
Answer: if you are aiming to get all the achievements then its a good time to get allot of the LEAGUE COMBAT achievements, other than that only thing i can think of is learning the different buildings, but you can do that vs AI anyway


Convert Data from fox pro table to sql server with c#

Question: I want write an win-form application to convert data from foxpro table into sql server.when i convert or transfer data in foxpro (with foxpro) my data is transfered very speed.but when i try to convert into sql server ,data is transfered very slow.how can i solve this problem.what technologies can i use? what tips can i use? please help me
Answer: You may want to look into SSIS.
In particular, this thread may help.
Answer: Did you know that there is a great tool that will mass transfer data from FoxPro straight into SQL Server. If the tables are not already in a DBC, you'll need to make a copy of the entire data set to a new folder, then add those tables to a DBC, then run this SQL Server Upsizing Wizard. Works with SQL Server 2005 and 2008.
http://vfpx.codeplex.com/wikipage?title=Sedna#SQLServerUpsizing


Does wire diameter affect the electricity bill?

Question: Does thickness of wire affect the electricity bill, for same load? Thicker Wire has lower resistive losses ,does this reduce power bill? Does a longer length of the same diameter wire increase bill by the same reasoning?
Comment: No chance you will see the difference in the bill. Just be careful, thin wires tend to burn under load.
Comment: In some places, you get charged a fixed distribution fee based on the size (rating) of your main breaker. In that sense, you'd actually pay _more_ for a "thicker wire" coming into your house. Other than that, the difference in your bill will be absolutely negligible.
Answer: You are REQUIRED by code to use specific wire sizes for specific loads and in no case can you use a wire that is rated for less than 125% of a continuous load. So if you follow the law, your wire heating is minimal. If you increase the wire size to be larger than necessary, it will technically reduce the losses, but at a much lower rate of increase in benefit so there is almost never a return on that investment. The energy savings may be only a few dollars per year, whereas the cost of the larger wire for an entire house may result in hundreds of dollars in increased cost. You would not likely live long enough for the energy savings to pay for itself.
Answer: "Power consumption drops when line resistance increases" is true when you have a resistive load, but not true when you have an active (regulating) load.  And it is not interesting to do so.
Let's assume that you have an apparatus that needs 100W to work.  For example your fridge that requires 100W to cool down to the set temperature level, or your computer that uses and adaptor and consumes 100W to do its work.
Suppose that your fridge or adaptor can run at 100V and also at 10V.  So at 100V the current would be 1A, and at 10V it would be 10A.
If the line voltage is 100V, then at 0Ohm of line resistance, the apparatus gets 100V and all delivered power is used by the apparatus.
If the line resistance is changed to 9 Ohms then there would be a 90V loss in the lines at 10A and the apparatus would get 10V and consume the 100W it needs.  However the loss in the lines would be 900W - 9 times the usefull power.
In practice you will not have such a big difference, but it will always work the same way.
For a constant "usefull" power, the lower the line resistance, the lower the consumption and the lower the bill.
If you accept the principle of getting less usefull power, the bill will be lower with higher line resistance, but you will also get a lower amount of light from light bulbs and the cost per useful Wh will be higher.
With (partially) inductive and capacitive loads, the computation is more complex but the line resistance will consume a lot with respect to the power actually consumed in the load.  This is why there is a lower limit on what is called the "Power factor", but that is another (related) story.
Answer: Power consumption drops when line resistance increases - it's a case of more resistance in series with the load takes less current and hence power is less. However, if you are boiling a kettle it will take longer to boil so, you maybe be taking less power but you are taking it for a longer time and so the power loss in the cable (due to \$I^2R\$) will add a bit on your bill.
Remember you are not billed on power but power x time i.e. energy.
Answer: Theoretically the thinner the wire and longer it is, the higher it's resistance and the overall load seen from the outside (power plant?). Some of the energy entering your house will just dissipate as heat on the wires and will not produce any useful work, while you are still billed for it. But practically it should be unnoticeable on your bill.
Comment: Higher resistance == higher power consumption??
Comment: Sorry, it's inaccurate indeed. Will try to rephrase.
Answer: Thinner wire will have higher resistance and will reduce current and voltage to the load. This will reduce power consumed but a higher percentage of that will be lost in the wires.

simulate this circuit – Schematic created using CircuitLab
Figure 1 (a) Low-resistance wire and (b) high resistance wire.
Figure 1 shows a 10 Ω load fed from 100 V supply. With zero wire resistance \$ I = \frac {V}{R} = \frac {100}{10} = 10~A \$. Power is given by \$ P = I^2R = 10^2 \cdot 10 = 1000~W \$.
(a) has 0.2Ω series resistance. \$ I = \frac {V}{R} = \frac {100}{10.2} = 9.8~A \$. Power in load is given by \$ P = I^2R = 9.8^2 \cdot 10 = 961~W \$. Power lost in wire is given by \$ P = I^2R = 9.8^2 \cdot 0.2 = 19.2~W \$. Overall, we've saved a little on our bill and lost 20 W to the wire.
(b) has 2Ω series resistance. \$ I = \frac {V}{R} = \frac {100}{12} = 8.33~A \$. Power in load is given by \$ P = I^2R = 8.3^2 \cdot 10 = 694~W \$. Power lost in wire is given by \$ P = I^2R = 8.3^2 \cdot 2 = 132~W \$. Overall, we've reduced our bill by 174 W (which means that the equipment won't run as intended) and lost 132 W to the wire.
<code>Summary table
R        Load power    Loss in wire    Total (billed)
0 Ω      1000 W          0 W           1000 W
0.2 Ω     961 W         19 W            982 W
2.0 Ω     694 W        132 W            826 W
</code>

For AC (Alternative Current)?

DC or AC. Same rules apply.
Comment: That assumes resistive loads.  Many loads today are constant-power loads. (PCs, TVs, anything running off of a switching supply.)  Probably  lighting and other resistive type loads will dominate, but not as much as they did years ago.


With what speed does a thermal front travel through a solid?

Question: The heat equation provides theoretical temperature distributions as a function of space and time. Take for example an infinitely long rod that is suddenly heated up at $t = 0$ at its end $x = 0$. The heat equation predicts that the thermal front (defined as the surface where the temperature changes from its initial temperature to a different temperature) moves away from $x = 0$ at a very high (almost infinite or infinite) speed. Does this match experimental observations, or is there a finite limit of this speed and if so, what is it? I imagine doing experiments near zero Kelvin to minimize "thermal noise".
Comment: Good question. Presumably it would move at about the speed of sound in the solid.
Comment: John Rennie is right. May be worth reading about "second sound" in superfluid liquid helium where great flows like a coherent wave, not diffusively. There the speed is the fastest in any terrestrial object and it is limited by the superfluid critical velocity. Most objects with conduct heat much more slowly, but this may give the right idea about the speed limits.
Comment: @KFGauss: Interesting. Second sounds also appears as a parameter in the "hyperbolic heat conduction" equation, which was proposed to upgrade the old parabolic heat equation, the latter being incompatible with relativity theory. Source: https://en.wikipedia.org/wiki/Relativistic_heat_conduction
Answer: I'm not aware of experimental results that answer your question, but I'll take a shot from a theoretical perspective.
Considering pure conduction, thermal energy is transferred through a material via a series of collisions between adjacent molecules. The finite limit of the speed at which these collisions could propagate through any material would be the speed of sound in that material, as that is the limit by which pressure waves (in this case, vibrations of thermally excited molecules) can travel.
However, the speed of sound is the speed at which the individual collisions/vibrations travel. Since this occurs in random directions, unlike a directed sound wave, a discrete heat front will travel slower than the speed of sound. If you define the thermal front as any tiny deviation from initial temperature, then the question simplifies from thermal front velocity to velocity of a vibration. In that case, it should move at the speed of sound, since a random vibration will propagate in every direction at sound speed and cause some infinitesimal temperature change as it goes through. However, if you define the thermal front by a specific temperature change, then you can easily calculate the propagation speed with the heat equation.
Answer: Short version:  the speed that the heat front moves will not be infinite, it will be equal to or less than the speed of light in the material. It will usually be much less than that.  The temperature of the heat front or just behind the front itself will determine roughly what regime of speed it will move at.  Pressure, density, composition of the material will also be important.
For your specific example of a rod, let's say it is at 20 K ("weird things" can happen at 0 K so we'll stay a bit above it), and the heat front is at 300 K.  Also assuming it is "normal" matter, not exotic, and it is not a metal.  In this temperature range, the thermal excitations in the heat front are the vibrations of atoms around their positions, and the speed of thermal energy transfer is the speed of sound on the solid.
For another example let's say the heat front is at 3000 K   The thermal excitations in the heat front are now vibrations of the atoms and motion of the electrons.  The energy of the latter will propagate through the material at the speed of light in the material.  Then there will be disequilibrium between the election temperature and the ion temperature which will resolve/propagate through the material.
To learn more you can search for "solid state physics heat conduction"


Is theodicy considered an open problem or a solved one, according to catholicism?

Question: Do catholics consider the problem of theodicy (Why God allows evil on this world) an open question, or is it "solved", seeing that so many people have tried to answer it. If it is not solved, what is the reason?
Comment: [Salvifici Doloris](http://www.vatican.va/holy_father/john_paul_ii/apost_letters/documents/hf_jp-ii_apl_11021984_salvifici-doloris_en.html) says it all!
Answer: Yes, it is an open question.
Yes, it is a "solved" question.
No, it is not an OR question.
The question you are asking "Is Theodicy an "open" question or a solved one?" is based on a faulty premise.  Namely, "open" does not mean "unanswered" or "unsolved" - it means one that is continually being investigated.  Thus, the opposite of an "open" problem isn't a "solved" one, but rather an unasked one.
At best, one can ask, "Is there a consensus position?" and the answer is "within a fairly broad spectrum, yes."  Yes, there are several "solutions" that theologians have posited that each have a significant consensus behind them.  Go find them - that's a book.
In order to highlight the faulty premise, let's use another question as an example: "Did God create the world ex nihilo or did it evolve?" As contentious as the question is, no one can say there are no "solutions" to it. Non-theists have posed several answers, Theists have posed several answers, and there are even significant groups that seek a "theistic evolution" answer in the middle.  Is it an open question? Sure! People wrestle with the origins question everyday. And is it a solved one? For many people, again, the answer is "Yes!"  They have weighed the evidence and personally picked the "solution" they feel is most compelling.
In theology, there are very, very few "closed" questions. John Dominic Crosson believes the literalness of the Resurrection to be an "open question," despite Paul's very obvious answer to "If there is no resurrection, then we are to be pitied above all men!"  The closest thing I can think of as a "closed" question is the literal existence of a King David. For many years, theologians thought it an "open question" as to whether or not David existed.  When archeologists found a steele mentioning the House of David, the evidence pretty thoroughly closed any further debate, because it is really hard to argue with tangible evidence to the contrary - and yet there are still some who claim it is "open"
Back to the "problem" of theodicy - or "Why does a Good God allow evil?"  There are several solutions to the problem - but that does not mean it is unasked.  It is thus open and solved at the same time.  And frankly, until there is no more evil, or until people see God as not good, it will always be asked. It will and must be open to new queries as long as man asks questions.
Answer: I'd like to post more or less a paraphrased pseudo-lazymans synthesis of Saint John Paul's Letter Salvici Doloris, which got me through a secular university's class on Job swimmingly.
First of all, the answer to the problem of evil is tied with the meaning of suffering.  If it weren't, it wouldn't matter whether it was evil or not, I think so Salvific Suffering (as the pope put it) is pretty much the answer.  But there is a lot more in the letter.
Pope John Paul II pointed to St. Paul's message that his suffering makes up what is lacking in Christ's to show that all suffering points towards salvation.  And Job, prefiguring Jesus, as well as the Suffering servant in Isaiah, who do nothing wrong and yet are tested and broken show that suffering is not merely retribution.  Jesus even says its sometimes rather arbitrary when talking about the tower that fell on some guys.  But, even of it arbitrary, its not outside God's will. It's just God's passive will in action.  The things he permits to happen by removing his hand (see Job 1 or 2) rather than my sticking his hand in the pot (see the 2nd to last Chapter of Job).
St. John Paul says the answer belongs to the class of problems tied up with the mystery of the incarnation (for God so loved the world...) and that the answer to it lies in Gods love rather than his wrath.  Because, even though stuff is tough, we are always under God's close and personal gaze, we are not in Hell, and because of it we should expect that any grace given, any supernatural truth that manifests itself in our lives, should be seen as a direct sign from God.
http://www.vatican.va/holy_father/john_paul_ii/apost_letters/documents/hf_jp-ii_apl_11021984_salvifici-doloris_en.html
Answer: It can never be fully solved because God is in the equation, and that is why it is also referred to as the mystery of evil.
This article, Evil | New Advent, states the problem as,

To what, then, is the evil of human life, physical and moral, to be
  attributed as its cause? But when the universe is considered as the
  work of an all-benevolent and all-powerful Creator, a fresh element is
  added to the problem. If God is all-benevolent, why did He cause or
  permit suffering? If He is all-Powerful, He can be under no necessity
  of creating or permitting it; and on the other hand, if He is under
  any such necessity, He cannot be all-powerful. Again, if God is
  absolutely good, and also omnipotent, how can He permit the existence
  of moral evil? We have to enquire, that is to say, how evil has come
  to exist, and what is its special relation to the Creator of the
  universe.

Herein is the difficulty, evil in the presence of an all-good, all-benevolent, all-knowing and all-powerful God, and from scripture, who himself hates evil.
The article continues

The solution of the problem has been attempted by three different
  methods.

Therefore answering the question, it is clear that the problem hasn't been fully solved though three solution have been proposed to attempt to solve it, and perhaps, the problem will never be fully solved at least this side of heaven.


How Do I Launch The SDK Emulator?

Question: How do I use the Android emulator? I made an AVD device in Eclipse, now im trying to launch it via command prompt. Android is the AVD name. I have also installed all the files for Android 4.2.2 in SDK manager. This is what I type into cmd: emulator -avd <android> -gpu on. It spits out, system cannot find the file specified.
Comment: By the way, this question is only marginally on topic for this site. This site is intended for questions about using Android devices. Questions about developing for Android belong on Stack Overflow. The emulator is a bit of a gray area, since it's usually used by developers, but non-developers could have some use for it too.
Answer: It sounds like the directory that emulator resides in isn't in your PATH. The path isn't updated by the installation of the SDK.
Answer: You shouldn't have <angle brackets> around the filename. They're used in examples to show you that you need to change that part of the command (to the AVD name you chose), and you shouldn't type them in.
Comment: Also a good possibility for causing the stated problem. I'm so used to that convention I didn't even consider this answer.
Comment: To be fair, I didn't think of your answer either until I read it :-)
Answer: You need to install AVD Manager inorder to run emulator...check for those and try running using the menu Tools|Android|AVD Manager after installing the necessary files and starting them...Wish u have success

How Do I Launch The SDK Emulator?


How to find the adware that shows pop up ads on my Android phone?

Question: If I have internet access, sporadically I see pop-up ads like the one below. I am trying to find the Android app causing them, but the recent button is not showing any app. Any tool I can use to find and remove the adware? I installed Kaspersky and it could not find the adware.
Comment: maybe check `/proc` directory
Answer: Two possibilities:

A background process (not an app, but started by an app) that is triggering ads. In this case you may not see the app itself in recent / running apps.
More likely, your ISP is triggering the ads. Airtel is known to have done this on and off (I've experienced it first hand), targeting specific circles and specific browsers. In my case, switching browsers solved it for sometime. If the ads are popping up when you're not browsing, you're situation is probably a bit different.
Comment: @defalt No, Android permissions model allows any app / process to draw in foreground without administrator / device administration rights.
This particular issue - of inserting ads that appear to be a part of a webpage is a more insiduous one. The ISP modifies the response (e.g., the html file being served) from the webserver without the permission of either party (the web server and the client browser).
Aside: There are some legitimate applications of this situation - e.g., CloudFlare where the webserver permits this to enhance its security.
Comment: Does an app require administrator permission to randomly display ads anywhere on the screen?
Answer: The majority of adware on the phone is designed to work around any administrative privileges. It does that by either, not needing admin rights to show certain information on the screen or attaching itself to an existing app that has permissions. If it is sophisticated or just merely annoying depends on the app.
My first advice would be to check your currently running processes and see if there is anything suspicious. If there is, attempt to kill the process and look if it might be tied to another app. After killing or uninstalling the apps, restart your phone and check if the problem persists. If it does, try to install a different anti-virus solution or see if now Kaspersky detects the issue.
If the problem persists, you can try tracking the app files or the process files that launch the said adware. I use ES Explorer for file viewing on the phone, but other alternatives should work just as good. (Unfortunately, if your phone is not rooted, there is a low chance that the files will be in a standard .apk format). Most likely the install might be named as a rogue an innocent app, like a game or utility app. See if you don't recognize something and try removing the files manually.
Lastly, if that doesn't help, you could try considering a factory reset, it is quite annoying and unfortunate, but it usually solves the issues.
Good luck! Hope this helps.


Double is to triple as halve is to ...?

Question: "Divide by three" is the only way I know to say this but I'm looking for a single word, if one exists.
Comment: Double is to triple as halve is to three fourths. Hah! I _rock_ the SATs
Comment: But a half is obviously singular, as it's one or the other half in question. With thirds, you could have one or two thirds, hence having to generally speak about a third or one third.
Comment: *[Is there a verb that fits in the pattern: quarter, third, halve (divide), ???, double, triple, quadruple. . .?](http://english.stackexchange.com/questions/80136/)* seems to imply it should be *third*, but I've never heard that verb usage.
Comment: Double is to triple as halve is to "cut into thirds."
Comment: You don't usually hear *third* used by itself like *half*. But I regularly hear *a third* or *one third*. For instance "I'll take a third of that za." That violates the single word restriction though.
Comment: [Related](https://english.stackexchange.com/q/613786/2085).
Answer: The OED defines third, verb, as such:

To divide (anything) into three equal parts; to reduce to one third of the number or bulk.

and it gives an example from 1874:

Such a course would have halved or thirded the number of our subscribers.

However, as choster pointed out in a comment, this seems to be rare.
I did an ngrams search which shows the relative changes in halved, thirded over time, and the usage of halved has increased while the usage of thirded has decreased. Similarly, the usage of thrice is also decreasing over time. This suggests to me that you should avoid using the word "thirded" despite it being a recognized usage.
Another answer provides the word "trisect". This word doesn't fit precisely in the collection "quartered, X, halved", but rather goes with "bisect, trisect, quadrisect". These words seem to me to be more technical in nature, used for geometry (e.g. Trisecting the Angle) or other mathematical or scientific contexts. Also, trisect can only be used for the sense "to cut into three equal parts" and not in the sense "to reduce to one third the size".
Interestingly halved is more popular than bisected but tristected is more popular than thirded.
Tl;dr:

If you want to fill the "quartered, halved, etc" group, use thirded.
If you want language people actually use, just say "cut into thirds" or "divided into three" or something similar.
If you need a current, single word, trisected is probably your best bet but it doesn't really fit the pattern nor satisfy all the meanings.
Comment: I upvoted all three answers I see, but this is the best one. I upvote it harder.
Comment: "What man/ Thirds his own worth?" *Two Noble Kinsmen, I, ii.*
Comment: @DCShannon - Exactly the problem I always get. I want to be able to vote for "this answer is useful", and have a separate vote, one per question, which says "this answer is best". But only the OP gets those. :( No fair!
Comment: thirded makeths me thpbfeel like Daffy Duck
Comment: Interestingly, the *present tense* version of "third" as a verb [has no results at all](https://books.google.com/ngrams/graph?content=third_*&year_start=1800&year_end=2000&corpus=15&smoothing=3&share=&direct_url=t2%3B%2Cthird_*%3B%2Cc0%3B%2Cs0%3B%3Bthird_ADJ%3B%2Cc0%3B%3Bthird_NOUN%3B%2Cc0%3B%3Bthird_ADV%3B%2Cc0%3B%3Bthird_NUM%3B%2Cc0).  I have no idea what to make of this.
Answer: Halve means two things:

to divide (something) into two equal parts: to divide (something) into halves
to reduce (something) to one half of the original amount or size

For the first, trisect is a good option, but in my experience the second meaning is more common, as in The budget for our project was halved.
You cannot equally say: *The budget for our project was trisected, as that sounds as if it was split into three, and shared among three parties.
Your best bet for the second meaning is probably: The budget for our project was reduced to a third.
(Actually, trisect already sounds a bit mathematical and formal; it doesn't sound like something you'd say in casual conversation. Split in three is better there.)
Comment: So, would you say that using "trisect" would be [cutting hair in four](https://en.wiktionary.org/wiki/couper_les_cheveux_en_quatre)?
Answer: I don't think there is a long-standing English word for dividing into three parts. Also, it looks like halve is from a word that originally  meant divide without carrying the specific meaning of divide in two:
https://en.wiktionary.org/wiki/half
https://en.wiktionary.org/wiki/Appendix:Proto-Germanic/halbaz
I would also note that neither halve nor half appears to bear any relationship to words for two or three. By contrast twice and thrice are words for two and three with endings that signify use as an adverb.

Double is to triple as halve is to ...?


timer library on Arduino DUE

Question: I'm trying to call an interrupt routine (let's call it handler() ) every second on my arduino DUE.
I thought the simplest way to do this was to use this library
but it seems that i don't know how to use it...

Comment: I guess you mean `start(1000000)`.
Comment: Serial print in an ISR is not recommended. Set a flag in the ISR and poll in the loop() instead.
Comment: yep............... this is embarassing ahahah

Do you have a guess on why wasn't it working with start(1000) ? is Serial.print too long to execute ?
Answer: You are trying to print 6 characters ("test\r\n"), i.e. 60 bits
including the start and stop bits, 1000 times per second. This requires
a baud rate of at least 60 kbit/s, yet the serial port has been
configured for only 9.6 kbit/s. Something has to give...
As Mikael Patel wrote in a comment, it is generally inadvisable to
Serial.print() from within an ISR. It is supposed to work, though, as
long as you don't fill the Serial output buffer. The problem is that
your code does fill the buffer, as it is writing to it faster than the
UART can empty it. Once the buffer is full, the program will
malfunction.
Comment: thank you ! I will definitly use a flag now

timer library on Arduino DUE


simple script to apply bootstrap pagination style in asp.net gridview

Question: is there any simple jquery script/plugin to apply bootstrap pagination style in asp.net gridview ? I've found some good tips about how to do this, like these links: here and here.
the only problem with these tips/solutions is we need to make a lot of changes to achieve the result and this is not preferable when you have large application and you want to transform it to bootstrap style. we need another solution. like a simple jquery script that can do the job without making lot changes to the current code.
Answer: I've made simple jquery script to apply the bootstrap pagination in asp.net gridview and I think it will be useful to share it here in stackoverflow.
source code of this script is hosted in github here.
usage is very simple:
-include the plugin js file in your asp.net page file:

-set gridview property:
PagerStyle-CssClass="bs-pagination"

that's is all you need to apply bootstrap pagination style in asp.net gridview.
check my blog for more info.
Edit:
about the problem when using gridview inside UpdatePanel, the reason of this problem is because "UpdatePanel completely replaces the contents of the update panel on an update. This means that those events we subscribed to are no longer subscribed because there are new elements in that update panel."
There is more than one solution to solve this problem:
Solution 1:
Use pageLoad() instead of $(document).ready. Modify the code like this:
function pageLoad() {
            $('.bs-pagination td table').each(function (index, obj) {
                convertToPagination(obj)
            });
        }

Solution2:
re-change the style after every update. We can do this by adding these lines to the bs.pagination.js file:
var prm = Sys.WebForms.PageRequestManager.getInstance();

prm.add_endRequest(function () {
    $('.bs-pagination td table').each(function (index, obj) {
        convertToPagination(obj)
    });
});

Comment: Thanks Issam Ali, the only problem I found is that after a postback the paging style is lost, even when I'm changing the property CssClass of the GridView on every page load..what am I missing?
Comment: yes that is correct, Gridview is inside an update panel. On my original comment I meant to say that when I do change to another page is when the style is lost..
Comment: @CesarDuran I've added two solutions for this problem, please re-check my answer.
Comment: I have added this solution to my blog too for others to make use of it. Thanks for the plugin @Issam
Comment: This only works for Bootstrap 3.x. For Bootstrap 4.x, you need to add more classes and structure it differently (see https://getbootstrap.com/docs/4.5/components/pagination/). How would you modify your bs.pagination.js to do this?


How do US Senate Elections Work?

Question: Do all the votes in a precinct go to the majority candidate (eg winner take all) and then the tallies added up by precinct? Or do they take the simple majority in the state?
Answer: The 17th Amendment to the United States Constitution established the popular election of United States Senators:

The Senate of the United States shall be composed of two Senators from each state, elected by the people thereof, for six years; and each Senator shall have one vote. The electors in each state shall have the qualifications requisite for electors of the most numerous branch of the state legislatures.

As the US Senate website explains:

The Seventeenth Amendment restates the first paragraph of Article I, section 3 of the Constitution and provides for the election of senators by replacing the phrase "chosen by the Legislature thereof" with "elected by the people thereof."

The candidate that receives the most votes in the state will win the Senate election.
Comment: Some states require a majority and have a second round vote of the top two vote getters in a runoff election if no candidates receives a majority, others require a mere plurality in the first round vote. The choice between these two rules is a matter of state law (other rules like instant runoff could also be adopted but have not been adopted).
Comment: Yep.  It is this amendment that tilted the balance of power between the States and Federal Government to the Federal Government.
Answer: It is a statewide tally of popular votes where the candidate with the plurality or majority of votes wins.

How do US Senate Elections Work?


Setting solid black in Photoshop

Question: I am trying to set a solid ( non 4 colour ) black in Photoshop. I am able to do this successfully; however if for instance I create a shape or a piece of text using my 100%K black all is well until I go back in to check the fill colour and then into the colour picker ... and without touching anything in there the colour has changed back into a 4 colour black. I cant see any way to make it stay as 0 0 0 100. I have been using this video as a reference and I can see this is is definitely possible, but I can't see what I am doing wrong. Is there some setting I'm unaware of. I'm using CS6.
Comment: First of all, are you use CMYK or RGB color space in your document?
Comment: @jamesy You're correct, but it would be more helpful all round if you'd say this in an answer so it can be voted up and accepted. :)
Comment: I think @Vnovak is right, and what your video doesn't show you — your document needs to be in CMYK. To validate this, in Photoshop for your document go to Image > Mode > CMYK Color

Now select the piece of text and check you've got solid black (0,0,0,100).
Comment: In addition. You can explore features of Channel Mixer. I really like how it works in CMYK mode for fine tuning.
Comment: @AlanGilbertson thanks for the reply. I just thought Vnovak's answer was 99% there and so I'd just add a finishing pointer. I will add it as answer though.
Answer: Your document needs to be in CMYK. To validate this in Photoshop for your document, go to Image > Mode > CMYK Color
Now select the piece of text and check you've got solid black (0,0,0,100) in the Select text color window, see image below:
Answer: I had same exact problem and took me forever to figure it out. In the color picker as the other person showed here, create your 100% black color and save as a swatch. Then highlight your text or shape and then click the swatch (for text), or fill it for a shape. It should be correct when you check with eye-dropper now - BUT, if there's any transparency to your element and if there's color underneath, it obviously won't appear as 100% black with your picker (took me a few minutes to realize that).
Comment: Can you show some screenshot to make your advice more understandable?

Setting solid black in Photoshop


WP function won't work on 404 template page

Question: I made a custom template of error 404 page.
In my header.php i use the_time() function to display current date.
There is some problem with 404 page because the_time() function won't work, not return any date.
Why is that happen?
Answer: the_time does not display the current date, it displays the date of the current post, and 404 pages do not have a current post.
For this reason, if it ever returned a date on the 404 page then it would be a bug.
Instead you should use the PHP functions that come with PHP to display the current date and time.
Comment: Thank you. I didn't read about this function properly. I just copied it from the WP developers documentation site and paste into my code. :) My fail and mistake.


General vs. Specific Tags

Question: I noticed a recent discussion about the use or deletion of the circuit tag. Is there value in using general tags in addition to specific tags that are often essentially sub-categories of the general tag? For instance in addition to the circuit tag, components, power, motor, transformer and capacitor are all tags that have more specific related tags.
Answer: There shouldn't be a circuit tag.  It is a meta tag that can apply to a huge number of questions on this site, and is thus useless.  All the others you mention are much narrower, with the possible exception of power.

General vs. Specific Tags


Text editor to open network files?

Question: Currently i am using Notepad++ for opening any text file in windows. but if u have multiple network file opened in the Noptepad++ and you switch to it , it got hanged for 10 to 20 sec.
Is this problem same in all editor?
please share your experience if any know any better one.
Answer: Most text editors will check to see if the open files have been modified since you loaded them. Over a network this can take some time (depending on the network).
However it looks like np++ is doing this on the UI thread and causing it to hang. Not all editors do this, so you might want to try others.
It is worth raising an issue on the np++ site as it should be fixable.

Text editor to open network files?


How do I get a deleted icon(widget) back in KDE Search & Launch?

Question: This has happened to me before, but I wasn't able to google a proper answer: I didn't have a habit of locking my widgets in KDE(Needless to say, I do now) and once in a blue moon I delete an icon from my Search & Launch strip that's positioned right below the task manager. I try launching a program, but sometimes I hit the "minus" sign and adios, icon! Is there a way to undelete it and get it back?
My screen looks kind of like this. I've scoured all the configuration files in the .kde directory in the hopes of finding a list of my current icons. So far-zip-o! I frequently back up my home directory so I might find my original list of icons there if only I knew where to look! Is there such a file?
This might be the most improper forum for such comments, but at times like these I dearly miss Windows!
Answer: Got it! Here's the file that I was looking for. If you have the same setup mentioned in this post, this might be useful to you: on line 531 of the <code>~/.kde/share/config/plasma-desktop-appletsrc</code> file is a list of your "stripwidget favoUrites". This is where your Search & Launch icons are stored(in logical order, of course), among all the other, useful settings!
I just popped into the backup file and found what I needed.
PS: If there's another way to go about the situation, ie. something lame like "taking the icon out of the trash and putting it back into the launcher"-please, do chime in!


What are some events/contests besides GSOC and ICPC that a first year undergrad Computer Science Engineering student should get involved with?

Question: I am currently in my first year (in Bangalore, India) and I'm eager to get into competitive coding and programs like GSOC. I'm not aware of too many besides the two in my title, though.  I'd love to learn about a few more that people take part in so that I can plan ahead.
Comment: If you could tell what you are looking for what sub-stream within programming and such I could Improve my answer.
Comment: Again the competition is for competitors wo are interested in the respective field, not the other way around.
Answer: There are lots of international programming competitions for undergraduate students .Here are some of the well known competitions(not mentioned by you).As you didn't mention what kind of competitions you were looking for, here are some general competitions. I have provided links to past problems so you can decide which fits you best.
1.Google Code Jam
Code Jam is Google's longest running global coding competition, where programmers of all levels put their skills to the test. Competitors work their way through a series of online algorithmic puzzles to earn a spot at the World Finals, all for a chance to win the championship title and $15,000.
How it works?
Register and compete in Code Jam's Online Qualification Round. It lasts 27 hours, but you won't need that long to earn the round's qualifying point minimum. Once you do, you'll advance to Online Round 1. They offer three sub-rounds for Round 1, and you can compete in as many as it takes for you to finish in the top 1,500 of any of these rounds.
The top 1,500 from each sub-round move on to compete for a spot in the top 1,000 contestants of Code Jam Online Round 2. These top 1,000 contestants advance to Online Round 3 and earn a limited edition Code Jam t-shirt.
The top 25 contestants from Code Jam Online Round 3 will compete in the Virtual World Finals for the championship title and a cash prize of $15,000.
Past Problems
2.Google Hash Code
Hash Code is a team programming competition, organized by Google, for students and professionals around the world. You pick your team and programming language and we pick an engineering problem for you to solve. This year's contest kicks off with an Online Qualification, where your team can compete virtually from wherever you'd like, alongside your virtual Hub. Top teams will then be invited to compete from our virtual World Finals.
How it works?
Once you've registered for the contest, you'll gain access to their Judge System. The Judge System is their online platform where you'll form your team, join a Hub, practice, and compete during the rounds. During the Online Qualification, you'll watch their YouTube livestream to be introduced to the problem, and your team will have four hours to solve an optimization problem using the programming language and tools of your choice. Teams can also compete alongside other teams in your area by joining a virtual Hub . The top teams from the Online Qualification will be invited to the virtual World Finals, where they'll work together to solve another challenge and compete for cash prizes.
Past Problems
3.Google Kickstart
Kick Start is a global online coding competition, consisting of three-hour rounds of a variety of algorithmic challenges designed by Google engineers. Participants can compete in one or all online rounds held throughout the year, and will have the opportunity to develop and grow their programming abilities while getting a glimpse into the technical skills needed for a career at Google (top participants may be invited to interview at Google).
How it works
Register for Kick Start once on the website and you'll be set up to compete in any round you'd like! When each round is open, the three-hour countdown begins and you'll compete on our website solving algorithmic and mathematical problems. Following the round, you can check out your rankings and the round analysis. If you were a top competitor, you may be contacted by Google for a chance to interview.
Last year's problems
4.Microsoft Imagine Cup
Reimagine technology solutions in a competition designed to help you make a difference in an online world. With four new competition categories—Earth, Education, Health, and Lifestyle—there's even more opportunity to share your tech ideas for some of today's most pressing global challenges. What could you create next?
Transform your ideas into action with a purpose-driven application using Microsoft Azure technology. You could win mentorship from Microsoft experts, the chance to showcase your work on a global stage, and other great prizes.
Get Inspired
5.facebook Hacker Cup
Hacker Cup is Facebook's annual open programming competition. Open to participants around the world, we invite you to apply problem-solving and algorithmic coding skills to advance through each year's online rounds, win prizes, and have a chance to make it to the global finals and win the grand prize.

Last year qualification round
Here are some other "competitions" and resources for practice:
1.Codeforces
Codeforces is a project joining people interested in and taking part in programming contests. On one hand, Codeforces is a social network dedicated to programming and programming contests. On the other hand, it is a platform where contests are held regularly, the participant's skills are reflected by their rating and the former contests can be used to prepare. Codeforces constantly develops and we plan to improve the platform to give the participants the opportunity to organize their own contests, filling the project with learning content, developing Codeforces as a training and learning platform.
It also provides contests and practice for ICPC.
Codeforces conducts different competitions for different rated coders regularly and you can participate in it depending on your Codeforces rating.
Competition structure
It may also conduct competitions with other companies and sponsors at times.
2.atcoder.jp
AtCoder is a programming contest website based in Japan.
There are three types of official contests in AtCoder:

AtCoder Grand Contest (AGC). This is our best contest. The problems will have high originality and require interesting observations.
AtCoder Regular Contest (ARC). The problems may be a bit typical compared to AGC problems, but still we think most of you can enjoy them and they are good for practice.
AtCoder Beginner Contest (ABC). This is mainly targeted for those who are new to competitive programming. The problems will be easy and educational.

3.SPOJ
SPOJ (Sphere Online Judge) is an online judge system with over 315,000 registered users and over 20,000 problems. Tasks are prepared by its community of problem setters or are taken from previous programming contests. SPOJ allows advanced users to organize contests under their own rules and also includes a forum where programmers can discuss how to solve a particular problem.
4.Project Euler
Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
The motivation for starting Project Euler, and its continuation, is to provide a platform for the inquiring mind to delve into unfamiliar areas and learn new concepts in a fun and recreational context.
The intended audience include students for whom the basic curriculum is not feeding their hunger to learn, adults whose background was not primarily mathematics but had an interest in things mathematical, and professionals who want to keep their problem solving and mathematics on the cutting edge.

P.S: There are a lot of other programming competitions but it all depends on your interests(including hackathons, A.I competitions, obscure code competitions etc.)
Answer: If you want to hone your broader skills, better look around for a open source project that you personally use and that interests you, and dig in. You'll learn how to program, how to test and debug, how to triage bug reports, how to work together with a motley crew of people from all over the world. You'll be able to show your work to potential employers (open source means nobody can forbid you showing the "secret sauce") later on.
Programming competitions, or solving problems to be checked by some online judge, trains you in how to come up with a smart algorithm and write it in your favorite language. Programming "in the large" is much, much more than this.


I stopped the server, but it pinged successfully, why?

Question: If I stop the apache server by systemctl stop httpd command, and if I ping it from command prompt (ping [server's IP address]), it pinged successfully, why ? I was created the server on Google Cloud Platform and then I stopped the server by run this command on shell. But when I ping it, it pinged successfully, why ?
Answer: You simply stopped the (Apache) service running on the (Google Cloud) server, not the server itself.
The ping demonstrated that the server is still running (and responding to pings). If you try to connect to port 80 then it should fail, confirming that the http service isn't running.
Comment: To make it more explicit: Apache HTTPd is not involved with ICMP Echo Requests ("Ping") at all.
Comment: @SanniddhaChakrabarti Richard's answer is entirely correct. The service that you run on your machine called Apache is only a web server. The machine that runs that web server is still running even after you terminate the Apache process. The machine's Operating System is what responds to ping requests, not apache. You also cannot "ping" an address with a port number. It's only used to determine if a machine is able to be reached, not if a specific port can be reached.
Comment: Terminology here is important - the terms "server" and "service" can usually be used interchangeably when referring to a piece of software but not a physical machine. You stopped the apache *service*, but the physical *server* machine is still running.
Comment: @SamForbis but if I try to reach the server by paste the IP address (http://IP_Address) in browser after stop the server it shows that **This site can't be reached**.
Comment: @Richard I tried to ping with port number (for example: `ping 22.214.171.124:80`). But it shows that `Ping request could not find host 184.108.40.206:80. Please check the name and try again`.


Is there a hook or trigger I can use to run a script when a new network connection is made?

Question: I have a 2014 retina Macbook Pro running Yosemite (10.10.3). I would like to run a script whenever a certain network interface is connected. Is there a hook or trigger I can use? I have a shell script that will perform the actions I want to perform, I just need to know how to trigger it.
Answer: There's a great utility called ControlPlane that allows you to trigger all kinds of events based on the context of system rules you set.
from the about page on the site: http://www.controlplaneapp.com

ControlPlane determines where you are or what you are doing based on a number of available evidence sources and then automatically reconfigures your Mac based on your preferences. Evidence sources can include your current location, visible WiFi networks, attached USB devices, running applications and more.
Comment: Thanks @rchrdg I'll look at Control Plane. What I am really looking for are the hooks or triggers that Control Plane utilizes.
Comment: Scott the entire app is open source and you can see the code on Github. Ill bet any specific hooks are easier to figure out if you review that.


Can I get online support from Canonical for improving performance?

Question: I am a real techno-tard. I need help optimizing my pc performance. Does Canonical offer any online support (such as online chat) to help people like me? I try to understand the posts but the language is just too technical for me to understand what to do to help myself.
Answer: Related to support ,
They have lot of subject based IRC Chat channels here as Chat support IRC

There are several Internet relay chat (IRC) channels on the freenode
  network (irc.freenode.net) used for Ubuntu support and development.
The best place to start for general help with Ubuntu or just to meet
  other Ubuntu users is #ubuntu or one of the many local language
  channels.
Specific channels are available for Ubuntu derivatives: #kubuntu,
  edubuntu, #lubuntu and #xubuntu and separate channels relate to development, bugs,local community (LoCo) teams, accessibility and
  documentation. Choose from the current list of channels .
When participating in Ubuntu IRC channels, please follow the code of
  conduct.

Plus - Gain enough Reputation to Chat on AU Chatroom itself.
Ubuntu also provides Paid support service, refer the Canonical Profesional Desktop Support Service page.
As far as PC performance goes , you can try asking the question in detail description as required. Example
How can I improve Ubuntu overall system performance?


Are AWOS data open for commercial use?

Question: I 've found out this great website where you can get historical AWOS weather data. I would like to use it for a commercial application but I don't find any data license. Do you know if the AWOS data are free to use for commercial purposes? Can you point me to a relevant link?
Comment: A bunch of random folks on the internet aren't going to know that. I'd recommend that you contact the fine folks at Iowa State and ask them.
Answer: Your website source from Iowa State University may or may not be open for commercial use. As Freeman has said in his comment, it would be difficult for anyone here on the internet not affiliated with the university to know. Your best opportunity to use their data is to contact the university directly. Their contact information is at the bottom of the link you provided.
Otherwise, your best option is to contact NOAA, the NCEP or the National Weather Service (basically, one and the same organization). They collect, consolidate, and archive the data at their source. That is where many commercial applications get their data.
Comment: @greghor - No problem. Unless Iowa State has done something to add a unique value added proposition to their data in order to make the user interface easier for commercial use, it might be better to go directly to the source of the information. After all Iowa State is getting their data (probably for free) from the NCEI, a division of NOAA. More than likely, so can you. Good luck.
Comment: Thanks for the reply. Really helpful. I was maybe thinking that it was commonly known in the aviation community if this data source is open or not, hence my question. Your suggestions and the one from FreeMan are very good, I will directly contact the organizations that you mentioned... and post the answer here once I get a reply :)
Comment: +1 on Dean's last comment.  This data is produced using taxpayer dollars, and is broadcast for public consumption.  You are wise to ask these questions, especially if another entity has gone to the trouble to collect and archive it, and especially if you intend to use it commercially, but you ought to be able to find a way to gather and use what is in the public domain.
Comment: The question has been closed but let me write here the answer I got from NOAA. In the general case, AWOS data can NOT be used for commercial purpose

Are AWOS data open for commercial use?


Unaccounted bandwidth usage in activity monitor

Question: Here is a screenshot of Activity Monitor network usage tab. Look at the Data Received total at bottom, and then look at the process amounts in the list on top. It doesn't add up.
Answer: As far as I'm aware, one value is 'since you opened the window to start logging' the other is 'since last boot'.
Example much, much greater discrepancy…
Comment: I see. If so, it makes Monitor a bit less useful than I thought it was :-P I'll check your answer as definitive unless someone contradicts you. Thanks.
Comment: I'm not certain the *exact* point from which the columnated data is derived. I leave Activity Monitor running all the time, it auto-launches at boot, yet I rarely look at this particular tab. So, the values you see above are for 4 days' uptime for the whole machine, with 10s of that tab being in view. If it were strictly from the time I opened the tab, you'd expect most of those values to be near nil, which they're not, even though they're small.


Choosing trend analysis period

Question: I'm having trouble figuring out whether its important to have a common period of time when performing hypothesis testing for trend analysis.  I have time-series data of varied length - some sites have 5 years of data while others have 20 or even 40 years of record. If I filter my data based on a common time period then I exclude a large part of my data set. I can understand that comparing the magnitude of trends isn't valid but does it matter when simply testing for monotonic change?
Answer: If you are not interested in making any cross-site comparison of magnitude of trends or in computing an overall trend across all sites, then you don't really need to have a common time record across sites. However, if you have many sites, you may still want to group sites according to whether they have (approximately) common support for reporting purposes. Then you can present trends together for all sites where the time record spans a common period like, say, 2001-2005 (5 years worth of data), 2001-2010 (10 years worth of data), etc.  This would make it easier for people to digest your results than if you report trends for ungrouped sites. In particular, if you choose a time span like 2001-2010 for your 10-year time series, you can group together trends which cover incomplete time records such as 2003-2009, 2002-2008, etc., by allowing a small fraction of data missingness in the incomplete time series relative to the complete record 2001-2010.
Answer: Thanks Isabella, its been very difficult finding information about comparing trends with different time periods. What you can and should not do.  I really wanted to use the entire data set for all data ranges because collectively I think they highlight some interesting patterns - namely, we detect a far greater number of trends using summer months compared with other seasons. Related or not, this corresponds with peak groundwater abstraction, and our records show groundwater use has "significantly increased over time, particularly over the summer seasons.
We have two main groundwater basins in our region and I expect one is experiencing greater declines than the other. Here, I will look to use the same time period for comparing the magnitude of trends. Unfortunately, it seems to take quite some time before we detect statistically significant trends so my comparison might be limited to a 'small' number of wells.
Thanks again,
Simon

Choosing trend analysis period


Find the closest point along a rectangle, given an another point and direction

Question: Given a rectangle, and a point with a vector direction towards the rectangle.
How can I find the closest point on the outside of that rectangle to the point in question?
Comment: They keyword to look up in Google is "AABB" (axis-aligned bounding box). If your "box" (rectangle) isn't axis-aligned yet, a simple transformation matrix - used on all items you care about, obviously - can be used first to change it into one.
Comment: Can you explain more what you're asking? I'm really not getting it.
Comment: Let's say I have a character facing an object (a rectangle), and I draw an imaginary line from that character to the rectangle. I would like to know how I can tell which point along the outside of the rectangular object the line is touching.
Answer: The rectangle has four sides.  Each side is a line segment.
Test each of the four sides for intersection with the ray.  Track the closest hit.
Here's some code to find out where on the segment the ray hits:
bool intersect(const ray& ray, const segment& segment,point& hit) {
    // where do we intersect this line?
    float t = ((ray.direction.x * ray.origin.y + ray.direction.y *
        (segment[0].x - ray.origin.x)) -
        (ray.direction.x * segment[1].y)) /
        (ray.direction.y * (segment[0].x + segment[1].x) -
        ray.direction.x * (segment[0].y + segment[1].y));
    if(t >= 0.0 && t<=1.0) { // in the segment
        hit = segment[0] + (segment[1]-segment[0]*t);  // lerp
        return true;
    }
    return false; // no hit
}

Answer: One technique which you could use is called "ray casting". It is commonly used for rendering graphics, but has other applications such as line-of-sight (as you are wanting to do) and path-finding. In general terms it works by finding the intersection of a ray and an object. In your example the ray is the vector for the character's direction.
A useful reference for ray/object intersections (and incidentally other object/object intersections) is www.realtimerendering.com/intersections.html (look under the references for ray/aabb and ray/obb).
Answer: If your box is axis aligned, you just have to clamp each coordinate axis to the box if the point is outside the box.
From RTCD pg 130:
// Do this for all 3 axes
if( point.x < min.x )  point.x = min.x ;
else if( point.x > max.x )  point.x = max.x ;

If you do this for x, y, z axes, then the point will be slammed to the nearest wall of the box, if it is outside the box to begin with.  if it is already inside the box, it will be left alone (where it is).
Answer: Well, you can use just linear algebra (analytic geometry, to be more specific) to solve this. It depends on how you modeled the rectangle.
Here's a general case:
http://paulbourke.net/geometry/lineline2d/