Is there a way to break down a signal into its respective phases (4 in this case) to be later analysed individually

Question: Is there a way to extract amplitude information but over 4 phases of the same frequency. That is, if the frequency is known could I extract amplitude peaks (between given values)in the time domain in lots of four, then separate them into "bins" or "arrays" ready for four individual FFTs?
Comment: From what I get out of your question, is it the overlap-add method that you are looking for?
Comment: Can you partition a signal in the time domain on whole sets of cycles and do DFTs on each?  Absolutely.  Is that what you are asking?
Answer: This would be trivial to do as simple decimation where each block increments its samples by $n+4$ with each block starting at sample 0,1,2,3 respectively. This is common with polyphase filter implementation and similar techniques to reduce the overall clock rate requirement for the processing (parallel processing).
For more details on both of those approaches see these posts:
How to implement Polyphase filter?
4-phase FIR LPF in MATLAB
Why does this parallel IIR filter have such a weird form?
Another straightforward and interesting approach specific to DFT processing is to use the Decimation In Time binning that is inherent in the (DIT) FFT algorithm itself.
See the block diagram below for further detail on how an 8 point DFT can be binned in time into 4 2 point DFTs, and specifically how to combine the resulting DFT outputs for any size data block.

Expanding this to any general block length of radix 4 length (a length that is a multiple of $2^4$ so that it can be divided into 4 identical blocks. For blocks not in increments of $2^4$ simply pad with zeros to get to this length), the input sequence is binned by two decimation in time operations each with a factor of 2, best demonstrated by example:
Consider a sequence $x[n]$ numbered 0, 1, 2, 3, 4, 5, 6, 7, ...
After one decimation in time by 2 we divide this into two groups, noted here as Group one and Group two:
Group one (even samples): 0, 2, 4, 6, 8, 10 ....
Group two (odd samples): 1, 3, 5, 7, 9, 11 ...
Each of these is then decimated by 2 to form each of the four separate sample blocks, as follows:
Group $x_1[n]$ (even samples of Group 1): 0, 4, 8, 12, ...
Group $x_2[n]$ (odd samples of Group 1): 2, 6, 8, 10, ...
Group $x_3[n]$ (even samples of Group 2): 1, 5, 7, 11, ...
Group $x_4[n]$ (odd samples of Group 2): 3, 7, 11, 13, ...
The reason for binning in this fashion is for the efficient recombining of the DFT results for each block using 2 point DFT "butterflies" as shown in the graphic above and further detailed below.
The four DFT results are combined to be equal to the DFT of the whole sequence as follows:
Given each $N/4$ length DFT result of the four groups as $X_1[k]$, $X_2[k]$, $X_3[k]$, and $X_4[k]$, where k is the integers from $0$ to $N/4-1$ representing the $N/4$ samples in each group:
Combine $X_1$ with $X_2$ to form a $N/2$ length DFT of the combined result. $k$ is incremented from $0$ to $N/2-1$ and due to the cyclical nature of the DFT, $X_1[N/4]=X_1[0]$, $X_1[N/4+1]=X_1[1]$ etc
$$X_{1,2}[k] = X_1[k]+ W_{N/2}^{k/2} X_2[k]$$
Similarly combine $X_3$ with $X_4$ to form a $N/2$ length DFT of the combined result.
$$X_{3,4}[k] = X_3[k]+ W_{N/2}^{k/2} X_4[k]$$
Finally combine the two resulting $N/2$ length DFT's to have the full $N$ length DFT result, here k is incremented from $0$ to $N-1$:
$$X[k] = X_{1,2}[k]+ W_{N}^{k} X_{3,4}[k]$$
The notation above $W_n^k$ are the "roots of unity", coined as "Twiddle Factors" by W.M Gentleman and G. Sande in their 1966 paper "Fast Fourier Transform - for fun and profit". The twiddle factors are simply convenient shorthand for the exponential expression below and represent phase rotations:
$$ W_N^k = e^{j2\pi/N}$$
Below shows a graphic example for N = 8


Configurator Stalled: How to force it to continue or quit?

Question: I connected an iPad to the Mac and ran configurator as usual, updating from a back up, however it stalled out at the "Restoring Default Data" step. I tried clicking the Stop button, and now it says it's "Canceling..." but it's still not doing anything. The iPad has the apple logo and a progress bar that's at about 80%. I was able to restore one of our other iPad without trouble after I pressed start, but I can't seem to accomplish anything. Can I unplug the iPad safely and "factory reset" it or something and then update from a back up?
Answer: Click Apple icon > Force Quit > confirm


How do I highlight parentheses whenever the cursor is inside?

Question: I'd like to highlight the enclosing parentheses whenever the cursor is inside them.  How do I do that in Vim?  The plugins that I can find with Google only highlight the match when the cursor is on the parenthesis itself, so I can't use those to solve my problem. I cannot figure out a good regular expression to use to account for nesting, so syntax highlighting definitions don't seem like a good approach.
Comment: The question is not asking for an offsite resource; it's asking how to solve a particular problem in Vim.  While the OP clearly hasn't done any research (as evidenced by their comment). That's a reason to downvote this question, but not a reason to close it.  If the OP were asking for a plugin, then it'd be an off topic question. Note that a plugin could be on-topic as an answer; and that's ok. It's the explicit asking for off-site resources that's the issue. [This question is a topic of meta discussion](http://meta.stackoverflow.com/questions/253055/b/253057#253057).
Comment: Are you searching for a plugin (such recommendations are off-topic here), or do you plan to implement such yourself? For the latter, what have you tried so far?!
Comment: This is not off-topic because it requires Vimscript, a programming language in itself. I am not specifically looking for plugin recommendations, but if someone provides, I'd appreciate it. I have not attempted to implement this myself because it requires pattern matching (regex?) for which I have zero knowledge and zero time to learn at the moment. Please reopen.
Comment: As far as closing because there isn't minimal information to diagnose the 'problem'. The OP hasn't asked for their code to be debugged (a problem to be fixed), they're asking for code to do something in VimScript. Downvote if you feel it doesn't show any research effort, but because this question doesn't have to do with already written code, that close reason doesn't really apply here.
Answer: Properly matching a pair of parentheses, with an arbitrary number of nested pairs inside, is not a solvable problem using regular expressions alone. Limiting the problem by limiting the number of allowed nestings results in very ugly regular expressions. So don't try going that route.
Luckily, Vim provides a searchpairpos() function to find such pairs accounting for nesting.
See the example in Vim at :help match-parens for an example on using searchpairpos() to highlight the parenthesis which matches the one under your cursor. You should be able to extend it to first search for the next parenthesis before or after the cursor to use as a starting point for the matched pair.
Comment: Thank you. You may have observed the negative atmosphere around this question. Without your answer it would never had occurred to me that I could use `searchpairpos()` to determine ranges of text enclosed in parentheses.


Emphasize background conversation in Audacity

Question: I have an audio file of a man talking and some conversation taking place in the background. I need to remove the main voice and emphasize the background conversation.
Someone can help me?
Answer: There is no generic recipe to remove a certain specific sound from a recording. It may or (most probably) may not be possible to do what you wish, as voices are voices and take more or less the same place in the frequency spectrum, so there is no "magical" way of isolating them. Depending on the specific case, it may me possible to attenuate the voice you don't want, probably at the expense of the overall quality and perceptibility of the rest. So it's a matter of experimentation and trying to get the best compromise.
It's hard to summarize an approach without going into lengthy technical explanations, and to be honest, I think it will be extremely difficult for someone without the proper training to do it, but even so it may be worthwhile to lay out the basic approach.
Basically, you need to identify frequencies in your recording that belong to the voice (ot other sounds) you want to eliminate, but not (as much as possible) to the voices you want to keep. Then you need to apply filters to remove these frequencies from the audio. Best results (or perhaps, not so bad results) may be achieved if this process is done specifically to different audio segments, depending on how the voices mix along the recording.
In practical terms I would suggest not to start with Audacity (as excellent tool as it is), but to a more specialized audio analisys tool. I recomend Sonic Visualizer as a free/open source tool widely used for audio analisys in research and academia (setting up Sonic Visualizer with the required plugins may be a bit of a project on it self, but there are instructions in the site and it's worth the trouble).
Load your audio into Sonic Analiser and perform a spectral analisys. The details of the resulting spectrogram will intirely depend on the specific audio, but here's an example of a spectrogram (probably from a piece of music):

Listen to your audio while looking at the spectrogram and see if you can visually correlate the sounds you want to eliminate with specific high energy presence in the spectrogram (i.e. the red lines). If so, find on the left hand scale the frequency (or frequency band) where it seems to fit mostly. Take notes of the time intervals and frequencies.
Then go to Audacity and apply Notch Filters for each time interval and frequency you previously identified. You'll probably need to repeat the process multiple times, refining the parameters. (There are probably tools that allow the analysis and filtering steps in the same tool, but I don't know them and, unless I'm mistaken, they are not publicly and/or freely available anyway)
Answer: Be prepared for the outcome to be that it is practically impossible.  You don't mention WHY you want to do this, or WHAT kind of result you need. Are you creating a sound-track for a video? Are you trying to understand what someone in the background is saying?  Do you need "broadcast quality"  Do you need "transcription quality"?  It is pretty unlikely that you can accomplish this with Audacity. There are software tools that cost as much as your car which are used by forensic professionals for applications like this.
Answer: ESSENTIALLY,
-you need to find the frequency that the main voice is talking in
-you need to find the frequency that the background voices are talking in
-increase dynamics of background frequencoes
-decrease dynamics of main frequencies
This is incredibly difficult as those frequencies will overlap and you will most likely be altering a RANGE of frequencies
There are probably plugins that can help you out a little bit but I would say this is mostly up to you having patience and taking this one step at a time.
Answer: There is no technology to separate dialogue from background dialogue sounding at the same time. Especially while the background dialog sounds lower than the main speaker. You better forget it and don't waste your time. The only thing you can do is to try to figure out what's been said in the recording as it is by adjusting your overall volume if need it.
Answer: I know this is a ¾-decade-old question and the person has likely given up by now, but there is a commonly used program that can raise background voices to at least come close to matching foreground voices, despite not being a forensic application. IIRC it was even able to do this back in 2016 when this question was asked.
The program is Adobe® Audition® and has gone through many versions. It was stand-alone for awhile, then part of Creative Suite® (Audition CS#), then Creative Cloud® (CC[year] until recent years, and now just [year]).
The trick is a built-in filter (no need to download any VSTs, etc.) that's not in an obvious place in most versions. In the latest Adobe Audition 2024, it's under "Effects" ⇒ "Amplitude and Compression". In earlier versions it may've been in different groupings including an Obsolete Filters grouping.
It's called "Speech Volume Leveler…," and it does exactly what its name implies: it levels the volume of speech in audio. The dialog gives you considerable control over how much leveling it does. It's available in the Effects Rack of newer versions, too, which means you can use it in real time, non-destructively, if your computer's fast enough.
Answer: There's no easy way to remove the voice, but one option is to put a super hard compressor on the recording so that both the background conversation and the main voice are loud.
Comment: I suppose it's worh experimenting, but the compressor is going to respond to the combined sounds, so although it may enhance the background when occurring alone, I gather when the main voice appears (and his compressed) the background is going to be further pushed into the backgound.


'Proof' that non-Abelian Berry phase vanishes identically

Question: For a degenerate system with Hamiltonian $H =H(\mathbf{R})$ and eigenstates $\left|n(\mathbf{R})\right\rangle$ the non-Abelian Berry connection is
$$A^{(mn)}_i=\mathrm{i}\left\langle m|\partial_in\right\rangle\tag{1}$$
and the non-Abelian Berry curvature is
$$
F_{ij} = \partial_i A_j-\partial_j A_i - \mathrm{i}\left[A_i, A_j\right]
$$
in matrix notation or, including the state indices:
$$
F_{ij}^{(mn)} = \partial_i A_j^{(mn)}-\partial_j A_i^{(mn)}-\mathrm{i}\sum_k\left( A_i^{(mk)}A_j^{(kn)}-A_j^{(mk)}A_i^{(kn)}\right).\tag{2}
$$
Substituting (1) into (2) gives
\begin{align}
F_{ij}^{(mn)} &=  \mathrm{i}\partial_i\left\langle m|\partial_jn\right\rangle- \mathrm{i}\partial_j\left\langle m|\partial_in\right\rangle -\mathrm{i}\sum_k\left(\mathrm{i}^2\left\langle m|\partial_i k\right\rangle\left\langle k|\partial_j n\right\rangle-\mathrm{i}^2\left\langle m|\partial_j k\right\rangle\left\langle k|\partial_i n\right\rangle\right)\\
&=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle+\mathrm{i}\left\langle  m|\partial_i\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle-\mathrm{i}\left\langle  m|\partial_j\partial_in\right\rangle +\mathrm{i}\langle m|\left(\sum_k|\partial_i k\rangle\langle k|\right)|\partial_j n\rangle-\mathrm{i}\langle m|\left(\sum_k|\partial_j k\rangle\langle k|\right)|\partial_i n\rangle\\
&=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle +\mathrm{i}\langle m|\left(-\sum_k| k\rangle\langle \partial_ik|\right)|\partial_j n\rangle-\mathrm{i}\langle m|\left(-\sum_k|k\rangle\langle \partial_jk|\right)|\partial_i n\rangle\\
&=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle -\mathrm{i}\sum_k \underbrace{\langle m| k\rangle}_{\delta_{mk}}(\langle \partial_ik|\partial_j n\rangle-\langle \partial_jk|\partial_i n\rangle)\\
&=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle- \mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle+\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle\\
&=0.
\end{align}
To go from the second to the third line I used
$$
0=\partial_i(1) = \partial_i\left(\sum_k |k\rangle\langle k| \right) = \sum_k |\partial_i k\rangle\langle k|+\sum_k |k\rangle\langle\partial_i k|.
$$
Clearly I have done something wrong as the Berry curvature is not zero in all cases. Please could someone point out my error?
Comment: I am not 100% sure about the resolution, but I am skeptical about the identity you use in the last line. Let's say the parameter of the curve you are following is $\lambda$. The state is in some degenerate subspace of the spectrum as it travels along a closed path. Now I agree you can choose an orthonormal basis $|k\rangle$ at some fixed $\lambda$ obeying the resolution of the identity. But are you sure that as you vary $\lambda$, these states remain an orthonormal basis? You need $\sum_k |k(\lambda) \rangle \langle k(\lambda) | =1$ to be true for all $\lambda$ for your identity to hold.
Comment: To be clear, it's not that the evolution will take you out of the subspace. You can always choose an orthonormal basis for every $\lambda$. My question is, if you take a set of states for a given $\lambda$, and then evolve those states adiabatically as you vary $\lambda$, will they remain an orthonormal basis?
Answer: Your calculation is correct. I believe the problem is that the non-Abelian Berry curvature is useful only when it is defined in a sub-Hilbert space and that subspace is not the same at different parameter R. In this case, the identity you mentioned in the end no longer equals to zero.
Comment: Do you have a source for any more details on this?
Comment: Sorry I'm not aware of any good references for this. But inspired by your question, I find it is more intriguing to write the non-Abelian Berry curvature in the following form $$\sum_{m,n}\Omega_{ij}^{(mn)}|m \rangle \langle n|=\hat{P}i(\partial_{i}\hat{P}\partial_{j}\hat{P}-\partial_{j}\hat{P}\partial_{i}\hat{P})\hat{P}$$, where $\hat{P}$ is the projection operator onto the sub-Hilbert space where the Berry curvature is defined.
Answer: To expand on Leon's answer, it may be helpful to think in analogy with the usual $U(1)$ gauge theory. We can think of the non-abelian Berry connection as an $SU(n)$ gauge field, where $n$ is the number of bands. Our gauge transformations are a change of basis by unitary matrix $U(k)$ and the connection and curvature transform as
$$
A' = U^\dagger A U + i U^\dagger\partial_k U\\
F' = U^\dagger F U
$$
respectively.
In matrix notation we may write the Berry connection $A = i M^\dagger \partial_k M$ where $M$ is a matrix of the eigenvectors,
$$
M_{ij} = |j\rangle_i.
$$
If $M$ is unitary, corresponding to the condition $\sum_k |k\rangle\langle k| = 1$, we may perform a change of basis which takes $A'\to0$ (explicitly we change basis by $M$), and therefore $F$ must be zero since it is a gauge-covariant quantity.
That is if we keep all bands, every Berry connection is "pure gauge".
However, if we consider a sub-space of the bands, then $M$ is instead a projector onto those bands $M = \sum_{k=1}^{n < N } |k\rangle\langle k|$ where $N$ is the total number of bands, and in general there is not guaranteed to be a change of basis which nullifies $A$.
Thus, $F$ can be non-zero in this subspace.
Comment: Thanks for adding more details to my vague answer. :)
Answer: Shouldn't it be ?
$$
F_{ij}^{(mn)} = \partial_i A_j^{(mn)}-\partial_j A_i^{(mn)}-\mathrm{i}\sum_k\left( A_i^{(mk)}A_j^{(kn)} \mathbf+ A_j^{(mk)}A_i^{(kn)}\right).\tag{2}
$$

‘Proof’ that non-Abelian Berry phase vanishes identically


Sum the infinite series $ \sum_{n=0}^\infty (2n^7 + n^6 + n^5 + 2n^2)/n! $

Question: What is
$$ \sum_{n=0}^\infty \frac{2n^7 + n^6 + n^5 + 2n^2}{n!} $$
Comment: Just a curiosity: I wrote a puzzle post some days ago, and it was closed in a flash... How comes this is still open?
Comment: Note that at least one of the two respondents did not realize that you already knew (or thought you knew) the answer to the question before you asked it.  It seems that just having the "puzzle" tag is not (yet) clear enough for that purpose.  I would appreciate it if in the future you would make it more clear when you are posing a problem rather than asking a question by calling attention to this in the question itself (or perhaps even the title).
Comment: Can you suggest some words that would clearly and politely indicate this?  Saying, "By the way, I know the answer" seems awkward; I was hoping that using the "puzzle" tag was sufficient.
Comment: @Pacciu: I don't know (FWIW, I didn't vote to close your question).  You may want to bring this up on the meta site.
Answer: From Dobiński's formula, $\sum \limits_{n=0}^\infty \frac{n^k}{n!} = eB_k$, where $B_k$ is the $k$th Bell number, the number of set partitions of a set of size $k$..
So the answer is $e(2B_7 + B_6 + B_5 + 2B_2) = e(2 \times 877 + 203 + 52 + 2 \times 2) = 2013e$.
Comment: Sigh, I was aiming at 2011e and overshot via a typo.  Cheers for the answer.
Answer: Here's a hint.  $$\sum_{n=0}^{\infty} \frac{n^k}{n!}=e B_k$$ where $B_k$ is the $k$-th Bell number (see Dobiński's formula for a proof).
Answer: You have that if:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
then:
$$
z \frac{\mathrm{d}}{\mathrm{d} z} A(z) = \sum_{n \ge 0} n a_n z^n
$$
In general, using $\mathrm{D}$ for the differentiation operator, and $p$ a polyomial:
$$
\sum_{n \ge 0} p(n) a_n z^n = p(z \mathrm{D}) A(z)
$$
So you are looking for:
$$
(2 (z \mathrm{D})^7 + (z \mathrm{D})^6 + (z \mathrm{D})^5 + 2 (z \mathrm{D})^2)
  \mathrm{e}^z
$$
at $z = 1$.


Dependency property in wpf/silverlight vs. javascript object property

Question: i am learning silverlight recently and find its difficult to understand the essence of the dependency property system. can I make a analogy between silverlight's dependency property and javascript's object property? both of them are implemented by a hashtable-like data structure  and can be add/remove at run time , am i right?
Answer: You are not right, the WPF depdency properties are not dynamic, objects have a set of depedency properties the way they store values may be the same (i think it's more of a dictionary structure) but you cannot just have a new property on the fly. However there is something called attached properties which allows for additinal properties.
I would suggest you read those articles i linked to, they should explain everything quite well.
Answer: Think of DependencyProperties as Pointers or Addresses. Instead of setting a property to a value, you are setting it to a an address which points to another value.
For example, instead of saying a TextBox's text is equal to SomeValue, you are saying it is pointing to a string value in the ViewModel. Or it is pointing to the Name property of some object. Or pointing to a specified ComboBox's SelectedText. Of course, you could always say it points directly to a string of "SomeValue"
<code><TextBox Text="{Binding SomeProperty}" />
<TextBox Text="{Binding SomeObject.SomeProperty}" />
<TextBox Text="{Binding ElementName=MyComboBox, Path=SelectedText}" />
<TextBox Text="SomeValue" />
</code>
Basically, it is a property that is dependent on some other value.


Galaxy S2 Messaging: persistent badge number

Question: I use a Galaxy S2, SPH-D710VMUB (Virgin Mobile) running Android 4.1.2
The default messaging app it comes with has this (very common) feature of showing a little red circle with the number of new text messages in it. When you read your messages the icon disappears and you are left with just the yellow envelope icon.

Lately it has been displaying "3" new messages whether I have new messages or not. It never goes away, always the little red circle and the "3" inside it.
My notifications tab at the top of the screen is properly updated, if I have no messages it does not display anything.
It doesn't seem to affect any of the other apps, notifications or functions of the phone. It's almost as if the icon simply never refreshed or something.
I have already restarted the phone, cleared RAM in Active Applications, etc.
Thanks for your time.
Answer:
Head to Settings
Scroll down to find Application manager
Scroll to the right until you reach the page All
Locate Badge-provider
Click on Clear data, if you cannot click clear data then first hit Force stop, then you should be able too.
Restart your phone

(Source)


What were the main reasons for creating iĉismo?

Question: I can understand why in the context of today's discussions on gender, and I see a need for it, but what were the main reasons for creating iĉismo? I would like to know who created it and why.
Comment: Uh, changing "I see a need for it" to "I see the need for it" is a pretty big edit. When we edit, we should try to keep the original intent of the question and not change it needlessly.
Answer: The default for certain core words is masculine. You can explicitly call out anything as feminine with -in-. This leads some people rather directly to thinking of an entity in any role as masculine unless it's explicitly called out. Conversely, it makes it difficult to specify that a person is male.
In real life, men are not the default, and women are not aberrations from the default. Both are equal options. We want a language that reflects that.
The Fundamento endorses ge- as a prefix to be used for groups of mixed gender. In other words, if you see doktorojn kaj doktorinojn, you can say that you saw gedoktorojn. If you need to go to see a doctor, you need to see a doktoron aŭ doktorinon, or unu de la gedoktoroj, or you can merely hope that your audience doesn't think you are speaking specifically about male doctors.
By reinterpreting all words as not specifying a gender and adding an explicit masculine affix that parallels the feminine affix, we fix this problem.
However, there is a large body of work with slightly conflicting usage. While this shift doesn't invalidate most usages of the masculine-or-epicene, it does invalidate some of them. This is a barrier to adoption.
Furthermore, since the rationale is to resolve a gender issue, a lot of people jump out of the woodwork to defend the status quo. They will say that any sexism is in the mind of the speaker, not in the language. That might be true, but in this case, the language certainly isn't helping, and that makes the language less useful for communication -- which is the entire reason it was built. But regardless, this is still a barrier to adoption.
Comment: I'd say the largest barrier to adoption is that most people learn the "regular" Esperanto, not "ri-Esperanto" and then becomes harder to change.
Comment: Because of the simpler morphology of Esperanto, it's easier to learn to use *ri* than English neopronouns, which are already pretty easy to learn. The *-iĉ-* affix is ever so slightly harder to learn than *ri*.
Comment: well if you learn _-in-_, you learn _-iĉ-_, there should be no confusion there if taught simultaneously. And if you learn _-ĝi-_ then _ri_ becomes irrelevant.
Answer: My understanding is that iĉismo is a lighter form of riismo, which has been around at least since the early 90's. Here is an early description viewable via the "Wayback Machine."
https://web.archive.org/web/20060924123828/http://www.rano.org/riismo2.html
There's a lot of information there. I thought this part was particularly relevant to your question:
Pro la evidenta analogeco "panjo - patrino : paĉjo - patriĉo" la sufikso -iĉ- estas multfoje kaj sendepende inventita de multaj diversaj homoj. Verŝajne ankaŭ la pronomo ri havas multajn inventintojn; se oni metas vokalon I post ĉiu konsonanto kaj forstrekas la radikojn, kiuj estas jam uzataj aŭ kiuj tro similas al alia pronomo, kiun oni intencas daŭre uzi, restas nur malmultaj ebloj.
Comment: This may be a "meta" issue - but the last paragraph above is a quote from a website. I considered editing the quote which was in the x-system, but opted not to in order to stay true to the document being quoted. It appears not everybody saw it that way.
Comment: Thank you for the answer! I don't really think it answers my question, though...
Comment: Could you clarify your question?
Comment: I wrote "I would like to know who created it and why." And I ask for the main reasons *why* it was created. The question about why the "icx" is another question, which you answered, actually. http://esperanto.stackexchange.com/questions/1141/kial-i%c4%89-por-viri%c4%89oj-en-la-i%c4%89isma-sistemo
Comment: What I do understand from your answer is: It is unclear who created it, and why they did.
Comment: I'm not sure how to clarify. I linked to a document, apparently signed by several people - in which they state that this is an idea which keeps coming up independently (and therefore does not have a "who" who created it). They put forward their reasons, and discuss common objections.
Comment: Cxio en ordas. Koran dankon por via respondo.
Answer: When Esperanto was created, animate roots that referred to people or animals were almost completely male or female. Unmarked roots were male and had to be marked to be female (koko "rooster", kokino "chicken"), or they were inherently female and weren't marked to be masculine as their meaning inherently referred to women (ex. damo "lady/queen", nimfo "nymph"), as is the case in the majority of European languages. As time progressed the majority of inherently male roots became gender neutral. This led to the desire to fully neutralize the remaining masculine roots (patro "father", reĝo "king") and itroduce a masculine suffix similar to -ino so that Esperanto would be completely neutral.
When it comes to movements like these, it is not always a single person who leads it. Even if a manifesto is written and the followers of the idea have a name for themselves (as is the case with riismo), this does not necessarily mean that they will have a leader or named founders. Often with affairs like these, a group of people may spread the idea anonymously, or people or groups of people may independently develop the concept and spread it anonymously. As to who created iĉismo, the best answer is that we don't know and that we may never know.
PS: This is a subjective evaluation of the situation, but if their anonymity is intentional, they may have desired that we more so consider whether or not their proposal has merit compared to other proposals, as opposed to the celebrity of Zamenhof and Sonja Lang if the affair succeeds, or the infamy of Louis de Beaufront if it failed.
Answer: Because it makes logical sence. Right after my first Esperanto-teacher had explained the female ending -ino to me (this was in 1984) , I immediately asked whether there was a male one too. Fortunately he admitted that there was an unofficial one, -iĉo.
Over the years I have found that the perceived lack of a male suffix is probably the most common reason for people who do like the idea of having a worldwide auxiliary language to not learn Esperanto.
Answer: Other than with Riismo (originating in Octobre 1976 in this publication), Iĉismo is more difficult to pinpoint. It seems to have been a proposition by a Per Hagemann in 1998 on Usenet in this post.
To clarify one misconception: Esperanto isn't "sexist by choice", but Zamenhof simply took the German system ("Freund/Freundin") as the Slavic alternative (adding an "a") did contradict his grammatical rules.
This asymmetry was raised by Louis Couturat in the late 19th century, and Zamenhof apparently liked the proposal for its beauty and it being desirable, so he embraced the default being a so-called Utrum Genus ("all of both", its negation "none of both" is "neutral").
To clarify another: "ri" is too near "li", and speakers from both China and Japan have a common pronunciation for both /l/ and /r/ (namely, a very hard /ł/ in between, that sounds a little more /r/ in Japanese and a little more /l/ in Mandarin). They cannot "learn to discriminate /l/ from /r/". There is an interesting YouTube video of a US-American wife of a Japanese husband where this is demonstrated.
So yes, phonetically, they can confuse "pri" and "pli" also ("płi"). That was why Schleyer didn't introduce /r/ into Volapük (Chinese mission was strong in the 1880s).
In most cases, this can be cleared by context. "łi", however, has the exactly same usage (3rd Singular Utrum/Masculinum), so if somebody hears "łi estas...", the context is no help to differentiate between "li" or "ri". This problem manifests only in "ri", which is why to my knowledge Riismo didn't catch on outside the Western world.
"Blueprints to Babel" derived "gi" from the utric præfix "ge-". In 1967, this was already derived independently by Manuel Halvelik who created a register for literary translation (Arĥaika Esperanto).
Comment: Some dialects of Chinese, including Mandarin, have both /l/ and rhotic phonemes; and standard Chinese has a phonetic process called erhua that produces rhotic sounds.


Are right handed neutrinos actually antineutrinos and vice versa?

Question: Is it experimentally ruled out that right-handed neutrinos are actually antineutrinos, and left-handed antineutrinos are neutrinos ?
Comment: A massive right handed neutrino turns left handed if you chase it fast so that the velocity turns around, and in the rest frame, such a particle would not be handed.
Comment: To deal with the issue that Ron brings up replace "right handed neutrino" with "neutrino having positive helicity" and "left handed neutrino" with "neutrino having negative helicity".
Answer: Short answer: Unknown
Slightly longer answer: the situation you describer would obtain if neutrinos were Majorana particles (and thus not Dirac particles). It is favored by theorists because it feeds into a nice explanation of why the neutrinos are so light by comparison to the other massive particles.
Experiments are underway that might settle the question by detecting so called "neutrino-less double beta decay" reactions which are forbidden for Dirac neutrinos but allowed for Majorana neutrinos.
June 2012: One of the neutrino-less double beta decay experiments, EXO, has announced modestly significant limit on the possible masses of the neutrinos if they are Majorana particles. This limit excludes some of the theoretically favored scenarios. Right now the significance is too low for a lot of fanfare, but this might be a the first indication of a surprise.
January 2013: KamLAND-Zen has reported a slight improvement on the EXO result as of the end of their first run, with purification of the working material planned before resuming data taking.
July 2013: EXO released another paper last month, setting a much beter measurement of the decay with neutrinos:
$$ T_{1/2}^{2\nu\beta\beta} = \left( 2.172 \pm 0.017\text{(stat)}\pm 0.060\text{(sys)} \right) \times 10^{21} \text{ years}\,,$$
but they do not update their limit for neutrino-less double beta decay, so still no definitive answer on Majorana/Dirac.
August 2013: GERDA has published a report on a low significance null result in the July 17 issue of Phys. Rev. Lett. (also as arXive:1307.4720). Like the EXO and KamnLAND-ZEN results this isn't a strong result by itself, but the combination of all three is starting to look pretty significant.
Comment: The short answer should be "theoretically yes, experimentally inconclusive", this is the content of the rest of the post.
Comment: The [seesaw mechanism](http://en.wikipedia.org/wiki/Seesaw_mechanism) - to which you appeal when you say the Majorana nature of Neutrinos may explain their tiny masses - needs right-handed neutrinos as separate fields from the left-handed ones. It is the right-handed neutrinos that then have the Majorana character. So it is wrong to say that right-handed neutrinos are simply anti-neutrinos. While right-handed neutrinos and right-handed anti-neutrinos would be indistinguishable, the difference can be made for the left-handed ones, as they are part of an $SU(2)$ isospinor.


Precooking sweet potato fries for quick warm/serve day of party

Question: I want to serve sweet potato fries at a party (80 guests) Is there a way to precook or partially cook them the day before, so I can just reheat them on the day of the party to serve warm and crisp?
Comment: It's almost standard procedure to soak and blanch white potato fries before the real cooking. I never realized how complicated making the perfect fry can be! Hopefully there's an expert that knows how white potatoes and sweeties differ.
Comment: We don't really answer questions about "health" topics. We can address whether it's possible and what method to use but whether or not it's "healthy" is outside our scope.
Answer: As user36802 noted, cooking fries normally requires blanching the fries beforehand, which will par-cook them.
You can also check out this Food Lab article; in it, Kenji Lopez suggests parboiling the fries, then drying them and frying them once, followed by cooling and freezing them overnight. His recipe continues to fry the fries a second time before serving, but you could theoretically reheat them in the oven instead. I would assume that most of the moisture has escaped the fries after the first fry, so ideally you wouldn't notice much of a texture difference between twice-fried and fried-then-baked fries, but I haven't tried that method myself.
Also good to note that his recipe is for regular-potato fries; however, the technique should work just as well for sweet potato fries (in fact, I'd assume that since sweet potato fries tend to be more moist than regular fries, any texture difference from frying then baking should be less noticeable).
Comment: I'd actually make kind of the opposite assumption for sweet potato fries; since they hang on to more moisture, I think they'd probably lose some of the initial crispness from the first frying, and would benefit more from a second round of frying immediately before serving. Looks like I'll have to make a big batch for testing. Shame.
Comment: Haha, we all appreciate your sacrifice. Let me know how the test turns out!


What is this leg accessory that Noelle wears?

Question: In Black Clover: Quartet Knight, Noelle wears an accessory on her legs.

What is that clothing thing on her legs called?
Answer: Aren't those just called leg warmers? Other characters like Hinamori Amu (Shugo Chara), Kagamine Rin and Len (Vocaloid), and Rock Lee (Naruto) are known to wear them.
Answer: In her case, they are Genie Leg warmers.
Not just leg warmers, because they are made of a lightweight transparent material.
Most probably Silk or Organza fabric.
You can find some examples of them on Etsy https://www.etsy.com/ca/listing/549549557/custom-color-organza-genie-leg-warmers
I do not believe they are historically accurate, or as easy to wear as Anime characters make it look.
More like something fantasy creators made to make Genies/Magical characters cuter/sexier.
A bit like super huge swords, weighting more than a car, that heroes can lift effortlessly with one hand.
They go for the look more than the realism.
Answer: As per cosplay site, it's listed as 足カバー.
In short, it's more of a "foot cover", as opposed to something like loose/knee-high socks seen as a set with school uniforms.
I believe their purpose is partially for utility to cover the leg, but at the same time allow someone to wear more fashionable shoes or sandals. But then again I don't know what fashionable, maybe it's just for being a cute play on the magic school uniform motif.


the ratio of training set, validation set and testing set in K-fold cross validation

Question: Suppose I have 100 samples, then I want to use 5-fold CV, is the ratio of training set: validation set: testing set is 84:16:16? Is the number of data in validation set should always equal to testing set?
Answer: For K-fold cross validation, you would not need a separate validation set. In your example of 100 samples, if you have train:test set ratio to be as 80:20 and follow the 5-fold CV algorithm, the train set (80 samples) will be split into five 16 sample subsets ($s_i$) as in $\{s_1, s_2, s_3, s_4, s_5\}$. At each iteration, the model will select any 4 subsets out of the 5 subsets, based on your selection criteria, and use the left-out set as the validation set.
Once the training is completed, you can test the performance of the final model with the held-out test set (20 samples) to get an unbiased estimate of the performance of the model.
It is not necessary to have same number of samples in validation set and test set.
Comment: Many thanks for the reply. But in the example, you seem treated the validation set as the test set, is it correct?
Comment: Yes, in K-fold cross validation, validation set plays the role of a test set within each iteration.
Comment: Thank you so much :)
Answer: To answer this we need to understand what is the purpose of the all kinds of sets:
-Training set -we use it for learning purpose.Our model learn from it
-Validation set-we use  to tune the parameters (for example to choose the number of hidden units in the NN).We do not learn from it.
-Test set -we use it to evaluate our model.We don`t learn from it
Many a times, people first split their dataset into 2 — Train and Test. After this, they keep aside the Test set, and randomly choose X% of their Train dataset to be the actual Train set and the remaining (100-X)% to be the Validation set. My personal preferences are 70-10-20,but it depends how much data and what kind of data do you have .
Comment: Hi, I always heard that train:val:test should be equal to 60:20:20 or 80:10:10 or 70:15:15. Sounds like the validation set and test set should be equal size, is that true?
Comment: @sherryxu -the train-test-validation split ratio is also quite specific to your use case and it is not necessary validation and test set to be the same .For more detail explanation please check https://towardsdatascience.com/train-validation-and-test-sets-72cb40cba9e7
Comment: Many thanks! I got that.:)


Can I claim UI benefits in NY if I moved few months ago?

Question: I started working in NY state in December '13, and for various reasons I'm currently unemployed. Can I still claim unemployment benefits, or 4 months are not enough for it?
Comment: You have a many unknowns in your question but here is a place to start: https://www.labor.ny.gov/ui/claimantinfo/beforeyouapplyfaq.shtm#0
Comment: Apply. Answer the questions truthfully. The worst they can say is no.
Comment: What did you do before December 2013? The time worked and the money earned in a previous state may also apply to your qualification situation. If you collected unemployment for that previous period that may also factor into your situation.
Answer: Please refer to the Claimant Handbook, pp 6-7, for New York Unemployment Insurance:

To qualify for benefits,   you must    meet    all three   of  the following   requirements:

You  must    have    worked  and been    paid    wages   for covered employment  in  at
  least two calendar quarters in  your    Basic   or  Alternate   Base    Period.
You  must    have    been    paid    at  least   $1,700  in  wages   in  one of  the calendar    quarters
  in  your    base    period. Effective   January 1,  2015,   you must    have    been    paid    at  least
  $1,900  in  wages   in  one of  the calendar    quarters    in  your    base    period.
The total   wages   paid    to  you in  your    Basic   or  Alternate   Base    Period  must    be  at
  least one and one-half times your high quarter wages.   Your    high    quarter
  wages   are the wages   paid    in  the calendar    quarter in  which   you were    paid    the
  most money. Exceptions: If  your    high    quarter wages   equal   $8,910  or  more,   your    wages   in  the other   base    period  quarters    must    total   at  least   one-half    of  $8,910, or  $4,455. Effective   the first   Monday  of  October 2014,   if  your    high    quarter wages   equal   $9,240  or  more,   your    wages   in  the other   base    period  quarters    must    total   at  least   one-half    of  $9,240, or  $4,620.

[Available in multiple languages.]
If you worked four months, then you worked in at least two quarters.  The Basic Base Period is the first four quarters of the five-quarter period immediately prior to your claim.  If your claim begins this quarter, then that means the period Jan-Dec 2013, so you won't have 2 worked quarters.  But under the Alternate Base Period, the last four quarters before the quarter your claim begins, you have Q1 '14 and Q4 '13.  So you have the two worked quarters.
You must have made at least $1,700 in a quarter.  And you must have made no more than 2/3 your wages from your highest paid quarter.  This latter rule may be tough for you if you were employed Dec to Apr, so you may be in a situation where 3/4 or more of your income from the Alternate Base Period came from first quarter of 2014.  The $8,910 rule does not help you out, even if you made that much in January through March, because you only have two worked quarters.
Work out the details based on your situation.  I suspect that you will not qualify because of the income distribution rule #3.


Imap search in gnus incorrectly returing empty group

Question: I currently have three email accounts I use and access with gnus. After
a recent upgrade to emacs 28.1 I lost the ability to search
the mail for one of these addresses. They are all initiated similarly with a
gnus-secondary-select-methods of 'nnimap. They all three retrieve mail
and allow viewing and sending. For two of them the new parsed
search queries work perfectly, and for one of them (microsoft exchange server
2016 imap 4) it always returns an
empty group. I thought it might be something funny with that server so I
logged on to it via telnet and did the search that way and verified it
correctly finds the correct messages. Turning gnus verbosity up to 10 does not reveal any error messages. Any idea what could be causing this idiosyncratic behavior? I also logged onto a different computer running emacs 27 and for the same email account imap based search works fine.
Answer: Sounds like a bug in Gnus, or a bug in the IMAP server you are using. You should report it to the Gnus maintainer by running M-x gnus-bug.
Comment: @grszkthfr The bug report was:  #55214 . The fix is to find the gnus-search file and replace "UID SEARCH CHARSET UTF-8 %s" with "UID SEARCH CHARSET US-ASCII %s" then recompile the source. Of course this way you have to repeat the action everytime you update your emacs version. Hopefully, they will figure out a longer term fix soon. I think they are working on it.
Comment: Reporting the bug led to me discovering that gnus is sending the imap command with the UTF-8 charset, but the MS Exchange server only suppports US-ASCII. You can change this manually in the gnus-search.el.gz file and recompile. Hopefully a better longer term fix will be implemented at some point.
Comment: That definitely sounds like a bug in Exchange :). Glad you figured it out though.


What is the name of this connector?

Question: A rectifier I salvaged from a broken oasta machine uses am odd plastic barrel connector. On the rectifier end, it looks like a banana pin housed in a translucent white plastic barrel. On the cable end, it looks like a tube. The diameters of the terminals are smaller than a standard banana connector.  Both terminals snap together on two ridges.
What is the name of this connector?
Comment: A regular bullet connector doesn't fir.  This one has a smaller diameter and locks in using the tabs.
Comment: Looks like a bullet connector to me.
Comment: Yeah, I thought that, but haven't seen this one.
Answer: It's called a custom connector.
common  Shell, and Pin Socket Contact for each side.
One of these pins may fit your dimensions when you find the single wire interlock shell
Comment: Thanks.  I found it.  It is made by Molex part # 1625-1PRT.  It is hideously expensive $16 for 1 at amazon. I guess I will use standard terminal connectors.  Kinda sucks because they are really cool and keeps things tidy.

What is the name of this connector?


Turning an iPad app into a universal app

Question: I've created an iPad app, and want to turn it into a universal app (i.e. one app that works for iPhone and iPad). The Apple docs seem to only talk about the "Upgrade Current Target for iPad" option. Is there an option to go the other way?
Answer: I don't know if there is a better way to do this, but I think I would start by making a new blank universal app, and then copy in your iPad code/nibs/plists/etc.
Comment: Brute force method - definitely one way to go. Hoping there's a better way than recreating the xcode project.
Answer: Open the project in XCode and in the Project menu click "Edit Active Target (...)".
Select "All Configurations" for Configuration.
Under the Deployment heading you'll find "Target Device Family". you can use the search box to locate it.
Edit and select iPhone/iPad to make the application universal.


How to grant certain apps root permission through ADB?

Question: I have an Android TV box (4.4.2, armv7), and I can login the device as root by ADB (local network). What I want to do is to give certain apps root permission. How can I do it?
Answer: You don't give certain apps root permission, but certain apps ask for root permission by executing su binary. It's the developer of the app who decides if the app would perform some tasks with root privileges or not. In order to let apps request root privileges, you need to root your device using e.g. Magisk, SuperSU etc. (whatever works on your device).
When you get root privileges on adb (e.g. by executing adb root or adb shell su) it runs adbd process on device as root (UID 0) user. So all commands which you run through adb have root privileges. To grant root access to apps you need a special su binary which is not shipped with Android.
The usual su binary (like on any UNIX-like OS) makes a transition to UID 0 by making setuid syscall to kernel. This is not possible on Android due to certain restrictions. The special su binary (the one provided by rooting solutions like Magisk) connects to a background daemon to get a root shell for the requesting app, where the app can execute commands with root privileges. However the app itself is still running with its own UID (assigned at the time of installation).
For more details see:

What special privileges "/system/xbin/su" does have w.r.t. root access?
How Magisk works?
How to manually root a phone?
Comment: I don't understand this answer. If you can `adb root` and you run have root privs, why can't you run the app? That doesn't make sense, you grant root access to anything root runs. Why can't someone with root access run an android app?
Comment: @punnpkin you need to root your device, using e.g. Magisk or SuperSU. The `su` binary already provided in Android cannot be used with SuperSU. They provide their own `su` binaries. That's what I tried to explain in the answer. For more details see linked answers.
Comment: @EvanCarroll because Android apps cannot make use of SUID bit. Please read "How Android restricts root access?" in the linked answers.
Comment: Thanks for ans, and I found there is a file named `su` which's location is `/system/xbin` and I run `chmod 777` to this file. Then I install supersu on the box but it remind "su binary occupied". So do I have a solution to solve it?
Comment: OK, I learned it, Actually, I'm just looking for a way to root my device without changing the rom. (Because it has some customized content.)


ADO NET Destination has failed to acquire the connection with the following error message: "Login failed for user

Question: Created a SSIS package, it is running fine when I run it on premise but when i try to execute it from sql server agent it gives me error for login failed for user, I am using SQL authentication to connect to the Azure database which are working fine on premise but erroring out on Job agent
This the Error I am getting:
<code>Server Execute Package Utility Version 14.0.1000.169 for
64-bit Copyright (C) 2017 Microsoft. All rights reserved. Started: 10:59:46 AM Error: 2019-06-07 10:59:47.71 Code: 0xC0208452
Source: Data Flow Task ADO NET Destination [2] Description: ADO NET Destination has failed to acquire the connection
 {F51F409E-2990-4C40-8DFF-0B078AF481D4} with the following error message: "Login failed for user 'SVCASQLDW'.".
 End Error Error: 2019-06-07 10:59:47.71 Code: 0xC0047017
Source: Data Flow Task SSIS.Pipeline Description: ADO NET Destination failed validation and returned error
 code 0xC0208452. End Error Error: 2019-06-07 10:59:47.71 Code: 0xC004700C
Source: Data Flow Task SSIS.Pipeline Description: One or more component failed validation. End Error Error: 2019-06-07 10:59:47.71
Code: 0xC0024107 Source: Data Flow Task Description:
There were errors during task validation. End Error DTExec: The package execution returned DTSER_FAILURE
  (1). Started: 10:59:46 AM Finished: 10:59:47
 AM Elapsed: 0.875 seconds. The
  package execution failed. The step failed.
</code>
Answer: You need to be sure that your connection string has password for your SQL Login or add configuration file for SSIS.


Email to contact group suddenly bouncing as "too many addresses"

Question: I have been sending for many years a daily message to a group of 199 names.  Yesterday, the group was rejected for "too many addresses".  I broke the group into 4 groups with 50 names.  The message is still bounced back.  How do I fix?
Comment: A solution might be to create a google group.
Comment: This may be related to the anti-spam measures announced by Google the other day along with their new postmaster tools.
Answer: From Message bounced due to sending limit - Gmail Help

Are you a Google Apps customer? Find out more about Google for Work
    sending limits or SMTP relay service sending
    limits.

In an effort to fight spam and prevent abuse, Google will temporarily
  disable your account if you send messages to more than 500 recipients
  or if you send a large number of undeliverable messages. If you use a
  POP or IMAP client (for example: Microsoft Outlook or Apple Mail), you
  may only send a message to 100 people at a time. Your account should
  be re-enabled within 24 hours.
If you communicate with the same group of people on a regular basis,
  you might be interested in Google Groups.
If you send a large number of undeliverable messages, we suggest
  verifying your contacts' email addresses. It's also important that
  everyone you are sending mail to is willing to receive it.
Learn more about
  best practices for sending a large amount of mail through Gmail.


Mentioning Chanukah in Al Hamichyah

Question: When Al Hamichyah is recited on a holiday or on the sabbath, it has a short passage inserted relevant to the day. However, there's no addition to Al Hamichyah for when it's recited on Chanukah: Chanukah is not mentioned at all in the blessing. Why not?
Answer: The Minchas Ani says that Al Hanisim when said in Birkas Hamazon, is considered an insert. It is not a special Beracha in the Birkas Hamazon. But, if we would add an insert in the Al Hamichya, people might come to think that Al Hanisim deserves its own Beracha.
See this link from Rabbi Eli Mansour for more detailed answer
Comment: Somone just told Rab Plagi not Chaim but his father said one of the reasons we Have Latkes on Chanukah is Because Al Hamichyah lacks a reference to Chanukah we Eat Shehakol foods because he thinks it Should.
Answer: A couple of answers from a couple of Soloveitchiks:
R' Chaim Soloveitchik explains that "beracha me'ein shalosh" (al hamichya) only includes those components of bentching which get a beracha. The additions that we add into al hamichya get a beracha to make up for them if they are forgotten in their proper place and remembered right after boneh yerushalayim before starting hatov vehametiv. This also explains the addition of the placement of the references in al hamichya - they are after "ובנה ירושלים עיר הקדש וכו", which corresponds to the conclusion of the 3rd beracha, and before כי אתה ה' טוב, which corresponds to the 4th beracha, as they are in place of the blessings that are made between the 3rd and 4th beracha.
R' Yosef Dov Soloveitchik explained that al hanisim is a modified version of nodeh l'cha, as opposed to yaaleh v'yavo and retzei which are addendums of an external nature. Al hanisim becomes part of the actual beracha of nodeh l'cha and therefore is included in the corresponding words of al hamichya which summarize that paragraph. As opposed to the other additions, which are external additions, are not represented without being explicitly included.
Answer: Mishna Berura OC 208, sk 9 in name of the GRA that even for Birkat Hamazon there is no need to mention the Al Hanisim,  for Meen Shalosh,  people have not the custom to  add (adding here is regarded as a stringency) so,  we cannot add it
In Bet Yosef we found in name of the Yerouchalmi to "mention some words linked to the event",  but Tosfot seems to think that this Yerouchalmi wasn't ruled for the halacha. Tosfot has an other explanation, that this mention was mandatory at the time of Chazal because they were using wine or other Erets Israel important fruits for whole meals. But nowadays we don't make meals with nothing except bread,  so,  to mention the event is less important We understand that the duty to mention the event is to mention it in the important meals. Diverse customs are mentioned in Rishonim.  The guideline is furnished by Rashba: "See what is going among people".
According to all poskim people did not mention Chanuka.
The last seems a weak explanation.  The first is more rational.  But following the both,  we see that  the fact that someone looks at any cost to mention Chanuka is not  right

Mentioning Chanukah in Al Hamichyah


Breit-Wigner Peak - Not quite a Lorentzian?

Question: The non-relativistic Breit-Wigner peak is given by:
$$\sigma_{fi}=\frac{\pi\hbar^2}{q^2} \frac{2j+1}{(2S_1+1)(2S_2+1)} \frac{\Gamma_i \Gamma_f}{(E-Mc^2)^2+\Gamma^2/4}$$
where $q$ is the center of mass momentum of one of the particles and $E$ is the center of mass energy. From what I am aware $q$ is a function of $E$ (changing the center of mass energy will change the center of mass momentum) typically $q\propto E$ so we would get:
$$\sigma_{fi}\propto \frac{1}{E^2} \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$
however in every source I can find (e.g. Martin, 2012 pg 27) it is stated that
$$\sigma_{fi}\propto  \frac{1}{(E-Mc^2)^2+\Gamma^2/4}$$
and is usually compared to been the same as a Lorentzian. My question is therefore what happens to the $E$ dependence of $q$ and why is it not considered?
Answer: You are right that the relativistic Breit-Wigner distribution isn't functionally quite the same as the Lorenzian distribution. However, the approximation that you described and which compares the relativistic distribution to the non-relativistic, Lorentzian one may be extremely accurate.
Well, it is extremely accurate whenever $Mc^2\gg \Gamma$. Note that the distribution is almost entirely concentrated in the interval
$$E=Mc^2 \pm {\rm few}\cdot \Gamma$$
because of the denominator that you kept in your last formula. But when $E$ belongs to this interval, $q$ may be well approximated by substituting $E=Mc^2$.
In other words, the relativistic Breit-Wigner distribution only "changes abruptly" in the same interval where most of the distribution is concentrated and when one accounts for all the abruptly changing factors in this region, the distribution becomes indistinguishable from the non-relativistic, Lorentzian one.
Answer: @LubosMotl answered the question already, I just wanna chip in some physical intuition for why $\Gamma$ is usually small compared to $Mc^2$. The Breit-Wigner-Formula applies to resonances, in fact only isolated resonances. In the case of overlapping peaks the more general Fano or shape resonances become relevant.


How to implement AppFabric Cache using WF and WCF

Question: I am new to Windows Server AppFabric Cache.
Does anyone know how to implement AppFabric cache using AppFabric hosting services and caching services with help of WF and WCF...?
Or is there any good and simple tutorial/ebooks for AppFabric implementation.....?
What is difference between AppFabric hosting and caching services..?
Our requirement is to convert an existing asp.net cache to AppFabric cache in order to reduce DB read/write overhead. Can we achieve this with limited code change...?
Thanks.
Answer: Here is a link to AppFabric product documentation:
http://msdn.microsoft.com/en-us/library/hh351389.aspx
Hey, have you also looked into NCache for your caching needs? NCache is a very feature rich
distributed caching solution for .NET applications and much more powerful than
AppFabric, especially in your case when you want to reduce the database overhead. Whereas, AppFabric has limited set of topologies (architectures), Ncache provides numerous topologies according to your application needs in terms of high-availability and fault-tolerance
I know that NCache provides all of ASP.NET Cache features like Cache Dependency and
SqlDependency which AppFabric does not.


Expected determinant of a random NxN matrix

Question: What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?
Comment: @Ben: The field is probably the reals.  The OP specified the distribution: i.i.d. uniform on 0 and 1.
Comment: OP stands for Original Poster.
Comment: I assume that is what Charles means by "on 0 and 1."  Anyway, as for the original question it might be more interesting to compute the variance.
Comment: What ring or field do the entries come from? What distribution on the entries are you assuming?
Comment: does googling for the words in the title not yield anything you can work with?
Comment: @Charles: That appears to be nonstandard notation, and I would interpret it as a matrix whose entries are each either 1 or 0, more commonly called a 0-1 matrix, a (0,1)-matrix, or a 0/1 matrix.
Comment: Sorry, I misread Charles Roque's comment. Never mind.
Answer: If $N \ge 2$, then the expected value is $0$ since interchanging two rows preserves the distribution but negates the determinant.
Answer: As everyone above has pointed out, the expected value is $0$.
I expect that the original poster might have wanted to know about how big the determinant is. A good way to approach this is to compute $\sqrt{E((\det A)^2)}$, so there will be no cancellation.
Now, $(\det A)^2$ is the sum over all pairs $v$ and $w$ of permutations in $S_n$ of
$$(-1)^{\ell(v) + \ell(w)} (1/2)^{2n-\# \{ i : v(i) = w(i) \}}$$
Group together pairs $(v,w)$ according to $u := w^{-1} v$. We want to compute
$$(n!) \sum_{u \in S_n} (-1)^{\ell(u)} (1/2)^{2n-\# (\mbox{Fixed points of }u)}$$
This is $(n!)^2/2^{2n}$ times the coefficient of $x^n$ in
$$e^{2x-x^2/2+x^3/3 - x^4/4 + \cdots} = e^x (1+x).$$
So $\sqrt{E((\det A)^2)}$ is
$$\sqrt{(n!)^2/2^{2n} \left(1/n! + 1/(n-1)! \right)} = \sqrt{(n+1)!}/ 2^n$$
Comment: There is another simple proof based on Cauchy Binet theorem. Turan published (in Chinese) a formula for the sum of the 4th power. (I think there are simpler proof but I dont know a reference). I am not aware of a formula for higher powers.
Comment: Previous version had many sign errors, I think they are fixed now.
Comment: I don't know what $\ell(v)$ stands for. I don't know what $i$ stands for in the sum over $u$.
Comment: $\ell(v)$ is the length of the permutation $v$. In this case, it only appears in exponents: $(-1)^{\ell(v)}$ is the sign of $v$.
Comment: I don't see an $i$ in the sum on $u$.
Comment: see (https://mathoverflow.net/q/210666)
Comment: There's a sum over $u$ in $S_n$, and the summand involves the number of fixed points of $i$.
Comment: Number of fixed points of $u$. Fixed, thanks.
Comment: @Gil Kalai which Turan's paper does contain the formula for E[(detA)^4]?
Comment: @Machinato See https://mathoverflow.net/questions/137377/a-simple-proof-for-a-theorem-of-szekeres-and-tur%c3%a1n
Answer: It is a little more convenient to work with random (-1,+1) matrices.  A little bit of Gaussian elimination shows that the determinant of a random n x n (-1,+1) matrix is $2^{n-1}$ times the determinant of a random n-1 x n-1 (0,1) matrix.  (Note, for instance, that Turan's calculation of the second moment ${\bf E} \det(A_n)^2$ is simpler for (-1,+1) matrices than for (0,1) matrices, it's just n!.  It is also clearer why the determinant is distributed symmetrically around the origin.)
The log $\log |\det(A_n)|$ of a (-1,+1) matrix is known to asymptotically be $\log \sqrt{n!} + O( \sqrt{n \log n} )$ with probability $1-o(1)$; see this paper of Vu and myself.  A more precise result should be that the logarithm is asymptotically normally distributed with mean $\log \sqrt{(n-1)!}$ and variance $2 \log n$.  This result was claimed by Girko; the proof is unfortunately not quite complete, but the result is still likely to be true.
Comment: Update: the central limit theorem for the log-determinant was worked out carefully by Nguyen and Vu, http://arxiv.org/abs/1112.0752
Comment: By multiplying rows and columns by signs, one can normalise the first row and column to be all 1s.  If one then subtracts the first row from all the others to make the first column (1,0,...,0), the bottom right n-1 x n-1 minor is now a random matrix of 0s and -2s, whose determinant is $(-2)^{n-1}$ times that of a random 0-1 matrix.  The various sign factors one incurs in this process can then be removed by interchanging some rows (assuming $n>2$).
Comment: How exactly was the Gaussian elimination carried over? I mean how can one get from (-1,+1) ran. matrix to another (0,1) ran. matrix
Answer: For some further results of this nature, see Exercise 5.64 of Enumerative Combinatorics, vol. 2. This exercise deals with the uniform distribution on (0,1)-matrices or $(-1,1)$-matrices, but the arguments can be carried over to other distributions where the matrix entries are i.i.d. The proofs are similar to the argument in David Speyer's comment.

5.64. a. [2+] Let $\mathcal D_n$ be the set of all $n\times n$ matrices of $+1$'s and $-1$'s. For $k\in\mathbb P$ let
  \begin{align*}
f_k(n)&= 2^{-n^2} \sum_{M\in\mathcal D_n} (\det M)^k \\
g_k(n)&= 2^{-n^2} \sum_{M\in\mathcal D_n} (\operatorname{per} M)^k,
\end{align*}
  where $\operatorname{per}$ denotes the permanent function defined by
  $$\operatorname{per}(m_{ij})= \sum_{n\in\mathfrak{S}_n} m_{1,\pi(1)} m_{2,\pi(2)} \dots m_{n,\pi(n)}.$$
  Find $f_k(n)$ and $g_k(n)$ explicitly when $k$ is odd or $k=2$.
b. [3-] Show that $f_4(n)=g_4(n)$, and show that
  $$\sum_{n\ge 0} f_4(n) \frac{x^n}{n!} = (1-x)^{-3} e^{-2x}. \tag{5.120}$$
  HINT. We have $$\sum_m (\det M)^4 = \sum_M \left(\sum_{\pi\in\mathfrak S_n} \pm m_{1,\pi(1)}\dots m_{n,\pi(n)}\right)^4.$$
  Interchange the order of summation and use Exercise 5.63.
c. [2+] Show that $f_{2k}(n)<g_{2k}(n)$ if $k\ge 3$ and $n\ge 3$.
d. [3-] Let $\mathcal D'_n$ be the set of all $n\times n$ 0-1 matrices. Let $f'_k(n)$ and $g'_k(n)$ be defined analogously to $f_k(n)$ and $g_k(n)$.
  Show that $f'_k(n)=2^{-kn} f_k(n+1)$. Show also that
  \begin{align*}
g'_1(n) &= 2^{-n} n!\\
g'_2(n) &= 4^n n!^2 \left(1+\frac1{1!}+\frac1{2!}+\dots+\frac1{n!}\right)
\end{align*}
Answer: Is it not zero whenever $n \geq 2$?  Let $A$ be a $n \times n$ permutation matrix with determinant $-1$ (which requires $n \geq 2$).  Then the uniform distribution of a random $n \times n$ $(0,1)$-matrix $X$ is the same as the distribution of $AX$.  The determinant is multiplicative, hence Det$(AX)=$Det$(A)$Det$(X)=-$Det$(X)$.  Hence the probability of Det$(X)=x$ is the same as the probability of Det$(X)=-x$.
Comment: That's somewhat of a Freudian slip (since I typically use permanents rather than determinants).  Thanks, I fixed it.
Comment: The permanent isn't multiplicative.
Answer: Unless I'm missing something, this also follows immediately from linearity and multiplicativity of expectation, treating each entry as independently $0-1$ with probability $1/2$.  Every permutation yields the same expected value in the sum, $\pm (1/2)^n$ depending on sign, and the number of even and odd permutations is identical (for $n \ge 2$, as noted above).
It's probably worth mentioning that an old result of Komlos shows that despite this, the probability the determinant is actually 0 is $o(1)$.
Answer: Miodrag Zivkovic has actually done a classification on small orders of 0-1 matrices by  rank and absolute determinant value.  You may be interested in the tables in his Arxiv paper http://arxiv.org/abs/math.CO/0511636 .
Gerhard "Ask Me About System Design" Paseman, 2010.01.26


Automorphism group of directed complete graph

Question: Given a directed complete graph on $n$ vertices, is there an efficient algorithm for computing its automorphism group? Is there a nontrivial upper bound on the order of its automorphism group? How about the isomorphism of two such graphs?
Answer: I'm guessing that by "directed complete graph" you want each edge directed in exactly one of the two possible ways.  If so, you have a tournament.  Automorphism and isomorphism for tournaments is potentially easier than for general directed graphs but nobody has proved that. One thing to note is that the automorphism group has odd order and is therefore solvable, which may facilitate group-theoretic approaches.
For an upper bound on the group size, take $n!$ and remove from it the highest power of 2, but probably that is way too large.  A better bound would be the answer to this stackoverflow question.
The best known algorithm needs time $n^{O(\log n)}$, which is better than the best known for general digraphs.
A paper which studies these questions is here.
Answer: As to an algorithm: There is Tarjan's algorithm for computing strongly connected components. That should certainly be a first step in every such algorithm because every automorphism will permute strongly connected components. In fact in a complete graph the components are linearly ordered so that any automorphism will in fact map all components onto themselves. Moreover: If $\Gamma = \Gamma_1 \sqcup \Gamma_2 \sqcup \ldots \sqcup \Gamma_k$ is the decomposition into components in descending order then by completeness all vertices in $\Gamma_i$ will be have an arrow pointing toward all vertices in $\Gamma_j$ if $i<j$. Therefore $Aut(\Gamma) = Aut(\Gamma_1) \times Aut(\Gamma_2) \times \ldots \times Aut(\Gamma_k)$ which reduces the problem to finding the automorphism group of a strongly connected complete graph.


Books for studying Dirac Operators, Atiyah-Singer Index Theorem, Heat Kernels

Question: I am interested in learning about Dirac operators, Heat Kernels and their role in Atiyah-Singer Index Theorem. From various sources (including this very helpful question), I have come to know of various references :
(i) Spin Geometry by Lawson & Michelsohn
(ii) Heat Kernels and Dirac Operators by Berline, Getzler and Vergne
(iii) Dirac operators and spectral geometry by Giampiero Esposito
(iv) Invariance Theory, the Heat Equation, and the Atiyah-Singer Index Theorem by Peter Gilkey
and (v) The laplacian on a Riemannian Manifold by Rosenberg,
however I am having difficulty deciding which one or two to study between these. I prefer reading books that start from basics but eventually cover the core aspects of the subject at fairly advanced level. Please advise which one or two of the above should I study such that the intersection of the selected books is minimal and the union is maximal. Any comments about the above-listed books will be very helpful.
My background is : Analysis (As covered in 'Principles of Mathematical Analysis' by Walter Rudin but not much of measure theory), Algbera ( 1 year grad level course based on Serge Lang), Manifold Theory and Differential Geometry (Differential forms, de-Rham theory, Riemannian metrics, Geodesics, Connections, Curvature, Vector bundles & Characteristic classes, Principle bundles ) but very little functional analysis (definition and basic properties of Banach and Hilbert spaces, and the four famous theorems) and almost no Algebraic Topology beyond the definition of fundamental group. By the time I plan to start studying ASIT, I am expecting to have studied Representation theory of Lie groups and Lie Algebras and may be a little bit of Clifford Algebras as well.Please advise what other prerequisites do I need to study the books listed above.In particular, do I need to learn more Functional Analysis and Algebraic Topology for this purpose ? I have studied the prefaces of the books I have listed but unfortunately the information about necessary prerequisites is not mentioned in most cases.
   Any other good references are welcome.
Note: I had earlier asked a related question here
Comment: You can try my course notes http://www3.nd.edu/~lnicolae/ind-thm.pdf  They're modeled after Roe's book but  I go at a slower pace.
Comment: I think you need to take a PDE course.
Comment: I think Taylor's second volume would be a good start.
Comment: What about "Elliptic operators, topology and asymptotic methods" by John Roe
Comment: @timur Thanks. Could you please specify which topics in PDE I need to learn and some good text which I could study for this ?
Comment: @EricO.Korman Thanks! I looked at it and it seems useful.
Answer: I think the best introduction is "Dirac Operators: Yesterday and Today", the proceedings of a summer school edited by Bourguignon, Branson, Chamseddine, Hijazi and Stanton. It does not cover heat kernels, but gives a good understanding of the other topics - and does start from basics; there is a (short) chapter on differentiable manifolds. Also I think a good understanding of the contents here is prerequisite to getting much out of, say, the Lawson-Michelsohn book.
Beyond the prerequisites you mentioned, I think the only thing necessary is a certain understanding of cohomology groups. To start learning about heat kernels, timur is probably right in saying you should first learn some PDE.
Answer: In have worked for a longer time with the mentioned books of S.Rosenberg (v), P.Gilkey (iv) and Berline, Getzler and Vergne (ii). Since my interests were more related to heat kernels than Dirac Operators I want to comment from this point of view.
The book of S.Rosenberg is excellent to start with. It deals with the basics though trying to touch topics are quite complicated, so it is very good to start diving in.
(ii) is a great source, too. But it's on a more advanced level. I would recommend keeping this book inside your mind but start reading later, if you feel very familiar with the basics - even I can't precisely define, what 'basics' mean.
Finally, (iv) is very often cited directly and belongs to the 'canonical' literature to the topics related to the heat kernel. Definitely worth reading - but don't expect do read it from end-to-end, I would recommend you to pick chapters you are immediately interested in, since it contains a wide range of topics.
edit: If you're seriously working on these topics, you'll be faced with functional analysis automatically, it is an important fundament. I also had no 'further' functional analysis education before, but for me it worked learning parts of it 'on the fly' while considering concrete problems. So don't hesitate if you never heard a special lecture. The book if Rosenberg e.g. does also contain very much of useful functional analytic material.
Comment: I think there definitely will be topics you don't know yet, but simply give it a try. Don't expect to read it 'straight'. Start reading a chapter you're curious of and switch to other sources if you start feeling lost. I would recommend you to start with the chapter about Pseudodifferential operators. It's also very useful to compare the mentioned sources while reading. Gilkey mostly handles a more general case of operators, so it is a good idea to compare with methods presented in the book of Rosenberg for the Laplacian on functions.
Comment: Thanks, this is very helpful. Just one more question : I have had just an undergrad course on Functional Analysis (basic definitions,examples and properties of Banach Spaces, Hilbert spaces, compact operators, and the four famous theorems), can I begin reading Gilkey's book straight away or there are topics in functional analysis which I should learn before starting the book ?
Answer: Another worthwhile book that hasn't been mentioned is Liviu Nicolaescu's Lectures on the Geometry of Manifolds which is freely available from his website (under papers'n stuff). While it doesn't cover the Atiyah-Singer Theorem itself, it does a great job of addressing some of the prerequisite material such as elliptic operators (including Dirac operators).
Answer: I have studied the whole book of Rosenberg and I can asure you that it is a very good choice to start with, I have studied some parts of the books of Gilkey and Lawson-Michelson too and I can asure you they are quite advanced, it would be hard to begin with them. I guess the book of Lawson-Michelson is better for you because it covers the classical proof of the Atiyah-Singer index theorem, which depends more on geometry, while Gilkey covers de analytic proof of the theorem, which depends very much on analysis. Anyway, if you prefer to choose the book of Gilkey, be sure to get the second edition, since it is fairly better than the first one.
Remark: The book of Gilkey starts in the context of elliptic complexes, which is more general than the context of Dirac operators and Dirac complexes


How to track user logins

Question: I need to track user logins so I know how often "UserA" is logging into my site.
Can anyone recommend a method to do this? Module or code?
Answer: I think you are looking for User Stats module.

Provides commonly requested user statistics for themers, IP address
  tracking and Views integration. Statistics are:

days registered;
join date;
days since last login;
days since last post;
post count;
login count;
user online/offline;
IP address;
Comment: @dibs You can try and check it yourself ;-)
Comment: Does it give each users logins by date?
Comment: No. It doesn't ;-)
Comment: then try this module : http://drupal.org/sandbox/AshishThakur/1614582
Answer: Just to close off this question I will share how I managed to track the user login times.
I simply wrote a module that added a table to the db and on login I would add the login/user event + date-time  to the table.
Then I was able to access how many times and at what time users were logging in. This was something I wasn't able to get from the User Stats module.
Comment: Could you post the code or some elaborate on how you did it? Not sure where to begin w/ doing the same.
Comment: @jay-ess which part don't you understand? Timestamp and UID are recorded to the custom database table on login. This info can then be used to analyse login counts and time patterns for users.
Comment: k. Will get working on which hooks, etc, and post some code if I can make it work. Thanks.
Comment: You're right, "User stats" is not a solution for your question. But maybe you want to have a look at my answer to [this question](http://drupal.stackexchange.com/questions/35484/how-to-grant-userpoints-to-users-who-perform-a-login-5-days-consecutively/211279#211279). If you'd change the "5" in it to eg 9999 (and possibly some other task configuration options), you might be close.
Answer: use User Track module

A simple module that tracks each time a user logs in and stores a
  historical record in your database. This data is then exposed to views
  so you can report on it if required.
Logs minimal data: user ID, and time logged in Ability to exclude
  users from tracking by role Hooks to allow fine-grained control over
  which logins get tracked Hooks to allow you to add information to the
  data stored Views integration

it's available on both Drupal 7, 8

How to track user logins


Wcf Friendly PagedList

Question: We have been working with various versions of the PagedList , however these pagedlists are not Wcf friendly. They do not serialize. Does anyone have a good way to get paged data across Wcf? (Including properties such as TotalRecords of the datasource.)
Answer: We worked out a solution to simplify the current PagedList to something that is more serializable and a little more geared to our solution. Instead of creating a custom List/Collection, we created a class that contains the list/collection.
Previously we were losing the custom properties of the pagedlist when it was deserialized on the client side. It still had all the items in the list, but it was missing the properties.
This is based on code by Troy Goode which I believe is based on some unreleased Microsoft MVC code. More info can be found here: http://www.squaredroot.com/2009/06/15/return-of-the-pagedlist/
<code> public class PagedList<T> : IPagedList<T>
    {
        public PagedList()
        {
        }

        public PagedList(IEnumerable<T> source, int pageIndex, int pageSize)
            : this(source == null ? new List<T>().AsQueryable() : source.AsQueryable(), pageIndex, pageSize)
        {
        }

        private PagedList(IQueryable<T> source, int pageIndex, int pageSize)
        {
            TotalRecords = source.Count();

            // add items to internal list
            if (TotalRecords > 0)
                if (pageIndex == 0)
                    Data = source.Take(pageSize).ToList();
                else
                    Data = source.Skip((pageIndex) * pageSize).Take(pageSize).ToList();
        }

        public int TotalRecords { get; set; }

        public IEnumerable<T> Data { get; set; }
    }
</code>
and then the actual implementation through the following extension:
<code>public static PagedList<T> ToPagedList<T>(this IEnumerable<T> source, int pageIndex, int pageSize)
{
    return new PagedList<T>(source, pageIndex, pageSize);
}
</code>
Comment: I've been trying to use the code in your solution to serialize the IPagedList results from my WCF service on the client but with no success, can you please help me out? I'm not sure if its ideal for me to create a new question to explain in problem in-depth.

Wcf Friendly PagedList


How do I block multiple users on MediaWiki?

Question: My wiki collected a lot of spam over the past few months, so I've blocked it to only registered users. Unfortunately I never turned off user registration.
Is it possible to block a lot of user accounts at once, or should I just restart?
Comment: Do you mean delete the user accounts?
Comment: That would work for me. I think MediaWiki may philosophically be designed to favor blocking over deletion.
Answer: The quickest and least invasive way to prevent a big list of registered users from making modifications is probably to do a bulk password change for the unwelcome users.
Assuming you're running a Unix-like, and that the list of folks you want to keep is relatively low:

Disable anonymous viewing and/or editing. (which it sounds like you've already done?)
Dump a list of users from the database to a file.

eg., for MySQL:
mysql -p -u root -B wikidb -e 'select user_name from user;' | tee user.list

Remove the users you want to keep from the user.list file.
Use the MediaWiki password command-line tool to change passwords to something random.

eg.
cd [wiki-docroot]/maintenance/
while read wikiuser; do
    echo $wikiuser;
    php ./changePassword.php $wikiuser BOGUS%PASSblahblah77;
done <user.list

Generating a different password for each user is left as an exercise for the reader. :-)
Comment: Yep, a nice list of command-line tools there.  Too bad there isn't a bulk password changer among them.
Comment: Interesting way to do it. It's very cool to see how flexible shell scripts can be. EDIT: found it, a list of command line tools is here: http://www.mediawiki.org/wiki/Manual:Maintenance_scripts
Answer: Connect to your mysql server:

mysql -h DBSERVER_HOST -u wikiuser wikidb -p

Enter your password at the prompt.
Once in mysql prompt, search for the user

mysql> select user_id,user_name from user;

+---------+------------+
| user_id | user_name  |
+---------+------------+
|       9 | spammer    |
|       7 | r2d2       |
|       1 | admin      |
+---------+------------+

Delete every "user_id" you need with:

mysql> delete from user where user_id in(7,9);


Topological Hochschild homology using equivariant orthogonal spectra

Question: In the Hesselholt-Madsen paper "On the K-theory of finite algebras over Witt vectors of perfect fields", the authors develop some results concerning the Topological Hochschild homology (THH) of functors with smash products.
They use the language of equivariant spectra in the sense of Lewis-May-Steinberger.
My question is if there is a more modern reformulation of this work using equivariant orthogonal spectra (or other (one-categorical) variants).
I am quite certain that using the results from Mandell, May, Schwede and Shipley, one can translate the results  from Hesselholt-Madsen to orthogonal  spectra. My Question then is if this has already been done. At this moment, I am not looking for an infinity categorical treatment.
To be more precise, I would like to see the construction of THH and topological cyclic homology using orthogonal spectra.
Preferably I would like to see the construction of the Tate spectral sequence as well, where for a  finite cyclic group $C$ and $T$ an equivariant $C$ spectrum, the Tate spectral sequence is
$$E^2_{r,s}=\hat{H}^{-r}(C,\pi_sT)\Rightarrow \pi_{r+s}\hat{\mathbb{H}}(C,T),$$
where the $E^2$ page is the ordinary Tate cohomology of $C$  and  $E^{\infty}$-page is the homotopy groups of the Tate spectrum.
Answer: This has a partial positive answer.
The theory of cyclotomic spectra using orthogonal spectra is developed in 'The homotopy theory of cyclotomic spectra' by Blumberg and Mandell. See https://arxiv.org/abs/1303.1694.
Moreover, Nikolaus and Scholze have an exposition of the Böckstedt construction of THH using orthogonal spectra in their paper 'On topological cyclic homology'.
See https://arxiv.org/abs/1707.01799.


web2py external libraries

Question: how can i import other external libraries in web2py? is it possible to
load up libs in the static file?
can somebody give me an example?
thanks
peter
Answer: If the library is shipped with python, then you can just use import as you would do in regular python script. You can place your import statements into your models, controllers and views, as well as your own python modules (stored in modules folder). For example, I often use traceback module to log stack traces in my controllers:
import traceback

def myaction():
    try:
        ...
    except Exception as exc:
        logging.error(traceback.format_exc())
        return dict(error=str(exc))

If the library is not shipped with python (for example, pyodbc), then you will have to install that library (using distutils or easy_install or pip) so that python can find it and run web2py from the source code: python web2py.py. Then you will be able to use regular import statement as described above. Before you do this make sure you installed the library properly: run python interpreter and type "import library_name". If you don't get any errors you are good to go.
If you have a third party python module or package, you can place it to modules folder and import it as shown below:
mymodule = local_import('module_name')

You can also force web2py to reload the module every time local_import is executed by setting reload option:
mymodule = local_import('module_name', reload=True)

See http://web2py.com/book/default/section/4/18?search=site-packages for more information.
Answer: In web2py you import external library as you normally do in Python
import module_name

or
from module_name import object_name

I am not sure what you mean by "in the static file"

web2py external libraries


Adding items to scrollview in storyboard (size inspector appears to be locked)

Question: I've added a scrollview with a content size that is larger than the screen size in my view controller using storyboard. Of course the purpose of the scrollview is to have content larger than the view that contains it. I would like to add buttons and label to the bottom of the scrollview graphically in storyboard, but I can't view  that part of the scrollview. How do I get there. I tried adjusting in the size inspector which worked for me when I first added the view a few weeks ago, but it don't let me change anything.
Answer: The only way I've been able to edit my content in a UIScrollView (and this dates back to way before the Storyboard, but I'm pretty sure it still applies) is to change the offset of the  x or y of your content frame.
Another option, depending on whether your content view is "static" (built in the .xib) or not, is to actually pull it outside of the scroll view, and set it as the content view programmatically in viewDidLoad or something.
Hope this helps, I can follow up with code examples if it makes sense and you need 'em.
EDIT (2/10/2012): Here's an example of how to hook up the larger view to your scroll view. In this example, the scroll view (scrollView) is set to the view controller's view property. In the .xib I've also added a UIView named largerView. The two are not hooked up in any way in the .xib file, they just both appear there.
Given...
@property (nonatomic, retain) IBOutlet UIView *largeView;
@property (nonatomic, retain) IBOutlet UIScrollView *scrollView;

In viewDidLoad: we just...
// add the largeView to the main scrollView
[scrollView addSubview:largeView];
[scrollView setContentSize:largeView.frame.size];

This is probably the simplest possible example though. Good luck!
Comment: I went back and looked at this again. I discovered the scroll view that has content larger than the view in storyboard was the top level view of the view controller. When configured like this you can't adjust the x and y offset in the size inspector. However, if a uiview is the top level and the scroll view is placed in that, then the content offset (x and y under "view" in the size inspector) can be adjusted exposing different areas of the scrollview in storyboard so they can be configured. Problem solved!
Comment: Code examples would be great! :)
Comment: Added a code example for hooking a view to a scroll view programmatically.
Answer: It's great that you solved that. If I had that problem I would create a view that has the same frame size as content size of the scroll view. Add all the controls I need (buttons, labels, textFields etc.) on this view at the positions I need. After that just added the view on scroll, make an IBOutlet of the scroll and set it's content size programmatically somewhere in my code, before using that scroll (setting it in viewDidLoad will be fine).
That's another solution of this problem.
Comment: I like that too. I'll give it a go next time I'm in that situation.
Comment: This is what I suggested also.
Answer: @T.J. & @livingtech:
Why don't you use embed property given in Interface Builder?
Here you are creating two views (UIView largeView and UIScrollView scrollView). So its kind of  wasting of memory for the same objects.

First add one more UIView in your nib with your height and width.
Add your controls wherever you want.  (let say largeView only)
Then use embed property with ScrollView. (let say myScrollView only)
Now add this myScrollView in your original view wherever you want with your custom height and width.
In code base just set myScrollView contentSize as per the largeView size.

This would be easier for the code reusability and expandability purpose. If in future you need to add more objects.
Comment: I'm not 100% sure I'm following you here, but I think this is still the same number of objects as what I suggested. I'm pretty sure you always need a `UIView` to be the content view of your `UIScrollView`.
Comment: the OP isn't using a nib. If T.J. were, you could drag a scroll view onto the nib canvas, resize it,etc so you could fit all the views you want in it, then resize it back and either add it to the subview in code, or, drag it back as a subview within the nib using the hierarchy view (you can't do it by dragging and dropping the actual view widgets in the canvas). But, this strategy doesn't appear to work within the storyboard editor.
Answer: You can work on iPad Full Screen size if your content won't exceed the screen size too much.


How can I access my Samba Server from a single windows PC using different usernames?

Question: I already setup Samba on Ubuntu Server. It's working fine, but I would like to have it accessed by any user using their username and password, in any windows PC that a user would choose to use. Please help me how to do that. Is that possible? Any help will be appreciated.
Comment: Thanks @utkuerd. Your answer game me clues and made me create more questions in my mind that forces mo to look for more answers. For that I thank you. I finally got it all working thanks to this post [here](http://www.jadota.com/2013/01/installing-samba4-on-ubuntu-12-04/comment-page-1/#comment-14503), I am now able to use a username with password to any PCs with Micro$#it Windows OS.
Answer: Samba has a built in user authentication mechanism. You can give users permission to access your shares and use smbpasswd to set their passwords. You can also synchronize user's login password with Samba password. Samba Server Configuration tool can help you on this, as illustrated here.
When you set up password for users, they'll be automatically required to enter it to access your shares from any computer that can access your Samba server. If your point is accessing with the same passwords they use on clients then you either need to set same passwords on clients and server, or you need a centralized authentication mechanism based on LDAP, Kerberos, etc.


ESP8266 as wifi Card?

Question: I am very new to arduino and electronics in general. Can i use the NodeMcu Esp8266 as one of those usb wifi dongles? I want to use it as a monitor mode capable wifi card, for my linux laptop.
Comment: Yes. No. Maybe. What you want to do?
Answer: No, you can't. All you have is a UART interface through a USB to UART adaptor. It can only ever be a UART.
The best you could get would be to write a program to run on the NodeMCU that did the monitoring for you (assuming you can get it into a suitable promiscuous mode) and report information back to you through the UART.

ESP8266 as wifi Card?


Application Open Triggers

Question: Are there any alternative ways to open an application quickly besides the 2 prominent ones, url extension and screen tap. Possibly using the volume button or homepage button to initiate the app? Android and Apple devices.
Comment: Are you asking this question as a user or a programmer?
Answer: I'm fairly certain that Apple doesn't allow you to change the functionality of their buttons from a programmer's standpoint. I think there are some private API's that you could use, though it doesn't matter if you intend on deploying your app because Apple would reject it on that basis. I'll try to find some documentation to back up Apple not allowing button functionality to change, and then I will edit my post.
EDIT:
See this question: Temporarily Lock or Disable iphone home button
They give examples of apps(Camera+) that change button functionality and were pulled from the store when these changes were discovered.
I'm afraid I can't speak for anything on Android, though I wouldn't be surprised if it was similar.
Comment: If you're considering this question closed, be sure to accept an answer :)

Application Open Triggers


Android Studio 2.1 freezes when Logcat starts

Question: Sometimes, when I use Android Studio 2.1 and build and deploy the app, Android Studio freezes, and looks like a still image.
It can't responde to useful user interaction: (though it does not change into Not responding state)
I can click:

Close - it does nothing
Minimize - it minimizes the IDE

The last build message is this:

Activating Logcat Tool window

If I kill adb.exe, nothing changes and adb.exe appears again after a few seconds.
What could I do about this? Should I report that?
Comment: Experiencing the same problem here. I guess it's a bug of Android Studio. Why not just report the issue?
Comment: Same here. Android Studio 2.1.3
Answer: I faced the same issue, found out that this was caused by the enormous filter in logcat and solved it in the following way:

Have your Android Studio closed.
Go to the {root project folder}/.idea/workspace.xml
Find the TOOL_WINDOW_CUSTOM_FILTER key and set its value to the empty string.
Start the Android studio again.
Answer: 1) Open TakManager and kill Android Studio.
2) Re Open Android Studio.
3) Go to File, invalidate and Restart Android Studio.
4) Click on invalidate and Restart.