Separating points of closed countable discrete subspace of a $T_4$ space with pairwise disjoint open sets

Question: Let $X$ be a $T_4$ space (i.e.it is $T_1$ and every pair of disjoint closed sets can be separated by disjoint open sets). Suppose that $D=\{d_n\}_{n\in\mathbb N}$ is a countable, closed and discrete subspace of $X$. Can we always find a family $\{U_n\}_{n\in\mathbb N}$ of pairwise disjoint open sets of $X$ such that $d_n\in U_n$ for each $n$? I know this is possible if $D$ is finite, but I would like to know if this happens as well if $D$ is infinite. This question is inspired by what happens with the subspace of integers on the real line.
Comment: It seems that $\{d_n\}$ and $D\setminus \{ d_n\}$ are both closed?
Comment: Clearly yes these are both closed.
Answer: It is true: the map $d_n\to n$ is continuous from $D$ to $\mathbb{R}$.
By the Tietze-Urysohn theorem it has a continuous extension $f:X\to\mathbb{R}$.
Now let $U_n=f^{-1}\bigl[(n-\frac13,n+\frac13)\bigr]$ for all $n$.
Answer: Yes, you can do it as follows, and it suffices that $D$ is discrete and $X$ is $T_3$.
First, chose open $U_1$ with $U_1\cap D = \{d_1\}$.
Let $V_1$ be such that $d_1\in V_1\subseteq \overline{V_1}\subseteq U_1$ which we can find since $X$ is regular.
Next, let $U_2$ be such that $U_2\cap D = \{d_2\}$ and $U_2\cap \overline{V_1} = \emptyset$. Again find $V_2$ from regularity with desired properties.
Continue this process to find a sequence of sets $V_n$ with $d_n\in V_n\subseteq \overline{V_n}\subseteq U_n$ and $U_{n+1}\cap (\bigcup_{i=1}^n \overline{V_i}) = \emptyset$.
Then $\{V_n:n\in \mathbb{N}\}$ is a pairwise disjoint family of open sets such that $d_n\in V_n$.


Change multiple polygons' area from square degrees to square meter in attribute table with QGIS

Question: I imported an ESRI personal geodatabase vector file (.mdb file) into QGIS, when I open the attribute table, it seems the the shape_area unit is in square degree.
I changed the project properties to a projection rather than WGS84.
But the units remain in square degree.
Is there a method that can convert all the data shape_area unit to square meters, then I can save it easily as csv file with square meters instead of square degree?
And should one be clever to choose the correct projection to a country to get the true areas or is there a universal projection?
The file contains administrative divisions.
Answer: Attribute table fields are static. This means that your shape_area field won't automatically update when you change the projection, or if you edit a polygon and its actual area changes.
Here's how to add a new area field in square meters:

Choose an equal-area projection that uses meters as units. Selecting the best projection depends on the size of the area of interest. ESRI provides some helpful information on this topic.
Save your vector layer as a new file with the projection you chose in step 1. QGIS can't create personal geodatabases, so you'll have to choose a different file type. Shapefile is a fairly safe choice if you're not sure.
Use the field calculator to add a new field for area in square meters. Use the expression <code>$area</code> to calculate area in the units of the layer's projection.

If you want a live-updating field, check the box next to "virtual field." A virtual field is only saved in the current QGIS project, so if you load the layer into a new project the virtual field won't be present. However, a regular field is good for most purposes.
NOTE: Different versions of QGIS handle unit selection for the $area calculation differently. As you can see in the screenshot here, in QGIS 2.18 the $area function uses the units set in project properties for the current project. Older versions of QGIS use the units of the layer's projection. It's a good idea to verify the calculated area of at least one polygon by comparing it to the known area. For example, if you have a country polygon, look up the area of that country on wikipedia.
Comment: Updating the existing field containing the degree area would work too.


What is the minimum number of locks on the cabinet that would satisfy these conditions?

Question: Eleven scientists want to have a cabinet built where they will keep some top secret work. They want multiple locks installed, with keys distributed in such a way that if any six scientists are present then they can open all the locks, but if only five are present then they cannot open all the locks. What is the minimum number of locks on the cabinet that would satisfy these conditions?
Answer: Let me show it in the general case as the numbers do not matter much.
We have a safe and $S$ scientists, we want $M$ scientists to not be able
to open the safe but more then $M$ scientists to be able to open it.
How many locks do we need to put on the safe?
Given a set of $M$ scientists out of the $S$ scientists (we call it an
$M$-subset), they are missing a key for some lock.
Moreover two such distinct subsets have a scientist not
common to both and thus their union has more then $M$ scientists.
If two such $M$-subsets are missing the same key then they're union
is missing that key, the union has more then $M$ scientists and thus
we have a contradiction.
We define a multimap as follows, for each lock we define its preimage to be the $M$-subset which is missing a key for that lock, as described above there cannot be more then one M-subset missing that key (A priori we may have locks which no $M$-subset is missing a key for, so nothing will map to them).
Now if some M-subset maps to more then one lock then we can throw away all but one lock, the M-subset will still not be able to open the safe because he is missing a lock.
So we end up with an injection from the collection of $M$-subsets to the set of locks.
Thus we have at least $N=\binom{S}{M}$ locks.
Now we ask if we need more locks then this. Suppose there are more then $N$ locks.
WLOG assume our injection from above maps to the first $N$ locks labeled
$1$ to $N$. So each $M$-subset is missing a key for one of the first $N$ locks.
Now consider the $(N+1)$-th lock which we call $L$, consider the collection
$\mathcal{C}$ of $M$-subsets not having the key for $L$.
Imagine we throw away lock $L$ and all its keys. Any subset of the scientists which
could open the safe before can still open it as we just removed a
lock. The $M$-subsets still cannot open the safe because they are missing
some key from the first N keys. Thus the $(N+1)$-th key is redundant,
By a similar argument any lock labeled with a number > $N$ is redundant.
Thus $\binom{S}{M}$ is the minimum sufficient number of locks.
In the case you mentioned we have $S=11$ and $M=5$.
Thus number of locks we need is $\binom{11}{5} = \frac{11!}{5!6!}=462$
Comment: Hi, I wrote how you would construct the injection. This gives you a map from the M-subsets to which key they do not have, to see which keys each scientist has, you need to look at all M-subsets which include this scientist and each of these gives you a key the scientist does not have. So you could in fact construct from this how the sets of keys for each scientist looks. In general for a scientist you have $\binom{S-1}{M-1}$ M-subsets which include him, this gives you the number of keys he does not have. Subtract this from the total number of keys to see how many keys he has.
Answer: I thought fiftyeight has a great and rigorous answer.
However, I think there is a shorter explanation/proof:
Given S #scientists and M size of subset - Since each subset of size M can't open all the locks, there should be just one lock (to impose minimality) that each such subset can't open. So we have a one-to-one / invertible mapping between subsets of size M and the locks. Number of subsets of size M is ${S \choose M}$. So the cardinality of set of locks is also ${S \choose M}$.
Comment: Can you elaborate on 'So we have a one-to-one/invertible mapping between subsets of sie $M$ and the locks'?
Comment: Sure.
Let's just mark [S choose M] as n.

We know that every M scientists can't open all locks. So let's look at a certain subset of size M, let's label it "M(1)"... So we can take a lock, and label that lock, L(1) and label it as the people that can't open it... that is M(1) == L(1), and say that no matter what, scientists from M(1) can't open L(1).
We continue labeling every subsets and locks.

There should be a one-to-one mapping from:
M(1) -> L(1)
M(2) -> L(2)
...
M(n) -> L(n)

So #locks = n = [S choose M]


Applying Korean Tourist visa from Nagpur or Bengaluru

Question: I will be travelling to South Korea on 31th MAY, 2023 for tourism purposes. Before travelling, I have to get my visa approved. So, I was looking for some service based out of Nagpur/Bengaluru which can assist me in the same. Usually, I apply through VFS global. However this time, the service is only available for Delhi & Kolkata as mentioned on their website https://visa.vfsglobal.com/ind/en/kor/attend-centre as well as here https://overseas.mofa.go.kr/in-en/wpge/m_2660/contents.do . I was wondering if anybody else  is facing the same problem and/or have a solution for this.
Answer: The embassy of the Republic of Korea in New Delhi, states :

Jurisdiction for Visa Submission
☞ To Embassy of the Republic of Korea in New Delhi
Arunachal Pradesh, Assam, Bihar, Chhattisgarh, Haryana, Himachal
Pradesh, Jharkhand, Manipur, Meghalaya, Mizoram, Nagaland, Odisha,
Punjab, Rajasthan, Sikkim, Tripura, Uttar Pradesh, Uttarakhand, West
Bengal, Union Territory (Delhi, Chandigarh, J&K, Ladakh)
☞ To Consulate General of the Republic of Korea in Mumbai
Maharashtra, Gujarat, Goa, Madhya Pradesh, Telangana, Union Territory
(Daman & Diu, Dadra & Nagar Haveli)
☞ To Consulate General of the Republic of Korea in Chennai
Andra Pradesh, Kerala, Karnataka, Tamil Nadu, Union Territory
(Lakshadweep, Puducherry)

They talk about the state you actually live in

If you live in any of the states under the Chennai consulate jurisdiction, the rules are as follows :

Visa applicants to collect tokens from ground floor of the consulate building at 9AM. 17 tokens for individuals and 8 tokens for
agents are issued on a first come first serve basis only at sharp 9AM.
NO EXCEPTIONS OR PRIOR APPOINTMENTS

Visa applicants without check list or documents mentioned on the checklist will not be allowed to submit their application

Those creating any issues/disturbance during token distribution or those submitting fake documents during application will be de barred
from further submissions

Please avoid calling for visa status enquiry. Visa status will be uploaded after 15 days from submission

Visa once approved can be collected from consulate between 3 to 4PM.

The first point means that only 17 individual applications will be allowed to be lodged on that day

If you live in any of the states under the Mumbai consulate jurisdiction, the applications are to be lodged at select travel agencies link to the relevant information

If you live in any of the other states, you indeed need to visit the VFSGlobal offices in New Delhi and Kolkata


What decides when a conceptual analysis is 'complete'?

Question: When an analysis of a concept is given like
"A bachelor is an unmarried male." How is it decided if the analysis is correct and complete? Is there any way we can 'check' an analysis?
Comment: The arm-chair philosopher decides and his colleagues check. You are onto something.
Comment: Are you assuming an Aristotelean notion of analysis where there is a right and a wrong way to analyze any given thing as opposed to just different degrees of usefulness?
Comment: @DavidGudeman I don't know what either of those notions are, I am just thinking about properties from a Platonist viewpoint
Comment: What would you say if my analysis included bats as a kind of bird because for my purposes all flying creatures should be grouped together? Would you say, that's fine if it's useful, or would you say that's wrong because bats are not, in fact, birds? The second answer means that you think there is something essential to conceptual analysis; the first answer means that you think conceptual analysis is just a pragmatic exercise that you have to match to your purposes, and it can be more or less useful for a purpose, but not right or wrong.
Comment: @DavidGudeman I would definitely say that there is something essential to the analysis
Answer: Short Answer
There is no consensus in regarding a normative framework for differentiating between 'complete' and 'incomplete' analysis of concepts. Some philosophers are highly skeptical of analysis and the linguistic turn entirely. In practice, philosophers largely rely on their intuitions or pragmatic factors on when to conclude an analysis, much in the same way a functional analysis might be conducted under Cummins' method of functional analysis.
Long Answer
Not only is there no universal framework or protocol for 'complete conceptual analysis', it's arguable that philosophy is essentially one, long continual analysis of concepts, while some philosophers argue against such a definition. Quine famously attacked the whole notion of analyticity in his Two Dogmas. In popular culture, philosophical discussion is often portrayed as an eternal rehashing of essentially contested concepts of which no genuine progress is ever made.
Furthermore, analysis is taken to be useful if it 'informative', that is produces some sort of new information. And yet, if an analysis is to be held 'correct', in some sense it needs to be complete, that is, all of the relationships it entails should be addressed. That the twin goals of informativeness and correctness are driving in different directions itself seems to be a problem. G.E. Moore considered such a matter to be a philosophical problem in its own right: the paradox of analysis. From WP:

A conceptual analysis is something like the definition of a word. However, unlike a standard dictionary definition (which may list examples or talk about related terms as well), a completely correct analysis of a concept in terms of others seems like it should have exactly the same meaning as the original concept. Thus, in order to be correct, the analysis should be able to be used in any context where the original concept is used, without changing the meaning of the discussion in context. Conceptual analyses of this sort are a major goal of analytic philosophy.

So, there is no one-size-fits-all approach to conceptual analysis. In fact, conceptual analysis, which is a mainstay of analytic philosophy, should provoke the idea that analytic philosophy is not the only style of philosophy. Many philosophers see, including analytic philosophers themselves, the incessant hair splitting of analysis to be a from of pedantry, one that has lost view of what is important in philosophy, such as the pursuit of eudaimonia.
Comment: Of course, within AI research, one might be tempted to apply a goal-oriented measure, such as the ability to have an algorithm complete a task. [Semantic frames](https://en.wikipedia.org/wiki/Frame_(artificial_intelligence)#Frame_language), for instance, are one practical data structure that can be evaluated for completeness in NLP or related tasks.


Mountain Lion does not show up under Purchased for ML pre-installed machine

Question: I recently purchased a new Macbook Air (2012) and it came with Mountain Lion pre-installed. I've noticed that Mountain Lion does not show up as 'Purchased' in the App Store and indeed when I search for the app, the store offers to sell it to me. Is this a bug or by design?
Comment: Was the Apple ID you used to register the device the same as the one you use on the store?
Comment: @DeepanshuUtkarsh Yes
Comment: I just checked and it's the same with me. Probably it's intentional, as pointed out by CanuckSkier.
Comment: Also, if you are running the App Store and on any tab except purchases one, and then hold the **option** key while selecting purchases - you might see Mountain Lion as downloadable. Sometimes the store will "hide" purchases like OS that are already "installed" at a same or newer version.
Comment: Can you technically just use another apple ID who may have previously updated to mountain lion and have proof of it in their ID?
Comment: I am having the same problem with the whole " recovery partition". It said that I haven't purchased it but this is my first Mac with it pre installed. So what I did was use my friends ID and now it is installing. But my question is, does their info sync onto my computer now or is that only for the software?
Answer: If Mountain Lion is pre-installed on your Mac, you won't normally get a redemption code to purchase the "upgrade" but that serial number will be checked if you have to do a bare install, and the installer will download because the Mac deserves to get Mountain Lion internet recovery install on a bare drive.
Basically, the license is tied with the hardware, not the Apple ID. However, some Macs purchased during the window surrounding the release can often take advantage of the Up To Date program which according to the web page is intended for Machines that shipped with Lion and not Mountain Lion.
The upgrade lets you install the software on all of your personal Macs whereas the machine license generally lets you just install it on the Mac that shipped with the OS.
Answer: It's actually hidden. If you open the app store and go to the "Store" menu and select "View My Account", it will tell you that you have "Hidden Purchases". Unhide and download.
Answer: I've been on the phone with Apple Support for almost two hours explaining this and why I would want an installer for the OS my computer shipped with. I was finally sent to the App Store support who then sent me back to Apple Support. Short answer is that nobody knows how to do it at Apple Support or the App Store.
Here's what I did. I attached a USB drive large enough for the Mountain Lion installer (8GB) to my 2012 MacBook Air that shipped with Mountain Lion. Then I booted up in Recovery mode (the only way we can download the OS that shipped with our books), told it to do a fresh install on my USB drive (you'll need an internet connection) then I unplugged the USB drive after it rebooted to install the OS. The result is the OS installer image in a folder called "Mac OS Install Data". I then copied that image to my book's drive, then used Disk Utility to burn the image to the USB drive. Now I have a USB bootable drive for installing Mountain Lion.
What a chore!
On a related note this is how I save app installers from the App Store:
Copy and paste this into the Terminal app while you're downloading an app:
sudo find /var/folders -mmin -1 -size +1024 -type f -exec ls -lAdh {} \;
I leave it to the UNIX gurus to come up with a better "find" line. You'll have to enter your password in order to search the directory where the file is being downloaded to.
This command searches in "/var/folders" and all of its subfolders for a file that's been updated in the last minute and is larger than 1MB (fairly safe assumptions for a downloading installer). Somewhere inside "/var/folders" is where the installer is downloaded.
Copy the folder from the Terminal window (don't copy the last bit...the [garble].pkg part. Your string should end in '/'.
Then in the Finder and press command-shift-g. Paste the folder string copied from Terminal and click "Go". You're going to want to copy that [garble].pkg file out of there (drag it to the Desktop. Keep a watch in the App Store as you'll need to start the copy AFTER the installer has downloaded and BEFORE it has finished installing!
What a chore!
Answer: I'm fairly certain this is by design. If it were "purchased", then you could download Mountain Lion on any other Mac that has your App Store credentials.
Comment: That sort of makes sense, but according to http://apple.stackexchange.com/questions/58303/will-i-be-able-to-re-download-mountain-lion-by-using-my-redemption-code-in-the-f if I had bought my machine with Lion installed I would be able to get ML via the App Store. Somewhat inconsistent...
Answer: If your computer shipped with 10.8 Mountain Lion you should DEFINITELY be able to see it as installed in the Mac App Store.
I would double check to make sure that you are actually running 10.8 Mountain Lion (click on the Apple in the top left hand corner of your screen, click on 'about this mac' and you should see a window that either says 'Version 10.7' or 'Version 10.8').
When I was selling Macs for a living we would sell Macs that had the prior operating system on them even though the new operating system was out.  In that circumstance you are expected to go to: http://www.apple.com/osx/uptodate/ to claim your free upgrade.  I would do so quickly since you only have 30 days from date of purchase.
If you have 10.8 Mountain Lion attached to an Apple ID you can install it on "any Macs you own or control for personal use." http://support.apple.com/kb/HT4461


How does Facetime compare to normal video calling?

Question: So, Apple is billing Facetime as though it is a revolutionary technology. I'm not going to ask you whether you think it is, as that is clearly a subjective question. What I will ask instead is how does it compare to normal video calling. Is there anything new that hasn't been done before?
Comment: What are you calling "normal video calling"? VoIP stuff like Skype, or the original 3G stuff that never took off? (IMHO because the networks overcharged for it)
Comment: "The original 3G stuff" wouldn't be a fair comparison because Facetime isn't 3G but only wifi.
Answer: Entirely new? No - not really, but the quality is quite acceptable (I'd say it's better than most Skype video calls I've had) and it's extremely painless to use. It's also at the present time far more limited in audience (only iPhone 4's, iPad 2's and Macs) than normal video calling.
You can think of it as video calling built right into the OS with software that doesn't require hoops (no log in required) to jump through to use it.
Answer: As for your question, Is there anything new that hasn't been done before, then one could be generous and say that the no-account-required and it-just-works is quite nice, as is the elegant use of front or rear cameras and the ability to switch between them, and the automatic landscape/portrait modes. That's quite slick. But it's not a competition-killer.
On the whole, Facetime is the typical Apple feature -- very polished, but a limited feature-set, and only available to users on the exact same platform (in this case, only iPhone 4's, iPad 2's and Macs). That alone limits the usefulness of the app quite a bit.
For general-purpose video calling, there are general-purpose apps like Skype which do it better, and which even work over 3G (if you're lucky), and which aren't limited to specific Apple-only devices.
I'd recommend Skype over Facetime any day.
(Notice I didn't comment on the revolutionary issue either.)
Comment: Note that Skype now does offer video-on-phone.
Comment: This would be more helpful if Skype actually offered video calling on the phone (and would actually be a viable alternative to FaceTime).
Comment: @Adam, you're right. I was confusing Skype with Fring which does offer video calling: http://www.fring.com/blog/?p=2276
Answer: You're obviously well aware of the long history of video calling implementations to ask this question, so

Having an idea is trivially easy.
Implementing it well is hard.
Implementing it so you don't get
sued is even harder.
Driving substantial user growth or taking over user share in adopting your implementation is the ultimate challenge.

Apple's hype and I would presume most non-hater's hope is that they are shooting for no less than #4 above. If you consider implementation and support costs of a video chat system as well as voice and picture quality, it's hard to say anyone is beating them at #2 either. One could easily see the desktop client catching up to Skype in short order but no-one is even close for mobile devices and two way video.


Method of image charges for a point charge and a non-grounded conducting plane

Question: I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential $V$.
Comment: @sammygerbil Yes I have and also I looked it up in several books, Jackson's Classical Electrodynamics, Intro. to electrodynamics (Griffiths), Foundations of Electromagnetic Theory (Reitz, Milford, Christy) the examples in all these books limit to "Grounded" planes,
Answer: The problem is addressed in Electrodynamics Fulvio Melia Example 2.1 page 39,
using Green function techniques. However, I cannot fully understand his final
solution. Let us define the plane held at a constant potential $V_0$ to be the $z$-$y$ plane. The charge $q$ is at the distance $a$, namely at $(a,0,0)$. The potential at large distances is zero. The solution (Eq. 2.54) is
$$V(x,y,z) = V_{image} (grounded) + V_0 f(x,y,z).$$
The first term is the usual image method solution, for the case where the conducting plane  is grounded, namely
$$V_{image}(grounded) = q\left[\left[(x-a)^2 + y^2 + z^2\right]^{-1/2} - \left[(x+a)^2 + y^2 + z^2\right]^{-1/2}\right],$$
which gives zero when $x=0$. In other words this is the solution of the Poisson equation, but the boundary conditions are not those required.
The second term, $V_0 f(x,y,z)$ is the solution of Laplace equation, and $f(x,y,z)$ is $V_0$ independent.
According to the text book Eq. (2.54) there
$$f(x,y,z) = (x/2 \pi) \int dz' \int dy' \left[x^2 + (y-y')^2 + (z-z')^2\right]^{-3/2}.$$
The integrals over the primed variables stretch from minus infinity to plus infinity and this function is V_0, q and the length scale a  independent.
The solution is mysterious for me. Namely, we can get rid of $y$ and $z$ in the integral (for any finite $y$ and $z$) by change of variables of integration. Then the integral can be solved (for $x>0$) and the solution is $f(x,y,z)= x/ x = 1$.  Well, that is wrong (or subtle) since that would mean that we only need to add to the usual image solution a constant $V_0$. This does not satisfy the boundary condition at large distances from the plane where the potential is zero.
Maybe the integral for the dimensionless function f(x,y,z) should be regularised somehow.
Since the image part of the solution, $V_{image}(grounded)$,  takes care of the charge in the Poisson equation, we can find a solution that is a linear combination of $V_{image}(grounded)$ and some other function, the latter solves the Laplace equation.   Then let the boundary at $x=L$ be grounded and we want to take $L\to \infty$ and on x=0 we have V_0 (this seems like a legitimate trick). The Laplace equation for this problem gives
$$V(x,y,z) =  -V_0 x/L + V_0,$$
and if we take $L\to \infty$ we get the result just mentioned (for any finite $x$) namely a constant (which does not satisfy the boundary condition at infinity). So I think the limit of a large system ($L\to \infty$) is creating a certain non-trivial problem, which ultimately is related to the Green function technique used by Fulvio Melia. Any ideas?
Of-course if f(x,y,z) is unity for any finite distance x,  this would mean that switching from grounded plane to non-grounded one
(compared to zero potential at infinity) has practically no effect. This hypothesis was postulated in some of the  previous discussions.
Comment: Please use MathJax for the equations
Comment: I think you wanted to write $dz'$ instead of $dx'$ in the equation that you call (2.54).
Comment: thanks Davide, corrected.
Answer: If you would like to define V at infinity as zero and the conducting plane as not grounded, you can also think of the solution as a superposition of two different elctrostatic cases: Take the fields expression of a single charge and a grounded plane, and sum this with the fields given off by a plane of fixed potential.
Comment: I don't understand why should V be equal to zero at z=inf. Because the electric potential due to an infinite charges plane increases by distance (since the electric field generated by it doesn't depend on distance)
Comment: This sounds logical. But what are the boundary conditions in this problem so we can check the answer? For example in the grounded version we could say at x=0 (where the inf. plane is located) electric potential has to be zero.
Comment: The important requirement for the boundary condition of a conductor is that the surface be equipotential. For example, say you have a a conducting plane fixed at V0 on the xy plane at z = 0 and a charged particle at z  = d above the plane. I would set the boundary conditions  to be V = V0 for z  = 0 and V = 0 for z = inf.
Answer: I also think that there would be a difference between the two situations. Lets take a non-grounded PEC plane and say a positive charge lies above it. The field created by the charge will induce negative charges to be stacked on the top (towards the positive charge) of the PEC, and since there is no creation nor destruction of charge, the same and opposite charge will be induced on the bottom surface of the PEC. In this way, treating the thickness of the PEC as infinitely small, both induced surface charges cancel each other over large distances, as seen by the positive charge. In this way, nothing changes in respect to the field generated around the positive charge (since the PEC as no total charge).
Though, if the PEC is grounded, it acts like an infinite source of charges that can be arranged in any way, dictated by the surrounding charges. So the positive charge induces an electric field on the upper surface of the grounded PEC, BUT no charge on the lower surface since charge must not be conserved in this situation (that is what ground means!). We are then left with a charge that as electric field lines pointing towards the PEC, all coming at right angles with the latter. This situation is exactly equal to a dipole made of the same positive charge, and a equal negative charge situated an equal distance normal to the PEC.
Answer: When we ground the conductor what we are doing is essentially removing the effect of positive charge induced. But if we do not ground it we have to take account of the positve charge induced too.
We can say the effect of induced negative charge(only) is roughly same as when the conductor was grounded so we assume a $-q$ charge at distance d. from surface(on which -ve charge is induced). If we assume charge distribution of $-q$ and $+q$ to be symmetrical we can similarly place $+q$ charge at a distance d from the surface where +ve charge is induced.
Note: This is just an approximation.
Answer: I'm responding to the May 13 2021 post by Eli Barkai -- which is a very nice post so many thanks to Eli for an insightful post. And many thanks to Alireza for bringing this nice question up. My perspective on his quandry: the INFINITELY large "conductive plane" that extends to infinity causes mathematical complexity because a) electrical charge can be harvested from an INFINITELY large number of locations along this plane, which thus acts as "grounding" because charge can be brought in from infinitely far away (but still along the plane), and b) any attempt to uniformly spread charge on the plane will cause voltage to be non-zero at infinite distances from the plane.
Therefore it seems this problem can only be formulated as 1) a non-infinitely large conducting plane that is non-grounded and can have its voltage modified, or 2) an infinitely large conductive plane that is grounded and cannot have its voltage modified.
For case 1, the large-yet-finite-sized conductive-foil will have the same charge distribution from the famous original "image method" solution PLUS it can be put to any additional voltage by spreading additional charge uniformly around it's surface.
For case 2, ANY attempt to modify the voltage of the infinite-plane will cause voltage at infinity to be nonzero thus invalidating the problem description.
Answer: If the problem you're trying to solve only contains one point charge and the conducting infinite plane at potential V, then there is no physical difference between the plane's potential being 0 (grounded) or +V, because the electric potential may be globally shifted by a constant value everywhere...
Comment: We actually know nothing about what happens on the other side of the plane. The image charge is a construction to help us write down an electric field that is perpendicular at the plane.
Comment: I don't feel quite right about this. Because in a grounded version we can assume electric field is zero on the other side of the plane but in this case one side of the plane has a positive charge and the other side has negative charge therefore there are electric fields on both sides due to this charges.
I'm not sure about this either though. Just a beginner in image charges method.


Use the normal model to approximate the binomial to determine the probability of at least 191 passengers showing up

Question: Because many passengers who make reservations do not show up, airlines often overbook flights (sell more than there are seats). A certain airplane holds 190 passengers. If the airline believes the rate of passenger no-shows is 9% and sells 208 tickets, is it likely they will not have enough seats and someone will get bumped?
Answer: $X$ - numbers of "shows". So, $X \in \mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) \geq 10$ we can make the following approximation, where the approximation was made in the last step
$$ P_{\mathrm{Bin}}(X \geq 191) = 1- P_{\mathrm{Bin}}(X \leq 190) = 1- P_{\mathrm{Bin}}(X < 191) \approx 1- \mathrm{N}(\frac{190.5-np}{\sqrt{np(1-p)}} ). $$


Is it fine to put 4 wood blocks around the crimson/corruption

Question: I am just about to kill the wall of flesh and i dont want the crimson/corruption to spread over my base or
jungle i am thinking that i can just put 4 wood blocks around to stop the corruption/crimson
Answer: Biomes can only spread within three tiles, so any 4-tile gap of resistant material will prevent spread. Wood is one such material. So is air, which is a lot faster and cheaper (though an air-only gap can be bridged by plant growth in some arrangements).
Bear in mind that large amounts of Corruption/Crimson and Hallow will generate upon entering hardmode, so trying to enclose the spawned biome beforehand is pointless. Smashing altars will also cause new evil to spawn. Enclosing the areas you want to protect (spawn, jungle) is a better strategy.
Answer: Yes, this will stop the corruption/crimson but bear in mind the hallow will spawn in hardmode, so this seems kind of pointless until hardmode. My recommendation is defeat the wall of flesh, see where the corruption/crimson/hallow is, and then put 4 blocks of wood/air to stop them spreading further.


Is there an algorithm to compress multiple images that are not much different apart?

Question: I have a set of 100+ images that only differ a few pixels. They also don't have many different colors. Is there a way to compress them together by taking advantage of that?
Comment: Lossless or lossy compression?
Answer: There's an algorithm: calculate the difference of each image to some reference image. Is there a ready made application? Probably not. One approach could be to combine the images to a very big image (by interleaving or placing them next to each other) and using png.
If this is for archiving purpose and you don't need a random access to them, you can concatenate them to zip / tar (with zero compression) and compress the whole thing. .bz2 algorithm (Burrows-Wheeler Transform) is able to search for similarities over much larger window than deflate of png. It's an order of ten megabytes vs. 10 kilobytes. If the images are large enough, the window size will limit the compression in both algorithms -- that would have to be combat by interleaving or delta compressing between each image.
The delta compression is utilized regularly in some video compressing applications and e.g. screen capture and virtual desktop applications where lossless compression is required.
Answer: String them together as a movie and use video compression.  That is exactly what video compression does.
Answer: Nothing ready-made. If you want lossless compression, you could store just a reference image and the delta images (difference of images at the pixel leve), each one encoded with PNG. But you must write yourself the delta transformation.
You could also take advantage of the delta compression mode of the MNG format, an extension to PNG for animation (each of your image would be an animation frame). But the format is not widely supported.
You could also take the same approach (one image = one video frame) and use any standard video format (MPEG) but this would be lossy.


How do I connect power to a light bulb?

Question: I can't seem to figure out how to connect my light bulb to the generator to make it turn on.  How do I do this?
Comment: There's so much potential for a title that would be awesome : How to I turn on a light bulb? How can a light bulb be turned on? Why isn't my light bulb on?
Comment: My idea isn't turning on this light bulb. :)
Comment: Keep in mind that if we favour quality over humour, replacing "light bulb" with 'all electronics' answers a lot of other questions, too.
Comment: @timelord in this case it didn't make sense as most electronics can be directly hooked up but light bulbs can't.  When I first asked this question it was because there was no obvious way of powering up a light bulb but I knew how to hook up a radio tower to a generator
Comment: @z, yikes. A few people disagree, this had already become the go do original for "how to i set up this electronic" duplicates. I suppose Ill have to go looking for them, now.
Answer: I'm going to try and answer this as clearly as I can as I have just worked out how to get everything working. This is how you connect power to objects in your settlement:

What you want to do first, as I realise you have already, is get the source of power, i.e. your generator
Next, couple the generator with one of the 'power pylons' from the 'connectors and switches tab'. Add a cable between the generator and the pylon
The next step is optional; add a 'switch', again from the 'connectors and switches' tab, and place it on a wall somewhere. Attach a wire between the pylon and the switch
Some lights, like the 'construction' lights, can be wired straight into the switch, and when you come out of workshop mode, you just flick the switch and they will come on
For other lights, TVs, etc you will need a 'power conduit', again from 'connectors and switches'. There are three types of conduit, there is one for ceiling, wall and floor, and depending on which of these surfaces your desired item is going on, you choose the appropriate conduit. e.g. if you want a wall mounted light, you add a wall mounted conduit
Wire your conduit to the switch or to your power source that you installed earlier, then, add lights and TVs to your hearts content. You don't need to wire them to anything

You must have copper in your (or your workshop's) inventory to attach a wire, although you can delete a previously-built wire for a refund of copper used.
Comment: nice and thorough answer, thanks.  That last bit about the conduit was extra useful.
Comment: you (and the game) should make it abundantly clear that when you're out of copper, you can't add wires.
Comment: You don't need the switch. Generator to conduit on the same wall does it. Might not even need the conduit if the generator is close enough.
Comment: Some lights like the construction lights can be hooked up straight to a generator or pylon, they would just be permanently on though
Comment: "If you want a wall mounted light, you add a wall mounted conduit" This is definitely not true, the conduit simply needs to be near the light. You can also use a pylon or a switch, but I've found the wall mounted switch to have a significantly shorter range.
Answer:
This picture should tell you how to connect power to a light bulb.
Comment: Sorry, PhilFox!  I had to add some text to get it to take the edit; without that, your picture wouldn't have appeared.
Comment: I tried this based on an Internet post but it didn't work for me on PS4, I had to use a conduit instead of the switch to get power radiated (yes I turned the switch on, and yes I moved the items close to the switch).
Comment: Just going to leave this here, in case any other user had the same question I did. [are image only answers acceptable - yes](http://meta.gaming.stackexchange.com/questions/7505/is-giving-just-a-picture-a-reasonable-answer); have previously seen these deleted as "link only".
Comment: @mxyzplk the switch works too, but its range is shorter than that of a conduit or a pylon
Answer: You need to connect the generator via anything in the Connectors & Switches section under Power in the Workshop build menu. Once you've got a switch or pylon in place, when you hover over it you have the option to Attach Wire:

Select this option and connect a wire to the generator and the light (or other object) and then flip the switch if required.

Additionally, some items appear to power just by being in the vicinity of power infrastructure (pylons, conduits, etc):
Answer: So after blowing through a lot of copper, as far as I can tell anything with out a direct power connection (lights, T.V.'s, ...) can be powered by being in a AOE proximity to either a switch, conduit or a pylon. Also a small generator seems to be able to power several lights (not just three).
Answer: You don't add a wire to the ceiling and wall lights.
Add a generator, and once running, put a light in close proximity to it. That's it. If you need it farther away you have to add a power conduit and run a wire from the generator.  Then place the light in proximity to the conduit. Usually within the same room.
Remember, generators are also rated for power. So if you've used 4 of 5 of the power units and add something that requires 2 units. It won't turn on until there is adequate power.
Answer: Use power conduits on the roof near a light bulb and power the conduit
Comment: Could you elaborate? Screenshots and such could be useful.
Answer: You could try connecting the generator to a pylon with a switch and then running the power with conduits? All light bulbs within that vicinity should work.
Alternatively, you could try placing a coupling capacitor (I think that was the name), near the object you want powered.
Comment: What category is the coupling capacitor under?  I can't seem to find it.  Some pictures would help a lot as well.
Comment: As I say, I think that was what it was called. When time allows (as I'm at work), I'll take a look for you.
Answer: Step 1: set up generator
Step 2: set up a switch appropriately with environment  (try to get it in between generator and conduit, which is coming up next) wire the switch to the generator.
Step 3: place conduit near desired area of electricity (inside or outside wall is acceptable) wire the conduit back to the generator.
Step 4: place light is desired area (near conduit)
Step 5: If the light(s) DIDNT already come on, try flipping the switch, or pressing A on the item in workshop mode.
I struggled, and this is what worked for me. Still haven't mastered the whole power/defense thing yet

How do I connect power to a light bulb?


Cannot start upstart job, error: Could not open input file

Question: I have an upstart configuration that worked before and now it shows the error:

Could not open input file: /var/local/test/run.php

but the file is there and I can normally run it with php /var/local/test/run.php command.
# Events
start on local-filesystems and net-device-up IFACE=eth0
stop on shutdown

# Automatically respawn
respawn
respawn limit 20 5

script
    exec php -f /var/local/test/run.php
end script

I tried chmod 777, but not helping.
What are my options?
Please help. Thank you!
UPDATE:
adding an instance variable:

instance $ID

made it work, The thing is that before it used the instance, but now I decided to change the script a bit, and removed it. So I was thinking, is there some cache or similar that I can delete (that does not get deleted with rebooting)?
UPDATE 2
I figured out that the script adds an end-line char at the end of command, so that is the reason it cannot find the path.
I tried surrounding it with quotes:
exec php -f "/var/local/test/run.php"

but it still starts the script with end-line char at the end of php script file path.
What can I do?
Answer: So, the alternative solution is to add some parameter to the script, so it "eats" the end-line char.

exec php -f /var/local/test/run.php 1


Use Scrapy to cut down on Piracy

Question: I am new to using Scrapy and I know very little of the Python Language. So far, I have installed Scrapy and gone through a few tutorials. After that, I have been trying to find a way to search many sites for the same data. My goal is to use Scrapy to find links to "posts" and links for a few search criteria. As an example, I would like to search site A, B, and C. Each site, I would like to see if they have a "post" about app name X, Y, and Z. If they have any "posts" on X, Y, Z. I would like it to grab the link to that post. If it would be easier... It can scan each post for our Company Name. Instead of X, Y, Z it would search the contents of each "post" for [Example Company name]. The reason that I am doing it this way is so that the JSON that is created just has links to the "posts" so that we can review them and contact the website if need be.
I am on Ubuntu 10.12 and I have been able to scrape the sites that we are wanting but I have not been able to narrow down the JSON to the needed info. So currently we are still having to go through hundreds of links, which is what we wanted to avoid by doing this. The reason that we are getting so many links is because all the tutorials that I have found are for scraping a specific HTML tag. I want it to search the tag to see if it contains any part of our App Titles or Package name.
Like this, it displays the post info but now we have to pick out the links from the json. Saves time but still not really what we are wanting. Part of that, I think is that I am not referencing or calling it correctly. Please give me any help that you can. I have spent hours trying to figure this out.
posts = hxs.select("//div[@class='post']")
        items = []
        for post in posts:
            item = ScrapySampleItem()
            item["title"] = post.select("div[@class='bodytext']/h2/a/text()").extract()
            item["link"] = post.select("div[@class='bodytext']/h2/a/@href").extract()
            item["content"] = post.select("div[@class='bodytext']/p/text()").extract()
            items.append(item)
        for item in items:
            yield item

I am wanting to use this to cut down on Piracy of our Android Apps. If I can have this go out and search the Piracy sites that we want, I can then email the Site or Hosting Company with all of the links that we want removed. Under Copy Right law, they have to comply but they require that we link them to every "post" that they infringe upon which is why App Developers normally do not mess with this kind of thing. They have hundreds of apps so finding the links to your apps takes many hours of work.
Thank you for any help you can offer in advance. You will be helping out many App Developers in the long run!
Grady
Answer: Your XPath selectors are absolute. They have to be relative to the previous selector (the .)
posts = hxs.select('//div[@class='post']')

for post in posts:
    item = ScrapySampleItem()
    item['title'] = post.select('.//div[@class="bodytext"]/h2/a/text()').extract()
    item['link'] = post.select('.//div[@class="bodytext"]/h2/a/@href').extract()
    item['content'] = post.select('.//div[@class="bodytext"]/p/text()').extract()

    yield item

Comment: I have removed the item['content'] as I am not wanting the content section. How do I tell it to search for X, Y, and Z though? Right now, it is grabbing all the links that are in the body/h2/a.

Thank you for getting me one step closer and fixing my code!

Use Scrapy to cut down on Piracy


Why is the red dot still present if there is nothing to review?

Question:
I am seeing this red dot beside Low Quality Posts, but when I click it, there is nothing inside to review. So why is it showing? Am I yet not eligible to review low quality posts fully?
Comment: This is a longstanding UX issue, but it is [working "as designed"](https://meta.stackexchange.com/questions/238698/) in the eyes of SE.
Answer: There are currently two items in the Low Quality Posts review queue (you could check that by opening another browser, or a private/incognito window, so that the site doesn't know you're logged in; another way is to use the Wayback Machine).
You can't review Low Quality Posts review items for posts you've already flagged, and if you skipped the item you won't see it again either. However, the red dot is calculated and cached on system level (it would be too expensive to do this on a user-by-user basis), that's why you do see the red dot.


How to organize a CTF?

Question: In the context of an IT event, my team would organize a CTF (Capture The Flag).
About the challenges (we're working on them). I want to have some references on the types and levels of difficulty that we find on CTF generally. (The target audience varies from students (beginner) to security experts and regulars participant in CTF).
Then for the deployment, do we should use a local server? Or a VPN?
In fact, we lack experience, and I need some advices/suggestions on all organizational aspects.
Thank you in advance.
D3_4dl1n3
Comment: If you have not participated in several CTF, I'm unsure if it's feasable to organize one.
Answer: You can find a lot of CTF event in this website as a reference about how they organize it, their CTF's type, and what kind of challenges offered.
For a more detailed and technical stuff about CTF challenges, I recommend you to read some CTF writeup by active CTF teams. As an example, Dragon Sector is an active team and usually make some writeup after finishing a CTF event, their blog is a good start to explore.
Answer: There are a lot of detailed accounts available of Capture the Flag at DEF CON which might provide the insight you're looking for, and you can try asking specific questions about organization in the DEF CON Community Forums. They do it up.
Answer: I have participated in Live and online CTF competitions. It depends what sort of CTF you would like to organize, and what sort of challenges there will be, I mainly categorize them into two categories.

Web Application
System/Network

Now if you want to organize a CTF involving web applications only that has vulnerabilities and users are required to exploit them and get the flag, you dont need to have a VPN setup. All you need to to use WAMP server and get it live.
If the CTF requires to have network access, then yes, you would need to setup a VPN , in
order to let the participants, get access to your local network and do their deeds.
From resources point of view, you dont need to have large resources, if you are not expecting large number of participants, you can use multiple virtual machines on the same box and have multiple vulnerable apps running.

How to organize a CTF?


Sea Routes shipment Dataset Caribbean South America, Free

Question: I would like to obtain gis data of cargo, container vessels as well as small cargo inter-islands ships and for an educational non commercial project/publication on food trade across the Caribbean Sea. Any information available.
Comment: It sounds like you are after [opendata.se].
Answer: You could use AIS data.
I'm not sure for the coverage of the caribbean sea and the temporal extent. IMO it is difficult to get historical data for free, but for a start you could work with AIS data from NOAA (..into the blue: Puerto Rico and Panama Channel). For the caribbean sea you will have a poor station AIS infastructure (in logging terms).
Commercial sites like vesselfinder.com, vesseltracker.com and marintraffic.com offer also historical databases and show more stations.

To mine the cargo (drought) and ship type info's you could parse/read AIS Message 1 and 5. In Message 5 you will find DST (Destination) and ETA (Estimated Time of arrival). The are several AIS decoder available in the web. Kurt Schwehr has a good repository at github.


RP Automatic Updates

Question: Does Raspberry Pi working with Raspbian make any automatic updates without our permission? If yes, How to disable it?
Comment: The way you phrased it makes it sound like Rasbpian is doing something bad.
Comment: This is not intended, but it should not do this.
Answer: In fact Raspbian does have daily upgrades enabled by default; to check:
systemctl status apt-daily-upgrade.timer

The bit to play attention to is in the second line of output.  Between the two semi-colons it will say enabled:
 Loaded: loaded (/lib/systemd/system/apt-daily-upgrade.timer; disabled; vendor preset: enabled)
                                                              ^^^^^^^^

Or disabled, as it does here. Whether it does anything or not, you can check the time it was last run with:
stat -c %z /var/lib/apt/daily-lock

If it does do something, there should be a record of that in /var/log/dpkg.log.
To disable this:
sudo systemctl mask apt-daily-upgrade
sudo systemctl mask apt-daily
sudo systemctl disable apt-daily-upgrade.timer
sudo systemctl disable apt-daily.timer

The reason for using mask is to prevent these from being re-enabled by some dependency.
Notice those are two separate services; they both run /usr/lib/apt/apt.systemd.daily, a shell script, with parameters install and update.
Comment: @Dougie, it's a sane default for an IoT device such as raspberry pi, one surelly wants to receive unnatended secrutiy patches to stay safe. And if you for whatever reason do not, well, it's easy to disable, nothing idiotic about that.
Comment: My system shows `Active: inactive (dead) since Sun 2019-12-29 06:58:54 AEDT; 5h 51min ago` but there is **NO evidence** that it has ever updated my OS.

On another Pi (which I last manually updated over a month ago) now shows `17 packages can be upgraded.` so while this service runs it doesn't seem to actually do anything.
Comment: @Milliways The line to check is the second one, `Loaded:`.  The `Active` line will always be more or less the same unless you happen to check while the update/whatever is actually in progress, since this is not a persistent service.  I've edited in a bit about this.
Answer: It does have that feature available, but it is not enabled by default.
https://wiki.debian.org/UnattendedUpgrades
Comment: You might want to check `systemctl status apt-daily-upgrade`... ;|

RP Automatic Updates


Exporting to WebGL

Question: I want to create models for website and interact with it. I chose Blender since it is open source. My intention is to create some model, say a table and export to webGL. How can I export the model using Blender in order to view it in a webpage? Thanks.
Answer: For Blender prior to version 2.6 you can use the Blender WebGL exporter wich exports to json..It supports native WebGL. Looks like the script is no longer maintained..
I think you should also have success with formats like obj, collada or md2.
Comment: The link says the script will not work for 2.6+. (https://code.google.com/p/blender-webgl-exporter/wiki/InstallScript) Have you used this for v2.74?
Comment: @Kaushik I see, the scripts is no longer maintained .. I looked into it and the is too much to repair to work in 2.6+.. Does your model use something specific to v.2.6+? Otherwise you should be able to export from older blender..
Comment: I am not specific about versions, want to give it a try.

Exporting to WebGL


Android as a commercial game platform

Question: I'm currently developing a small non-commercial game for Android which is based on OpenGL ES. Except for some minor issues (slow as crap emulator being the biggest) I've really enjoyed it so far and would love to continue developing for it.
The question is: What are your thoughts on the future of Android as an environemnt for the  game market? I read an article somewhere that said that the IPhone is the preferred platform for games, but since I don't own a Mac and won't bother with some complicated work around to get it to work on Windows I'm currently not interested in it.
So, what do you think? Is it worth spending alot of time (and potentially money) on a larger Android game project?
Comment: Just a tip on the slow as hell emulator: leave it open.  I've found that launching the emulator is by far the slowest part of developing for Android.  If you just minimize the emulator and relaunch your app, it takes only a fraction of the time to restart.
Answer: Short answer: it's too early to tell. Technically this platform is neither better nor worse for games developer than iPhone. Both have some support for 3d graphics, both have good support for multimedia - just the language differs ObjectiveC or Java.
The problem is not in the technical capabilities though, but in the popularity of the platform and purely business decision if Android is a good gaming platform. iPhone certainly is and many people are making money on it already. Will Android be as popular and will people be willing to buy the game for some money is a good question.
There is some difference Android and iPhone users and developers. Android is more open platform and more software is free (or even open source). For iPhone most apps cost 1$.
So the expectations in the audience is different. Android users expect more apps to be free and will (most likely) pay only when software you give them is really worth spending any money at all. iPhone users are willing to pay this $1 even to try an app out, after all they bought Apple product that was expensive to begin with.
Will that trend continue? Or will the audience types converge as more paid apps appear on Android market is an open question.
Finally -- there will be some millions of users of the Android platform eventually. So if you develop a kick ass game it will have its audience anyway. And if Android does not become a platform for games - that may even be beneficial for you, because you'll have no competition.
Comment: Absolutely agreed. The biggest factor will be the platform's popularity and the user perception of Android - would it be able to establish itself as a viable market for the game industry.

Future devices will be targeting that application niche, so you can assume that you won't have to worry about the abilities of the hardware (Snapdragon at 1 GHz sounds great).

P.S. Andy Rubin's words from http://www.reuters.com/article/internetNews/idUSTRE56U5XU20090801?pageNumber=2&virtualBrandChannel=0:
"You can start really thinking about serious gaming like you would on a Nintendo DS or a PSP handheld."
Answer: In my opinion it is worth it.
As @Marcin said, Android is more of a open platform then iPhone. And it's easier for developers to start with developing applications and putting them on the market.
But it means the market is populated with a lot of crap applications. And then, there are really shiny jewels too. They get good scores, are blogged/twitted about and are popular.
So, in my opinion, if you got an idea for a good game, go for it. If you put it on the market though, consider putting (at least a 'lite' version) in the free area (people in some countries are unable to even access the paid applications market).
the below part is completly my own opinion and you can simply skip it
There are some 'cool' games on mobile platforms that I love to play. As for me, there are few important things to watch for when developing such a game, that aren't this important in more 'traditional' gaming, on a console or PC:

it should be easy(fast!) to start and stop. If I play it at a bus, it could be just 1 stop. If it starts right away and then doesn't take 5 minutes to stop it, it's a +
one more thing about rapid start/stop. If I quit the game, then come back, give me a chance to continue where I stopped. No need to ask if I wanna save/load last game. Make it default, I can always start a new game if I want, can I?
controls - even that accelerometer-using games are fun at times, try to play it in a crowded place. like a bus. Touchscreen elements should be large enough so even the thick-fingers can use it. If they are - it's a +
for long time I was trying to realise why playing Bejeweled, or Puzzle Quest, was fun and enjoying, while many of their clones were simply irrytating. The diffrence was a tiny piece of user interface - in Bejeweled the pieces you got are both diffrent colors and shapes. They are much easier to "operate" (in a Bejeweled way) then, say, screen full of diffrent color triangles. If you make a game like this, make the diffrent pieces differ in many ways, not just one (diffrent shapes with same colors would suck even more; red triangles, yellow circles, blue squares and black skulls is a way to go - in my opinion)
it's a mobile device, connected to either WIFI or 3G network, you can use the internet to at least show/save highscores. But - remember - sometimes there's no network, we're in a diffrent country with roaming turned off or on the bottom of the ocean. Make the game work even there
Answer: The iPhone has the advantage of a huge market (approximately 10x the android one at the moment) and even better most of those users have already given apple their credit card so they are ready to buy. Getting android users to spend .99 is hard. Depending on the game, you can make decent money (50-100/day) off of ads. The first is likely to change with more android phones and more android carriers(Verizon would be huge for the US market)
Technically Android is a bit slower (Java versus Object C) but with the NDK you can get decent performance. The iPhone has better opengl support (1.1 and 2.0 on the GS) while android is more of a 1.0.
Answer: Thanks for the replies!
I myself believe that the marketplace will gain momentum once there are more Android phones on the market, though my biggest fear is that it'll end up like Xbox Live Indie Games were 90% of the games are of below avarage quality, which of course affects it's reputation in the gamer community.
One thing I don't understand is why there isn't some kind of pre-approval before letting the application or game onto the market place. I understand that they want the people not the company to decide what should be allowed to be uploaded, but then they could let the users be the reviewers so that there is atleast some quality checks before release.
Comment: users are revieweres - I always browse the reviews before installing anything :)
Answer: There is an interesting article about Android as a games platform on Gamasutra.
Developing Games for Android

Android as a commercial game platform


Do I need to keep apps on my Mac with iCloud

Question: I have a MacBook Air with limited disk space as well as an iPad and an iPhone. With iCloud, do I need to keep the apps in iTunes on the MacBook anymore? When I try to delete them all in iTunes by clicking on the Apps link on the left side of iTunes and then selecting all of the apps and right clicking delete, a message appears saying that the apps will also be deleted from my iPad and iPhone. Is this really the case with iCloud or is it safe to delete them from the MacBook and free up the disk space? I think these files are stored in a folder called Mobile Applications on the MacBook.
Answer: As Adam has said, you should disable syncing applications and yes it is safe.
However if an app gets pulled from the App Store by Apple, which you have purchased but do not have currently install on your iOS device, you will not be able to get it back from iCloud. Whereas if you have it saved on your computer in iTunes you can get the app back by syncing.
iCloud doesn't actually back up the apps on your iPhone, it only stores your data for that app, if your restore from iCloud then it will download the apps clean from the App Store and then insert your data from iCloud, so that it appears that the entire app has been backed up.
If they backed up the entire app, for every user of an app there would be that many copies of that app on iCloud which is hugely wasteful.
So don't delete apps that are no longer on the app store else you won't be able to get them back. And if you have an app that looks like it will get pulled shortly (e.g. the app has tethering hidden inside of it) keep it on your computer. But most apps you are fine deleting and just relying on iCloud.
Comment: What happens if I am backing up to my MacBook and I need to perform a restore? Will all of the current apps be restored with their data? I understand that any apps that have been pulled will not be restored.
Answer: Yes, it's safe to delete them. Just make sure you have syncing off (look for options at the bottom of your devices' summary pages in iTunes) and don't connect them to your MacBook while iTunes is open. To be safe, also go to the Apps tab (at the top, not the left) and untick Sync Applications.
But before you start deleting, in case of problems, I suggest you make sure iCloud is turned on in both your devices' settings, and perform a backup. This is all in Settings > iCloud.
Answer: No, after a few years or even months you can't download your apps from the App Store (iCloud) if they decide not to sell them anymore.
Comment: This is the exact opposite of the currently-accepted answer. Do you have proof? Can you cite sources?


Can we generate a microscopic image of light?

Question: Is there a way we can see light at a microscopic scale? In other words, is there an experimentally feasible way to see the structure of photons?
Comment: I edited protons into photons otherwise there is no sense to the question as is. If you want to discuss detection of protons you have to expand the question to make it understandable
Comment: Photons do interact with themselves interference patterns are one such example...
Answer: Photons are simply quanta (very small pieces) of energy - they have no physical size, and no physical 'structure'. This video is a good introduction to our modern understanding of what light is.
Answer: light is made of photons, not of protons or anything else. And as an electromagnetic wave, it's structure is often the structure of its interaction with the matter (typically made of dielectrics and conducters).
Then, it all depends what you accept as "seeing":

you can see it because it reach your captors (which are not exactly at the place of focus)
you can see it through how it affects something else (secondary emissions or transformation, etc)
in microscopy, you "see" by sending probes (usually photons, sometime electrons) which are modified in their transmission or by their reflection. I guess photons might slightly modify some electrons' flight, but probably not practical at usual energies.

Anyway it also depend what you expect to see: photons are traveling pretty fasts ! or due you want to see something like "the still surface of a fast river", like, steady caustics at some places due to the imperfect optics within the laser generator ?
Comment: Do you have a digital camera? Do the photons modify the electrons in its CCD or CMOS chip? Of course they do. How about your eyes? Do the electrons in your light receptor molecules get modified by light? Now, if you want "direct" modification of free electrons in vacuum by photons, are x-ray energies "usual"? They are in your doctor's office and the scattering at those energies is fairly strong.
Comment: Imaging something where it is is not the same thing that collecting it or its effects through collision after travelling. Measuring a 3D (/6D) light field on place via beams of electrons is probably less easy than making a CCD captor.
And by energy, I meant intensity, not wavelength: the question was on light, not the whole EM spectrum. :-)
Comment: Please read up on the photo effect and Compton scattering.
Comment: photo effect on a beam of electrons ? Compton scattering: same thing (maybe the inverse Compton, then). Are you sure you understood the question ?


How many valid users to test the software that has been developed?

Question: Anyone know, how many users are valid to use the prototype of the software that we have made? I have conducted research on the reference to the "Software Engineering A Practitioner's Approach book Roger S. Pressman". There is no mention for valid users of software testing?
Comment: There is no general answer for this question. It depends entirely upon your software.
Comment: ... and about two dozen other considerations.
Comment: Please read my warning comment below the answer you accepted against applying the results of one study on each and everything. I would also recommend to have a look into [5 worlds](https://www.joelonsoftware.com/2002/05/06/five-worlds/)  of Joel Spolsky and think about in which of those "worlds" you are.
Comment: This [table](https://softwareengineering.stackexchange.com/a/301540/134647) mentions some types of user tests at the end.  As others have said, though, it depends entirely on the nature of the software and the nature of the users.
Comment: For what purpose?
Answer: In Usability testing there is a sort of rule, based on an article by Jacob Nielsen,  that tells you that 5 users is usually enough to find most issues on a given system. Beyond 5, the new finds are almost negligible and you need to go to 15 users in total to be able to find all problems.
Granted, we are talking about usability and not functional testing, but given that any test on functionality has to follow a set of requirements to know if the system complies or not with the acceptance criteria (in usability you'll have some general heuristics), you can extrapolate the notion that you may not need a lot of users to find most issues within a system.
Of course, if you are testing a gigantic application, using more people could probably help you find bugs faster, but that doesn't necessarily mean that they'll find more.
Comment: "on a given system" - well, the reference is about web sites exclusively, and it seems to refer to the few systems of the kind and size the authors of that paper have used for [their study](https://dl.acm.org/citation.cfm?id=169166) in 1993. So I recommend against generalizing such statements to all kinds of software, especially against applying it to the OPs system of which we know nothing.
Comment: No, the studies done by Nielsen (and several others throught the years) are not limited to web sites, nor are they confined to any kind or type of system. Some authors do argue about the number, but between 5 or 10, it is still a small number of users what helps you find most usability issues. As i said, you can use this as a reference, not as a definitive answer as one might not be available.
Comment: And size is no issue: for either functional or usability testing, guidelines (or test cases) would be followed. This can limit the scope.
Answer: There is no fixed amount of valid users of a prototype. Because it's not about the number of users but their expectations.
Using prototype-grade software has many potential problems: It

will be buggy,
might lead to data loss, and
will change frequently.

However, having actual users interact with a prototype can be super valuable because that lets the developers see what is really required and whether the software works as expected. This requires some amount of cooperation between users and developers, e.g. the users should be able to send bug reports if they counter a problem.
So the number of users and the openness of a pre-release program needs to balance the potential risks of using prototype-grade software on one hand with the benefit of being able to deliver more useful software more quickly.
A few example scenarios:
A new accounting software is developed. One accounting team of seven people is trained with a prototype software in advance so that they can test it in practice. Risks are mitigated by only transitioning one team as a kind of pilot study. Also, the previous software will still be available if something goes wrong, so bugs won't be business-critical. Developers interview the accounting team members to learn about their experiences and needs.
This kind of approach is great for agile development as directly interacting with individual users lets you discover their real requirements more quickly.
A software as a service web app regularly releases new features. To test new features, they are first deployed to a beta server. Users can opt in to using the beta server. This allows the users to select their acceptable level of risk and acceptable rate of changes. Many organizations need to test new releases before they may be deployed widely. The developers get feedback via analytics, automatic crash reports, and an online support system. Here, the group of beta testers may include many thousands of people.
This kind of approach is useful where you already have concrete features in mind but need to test whether they work in practice, using statistical data. Well known examples of this strategy include the Windows Insider program or the Firefox Beta release channel. Firefox also used to have the Testpilot program where new experimental features could be installed in the browser, without being connected to the browser's release schedule. And here on Stack Exchange, the most recent site redesign was opt-in for a couple of weeks before it became the default.


Bi-polar input (AC) to be used in op-amp

Question: I am planning to amplify AC signal (without sacrificing dynamic range) using MCP602(Microchip), but the datasheet shows input voltage is from 2.7V to 6.0V. My question is can I use negative supply in the MCP602, instead of ground. Do dual supply op-amps are different from signal supply op-ams, or can we use both interchangeably.
Answer: Yes, but the difference between the positive and negative supply voltages must be between 2.7 and 6V.
You can also AC couple the input and use divider resistors to generate a voltage mid-way between V+ and V- without using a split supply.

simulate this circuit – Schematic created using CircuitLab
You could also use a resistor divider to generate Vcc/2 and connect the ground side of the AC input signal to that instead of using the AC coupling capacitor and R1-R2 divider.
Comment: In the schematic show, the dynamic rang of the input signal would suffer, as we can amplify the signal till Vcc/2. My question is can we use dual supply in single supply op-amp.
Comment: Every op amp is a dual supply one. However a single supply op amp often means it will work close to or right up to the negative rail so you can connect the negative supply to ground in a circuit. The opamp can't tell the difference between ground and negative- it will just see a different input common mode voltage.
Comment: The circuit does not affect dynamic range vs split supplies.  Either way, for the MSP1602, the linear output range is Vss + 100mV to Vdd - 100mV.  So, if you are using a 6V single ended power supply, your output can swing from 100mV (0V + 100mV) to 5.9V (6V - 100mV) for a total swing of 5.8V.  If you use a split supply of -3V and 3V, your output swing ends up -2.9V to 2.9V for a total swing of 5.8V.  As others have said, there is no difference between a single ended or split supply opamp except the techniques for ground referencing the input.


How to get Linear dependent columns indices in big dataset?

Question: I have a dataset which is a size of 50k x 280 i want to find linear dependent column indices so that i can eliminate them, rank of my matrix is 249 so what i understand that there are 31 columns are linearly dependent, i know a method to get directlty indices pf LI columns by sympy's rref() method but it is taking so long to get indices, as i know that we can also get LI columns by pearson coefficient but how to choose which columns are LI by looking at matrix of size 280 X 280 of P values.
Answer: Try sympy's rref() on a simple random sample of 1000 observations. This should speed up the process by 500x. The downside is that you may get some false positives, but I think that's pretty unlikely. If you do see 32 or 33 free variables, you can just try a different random sample and take the intersection.
You can't rely on pearson correlation coefficients since they'll only find pairwise relationships. If, for instance, column_1 = 3*column_2 + 4*column_3, you won't see that by looking at pearson correlation coefficients.


Can I neglect kinetic energy when modeling convective heat transfer for viscous, low mach airflow? Why or why not?

Question: I'm looking at convective heat transfer for air flow over a solid at Ma=0.2.  At this speed, it is plausible to model the flow as incompressible because of small density variations in the air.  Normally, when modeling compressible flow, we include the kinetic energy term in the energy equation.  But for viscous incompressible flow at this speed, is it necessary?  Can I neglect kinetic energy as a component of the energy equation due to the incompressibility assumption?
Answer: You should be fine to neglect kinetic energy from a thermal perspective.
The kinetic energy per unit volume will be $1/2 \rho u^2$, which is about 2800 $J/m^3$ for your conditions. Given that the volumetric heat capacity is ~ 1200 $J/m^3.K$ The most you'd get out of your KE is a few degrees K (if you're lucky), due to viscosity, rather than compression. Unless you're dealing with such a small $\Delta T$ between your gas and solid, that effect is probably unimportant.


Why does an LLC have to be filed with most or all states but not an LP?

Question: For example, in New Mexico, California, and Montana, one must create an LLC by filing through the state government.  But as far as I know, none of those states require the same for a Limited Partnership.  (Though if doing a certain type of business in the states, one probably has to register the LLC/LP but that's a different matter than what I'm asking.)
Comment: are you sure?  "Limited Partnership (LP)
To form an LP in California, go to bizfileOnline.sos.ca.gov, log in, select Register a Business under the Business Entities Tile, Certificate of Limited Partnership - CA LP and follow the prompts to complete and submit.  "  https://www.sos.ca.gov/business-programs/business-entities/starting-business/types#:~:text=To%20form%20an%20LP%20in,prompts%20to%20complete%20and%20submit.
Comment: I guess you're right about CA!
Answer: A limited partnership is a entity with one or more unlimited liability partners and one or more limited liability partners. This form of organization was popular, especially in the 1970s and 1980s in the oil and gas industry and in the real estate development industry, before it was possible to have partnership taxation with a fully limited liability entity (which became possible in the 1990s).
The limited partnership form of organization allowed limited liability investors to receive passthrough taxation, because there were other unlimited liability investors. Unlike limited liability partnerships (LLPs) and limited liability limited partnerships (LLLPs), some partners in a limited partnership (LP) have unlimited liability.
These days true limited partnerships that are not limited liability limited partnerships are used mostly in estate planning, and residually by not very up to date lawyers in the oil and gas industry and real estate development industries where they used to be the preferred form of organization. One of the main reasons that they are used now is that state law typically decrees that limited partners have no voting or management rights, which is attractive to many promoters and family business owners who want to bring in additional investment or transfer economic ownership rights without ceding control.
Historically, limited partnership organizational documents had to be filed in the real property records of the counties where the limited partnership owned real estate, because they were used predominantly for real estate investments and because unlimited liability general partnerships usually don't have to file organizational documents with the state because they are the default form of organization.
Most states have since changed their laws so that limited partnership have to file organizational documents with the state level Secretary of State (see, e.g., Colorado, California, New York, and Texas) rather than in county real property records. But a few states, by virtue of inertia, have not changed their laws and retained the historical requirement. Both the historic registration requirement in real property records and the modern requirement of Secretary of State filings flowed from the language of widely adopted model statutes for limited partnerships.


Can electric field pass through a charge particle so that the charge particle behind the one can get energy?

Question: Can an electric field pass through a charged particle? As the field is energetic it would get absorbed by charge as I think .  Let assume that there are two electrons one behind the other such that it seems like one electron in front is blocking the electric field to reach that electron behind. Would that charge be able to get that electric field?
Answer this question so that I can ask another question related to it.
Answer: Your point about "blocking" electric field is interesting in that what you suggest is similar in concept to electron shielding or screening, where outer shell electrons are partially shielded or "blocked" from the attractive force of the protons in the nucleus. This is actually caused by inner shell electrons, and the electric field of the nucleus is reduced.

Figure: Electron shielding is due to the blocking of valence shell electron attraction by the nucleus, due to the presence of inner-shell electrons. The interior electron cloud (light blue) shielding the outer electron of interest from the full attractive force of the nucleus. A larger shielding effect results in a decrease in ionization energy.

In an electric field, the field lines are not really "passing through" the charges as you have suggested, but they feel a force, and are accelerated by the field. But they don't really "absorb" the field since the field is unchanged before and after interaction with charged particles in the field (they do change their energy when interacting with the field).
The electric field is a region of space where a charged particle experiences a force at any point in that region. Assuming we ignore the electric fields of the individual charges$^1$ in the larger electric field $\vec E$, the location of one charge relative to another, or if we have "one behind the other", is irrelevant and both charges will still experience a force that depends on their location. The electric field (force per unit charge) is given by $$\vec E=\frac{Q}{4\pi \epsilon_0 r^2} \frac{\vec r}{\mid \vec r\mid}$$ where $Q$ is the charge causing the field and $\vec r$ is the location of the charges, and for simplicity assume that if we had individual charges $q$ in the field, then $Q\gg q$.
$^1$ Individual charges will have electric fields that affect each other, but here we are considering the effect of one large central charge on unit charges in that field.
Comment: You said "there is no electric field passing through" first of all electric fields do have a velocity which they will travel and they also have energy which is told in poynting theorem. So if the work dones by a source charge on a q charge is because of energy in the form of electric field released by it and absorbed by q. If the charge behind can still get energy which would mean that the energy was not completely absorbed by q or there is nothing like energy as physical thing to do with electric field
Comment: @PredakingAskboss see https://en.wikipedia.org/wiki/Electric_potential_energy
Comment: Do field have a real physical significance as it is said that they have energy stored which means they really exist and are not imaginary
Comment: they are a mathematical representation of the effect of a charge particle on charged particles that fits all data up to now.
Comment: Hi! You may find my question [here](https://physics.stackexchange.com/questions/647342/how-was-it-found-that-a-charge-produces-an-electric-field-rather-than-a-local) interesting @PredakingAskboss
Comment: @Predaking Askboss What do you mean by "electric fields do have a velocity"? Are you referring to the fact that a _change_ in an electric field moves at $c$? Is your original question about static electric fields or what?
Comment: It is not correct to think that the lines of force emanate with a velocity, and "strike" the charges in the same way say, a fluid flows and sweeps some object in its path.
Comment: Your explanation must fit the other requirements even if it is not shown in question. I am  not talking about any static or moving charge just want to know the real explanation that fits everywhere
Comment: If the force doesn't flow at a velocity then there are only two results of that first, there is a medium which helps in propagation of that force just like sound in air( but i don't know if space has any kind of medium). Second, The force isn't travelling instead it is instant which will then violate the speed of information
Comment: @joseph h   If the electrons in the inner shell were really shielding the outer shell from energy(E.M.F) given out by nucleus then, for this to happen the energy must be absorbed by electrons in that inner orbit and the electrons should accelerate out of orbit. And we know that when this happens one of the thing that can happen is that, electron will release some energy and will come down to its intial shell
Answer: I guess you are misunderstanding the concept of the electric field, let me help you.
The electric field is not like a fluid that collide with obstacles and change its propagation after it.
Any field (electric, gravitational or magnetic) are a region of space where at each point there is a single well-defined value of a given physical magnitude. This fields are originated by the presence of a charge in some region and the range of the field that they create vanish at infinite distances from the charge. This can be seen in the equation of the electric field created by a pointlike electric positive charge: $$\vec{E}=\frac{q}{4\pi \varepsilon_0 r^2}\vec{u_{r}}$$,
where $\varepsilon_0$ is the electrical permittivity in the vacum and $r$ the distance from the charge $q$.
Answering your question: no, a charge will never absorb the electric field because it is a magnitude that depends on the point $r$. Any charge placed at a non-infinite distance of the first charge will feel the electric field with a intensity given by the later equation.
Comment: The energy of the system can be defined when it exists more than one particle on it. If you have only one particle there only exists the field $\vec{E}$, or the potential $V$. Nevertheless when you have a system of more than one particle you can talk about forces $\vec{F}=q_2\vec{E}$, or the potential energy $E_p=q_2 V$. So, the energy in the electric field is not absorbed or emited by anything, the energy is a quantity defined due to the interaction of multibody systems on each point of space, and interacting particles can take advantage of this interactions to move through the field.
Comment: But electric field contains energy and if it is not absorbed and can apply force on other charge after being absorbed by some charge then we will get more energy output.
Comment: "The electric field is not like a fluid that collide with obstacles and change its propagation after it." Actually, if your fluid is inviscid & incompressible, then its field equation is Laplace's equation (same as for the electric field).
Comment: Thank you for pointing this out. I intented to make an analogy with a "real fluid"
Comment: First of all nothing like phyiscal fluid is described here by me. If someone says, "no, a charge will never absorb the electric field " then, i would ask," the electric fields strength helps us determine the amount of energy passing through a certain region in space. And when you provide that region with a charges (that is free to move), the charge will get accelerated and hence will also store some energy in its motion. This means that electric fields strength must decrease so as to ensure law of conservation of energy"
Answer: Current answers seem to focus on static fields, while the question seems to be about incident propagating fields. I'll try to address this (purely classically). Whether one thinks of electric fields as something corporeal or not, processes akin to what you describe happen constantly (if only partially) in conductors and dielectrics, but it's certainly not the case that a propagating electric field gets "absorbed" entirely whenever it encounters a charge, or else light could never propagate through the atmosphere.
If an electromagnetic wave were incident along an axis containing two free charges somehow arranged to be initially at rest the instant the incoming wave reaches the first charge, the response of the first would certainly impact the response of the second: the first accelerates in response to the varying field, causing it to generate its own radiation which interferes with the original wave, and one could think of the adjustment to the net propagating field as meaning the first charge partially "absorbed" the incident wave. But exactly how things play out would depend in a complicated way on the incoming wave-- its frequency, amplitude, and breadth. What can be said quite confidently is that it could not cancel the incoming field completely.
Remember that E&M is linear and hence has the superposition principle: one can think of the dynamics as the incident wave continuing on exactly as it would in the absence of the charges, plus the perturbation of whatever fields the charges generate as they move. Generally speaking, a single charge won't be able to muster much of a perturbation, so for the most part the wave would continue on on its merry way. Only with systems of charges responding en masse in an uninhibited manner can you get effective screening against propagating fields, as in a Faraday cage.
Comment: This dosen't work if i am right. Let me explain. Try to visualise the Coulombs law, it says that the to check the strength of electric field(which provides energy) extend outwards making  radially(showing spherical structure).  And to find the total strength on any of the test charge we draw a conical shape extended to the test charge BUT if you try to make re extended a conical structure from the charge thag is absorbing energy  you'll get a much smaller amount of energy reched to charge behind due to re-release.
Comment: I will make another  question showing that with an image
Comment: https://physics.stackexchange.com/questions/647963/do-electric-fields-really-contain-any-physical-form-of-energy-in-them  this is link for that what  i was talking
Comment: @PredakingAskboss It seems the scenario you have in mind is quite different from it seemed through my reading of this question (you weren't discussing propagating waves after all). I've answered that question in an attempt to clarify the misalignment of your image with standard E&M.
Answer: It depends:

A single unit of charge, located at fixed point in space will generate a field on its own, which will superimpose itself on a pre-existing field, hence it will influence the local field, but it will not have a shielding effect, it will merely add its own contribution to the global field.

For instance, when calculating the electric field resulting from a homogeneous spherical charge distribution (Maxwell $+$ Gauss), the contribution of the successive layers simply add up, hence the resulting field is the sum of all the elementary contributions, there is no shielding effect from a layer to the next.
In the preceding examples, however, all the masses are regarded as fixed in space, therefore the formulation of the problem is essentially static as if said charges were held in place by some magic glue.

In practice, however, charges can be found in insulators, conductors, semiconductors, plasmas.

In the case of conductors, the free electrons will tend to have a shielding effect as in the Faraday cage.

If a spherical (quasi-homogeneous) distribution of charge is build by stacking a large amount of charged concentric hollow spherical conductors (insulated from each other) of increasing diameter (like a Russian doll), every shell will shield the outer world from the charges contained in the other inner spheres it encloses and the only sphere contributing to the field outside the stack of spheres will be the contribution of the outermost shell.

In this last example, the shielding effect, however, is not due to the individual charges, as such, but to their ability to move freely inside each individual hollow sphere. This is a collective effect of the free electrons.

Another exemple if provided by Bragg diffraction: An electromagnetic wave (e.g. X-rays) will be reflected by a cristalline matrix (e.g.diamond) if the geometry meets the Bragg conditions. In this case again, it is some collective  behavior of the electrons of the atoms of the crystalline matrix which is responsible for the reflection, not the charges individually.
Answer: Yes, in the same way that the Earth's gravity can get to you through the chair you're sitting on.
There will also be an electric force between the two particles, dependent on their charge + distance, but the Coulomb forces all just add linearly for stationary charged particles, just like gravitational forces add linearly.
An electric field between two plates for example is created by large amounts of charged particles in each plate, many of them cancelling each other out (e.g. neutral atoms with equal numbers of each), but at least to a good approximation, everything just adds linearly.
(Electromagnetism and gravitation are different fundamental forces and behave differently in some ways, not just in terms of having different "charges" (electric vs. mass).  e.g. accelerating charges create radio waves somewhat differently from how accelerating masses create gravity waves.  But they're the same in this respect, and I think this is a useful analogy.)
Answer:
Can an electric field pass through a charged particle? … Would that charge be able to get that electric field?

Definitely, yes. Maxwell's equations are linear in the fields. This means that the principle of superposition applies and the field at any point is equal to the sum of the fields from all of the charges. Charges do not "block" the electromagnetic field.
Now, other answers have mentioned screening effects, and they are not wrong. Screening effects are not an example of the field being blocked, but rather an example of a field in the opposite direction being provided. For example, if you have a positive charge you get one field, and if you have a negative charge close to the positive one its field is almost opposite. So if you have both then they nearly cancel out and you are left with almost no field. Not because one absorbed the field of the other, but because they simply added together to make a small total field.

As the field is energetic it would get absorbed by charge as I think . Let assume that there are two electrons one behind the other such that it seems like one electron in front is blocking the electric field to reach that electron behind.

As in your previous question, to understand energy conservation in EM you must understand Poynting's theorem. For a charge to absorb energy from the field requires that $\vec E \cdot \vec J$ must be greater than 0. If the charges are all stationary, then $\vec J=0$ so no work is done. Only if the middle charge moves would some of the energy not be available to reach the last charge.