Left For Dead Characters

Question: Are the characters in Left For Dead 1 and 2 modeled after real-life people?
Answer: It depends on what you mean by "modeled after real-life people".
If you mean visually, then yes! Valve usually models the face of their characters based on some real-life person (probably a hired model or actor, but I can't confirm this). From Left4Dead Wikia, here is a list of the face models for each character (as well as the voice actor):

Game
Character
Face model
Voice actor

L4D1
Bill
Bernard Fouquet
Jim French

L4D1
Francis
Taylor Knox
Vince Valenzuela

L4D1
Louis
Sean Bennett
Earl Alexander

L4D1
Zoey
Sonja Kinski
Jen Taylor

L4D2
Coach
Darnell Rice
Chad Coleman

L4D2
Ellis
Jesy McKinney
Eric Ladin

L4D2
Nick
Taymour Ghazi
Hugh Dillon

L4D2
Rochelle
Shanola Hampton
Rochelle Aytes

The body is easier to model in 3D, and is probably based on sketches made by Valve artists. However, it is very difficult to accurately model a face, and that's why they use real-life models for faces. (but, of course, they fine-tune the modelled face to fit better to the character, like the G-Man from Half-Life series)
Comment: Valve often uses real people as models for their in-game characters.  Dr. Vance is modeled after a homeless person (http://half-life.wikia.com/wiki/Eli_Vance#Behind_the_scenes).  More: http://fc07.deviantart.net/fs38/f/2008/363/1/0/Half_Life_2_Character_Models_by_leebyrne.jpg
Answer: It appears that they are just stereotypes:
Wikipedia article

There are four playable human characters in the game: Francis, a stereotypical biker (voiced by Vince Valenzuela); Bill, a Vietnam Veteran (voiced by Jim French); Zoey, a university student (voiced by Jen Taylor); and Louis, a district Account Manager (voiced by Earl Alexander).

(my emphasis)
However, I wouldn't be surprised if each characters appearance wasn't based on a real-life person - but that person could be anyone, even one of the developers!

Left For Dead Characters


Probability for Magic Trick

Question: I had a probability problem to solve , but could not proceed further , we have a m identical decks having n cards , where each deck has n different cards . Now shuffle them and select n cards . Now a someone chooses a card from the new deck , I do not know the card but I also take card after he has put it back , what would be the probability I would pick the same card. So any hint or method to solve would be appreciated.
Answer: One suggestion would be to use conditional probability / the law of total probability.
If I understand the question correctly, we have m identical decks, each with n cards. We shuffle all m * n cards together, and now have a pile with n distinct card types, each with m copies. We then draw n cards randomly from these shuffled cards, and we want to compute the probability that two cards selected from this new deck are of the same type (call it type i). Note: My apologies for the lack of mathematical notation in the following explanation -- not super quick with LaTeX yet, and I figured you'd get the gist anyways.
Now, what is the probability of selecting two cards of type i if we selected 0 of type i in our initial draw of the new deck? What about if we had selected 1 of type i in our initial draw? What about two, and so on? Once you've calculated these conditional probabilities (there will be m of them, unless m > n, in which case there will be n), you can multiply each of them by the probability of actually drawing the specified number of cards of type i when drawing the new deck (can be calculated with some basic combinatorics ... or remembering a handy discrete distribution). You now have the joint probability of selecting two cards of the same type after drawing k cards of that type, where k runs from 0 to m for m < or = n, and 0 to n for m > n. Since each of the "draw" events are disjoint and their union covers the event space, we can use the law of total probability to calculate the desired probability (just add them up). For more information, take a look at the wikipedia article on the law of total probability.
Answer: Let $x$ denote the card drawn from the new deck. The probability that the new deck contains (exactly) $k$ copies of $x$ will be
$$
P(k)=\frac{\binom{m}{k}\binom{mn-m}{n-k}}{\binom{mn}{n}}
$$
The probability you seek will then be
$$
P=\tfrac{1}{n}\cdot P(1)+\tfrac{2}{n}\cdot P(2)+...+\tfrac{m}{n}\cdot P(m)
$$
Unless of course $m>n$ then you must truncate the above sum at $\frac{n}{n}\cdot P(n)$.
maybe the parameter mn-n should be mn-m

Probability for Magic Trick


Using MPFR (mpreal) with Eigen3

Question: I've seen on the Eigen3 website that there is an "unsupported" module for MPFR multi-precision data type. Does anyone know how reliable this is?
My main goal is not only to use mpreal, but complex < mpreal >.
Answer: It works fine so far for me and gives as far as I can tell the correct results.
However you will run into problems when trying to do complex as others have pointed out that complex only works with built in datatypes. Currently I don't know how to produce arbitrary precission complex numbers.

Using MPFR (mpreal) with Eigen3


How to calculate the distance between cluster center and datapoint in K-means

Question: I'm learning about the K-means model.
I'm going to detect the anomalies using the k-means for that I need to calculate the distance between centers and point how should I do that.
Comment: By pythagorean theorem, if your input are the coordinates of data points plus of the centres.
Comment: sorry, I didn't get it
Comment: (L2) norm(x-m) where m is the centre and x is the point. But this doesn't necessarily pick out real anomalies. For some clusters points 'far away' from the centre could be perfectly normal. You may do better using GMM and allowing each cluster its own covariance matrix.
Answer: You can just compute the Euclidean distance of any points within one cluster with respect to that cluster's centroid. For categorical variable, it is preferable to use distance as either 1 or 0 (means if 2 observations of same features are having same class then take distance for that feature as 1,else 0). To get the K-means cluster(assuming that you are using some language for clustering), K-means generally have a attribute like "cluster_centers_" (available in python)


Can the off-diagonal elements of Fisher information matrix be negative?

Question: The concept of Fisher information is new to me and as I understand the diagonal elements of the Fisher information matrix (FIM) are proportional to mean square error (to be precise the inverse of FIM). What is the interpretation of the off-diagonal elements of FIM and can these be negative?
Comment: The only constraint is that the matrix is semi-definite positive. Some elements may still be negative.
Answer: TL;DR: The information matrix can be diagonalized, in which case its diagonal elements are positive, whereas non-diagonal elements are zero.
A meaningful information matrix must be positive-definite, but there is no requirements that non-diagonal elements must be positive. That such elements are present indicates that the chosen variables are correlated, and one should exercise care when interpreting diagonal elements as mean square error. This interpretation is correct, if we diagonalize the information matrix, in which case the diagonal elements are positive, while non-diagonal elements are zero.
Answer: A counter-example: For a Beta $\mathcal B(\alpha,\beta)$ distribution, the Fisher information matrix on $(\alpha,\beta)$ is
$$I(\alpha,\beta) = \left(
\begin{matrix}
\psi'(\alpha) -\psi'(\alpha+\beta) & -\psi'(\alpha+\beta)\\
-\psi'(\alpha+\beta) & \psi'(\beta) -\psi'(\alpha+\beta)
\end{matrix}\right)$$
where $\psi(\cdot)$ is the digamma function and $\psi'(\cdot)>0$.
Answer: There is a link between Fisher information matrices and covariance matrices, see Demonstration and Interpretation between a Fisher matrix and its dual space which is covariance matrix  for details.  So asymptotically Fisher information can be interpreted as an inverse covariance matrix, that is, a precision matrix.  For derails see How to interpret an inverse covariance or precision matrix?  and especially Why does inversion of a covariance matrix yield partial correlations between random variables?.
Partial correlations can certainly be negative.


No internet access using USB tethering on Android phone

Question: I used to access the internet on my notebook by tethering it to my android mobile phone. However recently this stopped working. Even though the notebook was tethered to the phone there was no internet access. Changing phone settings did not help.
Answer: On googling I found that my mobile service provider recently disabled internet access over tethering. This can be fixed by the following the steps described here.

Enable USB debugging on your android phone (Settings > Developer options > Enable USB debugging).
Download adb.exe from here and follow the steps here.
Connect the phone to your computer using USB cable.
Open command window in the folder containing adb.exe.
Type: adb devices -l (this will list your device model).
Type: adb shell (to enter in your mobile's shell).
Type: settings put global tether_dun_required 0.
Reboot your phone and you're done!
Answer: After a lot of headache and trying everything, including Martin's answer (but I got no permissions error when I tried settings...), this solved the problem for me:
Go to Configurations > Wireless & Networks > Mobile Networks > Access Point Names, make a copy of the default selected APN (you probably need to input a copy all by hand after going to "New APN" in a menu), and then in the copy add ,dun to APN type (if it was default, make it default,dun).
Answer: I have resolved this issue by changing the APN in mobile. Go to your cellular network -> choose new APN with the same configuration of your selected one or create new APN with the same configuration of current one.


what can be used instead if captcha is insecure?

Question: As far as I read from here, the fact that captchas are not 100% secure. What can be used instead of captcha? As a programmer, what do you think? How to solve this issue?
Edit: thanks for all the answers.
Comment: See my answer on this question: http://stackoverflow.com/questions/2387496/how-to-prevent-robots-from-automatically-filling-up-a-form/2387686#2387686
Comment: Its "insecure",  unsecure isn't a word.
Answer: This is an unsolved problem, and will become more unsolved as time passes. The better the OCR tools get, the smaller the gap between humans and computers, and the harder it will be to tell them apart. Eventually, computers will be indistinguishable from humans, and then the game will be up.
If your server wants to make sure that a human is at the other end of a TCP pipe, there isn't a turing-test in existence that won't eventually be defeated (and there probably never will be one). CAPTCHA is doomed, it's just a matter of how soon.
Of course, that doesn't mean it's all over as far as human authentication is concerned. It just means that automated turing tests, as convenient as they are, won't be an effective way to achieve this for very much longer.
Comment: +1 I bet the next hurdle is facial or retinal scans. Most computers have webcams to do this already. The software side maybe a little behind.
Comment: "Eventually, computers will be indistinguishable from humans", yeaa .. I don't think so. They will get harder to distinguish for a computer maybe.
Comment: @hasen j, I think you are missing the point. We are not talking about a cyborg approaching a bank teller. We are talking about an unknown agent at the other end of a TCP pipe that simply has to respond with the right set of bits to convince another machine that it is a human sending those bits. I'm not saying this is an easy problem, but significant progress has been made, and CAPTCHA technology is already beginning to buckle under the strain.
Answer:
Captcha involving human reflexion (like calculation, really simple question, and the like).
Session tokens
randomly generated hidden input which requires to be null, on the server side generate a random identifier, keep it in a session for a while. If the input is filled and not null, then it might have been filled by a robot, do your users will fill an hidden input ?

I think it really depends on what you are trying to control over the use of captcha.
Comment: Then, generate two hidden input, one which needs to be filled but no others.
Comment: @Boris Guéry thanks, what do you mean by randomly generated hidden input which requires to be null?
Comment: Scripts tend to fill in every input field with some data, although when this prevention technique starts gaining traction, those automated scripts will simply ignore hidden inputs fields, while populating the rest of the form.
Comment: How were you going to fill that extra hidden input field? Ask the human to handcraft their HTTP POST?
Comment: @calmh.. the second input field can be linked to any of the visible input fields through JavaScript that has to be filled.
Comment: @anurag Yes, but Javascript is no problem for bots. It tends to get executed.
Answer: Further explanation of a suggestion made by Boris:

randomly generated hidden input which requires to be null

The idea is that your form contains several invisible inputs, their type should probably not be set to hidden, but they should be invisible to a human (e.g. set width or height to 0). The initial content of these fields should be empty. If a human fills out the form, the field will be empty, because the human cannot see the field in order to enter anything into it, but if a bot fills out the form the field will (possibly) not be empty, because bots usually just blindly enter something into every field.
Thus, you can distinguish between a bot and a human based on whether the content of this field is empty.
Comment: never thought of that, very clever!
Answer: Although captchas can be broken, Capthca's only add to security reCapthca is very good,  and a trained OCR like Tesseract is going to have very limited success in breaking it.   However,  there are outfits that use Human Computation  to break them for pennies.   But this makes attacks against your system more expensive,  and thats the best you can hope for.   Cryptography can be broken with brute-force.  All password hashes are breakable,  but we still use them because it makes it harder for the attacker.
Most of the "solutions" on this thread are "Security Though Obscurity"  and you should be wary of these quick fixes to a very complex problem.
Comment: @Berkay thank you,  that is a great complement. It isn't easy being on top (http://stackoverflow.com/questions/tagged?tagnames=security&sort=stats&pagesize=50).
Answer: Captcha's are used to determine that an actual human being is doing the request, not a machine. Captcha's and captcha-like systems will upgrade, and so will the technology to break them.
So how do you proof that you're talking to a human and not a computer? You could for instance require users to engage in a chat session and have small conversation. There's no AI nowadays that pass the turing test.
So the answer is, no system is perfect. Don't try to solve this issue, but try to find a way to reduce the impact of this.
Answer: In the long run government could run openid servers as digital passports for their citizens.
It would be a clean way to identify human beings and prevent sockpuppeting.
Comment: Am I the only person that hopes this never happens?
Comment: Although i agree i don't see how this solves the problem of bots.
Answer: At the moment on my website I opted for simple questions. Some questions I've used in the past:

What is two to the power of one?
What is 2+2? (this one was hacked though so don't use it)
What is the name of this website domain?
What is the sum of two and two?

Some other nice ones could be

type in 'stuff' to this box as a spam check
What does 1337 look like? (using only letters)
the current year is?
Comment: Why would anyone use this Turing test over reCapthca?
Comment: @The R the whole point of this question is alternatives to Captchas
Comment: I understand that,  but what you are describing is a capthca. or a "Completely automated Turing test"
Comment: @The R and how is my suggestion different from the other answers here?
Answer: The best way I can think of is using something unconventional, like a special hidden field that should be null (or another specific value) that robots will mess with.
If some robot maker adjusts his robot for your site, you'll have to quickly change the captcha to something different. It will (hopefully) take a good while before another robot maker adjusts his robot for your site.
Basically, it's a security through obscurity that has to constantly change to remain obscure.
This won't work very well if someone is specifically targeting your site.
Answer: Its just an idea, id used that in my application and works well
you can create a cookie on mouse movement with javascript or jquery and in server side check if cookie exist, because only humans have mouse, cookie can be created only by them
the cookie can be a timestamp or a token that can be validate
Answer: Gets the coordinates of the mouse, determine whether the coordinates have changed, you can determine whether it is a robot.
Then encrypt the coordinate data.
Answer: I realize that this is an older question. However, in 2024 the best CAPTCHA is a form that requires payment information. This may not be ideal for freemium services or unauthenticated input like a common contact us form.
The best way to ensure that your user is human is to require a payment method as it will be more expensive to spam your site.
Take a look at what Elon Musk / X / Twitter is doing. The verified accounts cost money so as to filter the bots.


What happens when the logbook of an aircraft is lost?

Question: If the logbook for an aircraft is lost, how does one replace it? Is the airplane still airworthy?
Comment: One thing I know for sure is that the resale value of that aircraft goes down pretty significantly.
Comment: you might want to start from here: http://www.coolinglaw.com/Content/UploadedFiles/Documents/Articles/Aircraft%20Logbooks%20.pdf
Comment: would photographs of each page of the logbook count? seems like it would be easy to just photograph every new page.
Answer: In order to state the plane is airworthy you must be able to show that all AD's have been complied with and that the aircraft is within annual (or 100 hour) inspection; these proofs are typically found in the logbook so no, without the logbook your aircraft is probably not airworthy. It may be possible to fly the plane legally in the US anyway if you receive authorization from your FSDO (which, I believe, would be a "ferry permit").
It's not legally required that your logbook be complete to the beginning of time, just that it shows the current requirements are met (along with the signatures/cert numbers of the AI's that are stating them to be met). However, an incomplete logbook is very important from a resale point of view. Not having complete logs will make a plane very difficult to sell and will certainly affect the price most purchasers would be willing to pay.
Answer: As mah pointed out, if you lose all your logbooks your aircraft/engine/propeller/avionics are no longer airworthy because they lack the documentation of required inspections, AD compliance, etc.
Everything is still "airworthy" in the sense that the aircraft doesn't know anything about paperwork and will happily fly, and getting a special flight ("ferry") permit from the FAA so you can take it to a mechanic to fix the paperwork will be pretty easy, but if you just go flying and something goes wrong the FAA and your insurance company will be most displeased about the lack of required paperwork.
In addition to the aircraft turning into a paperwork-pumpkin two things happen when you lose logbooks:

The value of the aircraft drops (I've seen aircraft selling for 30% less because of missing logs)
You have to "reconstruct" the aircraft's history and demonstrate compliance with all ADs and required inspections (an annual, an ELT check, a transponder check, a pitot/static check for IFR, etc...)

There's really not much you can do about #1 -- Logbooks are the history of the aircraft (and engine, propeller, avionics, etc.), and someone buying an aircraft is going to want to know that history.

How much a missing logbook matters is a matter of the buyer's personal opinion and what's missing. For example, I bought a plane missing Engine Log #1 (of 3). The engine has had two overhauls by well-known shops, one documented in Log #2 and one recently in Log #3, so my mechanic and I were not too concerned about the contents of Log #1.
  Had the first airframe log been missing (which documents the plane's life as a flight school trainer in Florida) there would have been a more substantial impact on the aircraft's value.

As far as #2 goes, "reconstructing" logs is usually pretty hard.
If you're the first owner, you've always used the same shop, and your mechanic keeps meticulous records they may be able to print you out a replacement log of everything that's ever been done to the aircraft and sign it off with no sweat.
If you're not the first owner but everyone who's owned the aircraft kept meticulous records (shop invoices, etc) you still have the history even if it's not in "logbook" format -- it won't satisfy the FAA, but it will make the process of inspecting the aircraft and determining its airworthiness a little easier for your mechanic.
If you're not in either of those situations your mechanic is going to have to give the aircraft a thorough going over (in addition to conducting an annual inspection they will need to verify that every AD has been complied with (or doesn't apply). That can be a very time-consuming (and expensive) process - particularly if there are many "inspection" ADs.
Other details about the aircraft's history can usually be reconstructed from your national aviation authority & accident investigators (in the USA you would do an N-number & Serial Number search in the NTSB aviation accidents database, and request the aircraft record from the FAA which will include things like FAA Form 337 for major repairs, alterations, and STCs.
Comment: Is '337 forms' a type of form, or a boatload of forms?
Comment: @cpast FAA Form 337 is the form used to report "major" repairs or alterations (such as rebuilding a wing after a ground handling incident).
Answer: It's important to know what an aircraft's logs are for - what they evidence, to understand why it's such a big deal when logbooks are lost.
To legally fly in the US (and UN/ICAO member territories), an aircraft must have a valid Type Certificate. This certificate signifies the aircraft's design meets legal requirements. Alterations, including significant repairs, require a Supplemental Type Certificate. STCs amend the original TC. STCs are filed with the aviation authority, and can be requested by anyone.
Modifications must be installed by a certificated A&P, aviation-speak for 'licensed' airframe and powerplant mechanic. Whenever a mechanic works on an airplane, they must make a logbook entry describing the work performed, the procedures followed, and the result (was the work accomplished successfully). The entry must include their certificate number.
The Type Certificate is valid only when accompanied by a logbook related to the airframe's serial number. So, every airplane must have an airframe logbook. If the airplane has engine(s), each engine must have its own logbook. If an engine has a propeller, the propeller gets its own logbook, too.
Each part installed in accordance with the STC installation document must be manufactured properly. This means it was made by an authorized manufacturer (PMA) and that manufacturer attests the part complies with the necessary specifications. Compliance is indicated by the 8130 form, commonly called a 'yellow tag'. They used to be born yellow, but given the age of most general aviation aircraft, most turn yellow in time.
Airplane designs are imperfect. These design flaws are usually found in mundane ways, but sometimes they make the news. When the flaw is moderate, the manufacturer issues a service bulletin, which is highly recommended, but not necessarily mandatory. If the imperfection is serious, the FAA issues an airworthiness directive, usually referring to the service bulletin. Airworthiness directives are mandatory, and must be addressed within a certain number of flight hours or some similar measure. These directives can require a one-time effort (e.g. replace a bolt with a differently spec'd bolt) or they can be recurring (inspect this part for cracks every 100 flight hours).
FAA regulations also require a thorough annual inspection of the aircraft. This must be performed by an extra-experienced A&P mechanic, one with 'inspection authorization'. The inspection must follow the aircraft's FAA approved maintenance manual. The results of the inspection are documented with entries in the various logbooks, including which revision of the manufacturer's inspection procedure was followed.
If the aircraft will be flying under instrument flight rules, the altimeter system must be tested every two years. This test must also be logged.
Although all of these logbooks are not required to be in the plane during flight, they must be made available for inspection if requested by an authorized person -- usually an FAA representative. Without these logbooks, the fed will issue a citation and possibly suspend the pilot's flying privileges.
It is interesting to note the pilot is ultimately responsible. The owner, as far as the FAA is concerned, has little responsibility. Although the owner may be sued in civil court, the pilot will receive the FAA's wrath.
So, if you lose your airplane logbooks, what do you do? You can request the type certificate and STCs. You must then attempt to reconstruct as many records as possible. You can review your emails and invoices. You can contact the places you purchased repair parts. However, if you cannot prove compliance with applicable airworthiness directives, you must re-perform the procedures required by the AD - like replacing the bolt, or inspecting the wing spar for cracks.
All of this will be costly, even by aviation standards. For old aircraft without historic value, it can be better to sell the plane for parts.


Was Manusmriti the constitution of Ram Rajya?

Question: In Sarga 18 of the Kishkindha Kanda of the Valmiki Ramayana, Rama explains to a dying Vali as to why He attacked him behind his back. Rama categorically explains all the questions put by Vali in the last chapter from the viewpoint of scriptures (Manusmriti) that lays down the principles of Sanatana Dharma. Vali finally concludes that there is someone superior to him and thus seeks refuge in Rama.
Answer: No. It is impossible. The Manusmṛiti mentions the fall of Kṣatriya groups that we know happens long after Treta Yuga. That, the Parāśara (possibly Pārāsara) Smṛiti saying it is for Satya Yuga and it mentioning the Sarasvati river as a landmark means it has to be talking about a future Satya Yuga. It makes no sense to apply to Rama Rajya.

शनकैस्तु क्रियालोपादिमाः क्षत्रियजातयः । वृषलत्वं गता लोके ब्राह्मणादर्शनेन च ॥ ४३ ॥
śanakaistu kriyālopādimāḥ kṣatriyajātayaḥ | vṛṣalatvaṃ gatā loke brāhmaṇādarśanena ca || 43 ||
But by the omission of the sacred rites, and also by their neglect of Brāhmaṇas, the following Kṣatriya castes have gradually sunk to the position of the low-born.—(43)

पौण्ड्रकाश्चौड्रद्रविडाः काम्बोजा यवनाः शकाः । पारदापह्लवाश्चीनाः किराता दरदाः खशाः ॥ ४४ ॥
pauṇḍrakāścauḍradraviḍāḥ kāmbojā yavanāḥ śakāḥ | pāradāpahlavāścīnāḥ kirātā daradāḥ khaśāḥ || 44 ||
The Puṇḍrakas, the Coḍas, the Draviḍas, the Kāmbojas, the Yavanas, the Śākas, the Pāradas, the Pahlavas, the Cīnas, the Kirātas, the Daradas and the Khaśas.—(44)

P.S. This timing is a good reason not to trust the Agni Purana's quoting of the laws of Manu from Pushkara. As the first Manu writing it means it would be highly unlikely he actually got emancipation as the common translation suggests as he would still be alive near the present.
https://www.wisdomlib.org/hinduism/book/manusmriti-with-the-commentary-of-medhatithi/d/doc201773.html
Comment: https://hinduism.stackexchange.com/a/49381/25960 this answer contains refutation of your claims regarding the Parashara Smriti statement (ps: it doesn't have any Agni Puran quote)
Comment: @जलन्धर Where are you getting your information from? I tried to find the Mahabharata quote and got nothing related. https://www.sacred-texts.com/hin/m03/index.htm https://en.krishnakosh.org/krishna/Mahabharata_Vana_Parva_Chapter_3:2
Comment: You tried searching in the prose translation you will find nothing there.....the prose translation u gave has different number arrangements.....however the site has the Sanskrit slokas if u can read Sanskrit try checking the verses who's references I gave. However I referred u to that post for countering the Parashara Smriti quoted u used as they already have been refuted kindly stick to the topic of Parashara Smriti first I will talk about the Mahabharata one as well
Comment: Ok so u searched a different site altogether saw just now. Do one thing try finding the verse by verse translation (Gita press one is available in pdf as well) then check for the references.
Comment: @जलन्धर What site are you talking about? How would reading the Sanskrit fix everything? The Mahabharata has different versions in Sanskrit too.
Answer: Short answer to your question is yes and there are more proves than just what you mentioned some references being direct and others indirect. Long answer with references waits ahead.
I) DASRATH RAJYA:
To know Ramrajya one must first know what was Dasrath Rajya as well and as the article progresses they shall understand why I have included Dasrath Rajya in this. The Valmiki Ramayan Bal Khand Sarga 6 describes the kingdom of Raja Dasrath stating that in his kingdom as a kingdom ruled as per the Dharma Shastras, it is stated the kingdom was full of scholars learned in the various scriptures, all the citizen law abiding and honest, crime rate was absolutely zero and there were no atheists and lustful people in the country. Special emphasis must be given on slok 5 of this sarga which states that "Raja Dasrath while dwelling in the city of Ayodhya protected the world, like Manu, the foremost protector of mankind" which is a clear reference to the fact that Manusmriti was the law book used in time of Raja Dasrath otherwise the comparison would not have been made, a fact which is proven in the The Valmiki Ramayan states in Bala Khand Sarga 70 sloka 16 that Ikshvaku was the founder of this ancient dynasty in which Shri Ram was born while it is stated in sloka 21 of this same Sarga that Vaivastava Manu was the father of Ikshvaku. It is later stated in the Valmiki Ramayana Ayodhya Khand Sarga 2 sloka 6 that King Dasrath who was father of Shri Ram ruled the kingdom as per the laws ordained by his forefathers which is a clear reference to the fact that the laws of Manu were used in his rule and the comparison stated earlier was made on this exact basis.
II) RAMRAJYA SWAROOP:
The Valmiki Ramayan describes Ramrajya in the Uttar Khand Sarga 99 sloka 11-14 as being full of prosperity where all creatures stayed in harmony, no one died before time, everyone was in perfect condition, there were no natural calamities and seasons came and went away on time without causing hardships or disasters.
The Ananda Ramayan states in Sara Khand Sarga 12 slok 117 that states Shri Ram throughout his life enacted Dharma and its doctrines and spread rules of the Vedas and describes Ramrajya in Yatra Khand Sarga 3 sloka 15-17 stating that in Ramrajya there was no obstruction in ruling the kingdom, everyone was happy and prosperous, there were no calamities and food was produced in abundance, crime rate was absolutely zero. Ramrajya has been described in its full glory in the Rajya Khand Sarga 15 however stating the exact statements here shall be too long for this Sarga has a total of 86 slokas hence only the important points not available in the other mentioned references shall be provided; this Sarga apart from attributing the same qualities to Ramrajya as stated in the other references already given mentions in slok 2-3 that the kingdom had an extremely strong army, in slok 16 that no person in the kingdom consumed any intoxicants, in slok 21-22 that the various forts and the capital was well protected with strategic defenses, in slok 33 that Shri Ram had an excellent spy network or in simple terms the best intelligence department, in slok 55 that there was no polygamy in his kingdom. In the Rajya Khand Sarga 18 slok 54 Shri Ram himself ordains that authority of Bramhans in matters of decision of dharma must be protected and ensured by all kings. There is yet another reference vital for this article in the Manohar Khand Sarga 4 slok 273 which declears Manu to be the authority on Dharma and hence in term directly bears witness to the fact that the Manusmriti was used in Ramrajya as is evident by viewing this verse in light of the other mentioned references which all unanimously states that in Ramrajya Dharma was strictly followed.
The Adbhut Ramayan describes Ramrajya in Adhay 27 sloka 8-9 stating that Shri Ram after coming back to Ayodhya after war against Ravan cleansed the earth of Adharmis and conducted many yajnas. This states though slightly indirectly that Dharma was strictly followed in Ramrajya.
The Adhyatma Ramayan describes Ramrajya in Chapter 4 verse 21-22 stating that in Ramrajya everyone was well nourished, happy and rich, the trees and fields produced food in abundance, everyone followed Dharma and the husbands and wives were strictly dedicated to each other with zero signs of adultery.
The Ramcharitmanas describes Ramrajya in Uttar Khand verses 20-23 stating that in Ramrajya there was no fear and disease of any kind, everyone followed their duties as per the scriptures and obtained happiness, there were no untimely deaths and natural disasters, husbands and wives were faithful to each other and there were absolutely no crime in the kingdom.
To put it all in short Ramrajya was the perfect period of governance where food was in abundance, everyone was prosperous, Dharma and scriptures was follow strictly, crime rate and adultery were non existent, there were no atheists and Manusmriti was the book of law used in this ideal kingdom.
III) THE COMMON LINK:
Why it was stated in the point II that to know Ramrajya first Dasrath Rajya must be known shall be visible in this point. As stated earlier in point II, Raja Dasrath followed and used Manusmriti to govern his kingdom. What must be known now is that as per the Valmiki Ramayana Ayodhya Khand Sarga 2 sloka 46 Shri Ram governed various matters using the same laws as that of his ancestors hence as Dasrath which in turn only solidifies the fact proved in the earlier point that Shri Ram used the Manusmriti to rule his kingdom. One must also be made aware of the fact that Swyambhu Manu who was the first of the 14 Manus was the composer of the Manusmriti and this Manusmriti was followed by the other Manus as per the Manusmriti Chapter 1 verse 63 which in turn when combined with point II shows again that what we call as Manusmriti which was written by Swyambhu Manu is the only Manusmriti and debunks reformist claims of some other documents being used.
It must also be noted that in the Manusmriti Chapter 7 which deals with conduct to be observed by kings states in verse 3 that when lawlessness and chaos was prevalent and all creatures were in terrible fear, Ishvar created king and monarchy to put an end to their misery and bring back order by establishing a peaceful and balanced society, stating in verses 14-15 that Ishvar created the institution of punishment to ensure an ideal condition prevails through its application which will prevent chaos and further stating in verse 31 that those who follow their duties as per command of the scriptures must not be punished, which in turn states than an ideal monarchy is governed as per scriptures and is crime free which in turn is the exact same description we find of the Ramrajya. The Manusmriti describes atheists as those who abuse and are against the Vedas in Chapter 2 verse 11 and states in Chapter 4 verse 61 that countries which has atheists and Dharma is not observed as per scriptures must be avoided which in turn naturally implies that the kingdom free from atheists and governed as per scriptures is stated to be ideal in view of Manu Maharaj. Yet another common link is the Valmiki Ramayan Ayodhya Khand Sarga 100 where Shri Ram while advising Bharat on how to govern kingdom gives advise strictly based on the Manusmriti, a fact than can be observed clearly by making a comparative study of the mentioned reference and the Manusmriti. Bramhans being authority in Dharma stated in the Ananda Ramayan Rajya Khand Sarga 18 slok 54 is also confirmed by the Manusmriti Chapter 1 verse 98. There are many more such comparative references that can be provided apart from direct evidence to prove that Manusmriti indeed was used as the Dharma Shastra in Ramrajya and for all who actually want Ramrajya, Manusmriti holds a very close position.
JAI VEERBHADRA
JAI SHANKARACHARYA
Credits of the writing and compilation belongs to: Jalandhar.
Comment: Wrong sources - Thumbs down. Modern Manusmriti 1.63 doesn't say author is manu. In fact, Manu is referred to in THIRD PERSON in today's Manusmriti. Ramji should have killed or punished Shabri if current Manusmriti was followed and we wouldn't have modern scriptures.............. Plus Manu's own children, grandchildren etc. changed Varnas repeatedly and all the time so modern Manusmriti could be bogus. Or if this is original, Rama discarded anti-Vedic, anti-Puranic, anti-Itihasa, anti-Tantric verses.
Comment: Also Shankaracharya or Smriti sect is against all sacred scriptures except Smritis, which have ZERO divine credibility as per Vedas.
Comment: Only a pure brahmin can make such a nice comment..... you are definitely not a shudra haha......... Shankaracharya has ZERO authority outside smriti sect
Comment: Cope harder hater all u can do is hate now go and do intercaste marriage with ur mum
Comment: @R.Kaushik did you read the text in original Sanskrit?
Answer: Yes only not current Manusmriti. Otherwise Lord would have killed or at least punished Shabri (low caste) for performing Tapasya. Instead he preferred her over other saints.
Current Manusmriti refers to Manu in third person and has caste laws which would have debarred not just Shabri, but also Ved Vyas, Jabala, Vishwamitra and many more than 100 people whose life we know details about from getting initiation!!

The decision is that—'one born to an Ārya from a Non-Ārya woman may be an Ārya in quality; but one born to a Non-ārya even from an Ārya woman is always Non-Ārya' (10.67)
The settled law is that both these are unfit for the sacraments (10.68)

So we would not have the 4 Vedas nor Upanishads nor Puranas nor Itihas if these laws were followed. We also know that Manu had progeny that repeatedly and freely changed their caste. So Lord Ram was quoting from the original smriti.
If he was quoting from current one, then he followed hierarchy of scriptures where Vedas, Puranas, Rahasyas, Itihasas etc. are more authoritative so he discarded contradicting verses.
It is also common knowledge that Manu had children, grandchildren etc. who repeatedly changed Varnas very freely. So again, it's totally impossible this was completely followed in Ram Rajya
Comment: Now while I don't think a lot (especially things people tend to like) of the Manu Smriti is true, I don't think it is unoriginal. Now I also think any original, true dharma shastra will send virtually all humans into a temper tantrum, so Manu is not being the worst here (at least).
Comment: Shabari did a different type of satvik tapasya, and that too on the orders of her Guru Matanga Muni. It is different from the ghora tapasya done by Shambuka for which he was punished.
Comment: Shabri's guru Matanga was also from the lower caste so he should've been killed or punished too if this bogus Manusmriti was being followed
Comment: Lol do u have any references for ur made up fake stories or all I can do is larp with ur tamsic Arya samaj theories? @rkaushik?


Joomla insert article in helloworld component

Question: I would like to insert an article inside my component, does someone have an example on how to do that?
I created Helloworld component , I need to show items of articles in site component , How to write code in model ,view, controller to show items?
Thank you
Comment: Your question is too broad. Have a look at other components that ship with Joomla, especially com_content, try to write some code on your own and come back if you have a specific problem.
Comment: Have you read through the Joomla documentation?
Comment: Yes but i need more documentaion!
Answer: Your question is too broad for any detailed response and we don't know where to start with an answer. But I will take a stab in the dark (despite it possibly leading to a Joomla horror movie...).
If you already have a MVC component going and want to pull data from other tables you should read up on database access in Joomla.
https://docs.joomla.org/Accessing_the_database_using_JDatabase
You are probably looking for data from these tables:
jos_sections
jos_categories
jos_content
I hope that helps, however I would like to suggest that you investigate a component creator tool to assist you in understanding Joomla MVC layout. I use Component Creator (https://www.component-creator.com/en/), but there are lots out there. https://stackoverflow.com/questions/4470272/component-builder-for-joomla
And if you have read the Joomla docs and still need more assistance you might try a Joomla book from Packt (https://www.packtpub.com/all?search=joomla).  I can recommend "Learning Joomla! 3 Extension Development-Third Edition". It has step by step directions for creating simple plugins and modules, as well as more complex components that are data driven.
HTH


Insufficient nominees for graduation election: What happens next

Question: Stack Exchange's position is that every site has a team of moderators to operate in our network. Currently, this community is being moderated by Jens Kubieziel and Alexey Vesnin, who have been doing an excellent job. In 2021, Tor left beta with 58 other sites. We attempted to hold the first graduation election in May of 2022 and then another in March of 2023 for your community.
While sites are in Beta, elections do not need to be competitive. However, once a site loses the Beta label, elections must be competitive to progress to the voting stage and be considered successful. We define an election as competitive when it has at least one more nominee than available moderator slots. In graduation elections, the Pro-Tem moderators need to nominate themselves again for consideration to continue as full site mods. So for a graduation election to be competitive, there need to be at minimum four nominees since there are three moderator slots.
Unfortunately, we only had one of the current moderators nominate themselves, while we require three nominations in total. Members of an SE site being willing to step up as moderators has been the primary threshold for keeping a site alive since we created this policy in January 2015. Without that, a site is in danger of being shut down. Today, I am posting on behalf of the Community Management team to discuss the next election for this site. In order for a future election to be successful, there need to be community members who are eligible to serve as moderators who are willing to step up to the plate.
Next Steps
We are going to recommend waiting until October 2023 before setting up another election for this site. We'll be looking for additional individuals to nominate themselves as moderators to join the team. This next election will help to bring those folks in. If we can't find at least four people willing to nominate themselves to be a moderator on this site by that time, the site is at risk of shutting down.
Do you have any questions about the election process or any concerns keeping you from nominating yourself as a moderator? Let us know, and we'll do our best to address them.
Comment: "We are going to recommend waiting until October 2023 before setting up another election for this site." It has been a while since October '23. You've already stated the risks. When will details about the next steps be announced?
Answer: The cancelled election in June 2022 had two candidates: Alexey and Jens. The cancelled election in March 2023 had only Alexey.
I wonder if there's a reason why so few of the active users here want to run in the election, especially considering that there's not much competition? Would at least three other people be (including or excluding Jens) be willing to at least consider running in the October 2023 election?
It would be sad to see this site go down, along with Alexey's diamond (he's one of my favorite network-wide diamond moderators). If anyone is interested in running for the election, maybe you can comment or answer here to provide a rough idea of the likelihood of this election meeting the requirements. Personally, I can't run for the election because I don't have enough points on the site.


How to detect Wolfenstein 3D secret area?

Question: It has been passed many years since the last time I played to Wolfenstein 3D. Is there a visual trick to detect secret area or should I try to press open on all walls?
Comment: Just a note that this game is available [for free now, online](http://wolfenstein.bethsoft.com/game_NA.php)
Comment: The free download is region-blocked though.
Comment: I've been enjoying this game recently. Great music for it's time, and enjoying Adrian Carmack's artwork.
Answer: No, there isn't.
I spent hours in 1992 staring at walls and listening to BJ going "uff ... uff ... uff" as a tapped the 'open' key. Now you can, too!
Comment: Nope, the easiest way is to strafe every wall mashing space in the vein hope that something will click.
Comment: So, to clarify your answer - there is no visual trick?
Comment: Since posting above I've played through the three Spear Of Destiny missions, and there *may* be a case for saying that if you are seeing rats on the floor there could be a secret wall around close by.
Answer: There is no visual trick to detect secret doors. The only way to find them is to try and "open" the wall as if it was a door. Sometimes the secret door is in some place which is stands out a bit (e.g. a wall painting, an alcove, the center of a wall etc.) but many times there's simply no indication at all.
The best way to find a secret door is to run diagonally into the wall - so that you stay touching the wall but constantly moving - and click the open key repeatedly.
However, keep in mind that:

The granularity of doors in Wolfenstein 3D is one "block". A secret door will never occupy less than that.
Secret doors always move two blocks backward (relative to the block's activated side) when activated. If, according to the map, it's impossible for the block to move two blocks backward, then there can be no secret door there.
Secret doors may, and do, appear inside areas which were originally accessed via a secret door themselves.
Answer: Well there are secret rooms and secret levels...
Secret Rooms
Although there are no strict guidelines to where a secret room might be, there are subtle hints that might help you find them. It helps to map out the level since the layout of the levels are accurate (no magical warped spaces), gaps between areas could hint at secret rooms.
As for secret room markers, first try the obvious places on the walls (e.g. Hitler paintings, Nazi banners, and other wall decorations).
Some secret panels seem to be in the corners of a room, while others can be found in the middle(point) of a wall (usually in a room).
W@WWWWWWWWWW@WWWWWWWWWW@W
@                       W
W                       W
W                       ]
W                       W
@                       W
W@WWWWWWWWWW@WWWWWWWWWW@W

Also, be on the look out objects placed against walls between sets of twos (such as plants or barrels).
WWWWWWWWWWWWWWWWWWWWWWWWW
Wo                      W
@                       W
Wo                      ]
W                       W
W               o o     W
WWWWWWWWWWWWWWWWW@WWWWWWW

Some secret panels can also be found by walking in a straight line from (either side) a object (such as an overhead light), or in narrow alcoves.
WWWWWWW@WWWWW....Corridor
W
W
[      O <-- Light
W
W
WWWWWWW@WWWWW....Corridor

Take note that certain secret rooms have secret rooms within themselves(!!) and sometimes the movable panels  of secret rooms can block off access to other secret rooms (this why you sometimes don't a get 100% secrets even though you've searched every nook and cranny).
Secret Levels
In every Episode, there is a secret (10th) level you can access by taking a secret elevator. There are pretty hard to find if you don't have a map to help you. It is usually accessible though a series of secret doors/panels within the secret rooms and usually located in the last room of that level (the room with the regular elevator).
Here is a list of what levels the secrets levels are on by Episode.
Episode 1: level 1
Episode 2: level 1
Episode 3: level 7 (This is the secret Pac-Man level! Crazy Nazis!)
Episode 4: level 3
Episode 5: level 5
Episode 6: level 3

Here is a list of all the secrets in the PC version.
Also everything you would typically want to know about Wolfenstein 3D.
Comment: Yes, if you want get 100% secrets.
Comment: "sometimes the movable panels of secret rooms can block off access to other secret rooms" ouch that's tough. Does it mean that sometimes I need to activate hidden area in the right order?
Answer: Usually they will not be unmarked along a flat wall, but they could be in a alcove (marked or unmarked) or be represented by a painting (or if I remeber right, maybe a suit of armor) on a flat wall.


Will 3/16 inch spiles significantly reduce the amount of sap I will collect?

Question: Made maple syrup last year for the first time ever, with a mix of 7/16 plastic spiles and a pack of 3/16 inch spiles I bought from Amazon.  Yes 3/16 inch, not a mistake, not 5/16 inch.  As last year was a very poor sugaring off year (even the local sugar bush had extremely low production) I do not know if my low collection amount was also due to the smaller sized spiles.  I used the larger spiles in my biggest trees.  I am tapping Manitoba maples (box elders) which don't generally get as big or live as long as other maples and I am only tapping trees that are in my yard so I do want to keep the damage to a minimum as all my trees are precious.  Should I switch to larger (5/16) spiles or will the 3/16 spiles be okay?  A small reduction in the amount I collect would be all right with me if it is better for the tree.
Comment: A smaller tube would presumably draw liquids from a smaller zone resulting in less production ; but worth a try. I used 5/8" metal tubing , but I only used one tube in a 12" diameter tree.  That worked well. I hung 2 gallon buckets on the metal tubes. I had silver maples  but I could only find information on sugar/black maples.  Interesting that you are getting results from box elders.
Comment: It is not sap , it is "juice. I found that the juice stops when the sap starts and the buds start growing. For awhile I was concerned that I would not be able to stop flow , but it  stops on its own ( in Silver Maples).
Answer: I don't know where your information on 2020 production is coming from, but the Maple News reports that production in the US (Vermont, New York, Maine) was a record, also in Quebec. The industry provides guidelines for how many spiles in a tree of a certain size, so you might want to get the recommendations for your area. A study by Wilmot, Perkins, Stowe and van den Berg concluded that the flow was pretty much the same from smaller spiles, the small spiles did less damage to the tree and were easier to work with.
Comment: There is a publication for everything! They call it "sap" , so I am wrong; but it is different from the sap in a leafed out tree.
Comment: Thanks for the info.  Western Manitoba is a looooong way away from Quebec and Vermont!  We had a very bad local weather pattern settle over us and no one around here got very much from their taps.


"no caching mode page found" error in Ubuntu without any drive partition

Question: [6.913785] sd 1:0:0:0 [sdb] no caching mode page found
[6.913885] sd 1:0:0:0: [sdb] assuming drve cache: write through
Busy box v1.22.1 (Ubuntu 1:1.22.0-15ubuntu1) built-in shell (ash)
Enter 'help' for a list of built-in commands
Initramfs

Comment: "without any drive partition" You must have a "/",
Comment: New install? Or did this message just appear? More info, please.
Comment: @Rinzwind I'm guessing the OP means that the block device is unpartitioned i.e. `/` (and everything under it) is on `/dev/sdb` rather than `/dev/sdb*n*`
Comment: No this is already installed and I think  it appeared bcoz of my mistake as I change the crypt lugs format to NTFS and yes it is unpartitioned pls help me what to do
Answer: I had the same issue before and I solved it according to the following steps:

I've created a USB key with the live of Mint distro and I booted the
machine with it.
List your partitions using the following command:

sudo fdisk -l /dev/sda

I got the following output where sda the only drive on my machine, notice that your linux partition is /dev/sda1
Disk /dev/sda: 232.9 GiB, 250059350016 bytes, 488397168 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa71bc45b

Device     Boot     Start       End   Sectors   Size Id Type
/dev/sda1  *         2048 480393215 480391168 229.1G 83 Linux
/dev/sda2       480395262 488396799   8001538   3.8G  5 Extended
/dev/sda5       480395264 488396799   8001536   3.8G 82 Linux swap / Solaris

Then, I executed the following command which rescue my disk again

sudo fsck -yv /dev/sda1

and I got the following result
fsck from util-linux 2.27.1
e2fsck 1.42.13 (17-May-2015)
/dev/sda1 contains a file system with errors, check forced.
Pass 1: Checking inodes, blocks, and sizes
Inodes that were part of a corrupted orphan linked list found.  Fix? yes

Inode 11275098 was part of the orphaned inode list.  FIXED.
Inode 11275099 was part of the orphaned inode list.  FIXED.
Inode 11275100 was part of the orphaned inode list.  FIXED.
Inode 11275101 was part of the orphaned inode list.  FIXED.
Inode 11280599 was part of the orphaned inode list.  FIXED.
Pass 2: Checking directory structure
Error reading block 51389518 (Attempt to read block from filesystem resulted in short read) while reading directory block.  Ignore error? yes

Force rewrite? yes

Directory inode 12855288, block #0, offset 0: directory corrupted
Salvage? yes

Missing '.' in directory inode 12855288.
Fix? yes

Setting filetype for entry '.' in ??? (12855288) to 2.
Missing '..' in directory inode 12855288.
Fix? yes

Setting filetype for entry '..' in ??? (12855288) to 2.
Pass 3: Checking directory connectivity
'..' in /var/lib/docker/aufs/mnt/efa5ec50cb7cda556e07082ce6f284bc89209ffc566a91212110072d83e73e91 (12855288) is <The NULL inode> (0), should be /var/lib/docker/aufs/mnt (12468994).
Fix? yes

Pass 4: Checking reference counts
Inode 2 ref count is 23, should be 24.  Fix? yes

Inode 12468994 ref count is 58, should be 57.  Fix? yes

Pass 5: Checking group summary information
Block bitmap differences:  -(45219840--45220206)
Fix? yes

Free blocks count wrong for group #1380 (31392, counted=31759).
Fix? yes

Free blocks count wrong (47147739, counted=46915808).
Fix? yes

Inode bitmap differences:  -(11275097--11275101) -11280599
Fix? yes

Free inodes count wrong for group #1376 (5, counted=10).
Fix? yes

Free inodes count wrong for group #1377 (7279, counted=7280).
Fix? yes

Free inodes count wrong (14328772, counted=14270663).
Fix? yes

/dev/sda1: ***** FILE SYSTEM WAS MODIFIED *****

      745273 inodes used (4.96%, out of 15015936)
        2132 non-contiguous files (0.3%)
         607 non-contiguous directories (0.1%)
             # of inodes with ind/dind/tind blocks: 0/0/0
             Extent depth histogram: 698409/204
    13133088 blocks used (21.87%, out of 60048896)
           0 bad blocks
           3 large files

      607356 regular files
       89743 directories
          56 character device files
          25 block device files
           0 fifos
       41620 links
       48079 symbolic links (46566 fast symbolic links)
           5 sockets
------------
      786884 files

Finally, my machine worked fine without re installation.


What's an easy way to supply power to a breadboard without using batteries?

Question: There are a number of kits available (for example sparkfun's Breadboard Power Supply USB - 5V/3.3V and Breadboard Power Supply 5V/3.3V or Adafruit's Adjustable breadboard power supply kit -  v1.0) but these all come as kits. What is a common and easy of-the-shelf way for a beginner to supply DC power to a breadboard without using batteries?
Answer: I have found these breadboard power supplies ($2.12 or so on ebay.com, shipped) very useful for the purpose described:

Jumper-selected 3.3 and 5 Volt simultaneous outputs from wall-wart, USB or 9 Volt battery.
They have a convenient 2.1 mm barrel jack port on board, besides a power button and an indicator LED. You can choose to have 3.3 Volts on one of the power rail rows of the breadboard, and 5 Volts on the other, something I find especially useful.
As a fellow breadboard user, you might also find two of my answers to other questions on this site, of interest:

On "Good Quality Breadboards"
On ""Starter Kit for electronics" - What to buy?"
Comment: @dumbledad  You're welcome. Also, it's always good to wait a day or so before accepting an answer, in case people from other time zones come up with something better.
Comment: +1 for the time zone tip - I'm usually so pleased I accept immediately!
Comment: @dumbledad I can understand the temptation :-) Of course, you can always come back and accept the answer tomorrow!
Answer: Wall wart power supples - cheap, plentiful, just be careful of their output voltage. Many are odd voltages or unregulated (the voltage changes with load) although many are quite usefully designed.
Alternatively, old PC power supplies - Dangerous prototypes do an ATX breakout board, either use one of those or read about how it works & use the information to power your circuit. PC PSU's typically have 12v, 5v, 3v3 available which gives you everything you should need to get started. Just beware that many require a small load (resistor/light bulb) to work properly (the ATX breakout adds this if required).
Comment: Either cut it off & solder offcut component legs on for easy insertion, or stick a barrel connector on the breadboard.
Comment: Thanks John. How do you connect the barrel connector on the end of the DC wall wart power supply to the breadboard?
Comment: An old phone charger will often do.
Answer: Really any basic power supply would do. For example, I use this:

simulate this circuit – Schematic created using CircuitLab
Transformer, rectifier, large capacitor and a variable voltage regulator. You can also split R1 in two (5K and 500R) for extra fine tuning. Gauge the potmeter so you can see the output voltage without measuring.
Comment: Thanks Camil, but I was more after "off-the-shelf" solutions rather than circuit diagrams. Simple instructions on how to connect a shop brought DC power supply to a breadboard would be perfect.
Comment: Build it one time and it is off-the-shelf ;)
Answer: I asked some colleagues too and got a couple of interesting answers.

Cut a USB power cable and solder on wires that can connect onto header pins in the breadboard. (Not off-the-shelf, but doesn't sound tricky either.)
Use a Gadgeteer power module. I have several .Net Gadgeteer kits so this one's easy for me.

Take the Gadgeteer power module (for example this one from GHI)

Run the Gadgeteer cable from it, but instead of hooking it up with the Gadgeteer mainboard, as one usually would, run it instead to a Gadgeteer extender (like this one) onto which header pins have been soldered.

(N.B. One manufacturer makes the extender board with the header pins already on, but I cannot find them at present.) The schematic of the power board should say which pins are 5v, 3.3v, and ground, or you can use a multimeter.
Of course this is of little use unless (like us) you have lots of Gadgeteer kit around the office!


Is there an agent that changes colour when exposed to air?

Question: And if so, does it come in a liquid form?
Thank you for your time
Comment: Need it be reversible?
Comment: It's too broad, there are lots of things that react with oxygen, changing color.
Answer: Many compounds change colour when exposed to 'air', however, this is usually due to the presence of either water or oxygen, rather than 'air'.
To give a few examples:
*when distilling a solvent such as THF over sodium/potassium, benzophenone is used as an indicator. In the absence of air, the indicator leaves the solution a deep purple colour, however if an excess of air is present, the indicator turns a clear/brown colour.
*drierite is an indicator often used in inert gas drying tubes. In its dry form, it is either deep blue or orange, however when wet it turns white/pale orange.
Several other indicators exist, however its worth noting that air is merely a mixture of multiple things, and as such any indicator will indicate for a specific component of air rather than air as a whole- with water and oxygen being the most common things to indicate for.


Iterated integral and integrability

Question: Hi everyone: Suppose $B_{1}$ and $B_{2}$ are balls in $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ (let's say for $m,n\geq2)$. Suppose that $f(x,y)$ is defined and measurable eveywhere. Beside,
$$0\leq \int_{B_{1}}f(x,y)dx\leq C$$
for all $y$ and $C$ is a constant. Does this imply that $f$ is integable, or integrable in the large sense (i.e., at least one of the positive or negative parts is integrable)? Thanks for your help.
Comment: You mean if $\int_{B_2} \int_{B_1} f(x,y) dx dy < \infty $ ?
Comment: No. I mean $\int_{B_{2}}\int_{B_{1}}\mid f(x,y)\mid d\lambda(x)d\lambda(y)<+\infty$. Or at least the double integral of $f^{+}$ (or $f^{-}$) is finite.
Comment: Yeah sorry I forgot, ofc. a function is called Lebesgue integrable when $\int |f| dx < \infty $ and not $\int f dx < \infty$.
Answer: This does not imply that $f$ is integrable.
Counter example
Consider $f(x,y) = \begin{cases}  \frac{xy(y^2 -x^2)}{(x^2 + y^2)} & \text{for } y \in (0,2] \\ 0 & else \end{cases}$ and $B_1= [0,1]$ and $B_2=[0,2]$.
Then
$$0 \leq \int_0^1 f(x,y) \, dx = \frac{y}{2( y^2 +1)^2} \leq 1 $$
for all $y \geq 0$ and for $y < 0$ the integral is 0. In short, $f$ fulfills your condition.
However, computation reveals  (see  here) that
$$ \int_0^2 \int_0^1 f(x,y) \, dx \, dy = \frac{1}{5} $$
and
$$ \int_0^1 \int_0^2 f(x,y) \, dy \, dx = - \frac{1}{20} $$
and since
$$ \int_0^2 \int_0^1 f(x,y) \, dx \, dy \neq \int_0^1 \int_0^2 f(x,y) \, dy \, dx .$$
Fubinis Theorem implies that $f$ is not integrable.

Iterated integral and integrability


Continuity of the electric potential due to a surface charge

Question: The Electric potential due to a charge distribution on a surface is:
$$\Phi \left ( x \right )=\int \frac{\sigma \left ( {x^{}}' \right )dx{}'}{\left \| x-x{}' \right \|}da.$$ I want to show that  it's a continuous function everywhere.
Comment: @Qmechanic I personally feel that Physics.SE is appropriate (not just because I was going to respond :))
Comment: Would http://math.stackexchange.com/ be a better home for this question?
Comment: You know what, actually I take it back depending perhaps on the level of rigor desired by the OP and his desire for accompanying physical remarks if any.
Comment: I think it might be one for [math.SE], since the fact that the function $\Phi(x)$ represents an electrical potential seems irrelevant to the problem.
Comment: possibly related:  http://physics.stackexchange.com/q/40919/
Answer: The answer is not in general. There exist pathological charge densities for which the electric potential has discontinuities. Fortunately, these can always be ruled out on physical grounds: they usually have finite amounts of opposite charge infinitely close to each other, which allows for arbitrarily high electric fields, and therefore divergent energy densities and infinite configuration energies.
How does this work? Take two conducting plates and put their edges in contact. I'm quite fond of using the north and south hemispheres of a given sphere, separated at the equator (which has the advantage of not introducing any extraneous infinities), but for simplicity let's use two semi-infinite planes, in the $x,z$ plane and separated by the $x$ axis.
Now, connect both plates to a voltage source and set one to a potential $+V_0$ and the other to $-V_0$. The potential at the $x$ axis is (of course!) discontinuous, by construction.
If that makes you uneasy, consider spreading uniform but opposite surface charges $\pm \sigma_0$ on the $\pm z$ plates. Fix some positive coordinate $z_0$. Then if you zoom in more and more towards the plates (i.e. as you take $x\rightarrow 0$) the plates loom larger and larger, and when $x\ll z_0$ the $-$ plate looks so far away that the local potential has to be positive, and bounded away from zero. Similarly, if $z_0<0$ the potential close enough to the plates must be negative. Thus you have points arbitrarily near the $x$ axis which have potentials bounded away from zero, both positive and negative, so the potential is discontinuous there.
So why is this not a problem? In these pathological examples you are putting a lot of negative charge, in conducting surfaces, next to a lot of negative charge. This assumes that you have separated them with an insulator of zero thickness which can withstand large (albeit finite) potential differences (so infinite electric fields). This is just not physical: any dielectric will break down or spark as you approach these conditions. As soon as you introduce a finite separation, the discontinuity goes away.
More generally the electrostatic potential can have discontinuities. These are most usually considered as surface discontinuities, in which case they are known as dipole layers, which you can think of as infinitely thin capacitors set at constant potential. (Therefore, they must have two layers of opposite but infinite charge, which is again not physical.)
A real dipole layer would be something like a lot of polar molecules, all pointing the same way normally to the surface. We know that the potential inside is not discontinuous and that the electric fields inside are large but finite, but if you're outside then all you see are two plane layers of charge which look for all the world like they are setting up a discontinuous potential, so you might as well put that into your math and assume it is infinitely thin.
Answer: Ok so let $\mathbf x_0$ be some point on the surface, and let $\mathbf n$ denote a normal to the surface at $\mathbf x_0$.  Consider the difference
\begin{align}
  \Phi(\mathbf x_0 + \epsilon\mathbf n) - \Phi(\mathbf x_0 - \epsilon\mathbf n)
&= \frac{1}{4\pi\epsilon_0}\int_S \sigma(\mathbf x')\left(\frac{1}{|\mathbf x_0+\epsilon\mathbf n-\mathbf x'|}-\frac{1}{|\mathbf x_0-\epsilon\mathbf n-\mathbf x'|}\right) da'
\end{align}
But recall the Taylor expansion
$$
  \frac{1}{|\mathbf x +\epsilon\mathbf a|} = \frac{1}{\mathbf x} + \epsilon\frac{\mathbf a\cdot\mathbf x}{|\mathbf x|^3} + \mathcal O(\epsilon^2)
$$
from which it follows that
\begin{align}
  \frac{1}{|\mathbf x_0+\epsilon\mathbf n-\mathbf x'|}-\frac{1}{|\mathbf x_0-\epsilon\mathbf n-\mathbf x'|}
&= 2\epsilon \frac{\mathbf n\cdot(\mathbf x_0-\mathbf x')}{|\mathbf x_0 - \mathbf x'|^3}  +\mathcal O(\epsilon^2)
\end{align}
and therefore
$$
  \Phi(\mathbf x_0 + \epsilon\mathbf n) - \Phi(\mathbf x_0 - \epsilon\mathbf n)
=\epsilon\left(\frac{1}{2\pi\epsilon_0}\int_S \sigma(\mathbf x')\frac{\mathbf n\cdot(\mathbf x_0-\mathbf x')}{|\mathbf x_0 - \mathbf x'|^3} da'\right) + \mathcal O(\epsilon^2)
$$
from which it follows that
$$
  \lim_{\epsilon\to 0}\Big(\Phi(\mathbf x_0+\epsilon\mathbf n)-\Phi(\mathbf x_0-\epsilon\mathbf n)\Big)=0
$$
So $\Phi$ is continuous at the surface charge.
Disclaimer. This is not a mathematically rigorous proof, but it's something that I think would satisfy most physicists.  I propose that if more rigor is desired then the question should be migrated to math.SE whose possibility was first suggested by user Qmechanic.
Comment: A quick comment on rigour. While this may not be quite in the language that would satisfy mathematicians, it is relatively easy to formalize it into a rigorous proof and keep the same spirit, provided it holds. However, the validity of the proof depends on the term in parenthesis to be finite, and that depends on $\sigma$. Pathological densities can look quite innocent: $\sigma=\pm1$ on the north (resp. south) pole of a sphere will misbehave.
Answer: An alternative, perhaps more physicsy argument is:
Since:

The electric field ought to be finite for a surface charge distribution e.g. because it is locally planar.
The potential is the line integral of the electric field.
The line integral of a function without singularities will be continuous.

the potential ought to be continuous. You may find the first claim to be lacking in rigor.
You are trying to solve this starting with the Green's function for the Poisson equation, which itself has a singularity. This is probably not the most natural method. As an alternative to the foregoing argument, you could look up the conditions that are necessary and/or sufficient for solutions of the Poisson equation to be continuous.
As an aside, I would note that what you are trying to prove in fact does not hold if you consider a point charge to be a special case of a surface charge distribution.


Validation Settings for column

Question: I'm trying to eliminate other users from seeing other users SS in list group. Is there a formula that I can write in the Validation settings for that Column?
Answer: OOTB SharePoint does not have away to secure a column in a list, I think that this is what you are trying to achieve.  There are third party products that provide this functionality, SharePoint Boost Column Level Security for example.  Another idea would be to have a second list with the SS in it, then you can permission that list more securely.

Validation Settings for column


T-test for percentage change

Question: Suppose I have some paired data (for example weight before and after), and I wish to find a confidence interval for the percentage increase in weight, can I simply apply the t-test to the percentage difference (A-B)/A or should I use another test?
Comment: are you intetested in confidence for mean or confidence intrrval of %age change? Moreover, it is not clear for what purpose t-test is desired ?
Comment: @subhashc.davar, I'm interested in the confidence interval for the %age change.
Comment: In addition to the answer below from @FrankHarrell, consider his [answer to this related question](http://stats.stackexchange.com/a/15795/28500); you might be better off modeling the weight after as a function of weight before  (with both weights perhaps in a log scale) along with your other covariates, and then expressing the final result in percentage terms if still you think that is appropriate. That will allow you to test whether weight behaves additively, proportionately, or as some mixture, as he notes.
Answer: Percent change is an improper measure because of asymmetry.  If you really think that weight operates proportionally (it usually operates as a mixture of additively and proportionately) then analyze log ratio, which is a symmetric measure.  Get a confidence interval for that using standard methods, then anti-log to get fold change and its asymmetric confidence interval.  This fold change is a ratio of medians (also a ratio of means in this case, I think).
Comment: @whuber I respectfully disagree.  Taking the arithmetic mean of percent change is not correct.  Think about one subject who increases from 1.0 to 2.0 and another starts at 2.0 and decreases to 1.0.  The mean % change is +25% which is misleading.
Comment: I agree that using logarithms is natural and convenient.  The "fold change" won't be a ratio of means--but it could be a close approximation.  When it is a close approximation, the asymmetry you refer to is likely mild to nonexistent.  Regardless, even if percent change is an "improper" measure, if it's the measure needed for the application, then perhaps finding a CI for it should not be dismissed out of hand.
Comment: What I was trying to say is that when a measure is improper on the scale for which it is being used, it should not be used .  The correct average of a 100% change and a 50% decrease is 0% change.  The math doesn't work.  For the lognormal case with equal $\sigma$, the fold change (ratio) is both the ratio of means and the ratio of medians.
Comment: If you're really interested in the percentage increase in something, it's hard to see how the math could not work.  There might be concerns about the distribution of observed variations from a (hypothetical, theoretical) true percentage increase, but that increase itself is well defined.  It's certainly good to warn naive users of the t-test about the likely failure of related confidence interval procedures when the observations cover a range as wide as -50% to + 100%. Nevertheless, there's nothing *inherently* wrong either with percentage increases or even with taking their arithmetic means.
Answer: Using normally distributed dataset one can run a t-test for both, the absolute and the percentage change. Have a look here
To keep things simple, if you apply a paired sample t-test for the previously computed %difference, i.e. $\frac{A-B}{A}*100$, then all you can do is a one-sample t-test which will test your new variable against a single value which I suspect would be 0 (0% change) using a one-sample t-test.
You could instead run a paired-samples t-test (assuming that the assumptions of a t-test hold in your dataset) and then express the absolute mean difference which will be computed by any softare running a t-test as a %
Comment: Correct me if I am wrong, but surely percentage change breaks normality, for example if my original value was 5, this can go to 0 which is -100% or to 10 which is +100%
Firstly -100% is the lowest possible drop, a normal distribution should not have a maximum or a minimum and secondly due to the nature of percentage change I have found in my datasets the difference between the mean and the upper quartile to be larger than the difference between the mean and the lower quartile.
Comment: Wouldn't one use the Delta method for such a CI ?

T-test for percentage change


Capacity of the Deletion Channel

Question: Consider a Binary Deletion Channel with a deletion probability p of 1/2 and the channel has no error correction coding at all and that any given message can only be sent once. I want to conjecture that the capacity for the aforementioned channel is zero ( channel capacity for the Deletion Channel is presently unknown), can anyone point to any known complexity results that would point me in the direction of proving or disproving my conjecture?
Comment: The capacity of the deletion channel $\text{Del}: X\to Y$ with deletion prob $p$ can be defined to be $\lim_n \frac1n \max_{P_{X^n}}I(X^n ; Y^n)$ where $I()$ is mutual information and $P_{X^n}$ is any possible distribution on $X^n$. This definition is independent of "error correction" and of "sending message only once". Please formulate properly what is the quantity you wish to get. Specifically, how the above two vague notions (marked in quotations) are defined.
Answer: The capacity of a deletion channel with deletion probability $p$ is still not fully understood, however it is known to be somewhere between $1-h(p)$ and $(1-p)/9$.
See a survey by Michael Mitzenmacher on deletion channels: A survey of results for deletion channels and related synchronization channels.
Comment: @WilliamHird Can you prperly define your restrictions? Channel's capacity is a property of the channel. If you use it in a strange way (e.g., without using error correction) you will not be able to achieve the capacity, but that doesn't change the capacity of the channel. Please clarify your question.
Comment: @WilliamHird it would be helpful if you gave a *formal* statement of what you are looking for.
Comment: Yes I have read that paper but my question is about  a deletion channel with the restrictions mentioned .
Comment: That's the point of my question, to try to prove that with no error correction and the restriction that any message can only be sent once, you can't reach the 1-h(p) and (1-p)/9 bound.  Is this bound mentioned in the paper for the "raw channel" or for a channel with error correction coding?
Comment: G:  Not trying to be cute here, but I thought my question was clear , maybe if you ask me specifically what you don't understand about the question as posed we can go from there.
Answer: This is an old question, but let me give a final response for anyone who comes across it.
Your restriction is that we are not allowed to do any coding. In other words, every source codeword is transmitted directly across the channel. Since there is no redundancy being added, the transmission rate is 1 bit per symbol. As mentioned by Ran, we know that such a rate is not possible due to the channel capacity upper bound of $1/9(1-d)$ for deletion probability $d$.
A more direct way to show that such a rate cannot be achieved is to consider the block error. For a block length $N$ approaching infinity, there is almost surely at least 1 deletion in the channel output. WLOG, consider an input sequence $x = (x_1,x_2,x_3\ldots,x_N)$ that becomes $y = (x_2,x_3\ldots,x_N)$ after deleting $x_1$. Looking at $y$, there are multiple possible input sequences, only one of which is the transmitted $x$. The best decoder is guaranteed to make an error with non-zero probability, and hence the block error of this symbol is non-zero. For multiple deletions, this block error is only going to increase towards 1.

Capacity of the Deletion Channel


How did Mad-Eye Moody survive in the enchanted trunk in Goblet of Fire?

Question: How did Alastor 'Mad Eye' Moody manage to survive for 9 months when he was trapped in the enchanted trunk in the 4th book?
Comment: One would assume that Barty Crouch Jr took care of him somehow.
Comment: It seems likely. Remember, he didn't kill moody. Instead, he just trapped him in a box.
Comment: O.o.... doesnt seem that likely since he is an auror,one who has gained an immense reputation for capture of dark wizards..
Answer: Barty Crouch Jr had to keep Moody alive, since he needed a source of hair/etc for the Polyjuice potion. There is no canon explanation, but it would be relatively easy for Crouch to ensure that Moody was fed. Dropping food and water into the trunk is a relatively easy matter. Moody was helpless in that box (no wand, unable to attack physically and too weak to do anything anyway), so Crouch wouldn't have been afraid of Moody. So I see nothing stopping him from dropping food/water into the trunk every now and then.
Addendum: As mentioned in StarPilot's comment, Barty would have had to keep Moody around for questioning anyway. Polyjuice potion just makes you look like the subject, you still need to do a convincing impersonation by acting like him/her. Barty doesn't know all about Moody, he'd need to question him if something came up.
Besides, Moody, being the paranoid guy he was (as well as being an Auror), may have had spells that would activate when he died. Aurors did dangerous things, and were hated by many, it was quite reasonable for the Ministry to want to know about an Auror's death (If someone is on an Auror-killing spree, they should be able to stop him/her before it becomes a spree). Add Moody's paranoia to this and it's very likely that he had such spells on himself.
Comment: Best zombie costume idea ever.
Comment: Barry Junior could also question or use magic to question Moody, to better impersonate him. Plus, if any magic was used to check if Moody was alive, it would show Moody was indeed alive. Moody as a paranoid person may have created and cast spells to alert his trusted friend that he was dead, and keeping him alive but helpless would prevent this.
Comment: @TheLollipopGuildmaster: He doesn't have a wand. And, one may need a live subject for Polyjuice potion. Polyjuice reflects the current state of your subject, so scars and all are mirrored. Which means that it is possible for death to break the potion
Comment: This answer is correct. [Potions did expire](http://scifi.stackexchange.com/questions/22822/is-there-canon-info-on-shelf-life-of-various-potions-in-harry-potter). And, corpse wouldn't survive much longer.
Comment: It never made sense to me that he kept him alive. He is a dangerous crazy old wizard hunter. And they never said the hair couldnt be taken from a corpse.
Comment: What about hair piling in advance? Also, I don't think there was an expiry date of polyjuice potion.
Comment: @StarPilot: Good point, edited, thanks :)
Answer: As far as I know, it hasn't be explicitly stated, but I believe the person being transformed into must be alive.  All the instances of Polyjuice potion being used that I could find was between people who were still alive.  Although that isn't evidence that you couldn't turn into someone who was dead, not that I could think of a reason to, or a part in HP where that would be beneficial.
It's also possible that he was kept alive because of the freshness of the ingredients or just in-case Crouch Jr ran out of "A bit of the person one wants to turn into."
Comment: Sorry, but your answer here is just dealing with *why* he was kept alive. The question asked was about *how* he survived or how he was kept alive.
Comment: @phantom42 The *how* is that Crouch Jr. kept him alive, most likely with charms and potions, so that he could continue to make Polyjuice Potions.
Comment: Sure, but the closest that @mcfuu's answer gets to that is the comment about the "freshness of the ingredients".


Bypass capacitor in 555ic

Question: What will happen if we don't bypass capacitor at pin 5 of 555 ic as an astable multivibrator.Will voltage/current changes in line affect on pulse width?will capacitor at pin 1 discharge late?
Comment: go play here ... http://tinyurl.com/y7ftysze ... look under the `circuits` in the menu bar
Answer: pin 5 of the 555 is "control voltage" a tap on the divider chain that sets the levels for setting and resetting the internal flip-flop.
without a capacitor there the thresholds will tract the supply voltage exactly ad a glitch on the supply (a brief spke or dip) would cause the threshold to move leading to early triggering of the flip-flop - so without the capacitor on pin 5 supply noise can cause your 555 to run fast.
Putting the capacitor there stiffens the voltage there with respect to ground, and if the timing capacitor is also ground referenced gives a large amount of immunity to supply glitches,
so to answer your questions:

Will voltage/current changes in line affect on pulse width?

yes absolutely, depending on the timing of the voltage changes they
may cause the pulses to begin and/or end early

Will capacitor at pin 1 discharge late?

Pin 1 is ground, it's unclear which capacitor you're asking about, because they all connect to ground.
pins 2 and 6 are the timing pins, the charge/discharge of the capacitor there can be triggered early by glitcing the power supply.
it can also be delayed if the glitch goes the opposite irection but that effect would be a smaller effect.
Comment: I don't understand, can you draw a picture?
Comment: Actually I didn't use 0.1cap betwn pin 5 and grnd.sometime my multivibrator was remaining in ON stage.it means relay was once energised,it wasn't getting de ernergise.hence  tOn=6sec was seems infinite.could it be due to absence of 0.01micro capacitor?
Answer: Depends on the Ground and VDD inductance.
The original circuit (Signetics) 555, in bipolar technology (schematic of original design shown on "Wikipedia 555 timer IC"),has current limiting in every path except the output driver.
All other paths are either resistor_limited or are actual current sources.
Be safe, be able to ignore that cross-conduction glitch of the output driver; use 0.1uF between VDD and GND pins.
Comment: OP is not asking about a VDD bypass capacitor, the terminology might not be the best, but pin 5 is definitely not VDD.

Bypass capacitor in 555ic


Does dark matter cause diffraction of light?

Question: If there is dark matter (Degrasse Tyson likes to call it "dark gravity"), would it diffract light?  If so, how could this be differentiated from gravitational lensing?  By spectroscopy?
Comment: Do you really mean diffraction, or do you mean refraction?
Answer: No, dark matter is electrically neutral, so it doesn't interact with electromagnetic waves.
Comment: ...magic dust...
Answer: If you are asking if light is deflected by the gravity of dark matter, the answer is a definite "yes":  that is why gravitational lensing occurs.
Often we think of diffraction as  being an effect that requires periodic structures or abrupt changes in the phase and/or amplitide of a wavefront,  but the meaning of diffraction is more general than that.  Diffraction can be described as propagation of a wavefront,  in situations where Huygens' principle produces more useful results than ray optics.  Anything that modifies different portions of a wavefront differently will result in diffraction, and the same is true regardless of the mechanism of modification - so it is true even if dark matter's gravity is responsible.
You also asked how diffraction by (the gravitational effects of) dark matter could be differentiated from gravitational lensing.  The simple answer is that there isn't a distinction between the two cases.
Perhaps more to the point is the fact that light following different paths around a clump of dark matter, even if the light arrives at the same place, won't form a static  interference pattern because the different paths are very unlikely to be the same length, and are sure to be changing continuously.  They can easily differ by light years when the dark matter clump is associated with a galaxy or cluster of galaxies.
Comment: The OP specifically says they *don't* want to talk about gravitational lensing.
Comment: True, but there really is not a hard distinction.  It is a bit like the distinction between what a refractive lens does to a (monochromatic) wavefront and what a holographic lens does to the same wavefront.  Downstream, there is no difference for most practical purposes.   Where the difference shows up is in path length dependence vs position on the lens.  It is not correct to say dark matter does not interact with light, because it obviously does-- via gravitation. The *mechanism* of interaction is irrelevant to diffraction; only the *pattern of interaction* matters.
Comment: I was thinking of "refraction" rather than "diffraction".   From the comments its seems to me that both are worth thinking about.  For example, gravity can change wavelength (as I remember, Einstein predicted that the sun lengthened it).  Perhaps dark matter or a galactic cluster responsible for lensing changes wavelength in an interesting and detectable manner as the light approaches and then recedes.


How to fix media test error after installing elementary os?

Question: I am new to elementary os.I was trying to dual boot elementary os and windows 10. So, my problem is that I installed elementary os by following the article from foss website. After I finished installing at the last step where I had to remove the pendrive I messed up. I removed pendrive before restarting the laptop. Now I can't even boot in to Windows or Linux.
Answer: Put the usb stick once again, power on machine hold F12, choose the usb stick to boot and try to install the Elementary OS again. You can select the existing partition with Elementary OS as a target partition.


How to setup KVM partitioning?

Question: I'm trying to install SolusVM/KVM on a server with one 500 GB HD and one 256 SSD. I am a complete beginner at Linux. SolusVM has a script on the site to download and install automatically, but how do I specify to install the script on the 500 GB HD and put the virtual machines on the SSD? And how should my partitioning look like? I want KVM isos and stuff on HD and virtual machines on SSD. Thanks a lot.
Answer: Just let the SolusVM install run, when you create the volume group, just create it on your SSD and define the volume group in the Administrator panel.
The ISO's are stored in /home/solusvm/xen/iso (/home/solusvm/kvm/iso is simply a symlink to /home/solusvm/xen/iso).
Comment: Thanks! But how should I partition the drives? And how do I install the SolusVM on the 500 GB HD?
Comment: @Richard Have a look at: http://docs.solusvm.com/installation#partitioning. You just install SolusVM as described in the installation Wiki **but** when you need to create the volumegroup, create it on the SSD ;-)

How to setup KVM partitioning?


Novel about a microscopic civilization living inside the Sun

Question: I read a novel a few years ago about a tribe of alien people living inside the Sun. They were surfing the electromagnetic waves, visiting a floating city, and traveling towards the center of the Sun to learn of their origins. It was pretty far out, but still felt somewhat plausible.
Comment: Hi, welcome to the site. In roughly which year did you read this and when do you think it might've been published? Also, do you recall anything about the cover?
Comment: Read maybe five years ago. It felt like a story out of the seventees or eightees, but cant be sure.
Answer: This is probably Stephen Baxter's Flux.
From Wikipedia:

Dura and her fellow human beings live inside a neutron star, as they have for generations. As long as they can remember, "glitches" (instabilities of the magnetic field inside the star caused by changes in the star's rotation) have happened from time to time. As the novel starts, the worst glitch that anyone can remember threatens to destroy her home. To survive, Dura must travel to a far-off city, and eventually outside of the star itself.
... the main race of the book, microscopic human-like beings who live within a neutron star, travel around using the magnetic field lines within that star.
Comment: This. The other possibility was _Dragon's Egg._
Comment: The title made me think of *Dragon's Egg*!  Fun story.  Now I'm going to have to check out *Flux*...


How can you create a hive mind race?

Question: I wanted to make a tech-based hive mind species in Stellaris, but when I went through the whole list of options, I could not find anything that made your species into a hive-mind other than the naming convention. I remember watching playthroughs of the game were the AI were generated as hive-minds, but is there any way to make your own? Does it need any of the DLC's as I don't have any of them?
Answer: It is in the Ethics circle - the icon in the exact middle. It is available only if have the Utopia DLC for the biological empire or "Synthetic Dawn" if you want a machine one:

Then on the right, under "Authority" you can select are you a machine or biological hive mind.
Comment: Is "Biological Constructs" from a mod (first screen shot)? I've never noticed it before.
Comment: @FrançoisAndrieux  it looks like it - I've taken those photos from the Google

How can you create a hive mind race?


How to root Android using CF-Auto Root Solution

Question: How to root using CF- auto root? How to use Odin with it?
Comment: Possible duplicate of [Do I need CFs auto root If I can install TWRP recovery?](https://android.stackexchange.com/questions/93135/do-i-need-cfs-auto-root-if-i-can-install-twrp-recovery)
Answer: When you decide to root your Android usually you have to pick a proper rooting solution next. Otherwise, you have to face a lot of problems. In the worst-case, you might have a bricked mobile phone or tablet instead of a rooted one. So in this article, I'm going to talk about one of the best rooting tools if you have a Samsung smartphone or a tablet. We are going to root it using CF Auto Root through download Odin.
Preparations before Root Android
So this an invention of the famous XDA developer Chainfire. The exploit is working for more than 300 common Android devices.  But today we are gonna focus on Samsung. If you want to work with this you better check the compatible list which is available in the XDA Developers forum. It usually unlocks the bootloader first. So make sure to keep backups unless you will lose your data. And the other special fact is that Samsung drivers are only compatible with Windows. So you should have a Windows PC for the task.

Android Rooting for Samsung

First, you have to make sure to install proper drivers. Check this
link out and you will be able to download them. You just have to extract the zip and launch the .exe.
The next step is to download the correct CF-Auto-Root file. There are
many. So to find your compatible version check the model Number and
the Android version. And then click the download.
Make a new folder and extract the downloaded zip file there. The file
should be located in the hard drive.
Then put your smartphone or the tablet into the download mode. Volume
down button, home button, and power button will help to do this.
Press and hold it simultaneously.
Now connect it with the windows computer using a USB cable.
Now browse to the folder that has extracted the content of the
downloaded zip file. On there you will see a .exe file with a name
ODIN. Right-click on that select to run it as an Administrator
Once Odin is running it will detect the device.
In the interface  "ID:COM" field should be turned to blue.

Click the AP button and keep the rest unchanged.
Again you have to browse to that exact same place that has extracted
the content and select CF-Auto-Root.rar.md5 file. Then click Open
When the log tab displays a message saying "Leave CS" click the
start.
Now the rooting process has been started.
When its finished the device will reboot back to Android.

Wrap Up
So this is an almighty process of rooting your Android Operating System using CF-Auto-Root and Samsung Odin. It is safe and pretty easy compared to some other tools. So if you want to check the root status you should download the root checker.


Does one make bracha when tasting food?

Question: Does one make bracha when tasting food? If one were to taste food and immediately spit it out, presumably he wouldn't have to make a bracha. But what if he/she wants to see if it goes down smoothly or if it is easy to swallow, would they make a bracha?
Answer: The Gemara in Berachos 14a says:

מטעמת אינה טעונה ברכה והשרוי בתענית טועם ואין בכך כלום. ועד כמה? ר' אמי ור' אסי טעמי עד שיעור רביעתא
Tasting does not need a blessing, and one fasting can taste and there is no problem. How much? R' Ami and R' Asi would taste up until a revi'is.

There is a dispute in the Rishonim about how to understand this Gemara. Rabbeinu Chananel (cited by Tosefos there) writes that it is only if one spits it out that he does not make a blessing (and the part about up to a revi'is is specifically with regards to a fast [Rosh 2:6]). The Rambam (hilchos berachos 1:2), as understood by the Beis Yosef, rules that one does not make a blessing upon tasting even when swallowing, up until the tasting is a revi'is.
The Shulchan Aruch, O.C. 210:2, cites both opinions, with the opinion of Rabbeinu Chananel as "there are some who say". The Rema applies the rule of ספק ברכות להקל, and therefore one does not make a blessing on tasting, even if one swallows it.
Answer: As context, note that when eating, there is no minimum amount necessary for a blessing, i.e., even if one eats a very small amount of food one must make the appropriate blessing.
However one does not make a blessing when tasting food. The concern is that there is a fine line before tasting and eating as the following two sources explain
from ou.org

If someone is tasting food to see what seasoning it may need, he
  certainly doesn't need to make a bracha if he spits it out. If he
  swallows it, however, it is questionable whether or not a bracha is
  required. On the one hand, he swallowed it, but on the other hand, his
  intention was not to eat! Therefore, when one is in this situation, he
  should intend to have the benefit of what he tastes as food, make a
  bracha, and swallow it.

from aish.com (see there for further sources)

A bracha is only made on food that is consumed. If you are merely
  tasting food (and then spitting it out), a bracha is not said.
What if you swallow a piece of food only for the purpose of tasting
  it? Is a bracha required, or not?

In this case, there are various rabbinic opinions. So if you want to    taste a food and swallow it, the best approach is to either:
Eat at least a kezayit (approx. 15 grams – 1 fluid ounce) or drink at    least a revi'it (approx. 98cc – 3.3 oz). In this case, a bracha
  is    certainly required.
Eat with the intention of enjoying the food as well, not only to    taste.
First say a bracha on another food, which is being eaten for    enjoyment. Or as we said before, simply spit it out instead of swallowing.
Comment: Thanks for the answer mbloch:) What would one do when tasting bread according that one should have in mind to eat (vs taste) it and then make a bracha? Would they have to wash, make hamotzei, and then not bench, because it is less than the d'rabanan shiur?
Comment: @DonCorleone best options are to taste/spit out with no bracha or to wash, hamotzi, eat enough, then birkat hamazon. Your scenario is also possible but please check with a qualified Rav first
Comment: It would seem that according to some one can eat a little in order to taste, without making a bracha.

Does one make bracha when tasting food?