Location of spelling dictionary in TeXShop on El Capitan

Question: I was doing a spell check in TeXShop 3.58 and accidentally clicked on "Learn" for a word that isn't a word ("subalfgebra", in case you're wondering).  I'd like to fix this by editing the dictionary manually but cannot figure out where TeXShop keeps its word list.  The fake word does not appear in ~/Library/Spelling/LocalDictionary, for example.  I have not installed any additional spell checker such as Cocoaspell or Excalibur.
Comment: I'm glad you got things sorted. Since this is something that others might encounter you should turn your comment into an answer. It's perfectly fine to answer (and accept) your own question. You can answer right away but I think you need to wait two days before accepting a self answer.
Comment: Does the misspelt word now show up as correctly spelled in both TeXShop and other Mac apps (e.g. Mail?).
Comment: Yes, it did.  Actually, I figured how to fix the problem - it wasn't TeXShop's fault.  I just had to restart AppleSpell.  For reference, here's what you do:

 (1) remove the fake word from ~/Library/Spelling/LocalDictionary

 (2) quit TeXShop

 (3) open up a Terminal window and type "killall AppleSpell"

 (4) restart TexShop.

For details: [https://discussions.apple.com/thread/6844365?start=0&tstart=0].  Thanks for asking the question that led to the solution
Answer: I discovered how to fix the problem.  Restart AppleSpell, as follows:

Remove the fake word from ~/Library/Spelling/LocalDictionary.
Quit TeXShop (and any other apps that are using AppleSpell).
Open a Terminal window and type killall AppleSpell
Restart TexShop.

See also: Discussion on Apple Support Forums.
Comment: I have the same problem and, like in your case, my fake word does not appear in  ~/Library/Spelling/LocalDictionary. So how can I remove it from that file if it is not there, as your suggested?


Where did the notion of 'dumbing down' first arise from?

Question: Can anyone tell me the origin of the expression 'dumbing down'? It seems increasingly difficult to trace the etymology of neologisms, but I'm curious whether anyone has any information on this phrase. Thank you.
Comment: A book by John Taylor Gatto titled _Dumbing Us Down_ appeared about 20 years ago and did a lot to popularize the term.  It criticized educational practices.
Answer: The OED's earliest citation is from the 'Los Angeles Sunday Times' in 1927:

Hollywood writers have a process called 'dumbing it down'.
Answer: The earliest Google Books search result to use the phrase "dumbing it down" is an article in Life magazine from 1932—and like the OED reference that Barrie England cites, it appears to involve Hollywood script writers:

It seems they are convinced that unless they make changes, their employers will not think they are working. The employers, on the other hand, have been sold (probably by script writers) on the idea that it is impossible for the general movie public to understand the contents of a novel unless the language is simplified. They call this "dumbing it down." Well, they can't fool me on this one. Any person of average intelligence could furnish the dialog for "Wayward." All they would need would be a copy of "Wild Beauty," a pair of scissors and a paste pot.

A Google Books search finds multiple instances of the phrase "dumb it down" between 1933 and 1939. The earliest is from Forum and Century, volume 90 (1933):

I can cheer too for the Hollywood gag men in conference on a comedy which has been revealed as too subtle, when they determine they must dumb it down. That phrase saves time and wearying gestures. And "switcheroo" has value in the state department as well as in the mouth of gag men.

From New York Times Saturday Review of Books and Art, volume 24 (1935):

If this little book could be made required reading for several categories of people, such as newspaper publishers and writers, producers and writers of plays, makers of movies, all those whose business it is to furnish the great masses of population, lowbrows and highbrows alike, with mental nourishment, it would be thrilling to watch its influence upon them. For it is the settled conviction of most of them that they have to "dumb it down" for the public, that the lower they push their standards of intellectual and emotional appeal, the greater will be their success

From the same issue of New York Times Saturday Review of Books and Art (though probably a different article):

But therein speaks merely the horror of the specialist at the iniquities sometimes perpetrated by crude writers who believe that to popularize a scientific work is to devitalize it and dumb it down to the lowest common denominator of intellect.

From The American Ecclesiastical Review (1936):

It was permissible to change the order of the clauses to avoid monotony, but I could not, as I valued my salary, add an expression of regret or a testimony to his magnificent character. When I had done that sort of thing for weeks I was eventually accepted as a newspaper man. The word "journalist", by the way, is a fighting word in any metropolitan news office. And the refrain, "Dumb it down! Dumb it down!" will ring in my ears, I think, forever.

The earliest instance of "dumbing down" without the "it" in Google Books search results comes from American Association of Group Workers, Group Work (1942):

As we become more competent, and clearer about and more certain of our scientific foundations, it may be that we will then be more ready and able to explain what we mean by "group work." Effective interpretation requires not so much a "dumbing down" of technical material as it does a common attempt on the part of the technician, the administrator, the board member, the citizen, and the consumer to seek out needs, evaluate services, and examine into their effectiveness.

It appears, then, that the original dumb-downers were people in the film industry, but that the phenomenon of dumbing down spread in short order to newspapers, popular science, and other sources of culture. The fact that Reader's Digest picked up and reprinted the 1933 Forum and Century article that same year suggests that Americans far removed from movie studios and newsrooms were exposed to the idea of "dumbing down" within a short time of the term's first known appearance.
Answer: According to Wikipedia the term originated in 1933 in screenplay writers' parlance.


Wifi thermostat

Question: Can I connect C-wire to the empty terminal(red arrow)
Answer: No, no, no!
That terminal is on the wrong side of the transformer!  It's a primary tap used when the furnace is hooked up to a circuit or service that is "two-thirds" of a three-phase system, and has nothing to do with your quest for the C wire connection point.
Instead, you want to look at the wires that connect at the top of the transformer (helpfully labeled "24VAC" with the upward pointing arrows) in your picture.  One of them will measure 0VAC or thereabouts to the R (red) wire going off to the thermostat (that's where the R wire comes from); the other will measure 24VAC or so to the R wire, so this second wire will be where you want to tap your C wire connection at.
Comment: Your second paragraph is extremely confusing. You make it sound like if the red wire from the transformer is the one going to the thermostat that you hook the C wire there.....when in fact that would be incorrect.

Wifi thermostat


WCF Named Pipes and Azure Web Roles

Question: I've spent much too long banging my head against a wall for this one.
Running Azure SDK 1.7 / Visual Studio 2010 Ultimate. I can't seem to run a WCF service over named pipes hosted by IIS/WAS. Any attempt to connect to my service in the compute emulator yields
The pipe endpoint 'net.pipe://localhost/services/MyService.svc' could not be found on your local machine
I was going to post my code, but it turns out that the sample code from this article http://blogs.msdn.com/b/tomholl/archive/2011/06/28/hosting-services-with-was-and-iis-on-windows-azure.aspx gives me the exact same error. All the code is available on that site.
As far as I can tell, the powershell tasks are executing successfully.  (Well, OK, I had to change the ExecutionPolicy from Unrestricted to RemoteSigned in the startup task, but after that it ran just fine.)  The NetPipeActivator service is running, and when I call Get-WebApplication I see net.pipe in the protocol list:
PS C:\> $WebRoleSite = (get-website "*webrole*").Name
PS C:\> Get-WebApplication -Site $WebRoleSite

Name             Application pool   Protocols    Physical Path
----             ----------------   ---------    -------------
WcfService1      4a16b147-f9ac-41d9 http,net.pip c:\Code\WasInAzure\WcfService1
                 -9543-0577da64fb9a e

What else could be going on?
Comment: Check if those threads help:
http://social.msdn.microsoft.com/Forums/en-US/wcf/thread/3a97c32d-07f1-40a9-ab8c-f1220c8feb76
http://social.msdn.microsoft.com/Forums/en-US/wcf/thread/55fae194-e1ea-4bd0-8c0a-dd924d4d4ac8
Answer: I solved it. Phew!
My issue was that I was hosting my WCF service from within the root website, not a child application. I had to add the following line to the powershell script to enable named pipes on the main website:
Set-ItemProperty "IIS:/Sites/$WebRoleSite" -Name EnabledProtocols 'http,net.pipe'

Answer: I also ran across this error while working with this sample code. In my case, I received the error when trying to replicate the Cloud Service, but not with the original Cloud Service project in the solution.
The key difference ended up being the osFamily attribute of the ServiceConfiguration element in the ServiceConfiguration.csfg file.
In the sample, it's set to "2:"
<ServiceConfiguration serviceName="WasInAzure" xmlns="http://schemas.microsoft.com/ServiceHosting/2008/10/ServiceConfiguration" osFamily="2" osVersion="*" schemaVersion="2015-04.2.6">

In the newer projects, it's set to "4:"
<ServiceConfiguration serviceName="WasInAzure" xmlns="http://schemas.microsoft.com/ServiceHosting/2008/10/ServiceConfiguration" osFamily="4" osVersion="*" schemaVersion="2015-04.2.6">

The change in OS breaks the PowerShell script. Here's an updated version of RoleStart.ps1, which should work on Server 2012 R2 (osFamily 4):
write-host "Begin RoleStart.ps1"
import-module WebAdministration

# Starting the listener service
install-windowsfeature -name AS-Named-Pipes
$listenerService = Get-WmiObject win32_service -filter "name='NetPipeActivator'"
$listenerService.ChangeStartMode("Manual")
$listenerService.StartService()

# Enable net.pipe bindings
$WebRoleSite = (Get-WebSite "*webrole*").Name
Get-WebApplication -Site $WebRoleSite | Foreach-Object { $site = "IIS:/Sites/$WebRoleSite" + $_.path; Set-ItemProperty $site -Name enabledProtocols 'http,net.pipe'}
New-ItemProperty "IIS:/Sites/$WebRoleSite" -name bindings -value @{protocol="net.pipe";bindingInformation="*"}

write-host "End RoleStart.ps1"

Summary of changes:

Added install-windowsfeature -name AS-NamedPipes, as the NetPipesActivator service wasn't present.
Updated EnabledProtocols to enabledProtocols. That one took some searching. I finally found a comment to a blog post that put me on the right path.

WCF Named Pipes and Azure Web Roles


Heating issue in Nissan Sunny Diesel

Question: I am driving Nissan Sunny Diesel in pahalgam Kashmir with a temperature range of -20 to 40. My car is heating slowly. I have changed the thermostat also. But after driving the car almost 30 kms the temperature gauge still shown -30 in dash board. Plz suggest.
Comment: Please explain whether the engine is not really heating up or the temperature gauge is just not reading properly.  If it's not heating up, then the problem is the thermostat.
Comment: With temperatures well below freezing, a new thermostat in place, and the engine not heating up, this may be a situation of blocking off all or a good portion of the radiator to allow the engine to warm up. Extreme cold temperatures can justify radiator blocking. This presumes correct antifreeze amount and the cooling system is maintained.
Answer: After 30km your engine and coolant should be warmed up. I'm sure your engine will be warm, what you need to do is check your radiator, be careful not to burn yourself doing it. A good test is whether your heater works after 30km, if it does you're probably okay, if it doesn't you have a more serious problem.
If your radiator is still cold at 30km and your heater is still blowing cold air then you probably have a problem somewhere in your cooling system:

Coolant: In extreme cold it could be that your anti-freeze isn't strong enough and your coolant has frozen. I had that happen to me, my coolant froze in the hose, which then exploded due to the pressure
System block: there could be an obstruction in the cooling system which is stopping it from circulating
Bad water pump: the coolant pump may have failed, or it's not being turned by the belt
Failed coolant thermostat: there's a valve that keeps your coolant circulating in your engine until it heats up to a certain level, if that fails then it will stay open or closed, if it stays closed then your radiator will never heat up

I'd start with checking your coolant and go from there. If your radiator is warm after 30km and your heater is blowing hot then it's one of these things:

Bad temperature sensor: there's an electrical device screwed into your radiator which reads the coolant temperature, it may have broken or the wiring to it is bad
Broken coolant gauge: it could be that the gauge on your instrument panel has gone bad and isn't registering changes in temperature
Answer: I'd lay money on it being a bad temp sensor ... typical for them to fail and show the low reading. These are easily tested, too. The output is linear. As heat is applied, the resistance goes up (or down, depending on the sensor). You can usually find the scale (temperature vs resistance) online for your specific vehicle. If not for your specific vehicle, manufacturers usually have the same scales over their line of vehicles, so if you can find it for one, you can find it for all.
Realistically, all you are truly looking for is for the resistance to change as temperature changes. Pull the temperature sender out and check the resistance. Heat the sender up and check the resistance again. It should have changed. If it doesn't, it needs to be replaced. You shouldn't have to heat it very much to see the difference ... even 10°C should see a change in resistance.

Heating issue in Nissan Sunny Diesel


Binance - Suspended

Question: I'm new to the site and tried searching but couldn't find anything related to Binance always saying "Network Congestion, Withdrawal Suspend" when trying to withdraw IOTA shares - any ideas what is going on with that?
Comment: I'm voting to close this question because it is focused of a temporary issue occurring these days. A more generic question about networking congestion issues "in general" could be better.
Answer: The volume being traded is at unprecedented levels today, and at the exact same time, the network is being hit with a spam attack. Therefore, it seems likely that binance is being overwhelmed in their network.
Comment: Thank you - it just seems that it has been "suspended" almost all the time since last week

Binance - Suspended


What species of berry is this?

Question: This is a berry bush outside my workplace. What species is it? I can easily get more pictures if that would be helpful.
Answer: Assuming that this had white flowers on it a few weeks ago, and assuming that you're in North America, I'd say that the shrub is a Chokeberry (Aronia melanocarpa). The shrub might be about 3 feet high and wide, unless trimmed by your workplace's maintenance people. If I'm right, then the berries are edible, but only if cooked first (too bitter otherwise - hence "choke" berry). They apparently make very good juice.
Comment: Wasn't in the office till recently, so unsure on the white flowers, but the rest of the description fits. I'll wait 24 hours and accept your answer if no one else chimes in. Thanks!
Comment: Later in the summer the berries should turn a very dark blue-black. Aronia also has excellent fall foliage (scarlet and orange, mostly).

What species of berry is this?


Did the US miltary ever reverse-engineer Goa'uld or other alien energy weapons?

Question: The SG- teams often bring back weapons like Jaffa staves and zat-guns, did they ever take any apart so they could learn to make energy weapons of their own?
Comment: they work at it, then the azgard give them their tech, so why bother.
Comment: It seemed like SG1 ended up using the stun-mode capable Gou'old weapons quite regularly at one point.
Answer: As Jack put it, Jaffa Staff weapons are weapons of terror. The US Military has no interest in terror. They already have superior weapons of war, they are made to kill your enemy. The p90 is more accurate at a farther range, and has proven better than Goa'uld weapons.
That said, all tech found is researched at the SGC and Area 51, as seen in multiple episodes. The US Air Force did reverse engineer the goa'uld Death Glider,  to create their first space worthy fighter, which quickly backfired due to Goa'uld anti reverse engineering counter measures. The staff weapons are uses for research in building better armor for the SG teams.
Comment: @Omegacron - There is a difference between terror and fear. Militaries do not want to terrorize their enemy, they want to be feared.
Comment: The US Military has no interest in terror. HA! Any decent military force has an interest in terror.
Comment: The recall device had nothing to do with reverse engineering but to enact revenge against a deserter or betrayer.
Answer: The answer is no. At first they lacked knowledge to do that - remember that everything Goa'uld is naquadah powered. Only after first Ancient Knowledge repository (Ep. "The fifth race") enough knowledge was gained to build working naquadah generator (but it still took years).
In the meantime the gunpowder weapons proved adequate to fight Goa'uld and Replicators. As another answer suggests - they were weapon of war, while Staffs and zat guns were weapons of terror. Also the gunpowder weapons were already available and people knew how to use it.
And then there was an enormous leap in tech base for Earth and Goa'uld weapons were deemed obsolete compared to Asgard/Ancient tech.
I did ask myself the same question: why not rework staff weapons into something better (that is: drop blast projector onto some sort of modern weapon stock), but it would require a lot of work, knowledge and of course power source. At the time every bit of naquadah was needed for research and power generation, later to power up fleet of ships, so I gather it was deemed uneconomical and with unreliable source of power for "ammunition" rather risky.
Also, there was a plasma (?) gun built completely using Earth tech (Ep. "Bounty")... It was a bit bulky...
Comment: Technically not "gunpowder" weapons. The term "ballistic" weapons would be a good way to distinguish guns from the alien energy weapons.
Comment: Technically correct, but that includes all sort of projectile weapons not gunpowder based. The Rail guns are quite late introduction and out of scope for this question... I am, however, convincible to changing that to "explosive propellant" to cover missiles.... How about that compromise?
Comment: I only say ballistic, as none of the Goa'uld or Asgard weapons are ballistic, and all of the Tau'ri weapons are (excepting various explosive and melee weapons).
Comment: Sorry, but you're incorrect. Goa'uld used bombs, which were energy explosive devices dropped by Al'kesh class. Bombs are ballistic weapons...
Comment: "excepting various explosive and melee weapons"
Comment: I would agree with you on principle, but as it's somewhat artificial distinction, I'll take exception to that. Grenades (and Goa'uld equivalent) and aerial bombs are used extensively in warfare, in contrast to melee weapons (which are mostly of duelling and ritual use). Additionally Earth employs guided missiles, which most certainly **are not** ballistic. And last but not the least: at the ranges the engagements were fought over, the projectile trajectory is basically flat, so "ballistic" would be more of a concept description rather than actual issue... So I'm keeping my classification. ;)


Covering number of the meager ideal

Question: I learnt from a paper that "Let cov(K) be the least cardinal k such that a perfect Polish space can be expressed as a union of k meager sets. (It does not matter which perfect Polish space is used to define cov(K) ) " . I don't know why cov(K)'s are equal for different perfect Polish spaces. I want to use this result in my own paper, but I fail to find any books contain this result, could anybody kindly to help me ?
Answer: This should be in standard texts on descriptive set theory, like Moschovakis's "Descriptive Set Theory" or Kechris's "Classical Descriptive Set Theory".  The basic idea is that any two perfect Polish spaces become homeomorphic after you remove (at most) countably many points.  Countably many points constitute a meager set and therefore don't affect cov(K).
Comment: Not `points' but `nowhere dense sets' I think:after you remove the boundaries of the memebrs of some countable base what is left is zero-dimenional and Polish and perfect and nowhere locally compact, hence homeomorphic to the space of irrationals. NB the plane minus a countable set and the line minus a countable set are not homeomorphic: the former is connectedm the latter is not.
Comment: Thanks for your help, i have understood the union of the boundaries of the memebrs of some countable base is meager, and i have found Alexanderov-Urysohn theorem which tell us baire space is unique up to zero-dimenional and Polish and perfect and nowhere locally compact, but I don't know why does a perfect polish space became a baire space with the union of boundaries removed ?
Comment: The starting space is Polish after removing those boundaries you have a G_delta-subset of that space, now apply the theorem that a G_delta-subset of a completely metrizable space is completely metrizable (using some other metric of course).
Comment: Oh, and to avoid confusion: `Baire space' (capitalized) is the topological product N^N, whereas `a baire space' is a space that is not meager in itelf, It seems to me that you confused the two but I might be wrong.
Comment: oh, it is my fault, the "b" should be capitalized. thanks again, i think i have understood.
Answer: Andreas gave an argument while I was typing, so let me give a different approach: The proof that the covering number is the least $\kappa$ for which ${\sf MA}_\kappa({\rm countable})$ fails is soft enough that it can be easily adapted to work for any Polish space you begin with. Details of this equivalence are, for example, in the excellent ``Set theory: on the structure of the real line'' by Tomek Bartoszynski and Haim Judah, that you will want to have as a reference anyway.
Comment: Thank you! I have found the charactorization of cov(M) for R^1 in that book, but it is not mentioned cov(M) for general perfect polish spaces, do you  mean the proof of the charactorizations also relevant to any perfect polish spaces ?

Covering number of the meager ideal


Vagrant install of CloudFoundry on AWS: Provider expected: aws Provider of box: virtualbox

Question: I am having an issue with the following guide: http://www.activestate.com/blog/2013/07/how-deploy-cloud-foundry-v2-aws-vagrant in installing CloudFoundry on AWS using Vagrant.
I have followed it down to the point where I have to deploy using the command: vagrant up --provider=aws
This prints the following output which I have had little luck debugging due to my lack of understanding of Vagrant:
Bringing machine 'cf-install' up with 'aws' provider...
[cf-install] Box 'precise64' was not found. Fetching box from specified URL for
the provider 'aws'. Note that if the URL does not have
a box for this provider, you should interrupt Vagrant now and add
the box yourself. Otherwise Vagrant will attempt to download the
full box prior to discovering this error.
Downloading or copying the box...
Extracting box...te: 49.7M/s, Estimated time remaining: 0:00:01)
An error occurred while executing multiple actions in parallel.
Any errors that occurred are shown below.
An error occurred while executing the action on the 'cf-install'
machine. Please handle this error then try again:
The box you attempted to add doesn't match the provider you specified.
Provider expected: aws
Provider of box: virtualbox
One thing I noticed is my Vagrantfile has aws provider configured as such:
config.vm.provider :aws do |aws, override|

override.vm.box_url = "http://files.vagrantup.com/precise64.box"

aws.access_key_id = "*****"
aws.secret_access_key = "****"
aws.keypair_name = "****"
aws.ami = "ami-23d9a94a"
aws.instance_type = "m1.large"
aws.region = "us-east-1"
aws.security_groups = ["open"]
aws.user_data = File.read('ec2-setup.sh')
override.ssh.username = "vagrant"
override.ssh.private_key_path = "******"
end

Which is using the same image (precise64.box) as Virtual box (seen below)
config.vm.provider :virtualbox do |v, override|
override.vm.box_url = "http://files.vagrantup.com/precise64.box"
v.customize ["modifyvm", :id, "--memory", 2 * 1024]
v.customize ["modifyvm", :id, "--cpus", "2"]
end

Has anyone experienced this error before or have any idea what could be causing it? Thank you.
Answer: Try this one:
vagrant box add dummy https://github.com/mitchellh/vagrant-aws/raw/master/dummy.box
Comment: The dummy.box specified here + the `ami` in the config will give you the actual base vm that you start with.


additivity of crossing number of composite knots

Question: Let $K_1 \# K_2$ denotes the connected sum of tow knots $K_1$ and $K_2$. The crossing number is the minimal number of crossings among all knot diagrams,  denoted by $c(K)$. It is conjectured that $c(K_1 \# K_2)=c(K_1) + c(K_2)$. Is this conjecture settled down or still open?   Is there any counterexample?.
Comment: «It i conjectured» by whom? Can you give references? That always make a question better!
Answer: It's an open problem in general (there are no counterexamples, to my knowledge), but consider these partial results:

$c(K_1 \# K_2) \le c(K_1) + c(K_2)$ for any two knots $K_1$, $K_2$.  To see this, suppose that you have diagrams for $K_1$ and $K_2$ with minimal number of crossings, say $n_1$ and $n_2$, respectively.  Their connect sum is a diagram for $K_1 \# K_2$ with $n_1 + n_2$ crossings, although this may not be minimal.
Assuming that the knots are alternating, this is a theorem of Kauffman, Murasugi, and Thistlethwaite using knot polynomials.  (See MR:0999974.)
There's a larger class of adequate knots that includes alternating knots for which the conjecture holds.
The conjecture holds for torus links.  (See arXiv:0303273)
There's a bound in the other direction of the form $c(K_1 \# K_2) \ge \frac{1}{N} \bigl( c(K_1) + c(K_2) \bigr)$ for some constant $N$.  In one result, $N = 152$, but this can presumably be improved.  Of course, $N = 1$ would prove the conjecture in full generality.  (See MR:2574742.)

This question is as old as knot theory and Tait's conjectures.
Comment: Thank you Sammy for your answer.  But really I wonder why the diagram of connected sum of two knots with $n_1 + n_2$ crossings is not minimal.  I see that we can not apply any of Reidemeister moves to the resulting diagram.
Comment: I, too, would like to know why the connected sum of two minimal diagrams is minimal **if it's true.**  It is *an open problem* (nobody knows if it's true).
Comment: Regarding your comment "I see that we can not apply any of Reidemeister moves to the resulting diagram":  you might be overestimating your intuition about Reid. moves.  There are some mildly counterintuitive truths lurking in the web of Reid. moves.  For instance, there exist diagrams of a knot (of the same number of crossings) that can **only** be obtained, one from the other, through a sequence of Reid. moves that temporarily increase the number of crossings.


How did Barney & Ted know that Robin was searching for that locket in Central Park?

Question: In Season 8 Episode 23/24, Robin looks for the locket in Central Park. She think to ask Barney & Ted to come to Central Park. But she doesn't tell anyone that she is Central Park. How can Ted & Barney reach that location.  I thought Robin didn't say that she was in Central Park
Comment: Maybe they just know like it's supposed to be romantic destiny / fate or something?
Answer: Keep in mind that this entire show is a flashback, representing the story that Ted is telling to his kids.  There are many things shown in the show that are factually incorrect, including but not limited to: people eating sandwiches when they were really smoking weed, everyone in the group being smokers even though it isn't ever mentioned except in one episode, and Lily and Marshall literally communicating through telepathy and, on at least one occasion, passing plot-relevant information to one another by doing so.
What we see on screen is what Ted tells his kids, not a full and proper accounting of what really happened.  The simplest explanation to your question is that Ted neglected to relate to his kids how he and Barney found out Robin was in Central Park.
Comment: I disagre..this is a perfectly acceptable answer. If it is not, please explain **why** not.
Comment: I agree with you, but this is not an answer. Please add a comment if it is not a perfect relevant answer. Thanks
Comment: @anvesh.veerelli: you cannot rely on the meticulous accuracy of a narrator who has already proven to be inaccurate.
Comment: This is the explanation of lot of things in that show. As narrator Ted is not very reliable.


Looking for chemicals to melt ice

Question: This might sound like science fiction but I am trying to learn if there is a chemical or material that can convert cold into heat. It also could be kinetic as well. An example would be placing a chemical in cold environment where it becomes warm. Thank you for your help.
Comment: Might [chemistry.se] be better suited for your question?
Comment: You can't convert a lack of energy into energy.
But you can maybe trigger a reaction that would let loose potential energy trapped inside of that matter.
Comment: Wiki page on the [2nd law of thermodynamics](https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Clausius_statement)
Comment: "*Looking for chemicals to melt ice*" Salt!
Answer: You would be looking for some type of exothermic reaction with the ice.
Salting the ice actually does this (slightly).  It also lowers the freezing temperature of water; which the the main reason it melts the ice.
With enough of a salt you could absolutely heat up ice.  If you wanted to use it to remove snow and ice you might run into other issues; for example it may add heat to the water; but if it's cold outside the water will cool down again once the reaction stops.
Also note that I don't mean only table salt (NaCl); this happens with many different salts; when they disassociate in water they release some energy from the bond (which is the exothermic reaction).
Comment: It seems counterintuitive that releasing energy from an already-frozen material would cause it to melt, but I guess what's actually happening is converting a pure substance below its freezing point to a solution above its freezing point, with an incidental and unrelated decrease in energy.

Looking for chemicals to melt ice


How do event listeners work?

Question: Do they repeatedly check for the condition and execute if the condition is met. Ex, how the OS knows exactly when a USB device is plugged in, or how MSN knows exactly when you get an email. How does this work?
Thanks
Answer: At the low level, the OS kernel "knows" when something happens, because the device in question sends the CPU a hardware interrupt.
So when, say a network packet arrives, the network controller sends an interrupt, and the OS kernel responds as appropriate.
At the program level, it works quite differently - most application programs run an "event loop", where they fetch a message (say, a message from the OS saying that "the mouse was clicked on this point in your application"), perform the appropriate actions in response to that, and then, listen for more messages. If there is no message, the OS sleeps the thread until it has a message to deliver.
Comment: +1 For clearly explaining the difference between interrupts at the hardware level and event listeners at the software level.
Answer: Take a look at Interrupts this should explain how the hardware initiates certain 'events'
Answer: Depends.
Often an event listener is registered with the object that generates the event.  When the event occurs, the object iterates through all listeners registered with it informing them of the event.  Have a look at the AWT/Swing event model in Java for example.

ow the OS knows exactly when it gets a USB

At a low level I suspect thats a hardware interupt (someone correct me if I'm mistaken) which is handled by the kernel/USB driver.  There's often higher level systems (e.g. DBUS) that listen for this and have event listeners listening to them.

or how MSN knows exactly when you get an email.

I suspect that's a simple case of polling the mail box (e.g. over POP3) every x seconds and checking the message count (could be wrong though).
Anon. explains the difference between hardware interrupts and software level event listening quite well.
Answer: Programs like email checkers will usually be running background services that query their email server every X period checking for new mail. There are other ways of doing it, but for software level events, it's almost certainly going to be something like that.

How do event listeners work?


security groups sharepoint 2016

Question: I recently installed SharePoint 2016 onto a VM and have a SharePoint site running. I am trying to use Active Directory Security Groups to give permissions to users. Some Security Groups are working as intended; I add the group to the designated SharePoint group and it gives the users the correct SharePoint permissions. However, some Security Groups are not working as intended; We add the group to the intended SPG and they are not allowed access: <code>"Sorry, this site hasn't been shared with you"</code>. If I manually add a user from the Security Group, they are given access to the site. Has anyone run into this issue before? We have tried creating a new test Security Group and the users in that group do not get permission when added to SP. Would manually adding them via Powershell be a work around? Is this an issue with Active Directory?
Answer: I would verify everything is properly updated after an access change. There can be delay between adding users to the AD group, adding the AD group to the SharePoint group, and the SharePoint group "knowing" who belongs to that AD group.
Also, caching at various stages in that process could be affecting you here as well. Close the browser completely before each test (ensure there are no tabs open).
If you wait, say, an hour, or a day, does it resolve itself? You shouldn't have to wait this long, but if it does resolve eventually, it's likely a timer/caching/sync/replication issue.
Comment: Other than waiting, is there something that can be done to force this?
Comment: For replication issues, force replication on your domain controller (if you have more than 1). Oops, hit enter... ok so... for other issues you need to run the timer service that correlates. I don't know offhand which one handles which, but if you dig around in central admin timer jobs you might find one that looks promising.
Comment: Also, try running a user profile sync. Incremental should be fine.
Comment: why would it work instantaneously if I add a single user as opposed to a security group?
Comment: So I believe the way it works is it caches the user "token" which contains a list of AD groups that user has access to for a certain period (several hours potentially). If you log in, the token is generated and cached, then you change access groups around, the old token is still being used. For direct user access it doesn't need to check a token, it just verifies the user object directly.

security groups sharepoint 2016


How efficient is the immigration service at Oakland?

Question: OAK doesn't handle too many international flights.  How efficient are CBP at processing arrivals for those it does handle?  I'm arriving as a U.S. citizen from Stockholm on a Monday evening (scheduled arrival 5 p.m.) and wondering how long I should expect to wait.
Comment: @MichaelHampton Quite sure.  ARN-OAK on Norwegian Air Shuttle.
Comment: 5-08-17. My flight arrived at 10:45 PM
By 11:15 I had my Bags and Was In Line For Inspections.
No Joke... I am writing this post while in line, it is currently 1:30am. Three people left. Over two Hours. This has to stop!
It is absolutely ridiculous the Oakland Airport Cannot Invest Im More People and equipment to get these inspections done promptly like EVERY other Country and Most US Cities! Really Oakland? I am Ashamed!
Comment: My experience as a non-US citizen travelling to the US is that it is worth every attempt to get off the plan and down to immigration as quickly as possible, so as to be as close to the front of the line as you can be. Your position in the line makes a huge difference to your wait time.
Answer: I've been going with Norwegian from Stockholm to Oakland a couple of times. It seems to take between 30 minutes and an hour to process all passengers from a Dreamliner. The immigration area has been completely empty when I arrive (as one of the first passengers leaving the airplane) so the waiting time is very dependent on your seating in the airplane. As a US citizen I would expect you to get through immigration in less than 15 minutes, as long as there are US citizens in line they keep one line exclusive for them and there are usually quite few US citizens on these flights.
If you have checked luggage be prepared to wait for an hour if you are unlucky. The first bags arrive quite fast but all bags doesn't fit on the belt so one or two persons manually removes the first round of bags from the belt to make space for more bags. This is very time consuming.
Answer: You can look up historical immigration wait times for any U.S. airport at the CBP web site.
From a quick look, it appears that average wait times at OAK vary dramatically, from as little as a few minutes to over an hour, with no obvious (to me) pattern. But it appears that most international arrivals at OAK are much later in the evening, so you may get lucky with a 5 pm arrival.
OAK doesn't have Automated Passport Control kiosks, though Global Entry is available if you are in that program.
Comment: As the figures don't take traveler's nationality into account, it's probably safe to say that US citizens and permanent residents are far more likely to be on the "few minutes" end of the spectrum, while foreigners who need to be fingerprinted and perhaps subjected to a more stringent interview will account for most of the "over an hour" data points.
Comment: @phoog There are separate queues, which they attempt to balance, but just not enough agents.
Comment: @phoog The times weren't equal, but the number of agents assigned to each queue was changed on the fly. I don't know what the parameters are, but they exist.
Comment: @AndrewLazarus I don't think they try to balance the queues.  I think they give priority to citizens and permanent residents.  I generally arrive at JFK or Newark and they typically have roughly the same number of desks serving each queue.  The citizen queue moves much more quickly than the non-citizen queue.  If they wanted everyone to have the same wait time on average, they wouldn't have separate queues.
Answer: I have come into OAK on the Norwegian Air flight from Oslo (not the one from Stockholm) and we were in line (i.e., queue) over 35 minutes. US Citizens. About one-third the way back the cabin. The experience had us thinking very strongly of paying for Global Entry.
The examination at itself took two minutes, only because we still had a European apple to discard.
Answer: FWIW: When I arrived at OAK on Feb 1st 2016 on the Norwegian flight from Stockholm my seat was about 2/3 the way back.  There was a single combined queue for all passengers except those with Global Entry.  Only 2 of the 5 CBP booths were staffed.  It took 55 minutes to get through and there were still probably 150 people behind me.
Answer: We went through Oakland Airport in the autumn of 2014 on the Norwegian airlines flight from Oslo. Clearly they were not setup for international travel - I hope things have improved now.
We were near the back of the plane (787) and there was just one long queue. We were American citizens and it took almost 2 hours to get through the line!! This was around 6 pm local time but in the wee hours of the morning Oslo time. Lots of Norwegian families with kids in line. An absurd wait and a shocking face for America compared to how efficient most European entry points are.
Make sure you go to the bathroom on the plane, as there are no bathrooms before the immigration inspection.
Answer: OAK, Norwegian out of Gatwick. Row 35. US citizens. It's been 40 minutes so far, probably another 10 to go. They've got APC kiosks going but it doesn't look like they save any time at all. There IS a bathroom back of the room.  Posting while waiting in line, after googling for info on the kiosks.
Comment: Update: agent thinks the kiosks help. There were two helping citizens and one doing visa holders. After baggage capital, there was ONE guy doing everyone. Big downside of the kiosks: the slips have to be collected by the customs dude.  No"nothing to declare" walk right through.
Answer: The experience was comically BAD for a single plane to be processed. It appears that they are not prepared to handle a full Norwegian Dreamliner.  There is poor signage to divide US citizens vs other, the roped lanes led us away from the automated kiosks so someone had to re-rope the lanes. One of the customs agents was having a bad day and began yelling at passengers, and in the meantime, because it was so slow, the luggage was stacked four high on the conveyor and was actually backing up into the belt. Then another agents takes a second look at the kiosk picture and the baggage goes through a tiny X-ray machine .. it all led to big queues and lots of frustration ... OAK not ready for prime time.


Encrypted data sharing in decentralised system

Question: Alice encrypts file using her public key and upload it to decentralised file store (some service). Bob buys access to the file. Is it possible to share decrypted file with Bob without having Alice's key?
Decentralised file store doesn't store any private keys, but it knows that Bob has access to the file (e. g. from smart contract).
Answer: If there is no limit on the cost of interactions,proxy re-encryption works for you.
Proxy re-encryption (PRE) is a type of public-key encryption (PKE) that allows a proxy entity to transform or re-encrypt data from one public key to another, without having access to the underlying plaintext or private keys.


There is an item stuck in my inventory in Minecraft!

Question: There is an Item in my inventory that stays in it's place and will not move no matter what. I am on a server so there is very little I can do about it. I am in creative mode. If I try to use the hotbar slot that it is in, I can't break blocks or interact with anything... can someone help me!?!?
Comment: Fill the other bar slots with materials and then switch to that item again. Then aim at another block you currently don't have in your bar and press the mouse wheel. This should replace the current material with the one you're looking at, if it's not already in another slot. If that doesn't work, what material do you actually have in that slot?
Answer: Is this a craftbukkit server with plugins? Sometimes, servers will pin a certain item to an inventory slot, so that those items can be used for special purposes. Since you mentioned that you are on a server, this is the only thing that I can think of.
If you haven't already, you can also try dying, as well as re-logging.
Answer: If you're playing on a server, chances are that's how it's meant to be. e.g. if the item is meant to be part of a GUI menu or open one.
Answer: I had some gold armor stuck in my inventory that I wanted to give to a friend but... for some reason, I couldn't get it out so I tried a whole bunch of stuff until I exited out of my inventory and accidentally swapped it with armor I already had on so if it's armor that's stuck just try this!
Comment: The OP is talking about the item being in the hotbar, so it's probably not a piece of armor.
Comment: Did the armor have curse of binding on it perhaps?


Book or novel where the spirits of kings and heros are enshrined in statues. When an enemy approches, unknown

Question: A book or novel where the spirits of kings and heroes are enshrined in statues. When an enemy approaches, they don't come to life, but the spirits transform common men in heroic soldiers.
Comment: You'll have a better chance at getting a good answer if you go to https://scifi.stackexchange.com/tags/story-identification/info and use the tips there to add more details to your question.
Answer: How about "The Sword of Welleran" by Lord Dunsany?
The city of Merimna was defended by six heroes: Welleran, Soorenard, Mommolek, Rollory, Akanax, and young Iraine.  When each died, his body was hidden in a canyon by his comrades, so that the city's enemies believed they still lived; and six lifelike statues were erected around the city.  Eventually some incident breaks the enemies' belief that the statues are alive.  In the emergency, a youth takes the sword of Welleran from its display-case ....


How long does it take for different types of textiles to decompose in a landfill?

Question: I've looked through the first five pages of Scholar search results to my request "landfill decomposition of textiles cotton" and found zero useful information. Is there any scientific research at all on how long does it take for different types of textiles to decompose in a landfill?
Answer: You are correct that it does seem difficult to find information on this topic, but I found a couple resources which may be of use.
From "Textile waste resource recovery: A case study of New York State's textile recycling system" (emphasis added):

When looking at the relative biodegradability of non-durable goods in landfills, there are three categories: Labile materials which are easily compostable and degrade fairly rapidly in about 5-10 years such as food scraps and other organic wastes; Resistant materials that are moderately degradable over 15-20 years; and Recalcitrant materials that degrade very slowly over 30-40 years in the landfill. Other outlying materials including plastics and metals hold their own label as Non-biodegradable. Within this classification system textiles are classified as Recalcitrant. However, since the mid-twentieth century there has been a rise in the use of man made synthetic fibers
created from oil-derived polymers. This means that many of the textiles that end up in landfills today possess some of the same properties as plastics and will therefore never degrade.

Unfortunately the reference for this information (Black 2008) is not included in the list of references, so that's a dead end. I tried contacting the author -- I will update if they get back to me.
Another potential resource is the chapter "Biodegradation Studies of Textiles and Clothing Products" from the book Roadmap to Sustainable Textiles and Clothing. Unfortunately this reference is behind a paywall, but based on the list of references it could prove to be a valuable resource.
Comment: @SergeyZolotarev I believe this book may have additional information: [Sustainable Fashion: Why Now?](https://www.google.com/books/edition/Sustainable_Fashion_Why_Now/oupAPAAACAAJ?hl=en).
Comment: Thanks for the effort
Comment: No scientific verification clothing decomposition rates, but one [apparel recycler](https://upparel.com.au/clothing-recycling/) claims ...*clothing waste takes up to 40 years to decompose, whilst shoes can take 1,000 years to break down*.
Answer: Landfill conditions are notoriously anaerobic, so ANYTHING in a landfill will not decompose barely at all. It may take 20-200 years or thousands. Fact is, few things decompose in landfills, organic waste anaerobically decomposes producing some methane and some residual organic hydrocarbons but plastics, last practically forever in an anaerobic environment. Better to reuse textiles wherever possible.

Cotton: as rags. I use old t-shirts as cleaning rages, when when no longer any good, I wash em one final time then recycle cotton. Better long term strategy is that cotton can be shredded and composted to feed mushrooms. Indoor cultivation of Volvariella volvacea otherwise known as Straw Mushrooms which are edible using shredded cotton fiber. As an organic material cotton can be safely burned and used ash as a source of fertilizer, compost amendment.
Fleece is made of polyester. Soda bottles, which can be sent to plastic recycling to make, more fleece or soda bottles.

The overall strategy is mitigate sending textiles to landfill in the first place.
Donate clothes to charitable donations groups.
Cotton (100%) can be bonfire fuel.
Synthetic fabrics, should be recycled.


How to remove part of a string value in the attribute table of a shapefile in PyQGIS

Question: I have a number of shapefiles with a field in the attribute table containing many different string values similar to this one: 'Waypoints_17-JUL-18Timothy waypoints.shp Waypoints_17-JUL-18Timothy waypoints.' The only difference in the values is the name. I would like to find a way in PyQGIS to remove everything after the first name, so before 'waypoints.shp...' and iterate through each of the fields to repeat this. Is it possible to do this in PyQGIS? I am new to PyQGIS so am unsure where to start but it should be a lot quicker than doing manually. I am using QGIS 3.0.0
Comment: Which QGIS version are you using? If you have a value `"Waypoints_17-JUL-18Timothy waypoints.shp Waypoints_17-JUL-18Timothy waypoints"`, you just want to instead have `"Waypoints_17-JUL-18Timothy"`?
Comment: @Joseph I am using QGIS 3.0.0. Yes, that is correct. However, some fileds have different names, e.g. 'John' instead of 'Timothy,' so I need to remove everything after this name / before the next section, which always starts with 'waypoints.shp...' as in the above example
Answer: Assuming you only want to keep the first part of the value (i.e. before the first space), you could try using the following:
<code>layer = iface.activeLayer()
with edit(layer):
    for feature in layer.getFeatures():
        new_value = feature['fieldName'].split(' ', 1)[0]
        feature.setAttribute(feature.fieldNameIndex('fieldName'), new_value)
        layer.updateFeature(feature)
</code>
Comment: Is activeLayer the top most layer in layer tree?
Comment: @BERA - No, it's the layer currently selected in the tree. So you would want to make sure you select the correct layer :)
Comment: @JClifford - Most welcome! Glad it helped :)


Number of ways to choose n objects from groups of indistinguishable objects

Question: Say I have 5 blue, 6 red, and 7 green balls. How many ways are there to choose 8 balls, without regard to order? My first thought was to just add up all the balls and choose 8 from the collection of 18, but I know that that'll overcount the possibilities, as I can get the same color combination of balls multiple ways by choosing different balls within the same color group. How would I correctly approach this problem?
Comment: So all the same color balls are the same?
Comment: Yep. All balls in the same color group are the same.
Answer: Any selection can be described as a triple $(b,r,g)$ where $0≤b≤5$, $0≤r≤6$, $0≤g≤7$ and $r+b+g=8$.
As the numbers involved are small, it's not difficult to work case by case.  We'll go through the possible values of $b$.
I.  $b=0$. Then $r≥1$ but any non-zero value of $r$ works so $\fbox 6$ cases.
II. $b=1$.  Now $r$ can be any value from $0$ to $6$ so $\fbox 7$ cases.
III.  $b=2$  Again we can use any value of $r$ so $\fbox 7$ cases.
IV.  $b=3$.  Now $0≤r≤5$ so $\fbox 6$ cases.
V.  $b=4$  Now $0≤r≤4$ so $\fbox 5$ cases.
VI.  $b=5$  Now $0≤r≤3$ so $\fbox 4$ cases.
Finally the answer is $$6+7+7+6+5+4=\fbox {35}$$
Note:  while the logic is fairly simple, case by case enumeration can be a bit error prone so I suggest checking carefully. As an alternate method you could multiply out the generating functions $$(1+x+x^2+x^3+x^4+x^5)(1+x+\cdots +x^6)(1+x+\cdots +x^7)$$ seeking the coefficient of $x^8$.
Answer: Let's say that we ignore the green balls for now. We have can have between 0 and 5 blue balls and 0 and 6 red balls. So the number of possible pairings of number blue and number red is: $$6 * 7 = 42$$ Now we have to eliminate combinations where we had more than 8 balls so we need to see how far above the target we can get by adding the number of blue (5) and red (6) balls and subtracting the total desired (8). Then we sum the integers from 1 to that answer. This gives us how many of our 42 scenarios are too high. $$5 + 6 - 8 = 3$$ $$\sum_{n=1}^{3} n = 6$$ $$42 - 6 = 36$$ Then we need to subtract off the bottom where the greens can't get us to 8 which only occurs when we have zero blue and zero red so: $$36 - 1 = 35$$ And thus you have 35 combinations.


Books on functional equations

Question: Could you help me please with some functional equations? I need some online books with exercises and some explanations. I'm interested in the Cauchy functional equation and Jensen functional equation.
Answer: You can find recent results of functional equations in the following references.

Marek Kuczma, An Introduction to the Theory of Functional Equations and Inequalities (Cauchy's Equation and Jensen's Inequality)  Second Edition, ISBN 978-3-7643-8748-8 Birkhäuser Verlag
P. Kannappan, Functional equations and inequalities with applications, ISBN 978-0-387-89491-1  Springer
Stefan Czerwik, Functional Equations and Inequalities in Several Variables, World Scientific, ISBN 981-02-4837-7
Answer: I know it is a bit late now but it is good for anyone looking it up:
I think the best introduction book to functional equations would be Engel's problem solving strategies. It is very easy for beginners to start with and would recommend it for anyone looking forward to studying this topic.
Answer: Nice are the handouts by Evan Chen, "Introduction to Functional Equations" and "Monsters".
Answer: I think mathlink is the best one. Usually, books about functional equations include old problems only(some of them doesn't even include problems made in 21th century...) But if you search 'functional equation' on Forum(which is in the 'Community' section), there are a lot of challenging problems, and you might see tons of amazing techniques in FE!!!
Here's the link : https://artofproblemsolving.com/
Comment: Additionally, techniques in functional equations are not 'theoretical'. I usually learned techniques by solving hard problems, and there are problems that require high-level technique.

Books on functional equations


Passing session variables from jsp to servlet filter

Question: I just wrote my first servlet filter and am already ahving problems. What I wanted to do was to make the filter handle authentication so that if a protected page was being requested, the filter would cut in and redirect to a login page. If the login succeeded, the filter would then redirect the user to the original page requested.
To give an example, one of the pages in question is a blog page. When a user clicks on "Leave a comment", he is redirected by the filter to a login page if he hasn't logged in already. Up until here, the filter does it's job. Then the user logs in and the username/password is checked. If they're ok, a session variable is set in the jsp and the page redirected to the comments page again. That request goes through the filter where I check the session variable. But no matter what, the session I get from the request seems to be a new one with no attributes set. I've tried calling request.getSession(false) and that seems to always return null. So not giving me the same session back that my jsp pages share.
I'm using tomcat 7.0.2. I don't know if this is tomcat specific or that there is just some very obvious thing that I'm missing.
Hope you cna help me.
Gisli
-- Update --
Here is my code. Guess that would be quite helpful :).
Basically, I have a login page, a blog page and a comments page. On the blog page, I have this link:
<a href="?page=blogg/komment.jsp?id=<%=blog.getID()%>"> Leave comment

In my filter, I have this:
public void doFilter(ServletRequest req, ServletResponse res,
        FilterChain chain) throws IOException, ServletException {
    // TODO Auto-generated method stub
HttpServletRequest request = (HttpServletRequest) req;
HttpServletResponse response = (HttpServletResponse)res;
HttpSession session = request.getSession(false);

Object userObj = null;
if( session != null ) {
    userObj = session.getAttribute( "username" );
}
String contextPath = request.getContextPath();
String queryString = request.getQueryString();
String url = request.getRequestURI();

String page = request.getParameter("page");

String user = "<user>";
if( userObj != null )
    user = userObj.toString();

if( page == null )
    page = "<page>";

if( userObj == null && page.indexOf( "blogg/komment.jsp" ) != -1 ) {
    session.setAttribute( "destPage", page );
    response.sendRedirect( "index.jsp?page=blogg/login.jsp" );
}

System.out.println( ", Time "
                    + new Date().toString() + " " + page + " " + user);

chain.doFilter(req, res);
}

So far, so good. This does redirect me to the login page. There, I do this:
    String user = request.getParameter( "username" );
    String pass = request.getParameter( "password" );

    if( blog.confirmUser( user, pass ) ) {
      session.setAttribute( "username", user );
      Object destPageObj = session.getAttribute( "destPage" );
      String destPage = "xxx";
      if( destPageObj != null )
          destPage = destPageObj.toString();
      response.sendRedirect( "index.jsp?page=" + destPage );

      %>
      User exists
<%
    }
    else {
%>
    User does not exist
<%
    }

The string "User exists" is never supposed to be printed out, but it always appears and I get the login page back.
This is what I don't understand. I set the session variable "username" so that when the next request hits the filter, that variable should contain a value. But it always seems to be empty. In fact, when I called getSession(false) in the filter, I always got null back. Even though the session variable is supposed to always exist in the JSP if it hasn't expried.
I have tried passing session variables between pages and that works perfectly. But to set a session variable in a page and read it back in the next redirect always fails.
I feel like there's something terribly obvious that's staring me in the face and I've just worked too long on this to see it!
Hope you guys can help!
-- Further update --
Too little space to write as a comment :). What I want to do is this. I wanted to make my jsp pages independent of the login page. I.e. I wanted the filter to act as intermediary. So that when I send a user to my comments page, I don't need to worry about if he's logged in or not. If he's not logged in, the filter should redirect him to a login page and then send him on to the original destination. My blog is not the only thing that will require authentication and I wanted to have this logic all in the same place.
I don't want to pass the username/password, as request parameters for security reasons. I want both to remain as session attributes.
I don't understand what you mean when you say that what I'm doing in the login page should be done in the filter. I need a page to ask the user for his username and password and I don't want to send those along with the request. I saw an example here, link which is exactly what I want to do. Except that this doesn't work for some reason because I can't seem to be able to access the same session where I stored the username.
If I've totally got it wrong, could you point me to a web page where this is done properly? I've spent hours googling and all I have found are examples that seem to work for everyone but me :).
Comment: Does the browser accept cookies? SHow us the code of the filter and the code setting the session attribute.
Comment: Did you get this to work? If so could you please post an update? I am trying to do the same thing and wondering if I'll run into the same issue as yourself...
Answer: I resolved this after a lot of googling and digging around.
One thing I failed to mention, was that I had Tomcat running behind a proxy. And the proxy was not delivering the session cookie. In my virtual host definition, I had this:
    <IfModule proxy_module>
        ProxyRequests Off
        ProxyPass / http://b6.is:8080/B6/
        ProxyPassReverse / http://b6.is:8080/B6/
    </IfModule>

So, everything is passed to the B6 webapp from the root.
All I had to do to translate the cookie path as well was to add this line:
        ProxyPassReverseCookiePath  /B6 /

Now it works perfectly.


Primary Gmail address

Question: Google made me create a new email address and then that became my primary. I want my original Gmail address and I want to get rid of this new one. I am afraid to delete anything because I do not know what it will do to my account or Android. Everything I read is so confusing. Just want my primary to be my original email not the new one.
Comment: You need to tell us more about how "Google made me create a new email address".
Answer: So you have an Android phone, maybe a new one? You had to register using a Google account, and you were tricked into registering a new one?
Add your normal Gmail account in the Settings menu. When both Gmail accounts are there, you can delete the first one. Your normal Gmail account will become the primary.

Primary Gmail address


How does one determine the effective temperature of a star from its spectrum?

Question: Determining effective temperature of a star is in general a non-trivial task. Simple reason for this is that we can only study the electromagnetic radiation from a star, but not the temperature directly. The complexity is due to the fact that the radiation is produced in stratified stellar atmospheres, which are only partially characterised by stellar temperature, but also by many other factors, such as stellar mass, elemental abundances, stellar rotation, etc. What is more, the temperature of atmospheres varies with depth, whereas effective temperature is just a number.
From the other hand, temperatures and magnitudes are the most important quantities, characterising stars.
So, the question: How exactly does one use the spectrum to extract the information about the temperature of a star? By temperature here I mean effective temperature, or even the temperature profile of the atmosphere.
Note: This a rather textbook question. I created it because I have encountered a good existing answer by @Carl, previously posted in a bit less textbook discussion How well can we in principle determine $T_{\textrm{eff}}$ of a star? . This question seems to be a much better place for the answer.
Answer: Temperature ($T_{eff}$) can be quite tricky to determine accurately as it interrelates to a number of other fundamental measurements.
Firstly, remember that the spectrum we observe from stars are pin-point, they give us the entire overall result and not a specific location or part of the star. We need to dissect the various parts to arrive at the fundamental parameters. We arrive at our results by iterating the values of the fundamental parameters until a model spectrum matches the true spectrum we observe. The issue is, like you say, the existence of a whole lot of uncertainties.
The first of these (although it doesn't have a large effect) is the Uncertainty Principle itself. This creates natural line broadening due to the emitted photon having a range of frequencies. The width of the line is determined by;
$$\Delta E \approx \frac{h}{T_{\text{decay}}}$$
where $\Delta E$ is the uncertainty in the energy,
$h$ is the Planck constant, and
$T_{\text{decay}}$ is the amount of time the electron stays in a high energy state before decaying.
Fundamental parameters
The rotation of the star causes a Doppler shift effect on the line spectra making it broaden. The faster the rotation, the broader (yet smaller) the line. Like the Uncertainty Principle, this is natural broadening as it doesn't impact the abundance of any particular element in the star.
Measuring the rotational velocity ($V_{\text{proj}}$) depends on both its axis of rotation and our line of sight to the star. Therefore, we use a combination of both velocity about the equator ($v_e$) and the star's polar inclination ($i$) to determine the projected radial velocity;
$$V_{\text{proj}} = v_e \sin i$$
Temperature ($T_{eff}$)impacts wavelength in such a way that higher temperatures impart higher random motions on the atoms. When these photons collide with an atom, they can cause the atom to become ionised, i.e. lose an electron. Different energy levels (and therefore temperature) will create different abundances at the various ionisation stages of atoms.
The temperature of the stellar photosphere decreases as we move away from the core. Therefore the line profile represents a range of temperatures. The wings of the line arise from deeper, hotter gas which displays a larger range of wavelengths due to increased motion. The higher the temperature, the broader the wings of the line profile ([Robinson 2007, pg 58][1]).
Here you can see the effect of various temperature values on the synthetic spectral line of FE I 6593 A. Red: $T_{eff}$ = 4000K; Black: $T_{eff}$ = 5217K; Blue: $T_{eff}$ = 6000K;
$T_{eff}$ on spectral lines" />
Microturbulence ($v_{\text{mic}}$) is the non-thermal localised random motion of the stellar atmosphere. It works in a similar way to temperature - an increase in the motion of atoms creates a wider range of wavelengths observed and therefore broader line profiles.
In strong lines, saturation can occur when there aren't any more photons to be absorbed. As microturbulence in these areas increases, it presents more opportunities for photons to be absorbed. This broadens the wings of the line profile increasing the overall strength of the line. We can use this fact to determine $v_{\text{mic}}$, by ensuring that the strength of the lines (equivalent width) have no correlation with their abundances.
Finally, surface gravity which is a function of the star's mass and size:
$$\log g = \log M - 2 \log R + 4.437$$
with $M, R$ being in solar units and $g$ in cgs.
A star with a higher mass but smaller radius will invariably be denser and under greater pressure. By definition, denser gas has a higher number of atoms per unit of area (abundance), leading to stronger spectral lines.
A gas under pressure provides more opportunities for free electrons to recombine with ionised atoms. For a given temperature, ionisation is expected to decrease with an increase of surface gravity, in turn increasing the abundance of atoms in the neutral or low ionisation states.
The measuring of $T_{eff}$
As we've seen, there are a number of ways in which a spectrum of a star can be altered. The one you're interested in is temperature. As temperature is interrelated to all the other fundamental parameters, we need to treat them together as a whole and tease out the value of $T_{eff}$.
We begin with a synthetic spectrum and modify its properties iteratively until it matches the shape of the star's spectrum. Adjustments of one parameter will invariably affect the others. The spectra will match when the temperature, surface gravity, and microturbulence values (amongst others) are correct. This is obviously very time consuming although programs exist to help.
Atmospheric properties can also be determined by other less time consuming means. Photometric colours can be used as a proxy for temperature, and absolute magnitudes for surface gravity. However, these determinations can suffer from inaccuracies due to interstellar extinction and are at best a close approximation.
[1] Robinson, K. 2007, Spectroscopy: The Key to the Stars (Springer)
Comment: This doesn't mention the most fundamental of assumptions. That the crude (usually plane parallel) single component atmosphere mdel adequately represents the real atmosphere of a star. $T_{eff}$ is a defined quantity in terms of luminosity and radius. The $T$ measured by spectroscopy is not $T_{eff}$, though many assume it to be, and is entirely model-dependent.
Comment: @RobJeffries, you're absolutely correct. Thanks for pointing that out. :)
Answer: There are many different ways to measure temperature of an astronomical object. Typically, effective temperature means just simply a blackbody temperature. However, the blackbody model is just the first-order approximation that we know it is inaccurate in many circumstances.
If you have a nice spectrum from a wide wavelength, your effective temperature might be better to be defined as the excitation temperature. However, what definition you should use really depends on what context you are in. Check this for a short summary: https://www.physics.byu.edu/faculty/christensen/Physics%20427/FTI/Measures%20of%20Temperature.htm
Comment: Thanks, Kornpob! Note, however, that photospheric temperature determined from the spectrum is the physical temperature of matter in the photosphere, and it is not derived from a black-body approximation. The latter is very common in photometry though.
Comment: Both paragraphs have problems. The effective temperature is $(L/4\pi R^2\sigma)^{0.25}$. Full stop. To measure it you need the luminosity and radius of the star. Fitting a spectrum can only yield some estimate of the effective temperature that is model dependent.
Comment: - I don't think you need a radius. You can set a multiplicative constant to scale fluxes as a fit parameter, together with the temperature. The radius will already be in side the constant.

- If the photosphere is optically thick, at the limit it is blackbody radiation.


Synonymize [di2] and [shimano-di2]

Question: Combine the tags for di2 and shimano-di2. There's no need for both.
Answer: I suggest merging di2 (24 questions) into shimano-di2 (36 questions) since the latter is more descriptive, and will show up in the tag search box when people start typing either "Shimano" or "Di2."
Comment: I also wrote up a quickie tag wiki for [tag:shimano-di2] since it didn't have any, and I'd hate that work to go to waste. :D
Answer: Done.
<code>Merge results
updating post history, 25 rows affected (pipe delimited)
updating posts, 24 rows affected  (pipe delimited)
updating PostTags associations, 15 rows affected
updating ForYou nudges, 0 rows affected
destroying 'di2': [di2] removed from (0 post, 0 history, 1 tag)
total: 1 tags deleted, 0 posts modified, 0 post histories modified
0 history records for type Question had empty tags; retagged to [untagged]
0 post records of type Question had empty tags; retagged to [untagged]
updating count for target tag, new count: 53
tag remapping of [shimano-di2] and [di2] complete!
remapping 0 synonyms
2 favorite and ignored tags remapped!
0 tracked tag badges were remapped!

Tag Synonym di2 -> shimano-di2 was approved!
</code>

Synonymize [di2] and [shimano-di2]


GEDCOM: how to declare two different individual records as the same person?

Question: If a person changes the gender or the name or both: how can I interlink the two individuals in GEDCOM to declare them as the same person?
Comment: Are you using a particular piece of software?
Comment: Gender change is addressed at https://genealogy.stackexchange.com/q/15847/6485
Answer: What you are attempting to do is a bad idea. Each person should have one GEDCOM ID.
One of the basic principles of genealogy is that you (almost always) refer to people by their birth name. How you show name changes will depend on the software. You can always handle it parenthetically if the software doesn't have a good method.
Example, this person Americanized their name: Heinrich (Henry) Schroeder (Shrader)
Name changes due to marriage should not be shown in the person's name. In other records it will be obvious that they either took their husbands name or they didn't.
Gender is more tricky:
If the person is living and the gender change is not public knowledge, I would not reveal it in my family tree.
If the person is deceased and had children: Genealogy is recording history, so you should record them as the gender they were when they were parents. Indicate a gender change some other way (depends on the SW). My software only allows a Male/Female pair to be parents, so I would be forced to show them with their original gender.
If the person is deceased and didn't have children: I would list them the way that they would want to be listed, which is almost certainly with their new gender. Indicating the gender change elsewhere would be optional.
Answer: The proper way to do this in GEDCOM is to use the ALIA tag.
GEDCOM 5.5.1 says:
<code>> +1 ALIA @<XREF:INDI>@      {0:M} p.25
>
> ALIA {ALIAS}:= An indicator to link different record descriptions of a
> person who may be the same person
</code>
However, a lot of programs use it improperly.  See:

The ALIA Tag of GEDCOM « Louis Kessler's Behold Blog
GEDCOM ALIA - tamurajones.net

Tamura Jones in his article recommends that you not use the ALIA tag at all, but instead you can use the ASSO record to link an INDI record to another INDI record.
e.g.:
<code>0 INDI
1 NAME Joe /Jacob/
1 ASSO @I30@
2 RELA same person
</code>
Comment: Many thanks! Will use both tags. ASSO for the short term solution and ALIA (bi-directionally) for better times (when applications get GEDCOM compliant).


Hypothesis test between means of week and weekend patterns of the same time series

Question: I have a data set with phone calls, that contains the caller id and call duration. I want to test the hypothesis that the call duration during the week is smaller than the call duration during the weekend. I was thinking about using the Welch t-test but as the same caller can do calls on week days and weekends, I believe that I can´t assume iid. The time series is autocorrelated, what complicates the matter. Would ANOVA help in this case?
Answer: I would not expect call length to be normally distributed, so I would not jump into a Welch's-t test. Try a non-parametric Mann-Whitney U-test. However, that latter test is less powerful than a paired test which goes to interpretation. That is, if the test is highly significant, then a paired test with the same $N$ would be more significant. Problem is, you have some $n<N$ that are paired, so your test will be less than optimally powerful. You could try a Wilcoxon signed-ranks-W test on the $n$ pairs if $n$ is a substantial fraction $N$, and of those two non-parametric tests one will be more significant than the other.
Which? You might need to run simulations to decide which is best, but if they are both highly significant, it would be overkill. Cannot say whether ARIMA is needed or not, ANOVA probably less needed. Cannot say more without more information.
Comment: Would the autocorrelation of the series invalidate the independent samples that M-W U needs? Imagining that there is one year of data with daily values, one would have 261 week days and 90 weekend days, so it's not a substantial fraction.
Comment: On the other hand, if each call is considered separatedely, the series may be considered independent. So Wicoxon should work.
Comment: Invalidate is probably too harsh a concept to be applicable. Autocorrelation of call length might reduce the power of M-W testing. Again, if M-W testing is highly significant then it doesn't matter. You might start by looking at [ARIMA](https://en.wikipedia.org/wiki/Autoregressive_integrated_moving_average) regression.


Do I need to buy premium on Xbox if I have it on PC?

Question: If I buy xbox 360 version of BF4 when I have it on PC with premium, will I have premium access on xbox or do I need to buy it separately on xbox?
Comment: I don't think so. PC and Xbox are 2 different platforms. Most devs don't go the extra mile to give you premium stuff for different platforms unless you bought them on each of them. And then there's EA. Expect them to not give a sh*t about their customers; that's what they do best.
Answer: No, you won't have premium on XBOX and will need to buy premium for your XBOX version as well.  Cross-platform use of premium access is not granted; your accounts are linked (through Battlelog) but not to the point of using the same access level in both.
This has been further discussed here.


Can FSI analysis be linear on the solid part?

Question: Is the FSI analysis always considered Nonlinear or the solid part analysis can be linear too?
Comment: Yes and yes. The structural model, the fluid model, and the interaction could all be (independently) linear or nonlinear. For example the acoustics of a flexible body and a gas is likely to be all linear (for small amplitudes) but modelling the free surface between a solid and a liquid with relative motion (e.g. the wake of a ship moving through water) is almost certainly nonlinear.
Comment: @alephzero while you and other users could answer my question , why does the webiste stupidly ask for more stupid details ?
Answer: IMHO this is a matter of convention more than anything else. However, I acknowledge that there are other more experienced and knowledgeable people in this forum (e.g. alephzero), that can offer a more scientific (and less based on experience) view on the subject.
My view is that Even if the solid part's constitutive equations are elastic, the interaction between fluid and solid is always time dependent, and has sinusoidal components of response even in the steady state. Therefore, I would consider it as non-linear dynamic.
However, there maybe some cases, where if you don't have any vibrations (not many for real FSI problems). In that case, some (I'm not one of them ) may chose to consider the problem as linear.


Why is the series named this way?

Question: I've been watching the anime since yesterday and couldn't help but wonder why it was called Gokukoku no Brynhildr or Brynhildr in the Darkness. So far I have yet to see any mention of Brynhildr; is this some person later on in the story? Or another name for the protagonist?
Comment: Maybe [Lynn Okamoto](http://myanimelist.net/people/2711/Lynn_Okamoto) just likes germanic mythology
Answer: Brynhildr in this series is  Kuroha Neko. Brynhildr is mighty female warrior of Germanic mythology and the chief of the Valkyries.
Neko's character almost portray the character of Brynhildr as mentioned in German myth. She is powerful than Valkyrie. She does not have any past memory. She gained her memory and became very powerful at the end in series.
Brynhildr in the Darkness trying to show the Neko in darkness or loss of memory. She only does not have past memory but also lose memory when she keep using her power.
Answer: Spoiler alert!

 It could be because one of the main antagonists is called Valkyria, and she's Neko's sister. Since Neko is even more powerful than Valkyria, Brynhildr could refer to her. Plus, Valkyria is almost identical to Neko, but with white hair instead of dark (darkness). Also, her memory is full of holes, so she's "in the dark" about her powers and background.
Answer: As confusing  as the explanation was by "MirrorOfTruth" given on 09-12-2014, I think shklim or shkler was trying to say the following: The refrence to Brynhildr, from the anime series "Brynhilder In The Darkness", refers to the female character Neko Kuroha. Brynhildr is a mighty female warrior and chief of the "Valkyries", which comes from German Mythology.
Neko's character almost portrays "Brynhildr" from said myth where as she is powerful like the Valkyrie, has no  past memories, and as she regains her memories she becomes much more powerful, like in the end of the series.
(I'm going to leave the last part out because it is the definition of the word "Spoiler") but lacking basic grammar skills.
Having said that, it is almost as damaging to the information given, as it is to the person lacking said skills.
Answer: Brynhildr (also spelled Brunhild, Brünnhilde, Brynhild) is a shieldmaiden and a valkyrie in Germanic mythology.
Best answer I could find.
http://en.wikipedia.org/wiki/Brynhildr
Comment: Could you elaborate on how this is connected to the series ?

Why is the series named this way?


Dependency of rate of diffusion on mean free path and velocity of molecules?

Question: What is the dependency of rate of diffusion ( Graham's law) on mean free path, molecular size , average velocity, collision frequency and pressure? I have searched everywhere on internet but just found that rate of diffusion is inversely proportional to square root of density or molecular mass.
Comment: A proper textbook on physical chemistry should explain this in detail. You might have better luck searching in one than on the internet
Answer: My reference is Concepts in Thermal Physics, Blundell and Blundell, Ch. 7, 2nd ed. Everything here is from there.
Consider a gas trapped in a rigid box. We begin by evaluating the molecular flux incident on one part of one wall of the box. Let there be $n$ particles per unit volume, so $n$ is the particle density of the gas in the box. Let $A$ be the area of a small patch of the wall, and let $\theta$ be the angle between the particle's direction of approach and the normal to the wall. Let $f(v)$ denote the speed distribution function of the particles, which we assume to be a Maxwell-Boltzmann distribution.
Consider all the molecules with speed between $v$ and $v+\text{d}v$ approaching the wall at an angle between $\theta$ and $\theta+\text{d}\theta$, which we write in shorthand as ($v,\theta$). There are $n$ particles per unit volume, and a fraction $f(v)\,\text{d}v$ of them are moving within the desired speed range. A further fraction $\sin\theta\,\text{d}\theta/2$ of them are moving in the desired direction, so the particle density for ($v$, $\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2.$$ (To derive the fraction $\sin\theta\,\text{d}\theta/2$, we consider a unit sphere with surface area $4\pi$. The surface area of the differential annulus is $2\pi\sin\theta$ (circumference) $\times\;\text{d}\theta$ (differential width), so the fraction of surface area occupied by the annulus is as stated.)
Now consider how many particles traveling at ($v,\theta$) hit the patch of wall in a time $\text{d}t$. These particles must be contained within a volume $$A\,\times\,v\,\text{d}t\cos\theta$$ (distance particle travels normal to the wall in a time $\text{d}t$), for all particles traveling at ($v,\theta$) must impinge on the wall within $\text{d}t$. Actually this area is slightly enlarged by a width $v\sin\theta\,\text{d}t$ uniformly, but this contribution to the area is of order $\text{d}t^2$ and vanishes in the small $\text{d}t$ limit.
All in all, then, the number of molecular collisions in $\text{d}t$ on a patch of wall of area $A$ at ($v,\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2\,\times\,Av\,\text{d}t\cos\theta.$$
The flux $\Phi$ is the rate of molecular collisions per unit area, so we divide by $A$ and $\text{d}t$, and integrate over all $v$ and $\theta$. This yields $$\Phi = \int_0^\infty\text{d}v\int_0^{\pi/2}\text{d}\theta\,nvf(v)\sin\theta/2\cos\theta = \frac{1}{4}n\langle v\rangle = \frac{p}{\sqrt{2\pi mk_BT}},$$ assuming an ideal gas and given the average speed of gases within the Maxwell-Boltzmann distribution.
By definition of the molecular flux, the rate of effusion is simply $\Phi A'$ for $A'$ the cross-sectional area of the hole out of which gas effuses, and the resulting dependencies can be read off from the equation for $\Phi$.

Interesting notes.

Due to the extra factor of $v$ in the integral for the molecular flux, the distribution of speeds of the effusing particles is $v^3e^{-\beta\epsilon}$ , rather than the standard $v^2e^{-\beta\epsilon}$. This extra factor of $v$ comes about because speedier molecules hit the walls of the box faster than do slower ones.
The hole through which particles effuse must be small, and the length scale that governs smallness is the mean free path. Make the hole too large, and the box eventually reaches equilibrium with the outside world. Consider a system of two boxes placed side by side, and a small hole between the neighboring walls of the boxes. For holes small enough that effusive behavior is observed, at equilibrium the molecular fluxes must balance and hence $$\frac{p_1}{\sqrt{T_1}} = \frac{p_2}{\sqrt{T_2}}.$$ On the other hand, if the holes are too large, then we simply have regular thermodynamic equilibrium, with $$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$$


Can Entities Have Their Own Shader In Deferred Rendering?

Question: Since you need to use the general g-buffer shader to fill g-buffer, how can entities have their own shaders, like giving a box normal mapping feature ? As I remember, you can't use two shaders at the same time. So there is a possible solution that every object will have their own renderers, but If I have to make each entity have their own renderers that also include g-buffer filling in their shaders, wouldn't there be unnecessary g-buffer shader code everywhere, extra variable passings to GPU ?
Answer: You can use any shader to fill the g-buffer, it doesn't have to be generic.  That's not the point of deferred, the point of deferred is when you come to do the lighting pass you combine the various render targets in the g-buffer with your lights; your lighting pass is confined to pixels on-screen (rather than shading pixels which may be occluded).  But it's quite agnostic as to how the individual render targets get filled, so in the case of normal mapping you can use a shader with normal mapping here, and output the result of it to your normals buffer.
Comment: So the point is, as long as there is something filling that gl_FragData[], i can use any shader ?
Comment: You could even use glDrawPixels if you were so inclined.


Cooking a 5 pound pork sirloin roast

Question: Put a 5 pound pork sirloin roast in oven at 175 degrees F for two hours. Then turned up oven to 350F for two hours. Is it safe to eat?
Comment: What does your thermometer say?
Comment: There is no way for us to tell you that your food is safe to consume... only you can tell by using a thermometer and comparing that internal temperature to the "safe" temperatures chart.
Comment: It's more about internal temperature than it is about the time and temp. Can't really help you.
Answer: The only way to tell if a pork roast has been cooked long enough is by taking an internal temperature with an instant-read thermometer in several places around the thickest portion of the roast.

The National Pork Board recommends cooking pork chops, roasts, and tenderloin to an internal temperature between 145° F. (medium rare) and 160° F. (medium), followed by a 3 minute rest. Since large cuts increase approximately 10° F while resting, remove them from the heat at 150° F. followed by a 10 minute rest.
Pork Cooking Times and Temperatures

Cooking a 5 pound pork sirloin roast


What is the reason behind marking forbidden pages as 404?

Question: This came to my mind after my previous question. Actually, most of the 404 pages I stumble into on SE are forbidden (403, eh?), not that they can't be found. Examples of such pages would be trying to access /admin despite not being a mod, or trying to access deleted chatroom, or somebody's inbox or profile editor, etc. Why do they throw 404s instead of 403s or something similar?
Comment: [Security through obscurity](https://en.wikipedia.org/wiki/Security_through_obscurity)
Comment: @DavidPostill Well this question relates purely to StackExchange sites, so by asking here on meta.stackexchange nicael *is* asking the website designers why it was designed this way.
Comment: @JonW I missed that bit. Let wait and see an answer from one of the developers ... ;)
Comment: Isn't this question "primarily opinion based?" ;) You can't possible know **why** unless you ask every website designer why they designed that way.
Comment: Well, isn't the reference a valid reason? @JonW
Comment: For the same reason on a site login form you don't specify which of password or username were incorrect, and just state "one or both" is wrong. Security through obscurity is usually a bad path to walk, but in this case it's a good idea as there is no weird setup, just a "not found" instead of "access denied". Also, for them, "admin" arguably *doesn't* exist.
Answer: "What is the reason behind marking forbidden pages as 404?"
Security through obscurity.
See Obscurity is a Valid Security Layer for nice explanation (as well as the citation below).

Source When Should You Return 404 Instead of 403 HTTP Status Code? by Max McCarty.

When talking about web application security, one common denominator
  that repeatedly comes up is the act of disclosing sensitive
  application data.
...
In my last post on Security Misconfiguration, part of the discussion
  was on properly handling error messages to ensure we don't expose
  sensitive data to our users.  But it was obvious that in certain
  circumstances we can inadvertently disclose information that a
  malicious user could use to their advantage just by returning the real
  HTTP status code.  One of those situations is when the resource is
  forbidden (403).
An authorized user has requested a forbidden resource in which they
  receive a HTTP 403 forbidden response to the request is a common
  scenario.  However, by returning the applicable and valid 403, we
  have also made it clear that the resource does exist.  The
  disclosure of the resource might only provide a piece of the profile
  puzzle a malicious user was assembling or it might actually directly
  provide the user opportunities.  For example, a valid case could be
  that a 3rd party web resource (.axd) had a known security flaw and
  could be taken advantage of through other side channels when knowing
  the resource exists.
Though, security through obscurity is never in of itself reliable security, it can be and should be used as part of any security
  in-depth approach.  We can leverage this approach when we determine
  that it would be better to not disclose the existence of a resource,
  but return 404 instead of 403 HTTP status code.  Unfortunately, if you
  have ever attempting to do so, you will have found it less than
  possible and far from easy.  Maybe you've even given up on the
  attempt.
...
Comment: When first starting in security, I think it's great to properly identify when "security through obscurity" is being used. However, once you get more comfortable in the field and you're confident that your system isn't RELIANT upon these techniques, it's okay to introduce some obscurity as more of a defense-in-depth mechanism. For example, ensure that your resource is actually secure and access is controlled. Afterwards, there's still something to be said about preventing attackers from even learning about the existence of said resource. The 404 over 403 is therefore allowed by RFC 2616.
Answer: Both existing answers say that this is a valid use of security through obscurity. Unfortunately, both miss the point of the original post: anyone attacking SE can easily determine with high confidence that all these 404s are in fact actually 403s in the same way the OP did … or just by reading this or similar posts, which are public. It would be silly indeed to suppose that someone attacking a highly visible site like SE would fail to dig up all the information (public and otherwise) that they can. And supposing that the designers of malware will stoutly cling to some standards-compliant, fully-generic attack code that has no hacky customizations for particular targets would be madness.
Therefore, this cannot be a valid security policy. It means nothing at all; it's just pointless obfuscation for the sake of it, or perhaps carelessness in setting up conformant response codes. And since it does make any legitimate generic clients that are standards-compliant a little bit less aware of the true situation, it's a bad idea, just like using tables for layout.
Breaking standards in a predictable, publicly-documented way to ensure security will only sacrifice standards-compliance to achieve no additional security at all.
Comment: This is the right answer. The existence of moderator tools, etc. *is not a secret*, but in fact is well-documented on Meta.SE for anyone who knows how to read, and even users who can't read can get around it by using a screen reader. The security through obscurity principle applies to *truly secret* resources. If SE (for example) tracks private banking information and doesn't want us to know that it does so, then yeah, they should return 404 when someone requests https://meta.stackexchange.com/banking.
Answer: This is an acceptable practice used by many major websites. For example, GitHub returns a 404 when trying to access a private repository that you don't have access to. Facebook will also return a 404 when attempting to access the profile of an individual that you aren't authorized to see. The intent is that you shouldn't be able to learn about the existence of a resource unless you have authorization to view it.
This doesn't break standards either. RFC 2616 states that for a 403 Forbidden status code, "If the server does not wish to make this information available to the client, the status code 404 (Not Found) can be used instead."
Comment: The difference is that Github and FB do not publicly list all existing private repository or profile names, so 404 is a plausible lie (usually it's correct, and you can't tell which is which). SE does list valid, consistent admin URLs, so the lie is pointlessly blatant.
Answer: Why not have a scenario? If you are a guard guarding something secret, would you say "Hey, it exists, but you can't get it!" or "Nope, dunno what that is" when someone wants it? Obviously the second! Same here. If you say the first, the person now knows the existence of the thing. The second does not let you know. So if we try to access something confidential about a user, giving 403 will give the existence of it away, and hackers know where to find it. For most of us, 404 means not there! So we don't know if it really exists or its just a lie to protect user data.


Ice-hockey penalty reasons

Question: Is there a list of all possible penalty reasons in ice-hockey/NHL?
Like High sticking, Holding, Hooking... but all of possible reasons. I would like to have the complete list for statistical purposes, like what could a referee officially write down for penalty reason on the match sheet.
Answer: Yes like almost all sports, there's a full list of penalties, otherwise referees wouldn't done what play is a foul or not. In hockey the penalties are split to minor (2 minute) or major (5 minute) penalties.
Here is the list of minor penalties:

Charging
Clipping
Closing hand on puck
Cross-checking
Delay of game
Elbowing
Embellishment
Goaltender interference
High-sticking
Holding
Holding the stick
Hooking
Illegal equipment
Illegal stick
Instigator
Kneeing
Leaving penalty bench too early
Leaving the crease (goalkeeper)
Participating in the play beyond the centre red line (goalkeeper)
Roughing
Slashing
Throwing puck towards opponent's goal (goalkeeper)
Throwing stick
Tripping
Unsportsmanlike conduct

And here is the list of major penalties:

Boarding
Butt-ending
Charging
Checking from behind
Clipping
Cross-checking
Elbowing
Fighting
Head-butting
Hooking
Illegal Check to the Head
Interference
Kneeing
Slashing
Spearing

Take a look at Wikipedia's article on hockey penalties for what each penalty is. There's also info on what results in a game misconduct, or 10 minute misconduct, or what results penalty on bench participants.
This is also a good resource talking about the penalties, with visualization for the signals.
Comment: Wikipedia is pretty badly inaccurate here. Some things which can be either minor or major penalties are missing from the minors list (e.g. boarding) and for statistical purposes you probably also want to be aware of the concept of double minors and goalkeeper penalties. See [reference tables 2-13 in the NHL Rulebook 2020-21](https://cms.nhl.bamgrid.com/images/assets/binary/319997074/binary-file/file.pdf).
Comment: Please don't use codeblocks for text that is not actual code. List formatting and tables are now well-established and available.

Ice-hockey penalty reasons


Why can't the Fajans' rule and lattice energy theory be used to explain the difference in thermal stability of magnesium and calcium carbonates?

Question: $\ce{CaCO3}$ is more thermally stable than $\ce{MgCO3}.$ I do understand that a larger cation would be able to stabilize a larger anion to a greater extent, but why can't we use Fajans' rule or lattice energy to explain?
Answer: (1) Thermal Stability:
$\ce{CaCO3(s) ->[\Delta] CaO(s) + CO2(g)}$
$\ce{MgCO3(s) ->[\Delta] MgO(s) + CO2(g)}$
Because of smaller $\ce{Mg^2+}$ ion, it will polarise the $\ce{CO3^2-}$ ion more than $\ce{Ca^2+}$ ion. So the removal of $\ce{CO2}$ will be easier. See this explanation.
Fajans' rule compares the covalent character in ionic compounds. For which, it compares the polarising power of cations if the anion is same. Hence, this rule in a way, explains the thermal stability of these two compounds.
(2) Fajans' Rule:
The covalent character is less in $\ce{CaCO3}$ than $\ce{MgCO3}$.
(3) Lattice Energy:
The lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions.
$\ce{CaCO3(s) ->[\Delta] Ca^2+(g) + CO3^2-(g)}$
$\ce{MgCO3(s) ->[\Delta] Mg^2+(g) + CO3^2-(g)}$
$\ce{MgCO3(s)}$ has more lattice energy but it is less thermally stable than $\ce{CaCO3(s)}$. Hence, it cannot explain the trend in thermal stability because of different products obtained in (1) and (3). More on this is here.