Unable to change country. Tried everything

Question: I had made some purchases on my daughter's US credit card initially when in US. So my location was tagged as US. Recently when I tried to download one Indian app from India, it was showing not available in your country.
So I removed my daughter's credit card, added my Indian card and also made a purchase on my card.
Still Google Play is not changing my country.
Pl help.
Comment: what app are you trying to install??
Comment: Reliance jio ----- my jio
Comment: Also Citibank in
Answer: The settings you have to change are at Google Wallet. Sign in with your account, go to Settings, and change "Home address" to your new address.
You might also have to change your phone number, and your default "Billing and delivery address" in the second tab.
So, there are TWO places where you have to update your address, and the settings do not usually take effect immediately.
From Google Groups :
You can update your home address whenever you want, including changing to a different country. If you're representing a company, enter your business address.

Sign into your Google Payments account at payments.google.com and click the Settings icon in the top corner. See screenshot:
Click Edit next to the "Home Address" listed and update the address (please note this is different from the "Address Book" which holds shipping addresses).
Open the Play Store and navigate to any item available for download.
Click to begin a download until you reach the "Accept and buy" screen (no need to complete the purchase)
Close the Play Store and clear data for the Google Play Store application (Settings > Apps > Google Play Store > Clear Data) or clear your browser cache
Re-open the Play Store. You should now see the Play Store that matches your default payment method's billing country.

Note that if you don't immediately see the Play Store that matches your default payment method's billing country, it may take a few hours for this to take effect.
If you haven't yet added a payment method to your account for the first time, add a card directly from the Play Store with a billing address that matches your intended country location. Then, follow steps 3 through 6 to show your intended country's Play Store.
Some things to note...
For legal reasons, you'll need to sign new Terms of Service if you change your home country.
1. If your Google Play balance is higher than these amounts, you won't be able to change your home country. If it's under those amounts and you change to a country with a different currency, you won't be able to see or use your remaining balance until you switch back to a country where that currency is used.
2. If you have a Payments Balance higher than USD $10, or a Checkout Merchant account, you won't be able to change your country.


Qt4Dotnet on Mac OS X

Question: I'm using Qt4Dotnet project in order to port application originally written in C# on Linux and Mac. Port to Linux hasn't taken much efforts and works fine. But Mac (10.4 Tiger) is a bit more stubborn.
The problem is: when I try to start my application it throws an exception. Exception states that com.trolltech.qt.QtJambi_LibraryInitializer is unable to find all necessary ibraries. QtJambi library initializer uses java.library.path VM environment variable. This variable includes current working directory. I put all necessary libraries in a working directory. When I try to run the application from MonoDevelop IDE, initializer is able to load one library, but the other libraries are 'missing':
An exception was thrown by the type initializer for com.trolltech.qt.QtJambi_LibraryInitializer ---> java.lang.RuntimeException: Loading library failed, progress so far:
No 'qtjambi-deployment.xml' found in classpath, loading libraries via 'java.library.path'
Loading library: 'libQtCore.4.dylib'...
 - using 'java.library.path'
 - ok, path was: /Users/chin/test/bin/Debug/libQtCore.4.dylib
Loading library: 'libqtjambi.jnilib'...
 - using 'java.library.path'

Both libQtCore.4.dylib and libqtjambi.jnilib are in the same directory. When I try to run it from the command prompt, the initializer is unable to load even libQtCore.4.dylib.
I'm using Qt4Dotnet v4.5.0 (currently the latest) with QtJambi v4.5.2 libraries. This might be the source of the problem, but I'm neither able to compile Qt4Dotnet v4.5.2 by myself nor to find QtJambi v4.5.0 libraries. Project's page states that some sort of patch should be applied to QtJambi's source code in order to be compatible with Mono framework, but this patch hasn't been released yet. Without this patch application crashes in a strange manner (other than library seek fault).
I must note that original QtJambi loads all necessary libraries perfectly, so it might be issues of IKVM compiler used to translate QtJambi into .Net library.
Any suggestions how can I overcome this problem?
Answer: I think I ran into the same problem earlier today.  As soon as I copied libikvm-native.dylib from my current Mono to my project's exe directory, it worked.  Frank reminded me to do that on the list here.

Qt4Dotnet on Mac OS X


Is it possible to protect my Windows XP users now that Microsoft is no longer releasing security updates?

Question: My office has about 20 PCs running Windows XP.  I got a message saying that Microsoft is no longer releasing any security patches for Windows XP.  However, we find that XP is very convenient for our work.  Please suggest some possible ways to keep Using Windows XP for my PCs without any risk.
Comment: You would have to security so tight your users would revolt. If you blocked all internet access.  You could sit with each user and have them tell you what to do and then **you** do only the safe parts of there request, 1 user at a time.  Hire a team to analyze attachments, or just be an attachment free email company and delete attachments automatically.  Basically, your company would have to be an IT prison camp.
Comment: I would suggest that the way you think about convenience may need to shift.  It may appear convenient, but the fact that you are guaranteed to have permanently open security holes with a 100% certainty suggests that it may not be as convenient as you think.
Answer: Disconnect from the Internet, don't allow any external devices to be connected to your network and physically guard your computers to make sure that nobody violates the no external devices rule, and you'd be fine.
Otherwise the only option is to upgrade.  Since security updates are no longer provided, any newly discovered vulnerabilities would remain unpatched and an attacker (either external or internal) would have an easy time of using one of those unpatched vulnerabilities to access your systems if they can connect to them in any way.  Keep in mind that a large portion of attacks originate from disgruntled internal workers and unpatched vulnerabilities will mean that tools will become easily available to let even a relatively unskilled user make use of complex vulnerabilities.
There is no way to secure against this cost effectively other than to upgrade to an OS that still receives patch support to counter security flaws as they are discovered.  Honestly, if you are concerned about properly securing your business, it was past time to upgrade from XP even prior to the end of life.  There are numerous performance and security enhancements in Windows 7 that rendered XP obsolete from a security perspective years before end of life was reached.
Comment: @Willamhendric - while it is a good idea to use security products, it will not protect your computer against a lot of threats.  Many security flaws in Windows, when exploited, can bypass the protection of third party security software because they can force the OS to do things.  Third party software works by looking for other user software that is causing issues and telling the OS to stop it.  Security flaws in the OS allow the attacker to be the OS.  So the third party security software would be asking the attacker to stop themselves. (That's oversimplified, but gets the basic idea.)
Comment: I am not bothering about cost. I am just looking for alternative ways to protect my system. Thanks for your reply and one more question Shall i use any system security products on my PC's? Will it gives complete protection similar to windows security patches?
Comment: @Willam hendric there is a way, but it would require researching security holes, reverse-engineering, and implementing a patch by yourself or paying for someone to do it for you.  Depending on the country you are in you may need at least two people (the US has laws that pretty much require two or more people).  This is not very cost effective as it could cost into the millions if not billions of dollars and you are not guaranteed to patch all holes.  But, is really the only way to keep it "secure" without support from Microsoft patches.
Comment: I don't know any people who provide security to my PC's with Windows XP. Shall i hire any security providing company? If so, please suggested some affordable security providers.
Comment: @WilliamHendric the reason I didn't mention what Travis mentioned is that there is no cost effective long term option.
Comment: @WiliamHendric As AJ Henderson mentioned there is no long-term feasible option.  I just mentioned what I mentioned to show that costs are way too high to keep XP especially for only 20 PCs it would be millions of dollars per PC.  If you are that attached to it maybe check out ReactOS it is still in the alpha stage and not suitable for use yet, but I say is worth keeping an eye on.  Your best option would be to upgrade to Windows 7 (I like it better than XP personally).
Comment: There are corporates with tens of thousands of XP machines, and while these typically go through a 3 or 5 year refresh (depending on industry) until they are replaced, extended support is possible - initially through MS, then trusted 3rd parties. It is **not** cost effective for a small organisation. As AJ says, just upgrade.
Answer: This depends on why you aren't upgrading.
The vast majority of security holes in Windows XP are in Internet Explorer.  If you can completely eliminate the use of IE to view external content (and this includes hidden uses such as reading HTML emails in Outlook), you can keep your computers reasonably secure from outside attack, at least in the short term.  This is harder than it sounds, though, since a great many applications embed IE as an HTML renderer.
If you aren't upgrading because some task requires a specific version of Internet Explorer, there's nothing you can do.
Comment: +1 for hidden uses such as reading HTML emails in Outlook
Comment: In response to the last paragraph: If your need for XP is only because of old IE versions, you should definitely try Enterprise Mode on newer versions of Windows, or simply using the compatibility  tags on your internal webpages. (You can tell it to simulate as far back as IE7 for particular pages. It's not perfect, but pretty good)
Comment: I down voted this answer because not using Internet Explorer will not protect your XP machine as asked in the original question. If a service in the machine becomes exploitable even simply being connected to the internet can compromise you. An example of this is the Blaster worm which affected the RPC service running on Windows XP and 2000.
Comment: @BradMetcalf, during the life of XP, Microsoft disabled or firewalled all Internet-listening services by default (Windows Firewall became active by default with SP2), leaving only user-assisted vectors (such as browsing a site in Internet Explorer).  A fresh install of XP SP3 that has been connected to the Internet isn't at risk of infection until the user starts using it, and IE (including hidden uses) is the primary source of user-assisted attack vectors.
Comment: @Mark I am actually seeing the opposite. I just nMap'd a fresh install of XP SP3 with updates and it revealed three service ports are open. 125 which is an RPC port, 445 which is a SMB port, and 3389 which is the remote desktop service port. Which would mean if either service becomes vulnerable the XP machine itself becomes vulnerable.
Answer: Windows XP is not secure and you do risk running into getting hacked. I do believe you can pay for XP support like the banks are doing. But it would be a much better switch if you jump to at least windows 7.
Comment: Microsoft doesn't offer an extended support agreement for XP officially. They only do it "under the counter" for very large and security-relevant customers like banks and governments. The official stance of Microsoft towards business customers is "stop your whining and upgrade".
Answer: Microsoft has a special support for XP for large companies, but I think it is expensive for smaller ones http://gcn.com/blogs/pulse/2013/09/xp-support-for-a-price.aspx .
There is one way how to receive XP updates until 2019, but it is not advised by Microsoft. The info is from this site http://www.sebijk.com/community/board15-other/board73-tutorials/2985-getting-xp-updates/ and I have to say that there is a warning on that page:
ATTENTION: Use at your own risk! These updates are not tested on a regular XP system and could damage your system.
Microsoft have released a warning, too.
You have to write to a text editor the next few lines of code:
<code>Windows Registry Editor Version 5.00

[HKEY_LOCAL_MACHINE\SYSTEM\WPA\PosReady]

"Installed"=dword:00000001
</code>
and save it as a .Reg file. You have to save it exactly as it is, with the blank lines.  Than just double click on it and allow it to save to Windows XP registry. I tried it and it worked, I began to receiving new updates. But it is not my only one PC so if something happens it is not such a big problem for me and I am not running a business on it.
Comment: This method will get you updates for the **Embedded Edition of Windows XP**, which is a very stripped-down version. These patches will be unlikely to fix any vulnerabilities in any components not included in the embedded edition.
Comment: This is one answer where we need some sort of extra formatting to indicate the danger. Yes, it might work (and it's an awesome find!), but gosh, I'd never implement this for a business. The risks are undefined. That said, I'm going to try it on a test system ...
Answer: There has been a deal with Microsoft and the UK Government that has enabled the government's Windows XP machines to be patched for another year - so if you want continued patching and support then this is the only secure way to go if you refuse to move from XP assuming you don't want to isolate your machines (but you will still be forced to upgrade after the period ends in a year). No general deal has yet been announced to make this available to private companies and end users.
Also, there have been hacks posted that allow updates to continue, such as this one. Whether the hack will allow you to receive the same updates is another matter, but it could provide short term protection to buy you a little more time to rollout an upgrade ASAP. As @elsadek notes in the comments this could result in broken or reduced functionality.
Comment: Microsoft has made a warning for the hack you mentioned, read this "..Windows XP customers also run a significant risk of functionality issues with their machines.." http://www.zdnet.com/registry-hack-enables-continued-updates-for-windows-xp-7000029851/
Comment: Yes - as with all hacks, never guaranteed to work.
Answer: I heard that Horizon Datasys (the makers of Rollback RX) is developing a freeware version of Rollback exclusively for XP users to stay protected.
Answer: I have a Windows XP machine that is being used due to some legacy software. In order to protect it- I put in multiple layer of protection. Av, Malwarebytes and monitor my network as well. Of course keep an updated back up.
Recently I have come across this freeware known as RollBack XP. This definitely helps in protecting Windows XP by allowing you to revert back to a previous state before the system crashed or got infection.
I am still testing the RollBack XP. So far so good.
More info on RollBack XP:http://www.horizondatasys.com/en/RollBack_XP.ihtml


Simple side line defense against English opening

Question: I'm looking for a simple plan to play against the English. Ideally a side line, even if it's objectively worse than main lines. I'm far from being a GM, so it won't matter a lot. I prefer that to a main line that English opening players will obviously know much better than me. Something like the Raphael variation (2.Nc3) against the Dutch. I like that kind of games, as well as Caro-Kann.
Comment: 1.c4 d5 can be a bit extreme, but should be playable. After 2.d4 is just a Queen's gambit, and after the more usual 2.cxd5 you can chose between Qxd5 or Nf6 or even the gambit with c6
Answer: The white opening move 1. c4 looks like an invitation to play a reversed Sicilian. For fun you can take up the challenge and play a reversed Grand Prix Attack!
So you would aim for a setup with e5, Bc5, f5, Nf6, O-O, etc. It's probably a good idea to make an escape square for the bishop by also playing a6 as soon as white tries a3 looking to follow up with b4 and c5 winning your bishop. An early d6 is another way of safeguarding your c5 bishop.
If white is not careful this can turn into something close to a classical Dutch where black has achieved the thematic break e5 on move 1.
This is an offbeat line which is reasonably sound and can give you some good attacking chances against an opponent who isn't familiar with the ideas. It also gets you away from the world of often stodgy standard lines.
Comment: What's the plan after 3.e3 and 4.d4?  It seems difficult to get the bishop to c5 or b4 and you end up in a not very Grand Prix position.
Comment: Is this really a side line?
Answer: The suggested answer is objectively good, but leads to heavily theoretical and concrete play. I would suggest to have a look at 1... b5!? First of all, it is a sideline many English players haven't studied properly, and secondly it is full of poison, leading to Benko-like positions, as the typical follow-up is 2. cxb5 a6 3. bxa6.
<code>[fen "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - "]
[Event ""]
[Site ""]
[Date ""]
[EventDate ""]
[Round ""]
[Result ""]
[White "A simple example"]
[Black "-"]
[ECO "A10"]
[WhiteElo "?"]
[BlackElo "?"]
[PlyCount ""]

1. c4 b5 2. cxb5 a6 3. bxa6 Bxa6 4. g3 d5 5. d4 Nf6 6. Nf3 e6 7. Bg2 c5
8. O-O Nbd7 9. Nc3 cxd4 10. Nxd4 Qb6
</code>
Answer: Im not sure what level of play to assume here. Voluntarily accepting an inferior line against a strong player does not seem like a good policy so I will assume that White is not very powerful. Such players often adopt 1.c4 because they prefer to avoid anything sharp and you may be able to exploit this. For example they will often play d3 rather than d4 even when d4 might be preferable. They are hoping to complete a solid development before having to calculate any tactics, and are frequently content to make safe but rather moves. This may be the sort of player that you are having difficulty with. It is hard to suggest specific lines of play because White may choose such a wide variety of setups. But you can turn that psychology against them.
For example, as well as a reversed Grand Prix, you can play other systems involving ..f5. I often play, in some order, g6, Bg7, e5, Ne7, d6, Nbc6, 0-0, f5, slowly building an attacking formation without creating any weaknesses. Naturally this has to be adapted
to what White actually does, but if they are content with slow development Black gets a nice position.


Android PowerManagerService code to handle Screen standby or Auto lock

Question: In Android Source code Android\mydroid\frameworks\base\services\java\com\android\server\ PowerManagerService.java which method I need to comment or change so that my Screen never goes in standby mode or gets Auto lock.
I tried creating a service which captures boot complete intent. Tried capturing in broadcast receiver still the lock screen comes and goes.
Do let me know how I can disable screen auto lock and standby after system bootup.
I will make changes in the java file and again build the source code.
Thanks in advance.
Answer: There is a method called doKeyguard which is called from systemReady in PhoneWindowManager. You can give that flow a try.


Show SQL table on SharePoint 2019

Question: We have some databases and tables on SQL 2017, we want to show one table content on SharePoint, is it possible?
Answer: Yes, it's possible. You could use Business Connectivity Services to access the data in SharePoint via external lists. External lists allow you to manage external SQL data using the SharePoint user interface.
Alternatively, you could create a bespoke application which returns the data via an API. Once you have that API, you can then create an SPFx web part that displays the data via the API.
Answer: To provide an idea: If you use Power BI, connect it to the database and display the table through Power BI. Then you could easily display the report in SharePoint using the Power BI web part.
By the way, Power BI needs to access the database through an on-premises data gateway.
Reference for Power BI to connect to SQL server: Refresh data from an on-premises SQL Server database
Reference for using Power BI Web Parts in SharePoint: Embed a report web part in SharePoint Online

If the answer is helpful, please click "√" on the left panel of the answer and kindly upvote it.

Show SQL table on SharePoint 2019


What could be a star-like object that shines at daytime for few seconds? What's the probability to see it?

Question: Some years ago I looked up to the blue sky, around 2 PM, and I saw, for 2 or 3 seconds, something like a star fading out.
Could it be a Supernova or another phenomena? Is it highly possible?
Comment: Unlikely to be a supernova! Not had one in our vicinity since the medieval times (perhaps, barring SN1987A)
Answer: Several satellites (but most notably Iridium) have large reflective panels.  If aligned with the sun properly, it can shine with sufficient brightness to be visible in the daytime.
An Iridium Flare usually lasts only a few seconds.  It sounds quite consistent with your description.
It can be hard to tell in a blue sky with no nearby references, but an Iridium satellite would also be moving while fading out.
There are sites that can show you upcoming flares for your area.  If you knew the exact date in the past, they could probably even "post-dict" what flares were available then and see if it matched.
Answer: I saw a similar "ufo" several decades ago, before there were Iridium satellites.  Possibly it was another type of satellite with large reflective panels.
Thinking it over, I later decided that if an airplane reflected the sunlight directly toward me at one moment of a turn, it might make a similar flash.


Seeking birth family of Michał (Michael) Gudzinowicz Praniewicz from Poland/Russia?

Question: after reaching out to some historical societies in Massachusettes, I learned that I was dealing with two exact people named "Michael Praniewicz" – sorry for all the confusion. Below you will see my edited, much more precise questions.
My main questions are as follows:

Where do you think Ryski, Russia is? This is the place of his birth listed on his naturalization (intent and decloration) papers. I cannot located any town called "Ryski, Russia".

Furthermore, I found this list of noble people of Vilnius in 19th century. Both of his surnames "Praniewicz" and "Gudzinowicz" – could this just be a coincidence, or should I write to the Lithuania archives to see if they have anything on his birth in Wilno/Vilnius? It sort of makes sense to me because I have been told that our family was a wealthy family. Our last name means "Son of a Laundered one" Lol - so we musta been clean/wealthy folks!?

http://genealogia.lt/pdfs/szlachta_wilenska.pdf

He came to New York on a ship called "Palatia" on July 20, 1900. Looking through ancestory at all of the names, and he does not come up having arrived on the Palatia. Any ideas? Is there a way to search deeper into the ship records?

Full name: Michał (Michael) Gudzinowicz Praniewicz
- Alternate spellings of surname: Prenewicz, Pronowicz?

birth date:

April 25th or 30th, 1871 or 1872 (listed on Naturalization papers July 9, 1912 and on death certificate June 24th, 1948)

birth place:

Ryski, Russia (listed on naturalization papers July 9, 1912) (Foreign residence St. Petersburg Russia)
Poland (listed on death certificate)

death date: June 20th, 1948 11:35am
death place: Allegheny, Pittsburg, Mercy Hospital
parent's names: Unknown, but they were Polish

Spouce:  Rozilla OR Rosalie OR Rose, Salachińsk OR Salaczińska OR Salacheeska OR Salawański OR Salatyński OR Sałacztńska

spouse's birth date:

1893 (1940 Census, enumerated in Pennsylvania)
09 Feb 1891 (united states Social Security death index)
OR 11 Feb 1890 (according to Family)

spouse's birth place: Poland (family says "between Poland and Russia)
spouse: death date: July 1976 (cannot find death record)
marriage date: unknown, cannot find marriage record. Around 1912?
Marriage place: unknown, I think Pennsylvania. cannot find marriage record. Family says she came to the US at 16 or 17 to live with her sister, met Michael when she worked as a waitress on the southside. He had been in Pennsylvania and owned property since at least 1906. He came into her restaurant and sat at the same booth and ordered coffee all the time. They liked talking because they understood the same language. Family said he was around 44 and she was 18 when they got married. Not sure if this is true.

religion: Catholic
If immigrated, residence in US: He stayed in Pennsylvania his whole life.
Comment: Are you sure that your Michael had 2 wives and lived in Pennsylvania & Massachusetts? You mention Bertha in the 1940 census (in Boston with a Michael) http://search.ancestry.com/cgi-bin/sse.dll?db=1940usfedcen&indiv=try&h=92096648 but Rose is also with a Michael in Upper Burrell PA in 1940 http://search.ancestry.com/cgi-bin/sse.dll?db=1940usfedcen&indiv=try&h=23360223 . I think you should reevaluate your evidence sort which records go with which of the (probably) 2 men. That may reduce some of the birth date & place variations.
Comment: Now that we have an answer that locates Ryski, perhaps we can break out the question about finding  Michał's parents into a separate question. Let us know if you need help.
Comment: Excellent!  Now I'm going to ask you to focus one more time and ask you to focus on question number three -- here's why.  1) We prefer to have ONE question in a question for the site.  It's okay to leave more than one question about the same family and to refer back to your previous questions.  2) You may get more information about the place you need to identify in your first question if you can find the passenger list 3) It's almost always better to start with what you know and to work backwards in small steps instead of making leaps. Your second question is a big leap.
Comment: I agree with @JanMurphy that it will be best to focus this question on 3. Perhaps ask 1 as a separate focussed question now but definitely wait until you have more evidence suggesting that spending time on 2 could be fruitful.
Comment: Hi, welcome to G&FH.SE!  Can we clarify something?  If you are looking at the 1940 Census, that is a US Federal Census, not a state census.  A better shorthand would be, "1940 Census, enumerated in [place name]".
Comment: I would suggest looking for other records that would help you see which records actually belong to your great-grandfather.  Looking for the records of any siblings of your great-grandparents might help you see what is going on.  It is not enough just to match a single person's name to know that two records belong to the same person, because many people can share the exact same name.  You need to look for other identifying factors.
Comment: Hi Jan and Bgwiehle. I am sorry for the confusion. After finding more information out, I have made edits above. This is as exact as I can figure out! You will see my three main questions above!
Answer: The Geographical dictionary of the Kingdom of Poland and Other Slavic Countries lists three different locations for Ryski. The first is a village in the powiat lidzki; the second is a small town in the powiat grodzienski; and the third is a nature reserve also in the powiat grodzienski.
If you search for Ryski at this site, it will show you all three locations.


Mid 2012 MacBook pro, firmware lock

Question: I was wondering if there was any way to remove a firmware password from a mid 2012 MacBook pro with USB or something. I just bought it off of Craigslist (sketchy i know but....) When it comes up all i get is a screen with blinking file with a "?" On it. And if i try and do recovery mode it goes to firmware lock, im wanting to get this unlocked and usable again but preferably without taking it halfway apart. Btw i have checked with Apple, unless i have the original proof of purchase (i don't), then they can't fix the issue for me. Any help is appreciated
Answer: You're sure that it is actually a Firmware lock? A flashing question mark just means there is no valid OS on a disk that the Mac can find.
Obviously, there would be little point in a firmware lock if it could be easily bypassed. I think your first port of call should be contacting the seller and asking them for the password, with a view to getting your money back, if possible.
You're sure you can't access the Firmware Password Utility when booting to Recovery?
Some models will remove a firmware lock if the amount of RAM is changed. That's pretty much the only option. It's fairly easy to open the case and pull out some RAM.
Comment: It doesn't boot to recovery. It asks for a password with a lock on the screen. No options. And i was hoping to not have to fight the bottom panel off if possible. But can if need be

Mid 2012 MacBook pro, firmware lock


Should neutral and hot wire be tied together if not in use

Question:
There was a fan attached, but I removed it because I don't need it. I'm not sure what to do with the wires, though. Should I tie the neutral and hot wires together or should I leave them as they are in the picture? Also, do I leave the ground wire just hanging there?
Answer: No, these should be capped off as your picture shows.  If they are still live, code requires that the ends be in an approved box with a lid, and not concealed behind drywall or plaster (i.e. so the next person can get to it.  It looks like the conduit that brings the wires to the metal box has lost its nut to keep it attached to the box;  you should try to replace that too, so that the individual wires don't become abraded and short across the edge of the hole in the box.
Answer: Nope. That would be a direct short and would either make some impressive sparks or trip a breaker and prevent its reset. Be sure the nuts are securely connected and leave them as they are.
Your misunderstanding probably comes from the notion that electricity requires a closed circuit. While true, home wiring is done in parallel, so that each device is given a clear path to and from the source (panel). By connecting hot and neutral together, you're closing the electrical path without a load. This results in very large current flow, and the breaker will trip.
Normally that ground wire would be connected to a metal housing like that. If the device is abandoned, though, it's probably not a concern.
Comment: Wouldn't hurt. No point in taping the ground, though. If anything I'd attach it to the metal housing with a sheet metal screw and call it a day.
Comment: Great, can I wrap a little bit of electrical tape around them (including the ground) just to be extra sure and safe the nuts stay on
Comment: @user1142130 Remember as long as those wires are hot they must remain accessible.  You can't just close that hole up with the wires capped. You said you don't need the fan, if your closing that hole you need to do something else.
Comment: @Tyson what exactly do you mean? Can't I just put the fan cover back on and call it a day? I don't want to cover the hole completely with drywall- if that's what you're saying....
Comment: @user1142130 yes you can put the fan cover back on, that leaves the wire "accessible".  Drywall, plaster, etc was in fact my concern.


Will ibles go to heaven at the end?

Question: I had a question like so we all know ibless will go to hell but we also know if you believe in Allah swt you will go to heaven after finishing his sins. Like ya he was given a very long life and he must have deed innumerable but if he believed don't you think he will come to heaven at the end? He himself said he still believes and fear him in Surah Al Anfal verse 48 (8:48) in which he says 'Surely I am quit of you for I behold that which you do not. Indeed I fear Allah, and Allah is stern in punishment.'so like won't he be in heaven after he finish his sentence?
Answer: Asalaam Alaikum
to give a short answer to your question, the answer is NO.
to fully understand the quran we must look at the tafsir. the verse which you have mentioned [8:48], is regarding what iblis said when an army of believers and an army of disbelievers met on the battle field. iblis saw what humans could not see, he saw jibreel (A.S) on the battle field with the believers and he was scared that he would be captured by jibreel (A.S) and shown to the disbelievers. if that happened the disbelievers would not obey him.
[refference : Tafsir Ibn 'Abbas and Tafsir al-Jalalayn]
and now to clarify a little more on the question about iblis forgiven and going to heaven, this following video has it explained very well.
https://youtu.be/1U8OnUrYKq4
may Allah grant you mercy and guide you to the straight path.
Allah knows best.

Will ibles go to heaven at the end?


Does applying a force to keep a body in uniform motion not considered as a force because there is no acceleration. (Friction included)

Question: Because I have studied that if there is no acceleration then there is no force.
Comment: "Because I have studied that if there is no acceleration then there is no force." ------> actually we call it as a case of no **net force**.
Answer: It is still a force.
Just because there is no acceleration doesn't make it any less of a force. The force just gets balanced out by a force in the opposite direction. The net force on the body is 0 though and hence no net acceleration.
So if you are doing a problem where the forces are balancing out, then (Most probably) taking the net force as 0 should do it.
Comment: So if the net force is zero, does it also imply that the work done is also zero? But we are definitely doing some work to keep the body moving.
Answer: If a body with mass $m\neq 0$ has not acceleration then the total acting force is null: $\vec{a}=\vec{0} \iff  \vec{F_{tot}}=m\vec{a}=\vec{0}$. It means that a force applied to keep a body in uniform motion, say $\vec{F_1} \neq \vec{0}$ has to be considered. In particular if the body moves in uniform motion it means that there has to be an other force acting to it, say $\vec{F_2} \neq \vec{0}$ and this other force has to be opposite to the fist: $\vec{F_2}=- \vec{F_1}$ such that $\vec{F_{tot}}= \vec{F_1} + \vec{F_2}= \vec{F_1}-\vec{F_1} =\vec{0} =m\vec{a} \iff \vec{a} =\vec{0}$. For example, if there is a friction force on a table and you want to move a book on it in horizontal direction in uniform motion, you have to act a force such that her sum with the friction has to be $\vec{0}$.
Answer: There is a net force. Suppose the body has speed $v$ and undergoes uniform circular motion in a circle of radius $r$. Then the net force is the centripetal force, namely $$\Sigma F=\dfrac{mv^2}{r}$$
The acceleration is therefore $a=v^2/r$. This acceleration is radial, and only changes the direction of velocity, and not the speed. Remember that if velocity (which is speed+direction) changes, then there is a net force. In uniform circular motion, speed does not change, but the velocity's direction definitely does.
Second, there can be a zero net force, but there are still forces applied. Consider a stationary block on a table. Of course gravity is acting on it, but so is the table's normal force. The reason the block does not move is because the force of gravity equals in magnitude the normal force such that the (vector sum) net force is zero.


How to use Cesiumjs without bingmap api key

Question: Can I use cesiumjs without bingmap api key. I need to create a viewer which should use my own wms layer and osm. Further i want to use it as a platform for geospatial analysis on my spatial data. As i know cesiumjs is a opensource tool but why it needs bingmap api key.
Answer: Cesium uses the Bing key for two purposes, so to get it to stop warning about the key you have to turn both of these off:

The default imagery layer is Bing Maps.
The Geocoder widget uses Bing API for geocoding.

The first one is obvious, and you already have a replacement lined up.  But the second one is the gotcha, most people overlook it.  You can construct the Cesium viewer with geocoder: false in the options object to turn it off.  Once you've done both of these things, the Bing key will no longer be in use and the warning message will disappear.


What series of books do you recommend before studying the Banach Tarski paradox?

Question: I want to learn Banach Tarski's Paradox but I only know Algebra (little really), so I decided to spend some time on this but I don't know where to start.
Comment: You don't need THAT much math, other than set theory, but honestly, it's hard to understand the intuition behind what the result means without a stronger mathematical foundation.
Comment: Group theory is probably helpful (you don't need too strong an understanding of group theory: a passing knowledge of what a free group is should suffice).
Comment: You might find it helpful to begin with [**The Pea and the Sun. A Mathematical Paradox**](https://www.amazon.com/dp/1568813279) by Leonard M. Wapner (2005), and then ask more specific background questions as you read through Wapner's book. By "more specific", I mean that in addition to specifying a particular topic (cardinality of sets, group theory, rotation matrices, etc.), by providing more specific information about what you do know and what you don't know about that topic.
Answer: I don't think you need too much in terms of prerequisites.
An excellent reference is

MR3616119. Tomkowicz, Grzegorz(PL-CEG2); Wagon, Stan(1-MACA-NDM). The Banach-Tarski paradox. Second edition. With a foreword by Jan Mycielski. Encyclopedia of Mathematics and its Applications, 163. Cambridge University Press, New York, 2016. xviii+348 pp. ISBN: 978-1-107-04259-9.

The book is essentially self-contained, and it may make more sense (be more efficient) to read it and look at  other sources as the need appears. You need some group theory, as the paradox depends on properties of free groups, but what is needed is explained in the book. You also need the axiom of choice, but the way it is used is fairly direct. I suppose one needs a little about the basics of measure theory, to understand why this is a "paradox", but this is also introduced in the book. (Deeper knowledge of these and other topics is required as the book goes on, but for the basics you can essentially just delve in.)
A nice feature of the book is that, while remaining user-friendly, it puts the paradox in context, relating it to other topics, some more advanced or even active areas of research (amenability, for instance). Another is that it is basically up-to-date. I know of only one (technical, but really neat) result that was proved after this edition came out (the solution by Marks and Unger of one of the open problems listed in the book: a constructive proof of Laczkovich's result).


Representation theory of locally compact groups

Question: My knowledge about representation theory of locally compact groups is rather scattered. As I got more interested with this subject, I would like to know some good references, where I could learn the basic facts of this theory. So I'm looking for the book (or books, if there is no one souce which can cover the material which I want to learn) which will contain:
1. The result that each irreducible representation of compact group is finite dimensional.
2. The theorem which states that for any locally compact group irreducible representations separate the points.
3. Discussion about how the topology on the set of $Irr G$ (classes of) irreducible representation is defined.
4. The answer to the question: if $G$ is compact then $Irr G$ is discrete and if $G$ is discrete then $Irr G$ is compact.
5. Definition of the Plancherel measure.
6. The theorem which states that for any compact group $G$ we have:
a) every representation of $G$ splits into direct sum of irreducible ones
b) every irreducible representation is contained in left regular representation.
I will be very grateful for any help.
Answer: A lot of this would fall under the subject of harmonic analysis; I learned a large part of what I know of it from Anton Deitmar's Principles of Harmonic Analysis. This covers the compact case completely (here, everything is essentially contained in the Peter--Weyl theorem), and roughly sketches the general Plancherel theorem. This is a pretty difficult topic, and the main reference for most of the proofs seems to still be Jacques Dixmier's $C^\ast$-algebras and Von Neumann Algebras. Unfortunately, I can speak from experience when I say that these are pretty difficult books to read, and that his treatment of the Plancherel theorem relies on huge amounts of material from both of them.
As far as I'm aware, in the non-compact case you can't really put a topology on $Irr\ G$ in a sensible way. This is where $C^*$-algebras come in; you instead look at the dual of the group's associated $C^*$-algebra which gives you something to work with. Then, in the compact case I've never personally seen anyone put a topology on this set -- that's not to say that it isn't done at all. Certainly though, in my experience I guess that we essentially put the discrete topology on it, which speculatively answers question 4. It's never occurred to me to try and work out how this fits in with the $C^*$-algebra approach until now, and it's now long enough since I read about this stuff that I doubt I could say anything sensible.
Let me quickly give a few brief answers to some of your simpler questions. There's a discussion on means of showing that irreducible reps of compact groups are finite dimensional here, including some ways of doing it without the Peter--Weyl theorem. The theorem that irreducible representations separate points is the Gelfand--Raikov theorem. I don't have my copy to hand, but I suspect that a proof is in Deitmar's book. If not, it's a very well-known result, so a reference shouldn't be too hard to turn up. Semisimplicity of representations should also be established during the proof of the Peter--Weyl theorem. That every irreducible representation is contained in the regular representation is Frobenius reciprocty (the regular representation is isomorphic the the representation obtained by inducing the trivial representation of the trivial subgroup to the entire subgroup).
That just leaves that definition of the Plancherel measure. As I mentioned above, this is in general extremely non-trivial. Here's an answer in the form of the Plancherel theorem, which at least gives a non-constructive definition:

Let $G$ be locally compact, second countable, unimodular and type 1 (a condition on the $C^*$-algebra of $G$), equipped with a choice $\mu$ of Haar measure. Then there exists a unique measure $\hat{\mu}$ on the $Irr\ G$, such that one has a direct integral decomposition
  $$L^2(G)\simeq\int_{V\in Irr\ G}^{\oplus}HS(V)\ d\hat{\mu}(V),$$
  where $HS(V)$ denotes the Hilbert--Schmidt space associated to V.
Comment: As far as I remeber the topology on unitary dual is called Fell topology. I found somewhere on mathoverflow its definition but still I would like to read further about this. Moreover, apparently there must be some topology on the unitary dual since I found for example the following equivalent definition of so called property T "trivial representation is an isolated point in the unitary dual"
Comment: Yes, the Fell topology is defined on the dual of the group's $C^*$-algebra. There's then a natural bijection between this algebra and the unitary dual of $G$, so I suppose you can define a topology on the unitary dual in this way. As far as I know, this is the only way in which the dual is usually topologized. However, to actually be able to say much at all about this (apart from in very simple examples, such as when $G$ is abelian) I think that you'd really need to learn about what's going on on the $C^*$-algebra side of things.
Comment: Ok, so it appears that this is just the same definition as for $C^*$-algebras. I have some familiarity with the notion of group $C^*$-algebra. If you have a discrete group $\Gamma$ then you can form two $C^*$-algebras $C^*_u(\Gamma)$ and $C^*_r(\Gamma)$ called universal and reduced (resp.) C*-algebras of group $\Gamma$. Both of them are unital therefore their spectra are compact (not necessarily Hausdorff) by a general theorem which can be found in Dixmier's book. The construction of $C^*$-algebra of group can be performed also for nondiscrete groups so the question whether the ...
Comment: ... obtained spectrum for a _compact_ group $G$ will be discrete is still meaningful and (at least for me) interesting. Note that in this more general setting $C^*$-algebras of the underlying group no longer have to be unital.
Comment: By the way, his name (second line) is [*Anton Deitmar*](http://www.math.uni-tuebingen.de/user/deitmar/), not "Dietmar".
Answer: Gerald  B. Folland, a course in abstract harmonic analysis, is the best source and contains the answers to your questins.


04 pontiac grand am wont start

Question: Everything is working fine, but just getting a single click sound when trying to start, could be security system but don't think security light is on. Any solutions? Jump start didn't work either.
Comment: Welcome to Motor Vehicle Maintenance & Repair!
Answer: If you are only getting a single click noise from the starter and a jump start doesn't make a difference, more than likely the solenoid on the start has gone bad or could be the brushes in the starter motor.
There is a thick copper washer at the back of the solenoid which allows power to be transferred from the positive cable of the battery directly to the starter. If this goes bad or gets to the point where it won't connect anymore, the starter will make behave like what you're talking.
Less likely, but still a concern, is if the brushes in the starter are worn out, they can no longer allow electricity to pass through and energize windings, which will also keep the starter motor from spinning.
Either way, I'd suggest your problem lies in the starter.
Comment: It was the stater, gave it a couple taps and it started right up, could it be that's it's going bad?
Comment: @user73107 - Absolutely.

04 pontiac grand am wont start


Integral of measurable function

Question: Suppose $E\subset \mathbb{R}^d$ is a measurable set with finite measure $m(E) < \infty$. Let $f_n: E \to \mathbb{R}$ be a sequence of nonnegative measurable functions such that $f_n(x) \to f(x)$ almost everywhere in $E$. Prove that $$\lim\limits_{n \to \infty}\int_E \frac{f_n(x)}{1+f_n(x)}dx = \int_E \frac{f(x)}{1+f(x)}dx.$$
I'm trying to prove this equality in preparation for an exam of real analysis. I would appreciate your help in solving this problem. Thank you.
Answer: Hint:
$$
\frac{f_n(x)}{1+f_n(x)} \leq 1,
$$
and since $m(E)<\infty$ we can pass to the limit (why?)
Comment: I believe we can pass to the limit by the bounded convergence theorem. Thanks for the hint.

Integral of measurable function


Third hand damages insulation

Question: My third hand damages wires insulation because it's holding them too tight. Any tips how to avoid this?
Comment: Put some heat-shrink on the clips?
Answer: Wrap the teeth in heat-shrink tubing or electrical tape. It will still hold just as tightly, but without the teeth, the pressure on the wire being held will be less.
Answer: I like to tightly wrap rubber bands around the back of the clips to take some of the pressure off. That way, you can adjust the pressure to your liking. Sometimes I use the alligator clips to hold a pair of cheap soldering heat sink clips and use those to actually hold more delicate components.
See the heat sink clip in this picture:
Answer: I have the same tool; I just bent the teeth on the clips inwards using a pair of pliers.
Comment: If the O.P. goes the heat shrink route, he'll be able to remove the heat shrink and restore the holder to its original state.  Bent teeth are more permanent.
Answer: You could also change out the alligator clips for a spring clip instead. I found some small clips similar to this at a hardware store (http://www.homedepot.com/p/HDX-3-4-in-Mini-Spring-Clamp-80001/100656782?MERCH=RV-_-RV_nav_plp_rr-1--NA--100656782-_-N). They still squish the insulation a bit, but at least they don't bite into it like an alligator clip does.

Third hand damages insulation


How to make "rm" require root privilege?

Question: I often misuse "rm" for "mv"! I hope to make "rm" a command requiring root privilege, like "apt-get". How to do that, please? My system is Ubuntu 10.10.
Comment: Create a shell alias for `rm`: `'rm -i'`
Comment: It's not the answer you want, but you should take care and think before pressing enter, especially around dangerous commands like `rm`
Comment: I agree with @CJBrew. And i think if you keep typing `rm` when you mean `mv`, and don't have backups or a versioning system (http://git-scm.com/) or at least a recycle bin or something (http://pages.stern.nyu.edu/~marriaga/software/libtrash/), you kind of deserve to lose files.
Comment: You could also alias rm to be "sudo rm"
Comment: This is off-topic for StackOverflow. It should be asked on http://serverfault.com or http://askubuntu.com
Answer: The best solution is to use rm -i
Answer: Definitely aliasing the interactive version of rm is the best solution, because you wouldn't want normal scripts to break when rm suddenly starts to behave differently. Just add alias rm="rm -i" to your ~/.bashrc and you'll get a question like rm: remove regular file 'filename'?, which you'll have to answer with y [ENTER] or n [ENTER].
I think it is better that aliasing sudo rm, because you are explicitly told which files are to be deleted, while with sudo you're only asked for a password like on dozen other occasions and before you know it you will approve the command sudo rm *. The command rm -i protects you from such (and less obvious) too greedy wildcard matches, because it asks you about every single file.
Answer: How about alias rm to 'sudo rm'? Every time you need to enter the password.
Comment: You hardly want to run `rm` with escalated privileges. I would go for something like `alias rm='sudo true && /bin/rm'` or just the good old `rm -i`. With `sudo` you get to run without the password for a while, whereas `rm -i` will always prompt.
Answer: You could alias rm to something like echo 'Do you really want to remove this?'
This will prevent you from using rm.
Then, if you really want to remove something you will need to type /bin/rm (to bypass the alias).
Answer: This is off the top of my head. I'm sure folks will correct edit if I have my syntax off or something.
Insert this script in your ~/bin and call it rm.
#!/bin/sh

# Run /bin/rm using all the arguments from the command line
/bin/rm $@

Make sure that ~/bin is first in your $PATH so your rm is found before /bin/rm.
Set that script to ownership root:your_group and permissions 760 so that you must sudo in order to execute rm and you will not need to be root to write or read the file. You can then also get around this script by simply using /bin/rm instead of rm, but you'll do that knowingly. Maybe you'll even get in the habit of specifying rm by absolute path and eventually will do without the little extra script.
If you aren't familiar with setting your own path variable, edit your .bash_profile with these lines (or edit if they already exist).
PATH=~/bin:$PATH
export PATH

You'll need to restart bash or use the command 'source .bash_profile' to bring your changes in to the current shell.
(Of course, rm -i is much simpler and less breaky than chmoding rm itself, but this answers the question as asked without breaking anything that isn't run by the user who already knows of the limitation.)
Answer: While you could use chmod to give only the root user execute permissions, this would likely create many problems on the system. A better choice would be to alias the command in your .bashrc file by adding the following line:
alias rm="rm -i"
This makes the command interactive and you will then be asked to confirm each deletion.
Thanks to other posters for pointing this out.
Comment: -1 This approach will wreck any system command which invokes `/bin/rm` behind the scenes (basically any shell script which uses temporary files, among others)! Go with the alias instead.
Comment: This is likely to break things.
Comment: `rm` needs to be usable by applications which created their own files, are owned by them and shouldn't need root priveleges to remove them. Use `rm -i`
Comment: Normally I wouldn't edit an accepted answer, but I've learned a lot over the years and having just come across this again today I'm a bit horrified about what I originally wrote. The original answer is still contained, but with a warning and the better suggestions, with thanks.
Comment: Exactly it is !


KNN classifier + cross validation

Question: how can I find the mean and standard deviation of error rate or accuracy of a k- fold cross validation performing K-nearest-neighbour classification model for each fold?
Answer: The mean and standard deviation of you metrics are calculated across results of all cross validation (CV) partitions. So, if you have 10 CV partitions with 10 repeats you will obtain 100 sets of metrics, which in turn are used to compute the mean and standard deviation of each metric. This is not limited to KNN but applies do all models used with CV, therefore this should also answer your other question.
Assuming you are using a software like R: this is computed by the software already, so no need to do this on your own. For the purpose of understanding, here's a minimal working example on how to calculate it by hand anyway:
<code>> library(caret)
> m <- train(iris[,1:4],
>            iris[,5],
>            method = 'knn',
>            tuneGrid = expand.grid(k=1),
>            trControl=trainControl(method='repeatedcv',
>                                   number=10,
>                                   repeats=10))
> print(m)
    [...]
    Resampling results

    Accuracy  Kappa  Accuracy SD  Kappa SD
    0.96      0.94   0.0454       0.0682

> head(m$resample) # performances for individual partitions
    Accuracy Kappa     Resample
    1 0.9333333   0.9 Fold01.Rep01
    2 1.0000000   1.0 Fold02.Rep01
    3 1.0000000   1.0 Fold03.Rep01
    4 1.0000000   1.0 Fold04.Rep01
    5 0.9333333   0.9 Fold05.Rep01
    6 1.0000000   1.0 Fold06.Rep01
    [...]

> print(apply(m$resample[,1:2], MAR=2, mean)) # calculate mean/sd yourself
    Accuracy    Kappa
        0.96     0.94

> print(apply(m$resample[,1:2], MAR=2, sd)) # calculate mean/sd yourself
    Accuracy      Kappa
    0.04544332 0.06816498
</code>

KNN classifier + cross validation


Windows 2012 R2 - RDP properties?

Question: Where did the Remote Desktop Services properties go in 2012 R2?  In 2008 R2, we had the RDP-TCP connections properties... where did all that go in 2012 R2?
Answer: It went to Servermanager -> Remote Desktop Services -> Collections -> $collectionname -> The "tasks" button on the properties field (upper left):
Comment: Yes, but this provides only about half of the properties that were available under RDP-TCP properties.  Missing are things like the port and the environment tab.  I can edit the registry if I want to change the port... but what about the Environment tab?  It provided a useful place to disable the Explorer shell for RemoteApp users.
Answer: I had this same issue, it appears that it is not longer a Administrative Tool for managing your server.  This will need be handled by using Group Policy.  To access it hit the Windows button and type "Edit Group Policy" to access the Local GPO.
The settings are located here:
Computer Configuration\Administrative Templates\Windows Components\Remote Desktop Services
Source: https://technet.microsoft.com/en-us/library/dd834779.aspx

Windows 2012 R2 - RDP properties?


can lyx remember the most used math commands?

Question: While using lyx, I see that it remembers my last used math command. For example, if I write \Phi, and then right after that if I begin to write \P, it will show me in the suggestion Phi so that I can just press Tab and complete it. But I would like it to remember my most used math commands in general and not just the last used one. For example, I never use \Game but I frequently use \Gamma. So, when I start writing \Ga, it won't pop-suggest \Gamma but instead, it pops with the suggestion \Game which I don't ever need. Is there a way for lyx to have just the most used math commands in it's auto-complete suggestions ?
Answer: No, LyX's math auto-completion is currently not smart enough. There is an enhancement request for what you would like here. If you would like, you should register on LyX's trac, and CC yourself in the issue, and add any comment to clarify the enhancement request. By doing that, you bring attention to the issue and it is more likely to catch an eye of a LyX developer who might decide to work on that issue. If you happen to have any C++ knowledge, patches would also be welcome.


How to reconcile the size differential between giants and non-giants in Norse mythology?

Question: In particular, is there any thinking on how things might work between Loki and Angrboða, who produced significant offspring? I'd also be interested in other examples of giant/non-giant matches.
From a more general perspective, I'm interested in how the size differential between the gods and giants is reconciled in stories where they interact, for instance in the third ríma of the Lokrur. Even if illusion were utilized to make Jörmungandr appear to be a cat, how can the world serpent fit into a hall?
Comment: Outside of mythology, see also https://scifi.stackexchange.com/q/27772/4918 "In the potterverse, how do giants procreate with humans?"
Comment: @b_jonas lol. If this is a subject you have any interest in, I'd highly suggest Gene Wolfe's [The Wizard Knight](https://en.wikipedia.org/wiki/The_Wizard_Knight).  He gets into the subject of human/giant intimate relations in some detail.  *Quite salty!*
Comment: My answer there should make it clear that I'm not interested to know the intimate details. I know just enough to be almost sure such a sexual relationship is possible. (Whether they can procreate this way is a different matter.)
Comment: @b_jonas Wolfe is always worth reading in any case, and that subject constitutes only an aside in a very cool story that involves an excellent take on the division of the universe into connected world/realms, such as in the Norse mythology.  *([Surt](https://en.wikipedia.org/wiki/Surtr) also make an appearance;)*
Answer: Well, let's start with the obvious: where the jotuns in fact, giants?
(I will be using "jotuns" for the group throughout, to have avoid confusion).
This seems likely, on at least some level; etymologically, the word means "(great) eater", so it seems like that they were always associated with large size.
Now, the source which probably is best for size comparisons between giants and gods is the story in Gylfaginning about Thor's journey to Útgarða-Loki, which he does in company of (ordinary) Loki and Tjalfi and his human servants Þjálfi and Röskva. On their way, they meet up with the jotun Skrymir, which is much, much larger than them: they in fact mistake his gauntlet for a house. This indicates that jotuns other than Ymir can be really gigantic. It also does indicate that there is no significant difference in size between humans and gods.
When they arrive at Útgarða-Loki's castle, it is also enourmous. Even Thor can not open the gate, but they manage to squeeze in through the trellis. However, when they meet the inhabitants, it is only noted that they are "big enough". Finally, as Thor is asked to wrestle with Útgarða-Loki's "grandmother", she can not have appeared to be so much larger than him that it would be impossible.
It should be noted that Útgarða-Loki is a master of illusion, so that while the sizes of the giants where probably not surprising to the party, we can not take the size of anything else for granted.
As for other pairings of gods and giants, we should first note that all the Aesir are descended from jotuns, many born by jotun mothers (including Thor). Loki is in fact seen as a jotun. A few notable examples are Njord and Skaldi, Frey and Gerðr, Thor and Járnsaxa, and Odin and Gunnlöð.
Sources

Snorri, Gylfaginning, in translation by Karl G. Johansson and Mats Malm
SAOB, for the etymology of giant.
Gro Steinsland, Fornnordisk religion, for double checking details.
Answer: Norse mythology doesn't, outside of a few specific instances, actually specify the size of the giants (or the Aesir, for that matter) as far as I'm aware.
As far as other giant/non-giant matches, there's Skadi and Njord, Borr and Bestla, Freyr and Gerd.
As far as other size differentials, this varies but story, but in general, Norse mythology is rife with magic and shape-shifting that could explain most examples.
Comment: This holds with my take on the matter--never in my reading have I come across this as issue, which leads me to believe it wasn't an issue for the original audience.  Thanks for contributing and welcome to Mythology!
Comment: In the story of Hadingus, his nanny [Hardgrepir](https://en.wikipedia.org/wiki/Har%C3%B0greipr) is a giant, who later becomes his lover. He objects that she's larger than him, but she tells him that she can [adjust her size](http://www.sacred-texts.com/neu/saxo/saxo01.htm) to make things work: So that's one possibility...
Answer: Giants are not beholden to our rules of geometry
Myths and fairy tales include all kinds of impossible things like mermaids turning into sea foam, women selling their eyes to rivers, giant castles built on the clouds, scarecrows spontaneously coming to life with speech capability, men surviving dismemberment by replacing their entire body with tin, and household objects transforming into geographic features. Absolutely none of the rules we take for granted in reality will apply here, and that is just as terrifying and wondrous as it sounds.
The inconsistent size of giants in myths and fairy tales makes them difficult to put a handle on. Nay, they defy geometry entirely. They may be described as arbitrarily large, but this size never inconveniences them when they want to interact non-destructively with anything smaller. Actually changing size is virtually never mentioned, and would not make sense in many examples where the giants perform multiple tasks at once that each require different sizes. The most extreme and telling convention I remember would be that a giant can have a wife who fits in the palm of his hand yet experience no difficulty having kids (see the book Monsters of Greek Mythology by Bernard Evslin, section "Amycus" for a clear-cut example).
The 1952 Jack and the Beanstalk movie, due what I assume were technical limitations or a stylistic choice, depicted the giant as being a tall human who had the capabilities of something vastly larger. He stole the food in the land thereby leading to mass starvation, his visits were accompanied by thunder and eclipsing the sun, and his footprints and shadow were comically huge. Although it may come off as silly and unbelievable to modern audiences what with our tendency to overthink the logic of fantasy worlds and our desensitization to expensive CGI spectacle (plus the movie was panned by critics), this movie does convey a visual analogue to how giants were idiosyncratically treated in myths and fairy tales.
In fact, the same thing applies to human heroes in myths and fairy tales. They can wrestle with giants many times their size despite that being physically impossible. For that matter, heroes can do other physically impossible stunts like splitting mountains in twain, swallowing the ocean and running over multiple different geographies in the span of one scene.
The only logical explanation I can think of is that giants and heroes are simply following non-Euclidean geometry. Space and time simply hold different meanings to them compared to us mere mortals. Naturally, this is difficult for human minds to imagine and nigh-impossible to portray in a visual medium. The movie The Ritual is probably one of the few examples that pulls this off, and predictably it is a horror movie rather than a traditional fantasy romp.
Comment: I think you're right on the money here. Fables and legends, which make up mythology, are inconsistent because they're invented stories. None of it really happened, and original audiences, perhaps excepting some unusual duck, expected it to "make sense" upon scrutiny. These stories are beyond scrutiny!


Removing GRUB from windows system after uninstalling Ubuntu from Windows 7

Question: I have installed Ubuntu alongside Windows 7 (no separate partitions, I selected inside windows option).
I want to uninstall Ubuntu but last time I tried it and the Ubuntu is released but the GRUB remains. How do I remove GRUB?
Answer: You need Windows 7 installation disk, boot it, choose repair option when you are asked to menu.
It will check automatically for start up problems. Let it check it.

It will not find any problem with start up.
It will allow to choose options. Choose command prompt.
In cmd you have to type just two following commands and grub will be removed.
bootrec /fixmbr
bootrec /fixboot

Close cmd and restart your machine.
Answer: Here's how.

Download and burn the Super Grub Disk ISO to a CD.
Boot from it.
Select the Choose Language & HELP  :-)) entry.
Select the English entry.
Select the Windows entry.
Select the Fix Boot of Windows entry.
Reboot once the process is complete.

No more Grub! :)
Comment: Better get Rescatux from here http://www.supergrubdisk.org/category/download/rescatuxdownloads/rescatux-stable/ View: https://www.youtube.com/watch?v=kpJoNHjiOOs
Comment: The link in this answer is broken.
Comment: Rescatux is now the way to go, or Boot-Repair.. This answer was on point when it was written, three and a half years ago. :)
Answer:
Put windows 7 disk and boot from it.
Choose to repair windows now.
It will ask to repair, select yes.  After restart if grub is still there, follow above    step.
Select windows partition found by the recovery manager and click Next
Now choose startup repair.

This internally uses command fixboot and fixmbr. You can google for more details.


Replacing two toggle switches that are linked with the same grounding wire

Question: The bathroom light switches are wired to independent red and black switches but are connected together by a copper grounding wire that is looped around one screw and then terminates on the second toggle. Can I take both switches out and just duplicate the setup with the new toggles? How do I know I bought the right switches?
Answer: If the switches have the same number of screws you should be all set. Loop the ground as it is now you should be OK. If there are three screws on one or more of the switches look for the one screw that is identified, the screw will be a different color. Be sure to take pictures before you remove anything.


CakePHP Auth->login not generating query for request data

Question: I have the Auth component working perfectly for an application running Cake 2.4.1, and I need to copy it to another application that is also running Cake 2.4.1. I copied over everything I could think of that has to do with Auth stuff, but it just does not work in the second application. It refuses to log me in. The call to $this->Auth->login() fails even though it gets exactly the same data as the application that works.
Both applications are using exactly the same database and user login.
Both applications get exactly the same request data in the POST request.
First of all, here is my setup.
In AppController.php:
public $uses = array ('User');
public $components = array    ( 'Session',
                                'CPRACL.CPRACL' => array (),
                                'Auth' => array (
//                                        'loginAction' => array ('controller' => 'users', 'action' => 'login'),
//                                        'logoutRedirect' => '/users/login',
//                                        'loginRedirect' => '/',
                                        'authenticate' => array (
                                            'Form' => array (
                                                'fields' => array ('username' => 'email_address'),
                                                'scope' => array ('User.active' => 1)
                                            )
                                        ),
                                        'authorize' => array ('Controller'),
//                                        'unauthorizedRedirect' => '/' // when going to a page the user is not authorized to go to
                                    )
                                );

The commented out lines use the defaults, but they are included for reference to their default values. This is exactly the same in both applications. CPRACL is a plugin component that I am using instead of Cake's ACL component, but it fails long before it gets to any of that.
Also in AppController.php:
public function beforeFilter ()
{
    $login_user = $this->User->find ('first', array ('conditions' => array ('id' => $this->Auth->user ('id'))));
    if (!empty ($login_user))
        $this->set ('login_user', $login_user ['User']);
}

public function isAuthorized ($user = null)
{
    // This never gets called in the application that does not work, but it does get called in the one that works.
}

In both applications, it sets up the Auth component as defined in AppController.php, then it calls beforeFilter to check to see if the user is already logged in. This is a convenience so that when they get redirected to a different page, the application can display the username of the user who is logged in. It works fine and I don't think it is causing any problems.
The next thing that happens is it calls UsersController::login()
    public function login ()
    {
        $redirect_url = $this->Auth->redirectUrl ();
        if ($this->Auth->loggedIn ())
            return $this->redirect ($redirect_url);

//die (print_r ($this->request->data, true));
        if ($this->request->is ('post'))
        {
//die (Debugger::dump ($this->Auth));
            if ($this->Auth->login ())
                return $this->redirect ($redirect_url);
            else
                return $this->redirect ($this->Auth->loginAction);
        }
    }

I used the two commented out lines to compare what is happening between the two applications. The two applications get exactly the same data in the post. For the dump of the Auth object, here is what I get:
object(AuthComponent) {
    components => array(
        (int) 0 => 'Session',
        (int) 1 => 'RequestHandler'
    )
    authenticate => array(
        'Form' => array(
            [maximum depth reached]
        )
    )
    authorize => array(
        (int) 0 => 'Controller'
    )
    ajaxLogin => null
    flash => array(
        'element' => 'default',
        'key' => 'auth',
        'params' => array([maximum depth reached])
    )
    loginAction => array(
        'controller' => 'users',
        'action' => 'login',
        'plugin' => null
    )
    loginRedirect => null
    logoutRedirect => array(
        'controller' => 'users',
        'action' => 'login',
        'plugin' => null
    )
    authError => 'You are not authorized to access that location.'
    unauthorizedRedirect => true
    allowedActions => array(
        (int) 0 => 'register',
        (int) 1 => 'login',
        (int) 2 => 'logout'
    )
    request => object(CakeRequest) {}
    response => object(CakeResponse) {}
    settings => array(
        'authenticate' => array(
            [maximum depth reached]
        ),
        'authorize' => array(
            [maximum depth reached]
        )
    )
    Session => object(SessionComponent) {}
    [protected] _authenticateObjects => array()
    [protected] _authorizeObjects => array()
    [protected] _user => array()
    [protected] _methods => array(
        (int) 0 => 'index',
        (int) 1 => 'view',
        (int) 2 => 'add',
        (int) 3 => 'register',
        (int) 4 => 'login',
        (int) 5 => 'logout',
        (int) 6 => 'toggleactive',
        (int) 7 => 'changePassword',
        (int) 8 => 'passwordHash',
        (int) 10 => 'edit_pwd',
        (int) 11 => 'edit_your_pwd',
        (int) 12 => 'isAuthorized'
    )
    [protected] _Collection => object(ComponentCollection) {}
    [protected] _componentMap => array(
        'Session' => array(
            [maximum depth reached]
        ),
        'RequestHandler' => array(
            [maximum depth reached]
        )
    )
}

That is what both applications return from the Auth object dump when I send the login request. It says the user is not authorized to access that location, but I think that is because it hasn't actually logged the user in yet, so I don't think the dump of that Auth object is of much value, but I included it here anyway.
But here is the most important part because it is the only thing that is obviously different between the two applications. When I perform the login on the working application, it logs me in successfully and my layout (which displays the most recent query) shows this query. Notice the id field. The id field is correct in the query and it works.
SELECT `User`.`id`, `User`.`user_id`, `User`.`email_address`, `User`.`first_name`, `User`.`last_name`, `User`.`password`, `User`.`active`, `User`.`created`, `User`.`modified` FROM `brian_cake_test`.`users` AS `User` WHERE `id` = '529f9a8c-3460-46a2-a97c-6a6242a26b88' LIMIT 1

But when I do exactly the same thing for the application that does not work, everything happens exactly the same way except that it redirects me back to the login page, which it should do if it fails to login, and it displays this query in my layout.
SELECT `User`.`id`, `User`.`user_id`, `User`.`email_address`, `User`.`first_name`, `User`.`last_name`, `User`.`password`, `User`.`active`, `User`.`created`, `User`.`modified` FROM `brian_cake_test`.`users` AS `User` WHERE `id` IS NULL LIMIT 1

Notice how the id is NULL in the second query. I have no idea why this is happening.
My User.php model is just an empty class, exactly the same in both application, but one works perfectly and the other does not.
<?php
App::uses ('AppModel', 'Model');
class User extends AppModel
{
    public $name = 'User';
}

I have racked my brain trying to figure out what is different between these two applications, and I can't think of anything, and nothing I try will make it work. I don't want to post hundreds of lines of code when I know that 99% of it has nothing to do with this problem, but I have no idea why it works in one but not the other. I will gladly post more information if you think you might know what the problem is. Any help would be much appreciated. This is very important to me.
Answer: Actually if its exactly the same code and DB it shouldn't be a reason not to work.
Have you tried clearing the cache folders? Those files sometimes mess up the app.
Comment: I did try clearing the cache, but it did not solve the problem and I never figured out why it wasn't working. There must be something configured wrong somewhere, or a conflict, and I never figured it out. I wound up copying the working application to a new directory and tediously moving all the functionality into it. It took me a whole day, but all the features seem to work now. Unfortunately, I have no information to offer about what was causing the problem.


Restoring backup from a stolen iPhone

Question: I am trying to restore my backup but it is asking to turn off "Find my iPhone" on the old device, which was stolen. How can I restore my backup?
Comment: What does a missing / stolen device have to do with restoring a backup onto a device you have in your hand? Are you really asking how to detach the current device from iCloud? Whether other devices are locked to iCloud doesn't affect a new device restore - just the status of the device in question matters. (basically restoring a backup **from a stolen phone** has zero hoops to jump through - restoring backup **to a stolen phone** requires you to get the owner to release the reservation).
Answer: Well I guess this might help you : iCloud: Remove your device from Find My iPhone
See :
Remove an iOS device you no longer have
If you no longer have the iOS device because you gave it away or sold it, you need to remotely erase it before you can remove it.

Sign in to icloud.com/#find with your Apple ID (the one you use with iCloud). If you're using another iCloud app, click the app's name at the top of the iCloud.com window, then click Find My iPhone.

Click All Devices, then select the device.

Click Erase device, then enter your Apple ID password. Because the device isn't lost, don't enter a phone number or message.
If the device is offline, the remote erase begins the next time it's online. You'll receive an email when the device is erased.

When the device is erased, click Remove from Account.
All your content is erased and someone else can now activate the device.
Comment: I'm doing exactly this, however, it is still saying to switch off find my iphone!

Restoring backup from a stolen iPhone


Changed base URL in Magento 1.7 and can't log in now

Question: I changed my base URL from a local IP address to a hostname. Magento loads everything fine but when I attempt to login to the admin section the page refreshes without an error. If I type in the wrong password then the page refreshes and displays invalid credentials message.
Answer: It looks like this is the classic cookie error. Remove all the cookies from the previous domain or try to login with another browser/incognito mode. Also remember to change the cookie path in your DB. Check this link to achieve that.
Answer: I've been pulling my hair on this one a bunch of times for different reasons (Admin login loop after changing BASE URL / SECURE URL settings)
This allows me to log in everytime, fix whatever I did wrong, clear all cache, sessions, etc & undo the changes:
go to app/code/core/Mage/Core/Model/Session/Abstract/Varien.php and comment out those 3 lines (in 1.7 it's from line 89 to 91)
    $cookieParams = array(
        'lifetime' => $cookie->getLifetime(),
        'path'     => $cookie->getPath()
//      'domain'   => $cookie->getConfigDomain(),
//      'secure'   => $cookie->isSecure(),
//      'httponly' => $cookie->getHttponly()
    );

Comment: This worked perfectly fine for me.
Answer: Check if the cache isn't in /tmp/magento instead of [magento root]/var/cache/
Check the permissions, user and group of [magento base dir]/cache folder.
If this is it, fix it and clear cache from dir var/cache.
Answer: Log into your database with a Mysql Client (I recommend Mysql Workbench), and issue this query:
select * from core_config_data where path like '%base_url%';

You will get 2 rows for that query (a secure (for https:// access ideally) and insecure URL), what is the value of value column ?


Mining In Remix IDE

Question: Normally in MIST, we have to mine by miner.start() to deploy the smart contract. But in Remix we don't need to mine. Can someone tell me why is that ?
Answer: Mist starts a geth node, if not already running on your machine. If you are working with main-net, you need to worry about mining. Just send your tx and miners already out there will mine your tx.
But if you are working with private net (setting up your local PoW blockchain). You yourseld need to mine each tx you send. This is irrespective of whether you are using remix or Mist.
If you are using Javascript VM as your environment in the remix. You need not to worry about mining. But if you are connecting remix to custom web3, you need to use same command miner.start(theradNo) on your geth to mine the txs.

Mining In Remix IDE


Slot size bound for chaining

Question: This is a question from CLRS
Q) Suppose that we have a hash table with n slots, with collisions resolved by chain-ing, and suppose that $n$ keys are inserted into the table. Each key is equally likely to be hashed to each slot. Let $M$ be the maximum number of keys in any slot after all the keys have been inserted.
a) Argue that the probability Q$_{k}$ that exactly $k$ keys hash to a particular slot is given by
Q$_{k}$ = ($\frac{1}{n})$$^{k}$(1-$\frac{1}{n}$)$^{n-k}$ $\binom{n}{k}$.
Which is pretty straight forward as hashing of a particular key at any slot is binomial trial with prob. of success = ($\frac{1}{n}$) and prob. of failure = (1-$\frac{1}{n}$) . So according to binomial trial
probability that exactly k keys hash to a particular slot is
Q$_{k}$ = ($\frac{1}{n})$$^{k}$(1-$\frac{1}{n}$)$^{n-k}$ $\binom{n}{k}$.
b) Let P$_{k}$ be the probability that $M = k$, that is, the probability that the slot containing the most keys contains $k$ keys. Show that $P_{k} \leq nQ_{k}$.
Which is, that a particular slot contains exactly $k$ keys and other slots less than $k$ keys.
Pr (a slot contains less than $k$ keys)= It contains 0 keys or It contains 1 key or It contains 2 keys........ or it contains k-1 keys
Given by
($\frac{1}{n})$$^{0}$(1-$\frac{1}{n}$)$^{n-0}$ $\binom{n}{0}$ + ($\frac{1}{n})$$^{1}$(1-$\frac{1}{n}$)$^{n-1}$ $\binom{n}{1}$+......+ ($\frac{1}{n})$$^{k-1}$(1-$\frac{1}{n}$)$^{n-(k-1)}$ $\binom{n}{k-1}$.
So $P_{k}$ = $Q_{k}$ . Pr (a slot contains less than $k$ keys)
           = $\left(\frac{1}{n}\right)^k\left(1-\frac{1}{n}\right)^{n-k}\binom{n}{k}\times\Bigg(\left(\frac{1}{n}\right)^{0}\left(1-\frac{1}{n}\right)^{n-0} \binom{n}{0} + \left(\frac{1}{n}\right)^{1}\left(1-\frac{1}{n}\right)^{n-1}\binom{n}{1}+\cdots+ \left(\frac{1}{n}\right)^{k-1}\left(1-\frac{1}{n}\right)^{n-(k-1)}\binom{n}{k-1}\Bigg)$
My question is that how do i converge this. ?
Answer: To solve the task you should consider events $E_i$ which is that $i-$slot has $k$ elements hashed into and all other slots have less elements. Thus, $Pr(E_i) \le Q_k.$
From above $Pr(M=k) \le \Sigma_{i=1}^nE_i = nE_i \le nQ_k.$

Slot size bound for chaining


How to read some maths expression in English?

Question:
Hi there! I am learning English and I wonder how to read this maths expression in English.
Comment: can you identify a particular issue?  Can you read this in your native language? Are there any simpler expressions that you can read, can you read any part of this expression?  Please show us some details!
Comment: yeah, I can read this expression in Russian.
Comment: great.... now translate part by part until you get stuck....  so the first thing is "a"  Can you say the name of that letter in English? Yes? Great! The next part is  "²"  Can you say that in English?  No? Okay now you have a specfic question to ask.  This is what I mean by "can you identify the particular issue" and "Please show us some details".
Comment: Скобка открывается, а в квадрате минус двадцать пять поделить на на а плюс три, скобка закрывается умножить, на скобка открывается, один поделить на а в квадрате плюс пять а, скобка закрывается, минус, скобка открывается, а плюс пять поделить на а в квадрате минус 3 а, скобка закрывается
Comment: Here is a translation of this expression into Russian
Comment: But it is better to show the expression to your companion instead of reading it, because he can understand it not correctly
Answer: This expression is difficult to say unambiguously in English (or probably any natural language); mathematical notation is an artificial language designed to be exact within a much more limited domain than is covered by natural language. There is nothing in English like the rule of construing mathematical notation called PEMDAS in US schools. Of course, any mathematical expression can be stated exactly in English if you are very wordy. For the relatively simple expression shown as your example, here goes a rendering that is probably about as concise as possible:

The fraction a squared minus twenty five all over a plus three plus the fraction one over a squared plus five a minus the fraction a plus five all over a squared minus three a.

That will probably be interpreted correctly by most numerate speakers of American English if you are careful to indicate that the numerators of two of the fractions are a difference and a sum respectively. If you want to be absolutely sure of being understood by someone with mathematical training, you can combine the terms fraction, sum, difference, numerator, denominator, minuend, and subtrahend to duplicate the mathematical notation exactly, but the result will be a long paragraph. For example, you can describe

the first fraction, which has a numerator that is a difference, with a minuend of a squared and a subtrahend of twenty five, and with a denominator that is the sum of a and three

Thus, you can describe exactly each fraction in detail that way and put them together as

The expression is a difference. The minuend of the entire expression is the sum of two fractions. The first fraction in that sum is __. The second fraction in that sum is __. The subtrahend of the entire expression is the fraction __.

To conclude, brief descriptions in English of anything more than very simple mathematical expressions have a material probability of being at least somewhat ambiguous. Use the universal language of mathematical notation if you want to avoid any possibility of ambiguity.


Why I can't change font in Label, but in Button i can (Tkinter.ttk)

Question: Well, as in topic, why I can't change font in Label using ttkbootstrap, but in Buttons everything works fine?
And second question, is there some documentation that lists all of the things that I can change in ttk stylesheet? Like eg background color (as in ttkbootstrap is somehow done), because everywhere I searched, was mentioned 'bout background which only changed  a "frame" of a button.
Here's a problematic code:
import tkinter as tk
from ttkbootstrap import Style as StyleBs
import tkinter.ttk as ttk

cfg = {
"args_label" : {
    "style" : "TLabel",
    "text" : "Label 12345",
    },
"args_button" : {
    "style" : "TButton",
    "text" : "Button 12345",
    },
}

if __name__ ==  "__main__" :
    root = tk.Tk()
    style = StyleBs("darkly")

    style.configure('TButton', font=('Times New Roman', 21), foreground = "red")  # foreground is changed, font too
    style.configure('TLabel', font=('Times New Roman', 21), foreground = "red")   # foreground is changed, but font is not

    button = ttk.Button(root, **cfg["args_button"], ).grid(row=0, column = 0)
    label = ttk.Label(root, **cfg["args_label"], ).grid(row=1, column = 0)

    root.mainloop()

Answer: The following code worked for me, if you import Style from tkinter instead of  ttk.bootstrap you may have an easier time.
I'm not sure why the button and label text, as well the style names were in a dictionary, that's new on me.
By removing the dictionary and simply placing the desired text and style name inside the button and label builds, it'll meet your requirements.
from tkinter import Tk, Button, Label, ttk
from tkinter.ttk import Style

root = Tk()

#Assigning variable to Style import
style = ttk.Style()

#Building Button and Label in window
button = ttk.Button(root, text='Button 12345', style = 'ButtonStyle.TButton').grid(row=0, column = 0)
label = ttk.Label(root, text='Label 12345', style = 'LabelStyle.TLabel').grid(row=1, column = 0)

#Configuring style to be used by window
style.configure('ButtonStyle.TButton', font=('Times New Roman', 21), foreground='red')
style.configure('LabelStyle.TLabel', font = ('Times New Roman', 21), foreground = 'red')

root.mainloop()

Comment: Well, I wanted to use ttk.bootstrap cause it just looks nice and it's easier than making hole new style on my own (I tried and it wasn't as easy as I thought). I used a dictionary to contain all the arguments in one place, so I wouldn't have to search each button's or label's args in code to change it. I know it will work when I use just ttk and not a bootstrap but I am currently using bootstrap in my project.  But thank you 4 help <3
Comment: ahh I see what you mean, didn't consider that approach. I'll admit ttk.bootstrap isn't something I'm entirely clued up on either. I'd be interested to see your final code! hope it goes well.
Comment: Yup you can see it on my github https://github.com/blackcaer/calculator-tkinter-ttk.
Nothing special at all, just learning some gui to make a "frontend" to my future apps


db access via interactive tags map

Question: I want to create an interactive tags map, kinda like 'visual thesaurus' (http://www.visualthesaurus.com/), and use it as an a access portal to my DB (display tagged data in a separate area), what do you think would be the best technology to use?
I need something that would be SEO friendly of course, easy to use (user wise), and easy to program...
Any thoughts?
tnx
Liat.
Answer: you talking about the Graph display?
http://www.jayday.org/thesaurus.jpg

db access via interactive tags map


Brightcove Video player not working on Internet explorer 8

Question: I am using Brightcove Video player to my web application. The requirement of this this player is html5 or flash player. Now in I.E. 8 we all know that it partially support html5. I would like to know is there any plugin available which get installed to the client browser before playing the video. If any support related to it let me know in details please note i do not want that browser need to install the flash player.
Comment: The player is likely to fall back to Flash in IE8. Is Flash installed?
Answer: Brightcove delivers Flash to ALL desktop browsers, even those that support HTML5. Brightcove's HTML5 video delivery is reserved for mobile browsers and then only iOS and Android are supported. This was the answer I was given a few months back anyhow.
In any case, IE8 does not support HTML5 video as far as I know, at least not without a shim or some other hack.
Answer: I would like to add something to this response.
As long as you are using their Smart Player API - you are more or less safe for any backward compatibility and as mentioned by Brian - you will receive Flash on desktop UNLESS you force html5 (which is not recommended at all) or you are only using their video api to retrieve the video.
Other than that - he is correct that HTML5 auto switch on mobile only (and HLS live stream delivery on iOS).


Alternative ways to backup Outlook 2010 emails if archive .pst is prohibited

Question: Is there any other tool or work-around in Outlook to backup emails if the exchange server is configured to prohibit archive emails in .pst? I am not sure what would happen if ignoring the rule and keep archiving emails to .pst. Maybe it won't work at all, or exchange server might still can delete the old emails even it's already archived. So I would like to find a more reliable solution.
Answer: Not sure how the serve is limiting PSTs. An administrator can prohibit use of PST files in Outlook or disable auto-archive to PST or disable use of an Archive mailbox. In any case, you may still be able to export to some extent.
From Outlook click File -> Open & Export -> Import/Export to start the export wizard. You can try to export to PST. Worst case would be exporting to CSV. It's not going to maintain full integrity (formatting, for example) when you import from CSV later. If there were just a few item you wanted to save you could just drag them to the desktop and save as individual items.


Selective ultrafilter and bijective mapping

Question: For arbitrary (selective) ultrafilter $\mathcal{F}$ does there exist bijection $\phi:[\omega]^2\to\omega$ with the property: $\forall B\in\mathcal{F} : \phi([B]^2)\in\mathcal{F}$ ?
Answer: No, this fails not only for selective ultrafilters but for all non-principal ultrafilters $\mathcal F$ on  $\omega$.
The main ingredient in the proof is the theorem that, if an ultrafilter $\mathcal U$ on a set $X$ is sent to itself by some map $g:X\to X$ (meaning that $\mathcal U=g(\mathcal U):=\{A\subseteq X:g^{-1}(A)\in\mathcal U\}$), then the set of points in $X$ fixed by $g$ has to be in $\mathcal U$. This easily implies that, if $\phi:X\to Y$ is a bijection and $g:X\to Y$ is an arbitrary map and $\mathcal U$ is an ultrafilter on $X$ with $\phi(\mathcal U)=g(\mathcal U)$, then $\{x\in X:\phi(x)=g(x)\}\in\mathcal U$.
In your situation, suppose toward a contradiction that $\phi$ were a bijection as in the question. Let $\mathcal G$ be the filter on $[\omega]^2$ generated by the sets $[B]^2$ for $B\in\mathcal F$. ($\mathcal G$ may or may not be an ultrafilter, depending on whether or not $\mathcal F$ is selective.) Your requirement on $\phi$ implies that each set in $\mathcal G$ meets each set of the form $\phi^{-1}(A)$ for $A\in\mathcal F$.  Since the intersection of two sets of the form $[B]^2$ for $B\in\mathcal F$ is again a set of that form, and since the intersection of two sets of the form $\phi^{-1}(A)$ for $A\in\mathcal F$ is again a set of that form, it follows that the sets of these two forms generate a proper filter. Let $\mathcal U$ be an ultrafilter on $[\omega]^2$ extending that filter. So we have $[B]^2\in\mathcal U$ and $\phi^{-1}(A)\in\mathcal U$ for all $A,B\in\mathcal F$.
It follows that $\phi(\mathcal U)=\mathcal F$. It also follows that $g(\mathcal U)=\mathcal F$, where $g:[\omega]^2\to\omega$ is the function sending each pair $\{x<y\}\in[\omega]^2$ to its smaller member $x$. By the theorem cited above, we have that the set $E=\{\{x<y\}\in[\omega]^2:x=\phi(\{x,y\})\}$ is in $\mathcal U$. Since $\phi$ is one-to-one, $E$ contains, for any $x\in\omega$, at most one pair whose smaller element is $x$. Thus, there is a function $f:\omega\to\omega$ such that $E\subseteq\{\{x,f(x)\}:x\in\omega\}$. Note that $x$ here is the smaller element (the one given by $g$) in the pair $\{x,f(x)\}$, and so we have  $x<f(x)$ for all $x$.
Consider any two sets $B,C\in\mathcal F$. Since $[B\cap C]^2$ and $E$ are both in $\mathcal U$, they have a nonempty intersection. In particular, there is some $x\in C$ with $f(x)\in B$, i.e., with $x\in f^{-1}(B)$. I've just shown that $f^{-1}(B)$ intersects every set $C\in\mathcal F$; since $\mathcal F$ is an ultrafilter, it follows that $f^{-1}(B)\in\mathcal F$. And since this holds for all $B\in\mathcal F$, we have $f(\mathcal F)=\mathcal F$. By the theorem cited at the beginning, $f(x)=x$ for $\mathcal F$-almost all $x$. This contradicts the fact that $x<f(x)$ for all $x$.
Comment: The theorem used in the solution is due to Katetov (1967), you can find the short paper where it appeared through the link: https://dml.cz/bitstream/handle/10338.dmlcz/105124/CommentatMathUnivCarol_008-1967-3_6.pdf.  You can also find Solovay's rendition of Katetov's theorem through the link: https://math.berkeley.edu/~solovay/Preprints/Rudin_Keisler.pdf
Comment: Thank you for your answer. Where can I read about the theorem you mention in beginning?
Comment: @AliEnayat: Thank you. I got the links. But what about reference of  Solovay's rendition which could be cited in article ?
Comment: @ar.grig To my knowledge, Solovay's rendition has never appeared in print; his argument is "essentially the same" as Katetov's.  As pointed out in the following source, Katetov's theorem can be viewed as a special case of the Bruijn-Erdős theorem: https://dml.cz/bitstream/handle/10338.dmlcz/126182/MathBohem_122-1997-1_9.pdf