Why my Feature Class created with the tool "create feature from XY table" is outside the Feature Dataset?

Question: I'm using ArcGIS 10.2.2 and I'm trying to create a Feature Class inside a Feature Dataset from a table opened through OLE-DB connection using the tool "Create Feature Class from XY Table".
The output Feature Dataset Coordinate System is the same than the Feature Class I want to create (WGS 1984), and , when I run the tool, I don't get any error message.
But the output feature class is created OUTSIDE the Feature Dataset!!
(Still inside the File Geodatabase, but outside the desired Feature Dataset).
Does anyone have an idea why this is happening and how I could avoid this problem ?
Answer: When creating a feature class from a table (or a text file), you cannot export directly into a feature dataset; this is just an existing limitation of the software. but there could be a good reason behind it.
The recommended workaround is to use the Make XY Event Layer GP tool first and then exporting the event layer into a feature class within the feature dataset; this is supported.


Is it faster to have two smaller hard drives or one big one?

Question: Assuming that the drives all have the same RPM and overall same specs except for storage space, would it be faster to install two hard drives together or just one larger one in a computer?
Answer: Yes. Two drives are faster.
As someone already said, you can use RAID 0. This will give you almost double the speed for anything larger than a small file, where small depend on how your raid is formatted. This makes both drives appear as one large, so there is no extra steps to get improved performance once the RAID is setup. The one disadvantage of RAID 0 is that if ONE drive fails, you lose ALL your data.
You can also simply use both drives separately. In this case the performance advantage will depend how they are used. To get maximum advantage you will need to ensure both drives get used simultaneously. If you run two or more disk I/O bound applications, that is relatively easy. If not, you can separate the OS, Swap (if used), applications or data. What to do really depends on the types of applications you use.
Note, do not confuse drive and partitions. Even though Windows presents them as different drives, using two partitions can only slow things down because when they are used simultaneously, the head has to move from one partition to the other since they are separated by cylinders. This does not apply to SSDs though.
Comment: Well, I have not tried most motherboards :) I did manage to get within 15% of the theoretical maximum with a plenty of them as part of a previous position I held.
Comment: I agree with the answer, however factors such as cache sizes and read/write/seek times/etc will factor in the overall performance of the test.  2 *slow* drives might not be as fast as 1 fast drive.
Answer: I had Raid 0 for a while, but noticed only a slight overall improvement. I was disappointed with the whole idea and got SDD for my system and program files, and my conclusion is that no other affordable computer upgrade will gain you so much performance, before I got SSD I really didn't realize how much of bottle neck my HDD was.
Back to the original topic, depending on how you would use you computer, it might be faster if you have two hard drives connected to different controllers on Motherboard, if it has more then one. Sata ports for the same controller usually marked with same color eg. my motherboard has 4 orange ports controlled by Intel chip and 2 red port controlled by J-Micron chip. Connecting both hard drives to the same controller will reduce performance, in my tests, writing speed of a file copy from one HDD to another dropped more then twice compare to setup when hard drives where connected to different controllers.
Comment: RAID performance factors heavily on the controller.  Software RAID will give minimal performance benefits, while hardware RAID is considerably faster.  Also, note that some motherboards with integrated RAID controllers use low end chips with poor performance.
Comment: Good point about the controllers.  My experience is mostly with "big iron" systems where the controllers can generally handle multiple simultaneous transfers, but such may not be the case some with PC controllers.
Answer: It can be useful to have two drives because you can take advantage of parallelisation. For example, you can copy a file from one drive to another drive much faster than copying a file to another location on the same drive.
This parallelisation is also what improves read and write performance with RAID 0, and read performance with RAID 1.
Comment: Thanks! Are there any downsides to performance with using two drives then?
Answer: Two drives will be faster if both drives are often used simultaneously.  Eg, in Windows you can page off of one drive and put your files on the other, or put system files on one and user files on the other.  Depending on your usage patterns this may cause both drives to be busy simultaneously a reasonable amount of the time.
However, two drives won't significantly improve performance (but won't do any harm) if you simply put a random chunk of your file system on the second drive -- it takes some conscious planning.
(The two exception to this are IBM System i (or whatever it's called these days) and some RAID systems.  System i spreads (almost) all data out over all drives more or less uniformly, splitting single files across multiple drives, so that drives tend to be equally busy.  RAID systems similarly spread individual files across multiple drives, but whether this improves performance (or instead makes it worse) depends on the particular RAID scheme.)
It used to be that average drive access speed got worse with larger drives, providing another motivation to use multiple smaller drives vs one large one.  I'm not sure that this is so true anymore, though.  (Haven't looked at drive performance specs recently.)


How do I delete these unclickable shapes in Illustrator?

Question: There are hidden shapes in the artwork and when I hover over the shape they appear but they are not clickable to delete. How do I delete these four, large rectangles?
Comment: Check at first is there a clipping mask or several of them. This can be seen in the layers panel or as well if Object > Clipping mask > Release is selectable. Clipping mask can be seen as a ghost when the mouse hovers on it. Releasing them reveals do they anything useful. They can be inserted to hide some unwanted parts or they appear without asking when one uses Illustrator for Images which are made with different, probably newer version.
Answer: The rectangles would indicate a Mesh Object. Possibly an Envelope Mesh which is used to distort artwork into various shapes.
You can...

Simply hit the delete key twice to remove the mesh object. This will also remove the artwork within the mesh.
Select the mesh and choose Object > Envelope Distort > Edit Contents. This will allow you to alter what is inside the mesh without removing the mesh.

For more information on Envelope Meshes, see the Adobe Help files linked to below:

Adobe Help: Envelope Meshes


Why did Voldemort need Hagrid's help to get past Fluffy?

Question: In the Sorcerer's Stone, Voldemort, acting through Quirrell, gets Hagrid drunk and "loses" a dragon's egg to him in order to find out how to get past Fluffy. That seems like incredible overkill. Surely Voldemort can teach Quirrell a sleeping spell, or just have Quirrell kill the dog outright. Not to mention the fact that dragons' eggs are incredibly rare and hard to get a hold of.
Comment: Book 1 would have been called "***Harry Potter and the giant dog that's immune to Avada Kedavra that ate Voldemort***"
Comment: An interesting piece of fan fiction showed alternate methods of getting past the blockages. In this case, the idea was to get past fluffy without leaving tracks or to deliberately leave Fluffy to stop anyone trailing him. Otherwise he could have incinerated Fluffy, the plants, the troll, and the door.
Comment: My thinking is that Voldemort was not yet strong enough to use magic; he was impelling Quirrell by force of personality and was able to inflict pain (in the form of headaches). It was easier to have Quirrell dupe Hagrid into giving the way to pass Fluffy than to expend energy on the 'dog'. There would still be other reasons to use magic or expend energy once passed Fluffy, so he (Quirrell + Voldemort) needed to marshal his resources.
Comment: @sabbahillel's comment about how `the idea was to get past fluffy without leaving tracks or to deliberately leave Fluffy to stop anyone trailing him` is on the right track. Voldy would have needed to get in and out past the traps on multiple occasions. Why go in nuking everything one time, when you can slowly work your way in whenever you want? One thing Voldemort is is calculated and thoughtful.
Comment: Seems to me like Voldemort/Quirrell would want to keep this plan hidden until Voldemort was actually strong enough to reveal anything. If Fluffy was hurt or killed, it would set off some red flags that the stone was compromised. In the case that the stone doesn't immediately revitalize Voldemort (which I see as likely to be the case) he would be incredibly vulnerable, and would not want to be hunted down in such a state. This assumes of course, that Harry and friends didn't figure the plan out.
Comment: Yeah, Quirrell could have simply levitated Fluffy and then opened the trapdoor.. I would have simply cut the chain to make it free and then simply wait for it to go outside.
Answer: Fluffy being clearly alive and alert reduces suspicion that the Stone was in danger.
Fluffy was the first defense, and therefore also the one everyone would notice being incapacitated. Even Harry kept checking on Fluffy to know the status of the Stone.

"Every time they passed the third-floor corridor, Harry, Ron and Hermione would press their ears to the door to check that Fluffy was still growling inside. Snape was sweeping about in his usual bad temper, which surely meant that the Stone was still safe." - Harry Potter and the Philosopher's Stone, Chapter 14, (Norbert the Norwegian Ridgeback)

They were also listening for him growling, which meant a spell that didn't kill him but resulted in him being quiet would still cause suspicion.

"Yet the days crept by and there could be no doubt that Fluffy was still alive and well behind the locked door." - Harry Potter and the Philosopher's Stone, Chapter 16, (Through the Trapdoor)

If he suddenly got quiet, it would alert Harry and presumably also the staff that the Stone was in more serious danger. Leaving Fluffy meant not raising suspicion. It would be especially useful if it took multiple attempts to get to the Stone, and we do know of at least one other attempt.

"You're too nosy to live, Potter. Scurrying around the school at Hallowe'en like that, for all I knew you'd seen me coming to look at what was guarding the Stone." - Harry Potter and the Philosopher's Stone, Chapter 17, (The Man with Two Faces)

If Fluffy had been killed or seemed incapacitated after that, it's likely the Hogwarts staff would have increased their defenses or even moved the Stone entirely. As it was, Fluffy was alert enough to maul Snape's leg, and Snape was probably the only staff member that suspected what was going on.
Even presuming his success, it would be highly beneficial to Lord Voldemort if he was able to get the Stone without anyone noticing, and have the Hogwarts staff diligently guarding nothing, unaware he would be strengthened by the Elixir of Life.
As it is, the only one of the defenses that was already handled before Harry got to it was the troll, which was further down the line of defense, and therefore would be less noticeable.
Many magical creatures are resistant to magic, and Fluffy might be as well.
There isn't any more information about three-headed dogs specifically, either in the seven books or in "Fantastic Beasts and Where to Find Them" (either version of the book), but from what we know about other magical creatures, it's quite possible that they might be resistant to magic.  Dragons, Graphorns, and Manticores are all resistant to magic. Casting a sleeping spell on them would likely be difficult or even close to impossible.
Many magical creatures are very difficult to fight even if they're not really resistant to magic.
It's never mentioned in "Fantastic Beasts and Where to Find Them" if Acromantulas, Chimaeras, Erumpents, Griffins, Nundus, Quintapeds, trolls, and yetis are actually resistant to magic. However, they are all highly dangerous, and some require either very powerful magic or multiple wizards working together to subdue them. It's unknown exactly how hard it would be to fight Fluffy using magic, but Hagrid does imply that Fluffy is harder to take care of than a dragon would be.

"So I told him, after Fluffy, a dragon would be easy …" - Harry Potter and the Philosopher's Stone, Chapter 16, (Through the Trapdoor)

Hagrid often underestimates the danger of magical creatures, but this gives us at least a good comparative sense of how dangerous Fluffy might be.
Lord Voldemort was still weak, not at his full strength, and Quirrell was weakening as well.
Lord Voldemort was in a fairly weak state when he was possessing Quirrell. Although he had a servant to possess, and was strengthened somewhat by unicorn blood, he was still a wraith and nowhere near his most powerful. Quirrell even doubts that he's strong enough to be revealed to face Harry.

"Let me speak to him … face to face …'
'Master, you are not strong enough!'
'I have strength enough … for this …" - Harry Potter and the Philosopher's Stone, Chapter 17, (The Man with Two Faces)

Lord Voldemort even says himself to Harry that he's been greatly weakened, and needs the Stone to regain a body, and with it his own power.

"See what I have become?' the face said. 'Mere shadow and vapour … I have form only when I can share another's body … but there have always been those willing to let me into their hearts and minds … Unicorn blood has strengthened me, these past weeks … you saw faithful Quirrell drinking it for me in the Forest … and once I have the Elixir of Life, I will be able to create a body of my own …" - Harry Potter and the Philosopher's Stone, Chapter 17, (The Man with Two Faces)

Quirrell was also likely getting progressively weaker due to being possessed, as the animals that the Dark Lord possessed did.

"I sometimes inhabited animals – snakes, of course, being my preference – but I was little better off inside them than as pure spirit, for their bodies were ill-adapted to perform magic … and my possession of them shortened their lives; none of them lasted long …" - Harry Potter and the Goblet of Fire, Chapter 33 (The Death Eaters)

We see evidence that Quirrell is indeed growing weaker in this way.

"Quirrell, however, must have been braver than they'd thought. In the weeks that followed he did seem to be getting paler and thinner, but it didn't look as though he'd cracked yet." - Harry Potter and the Philosopher's Stone, Chapter 14, (Norbert the Norwegian Ridgeback)

It was comparatively easier to get information from Hagrid than try fighting Fluffy without it.
Considering all factors, it would be much easier and more logical for Lord Voldemort to get Hagrid to accidentally mention how to get past Fluffy than for him and Quirrell to fight Fluffy without this knowledge. Getting information from Hagrid is very easy, and playing a harp is much less effort than spells which may have little effect anyway.
Comment: A very good answer, uses direct quotes to back up the most likely in-universe reasons! Well done bella!
Comment: If I recall correctly, the troll was merely knocked out so that too would've been alive and operational should anyone call to check on the stone. In fact, given that there only seemed to be a limited quantity of potion to access the final stages it's unlikely that anyone would have felt the need to proceed any further than the troll room. I agree with the others, very good answer!
Comment: Great answer!  This should be the accepted answer, imo :)
Comment: @TheDarkLord Yes, the troll was knocked out. It wasn't conscious when Harry had to pass it, but it would have woken back up. I agree, it doesn't seem likely that anyone checking on the Stone would have gone all the way through the stages. Thanks a lot, I'm glad you liked it! :)
Comment: Are those quotes from the UK edition?
Comment: @JohnP I think so, the cover of the first one in the series says "Philosopher's Stone", rather than "Sorcerer's", and it's a matching set.
Comment: @Bellatrix - That makes sense, because I didn't recognize some of those quotes being in the US edition.
Answer: There is no direct reference in the books, so all in-universe explanations are my assumptions and deductions.
Out-of-universe reason: protection of the Stone was meant to be a quest for Harry and his friends, as well as for Voldemort. Also, there was a need to expose a particular trait in Hagrid's character - love for dangerous beasts.
In-universe reasons: - Knowing Hargid, I would assume that Fluffy would be tough to disable or kill with magical methods, like Hagrid himself. Also, Quirrell was not as good at using magic as Voldemort, and Voldemort was weak, relying on Unicorn blood to keep his strength. He needed his strength for a possible encounter with someone possibly more powerful than Fluffy. If Voldemort/Quirrell failed to kill or disable Fluffy in one strike, he would put up a fight. The whole castle would be alert, which was not something Voldemort wanted.
Answer: Consider these facts:

Voldemort was weak - Having not regained his body yet, Voldemort was simply a disembodied face on the back of Quirrel's head. He was not capable of using his own powerful magic to neutralise the dog somehow.
Voldemort's host body was Quirrel - We don't know for sure whether Quirrel, who was providing Voldemort's only means of getting past Fluffy by being his host body, would be capable of using magic to stop Fluffy. The dog was generally portrayed as being very frightening and powerful, and Quirrel was generally portrayed as being a weak, stuttering and timid man. Therefore I think evidence points to Quirrel's magic not being strong enough to stop Fluffy, even if Voldemort instructed him on potential powerful spells.
Voldemort needed to find out how to stop Fluffy in a way that did not draw attention to himself - Voldemort couldn't do anything that would arouse Hagrid's suspicion about Quirrel, in case of a follow up investigation at Hogwarts. Voldemort was planning on getting the Philosopher's stone, and the best way of doing that would seem to be quietly sneaking into the third floor corridoor without causing anyone any reason to suspect until the moment he enacted his plan. Doing so by having Quirrel pretend, as a fellow Hogwarts staff member, to socialise with Hagrid, get drunk together and lead Hagrid to spill the secret, would not be a bad way of doing so, as Hagrid would probably be slightly embarassed by the encounter but otherwise not think much of it.

With these considerations I think the method Voldemort used to get the key to stopping Fluffy made sense, and was not necessarily overkill.
Comment: As a counter to your first two points, remember that Quirrel was able to break into Gringotts before the start of the school year. I agree with you on point 3.
Answer: I think that the Dog has also some sort of Protection in his blood. Like a Giant or a Dragon. This means you can't just kill Fluffy. In Harry Potter and the Goblet of Fire, they couldn't set the dragons to sleep with only one spell. They did it with 3-5 people.
In the Harry Potter Wiki
there are some paragraphes about spell resistance.  In there is mentioned that Dragons have some resistance because of the Skin. And also in the book Harry Potter and the Prisoner of Azkaban, where some Wizards try to stop Hagrid.

Ron: "Hagrid did well, didn't he? How come all the spells bounced off him?"
  Hermione: "It'll be his giant blood. It's very hard to Stun a giant, they're like trolls, really tough... but poor Professor McGonagall... four Stunners straight in the chest and she's not exactly young, is she?"
Ron and Hermione discuss Hagrid's surviving multiple Stunning Spells
Comment: Do you have any sources to support your belief??


Compiling with outdated compiler

Question: In my contracts I've been placing pragma solidity ^0.4.2; Without knowing my compiler is outdated at 0.4.16! Interestingly, all contracts still compile and are able to migrate to testrpc. Still, I would like to make sure, is it possible this could introduce unseen errors in the future?
Answer: Actually, 0.4.2 is older than 0.4.16. Versions follow consecutive numbering within smaller releases.
The latest as of now is 0.4.19.
Comment: Ok wow! Then V 0.4.2 significantly predates 0.4.16 then.. Back to my question, any possible ill effects from incorrect compiler declaration?
Comment: Yes, you are using a compiler version that doesn't have the features and security improvements that have been made during the past year. You should try to use the latest stable version. Changes made to new versions don't affect what has already been deployed as those will compile to the specified compiler version.

Compiling with outdated compiler


Logging in via Pokemon trainer club vs Google account

Question: My son's account was set up with trainer club. Tab did a software update overnight that logged him out. Trainer club wouldn't accept password so he selected the Google login option but it wants to create a new character.  Will he lose all his progress if we move forward? We restarted the tab but the Pokémon app is still in the midst of picking a new character and doesn't give us the  trainer account option.
Comment: It doesn't matter if you create a new account with Google, it won't erase the trainer club account.
Comment: As Patrat has said in the answers below, the PTC servers are experiencing some trouble over the past few hours. Try checking this website for the PTC login status: http://ispokemongodownornot.com Once it's stable, you should try logging back in again.
Comment: Something that I noticed that might be your problem. The usernames for PTC are case sensitive this is unlike a lot of login systems.
Answer: Characters and progress are tied to accounts, so if your son wants to hold onto his, he should log in with the Pokemon Trainer Club as before.
If the app isn't doing this even after a restart, the best option is probably to uninstall the app and download it again from the Google Play store. This won't harm his progress.
Comment: Thank you! I considered this but was afraid to lose it and that would be a catastrophe  in my house! :)
Comment: Yeah, I get that feeling :) If you found the answer helpful, please consider accepting/upvoting it :)
Answer: Also, pokemon servers are highly unstable at the moment.  This doesn't affect google accounts but for those logging in with a pokemon trainer club account the odds are that it's the servers.  Try again Later before giving up.
Comment: @KRyan Just because the pokemon servers are in a worse state doesn't mean the general servers have no problems. Google accounts may not always be working but pokemon trainer club accounts apparently have it even worse.
Comment: Doesn't affect Google accounts? I can't log in with my Google account pretty often.
Answer: All the game data is stored on server side, you can log in and out with as many accounts/game progresses as you wish from the same device and everything will keep working fine. When you logged in with a google account you loaded a new (empty) profile and game progress in which nothing had been done yet. As soon as you log in with your old account again, the old game progress will be restored. This feature also allowes to keep the game progress if you buy a new device or want to play on a device of a friend.
If sometimes the login fails with a message like "Authentification failed. Please try again." repeatedly, don't panic, it's probably just the Pokemon Go servers being down again. Wait a few hours or until the next day and they'll be working again.


Hydrophilicity and polarity

Question: If any given molecule is polar, and is capable of hydrogen bonding, I can be sure that it's hydrophilic.
However, are those two the only conditions that make a molecule hydrophilic?
Comment: Is this a biology/biochemistry question?
Comment: its a common term used in biochemistry
Comment: As phrased right now it's a chemistry question.  Maybe if it asked about specific biological molecules instead?
Answer: I think the answer is no.
The definition of polarity is basically that a molecule has a dipole, and as water has a very large dipole, if a molecule will mix with water readily, its usually pretty polar.
Water also can share its hydrogens in hydrogen bonding, which is an actual shared covalent bond where a hydrogen atom is shared between two acceptors.  The hydrogen bond is fairly weak (~ 0.5 kcal/mol) but water does a lot of hydrogen bonding.
So hydrogen bonding and polarity are often a shared characteristic of an molecule, but are not the same thing.
Ions are quite hydrophilic but do not need to be hydrogen bond donors or acceptors.   Chloride  (Cl - ) is not a hydrogen bonding atom - its so acidic that it exists nearly entirely in the Cl- form in water solution. But its definitely electro negative and really soluable in water. Are Ions polar though?  Chloride is symmetrical and so its not really polar.
Still is there a polar molecule which is neither a hydrogen bond donor or acceptor?  Acetone and the cyanide ion (CN-) are a polar molecules which have no hydrogen bond donor, but can accept them.
I think sulfate is an example of a polar molecule which does not form hydrogen bonds.
Sulfate (SO4 -- )  has four S-O bonds arranged in a tetrahedron but each of them is highly polar - with 90% of the negative charge on the oxygen.  But like the Chloride ion its so acidic that it much prefers its free ionic state.  It would have very little hydrogen bonding character even when its in water.  Sulfate can form hydrogen bonds, but I think they will only be stable when there are several of them at once.  Sulfate binding protein has seven hydrogen bonds when it binds sulfate, using the chelation effect to make the binding strong enough to be stable.

Hydrophilicity and polarity


what scanning tools are unlikely to set off network IDS?

Question: What network scanning tools have a stealth option to avoid detection?
Comment: please read the faq on how to ask questions. Most of yours so far have taken work to make them even vaguely suitable here.
Comment: The most correct answer to this question would be several conference presentations. Are there any specific situations you're interested in? Specific IDS implementations or configurations you're worried about? These are the kinds of questions that make it so hard to succinctly answer tuis.
Answer: As with most things - it isn't about the tool, it is how you use it.
Take nmap for example.
Beginners to the tool or script kiddies might run scan using the default option.

nmap 192.168.1.0/24

The default behavior for nmap without specifying any addition flags would be to perform a syn scan, so the above query would be equal to.

nmap -sS 192.168.1.0/24

Syn scans are reasonably stealthy. However, most modern firewalls/IDS should be able to detect syn packets as it is a very common scan technique.
Slightly better but still inexperienced users might perform even noiser scans using nmap.

nmap -sV 192.168.1.0/24
nmap -A 192.168.1.0/24

An nmap scan triggered with those two scan flags are VERY noisy. In addition to sending a variety of packet types and grabbing banner information, nmap will use the nmap scripting engine to perform additional tasks such as bruteforcing the SNMP community string if applicable. Such scans will be easily detected.
More experienced users can perform more stealthy scan types by specifying certain flags. nmap is a very powerful tool as it allows you to set different flags in a TCP packet to evade firewall rules. ACK & FIN are two very commonly used techniques to evade firewalls.
Other advanced techniques like idle scan can also be performed using nmap. It is a very versatile network scanner that can be very silent in the hands of an advanced user.
Some tips
Do not perform syn scans or version scans on entire subnets unless needed. It is slow and noisy. Instead, use the ping scan option to find out which hosts are active on a subnet.

nmap -sn 192.168.1.0/24

This option would return a list of active host IP and MAC addresses in the subnet.
Once you have obtained a list of active hosts, selectively scan hosts which you think are interesting. For example, the IP address 192.168.1.1 seems interesting to me, I might want to know more about it. Perform a syn scan.

nmap -sS 192.168.1.1

By default, nmap scans for the most popular thousand or so ports according to a list compiled by the author. This is good enough for most purposes. However, there might be some scenarios where you might want to specify specific ports. You can do that using the -p flag.

nmap -sS -p 80 192.168.1.0/24

This scan would send syn packets to the entire subnet, searching for active IP addresses with port 80 open. This is a useful scan for determining which host might have an active HTTP server.

nmap -sV 192.168.1.1

The -sV flag specifies the version scan option. This option instructs nmap to send a wide variety of packets to the hosts and scanning the headers of the packets returned to check for the exact version of a service the host is running.
This is a very powerful but noisy version. Use with caution.

nmap -A 192.168.1.1

The -A flag instructs nmap to use an aggressive scan. This is one step up from the version scan and utilizes the nmap scripting engine. This is very noisy.
nmap is a highly powerful tool. If you want to learn more, i highly suggest you pick up the book written by its author.
Comment: I was just dealing with network IPS and client-side active defenses: what worked best for me was using nmap -sS, -sC -f, T2 or 3, --osscan-limit, --version-light, and maybe -iR.
Answer: I'd actually say that the best way to avoid detection from IDS is not to use scanners at all.  Most scanning tools have signatures which can be recognised by IDS, so can be incorporated into their rulebases.  Some (like nmap) have options to slow down the scan or scan in a more random fashion to try and avoid detection, but whether IDS trips here will depend on how it's configured.
There are two types of traffic which I'd expect to bypass IDS.

Legitimate traffic - For example connections to ports 80/443 on a web server. if you're looking at an external network then trying to manually connect to systems on web ports is unlikely to trip IDS unless you try to do it a lot very quickly.
Obvious Junk Traffic - Again on external networks, to avoid getting overwhelmed with false positives, the IDS might be tuned to discard specific scan types (e.g. scans for port 445/TCP) .  Whilst those aren't likely to be successful, you could use a scan for that port to try and establish live hosts in a range and you might bypass the IDS as it'll be taken as one of the usual scans that every system on the Internet gets..
Answer: She can use standard, non-intrusive scanners like Nmap. It should be adjusted the way that it's slow enough to not trigger alarms. The information gathered is:

Active hosts
Operating system version
Running services (ports)
Banners / versions of these services

Also, you can perform non-invasive Nessus scan, the way that it's not trying to run exploits or authenticate, but just pings, connects to ports and check banners. The key elements:

Slow scan, maybe from multiple ip numbers
Do not attempt to login, but just read the banner

In the easiest scenario, is discovering their subnet, and then visiting via the browser each host. It has high efficiency and cause little alarming.
Answer: You could use nmap with tortunnel  and proxychains. But scanning through tor is very, very slow.
And remember to create a rule that drops packets form your real ip so your real IP address is not disclosed.
(if your are on linux iptables -A OUTPUT --dest "target" -j DROP)
Comment: Abusing Tor for this purpose is unfriendly behavior and an abuse of the Tor network.  Please don't recommend that people use it for port-scanning.  Tor is a useful service; people who abuse it ruin things for everyone.
Comment: @D.W. +1 by me. I agree with what you saying , but as you say there are people who abuse the network, so someone could use the above method to see how thing work and then find a way to make the IDS to block when someone is scanning him from tor.


Configure Tor Browser to run over SSH SOCKS proxy?

Question: I'd like to proxy all of my Tor Browser traffic through my home network remotely over SSH. If I have SSH access to a computer in my home network and a Tor Browser bundle here, how can I set up the tunnel and the browser to run remotely over a SOCKS tunnel?
Answer: You can configure it so that it's:
you > tor network > ssh proxy server > internet

By using torsocks
torsocks ssh -D 8080 user@IP

I suppose you could configure it as you > proxy server > tor > internet, by having the traffic on your proxy server go through tor, but I'm not sure how, or even if it's possible.


Elementary collapses implies same homotopy type

Question: Let $\Delta$ be a simplicial complex, and suppose that $\sigma \in \Delta$ is a proper face of exactly one maximal simplex $\tau \in \Delta$. A simplicial collapse of $\Delta$ is the removal of all simplices $\gamma$ s.t. $\sigma \subseteq \gamma \subseteq \tau.$. It is well known that a sequence of collapses yields a strong deformation retraction,
in particular, a homotopy equivalence. I have an intuitive geometric feeling of how to prove this fact, but I don't know how we can construct this deformation retract map. Can anyone give the idea for the deformation retract map?
Answer: Such $\sigma$'s are called free faces. You can think of these collapses as "pushing in" the free face. For example, if your simplicial complex is a triangle $\Delta^2$, then any edge $\sigma$ is a free face as it is contained in the unique $2$-simplex $\tau$. Pushing in $\sigma$ through $\tau$ collapses those two simplices onto the remaining two edges. This is a strong deformation retract as all other faces remain fixed through this process.
Comment: It may be that the OP recognises this visually but would like a precise description or construction of this function


limit spatial extent of arcpy cursor

Question: How do I limit spatial extent of features that is returned by arcpy.da.SearchCursor?  Can I use arcpy.env.extent?  Or somehow come up with sql statement to specify spatial relationship?   Or do I have to arcpy.SelectLayerByLocation_management to do intersection of my area of interest.
EDIT:
Here is what I want to accomplish.  Want to visit features from a feature class within a set extent.  I was hoping that second use of arcpy.da.SearchCursor respects arcpy.env.extent, but apparently it ignores it.
How can I do this (visit spatial subset of features)?  I want to have method with decent efficiency (e.g. avoid copying of features).
<code>import arcpy
def proc():

    # I/O files
    fo = open('c:/users/yosuke/documents/log.txt','w')
    fc = arcpy.CopyFeatures_management('c:/Users/Yosuke/documents/ArcGIS/polytical_boundaries.mdb/tl_2010_us_state10', r'in_memory\fc')

    # grab everything, works fine
    arcpy.env.extent = 'MAXOF'
    n = int(arcpy.GetCount_management(fc).getOutput(0))
    m = len(list(arcpy.da.SearchCursor(fc, ('SHAPE@',))))
    extent = arcpy.Describe(fc).extent
    # got 52 from both, as expected
    fo.write('n: %s\n' % n)
    fo.write('m: %s\n' % m)
    fo.write('extent: %s\n' % extent)
    fo.write('\n')

    # want to grab only western USA
    arcpy.env.extent = arcpy.Extent(-120, 0, -90, 80)
    # this counts 27 western states, I suppose.
    n = int(arcpy.GetCount_management(fc).getOutput(0))
    # this still marches through all states, I want this to be spatial subset of USA
    m = len(list(arcpy.da.SearchCursor(fc, ('SHAPE@',))))
    extent = arcpy.Describe(fc).extent
    fo.write('n: %s\n' % n)
    fo.write('m: %s\n' % m)
    fo.write('extent: %s\n' % extent)
    fo.write('\n')

    # For now I am making copy of feature class
    # turned out that this is not very bad for my application, other processing
    # tasks I am applying on cursor rows are more computationally expensive.
    # I still want to know if this is standard way to accomplish this kind of
    # tasks (subset cursor by extent)
    fc2 = arcpy.CopyFeatures_management(r'in_memory\fc', r'in_memory\fc2')
    n = int(arcpy.GetCount_management(fc2).getOutput(0))
    m = len(list(arcpy.da.SearchCursor(fc2, ('SHAPE@',))))
    extent = arcpy.Describe(fc2).extent
    fo.write('n: %s\n' % n)
    fo.write('m: %s\n' % m)
    fo.write('extent: %s\n' % extent)
    fo.write('\n')
</code>
Answer: From what I have tested the extent does not limit the search cursors spatially..When using tools like "GetCount_Management" etc. it is limited but it appears that Cursors do not follow the extent
Answer: I haven't tried this out so I may be incorrect but I think the search cursor will honor the env.extent based on these notes in ArcGIS help.
Output Extent

extent (Read and Write)
Tools that honor the Output Extent environment will only process
  features or rasters that fall within the extent specified in this
  setting.

Cursors

Note:   Cursors honor layer/table view definition queries and
  selections. The cursor object only contains the rows that would be
  used by any geoprocessing tool during an operation.

I'm taking that last sentence to mean if a geoprocessing tool will only process features that fall into the defined extent than cursors will have the same behavior.
Comment: Thank you, I also thought that arcpy.env.extent does what I wanted, but I am not sure about the behavior of my script.  I should simply say "arcpy.env.extent = arcpy.Extent(a,b,c,d)" then "for row in arcpy.da.SearchCursor(lyr,['SHAPE@',]):" should let me iterate he geometry, correct?  I try post my code see why it's not behaving what I exptect

limit spatial extent of arcpy cursor


how to implement Step definitios from gherkin feature files

Question: I'm new to GHERKIN-CUCUMBER. so please anyone guide me to find solution to my problem.
"How to generate or implement Step definitions (from feature files) programmatically in JAVA?"
Answer: There are multiple tutorials you can follow on the net.
I wrote one a couple of years back that may be a starting point http://www.thinkcode.se/blog/2014/05/29/cucumberjvm-hello-world The versions are outdated, but it will work, And when you get it to work, update the versions of the dependencies.
Another source of malformation may be the getting statred project supplied by Cucumber, https://github.com/cucumber/cucumber-java-skeleton
I would advice not to look to hard for anything that generates the steps. Cucumber will suggest code snippets for missing steps. They are a good start but they will need some attention.
Comment: Thank you for your answer Thomas.


Academic Conferences: Are "program committee" members always present at the actual conference?

Question: For example, this conference (https://www.monash.edu/tenor2019/committees) is being held in Melbourne, Australia; but the Program Committee consists of over 40 members from interstate and overseas. Would it be expected for all of these committee members to be present at the conference, or could their roles be carried out without actually attending?
Comment: Most will be actually involved but not necessarily all
Answer: Program committee members are not expected to be present at conferences. They are responsible for reviewing submissions (before the conference) and helping to select submissions to be presented at the conference. Nonetheless, program committee members may attend, possibly even to present a submission that was selected (by other members of the program committee). Beyond the program committee, other committee members may be present, indeed, some must be present.


In cavity optomechanics, what does it mean that a cavity mode is driven by a classical field?

Question: What do we mean by the statement when we say that a cavity mode is driven by a classical field or classically controlled field, in Cavity Optomechanics?
Answer: It means that one or multiple channel modes (an external mode in the Jaynes-Cummings Model of CQED) of the cavity are driven by a coherent state. Here "driven by a coherent state" is to be understood within the Input-Output formalism: The input operator is in a coherent state.
It turns out that for a coherent state the quantum stochastic equations of motion can be replaced by an effective Hamiltonian that is equivalent to a classical driving field Hamiltonian.


Could Nmap scan switches/routers

Question: Nmap can scan and sometimes successfully detect the running OS in the remote host. However, can nmap scan routers and switches? Most of them use an embedded system.
Comment: Why not just try it at home? At my place, the TP-LINK wireless router is the noisiest node of the network, leaking all everything about it's firmware :(
Answer: Yes, provided the devices are reachable on the network they can be scanned.
The accuracy of the results (e.g. fingerprinting) is dependant on the platform, software version, running services and configuration.
Example:
<code>$ nmap -A -T4 10.1.1.1

Nmap scan report for 10.1.1.1
Host is up (0.020s latency).
Not shown: 999 closed ports
PORT   STATE SERVICE VERSION
23/tcp open  telnet  Cisco router
Service Info: OS: IOS; Device: router

Service detection performed. Please report any incorrect results at http://nmap.org/submit/ .
Nmap done: 1 IP address (1 host up) scanned in 2.21 seconds
</code>
Answer: Don't forget other network protocols, routing protocols, Cisco Discovery Protocol, SNMP, etc.., there's a ton of information on these types of devices and "Yes" you definitely can identify them and their OS versions using nmap. Don't forget about nmap's scripting options either, you could write a .NSE just for this if you wanted (there may already be one for that matter). SNMP and CDP will give you the exact OS version information if you have access to them. Even SSL certificates on these devices may leak versioning information. Hope this helps.
Answer: This command helped me get info about a Cisco router:
<code>$ nmap -O -v 172.22.253.1
</code>
Output:
<code>c:\nmap>nmap -O -v 172.22.253.1
Starting Nmap 7.70 ( https://nmap.org ) at 2019-10-18 11:51 W. Europe Daylight T
ime
Initiating Ping Scan at 11:51
Scanning 172.22.253.1 [4 ports]
Completed Ping Scan at 11:51, 1.77s elapsed (1 total hosts)
Initiating Parallel DNS resolution of 1 host. at 11:51
Completed Parallel DNS resolution of 1 host. at 11:51, 0.01s elapsed
Initiating SYN Stealth Scan at 11:51
Scanning 172.22.253.1 [1000 ports]
Discovered open port 1720/tcp on 172.22.253.1
Discovered open port 22/tcp on 172.22.253.1
Increasing send delay for 172.22.253.1 from 0 to 5 due to 40 out of 132 dropped
probes since last increase.
Completed SYN Stealth Scan at 11:51, 12.17s elapsed (1000 total ports)
Initiating OS detection (try #1) against 172.22.253.1
Nmap scan report for 172.22.253.1
Host is up (0.00s latency).
Not shown: 992 closed ports
PORT     STATE    SERVICE
22/tcp   open     ssh
25/tcp   filtered smtp
1720/tcp open     h323q931
6666/tcp filtered irc
6667/tcp filtered irc
6668/tcp filtered irc
6669/tcp filtered irc
7000/tcp filtered afs3-fileserver
Device type: router
Running: Cisco IOS 12.X
OS CPE: cpe:/h:cisco:2811_router cpe:/o:cisco:ios:12.x
OS details: Cisco 2811 router (IOS 12.X)
TCP Sequence Prediction: Difficulty=262 (Good luck!)
IP ID Sequence Generation: Randomized

Read data files from: c:\nmap
OS detection performed. Please report any incorrect results at https://nmap.org/
submit/ .
Nmap done: 1 IP address (1 host up) scanned in 32.82 seconds
           Raw packets sent: 1153 (52.522KB) | Rcvd: 1014 (40.914KB)
</code>
Answer: I'm not so sure you can get switches infos the same way you get them from routers.
Switches (layer 2) are not routers (layer 3), but routers need to be also switches (layer 2) to route the traffic; They intercept and manage the packet to be vehicolated trough the net, while switches don't manipulate traffic; They only switch between MAC addresses of different NICs.
Every time you get a port opened or an OS fingerprinting, that's because all the traffic sent and received passes through layer 3.
If you put a switch between two nodes and sniff MAC addresses, you'll always see only two MACs: source and destination.
You will never see the MAC address of the switch port the two nodes are connected to, because the switch is a transparent device.
Comment: Switches can have IPs, too ... And you bypass the info offered by the accepted answer. You can get detailed info from a switch.

Could Nmap scan switches/routers


What building boosts Saka Horse Archers?

Question: Are Saka Horse Archers boosted by a Barracks or a Stable?
Comment: @Vemonus I think they are asking what I answered, what type of unit is the unique unit classified as that would determine what type of building you need in your encampments to boost their XP when producing them.
Comment: Are you asking which one you should choose?
Answer: The Saka Horse Archer is a ranged unit, not cavalry. This is noted in it's wiki page as I linked and it is also noted that the unit upgrades directly into the field cannon, not another cavalry.
Although it is researched under Horseback Riding, and benefits from the "Make 1 Get 1" bonus of Scythia for light cavalry and horse archer units, they do not actually require horses to build, and they are not actual cavalry units.
So to answer your question specifically, you will need a barracks to boost the units XP since they are for melee and ranged units, and a stable is only used for cavalry units.


What's the difference between Tags and Workspaces in awesome?

Question: I'm trying out awesome window manager, and I'm still not getting the hang of it. I don't quite understand the difference between tags and workspaces. I would appreciate a good explanation of how tags work.
Comment: Describe what you think is a workspace in awesome. AFAICT, tags *are* workspaces here. They're just called tags to perhaps better convey the nature of their operation, as in one tag has multiple windows, but it's also possible to tag a window with multiple tags.
Answer: Tags are a superset of workspaces. Workspaces are a subset of tags. Each client (window) has an associated list of tags, so you can display it in multiple "workspaces".
From the documentation : "Windows are grouped by tags in awesome. Each window can be tagged with one or more tags. Selecting certain tags displays all windows with these tags."
You have three clients, Firefox, Spotify, and Terminal. Firefox lists tag 1, Spotify lists tag 2, and Terminal lists tags 1 and 2. So when you are on tag 1, Firefox and Terminal are displayed. If you switch to tag 2, Spotify and Terminal are displayed. That's the same thing as workspaces, but you can display a client on multiple tags. You can even display several tags at the same time, so in this case, you'd have Firefox, Spotify and Terminal displayed at the same time.
Hope that helps, have fun with Awesome !
Comment: I would go so far as to say workspace is an obsolete concept. You can use tags to emulate workspaces by limiting the way you use them. :)


Hide attribute added by Manage attribute

Question: In project I have added an attribute Status in Dropdown. I want to hide that attribute on adding product page. While Managing Attribute there is no such configuration to hide Attribute. Is there any way to hide attribute on product page ?
Thanks in advance.
Answer: If I understand your question correctly, you don't want that attribute to be displayed on the Add/Edit Product page from admin, right?
If so, that's not possible by default on Magento.
If you want it to be displayed only on certain products you can try attribute sets. Adding that specific attribute to a specific attribute set will make it visible only to the products that have that attribute set.
But, if you really want to hide it you could try this in a setup script:
<code> $installer->updateAttribute('catalog_product', $attribute_code, 'is_visible', '0');
</code>


Why are .so files executable?

Question:
Possible Duplicate:
Why are shared libraries executable?

Why do the .so files in /lib have permission 755 and not 644 in Linux? That seems strange
According to http://www.gossamer-threads.com/lists/gentoo/user/231943, it seems for old glibc. On my embedded system /lib/libc.so.6 it works even permission is 644.
Comment: I've updated my answer.
Answer: I suspect it's probably just for historical reasons.
BlueBomber's answer is probably historically correct, but it's not actually necessary for shared objects to be executable.
On my Ubuntu system, they aren't. Of the 30 /lib/*.so* and 600 /usr/lib/*.so* files, only one has execute permissions, and that's probably just a glitch.
Execute permission permits a file to be executed via one of the exec*() functions; shared object files contain executable code, but they're not executed in that way.
On the other hand, on a CentOS 5.7 system I have access to, those files are executable; the same is true on a SPARC Solaris 9 system. (It would be interesting to try turning off executable permissions on some of those files to see if it breaks anything, but I'm not able to do so.)
(What Linux distribution are you using?)
UPDATE:
This answer to this question shows an example (HP-UX) of a system where the execute bit is actually required. That doesn't appear to be the case on Linux, where some distributions set the execute bit (probably out of historical inertia) and others don't. Or maybe some Linuxes actually require it.
Another data point: On my Ubuntu system, I just tried creating my own shared object file. The generated "libfoo.so" file was created with execute permissions, but if I manually chmod -x it, the program that uses it still works.
In any case, setting execute permission on *.so files is Mostly Harmless (and certainly less annoying that, say, setting execute permission on source files).
UPDATE 2:
As fwyzard points out in a comment, some *.so files can actually be executed. For example, on my current system, executing /lib/x86_64-linux-gnu/libc-2.27.so prints version information for the GNU C library.  /lib/x86_64-linux-gnu/libpthread-2.27.so behaves similarly.
Comment: AFAIK, only libc.so.6 has a use for the execute bit, in that you can actually execute it: running /lib/x86_64-linux-gnu/libc.so.6 will print its version information, e.g. "GNU C Library (Ubuntu GLIBC 2.27-3ubuntu1) stable release version 2.27."
Comment: Another use case. You can use `/lib64/ld-linux-x86-64.so.2` to execute ELF executables that do not have the execute permission bit set. e.g., `cp /bin/ls /tmp/ls; chmod 0600 /tmp/ls; /lib64/ld-linux-x86-64.so.2 /tmp/ls`.
Answer: If I understand your question, you are asking why shared object files have execute permissions, and that's because those are library files intended to be linked, and the code they contain is intended to be executed.
Reference:

http://www.gossamer-threads.com/lists/gentoo/user/231943

Why are .so files executable?


How to enable Internet Connection Sharing via a script?

Question: I have two NIC ports [port1 & port2] and want to enable Internet Connection Sharing on port1 through either C# or batch script:

I have only been able to find how to start the ICS service, not to enable the option:
Answer:
On earlier OS versions, netsh routing could be used, but this no longer works in Win7.

Have a look at this Powershell snip, otherwise StackOverflow likely has a C# example:
# Register the HNetCfg library (once)
  RegSvr32 "hnetcfg.dll"

# Create a NetSharingManager object
  $m = New-Object -ComObject HNetCfg.HNetShare

# List connections
  $m.EnumEveryConnection |% { $m.NetConnectionProps.Invoke($_) }

# Find connection
  $c = $m.EnumEveryConnection |? { $m.NetConnectionProps.Invoke($_).Name -eq "Ethernet" }

# Get sharing configuration
  $config = $m.INetSharingConfigurationForINetConnection.Invoke($c)

# See if sharing is enabled
  Write-Output $config.SharingEnabled

# See the role of connection in sharing, only meaningful if SharingEnabled is True
  # 0: Public || 1: Private
  Write-Output $config.SharingType

# Enable sharing
  # 0: Public || 1: Private
  $config.EnableSharing(0)

# Disable sharing
  $config.DisableSharing()

How to enable Internet Connection Sharing using command line?
Comment: not wokring on win8.1


Why is System.Environment.MachineName value uppercased?

Question: My machine name is lowercased (i see that in Advanced system settings dialog, Computer Name tab) but System.Environment.MachineName reports it uppercased. Why is that? This is a real problem for me because from my tests PrincipalPermissionAttribute performs case sensitive comparison for role names (i map my custom roles to Windows groups and my environment is non-domain). Any advise?
Answer: The source for Environment.MachineName for .NET 4.7.1 is here: https://referencesource.microsoft.com/#mscorlib/system/environment.cs,be0b5c103d248dce
It p/invokes GetComputerName as seen here: https://referencesource.microsoft.com/#mscorlib/microsoft/win32/win32native.cs,0c7d7f4f83d4ddd0
Here is the GetComputerName function: https://msdn.microsoft.com/en-us/library/windows/desktop/ms724295(v=vs.85).aspx, which states:

GetComputerName retrieves only the NetBIOS name of the local computer.
  To retrieve the DNS host name, DNS domain name, or the fully qualified
  DNS name, call the GetComputerNameEx function.

The MSDN for Computer Names, https://msdn.microsoft.com/en-us/library/ms724220(VS.85).aspx, states:

NetBIOS names consist of up to 15 bytes of OEM characters including
  letters, digits, hyphens, and periods. Some characters are specific to
  the character set. NetBIOS names are typically represented in the OEM
  character set. The OEM character set depends on the locale. Some OEM
  character sets represent certain characters as two bytes. NetBIOS
  names, by convention, are represented in uppercase where the
  translation algorithm from lowercase to uppercase is OEM character set
  dependent.

So, NetBIOS names are upper case by convention and System.Environment.MachineName returns the system's NetBIOS name.
Comment: Thanks a lot for this breakdown
Answer: Use Dns.GetHostName instead, that should return it with the correct case (at least it does on my computer).
Comment: If anyone is interested (as I was) [here's the native method](https://msdn.microsoft.com/en-us/library/windows/desktop/ms738527(v=vs.85).aspx) that is called to get the result.  Has some details about the process that may be relevant if you plan on using this.
Comment: Thanks, worked for me! Is it possible that Dns.GetHostName() may return a name different from NETBIOS name ('cause i use it to construct local Windows group names)?
Comment: @User: Well, `GetHostName` will return the DNS hostname, but I can't remember how that relates to the netbios name.
Answer: According to this MSDN article, its case - insensitive
http://msdn.microsoft.com/en-us/library/ms724220(VS.85).aspx
In which scenario, it is doing case sensitive comparison ?
Comment: I've just retested it and can confirm that if i have role as "MYBOX\SupportPersonnel" PrincipalPermission throws the exception but in case of "mybox\SupportPersonnel" everything works as expected. My environment is Windows 7 x64.


Illustrative example of a quality post in FAQ or elsewhere

Question: Just an opinion with insights as a bilingual. I think the FAQ or another area on SO could benefit from a clear illustrative example of what constitutes a quality question in terms of structure. I mean a newcomer who has a A1/A2 (or even B1 CEFR?) level of English won't necessarily catch on to mentions of "quality standards" or format.
Comment: According to [data.se], there are over two *hundred* **thousand** open questions with a score of 5 or better. If those don't provide sufficient example as to 'what constitutes a quality question', I'm not sure what would.
Comment: We have editing badges for those who like to fix these types of questions. A sample boiler plate question isn't going to solve much and there is already a definitive list of what people should and shouldn't do with their questions.
Comment: People are just as unlikely to read and learn from such an example as they are to read the various SE FAQs or blog posts like [this one](http://dba.blogoverflow.com/2012/06/help-us-help-you/), [this one](http://msmvps.com/blogs/jon_skeet/archive/2010/08/29/writing-the-perfect-question.aspx) or [this one](http://catb.org/esr/faqs/smart-questions.html). What makes you think this example post would be a raging success in comparison? How would you make such a post that caters to PHP, MySQL, SQL Server, Rexx, VBScript and Oracle?
Answer: Well, it really depends on the nature of the question, and it isn't just about language to me.
A good question gives as much information as possible about the environment, hardware and other such things.
A good answer is clear and stands on his own merits.
On SU, we've fixed up answers in Russian and found that had it been in English, it would have been a pretty good answer. Bad language quality is fixable. A link only answer might be fixable. Garbage is garbage, and is quite obvious. I don't think its an English issue. It's an effort and overall quality of information problem. Giving a example of a good question or answer linguistically won't help a question with insufficient detail, or a rant in perfect English as an answer.
Comment: As I understood the OP, he thinks (correctly, in my opinion) that some expressions used by native speakers ("quality standard") are difficult to grasp, and an example would be easier to understand.
Comment: I'd agree up to a certain point. But wouldn't it be possible to put an illustrative example (a picture is worth a thousand words) along with some structure guidelines ("situation, problem, attempts" or "context, question") or tips ("go to the point", "imagine you are reading your own question"...). These might seem obvious reflexes for some but they are far from acquired for a good portion of people including myself. A post may seem like garbage and a lack of effort on behalf of a poster but I'm sure you've lived a situation personally where form didn't communicate your intent properly.
Comment: @Monolo Spot on. It can't really hurt to have an illustration to add to the FAQ or elsewhere. I mean a lot of the programmers have learned by example. I am aware that a few visitors are content with the "gimme the codez" approach but lets avoid generalization :)
Comment: @JamesPoulson that reminds me of an old proposal of mine: http://meta.stackexchange.com/questions/481/whats-the-single-biggest-barrier-to-entry-on-stack-overflow/521#521


Neural Network: Approaching the input layer with object abstraction

Question: I have recently completed a course on Machine Learning and am now attempting to apply my knowledge by creating a neural network that attempts to, given a series of $n$ candlesticks (stock market data), predict what the next candlestick might be. I have over 1 million candlesticks available as a training set.
Each candle contains five variables that help shape it: Open Price, High, Low, Close Price, and Volume. In addition to the $n$ candlesticks I feed the network, my goal is to have it output a candle with these same variables filled in. How can I best approach this?
Let's assume I set up the network to make a prediction based on $n=10$ previous candles. Would I be required to have $10*5=50$ input neurons, five for each variable spanning a total of $10$ candles? In class we made simple 2-input, 1-output networks and didn't spend much time on creating an optimal network layout, so using that many features seems like a foreign concept to me.
It also raises the question of efficiency. If I train with backpropagation, which is very computationally expensive, I feel like having a total of $50$ input features would be very inefficient to train, especially considering the size of my training set (and this issue would likely persist even if I set aside 20% for cross-validation). Would it then be slightly more optimal to have separate networks for each variable, so that each network has only $10$ inputs and one output?
Edit: For a bit more clarity, here's an explaination of a candlestick.
Say I'm looking at the 1-minute candle chart. This means that each candle on the graph represents one minute of stock price history. As soon as a minute passes, a new candle is created representing the new minute, and it's open value is whatever the last traded price was. So if the last trade was for \$150, the "open" for this candle is \$150.
Then lets say 10 trades take place that drive the price down to \$140. \$140 is so far the lowest price point for this minute/candle, so the open stays \$150, the high stays \$150, the volume increases by however many shares were traded, and the low is set to \$140. The candle is red because the current price is now below the open price.
Then, let's say another 10 trades occur and drive the price up to \$160. Now the high is \$160, the low is still \$140, the open is \$150, and the volume has yet again increased by however many shares were traded. Now that the current price is above the open price, the candle is green, and there is a thin line on the bottom of the candle showing that the price dipped to that point, but recovered.
Now 12:02 rolls by, and the candle closes at \$155. There are thin lines on both the top and the bottom representing where the price had been in that timeframe, and the solid body of the candle represents where it opened and closed at.
Here's a picture of what it might look like:

Granted, this is a 1-day candle chart, so each candle represents the passage of 24 hours. And as an example, the very first candle is red, so we can assume that it started close to 2708, dipped down to 2705ish, and recovered to 2707 before the candle closed and a new day started
Comment: Please register &/or merge your accounts (you can find information on how to do
this in the **My Account** section of our [help]), then you will be able to
edit & comment on your own question.
Answer: This is likely a a good candidate for LSTMs. To your point, it seems excessive to have 50 inputs, 5 for each previous candle. With an LSTM, your inputs are the hidden state from before, plus the current candle. The simplest thing you can try is an LSTM network, where the final output passes through 5 additional neurons which predict the individual features of the candle.


why does current like circulating?

Question:
why  does the current in R2 remain always zero regardless the value of other parameters(V1,V2,R1,R3)?)
Comment: Note that there is no need to use Paint to draw schematics. There is a button on the editor toolbar that has a proper schematic editor.
Comment: because of a potential difference. answered q in title...
Answer: Let me turn the question around. "How much current do you think should flow through R2?"
No current flows because there is no potential difference across R2.
Another way of looking at it is that if current did flow somehow that there would be an imbalance of charge which would cause current to flow the other way. Imagine that (conventional) current flowed from left to right through R2 then the left side would take on a negative charge and the right side would take on a positive charge. That would create a potential difference or voltage between the two sides so the current would flow from + to -, that is, from right to left. The result would be that the voltage and current would fall to zero.

A water analogy might help. The two voltage sources in your diagram would be replaced with circulation pumps. Water will be pumped around the V1 R1 loop and around the V2 R3 loop. Water can't go from one loop to the other as there is no return path.
Answer: This is more a physics question, not normal electronics.
OP question is probably caused by a common physics misconception:  "Current" is not conserved.  Current isn't substance-like.   It does not flow, instead it just appears and vanishes.  (Wave a magnet near copper, and current springs into being from nothing, with no source or sink.)
On the other hand, charge is conserved.  Charge is substance-like.   Wave a magnet near copper plates, and the sea of mobile electrons within the surface will begin moving, and a current will appear inside the metal.  Where is this charge?  All metals are always full of electrons.
And  ...if charge is like water, then electric current is like water-motion. In other words, the phrase "flow of watermotion" is wrong. The correct phrase is simply flow of water.  Water moves from place to place, while "watermotion" just appears and vanishes.  (What flows in rivers?  "Current?")   And in E&M physics, the phrase "flow of current" is just as wrong.  Electric current is itself the flow: it's a flow of charge.  So, the correct phrase is simply "flow of charge."  In circuits, charge flows from place to place, while current just appears or vanishes.  Everything becomes simple if we strike out "flow of current" everywhere, and instead speak of "charge flows."  Like below:

"Since a current is a flow of charge, the common expression 'flow of current' should be avoided, since literally it means 'flow of, flow
of charge.'" - MODERN COLLEGE PHYSICS, Richards, Sears, Wehr,
Zemanski

Still we can think like an EE rather than a physicist.  In the OP schematic, if R2 is always full of charges, and there is a flow of charge inside R2, then the charge has no return path, and it must build up in the two circuit-loops shown above.  Treat them as two capacitor plates.  We add a capacitor "C1" (1pf?)  Perhaps place it between R1 and R3.
Now, if there is 1 amp in R2, then the potential between the two above loops will increase by 1,000,000,000,000 volts per second!  (One ampere through one picofarad.)  That's not going to happen, so what if there were only 1 microamp in R2?  Then the voltage between the two loops will only increase by a megavolt per second.  Big nope.
Final answer: in a real circuit, for DC supplies during turn-on, perhaps a few nanoamps briefly appear in R2.  It depends on the exact points where parasitic 1pF C1 is connected.  But then it dies away almost instantly.  After that, the flow inside R2 is zero, since there's a C1 picofarad capacitor in series with R2.
The OP question is for a DC circuit.  What if the power supplies were AC?   In that case we'd see a small flow in R2, where the return-path for the complete circuit was the invisible capacitor.  The closed circuit goes through the space between the two loops.  (Better add even more low-value parasitic capacitors, between every circuit node.)

The answer is clear: charges in DC circuits can only flow in loops, since otherwise we'd get billions of volts appearing.  To eliminate any need for million-volt power supplies, our circuitry must resemble flywheels made out of charge: closed rings.  "Complete circuits."  Or instead, if your flashlight batteries were rated in billions of teravolts, then perhaps you could create some milliamps in open loops for awhile, until the picofarads of space-capacitance became filled with enormous e-fields.   In the real world we'd just get lightning bolts.   To work with such teravolt D-cells or tiny hand-held Wardenclyffe towers, either use them in hard vacuum high earth orbit, or embed it all in solid quartz!
:)
too long didn't read; wires are always full of electrons, so circuits are like flywheels made of electrons.  Cutting a circuit is like applying a brake to the flywheel.  That gives us the closed-circuits law.  In above OP circuit, we have two flywheels, but R2 isn't part of either one.  To cause a continuous current inside R2, R2 has to be one part of a wheel made of movable charges.
Comment: Judging from OP having to ask this question in the first place, he probably isn't very knowledgeable. Arguing semantics and mentioning parasitics in the context of what's likely intended to be an ideal circuit isn't helpful (although entertaining to do among EEs or physicists), sorry. You're right that "flow of current" is redundant, but that's not what was asked about.
Comment: @SwapnilSaha  glad you liked it!  Engineers tend to hate this stuff, since it's correct physics, yet exposes such problems as "current carriers" and "conservation of current." Neither one actually exists.  Physics calls them charge carriers, and the law called Charge Conservation not "current" conservation.  Physics always trumps double-E. But double-E fights back. Removing such confusing mistakes only happens over decades. (Well, unless internet memes are harnessed. If errors are like textbook-disease, then we design and release a bacteriophage that finds and cures the disease unsupervised.)
Comment: @jms aha, *needs a TLDR.*  I go by my own experience: I was just like the OP until I learned that circuits are full of charges, and they behave like flywheels made of electrons.  I'd been taught that circuits are hollow, and electrons are "particles of current" supplied by the battery, and that "current flows."  I instantly understand this circuit if I first know that "wires are stuffed full of coulombs," and also that current is just charge-flow, coulombs per second. We do see these concepts in EE texts, but buried in math.  They should be taught right at the start, in grade school.

why does current like circulating?


Is it OK to use UK Visitor Visa for multiple entries with different reasons

Question: My girlfriend had applied a standard visitor visa last year and her stated reason was tourism. Is it OK for her to visit me now, as it's now for a different reason?
Comment: what is the reason now? It could continue to be tourism. She is not going to work or stay in the country...
Comment: Is it a multiple entry visa she already used once last year and now wants to use again for another visit, or a single entry visa she applied for with one reason but now wants to use for something else?
Comment: Were you her boyfriend at the time of the Visa application? if so was she honest about it?
Comment: @PeterGreen  Thanks for your reply! I was her boyfriend at the time. She didn't state she wanted to visit me because at that time her college arranged a trip to UK. I wasn't in UK at that time yet.
Comment: @MarcelP. The reason I want to check this is because she might end up applying for the Tier  dependant visa in the future (I'm current holding Tier 2 working visa in UK).
Comment: @ZachLipton It's a multiple entry visa now want to be used again for another visit.
Answer: With a Standard Visitor Visa, she can come and go and do any activity listed in Appendix 3 of the Immigration Rules Appendix V visitor rules. This holds true even if the stated purpose on the application now differs from the current purpose. Among them, you'll see

Tourism and leisure
  3 A visitor may visit friends and family and / or come to the UK for a holiday.

Of particular relevance, as this question is frequently asked on this site, @GayotFow explained that:

On 24 April 2015 all existing visitor visas, including family visitor, business visitor, student visitor, child visitor, and others were automatically converted to the 'standard visitor visa'.
[ ]
  The controlling link if you are interested is the Immigration Act 2014.

For additional reference: Can I travel to the UK for business on a C visit visa?


deval function in matlab

Question: I can't understand how the deval function works in matlab, it's hard for me to understand the code, are there any books or descriptions of numerical methods where I can read about numerical algorithms, how to evaluate solutions of differential equations at the right points?
I need to implement this feature myself.
Thanks for the help!
Comment: What exactly you do not understand? This function just evaluates the solution of a differential equation at some desired points. This is basically some interpolation. You will have to be more precise about what you want to do.
Answer: The solvers in last mode mentioned in the documentation of them produce as part of the returned object a "dense output" solution, that is, a piecewise polynomial interpolation, represented by the coefficients for every segment. Many of the standard solvers can produce this interpolation directly from the values of the ODE function produced during the solver step, or with a few additional evaluations.
So deval has just to determine the correct segment, the offset inside that segment, and evaluate the polynomial.

deval function in matlab


"Unicorn": what other words have this "cornus" etymology?

Question: "Unicorn" comes from the French and late Latin, with the "cornus" part meaning "horn". I am wondering what other English words share this root. I could think of "rhinoceros". Can you think of something (or multiple somethings) else?
Comment: Rhinoceros? Sure, it's the same PIE root, but from Greek rather than Latin.
Comment: The ancient Biblical Hebrew word for horn (of an animal) is "keren."
Anyone else think that may be where the Latin connection comes from?
Comment: @Faith Schames - An interesting hint. In my view there might be a connection. But there is not much to be found on the Internet about possible relations between semitic and IE words.
Comment: @FaithSchames It has been suggested and is reasonably commonly held that the Proto-Indo-European root was borrowed either to or from Proto-Semitic at some point (similar to the number seven, PIE _*septḿ_, which is almost universally considered a loan word from Semitic). Unlike with 'seven', though, there's nothing unusual about the root in either language, so we don't know which way the borrowing went.
Answer: The Latin word for horn is cornu, stem cornu- (with null-inflection in the nominative case). Note that Latin cornus, "cornel/dogwood", comes from a different Proto-Indo-European root and is not related. Rhinoceros comes from Greek keras, horn.
Both Latin cornu and Greek keras come from the same PIE root *ḱer- (very frequently, and seemingly somewhat at random, expanded to *ḱerh₂- ), which meant something like "horn, head" (note that there appear to be other PIE roots *ker- that are not related).
The examples Snumpy and Trideceth gave all come from Latin cornu-. The following words come from the PIE root *ḱer- through Latin, but not from Latin cornu-:

cervix
cerebral
triceratops
rhinoceros

The word horn comes from the same PIE root, but not through Latin: the /k/ sound was lenited to /x/ or /h/ in Proto-Germanic, as Colin Fine said, which is why we have /h/ now, just as in other Germanic languages.
Comment: The most obvious seems to be "cornet".
Comment: +1 for a very good answer. But a nitpick: /k/ did not "disappear" in Germanic (the way /p/ really did disappear in Celtic, for example) but was lenited to /x/ or /h/, just as /p/ was lenited to /f/.
Comment: To be exact, the PIE root is _*ḱer-_ (very frequently, and seemingly somewhat at random, expanded to _*ḱerh₂-_), rather than _*ker-_.
Comment: @ColinFine: Ack, I knew someone would catch me! At first I had an even worse expression. Then I changed it to this, but I knew I should have looked up what happened exactly to the k, apart from the fact that it is now an h; I knew this was weak, but I was too lazy to look it up. I will incorporate your comment into my answer, if you don't mind. I figured "disappeared" might be taken as "it didn't exist after a certain time, when something else came instead of the k"; but I know that isn't the right way to say it.
Answer: cornucopia
corneous
corner
tricorn
and, believe it or not, corn
Comment: Presumably you mean a *corn* on the foot but not as in grain.  [Different roots](http://www.etymonline.com/index.php?term=corn)
Answer: Previous answers listed cervix, cerebral, triceratops, rhinoceros, cornea, Capricorn, cornicen, corniform, cornucopia, corneous, corner, tricorn, corn (on the foot), and cornicle as being from the same PIE root. There are others, including hirn, keratin, kerato- and anything formed therefrom, cerebro- and anything formed therefrom, many (but not all) other corn- and -corn words, and, possibly, Cornelia. Unrelated are cerul- words, ceremony, karma (related to ceremony, though), hurl, and, as someone else has mentioned, corn (grain).
Answer:
cornea
  Capricorn
  cornucen
  corniform
  cornicle

There are many medical terms as well, which I'll not bother mentioning.
Answer: bicorn and tricorn (as in 18th century hats)


how to get all records in batch job?

Question: How to get all record in batch? my requirement need to get all records from a object. but, SFDC batch size limit is 2000, even if i use iterable batch still can not get all records. If i use Database.Stateful(), it will trigger to "instacne too big" or "string to big". For example, my object records more 30K, i need to get all and compare all record base on some condition. Searching google, looks like can not get all record. So, would you mind give me a solution for me?
Comment: The batch job reads all the records, in execution batches of 2000 per execution of the "execute" method. Why do you think you need to have all the records available at once?
Comment: I need to compare all result and de-duplicate(meger) record. if 2000 per execution, I still can not get all.
Comment: You dont need to compare the whole set at once. Take each batch and compare against the db.
Answer: In the start() method you can query all the records of your object (up to 50 million providing you use the QueryLocator rather than an Iterable). The 2000 limit is the number of objects sent per batch. For example, if your start() method returns 50 000 objects, you will have 25 batches (25 * 2000).
So you can do like this :
global class MyBatchClass implements Database.Batchable<sObject>{

   global Database.QueryLocator start(Database.BatchableContext BC){
      String query = 'select id from Account'; // Example to get all your Accounts

      return Database.getQueryLocator(query);
   }

   global void execute(Database.BatchableContext BC, List<sObject> scope){
       ...
    }

   global void finish(Database.BatchableContext BC){
       ...
    }

http://www.salesforce.com/us/developer/docs/apexcode/Content/apex_batch_interface.htm
Comment: Don't you mean 25 batches?

how to get all records in batch job?


Early Rose Rosette Disease?

Question:

I have several different rose bushes that I am concerned about. The buds and leaves are unusual shapes. They don't have the classic extra thorns or red leaves of RRD.  This just showed up in late August when the Japanese beetles were ravaging my roses. I was spraying with pesticides because I was desperate. I have not sprayed since mid-August.
Answer: Early Rose Rosette Disease, also known as Rose Rosette Virus (RRV), is a viral disease that affects roses. It is caused by the Rose Rosette Virus, which is transmitted by microscopic eriophyid mites. Here is some information about Early Rose Rosette Disease:
Symptoms:

Abnormal growth: Infected roses may exhibit excessive thorniness and the development of multiple, thin stems emerging from a single bud (witches' broom).
Red discoloration: New growth and stems may display a reddish or purplish coloration.
Deformed leaves: Leaves on infected plants may be distorted, elongated, and have a mottled or yellowish appearance.
Excessive thorns: Infected roses may produce an abundance of thorns, more than what is typically seen in healthy plants.
Stunted growth: Affected plants may experience reduced vigor and fail to grow to their full potential.

Control and Management:

Early detection: Regularly inspect your rose plants for any signs of the disease. If you notice symptoms, take immediate action to prevent the spread to other plants.
Pruning: Remove and destroy any infected plant material, including stems, buds, and roots. Be sure to disinfect your pruning tools between cuts to avoid spreading the virus.
Mite control: Since eriophyid mites are responsible for transmitting the virus, managing mite populations is crucial. Use miticides recommended for eriophyid mites, following the instructions provided by the manufacturer.
Quarantine: If you have an infected rose, consider isolating it from healthy plants to prevent the spread of the disease.
Resistant varieties: When selecting new roses for your garden, choose varieties that are known to be resistant to Rose Rosette Virus. Resistant varieties are less likely to develop severe symptoms if infected.
Monitor and remove wild roses: Wild roses can act as a reservoir for the disease. If you have wild roses growing in or near your garden, monitor them closely and remove any infected plants to prevent the spread of the virus.

It's important to note that there is no cure for Rose Rosette Virus. If the disease becomes widespread or severe, it may be necessary to remove and destroy infected plants to protect nearby healthy roses. Consulting with a local horticulturist or plant disease specialist can provide more specific advice for managing Early Rose Rosette Disease in your area.
Comment: Sorry it also sounds like chatgpt.
Answer: IMO this does look like early stages of rose rosette. I see you posted this 8 months ago - what did you end up doing?

Early Rose Rosette Disease?


Mobile data connection lost on iPhone 5S

Question: I have iPhone 5S with 10.1.1 iOS and sometimes happens to me that I loose mobile data connection. In other words in top left corner on my screen there is provider (operator) and right next to it I usually have LTE. This LTE disappears from unknown reason (even in places that are 100% covered by LTE, but that is ok, I can accept it) but the problem is that it doesn't come back until I restart iPhone completely. I tried to restart data connection or some other things, but nothing worked. Only turning off and on again completely. Has anyone experienced this or have suggestions? Many thanks
Comment: Have you tried resetting network settings and checking for a carrier update?
Answer: This is probably not a problem with your phone. It is up to the carrier to decide which individual handsets are actually allowed on LTE on the current cell tower. So you might just be experiencing that your carrier hasn't got enough capacity to allow everyone on LTE, so they put you on a slower network.
So even though a coverage map tells you that LTE is available - it doesn't necessarily mean it is available to everyone.


HighlightGraph Problem

Question: Bug introduced in 9.0 or earlier and fixed in 11.3

HighlightGraph does not work.
HighlightGraph[Graph[{{0, 0}, {0, 1}, {1, 0}, {1, 1}}, {{0, 0} \[UndirectedEdge] {0, 1}}], {{0, 0}, {0, 1}}]

In Mathematica 9 I see:
Comment: You should report the problem to Wolfram Support.  I can reproduce it in 10.4.1.
Comment: Maybe: `HighlightGraphR[g_, vertices_] :=
 HighlightGraph[
  VertexReplace[g,
   Thread[VertexList[g] -> ToString /@ VertexList[g] ]],
  ToString /@ vertices]`
Comment: a simpler example : `HighlightGraph[Graph[{{1}, {2}}, {{1} <-> {2}}], {{1}, {2}}]`
Answer: We can reproduce the problem in a simpler example:
g = Graph[{{1}, {2}}, {{1} <-> {2}}]

HighlightGraph[g, {{1}, {2}}]

HighlightGraph invokes the function GraphComputation`GraphHighlightDump`vertexEdgeExtract to get the list of vertices and edges to be highlighted. For the example above, this function returns {1,2} as the list of vertices to be highlighted,
GraphComputation`GraphHighlightDump`vertexEdgeExtract[{{1}, {2}}, g]

{{1, 2}, {}}

not {{1},{2}}. Since 1 and 2 are not in VertexList[g] nothing is highlighted.
The culprit lies deeper in a function called by GraphComputation`GraphHighlightDump`vertexEdgeExtract, namely, in the function GraphComputation`GraphHighlightDump`veExtract. The code for this function is
GraphComputation`GraphHighlightDump`veExtract[
   GraphComputation`GraphHighlightDump`set_,
   GraphComputation`GraphHighlightDump`g_,
   GraphComputation`GraphHighlightDump`vbag_,
   GraphComputation`GraphHighlightDump`ebag_] /;
  VertexQ[GraphComputation`GraphHighlightDump`g,
   GraphComputation`GraphHighlightDump`set] :=
 Internal`StuffBag[GraphComputation`GraphHighlightDump`vbag,
  GraphComputation`GraphHighlightDump`set, 1]

If we remove the 1 in the last line, i.e., change the definition of veExtract to
GraphComputation`GraphHighlightDump`veExtract[
   GraphComputation`GraphHighlightDump`set_,
   GraphComputation`GraphHighlightDump`g_,
   GraphComputation`GraphHighlightDump`vbag_,
   GraphComputation`GraphHighlightDump`ebag_] /;
  VertexQ[GraphComputation`GraphHighlightDump`g,
   GraphComputation`GraphHighlightDump`set] :=
 Internal`StuffBag[GraphComputation`GraphHighlightDump`vbag,
  GraphComputation`GraphHighlightDump`set]

then we get the correct vertex list to be highlighted:
GraphComputation`GraphHighlightDump`vertexEdgeExtract[{{1}, {2}}, g]

{{{1}, {2}}, {}}

TO DO: I couldn't figure out how/where to make the appropriate changes to get HighlightGraph working as expected.
Notes: An example to see how the third argument of Internal`StuffBag is working:
bag1 = Internal`Bag[];
bag2 = Internal`Bag[];
Internal`StuffBag[bag1, #, 1] & /@ {{1}, {2}};
Internal`StuffBag[bag2, #] & /@ {{1}, {2}};

Internal`BagPart[bag1, All]

{1, 2}

Internal`BagPart[bag2, All]

{{1}, {2}}
Comment: which version of Mathematica are you using? I got ``In[7]:= GraphComputation`GraphHighlightDump`vertexEdgeExtract[{{1}, {2}}, g]

Out[7]= {{{1}, {2}}, {}}``
Comment: @halmir, i am using version 9 (windows 10)
Answer: This is clearly a bug, which appears when the the vertex names are lists.  You should report it to Wolfram Support: http://support.wolfram.com/
I can reproduce it with version 10.4.1.
Proof that lists as vertex names are reasonable: some functions return such graphs.  Example:
pts = RandomInteger[{1, 5}, {10, 2}];
g = NearestNeighborGraph[pts]

VertexList[g]
(* {{2, 5}, {5, 2}, {5, 5}, {4, 5}, {3, 3}, {5, 1}, {1, 3}} *)

HighlightGraph[g, Take[pts, 2]] (* Doesn't highlight *)

HighlightGraph[g, {Take[pts, 2]}] (* Also doesn't highlight, even though this syntax is unambiguous *)

Answer: Okay, forget what I wrote in the other post. It took quite a while but I think I finally found the bug. That would not have been possible without the great spelunking tool PrintDefinition. The issue is within the function GraphComputation`GraphHighlightDump`setHighlight:
Needs["GeneralUtilities`"]
HighlightGraph;
PrintDefinitions[GraphComputation`GraphHighlightDump`setHighlight];

setHighlight[g_, i_] /; VertexQ[g, i] :=

  (PropertyValue[{g, i}, GraphHighlight] = True;);

setHighlight[g_, i_List] := Scan[setHighlight[g, #1] &, i];

setHighlight[g_, i_] /; EdgeQ[g, i] :=

  (PropertyValue[{g, i}, GraphHighlight] = True;);

Attributes[setHighlightProperties] := {HoldAll};

Let's consider kglr's toy model:
g = Graph[{{1}, {2}}, {{1} <-> {2}}];
data = {{1}, {2}};
HighlightGraph[g, data]

After calling HighlightMesh[g,data] (after traveling through a maze of calls to many other functions), we arrive at the call setHighlight[g_, {{{1},{2}},{}}] (here, {} represents the empty set of edges to be highlighted).
This activates the second  SetDelayed for setHighlight and so we start to scan through the set {{{1},{2}},{}}.
At some instance, we arrive at the call
PropertyValue[{g, {1}}, GraphHighlight] = True

and although there is no error message, it fails. We also cannot retrieve any property values with PropertyValue[{g, {1}}, <prop>], even if we try to set them beforehand (e.g. with the VertexLabel option). So, this is pretty broken. Note that PropertyValue is a kernel function and we can do nothing about it. However, we can prevent that these calls are made by making the following redefinition which has essentially the same effect (modulo a deep copy of the graph). So, this is my proposition for a fix:
Unprotect[HighlightGraph];

HighlightGraph =.

HighlightGraph := (
   System`Dump`AutoLoad[
     Hold[HighlightGraph],
     Hold[HighlightGraph],
     "Network`GraphHighlight`"
     ] /; System`Dump`TestLoad;

   GraphComputation`GraphHighlightDump`setHighlight[g_, i_] /;
   VertexQ[g, i] := Block[{gh},
     gh = GraphHighlight /. Options[g];
     If[Length[gh] == 0, gh = {}];
     (*This is a very slow workaround, only meant as fallback.
     Should be avoided at all cost!*)
     g = Graph[g, GraphHighlight -> Join[gh, {i}]];
     ];

   GraphComputation`GraphHighlightDump`setHighlight[g_, i_List] :=
    Block[{gh},
     With[{vlist = Select[i, VertexQ[g, #] &]},
      gh = GraphHighlight /. Options[g];
      If[Length[gh] == 0, gh = {}];
      (*This enforces a deep copy (?) of the graph but does it only once for all vertices detected.*)

      g = Graph[g, GraphHighlight -> Join[gh, vlist]];
      (*Proceed as usual for all remaining data.*)
      Scan[
       GraphComputation`GraphHighlightDump`setHighlight[g, #] &,
       Complement[i, vlist]
       ];
      ]
     ];

   GraphComputation`GraphHighlightDump`setHighlight[
    g_,
    GraphComputation`GraphHighlightDump`iSubgraph[v_, e_]
    ] := (
     (*Parse vertex and edge list; do not use Scan in order to avoid the call to setHighlight[g, i]/;VertexQ[g,i]:=...*)
     GraphComputation`GraphHighlightDump`setHighlight[g, Join[v, e]]
     );

   GraphComputation`GraphHighlightDump`setGroupProperty[g_, i_, color_] /; VertexQ[g, i] := Block[{gh, style},
     {gh, style} = {GraphHighlight, GraphHighlightStyle} /. Options[g];
     If[Length[gh] == 0, gh = {}];
     If[Length[style] == 0, style = {}];
     (*This is a very slow workaround, only meant as fallback. Should be  avoided at all cost!*)
     g = Graph[g,
       GraphHighlight -> Join[gh, {i}],
       GraphHighlightStyle ->
         Join[style, {i -> {VertexStyle -> color}}];
       ];];

   GraphComputation`GraphHighlightDump`setGroupProperty[g_, i_List, color_] := Block[{gh, style},
     With[{vlist = Select[i, VertexQ[g, #] &]},
      {gh, style} = {GraphHighlight, GraphHighlightStyle} /. Options[g];
      If[Length[gh] == 0, gh = {}];
      If[Length[style] == 0, style = {}];
      (*This enforces a deep copy (?) of the graph but does it only once for all vertices detected.*)
      g = Graph[g,
        GraphHighlight -> Join[gh, vlist],
        GraphHighlightStyle ->
         Join[style, Table[j -> {VertexStyle -> color}, {j, vlist}]]
        ];
      (*Proceed as usual for all remaining data.*)
      Scan[
       GraphComputation`GraphHighlightDump`setGroupProperty[g, #, color] &,
       Complement[i, vlist]
       ];
      ]
     ];

   GraphComputation`GraphHighlightDump`setGroupProperty[
    g_,
    GraphComputation`GraphHighlightDump`iSubgraph[v_, e_],
    color_
    ] := (
     (*Parse vertex and edge list; do not use Scan in order to avoid the call to setGroupProperty[g,i]/;VertexQ[g,i]:=...*)
     GraphComputation`GraphHighlightDump`setGroupProperty[g, Join[v, e], color]
     );

   (*Making sure that kglr's bugfix is available in older versions, too.*)
   GraphComputation`GraphHighlightDump`veExtract[set_, g_, vbag_, ebag_] /; VertexQ[g, set] := (
    Internal`StuffBag[vbag, set]
    );

   (*Not sure whether this is set correctly in older version.*)
   GraphComputation`GraphHighlightDump`veExtract[
    GraphComputation`GraphHighlightDump`iSubgraph[v_, e_], g_, vbag_, ebag_] := (
    Internal`StuffBag[ebag, e, 1];
    Internal`StuffBag[vbag, v, 1]
    );

   HighlightGraph
   );

Protect[HighlightGraph];

This seems to work, at least for the examples from the other post. Note that I am using version MMA 11.0.1 and that this version does not contain the bug in veExtract that kglr found.
Of course, one can enforce to break it by introducing ambiguities like these:
g = Graph[{1, 2, {1, 2}, {{1, 2}}}, {UndirectedEdge[1, 2],
    UndirectedEdge[1, {1, 2}]}, VertexLabels -> "Name"];
HighlightGraph[g, {{{1, 2}}, 1, 2, UndirectedEdge[1, 2]}]

So, introducing a Vertex primitive might still be a good idea.
Improvements and suggestions are welcome!
Update

Added a fix for a case that {1} and not 1 is the only vertex to
be highlighted.
Reformulated the fix such that it gets applied automatically after autoload.
Fixed the issue for Subgraph.
Defined new rules for setGroupProperty  in order to fix the issue for lists of list.

Here is a demonstration of the newest changes:
n = 24;
m = Floor[n/3];
SeedRandom[123];
H = RandomGraph[{n, 2 n}, VertexLabels -> "Name"];
rules = Normal@AssociationThread[Transpose[{Range[n]}], Range[n]];
G = VertexReplace[H, Reverse /@ rules]; // AbsoluteTiming
v1 = DeleteDuplicates@RandomChoice[VertexList[G], m];
v2 = Complement[DeleteDuplicates@RandomChoice[VertexList[G], m], v1];
HighlightGraph[G, {Subgraph[G, v1], Subgraph[G, v2]}]

The original version leads to:

And the fixed version from above produces:
Comment: @Szabolcs Thank you very much for the flowers! Can we expect that the bugfix will be made part of IGraph/M?
Answer: I am afraid, the only unambiguous solution would be to introduce a keyword for vertices, e.g. Vertex. Compare that to the need for Key when addressing associations whose keys contain integers. Let's see what we can do.
Fortunately, HighlightGraph is Protected and ReadProtected but not Locked, so we can mess around with it. We can see the code with
Needs["GeneralUtilities`"]
PrintDefinitions[HighlightGraph];

Warning
People who have the tendency to get nervous when unprotecting symbols: Stop. Reading. Now.
The code
For some weird reason I had to Clear HighlightGraph first, in order to overwrite its definitions. Here is what I do:
Vertex::usage =
  "\!\(Vertex[\*
    StyleBox[\"vertex\", \"TI\"]]\) represents a \
vertex used to access vertices in a graph.";
Protect[Vertex];

HighlightGraph;
Unprotect[HighlightGraph];
ClearAttributes[HighlightGraph, ReadProtected];
Clear[HighlightGraph];

Begin["GraphComputation`GraphHighlightDump`"];

HighlightGraph[graph_, None, opts : OptionsPattern[]] :=
  HighlightGraph[graph, {}, opts];

HighlightGraph[graph_?GraphQ, {}, opts : OptionsPattern[]] :=
  Block[{res}, res = Flatten[{opts}];
   GraphComputation`CloneGraph[GraphComputation`GraphHighlightDump`graph, res]
   ];

HighlightGraph[e_List, {}, opts : OptionsPattern[]] :=
  Block[{res, graph},
   graph = Graph[e];
   (
     res = Flatten[{opts}];
     GraphComputation`CloneGraph[graph, res]
     ) /; GraphQ[graph]];

HighlightGraph[graph_?GraphQ, data_, opts : OptionsPattern[]] :=
  Block[{res, nopts, dopts, n, vertexlookup, vertexreverselookup,
    edgelookup, edgereverselookup, g, h, gdata},
   nopts = FilterRules[Flatten[{opts}], Options[HighlightGraph]];
   dopts = Complement[Options[HighlightGraph], Options[Graph]];
   If[dopts =!= {}, nopts = Join[nopts, dopts]];
   res = If[FirstPosition[data, Vertex] === Missing["NotFound"],
     GraphComputation`GraphHighlightDump`highlightGraph[graph, data, nopts]
     ,
     n = VertexCount[graph];
     vertexlookup = AssociationThread[VertexList[graph], Range[n]];
     vertexreverselookup =
      AssociationThread[Range[n], VertexList[graph]];
     With[{edges = EdgeList[graph] /. vertexlookup},
      edgelookup = AssociationThread[EdgeList[graph], edges];
      edgereverselookup =  AssociationThread[edges, EdgeList[graph]];
      ];
     g = AdjacencyGraph[AdjacencyMatrix[graph]];
     gdata = data /. KeyMap[Vertex, vertexlookup] /. edgelookup;
     h = GraphComputation`GraphHighlightDump`highlightGraph[g, gdata, nopts];
     Graph[graph,
      Join[{GraphHighlight -> (GraphHighlight /. Options[h] /. vertexreverselookup)},
      FilterRules[Options[h], Except[GraphHighlight]]]]
     ];
   res /; res =!= $Failed];

HighlightGraph[e_List, data_, opts : OptionsPattern[]] :=
 Block[{res, h, nopts, dopts, vlist, elist, graph},
  graph = Graph[e];
  (
    res = HighlightGraph[graph, data, opts];
    res /; res =!= $Failed
    ) /; GraphQ[graph]]

HighlightGraph[expr___] := (messageHighlightGraph[expr]; Null /; False);

End[];
Protect[HighlightGraph];

The main SetDelayed is the one which is defined for the pattern HighlightGraph[graph_?GraphQ, data_, opts : OptionsPattern[]]. This is where we try to slip in. What I do is the following: The original HighlightMesh has no problems with integer vertices. So, I merely replace all the vertices by integers (that's where we need the wrapper Vertex around vertices appearing in data. Afterwards I run the usual backend of HighlightGraph, grab the options of its output, transform that back to the original vertices and append it to the original graph. That's it.
Some examples
A usage example for the OP's example:
graph = Graph[{{0, 0}, {0, 1}, {1, 0}, {1, 1}}, {{0, 0} <-> {0, 1}},
   VertexLabels -> "Name"];
HighlightGraph[graph, {Vertex /@ {{0, 0}, {0, 1}}}]

And for kglr's example:
g = Graph[{{1}, {2}}, {{1} <-> {2}}];
HighlightGraph[g, Vertex /@ {{1}, {2}}]

Apart from that, I haven't tested it too thoroughly. Maybe this won't work in all cases or break something else. So, better not run any nukular power plant with it. And maybe this is not very performant (quite many replacements that have to be made).
Edit
Added detection for occurence of the symbol Vertex. If not found, the standard implementation is used.
Comment: While it is true that `HighlightGraph[g, {1,2}]` and even `HighlightGraph[g, {{1,2}}]` are ambiguous (is `{1,2}` a vertex or a list of vertices?), `HighlightGraph[g, {{{1,2}}}]` is unambiguous. I actually didn't realize that this syntax is not documented: `HighlightGraph[CycleGraph[10], {{1, 2, 3}, {4, 5, 6}}]`.  The situation is a bit similar to `Extract`: in `Extract[list, 1]`, `1` is an index. In `Extract[list, {1,2}]` we have a single position spec instead of two indices. in `Extract[list, {{1,2},{1,3}}]` we have two position specs.
Comment: What I am trying to say is that I consider failure acceptable when the second argument is a vertex or a list of vertices (for vertices with the head `List`). It should be possible to disambiguate these by providing a list of lists of objects as the second argument of `HighlightGraph`.
Answer: Faced the same issue on Mathematica 11.0
Workaround started with @ZYX's comment extended to PathGraph too for DirectedEdges. However this can be further extended for Optional Parameters
HighlightGraphR[g_, vertices_] := Block[{vt, gt},
    vt = ToString /@ vertices;
    gt = VertexReplace[g, Thread[VertexList[g] -> ToString /@ VertexList[g]]];
    HighlightGraph[gt, PathGraph[vt, DirectedEdges->True]]
]

HighlightGraphR[g, v];

HighlightGraph Problem