If someone screenshots a photo you posted to social media and reposts it is that against any laws?

Question:
The account centers around taking screenshots of people's photos and pairing them with another person as a "couple".
I don't know about the other pictures but mine did not have consent.
This is on Instagram.
I only found out when a friend texted me.
I asked twice that it be taken down and both times I was refused.
Comment: What is the jurisdiction? e.g. country/state/territory/etc/
Answer: I depends on who took the picture: that is the person who controls what can be done with it. For example, if you took the picture, you own the copyright in the picture. If some other person took the picture and you are just in the picture, that person owns the copyright, and you need their permission to post it on Instagram. Taking a screenshot is copying, and all copying requires the permission of the copyright owner.
Things are very cut-and-dried when you put things up on a personal web page, but less clear when you use some social media platform, because then you have to navigate their terms of service to see what you have legally done by posting some content. There is generally (i.e. universally, with any serious SM platform) some license that you grant to the platform, and they also grant users some permissions. You simply have to work your way through all of the documentation that seems to be about policies etc. to see what you can and can't do.
Instagram has this terms of use document. On the matter of copyright, your starting point is "Permissions You Give to Us". Their summary statement is that "We do not claim ownership of your content, but you grant us a license to use it". More specifically

you hereby grant to us a non-exclusive, royalty-free, transferable,
  sub-licensable, worldwide license to host, use, distribute, modify,
  run, copy, publicly perform or display, translate, and create
  derivative works of your content (consistent with your privacy and
  application settings). You can end this license anytime by deleting
  your content or account. However, content will continue to appear if
  you shared it with others and they have not deleted it.

You have not, by licensing the content to Instagram, granted any third party a license to copy. Right below this, they start to address third parties (by putting "you" in the position of the third party), and they state that they can take action against you (anyone) who violates the rules or "if you repeatedly infringe other people's intellectual property rights". They do not at any point say "It's okay to copy stuff from another person's account without their permission", though they do not directly say that this is forbidden (see their section on copyright) – they come pretty close by saying "It's possible to infringe someone else's copyright, even if you don't intend to do so. In most cases, you shouldn't use someone else's copyrighted work if you don't have permission". They are not trying to clearly define what constitutes "fair use", because there isn't a clear answer to whether a given use is "fair use", except in a very small number of cases.
On the face of it, this is both copyright infringement and, if the other screenshots are also without permission, a violation of the terms of service. The legal question is answered by filing a lawsuit. Theoretically there is an Instagram-internal method for complaining about copyright infringement – it is probably only visible to logged in users.
Comment: isn't also photographs of people subject to things like model release? not sure, but isn't it another different area of law which could apply?
Comment: @Gnudiff Sometimes "yes", sometimes "only if the photos are used to promote something other than the photos", sometimes "no". https://commons.wikimedia.org/wiki/Commons:Country_specific_consent_requirements


Should the bullet trajectory calculated by server or by client?

Question: In a FPS game like Battlefield or Arma, when playing online, should the bullet trajectory be calculated by the server or by the client?
Comment: I would say that both. Client calculates it, so the gameplay is smooth, and every n-th frame everything is synchronised with the server.
Answer: Always assume the client is a lying, cheating, bastard.
The client is responsible for:

Receiving input from the player (and sending commands to the server, which validates)
Rendering the known gamestate

The client is in no way allowed to calculate the gamestate except as client-side interpolation for smooth animations.
Comment: That header?  I've made that statement a dozen times this year already.  It never ceases to be accurate and straight to the point.
Comment: +1 for making me laugh. You touched just the right parts of this problem.
Comment: hahaha, I agree 100% , have some more internet points!
Comment: @Jon Yay internet points! I always wanted those!
Answer: Note that most of us cannot answers specific questions about either of the games you mentioned. We aren't the authors of that code and can't reasonably know exactly how it was built.
Generally speaking, there are multiple approaches.
One approach is that the server is authoritative on everything. When a client wants to fire a weapon a round-trip to the server is required. The server tells the client which trajectory to use. This approach adds significant latency to firing. Games that use this approach might hide the latency by animation or a smoke particle effect of the like so that the player doesn't realize that the client is waiting on the server (on well-behaved networks).
Another more complicated approach does firing as part of client-side prediction. The client fires the weapon. It tells the server that it fired the weapon along with information about the specific time of the fire request. The server is then able to simulate the same firing trajectory (or very close to). The server's fire message, when received by the client, is used to adjust the projectile (if necessary). Note that the server should still be in charge of determining what the projectile hits and what the results are. The weakness of this approach is that, in some situations, the client can see a firing effect that is wildly different than what actually happens. However, in most cases the client should see something close to what happens, and the firing will feel instant, which makes this approach "feel" good on low-latency connections.
These are not the only two approaches, nor did I describe the most advanced approaches. Networking games is really hard. There's a good reason that network programmers are some of the best paid engineers in the game industry.
Answer: More than likely, it works like this (major speculations):

Client sends Shoot command to server, along with parameters such as
position, direction, velocity, type, etc...
Server accepts command and broadcasts to all players on next frame.
Clients render the shot.
If shot hits a player, server broadcasts kill.

Can't tell for sure though,  as I don't have access to the source code.
Comment: The client only needs to send the "shoot" command; in theory, the server already knows where the client is (position, direction, etc).
Comment: Yeah, I did put this as my original #2... but then I figured, that could very well depend on the design.  For example, a grenade wouldn't immediately know if it hit a person.  Personally, I think this question is pretty much worthless in itself -- You would need to tune it to what you are trying to accomplish to get something out of it.  As it stands now, all we can provide is a general idea of how it works.
Answer: One thing a lot of people say here and is pretty fangerous is to have the player send information about his state when he shoots. The server is a machine, he can't decide who cheats and who doesn't.
"Yes, I really did stand in front of him for 30 seconds without he noticing it, then I shot him in his eye while he was running around in a thight place, looking down...while crouching...on a ledge."
You need to store information about the playsrs server side.
What a client should do

Draw the scene
Talk with the server, sending him his current position, and receiving the positions of other players
Handle keyboard and mouse control.

What a server should do

Have a nice, clean command window.
Manage collision detection
Speak with the client

Servers tend to have a better CPU, than clients, so they can manage more complex things, like collision detection, or player movement without having a hard time.


Why can't I auto-focus on semi-close subjects with my new Nikon 50-300?

Question: I just got a new lens (Nikon DX 55-300) for my Nikon 3300 camera. I am just a beginner, however when I try to focus and capture  something semi- close by (in the same room) it does not allow me to focus or shoot the image. The only way to capture the image is if I flip the manual focus switch on the side. Why is this? Does it have to do with aparture or ISO. Also, I am taking a trip to South Africa and  need all the beginner help I  can get in order to learn my camera quickly.
Comment: Wonder if problem solutions ala "consider getting a diopter, a full macro lens, or extension tubes (not if the lens is internal focusing!)" make good or off topic answers here? Technically, they asked "why does it not work", not "how to make it work" ...
Comment: @rackandboneman I think it's good to mention solutions as long as the "why" is also answered. But there are also other QAs that discuss such solutions that might be worth linking to.
Answer: This lens is advertised as having a minimum focus distance of 1.4m. That means it can't focus on subjects closer than that distance — about 4.5 feet. That might be some of your problem.
It is also likely that indoors is darker than you think it is — sunlight is many times more light than the typical interior. The lens may have trouble focusing in low-light conditions.
Many cameras have what's called "focus priority" shutter modes, where the camera won't click the frame until it's decided that the subject is in focus — and if it can't get focus, it won't let you take a picture. This is the default on your Nikon D3300. (Of course, disabling that mode doesn't solve the focus problem, but it does explain why you can get pictures in manual mode.)
Comment: Worth adding that Nikon bodies come with default settings that do not release the shutter until the autofocus has locked on.  OP was wondering why they had to switch to manual focus to get it to take a photo - it's because in manual focus mode the camera will always take the shot, regardless of whether it thinks the image is in focus or not.  There are autofocus options in the camera, however, which will autofocus with a half-click and release the shutter on a full click without waiting for focus confirmation (ie: the photographer has to decide whether the autofocus has done its job or not).
Comment: What are you trying to focus on? Phase detect focus systems (viewfinder focus) needs contrast to focus. They cannot focus on an all white, black or any color plane. Aim at the line between black and white, and there is no problem.
Comment: Thank you all for this advice. I have found it is a little better outside than inside, however not the best close up.
Answer: What other lenses do you have? What subjects do you plan to photograph? If there is a mismatch between your intended subject matter and lens choice, you should consider getting a more suitable lens. Since you "just got" this lens, you may still be within the return window.

As mattdm mentions, the lens is likely unable to focus closely because of its minimum focusing distance. Although you are able to fire the shutter with the lens in manual mode, the images you capture may not be optimal. If there is sufficient light, stopping down the aperture can widen the depth of field and help make closer objects appear more in-focus.
You may be able to switch the camera to "release priority" mode, where the shutter will fire regardless of whether focus has locked. However, this will likely increase the number of out-of-focus shots you will obtain at all distances.
As tetsujin had formerly noted, the lens has a slow F4.5 max aperture. That would make it a good daylight lens with marginal indoor performance (which mattdm also notes). Attempting to adjust the aperture in this case would be unhelpful because the camera likely focuses with the aperture wide open regardless of the setting.
You can try adding more light, as mattdm comments. This could be in the form of a focus-assist or constant light source. A speedlight without focus assist would not be helpful.
A controversial source mentions that the lens' autofocus is slow. The camera may be giving up on attaining focus before the lens is able to complete its focusing routine. Adding a focus assist light may help. Otherwise, there isn't much you can do besides get a different lens or camera body.
Comment: My other lense is the one it came with the Nikon DX VR 18-55m . I plan to take photos of landscape, people, animals on the safari. Thanks for all your help in this comment.


Gnucash: How to find transactions of a specific amount?

Question: I've got a Gnucash credit card account in which I'd like to find all transactions of a specific amount. The Find window doesn't have the Charge Amount field in it, and I don't see any relevant information in Help. Is this something that I can do in Gnucash?
Answer:
The Find window doesn't have the Charge Amount field in it

It's the "Value" field.
Comment: Aha! "Value" and "charge amount" don't correspond very closely in my head, so that one never occurred to me. Thanks.
Answer: As another option, you can go to "View:Sort By..." and choose "Amount".  This should make it easy to find those transactions in the list.
Note that payments will be listed separately from charges, as they are effectively negative and positive amounts.
Comment: A creative solution. I like it.
Comment: This is good for situations where you don't know the exact amount.


How is 2D spectroscopy able to show quantum coherent transport through networks?

Question: For wave-like (quantum coherent) energy transfer in networks (eg. propagation of excitation in photosynthetic protein complexes of algae or FMO complex in plants) 2D electronic (photon echo?) spectroscopy is used (eg. in this article). I have read an article (unfortunately I can't cite it here) about how it works, but I didn't understand it. I know that the off-diagonal peaks in the spectrum mean some kind of coherence in the network but I don't know why. I also don't know what kind of coherence it means. What are meant by electronic eigenstates? What are the so called 'quantum beatings'? It would be very much appreciated if somebody would be so kind as to make these thing clear.
Comment: Why can't you *cite* it? You can generate an academic citation for anything, even personal communications.
Comment: I don't understand why you can't cite the article here.  You don't remember the title or you don't have permission?
Comment: The second. I could cite the abstract, but I don't see much point in it. It's G. R. Fleming's "Two Dimensional Electronic Spectroscopy of Molecular Complexes" by the way.
Comment: @GeoffHutchison I can cite it, but there's no point since it is not open acces, but since two of you have asked, here you go: M. Cho, T. Brixner, I. Stiopkin, H. Vaswani, and G. R. Fleming, "Two dimensional
electronic spectroscopy of molecular complexes", Journal of the Chinese Chemical
Society, vol. 53, no. 1, pp. 15–24, Sep. 25, 2013. doi: 10.1002/jccs.200600002.
Answer: I'm going to focus on one question here. You really have several and it'd be better if you asked these as separate questions on the site.
First off, 2D spectroscopy involves using multiple pulses, just like multi-dimensional NMR experiments. There are a variety of techniques but in all cases, you're sending some sort of pump pulse which excites the system (e.g., electronic or vibrational levels) and then probing the system over time.
The resulting data is inherently two-dimensional. You know the absorption of the system along one axis, and the emission (or second absorption) along another axis. Often, you can watch this 2D data over time, giving you even more information.
Now how does this show coherence? Let's say you have some complex system. You take a UV/Vis electronic absorption spectra and you see two peaks.

Now normally, you'd think of this system as having two distinct absorptions. Maybe it's a metal complex and you see a ligand-metal or metal-ligand charge transfer band and a d-d transition at longer wavelength.
But perhaps instead, some of the intensity in a peak comes from energy transfer. This can particularly be the case in biology or materials systems, where there are multiple photosynthetic reaction centers with ever-so-slightly different energies. So you excite one center, but then energy dumps into another one (e.g., at lower energy).
By using two-dimensional spectroscopy, that will show up as an off-diagonal peak. You'll excite at one wavelength, but it fluoresces at another one!

That's an example of coupling between oscillators (either vibrational or electronic). Now there are multiple ways that energy transfer could occur. Usually, this happens essentially as a diffusive random walk - all phase information is lost.
In some cases, the wave packet transfers from one state to another, but retains its phase. That's coherent.
In 2D photon echo spectra, you'll see oscillations in the peak heights when coherent transfer occurs, thus "beats."
Comment: For applications of 2DES to photosynthesis, I highly recommend ["Two-dimensional electronic spectroscopy and photosynthesis: Fundamentals and applications to photosynthetic light-harvesting" *Chemical Physics* 386 (**2011**) 1–22.](http://dx.doi.org/10.1016/j.chemphys.2011.04.025)


Book recommendations - Topological Insulators for dummies

Question: Is there a pedagogical explanation of what is a topological insulator for those that do not even know what the Berry phase is but have a basic understanding of quantum mechanics and solid state physics?
Comment: If you don't know what a Berry phase is but what to learn what it is, I recommend Bernevig and Hughes' book. It is not a fast explanation, but a very clear and comprehensive one. Quite some typos though. If you want something even faster, Qi and Zhang's review paper or Hasan and Kane's review paper.
Answer: I think the clear answer is A Short Course on Topological Insulators. There's both a textbook and an arXiv version. I think it's notable for assuming the absolute minimum knowledge and building up from there. It teaches the adiabatic theorem, Berry phase, Chern number, etc all from scratch. It is also notable in being the only textbook on the subject I've seen that doesn't use the second-quantization formalism. The only background necessary is that you understand quantum mechanics, and have seen tight-binding models before.
Answer: I did not start too long ago, but when I did, I used the following resources in the given order. I used the chapters about adiabatic evolution and Berry phase in Griffith's book (they are very short!), then used both this website 2 and this review 3 hand-in-hand for a basic exposition of topological insulators (maybe the section about the Kene-Mele model might be overkill for a very basic overview). These two reads gave me the basic idea, but I still had trouble thinking of things in the bigger picture physically. Then I started working through the book by Shun-Quin 4, along with other lecture slides from the internet (like 5, which has great figures and was written by Haldane, a pioneer in the area). Perhaps, if you are willing to spend the time, starting with Shun-Qing's book might be better for a comprehensive understanding; it starts with most things like Dirac dispersion relations that most other materials take for granted, and helped me fit topological insulators into the general context within the Hall effect.

Introduction to Quantum Mechanics 2nd edition, by Griffiths
Topology in Condensed Matter website
An Introduction to Topological Insulators, by Fruchart and Carpentier
Topological Insulators — Dirac Equation in Condensed Matters, by Shun-Qing Shen
Topology and geometry on condensed matter physics, by Haldane
A Short Course in Topological Insulators, by Asbóth, Oroszlány, and Pályi

Hope this helps!
Comment: I found the resource that Jahan recommended only last week, and it was pretty good!
Answer: A more mathematical approach aimed at physicists (and to really understand topological insulators, you do need quite advanced mathematics) is the book by Emil Prodan and Hermann Schulz-Baldes titled "Bulk and Boundary Invariants for Complex Topological Insulators: From K-Theory to Physics". Strictly speaking, they only cover two of the 10 Altland-Zirnbauer classes, but they do explain and later define what K-theory is, how to derive and understand bulk-boundary correspondences and so forth. They also give a good account of the state of the art at the time of writing (circa 2015) when it comes to experiments.
Prodan also collaborates with experimentalists, so in my mind much of the book is written in a style that is amenable to physicists.
Answer: On top of the excellent recommendations mentioned above, I'd recommend Berry Phases in Electronic Structure Theory by David Vanderbilt. The book is pedagogically written. It starts with simple explanations in terms of polarization in 1D systems and slowly builds up to introduce the concept of Berry Phase and finally talks about topological insulators.
Another great aspect about the book is that the book gives you simple tight-binding codes to reproduce some of the figures in the book using pythTB package. This was super useful to me in order to visualize the concepts mentioned in the book.
I hope this helps.
Comment: This book also has a fabulous discussion of Bloch functions and the various normalization conventions.


Why aren't electrons waves by "default"?

Question: I was reading "The holographic universe" by Michael Talbot and it said that most scientists believe, and there is proof of the fact that an electron is only a particle when we are observing it. Now, when an electron is zooming around an atom it is not being observed (Usually) So an electron on, say, the couch I am sitting on should be a wave instead of a particle, but we are taught from a young age that an electron is a particle zooming around the nucleous of an atom. So why is an electron in an atom a particle and not a wave?
Comment: Quantum objects aren't "waves" or "particles", they're *quantum states*. Thus, this question isn't really meaningful. See also e.g. [Is the wave-particle duality a real duality?](http://physics.stackexchange.com/q/46237/50583)
Comment: @ACuriousMind funny how although this question "isn't really meaningful" it's also one of the most interesting questions in physics and one of the most often asked and closed, despite the lack of any really good answers to it :-)
Comment: Don't close as unclear. The OP is basically asking why we concentionally treat fermions as particles (and implicitly we we conventionally treat gauge bosons as waves - well, photons as waves :-). That is, why do electrons appear particle like more often than they appear wave like?
Comment: @DanielSank: are you volunteering yourself?
Comment: Someone should write a really good explanation of the *experimental* reasons we sometimes think of electrons as particles and sometimes as waves, *and* then discuss the role of measurement, entanglement, and decoherence in the transition from wave-like to particle-like behavior. Wouldn't that be nice...
Comment: @Kyle Well, I think that ACouriousMind and I are the most vocal proponents and we'd be the logical candidates. Alas, the real world is commanding a lot of my time just now.
Comment: Nico, if I understand you correctly your question is basically the same as [Fermion vs. Bosons and particle vs. wave: is there a link?](http://physics.stackexchange.com/questions/76965/fermion-vs-bosons-and-particle-vs-wave-is-there-a-link), though I suspect this question may be posed a bit too technically for you.
Comment: NicoA: *"... there is proof of the fact that an electron is only a particle when we are observing it."* How can you prove electron is not a particle when you are not observing it, since you are ... not observing it?
Comment: @ACuriousMind: "Quantum objects aren't "waves" or "particles", they're quantum states." Yet another state of ... matter? Or just a dead end?
Answer: In quantum mechanics, things are not "particles" or "waves" - they may behave like both, or like neither. But a quantum object "is" neither of those - it is a quantum state, usually described as a vector in a Hilbert space.
The Bohr model of the electron orbiting the atom is false (for one inconsistency, that of moving charges classically radiating, see this question) - there is no orbiting in the classical sense going on at this scale.
Quantum objects don't really "become particles" when we observe them, but they are forced into a state that is sharply localized, usually, and this is what we intuitively would call a particle - a small blob with a more or less definite position in space.
Conversely, their wave-like characteristics are more pronounced when their position is not definite, but rather their momentum (by the uncertainty relations, these two states are mutually exclusive at scales where $\hbar$ is large).
It is very difficult to cast the classical concepts of "wave" and "particle" into precise enough terms to discuss them in quantum mechanical settings, and it doesn't help that there is no consensus about the measurement problem. The quote you read that "an electron is only a particle when we are observing it" is probably meant to mean that most of our measurement procedures force the electron into a sharply localized state, but that "usually", i.e. for instance when undisturbed in an atom, it is in a very smeared out state, an atomic orbital, and has no position to speak of (and hence can't be talked about as a classical particle).
Comment: Good way to answer this. I will add that if you look at the electron in the ground state of a Hydrogen atom, it has $\ell = 0$ hence $\langle L^2\rangle = {\hbar}^2 \ell(\ell+1) = 0$; it's not "orbiting" at all in the conventional sense and its only angular-momentum is spin-angular-momentum: so the inconsistency with the classical picture can at times be 100%.
Answer: Who says electrons are not waves in the atom?
Discard your bygone, quaint, outdated, perplexing, stupifying, nonsense idea that electrons move in a solar-system sort of orbits. That model(attributed to Rutherford) was an attempt to understand the atomic arena of Nature; unfortunately it was far from being correct as it had many in-built flaws.
Then the age of Quantum Mechanics began. That ushered in a new perspective to look at the phenomenon at the atomic scale which not at all follow classical determinism.
Everything at the quantum level is ruled by probability-uncertainty. The probability is not due to our ignorance but it is rather fundamental. The matter of irony is that despite being deterministic, classical physics is but approximation of Quantum Mechanics(see complementarity Principle).
Caution:Electrons are not waves.
They do, however, behave as wave. Waves of what? Waves of probability. These waves are described by $\Psi(x,y,z,t)$, a complex-function known as wavefunction. It tells you all the possible states the electron may be at now. Since there is uncertainty in the present, we cannot predict the future by certainty. This wave is spread in the whole space. The probability of finding the electron at a certain point at a certain time is the square of the amplitude of the wavefunction there i.e. $\Psi\cdot \Psi^*$, where $\Psi^*$ is the complex conjugate.
Caution: There is a huge difference between the probability of the event & the event itself; Probability wave of electron doesn't mean that it is smeared throughout the space; it can be at any place but we are not sure where it is. The probability wave just provides the probability of finding the electron at a certain place at a certain time.
Answer: I think the default that electron is a particle is taught us first because of historical reason. You have to understand the newtonian universe before the quantum mechanics; else it would be too complicated.
The atomic model was first made with electron on discrete observable routes. The next step was the thinking that they have to be stable and quantized so that they are not emitting photons as default. So the orbital model follows.
So after this with quantum mechanics it turned out to be, that everything has the wave particle duality and these behave also like waves, but the de broglie wavelength is mostly so small that it isn't measurable. But scientists proved that the double slit experiments also works with carbon molecules.
It would be nice if the kids are learning all this, but also in maths you don't start with unknown variables; you make them first understand actions like addition, multiplication etc. Afterwards you can build up your system step by step. But our brain just needs first a simple model before it can imagine more complicated things.
Also the quantum mechanics is not the least and best theory. I think the quantum field theory adds a bit more to the states of the quantum mechanics since there are fundamental problems when combining it with relativity.
But, you don't can add up just the theory; you have to find new ones and an interesting question is, if we find them be seeking valid equation and put them together. In "Emperor's Mind" Roger Penrose makes a good approximation with Goedel theory. Goedel actually proved that not everything which is valid is proofable so you have to assume that some of your things are true and build your universe upon it. But at the end every theory is either disproved or not already disproved.
Hopes this helps.
Greetings
Comment: In your first sentence, don't you mean to say we are first taught that electrons are particles, not waves.


For king of the hill, do you need to do tests yourself?

Question: I have an awesome idea for King of the Hill, but what if you can't do tests yourself due to time and system constraints. Would providing a test program be enough, or do you have to do the tests yourself?
Comment: What makes you think that your resources won't cut it? If you start getting into a situation where running the tourney would take more than overnight on a basic consumer machine you're talking about a *big* job. Modern computers are ridiculously powerful.
Comment: Time constraints? Writing a good question is going to take longer than testing the results.
Answer: If you post it in the sandbox explicitly saying that, someone might volunteer to test if they're interested enough. You can take them up on the offer if you wish, but remember that if you post it, it's still your responsibility to award a winner. If something happens and the volunteer can't do it, you need a backup plan.
I certainly wouldn't expect each user to test themselves. While having a program to test with is fine, they can just report whatever they want. In fact, according to my tests, I'm undefeated in the BattleBots Tournament right now. Of course, I may or may not have tested against all entries, but my statment remains true. I'm not saying all (or even most) users would outright lie, but even innocent mistakes could be detrimental to the post in general. Besides, people do crazy things to win at contests, especially when you don't win much of actual value from them. Go figure, humans are weird.
Bottom line: Unless the testing is done by you or someone you trust, you really have no way of knowing for sure who won. Bickering will ensue, hordes of goblins will descend from the mountainside, and chaos shall reign eternal.
Comment: Not to mention that in one previous king-of-the-hill there was some weird irreproducibility issue.
Comment: This brings up another point...How does an asker reward a volunteer tester?
Comment: @DavidWilkins My definition of *volunteer* is "to *freely* offer to do something". I think if you're only doing it for a possible reward, you're not really *volunteering*.
Comment: @Geobits Poor choice of words on my part...Recognize would be more what I mean....Or are the volunteers destined to be eternal unsung heroes of PPCG?
Comment: @DavidWilkins A bounty on a good answer of the volunteer is possible. Personally, if I were going to do the testing, it would be because the challenge sounds interesting and I want to compete, not because I want recognition for doing the testing. Honestly, though, I think dmckee's comment on the question is a good one, and *probably* a better answer than this.
Answer: I think if you don't have one central 'authority' doing the scoring runs, then all the contestants are going to have to install all the compilation/execution environments for all their opponents on their machine to check that their bot has a chance. Not everyone is prepared to do that so you might alienate some people who might otherwise compete.
Answer: I agree with Gareth that ultimately someone has to take on the burden of installing all the necessary build environments and running the tests, or else in the absence of someone it becomes everyone.
The idea of providing some incentive to someone to volunteer to host a KotH contest is nice in theory, but I wonder in practice how many willing volunteers would not simply busy themselves with hosting KotH contests they dreamed up themselves?
There is another way: create KotH contests structured around existing test beds. The Internet is full of game servers that accept bots for play. I know that Go and Backgammon are definitely options as I've played against bots on KGS and FIBS, and there are bound to be many others. There's obviously a trust aspect any time you enable a user to participate in their own testing, as Geobits said, and I am sure you'd get someone running a top-rated bot program not of their own making now and then. But with the proper limitations (e.g. Best Go bot in 200 tokens of atomic code golf) it could be made pretty obvious that someone was cheating.
The upside here is that most of these servers come with built-in rating systems, so the contest management becomes something as simple as tracking and reporting the bots' elo ratings during the contest.
Having given this fine suggestion, I do join Geobits in suggesting that you give some thought to dmckee's comment. The overhead to manage one of these contests is probably not as great as you imagine. Even if you are paranoid about foreign code on your system, virtual machines plus package managers like apt & homebrew make setting up an environment not too big a headache, and you can easily blitz it all when you're done.
I think I just talked myself into hosting one myself soon.  :)


Heartbleed: revoke or replace ssl certificate?

Question: I have a free ssl certificate issued just before Heartbleed was found. Now my CA wants me to pay 25$ to revoke it.
Should I pay or it's just enough to create a new certificate from another CA an replace existing?
Considering there are no real users of my site (the site is still in development and nobody has my old certificate installed in his browser yet)?
Comment: I think you're talking about the StartSSL (StartCom) root certificate? I personally removed it from my trust store, as they'd rather make money out of their users instead of protecting the end-users. Oh well. :P
Comment: Yes, it's StartSSL. I understand they need to make money, but... I feel that my hands are tied in this situation and I don't like it.
Answer: If you did not use your certificate, there is no need to of revoke it or re-issue. Seems to be CA is making more money with Heart Bleed.
If you have used certificate following things could be exposed with exploiting the HeartBleed.

long-term server private keys
TLS session keys
confidential data like passwords
session ticket keys.
Answer: I think there are a couple of misconceptions.
First of all, browsers request and check the certificate on the fly. They do not store it. This can be done manually, but it's neither common nor useful in standard cases.
If anybody has managed to obtain your certificate and the corresponding private key, they now "own" a perfectly valid certificate for your site and can use it whenever they want. It doesn't matter that you've used the certificate only for testing and plan to buy a new one. Browsers aren't aware of this. They will accept either certificate – well, unless your CA explicitly revokes the old one.
So merely buying a new certificate doesn't solve the problem. It doesn't change the fact that there's a potentially compromised certificate which could be used for a man-in-the-middle attack against the site.
It's certainly annoying to pay extra money for the revocation, but there isn't really another option. You should choose a fairer CA next time, though. ;-)


Why is the voltage wave magnitude constant along a transmission line that is perfectly matched to a load?

Question: I think I understand reflected waves and standing waves on transmission lines, but why is a traveling wave voltage magnitude constant on a perfectly matched  transmission line? Why doesn't have peaks? I would expect to a decrease to zero in between peaks racing by what ever point is probed, dependent about the phase velocity. I get that there isn't a reflected wave for a perfect match.
Answer: You understand that for a standing wave in a T-line, the amplitude depends where you are on that T-line.
Now the wave is not standing but travels. So pick a point on the T-line and you see the waves passing by. They all have the same maximum amplitude, right ?
I could move to the left or right on the T-line but I would still see the same as the waves are all the same and they keep passing by.
Why would there be any amplitude peaks ? For that to happen the wave and its amplitude has to be influenced by something. For the standing wave it is the reflected wave combining with the original wave resulting in a combined wave which appears to stand still.
Note that a wave cannot stand still, it is just the vector-sum of two (or more) waves which appear like the wave stands still.
A T-line which is perfectly matched looks like an infinitely long T-line to a wave that is travelling in it. The wave experiences no outside influence/interference because there is none. So it will have and continues to have a constant amplitude.
At the end of the T-line the wave's energy is dissipated in the load (assuming a resistive load), the wave is no more, it is converted into heat. So it cannot reflect and it will not interfere with the other waves.
Comment: No the amplitude stays the same, the amplitude is multiplied by a sinewave and the sinewave varies between -1, 0, +1 but the amplitude remains constant ! Perhaps you're assuming that when at a certain point you measure 0.5 that then the amplitude must be 0.5 but that isn't so. The amplitude is not the **momentary** value at one point but rather the **maximum** of the wave. The amplitude can be 1 and the sine 0.5, these multiply and give 0.5.
Comment: As the waves pass by won't their amplitude decrease then increase relative to a certain physical point? Otherwise isn't the signal DC? I.e won't the voltage a point change continuously? And therefore change continuously along the line?
Comment: https://www.youtube.com/watch?v=mh3o8gUu4AE

In this video the dude is using an animation to describe standing waves, it shows a traveling wave with reference points that rise and fall as the wave travels through
Comment: So are you saying i am confusing instantaneous magnitude with amplitude?
Comment: yes...at any point the amplitude is what it is along the sinewave at that time, but the amplitude of the sine wave is the full amplitude is traverses over a period (or half a period).   for ex. v= A*sin(wt).    A is the amplitude, reaching its peak when sin(wt)=1, so the amplitude of the sine wave is A, even though at any time t it may be less than that as the oscillation completes a full cycle w=2 *pi.
Answer: If the transmission line is perfectly matched (or its length is infinite), there is no reflection.
Therefore the voltage at any point x, and any time t is (assuming losesless line):
$$V(x,t) =A\cdot \sin{(\phi + 2\cdot\ \pi\ \cdot ( f\cdot t - \frac{x}{\lambda}))} $$
where λ is the wavelength, and Φ is the phase.
If you choose any x point, you can rewrite the voltage as:
$$V(t) =A\cdot \sin{(\theta + 2\cdot\ \pi\ \cdot f\cdot t )} $$
As you can see, the amplitude of that sinewave is A, regardless the particular x.
Comment: thanks for your answer but my question isn't really mathematical, it's more conceptual
Comment: Ok. Assume there is no reflection, so we don't have to deal with stationary waves. Assume also that the line is loseless. In ANY point you'll get an oscillating wave, but it's amplitude will remain constant, i.e. it will oscillate from +A to -A. Think of a traveling sine wave: if you  stop at any point, and you measure the instantaneous value, you'll measure of course a sinewave, but the amplitude (A) won't change. And A will be constant regardless the particular position you're considering.


Design PI controller using LM2743 analogue controller

Question: I'm final year student currently designing a buck converter with an analogue controller (LM2743). What my supervisor told me is to used a PI controller using bode plot. My design spec:
1. Vin=12V
2. Vout=5V
3. Pout=10Watt
4. Fsw=100KHZ
From there I obtained my inductor (48.6uH), output capacitor (15uF) an load resistor at (2.5ohm). From there my transfer function is:

From there I'm quite stuck on how to do it. Where I know the T.F for PI controller is Gc(s)=Kp+(ki/s). After that, lets say I've managed top design the PI controller. From datasheet LM2743, there are few components need to be connected to the main controller which is pin FB and EAO.

Its mentioned in the data sheet that the calculation can be made using the equatio given but didn't tell on how to get the Fz1, Fz2, Fp1 and Fp2 where these value that can decide the components.

LM273 datasheet
Comment: I used to learned how to design PI controller using damping factor and freq using a P factor point, but my transfer function got imaginary number so thats why my SV asked to design it using frequency response.
Answer: Although P feedback is stable and simple, PI feedback reduces gain error thus better load regulation but increases overshoot. While PID offers a better compromise. The RC filters in series are  Lead-Lag compensation being a  limited D frequency range.
The design must start with tolerance specs for each source of error: load limits, overshoot transient and time; load regulation error ;input sensitivity, and temperture rise worst case, for each range of expected variables.

It is a good practice to get used to doing things this way by learning how formal PSU's summary specs are done and follow these definitions.  An ATX spec might be >10 pages and have a dozen parameters that you can summarize including environment specs.

The lead-lag filter is described with the phase-gain margin of the response in the App. Note.
Integrators reduce ripple but decrease stability.
Differentiators, increase stability and step response time but add ripple.
A Lead-Lag filter usually adds enough derivative gain just before unity gain to boost the phase margin. It may only be a 10dB boost in gain in the HF range just before unity gain.
The basic process of choosing these filters requires the entire loop response in order to apply asymptotic lines with the understanding that the breakpoint applies a 45 deg. phase shift and goes to 0 and 90 deg exceeding a +/- decade span.  You can find a full step-by-step description elewhere, such as your Control Systems textbook.
Generally when you do this on a regular basis, you will rely on a tool to visualize the results in sumulation.
Although this one, I did for you, (Falstad) although no phase, it is interactive  and you can edit any components with a Bode plot.  It can be exported and do DSO time plots as well.
You can use Vspice or any others you have for Bode Analysis or Nyquist analysis.
You might also look at my choice of a  more efficient TI regulator in WebBench.


BIP91 Activation Period

Question: I understand that BIP91 needs 269/336 signaled blocks.
My question is is it a rolling window of any 336 blocks or are there activation periods of 336 blocks that start and end at predetermined block numbers?
Answer: It looks like it is actually periods of 336 blocks instead of a rolling window. It is like what BIP 9 specifies, but uses 336 blocks as the window and 269 blocks as the threshold. The windows begin when a block's height is a multiple of 336.
Comment: Is there any particular reason why it is implemented in this way instead of rolling fashion?
Comment: Yes, this is inherited from BIP9. The reasons there are to be more conservative (there is randomness involved, if you give every block a chance to activate, each of those increases the probability that it activates when in reality the threshold is not reached), giving people less sleepless nights (knowing in advance when something can activate is much nicer than having to check at every block), and to make the behaviour more analysable (each attempt now is a statistically independent sample).

BIP91 Activation Period


Building a 1 layer thick brick wall over drywall

Question: Looking to lay a brick wall over drywall in my basement.  The floor is a concrete slab which covers my entire basement.  I am looking to see what issues or concerns I should have laying a 1 layer brick wall from floor to ceiling. I am concerned the wall would topple over.  How can I lag this in?
Comment: I'd be very concerned about such a thin masonry layer over such a flexible substrate. Either use full brick or use a fake brick that's not so heavy and brittle.
Answer: Assuming the bricks would be adequately supported from below, you would use brick ties (attached to wall studs on one end, and embedded in mortar on the other) to give your brick wall vertical stability.
If you want the look of brick wall with less effort, you can just install brick-look tiles on a suitable backer board onto your existing wall.


Bandwidth saturation due to Googlebot search for spam pages

Question: I recently had to update the hosting plan of a WordPress site I run due to bandwidth excess, which was already quite high.
I've installed a security plugin that keeps log of 404 errors. From here I see a large amount of accesses to spam pages from an IP that results being from a Googlebot.
I've disallowed all bots to crawl my pages and that seems to stop those bulk access to non existing pages. Of course this is a temporary solution to avoid that my site is being blocked. How can I stop Googlebot from trying to access only those pages?
Answer: If the URLs they are requesting follow some patterns, I think it is a good idea to block those pages. Based on your comment above, in your robots.txt you could add:
User-agent: *
Disallow: /data/

That will work as long as you don't have any real URLs beginning with data.
The second thing you may want to do is decrease Google's crawl rate. You can do this in Google Webmaster Tools.
There is also a Crawl-delay setting in robots.txt - Google ignores this but it is useful for other search engines. I'd advise not setting it higher than 4.
Comment: Yes, use Webmaster tools. Add your site to Google Webmaster Tools and set your preferred crawl rate, please see https://support.google.com/webmasters/answer/48620?hl=en
Answer: 404 errors are a part of business. They can come from poorly made links on other sites, current link on your own site, or from pages that have been removed without a current existing link.
You do not want to block the search engines from accessing pages that do not exist unless you do not want to be in the search engine at all.
If you are blocking Google, for example, using a robots.txt file, you will be delisted. You are saying to Google, I do not want what you are offering.
You do want to allow 404 errors to occur. How else can Google know these pages are gone? You can replace a 404 error with a 410 which would shorten the process. A 404 means temporarily gone while a 410 means gone. Google will try and access pages that return a 404 error for a period of time for a certain number of accesses before it determines the page is truly gone. If you return a 410 error, Google will stop trying to access the page after one access. But keep in mind that any search engine may try again if they find a link on another site just in case.
Comment: I know that blocking googlebot is not a good idea, but it's a temporary solution to avoid bandwidth saturation. Googlebot is trying to access pages that NEVER existed on my site (spam pages due to hacking of site, i.e.
data/annonce-rencontre-homme-serieux-poissy-27062dd-1.html) in means of one per minute, thus quickly saturating my bandwidth. What I wanted to know is how to stop googlebot from pointing to these spam pages...
Comment: @RobertoPravisani By letting Google do its work. Otherwise, it will not stop. These links are out there and no matter what you do, Google will chase them down thinking they are valid pages. The only way for Google to know they are invalid is to attempt to fetch them. If you block Google, you are causing more damage than if you just let it go. If one spam access per minute is eating your bandwidth, I wonder how little bandwidth you have??


What are some possible motivations for Peirce's use of only one operator(his NAND) to recreate the "and", "or" and "not" operators?

Question: It was found in Peirce's unpublished papers that he defined the NAND operator and discovered that all classical logical operators could be replaced by combinations of NAND.
What are some possible motivations for Peirce's use of only one operator(his NAND) to recreate the "and", "or" and "not" operators?
Comment: The issue about "minimization" of concepts/symbols/axioms is a driver for science and mathematics (as well as philosophy) development.
Comment: If you are studying logic, as opposed to using logic, then it is helpful to reduce it to its simplest form. If you are trying to prove something general about logical connectives, then the fewer there are the better. Also, it has practical consequences: electronic circuits often consist of a bank of NAND gates. It is useful not to have to create separate gates for different connectives.
Comment: It meant that he only had to have one type of IC in his parts bin.
Comment: Yes but there may have been more specific motivational forces guiding Peirce's efforts at minimization of operators. Perhaps De Morgan's laws or some remark by earlier logicians that 2 of the classical operators suffice to express all 3 if one of the 2 is  "not"
Answer: See Charles Sanders Peirce, Collected Papers : Volume 4. The Simplest Mathematics (1933), page 13:

[4.12] A Boolian Algebra with One Constant [untitled paper c.1880]

Every logical notation hitherto proposed has un unnecessary number of signs. It is by means of this excess that the calculus is rendered easy to use [...]; at the same time, the number of primary formulae is thus greatly multiplied, those signifying facts of logic being very few in comparison with those which merely define the notation. [...] The apparatus of the Boolian calculus consists of the signs, [...]. In palce of these seven signs, I propose to use a single one.

Two propositions written in a pair are considered to be both denied.

And see [4.264] [from The Simplest Mathematics (1902)], page 215:

for example x⋏y [a sort of "harpoon" symbol] signifies that x is f and y is f. [...] With these two signs, the vinculum (parentheses, braces) and the sign ⋏ , which I will call ampheck (from Ancient Greek ἀμφήκης, "cutting both ways") all assertions as to the values of quantities can be expressed.
Comment: I was hoping that you would post the descritpion of some attempt to minimize the operators before Peirce.
Answer: Actually, my slightly snide comment above may be close to an answer.  All possible logical operations can be produced with some combinations of NAND operations, meaning that only that one operator needs to be studied.  While I have no idea how this would "simplify" philosophical problems, it would at least regularize the notation and perhaps make contemplation easier for folks with a "mechanical" thought process.
This certainly made early digital electronics simpler to deal with.


Did Karl Marx believe there is an innate human nature and does it matter in the light of contemporary biology?

Question: Marx stressed the historical specificity of social practices and institutions. Did he think that human nature was similarly historically specific and aren't his views only of antiquarian interest in view of the findings of modern biology ?
Comment: Wikipedia answers this question at length in [Marx's theory of human nature](https://en.wikipedia.org/wiki/Marx%27s_theory_of_human_nature):"*Man is directly a natural being. As a natural being and as a living natural being he is on the one hand endowed with natural powers, vital powers – he is an active natural being. These forces exist in him as tendencies and abilities – as instincts*"; "*producers change, too... develop themselves in production, transform themselves, develop new powers and ideas*". But of course he did not use terms like "innate knowledge" or "environmental conditioning".
Comment: He basically took Feuerbach's (Christian) anthropology as a given in his early days but never made anything else explicit even later from what I've heard from experts on Marx. Alienation is not really conceivable without a thick concept of what humans should be like "naturally", is it?
Comment: RE: "Feuerbach's (Christian) anthropology" -- wasn't Feuerbach an outspoken critic of christianity?
Comment: "In the first part [of "Essence of Christianity"] Feuerbach considers religion "in its agreement with the human essence" (WC 75), arguing that, when purportedly theological claims are understood in their proper sense, they are recognized as expressing anthropological, rather than theological, truths" ([see SEP](https://plato.stanford.edu/entries/ludwig-feuerbach/#CritChri)). So the point is rather that religion co-opted the essentially philosophical/anthropological content. This is why on one hand, he was critical of Christianity (Theology) but on the other, adapted much of its content.
Comment: Here is Erich Fromm writing on this question:  https://www.marxists.org/archive/fromm/works/1961/man/ch04.htm
Answer: ▻ MARX AND CONTEMPORARY BIOLOGY
Marx knew nothing of the findings of present-day evolutionary psychology, neurophysiology, neurochemistry and ethology, and of the empirically-based information they provide about human nature. So much is obvious.
But we need to be aware of his normative stance towards human nature. He did not offer, or claim to put forward, a scientifically informed view - a view up to speed with the 21st century - of the nature of human beings. He would not have wanted to contradict scientific findings but he was not in competition with them. When he conceptualised human nature he did so from the standpoint of an emancipatory revolutionary. That at any rate is how he saw matters. It is not evident that 21st-century biology undermines his ideas about human nature.
'Innate' would not be my preferred term. Marx did not think in terms of a primordial, pre-evolutionarily fixed human nature. He could readily agree that, as we have evolved, this is the nature we have. It is not artificial or contingent on social conditions.
From his limited, or at any rate specific, perspective he had views about the conditions in which human beings could flourish given their 'species-being'. Okay, jargon : but not empty or pretentious jargon. I don't intend a defence of Marx, only a clarification.
▻ HUMAN NATURE - SPECIES-BEING
Marx's view is that we have 'species-being' (Gattungswesen). This is another way of saying that we have an innate human nature - that there is a human essence. He rejects the Christian view  that work or labour is 'the curse of Adam' which ideally we would be well rid of. His view is also that under capitalism we are exploited and alienated workers at worst and aspirant consumers at best. I am only outliving my understanding of what I think he believes.
Marx is definite that we are active, creative, productive beings by nature; and that we can achieve our deepest satisfaction from developing and exercising skills, overriding obstacles, and otherwise structure the world. Work or labour does not simply produce a result that can be consumed; it transforms the labourer as action, creation, and production draw out his or her 'human freedom'. The world that is right for our species-being is one of freely-chosen creative activity.
▻ CONDITIONS FOR HUMAN FLOURISHING
That freedom is not complete until work or labour takes places under conditions quite different from those that prevail. What Marx has centrally against capitalism is not its exploitation (Ausbeutung) the worker (though he thinks it does exploit, through the extraction of surplus-value or Mehrwert, i.e. selling the products of labour above their cost of production without benefit to the worker) or that it induces alienation (Veräußerung), estrangement of the worker from what he or she produces (which is not his or her own). It's rather that capitalism treats the worker as someone to be kept alive, to reproduce further labour, and to consume. The capitalist form of social organisation has no aim or incentive to create the conditions to enable - to repeat the phrase -
freely chosen creative activity.
▻ REVIEW
This is (to repeat) simply an answer to the question, not an endorsement of Marx's views or a critique. In my view Marx did believe that there is an innate human nature; I have tried to spell out what he believed it was. His views were not science but it is not clear that, if empirical contemporary biology cannot verify or confirm his normative view of human nature, it does anything to contradict it.
▻ REFERENCES
K. Marx, 'Economic and Philosophic Manuscripts of 1844', tr. M. Miligan, NY : Dover, 2007.
K. Marx & F. Engels, 'The German Ideology', 1845/6, London : Lawrence & Wishart, 2004.
Norman Geras, 'Marx and Human Nature', London : Verso, 1983.


If a server only opens port 22 and 80, do we only have those two ways to hack it?

Question: If a Linux server only open SSH port 22 and HTTP port 80, must we go through one of these two ports to hack into server from the internet?
Comment: If you only have these two ports open, then you can only communicate with those two, yes.
Comment: Depends on how you define "open" - perhaps as "replies with either SYN/ACK or RST to a SYN"? In that case, it might react to other ports (or non TCP traffic) in vulnerable ways though perhaps not by replies per se.
Comment: Are you talking about which ports are open on the server itself? Or are you talking about what ports the router forwards? Or what exactly?
Comment: As far as remote attacks are concerned then sure. Whether or not you're safe heavily depends on which programs are listening on those ports and whether they are properly secured.
Comment: You probably also want 443.
Answer: Not really. I'd say it depends on your threat model. There might be other threats that don't need to use those ports in order to compromise your server. The first example that I can think of right now is a supply-chain attack. When you update any software on your server, if the updated software has been compromised by a supply-chain attack, your server will get infected. Or if you install example-program by mistake instead of example_program (note the hyphen instead of the underscore), and example-program was malicious and had been given that name on purpose to confuse you, then your server will be compromised. I think something like this happened recently... oh, yesterday (Bitcoin stealing apps in Ruby repository). Other examples? Maybe some MITM in the outgoing connections from your server. Then let's not forget about phishing, or anything involving social engineer.
So to be precise, if you asked me "in general, can I only be hacked by a remote threat through open ports?", my answer would be no. Whether some threats are likely or not though, depends on your threat model, which in turn depends on what your server does, how you are managing it, who you are, etc.
Comment: @onurcanbektas Yup. You can check for yourself (you can run `netstat -a` to see active connections, and most will have some high port number with the local address). Port 80 being assigned to HTTP is for destination. So when you want to connect to server example.com over HTTP, you connect to example.com at port 80, but the connection can come from any port. And given that low port numbers are generally assigned, high ports are usually used for outgoing connections.
Comment: @onurcanbektas Nope, open ports generally mean listening ports, so only incoming connections use those. Outgoing connections generally use a random port (with some restrictions). Unless by "open ports" you mean that everything else is blocked by a firewall. But in that case the server probably wouldn't be able to make outgoing connections, such as one needed to download the update mentioned in this answer.
Comment: @onurcanbektas just for completeness, not *any* random port. You can't use ports already in use (e.g. port 80 in this case), and usually OSs restrict you to ports above 1024 or above a higher number (the exact number depends on the OS, and you can probably change this with some effort).
Comment: But, to be clear, even if the server compromised with such an attack, the incoming / outgoing communication between the server and the attacker would be through either port 22 or port 80, right?
Comment: @Tomeamis So, unless it is blocked by a firewall, an application can use any random port for outgoing connects?
Comment: @Tomeamis oh I see; thanks for the answer.
Answer: No.
There are multiple things which can be attacked on a target computer, and a serving application (httpd or sshd for example) is just one of those things.
Remember, there is an entire network stack between the physical network port on the network card and the application handling the actual traffic (ie sshd) - in this stack there includes things like kernel functionality such as firewalls, network drivers etc all of which can be attacked separately to the handling application.
See the number of Linux kernel remote execution vulnerabilities highlighted here which require no handling application to exploit, and instead allow an attacker to execute code just by crafting a bad network packet.
Of course, its easier to attack the application rather than the kernel, because the kernel tends to be much more heavily inspected.
Answer: An attempt to list some ways to hack into a server without using http or ssh:

Using a vulnerability in Management Engine
Using a bug in a network board firmware or driver
Using something not very secure in the virtualization platform, having a legitimate access to (or hacking into) a neighbouring virtual machine
Exploiting some bug in the IP or TCP driver in the OS
Making use of some network interaction where your server acts as a client (DNS queries, automatic updates, database access), spoofing or hacking into the legitimate server of these services.

Example for 4. :
Long ago, there was a ping of death attack exploiting a bug in the IP layer driver, no open port needed at all.
Comment: haha yeah back when i waz a scr1pt k1d i once shot dead a win95 client in japan.. sorry
Answer: Closing ports is one very early line of defence.
When a port is open, there will be some piece of software running which handled the data entering on that port. That software can have bugs that allow an attack to succeed. If you open 100 ports, there are 100 pieces of software that are potentially insecure. With two open ports only, there are only two pieces of software that are potentially vulnerable. Obviously making sure that 100 pieces are not vulnerable is much harder than two pieces.
But an attacker can try to get in through another route. Your server should receive blocks of data containing a port number, and your software should direct the block to the right port or throw it away, possibly log it. If the software sending incoming blocks to ports has bugs, an attacker might exploit those bugs, and such an exploit could be independent of which ports are open.
Comment: "If you open 100 ports, there are 100 pieces of software that are potentially insecure. " <-- Why do you assume a 1-on-1 mapping between ports and "pieces of software"? A single program can easily listen on a hundred different ports. Sure, that would be a weird thing to do, but there are many programs that listen on more than one port at least. (And the converse is true too, it's possible to share a single listening port between multiple processes.)
Comment: This is inaccurate. Ports that are not closed are not open: https://security.stackexchange.com/questions/96568/open-ports-with-no-services-bound-to-them
Comment: @PeterW. What are you trying to say? Nobody even made any claim regarding filtered vs unfiltered ports. An open port is not merely an unfiltered port, it's a port that you can actually successfully connect() to.
Answer: If the server has only ssh and http services open, then you will have to exploit possible vulnerabilities related to one or both services, to hack into the server.
One other way would be physical access to the server keyboard, but I think, you meant just attacking through network. (remember to scan all 1-65535 ports)
Comment: Remember that theres an entire network stack between the physical network port and the receiving application, including kernel functionality and drivers that can all be attacked as well.  This can be done even on closed ports - just because nothing is listening on those ports doesnt mean the system isnt handling them, for example in the firewall rules system etc.  Ample opportunity there for an attack vector.
Comment: -1. OP, you are encouraged to wait at least a day or two before accepting an answer, to give the community and yourself a chance to judge the quality of a selection of answers before choosing one (if any at all!).
Comment: The only answer so far to mention physical attack.
Comment: @mckenzm True, but the question says _from the internet_ which seems to exclude physical attacks.
Comment: I'll pay that, it precludes even compromised appliance on the local network. (Peers).


distinguishable particles' Hamiltonian

Question: Let us consider a classical Hamiltonian of a many body system
\begin{equation*}
H = \sum_{j=1}^N\frac{p_j^2}{2m}+V(\mathbf q)
\end{equation*}
and let us pass to quantum dynamics by promoting the positions and momenta to operators:
\begin{equation*}
q_j\rightarrow \hat q_j \;\;\; p_j\rightarrow \hat p_j
\end{equation*}
where the index label still refers to THE particle $j$.
Additionally, we could define a bosonic Fock space via the creation and destruction operators (by taking $\hbar\equiv 1$)
\begin{equation*}
a^\dagger_j=\frac{\hat q_j+\hat p_j}{\sqrt 2} \;\;\;
a_j = \frac {\hat q_j+i\hat p_j}{\sqrt 2}\hspace{8mm} j=1\cdots N
\end{equation*}
creating/destroyting A particle in A STATE $j$. To me the fact that particles are seen as indistinguishable once a Fock space is build appears natural. In particular the label $j$ does not refer anymore to a particle but rather to a state. Instead, if we stick to the quantum dynamical system $\hat H(\mathbf{\hat q}, \mathbf{\hat p})$, is it true that particles are still distinguishable? In particular, I think that in the calculation of the canonical partition function
\begin{equation*}
Z(\beta) = \int_{\mathbb R^N} \mathrm d \mathbf q \langle \mathbf q|e^{-\beta \hat H(\mathbf {\hat q}, \mathbf {\hat p})}|\mathbf q\rangle
\end{equation*}
the states in the braket doesn't need to be symmetrized (as we would do for a trace in the Fock space) and that we can stick to the tensor product of the single particle Hilbert spaces:
\begin{equation*}
|\mathbf q \rangle = |q_1\rangle \otimes\cdots \otimes |q_N\rangle
\end{equation*}
Is my line of arguments correct?
Comment: Careful with your definition of particle and state. For example if $V(q)=\sum_\limits{j=1}^{N}V_j(q_j)$. Then an excitation by $a^\dagger_j$ can still be distinguished from one by $a^\dagger_{i\neq j}$
Comment: I agree, so I guess in case of a one body potential as the one you mentioned we would have to rewrite the interaction in Fock space as $\hat V = \sum_{rs} c^\dagger_rc_s$ (by assuming $V_j=v \;\;\forall j$) which is indistinguishible. However I think in the coordinates representation $\mathbf q$ no such procedure is needed. Is it true?
Comment: I think I see what you mean. The excitation $a^\dagger_j$ for a given j are indistinguishable. As for the $q_j$ between themselves, that's up to you. You have to choose either $[q_j,p_k] = i\hbar \text{ or } 0$ for $k\neq j$.
Answer: This seems to be a rather confused question. Given a general interaction potential $V(\mathbf{q}_i)$, your operators $a_j^\dagger$ and $a_j$ simply aren't creation or annihilation operators, in the naive sense. Their commutators with the Hamiltonian are much more complicated.
Assuming that $V(\mathbf{q}_i)$ has the very special form
$$V(\mathbf{q}_i) = \frac12 \sum_i k_i q_i^2$$
then you indeed do get a set of $N$ independent harmonic oscillators. In this case, your conclusions are still confused because you're describing the same state in two incompatible ways.
For simplicity, let's restrict to $N = 3$ and consider the state
$$|2, 1, 0 \rangle = a_1^\dagger a_1^\dagger a_2^\dagger |0 \rangle.$$
We can describe this state in two ways: either "there are three harmonic oscillators; the first harmonic oscillator is in the second excited state, and..." or "there are three units of excitation; two are in the first harmonic oscillator, and...".
The source of your confusion is that you're calling the harmonic oscillators themselves the particles in the first case, but then switching to calling the units of excitation the particles in the second case. Of course, "particle" is just a word which you can freely attach to either concept, with one or the other being more useful depending on the situation, but it shouldn't surprise you that the properties of the particles change if you change what the word "particle" means halfway through.

To hopefully clear some of the confusion, let's do an explicit example. Let $N = 3$ and consider the state $a_1^\dagger a_3^\dagger |0 \rangle$.

In the original picture, the Hilbert space is $\mathcal{H}_1^{\otimes 3}$ where the tensor product is not symmetrized. We describe this state in words as "there are three particles; the first particle is in the first excited state, the second is in the ground state, and the third is in the first excited state".
Next, we may switch to a completely different description of the states. We rename "particles" to "modes" and rename "particle in $n^{\text{th}}$ excited state" to "$n$ particles in the mode". Hence to describe the state we say "there are two particles; the first particle is in the first mode, and the second is in the third mode". We can write this state as $|1, 3 \rangle$ or as $|3, 1 \rangle$.
Since the "particles" are identical, the states $|1, 3 \rangle$ and $|3, 1 \rangle$ are both the same state, namely $a_1^\dagger a_3^\dagger |0 \rangle$. So we must work with a symmetrized Fock space.

To say the distinction yet another way: suppose I own $3$ sheep among a large herd, and to pick them out I give them identical bells. In the first langauge, the particles are the sheep, and I specify the state by saying where they are. In the second language, the particles are the bells, and I specify the state by saying what sheep each bell is on. Since the bells are identical, I then have to quotient out by permutations of the bells. But that doesn't mean the sheep have become identical.
Comment: I think to the labels assigned to $a_j$ and $a_j^\dagger$ as states because a Fock state naturally constructed via them $|n_1...n_N\rangle\propto(a_1)^{n_1}\cdots(a_N)^{n_N}$ reads as the state where $n_j$ bosons are in the state $n_j$. Instead, starting from the classical Hamiltonian H, $\hat q_j$'s and $\hat p_j$'s are the direct quantum correspondent of physical particles.
Comment: I can rephrase my question asking whether I need to fix bosonic indistinguishability already at the level of the operators $\hat q_j$'s and $\hat p_j$'s or I can 'postpone' the creation of the symmetrized Fock space once I define $a_j^\dagger$ and $a_j$.
Comment: @Graz Again, you're using two totally different definitions of the word "particle", which is leading to all the confusion. For example, consider the state $a_1^\dagger |0 \rangle$ for $N = 3$ harmonic oscillators. In your original language, there are $3$ particles. In your new language, there is $1$ particle. It's the same state, you've just completely changed the words you use to describe it.
Comment: I see what you mean. Though is it right that the Hilbert spaces for the two set of coordinates are interpreted differently? In particular that for the canonical coordinates $\mathcal H = \mathcal H_1^{\otimes N}$ (where $\mathcal H_1$ is a one-particle Hilbert space), while for the creation destruction operators we have to work in a Fock space? This would imply that to take traces in the case of $(\mathbf q,\mathbf p)$ we can simply consider a factorized basis, while for creation/destruction operators a complete set of orthonormal vectors is symmetrized
Comment: @Graz I edited with some more explicit explanation, does that help?
Comment: Great explanation thanks, very helpful in resolving my confusion on the issue

distinguishable particles' Hamiltonian


Separated servers or virtualization, which is more efficient?

Question: My company is hosting a few separate, but related, moderately hit, web sites.  Accordingly, a production database server, staging database server, production web server, staging web server, etc are needed.  My question is, should we invest in physically separate servers for each of our needs, or should we put that money together and invest in a much higher end server and virtualize all of the aforementioned servers?  Which route would you guys decide on, and why?
Comment: while not _directly_ answering your question, much of the why could be found in the answers to http://stackoverflow.com/questions/9749/what-kind-of-servers-did-you-virtualize-lately
Answer: That depends on a lot of things, here are the main considerations.
If you have a lot of servers with low to moderate usage, virtualization should generally save you money on hardware, power, and floorspace. There is a tipping point, however, based on the overhead of the VM layer itself. Honestly, you will have to experiment to find the right cost/performance balance on this. I am sure the VM vendors would be happy to help you with the math.
The downside is that virtualization creates a single point of failure. If that box fails, you have downtime for all of your servers. Having them separate makes it far less likely for everything to take a dive at once.
Answer: You certainly want physical separation between the development and the production servers. You shouldn't ever have to worry that something you do in dev could bring down the machine on which production is hosted. And, there are some problems in development that really require either a hard reset of the physical machine or a ludicrous work-around to avoid a hard reset.
As for production web server and production database, you're not really introducing any new points of failure by virtualizing them on the same machine, particularly if you can colocate a static version of the site on another server. For any modern web site of even moderate complexity, database failure is web site failure anyway.
Comment: Additionally, using virtual servers allows you to easily take and restore snapshots if you do boink the dev (or production!) environments, without having to rebuild OS, reinstall server software, reinstall apps, etc, etc.
Answer: From my experience, for low or moderate usage a VM is the way to go - if you get just one very powerful server instead on several moderately powerful servers you save money, power and space and make the application faster at the same time because it's running on faster hardware.
A VM also have same another nice advantages, if the server hardware fails you can load the same VM on different physical hardware and continue running like nothing happened (you do have full backups, don't you?) and you can take a copy of the actual production server and run it in isolation on a development machine to debug those annoying bug that only appear in production.
But I would put the development (and maybe testing) servers on a different physical machine than production, you need to make sure no matter what stupid mistake you made in development it wouldn't take down the production server.


How do you know if you're on a 10 meter band?

Question: I am just getting into ham radio, and I find it very interesting, but to make phone calls, you have to be on a 10 meter band. Does anyone know how to infer the band you're on is 10 meter? Thank you very much.
Comment: I think Jon is talking about phone privileges for Technicians in the US, which are available from 28.3 MHz - 28.5 MHz only.  I'll edit my answer to explain how "phone" ≠ "phone calls".
Comment: I forgot to mention yesterday, hello Jon and welcome to ham.stackexchange.com!
Comment: Are you thinking about a repeater's telephone autopatch feature? I thought those went out after cell phones came into common use. In any case, back in the day most everyone used 2 meters for that. Perhaps there are/were some in the FM portion of 10m.
Comment: @rclocher3 Could be, but he did write "make phone calls".
Answer: Frequency and wavelength of a wave are related: the frequency times the wavelength equals the speed of the wave, in this case the speed of light.  For HF, the frequency in megahertz times the wavelength in meters equals 300.  Modern radios display the frequency.  Bands are referred to by their approximate wavelength for historical reasons.
If the frequency displayed on your radio is between 28 MHz and 29.7 MHz, then you're on the 10m band.
For hams in the US, the ARRL's band charts come in handy for remembering what band is where.
By the way Jon, I think ham radio jargon may have gotten in the way.  In the US, Technician-class licensees have "phone" privileges between 28.3 MHz and 28.5 MHz only.  But "phone" here means radiotelephone, which means any analog mode used to carry voice sounds, such as AM, FM, or SSB.  So ham radio "phone" isn't the same as telephone.  In other words, one can't ordinarily use a ham radio to place a telephone call.  There is such a thing as an autopatch, which is a device to connect a radio to a landline telephone.  So if you're talking on the radio to someone with an autopatch, the other operator could make a telephone call for you and connect the radio to the call.  But autopatches are rare these days.
Comment: Great answer. Love that it is clear ànd welcoming, a better introduction to the ham community than lots of hams give to newcomers!
Answer: If you are a new ham, you need to go to the https://arrl.org web site and get a listing of the ham bands that they publish. You can find them on their site by searching for ham bands.
Print out one of the listings; the 8.5 by 11 inch chart is in color. They also have some that are able to be printed in black and white. On those charts, it lists that you with a Technician class license can transmit Phone from 28.30 to 28.50 MHz in the ten meter band.
I would also recommend you search for local clubs nearby you and see if they have a web site and try to join. Local hams usually enjoy helping new hams get started in this hobby.
Comment: Hello Lewis, and welcome to this site! :-) +1 for suggesting that the OP try and get help from a local ham or amateur radio club.


How can I confirm to whom the bitcoin address I sent a payment too?

Question: Where can I find out what company is associated to the address I sent a payment too?
Answer: You usually cannot.
Bitcoin addresses are not explicitly linked to companies or people, unless they decide to publicize that they own an address (or receive funds from it via a third-party custodian).
There is no directory or listing of who owns which address, although some companies and explorers try to tag major addresses to exchanges etc.
You can attempt to follow the BTC from that address and see if it touches a known address. That might give you a lead as to which company is being used to store/cash out Bitcoin by the company you made the payment to. However, if the company you paid has their own system and doesn't cash out frequently, you won't really find anything actionable.
Answer: In Bitcoin, you need to verify that you're sending to the right address before sending. Usually this works by the recipient of the payment sending you a payment request with a destination address. If you're unsure whether the source is authentic, verify your information on a second channel.


Is it ok to store peanut oil in the container it came in after being used?

Question: I am starting to get into frying again.  I was wondering if it is ok to store the oil after it has cooled down in its original container or if there is a better option for storage.
Comment: I typically gather up the oil after it's cooled down, and pour it bit-by-bit through a funnel + coffee filter into it's original container.
Comment: It would be helpful to know more about the actual container. Is it metal, plastic or glass? A link to the actual product (in the same container) would be great.
Comment: Why would it not be safe? What are you concerned about?
Answer: If the container is labelled with one of the standard logos that mark an unconditionally food safe container, there should be no concern: Used peanut oil is still food, so a container specified to deal with food should deal with it too.
If not, assume that the container has only been designed to be safe with the exact, unaltered ingredient shipped in it (oil and plastic chemistry can get very complicated, some edible oils WILL dissolve some plastics).
The best option will always be an airtight, food safe, glass, stored in the dark container well filled.


Phytomass and photosynthesis

Question: I have been introduced to the term "phytomass" and I was wondering if phytomass can be both photosynthetic and non-photosynthetic?
Answer: "Biomass is defined as live material, necromass as dead, attached material.
Biomass + necromass form the "standing crop" or phytomass." (from https://link.springer.com/chapter/10.1007%2F978-3-642-68786-0_22)

But even if we were only talking about biomass, there is only a loose correlation between that and photosynthetic activity -  think about single-celled algae, where 'every' cell is photosynthesizing, versus trees, where most cells are non-photosynthesizing, and there may even be times where none are (in trees that loose their foliage in winter).
Comment: Oh I see. Is true though that phytomass has both photosynthesising and non-photosynthesising components (with the majority being non-photosynthesising)?
Comment: That depends on what those 'components' are - if you are counting cells, the answer is a clear yes, if you are counting organisms the answer would be no, and if talk about mass, i don't know.  - 'Phytomass' and 'biomass' are both in need of a qualification if you are going to use those terms - 'biomass' might also be the sum of all living organisms, and is also sometimes used for some types of harvest (plant matter for instance), where the 'bio' only refers to the formerly living state of the matter...
Comment: .. also note that in my answer - citation the phytomass only comprises 'attached' dead material, whereas sometimes it can also comprise all dead former plant material.

Phytomass and photosynthesis


Should "ads" and "advertising" tags be merged?

Question:
advertising × 184 - the practice of producing text, images, or other media intended to influence others' purchasing decisions.
ads × 60 - Short for advertisements. Visual elements added by a webmaster on a web site or web page to earn money.
Comment: Seeing no objections, I'm going ahead with this tag merge.
Answer: These tags mean basically the same thing and should be merged.  The canonical tag should be "advertising" because it is used more and because "ads" is a short form of it.
Answer: Missed this one...
My preference would have been to leave advertising, which is the practice, and ads, which are the elements, as separate tags.
Not sure if this can be reversed at this point however.


Guarding Eyes for Noahides (Shemirat Eynayim)?

Question: Are Noahides required to guard their eyes (Shemirat Eynayim) as Jews are? Is there some information in Jewish religious literature addressing this question?
Answer: The concept of guarding the eyes is set up to prevent one from forbidden relationships. The list for a Jew is enumerated here. The Talmud Yerushalmi Brachot 1:5 states that the eyes [and heart] cause one to sin (in the right context). In Nedarim 20a it also states that one who looks at forbidden relationship will eventually come to commit a sin.
Recounted by Rambam in
Repentance 4:4:

He who wantonly stares at the opposite sex, for he thinks that it is of no matter, saying: "Have I stilled an urge, or even been near her"? whereas he remains oblivious to the fact that purposeful eye-staring is a great iniquity and begets actual unchastity, even as it is said: "And that ye seek not after your own heart and your own eyes" (Num. 15.39).

Non-Jews however, do not have a lengthy list of prohibited relationships; just six.
Rambam, Kings 9:5:

There are six illicit sexual relations forbidden to a gentile:
  a) his mother;
  b) his father's wife;
  c) a married woman;
  d) his maternal sister;
  e) a male;
  f) an animal.

It comes out that a form of guarding the eyes may need to be implemented, should such issues arise. But surely any method could work.
Comment: Thank you Dr. Shmuel, you fully answered my question!
Answer: One of the primary prohibitions in looking at illicit material is that it causes improper thoughts.

מכאן א"ר פינחס בן יאיר: אל יהרהר אדם ביום ויבא לידי טומאה בלילה - כתובות מ״ו

The prohibition (see Igros Moshe E"H 1 §69) of improper thoughts is because it can cause הוצאת זרע לבטלה [wasting seed].
Whether or not Noahides have the prohibition of wasting seed would depend on the reasoning for the prohibition: If wasting seed is prohibited because of P'ru Urevu (Tosfos Sanhedrin 59b), a Noahide would not be forbidden (because a Noahide is not commanded to procreate [Tos. ad loc]. See, however, Tosfos Chagiga 2b, who implies that a Noahide is commanded in Pru Urevu, in accordance with the Sheiltos 165 and the pashtus of Yevamos 62).
If, however, the prohibition is from Lo Sinaf (או״ז א-קכ״ד, סמ״ק רצ״ב), a Noahide would indeed be prohibited from wasting seed.