No Limits Humans - Physical Side

Question: In a world I'm creating, there is a sect of humans that hunts monsters in secret.  All of them have magic "cores" installed on their person.  These cores can store massive amounts of neutral biological matter (think enough to rebuild a person) as well as convert that mass into whatever tissue type needed and replace damaged cells (and repair DNA).  The very obvious affect of these cores is practical immortality.  But if they were fully (consciously and subconsciously) aware of the fact that they didn't have to hold back, could they be faster or stronger than a normal human?
Comment: Unfortunately no, the volume and types of muscle and skeletal tissues didn't change remember you are healing or replacing cells not augmenting or mutating these cell. Yes since you can recover from any fatal wound in minutes you don't need to hold back but are capped by your genes and no gamma ray don't turn you into hulk!
Comment: enough matter to rebuild a person is roughly person-sized (or at least person weight, if it was compressed). Unless the magic also transfers this matter to a different dimension (or whatever) they're going to be waling around with quite a hump!
Comment: The only way I see for them to be stronger and faster, etc. is if they're born with this *core* (resp. they get it implanted quite early in their life) and hence grow up with this *potential*
Comment: @dot_Sp0T That's why I included the fact they are aware on all levels what they can do ;P
Answer: Yes, they could be faster and stronger than a normal human, but not by a whole lot. The mind limits what it asks of the body to ensure that it does not injure the body, but a lot of this tendency gets conditioned out of elite athletes (weightlifters in particular) during the training process. From an article in Scientific American:

Vladimir Zatsiorsky, a professor of kinesiology at Penn State who has extensively studied the biomechanics of weightlifting, draws the distinction between the force that our muscles are able to theoretically apply, which he calls "absolute strength," and the maximum force that they can generate through the conscious exertion of will, which he calls "maximal strength." An ordinary person, he has found, can only summon about 65 percent of their absolute power in a training session, while a trained weightlifter can exceed 80 percent.

So you take a "no limits" weight lifter and a regular weight lifter of identical absolute strength. "No Limits" would be able to lift about 250 lbs while "Regular Lifter" would be able to manage a little over 200 lbs. No Limits would probably be in a lot of pain until the muscles they wrecked were repaired, while Regular Lifter would just have tired muscles.
Fatigue also appears to be another such limit. Your "no limits" people would be able to walk and run for ages, given that their "cores" would be constantly repairing the damage being slowly done to their bodies. They would also be able to forgo sleep without experiencing permanent brain damage. But the healing might take a while if the relative speeds of natural healing processes apply, however accelerated they may be. (Brains are really slow and hard to heal.) And they would probably still get dingy and hallucinate due to the toxins building up in their brains.
Comment: This could be interesting to explore in plot, mind - effectively, the core removes the physical limits, so it's all in the mind. It'd be kinda like the Matrix - when training Neo, Morpheus makes the point that he can do whatever he believes he can; he's only tired because on some level he believes he should be after such exertion ("you think that's air you're breathing?" and suddenly he stops panting). Someone new to a core would still hold back for fear of injury, like a normal person, while someone more experienced would push harder because they've learned that they can.
Comment: would they still be able to lift something with 'wrecked muscles' - unless healing was instantaneous, the ability to lift would not be much greater than the weightlifter.
Answer: Going from ryepdx's answer which was great, I think someone who is going to have their body handle the damage done to it, can push it a lot harder than the 'No Limits' weight lifter.  This will likely take months or years of training to be able to push themselves past the normal limits and even into limits that should kill them if they weren't 'special'.
Speed is going to have limits, these are going to be physical limitations that the body just isn't designed to do, however, they might take a lot less training to be able to sprint at max speed for much longer periods, and once you trained your body to not flinch when you hit something as hard as you can, because the damage will be repaired quickly you are going to be much more likely to perform amazing acrobatics and feats of strength that might break bones.
Basically once they have retrained fear responses and the bodies natural break limits many amazing things can be accomplished, but they are still going to be within what the human body can physically take.
Do they have Wolverine style healing?  Then Wolverine is an excellent model for this (after he lost his adamantium skeleton).
Comment: Another thing to consider: damage and repair is how the human body makes itself stronger. Would these magic cores then allow them to fit a few months of training into just a few days, since there'd be no recovery period?
Comment: Yes, the healing is Wolverine speed (or faster in a non-fatal wound) so that is an excellent comparison, but I never heard he lost the skeleton, when did this happen?
Comment: @HadesHerald Magneto ripped the metal out of his body.  Almost killed him.  I think it was around 15 years ago.
Comment: @ryepdx with wolverine's healing factor, I would say that is a yes.  Even more, they might be able to build muscle during an actual fight.
Comment: @bowlturner, I imagine those are going to be some extremely bulky humans then!
Comment: Yes, I would imagine, the question would be how fast do they 'go back to normal;?
Answer: One that you might not have considered - how do they deal with pain? That'd be a major limiter on the extra speed and strength.
Just because something's going to be fixed, doesn't mean its not going to REALLY hurt if you break it. A friend of mine wrote a short story that dealt with this quite cleverly. Just because you're invincible, doesn't mean you're going to throw yourself head on into danger with little regard to the consequences.
It could potentially be assumed that they'd be the peak of physical fitness because they would be able to continuously train without worrying about permanent injury so they could conceivably be pretty effective fighters, but not necessarily super-human.

No Limits Humans - Physical Side


Change Sharepoint field by custom

Question: I was write custome sharepoint field that extend functionality for select users, but now i need change default user field to my field type. I find User in fldtypes.xml and change FieldTypeClass element but if I try to add new entry in list I see older (default) field.
Answer: You should never, ever, ever edit any files in the SharePoint 14 or 15 hive. If you do that, a few things happen:

The next time you deploy a hotfix or service pack your files may be overwritten and your custom code will be gone and things will stop working
Your farm will be in an unsupported state and Microsoft will not give you any support if you have any issues in your farm
You will unleash all the SharePoint deamons and all the ghosted files will haunt you for the rest of your live

There are a few exceptions and these may be the web.config files in some folders like the lauouts and securitytoken folders.
That said, what you should do is:
 1. Deploy your custom field in a new fldtypes_something.xml file via a wsp solution
 2. Create a new site column that uses your custom field
 3. Create a new content type that uses your site column
 4. Create a new list definition
 5. Create a list instance using that list definition
This way you'll be using your custom field in a supported way. And following the SharePoint development guidelines provided by Microsoft.
Comment: Thanks for answer. How I can replace standard user field wich my custom in the earlier created lists?
Comment: You can add the new custom field to the existing lists, copy data from the old field to the new field using a simple PowerShell script and then delete the old field.
Comment: If you have many lists where you want to do this, you can script all three steps in a single PowerShell script.

Change Sharepoint field by custom


How can I speculatively invest $1000 in (micro)bitcoin?

Question: How can I invest $1000 it micro-bit-coins, or whatever they are called?
Comment: this is better suited for the bitcoin stackexchange, found here: https://bitcoin.stackexchange.com/
Comment: Are you asking how to buy bitcoin, or are you asking how to make money at it?
Comment: In addition to the buying it directly (as answered below), you can buy futures or other instruments (such as spread bets) that track its price, and these products can be bought from more reputable (and regulated) institutions than Bitcoin exchanges.
But as others have said, be aware that this is high risk, and probably not advisable if you don't know much about it.
Answer: Find a (reputable) exchange that accepts business from citizens of the country you re located in and purchase through them.
Be advised though that fear of missing out and investing in something you dont even know how is called (let alone understand) can be a very bad practice and sets you up for losses.
Answer:
Acquire $1000 you can afford to lose.
Meditate on the transience of material wealth until you acquire a mindstate where you are not bothered by losing $1000 to a gamble.
Download, install and run a bitcoin wallet software. Alternatively, use an online wallet service and hope they won't scam you.
Wait until the software synchronized with the blockchain. This can take a while, because it's several GB by now. This is not necessary with an online wallet service because they already did that for their other users.
Use the software/service to create a new wallet (which is just two long codes: Your public wallet address and your secret wallet key).
Create a reliable and secure backup of your wallet key. When you lose it, there is no way to recover your bitcoins. When someone else finds it, they can steal your bitcoins.
Go to the website of a Bitcoin exchange. Pay $1000 in exchange for a bitcoin payment to your wallet address.
Wait until the software says that the transaction arrived in your wallet (can take a while).
Monitor the bitcoin chart until you decide it's time to cash in / rescue what you still can.
Sell your bitcoins to a bitcoin exchange in exchange for real currency (does not need to be the same exchange you bought from)
Comment: Perhaps you provide a ballpark on how long "a while" is for #4 and #8.


Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$

Question: I have a few students that are having trouble understanding that
$$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$
specifically that
$$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} = 2n+1.$$
I've tried explaining it a few different ways, but it didn't seem to go over very well.  What would be a good way to explain this?
Comment: Perhaps demonstrate it for low values of $n$, and then sneak in an induction proof?
Comment: From what you wrote, one thing that I notice that the students would find confusing is that the products end at different values.  Perhaps write out the product (or have the students figure out) the last three factors of both the numerator and denominator.  Then, they can see some cancellation.  If this is an intro to proof class, they might appreciate learning product notation at this point; with product notation, it becomes somewhat clearer.
Comment: Maybe put $x_n=1·3·5···(2n-1)$ and write $\frac{x_n(2n+1)}{x_n}$
Comment: Perhaps point out that $$2n-1= 2(n-1)+1$$ $$2n-3=2(n-2)+1$$ etc..
Answer: I would suggest including more terms at the end of the products. That makes it easier to see the pattern and to see why it's just $(2n+1)$ that's left at the end.
$$
\require{cancel}
\begin{align}
\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} &=
\frac{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)(2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)\quad\quad\quad} \\
&=\frac{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}(2n+1)}{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}\quad\quad\quad} \\
&=2n+1
\end{align}$$
It might also simply be a matter of spacing: students might get confused by seeing the $2n+1$ directly above the $2n-1$. I've tried to format the above to emphasize that the $2n+1$ is an extra factor that doesn't appear in the bottom.
Finally, it could also be helpful to pick a small $n$ and work out the product concretely.
Answer: It might have helped if you did the equation in reverse, but it seems like they're not seeing the general pattern, namely, that the $j$th factor in $1 \cdot 3\cdots \cdot (2n + 1)$ is $2j - 1$. If you could explain that, then they'll understand that the $n$th factor of the sequence is $2n - 1$, and the $n+1$st factor is $2(n+1) - 1 = 2n + 1$.
Answer: A picture is worth a thousand words.
Just give an example with a sufficiently large value of $n$, nothing works better.
Suppose $n=10$, then:

$1\cdot3\cdot5\cdot\ldots\cdot(2n+1)=$
$1\cdot3\cdot5\cdot\ldots\cdot21=$
$1\cdot3\cdot5\cdot\ldots\cdot19\cdot21=$
$1\cdot3\cdot5\cdot\ldots\cdot(2n-1)\cdot(2n+1)$

Numeric examples are also very useful when approaching series/sequences related problem, as they allow you to understand the behavior of the expression at hand before trying to prove it.
So this technique might also help your students with more advanced tasks in the future...
Answer: I would suggest put some various number
  for example n=3 , and force them to write numbers in two rows
  then simplify
  $$n=3 \to 2n+1=2(3)+1=7 \to18.104.22.168 \\2n-1=2(3)-1=5 \to 1.3.5\\ \frac{1.3.5...(2n+1)}{1.3.5...(2n-1)}=\frac{184.108.40.206}{1.3.5}$$


Login form: User ID and Password in same page vs User ID in the first page and password in the next

Question: Some major sites like Google and Amazon take 2-page approach for signin. They first ask for ID and in the next page ask for password.
What is the advantage in this against single page signin process?
Is this because of security, technical, usability or any other reason?
Comment: I know that this plays havoc with several auto-fill extensions, so it might have something to do with hiding the connection between the two inputs for key-loggers
Answer: My guess is that the cybersecurity folks have more say in that case than the UX people.
Transitioning from click to keyboard to mouse takes a lot of mental energy and doing so multiple times separated by a page reload is a good argument against that practice from a UX standpoint.
However, I would guess that companies like Amazon and Google are doing a lot of "work" behind the scenes during that screen refresh. I would be checking to see if this is a valid email address. If the device location seems normal. And so on. All while blocking form fill scripts and brute force user/pass combo attacks.
Since the sites in your example are big players with a lot to lose from a security perspective, and because users of those sites want to access making them willing to endure a bit of pain, UX kind of loses the business case.


Straight lines and longest distance

Question: In Hartle's book 'Introduction to General Relativity', he says that straight-line paths between two time-like separated points are the longest paths. He uses this in the case of Twin paradox, where the moving observer does not have a straight path between the same points, hence its 'distance' or the proper time is shorter. This is understood clearly using the time-dilation formula, but how are straight-line in non-Euclidean space longer? Any geometrical or mathematical explanation would do.
Comment: This is more a problem of Lorentzian vs Riemannian as opposed to Euclidean or not Euclidean but still Riemannian.
Comment: Hi Maryam. There is some discussion of this in [What is time dilation really?](http://physics.stackexchange.com/q/241772/) and [What is the proper way to explain the twin paradox?](http://physics.stackexchange.com/q/242043/) though neither are a direct answer to your question.
Comment: Quanta repeated the longest time bit, but without specifying timelike.. (!). They just write about two points in space-time, so naturally it made no sense to me. This question was the answer I needed :)
Answer: The "distance" between two points(events) in space-time is given by,
$ds^2 = dt^2-d\bar{x}^2$
Where $d\bar{x}^2$ is the spatial part of the space-time. Now moving in a timelike straight line you can always choose a co-ordinate such that this spatial part is zero. And then in that co-ordinate $ds^2 = dt^2$. Now for any other path between this two events the extra term $d\bar{x}^2$ comes with a negative sign and therefor decreases the distance. Thus the longest distance will be straight line one.
Answer: The "length" is the Lorentzian length: For a world line $\gamma$ with velocity $v^\mu$, we have $L(\gamma)=\int \sqrt{-g_{\mu\nu}v^\mu v^\nu}\,dt$. A geodesic is a critical point of the length functional. To see that a geodesic (straight line) is not a minimum, we need to show that there is a nearby curve with shorter length. This construction is standard: We can perturb $\gamma$ slightly to make it "almost spacelike", which decreases the total length since $g_{\mu\nu}v^\mu v^\nu$ becomes arbitrarily close to $0$. To do this, we can zig-zag a curve very close to $\gamma$ so that each "straight" part is very close to being null. This clearly fails in the Euclidean/Riemannian case. If there is no conjucate point along the geodesic, then it is in fact a maximum of length.

Straight lines and longest distance


How to Mention the Use of Pseudonym when Cold Pitching

Question: I'm looking to gain experience writing for blogs. To do this, I plan on cold pitching a list of bloggers with guest post ideas. Only, I'd like to publish the posts under a pen name. I don't want the pitches to get rejected because of that slightly unusual requirement, either.
I have no experience writing under that name and very little experience otherwise.
When should that I mention that I'd like to use a pseudonym? The second email, once they'd responded to my pitch? Or at a different time?
When, where and how would you recommend I bring it up?
Answer: It really is not that uncommon in an online world to use a pseudonym. You're doing it right now, in fact.
If you feel it is important to get settled early, though; still don't mention in the initial cold pitch. It is presumptuous, and it really does not relate to the content you are pitching, hence distracting from it.
Normally, you'd enter some sort of contract--or at least a work (for hire) agreement--and in the negotiation of that is when things like pseudonyms should first come up. If you don't expect to be compensated at all, it is also not entirely unreasonable to not treat it as a 'pseudonym' at all in your interactions, and to just sign your emails with it.
Comment: @user613 If you do not think you will want to keep using a pseudonym for your whole career, it may be best to avoid that, yes. If sums of money start exchanging bank-to-bank, same thing.
Comment: Is it a bad idea for future business with that person if I pretend a pseudonym is my real name? I mean, the other person might assume that the pseudonym is my real name, which could make me seem deceptive if I have to tell them later on that in fact, it's a pseudonym.


Can the study of the quantum information structure in QFT with holographic duals be relevant to string theory?

Question: I'm interested in characterizing the behaviour of measures of quantum information in strongly correlated quantum field theories which admit a gravity dual description, e.g through AdS/CFT duality. In a more broad sense, my question points towards the following: in my limited experience, I use to accept that a duality means that we don't have any right to give a physical meaning to one part of the duality and use the other part as a mere computational tool. I would like to be illuminated by some experts.
Comment: This is an interesting topic, and there may be a few things to be said. But, in the present form it is not clear to me what you are asking precisely.
Comment: Thanks for your interest @Moshe R. I acknowledge that I wasn't very precise in my question. What I meant is the opposite to something like, what can AdS/CFT duality can tell us about condensed matter? My question should better be rephrased into: could the study of quantum entanglement in strongly coupled field theories (e.g-critical quantum many body systems-) with holographic duals tell us something interesting about string theory?
Answer: This is an interesting question, which quite a few people (myself included) find fascinating, and by and large still remains to be explored. To my knowledge, the farthest this idea has gotten so far is in the work of my colleague, Mark van Raamsdonk: Comments on quantum gravity and entanglement and the essay Building up spacetime with quantum entanglement which won the GRF essay competition last year. If you have background in many body systems and their description in terms of quantum information, your input on this set of issues could be valuable.
Comment: Thanks again @Moshe. These two papers are really exciting. I hope to get more input on those issues soon.
Comment: I will also post something on this issue soon.
Answer: The holographic principle in general gives an equivalency between gravity in the interior of a space and the quantum fields on the boundary.  So in the case of the $AdS_n$ its boundary $\partial AdS_n~\simeq~CFT_{n-1}$ is equivalent to a conformal field of one dimension lower.  This is a form of the black hole holographic principle, where the field configuration of a black hole is defined by strings on the horizon.  This means the number of pieces of information required to describe physics in dimensions $N$ is reduced to $N~-~1$ dimensional description.
Comment: @Daniel Mods are not here to judge quality of an answer; if you think it is wrong, downvote and/or (better) add a comment explaining what is the problem.


Is it allowed to ask whether there is a synonym of a specific Russian expression\idiom in English?

Question: I have a question regarding specific saying in Russian. I'd like to know if there is a saying\expression in English which has a similar meaning. Is such question fits this site?
Answer: Edit: I've just noticed that in the definition phase, these questions were highly upvoted.
Technically we should consider them on-topic, if they're the same kind you were referring to (I think so). In order to deem them as off topic, a Meta question should be proposed to ask the community opinion.

Initially I was in favor, and actually I'm not against, but this would be easier for the users in EL&U perhaps? I think that here for example, you could post the opposite: English to Russian saying correspondence.
Remember that if you ask it on one site, you theoretically cannot post the same question on another site. One of them will be deleted/migrated and merged with the other one. So choose the best site for your question, don't force it.
Comment: Thank you for your answer. I just wonder how many Russian speakers are there on EL&U and what are the chances that one of them will see my question and actually know the answer... That's the reason why I think it'll best asking it here rather than EL&U..
Comment: There are some there... But the fact is that you need to look for those that know best the "target" language.
Comment: @EugeneS See my edit.
Answer: This Q&A is not for questions about English that are targeted to Russian speaking audience. Therefore, the type of questions you're talking about should not be welcome on RLU. They must be welcome on ELU, with the caveat that you should not only post the Russian word (so that Russian speakers can see it) but also explain it thoroughly so that others can answer your question as well.
Comment: See my edit, Armen.


I want to connect my iMac audio port to both a set of headphones with a mic and a set of external speakers

Question: I have a 2013 iMac with a 3.5mm headphone jack that provides stereo out and mic input in, I would like to connect this single port to my headphones that have an inline mic so I can listen to music/take phone calls/skype through my computer, or output from the comuter to the speakers for when I want to watch a movie or broadcast music around my apartment. I think I need an audio splitter/amplifier, but I am not sure how they handle input from a mic coming back through one of the outputs. It seems like there should be a simple solution but I cannot find one. Just a quick recap(because sometimes I am not clear) I have 1 3.5mm jack on my iMac (that supplies stereo out and mic in) and 2 3.5mm connections, 1 is a simple stereo connection(the external speakers) and the other is stereo + mic on a single 3.5mm connector(like iphone headphones, I don't know that this is called, and I know sometimes the audio and mic are split into 2 different connectors, but mine isn't). I don't necessarily need the mic to be connected to the speakers, and I don't plan to use both the speakers and headset at the same time. I am looking for cheap recommendations or better search terms to use when looking, or specific places/websites to look.
Comment: walloftext TL:DR
Answer: I know one thing you could do is use a splitter.  They should be quite cheap(I got 2 for a dollar).  Connect the headphone to one end and speakers to the other. If you have headphones with dedicated mic port (like mine) then use a splitter for the headphones and connect that to the 2nd splitter . This works for me and hopefully for you too.
Happy listening...
Comment: I have tried that, but it was too quiet, I am hoping to find a splitter with an amplifier to increase the signal


Photoelectric effect

Question: Why the photoelectric effect is observed only for metals and not for non-metals?
Isn't it possible for the photon to release an electron from a non-metal surface?
Answer: Conduction electrons in metals lie in the conduction band, they have a very low binding energy compared to electrons in an insulator, for instance. It thus requires a low amount of energy to excite them and have a photoelectric effect.
You can see on this article showing original data from Milikan, that the minimum amount of energy per photon required to have a photoelectric effect with sodium is of order of $\Delta E_{\text{conductor}} \simeq 1,8$ eV.
In insulators, the highest-energy electrons lie in the valence band and you need a huge amount of energy to excite them : it is harder to have a photoelectric effect in this case. In the case of insulators, this energy is of order of magnitude of the first ionization energy of the atom or molecule which the solid is made of.
You can see here a table giving in eV the first ionization energy of each atom. (It's in french, couldn't find the same table on English wikipedia). You see that the energy required is of order of $\Delta E_{\text{insulator}} \simeq 10$ eV for most insulating pure compounds. It is thus possible to have a photoelectric effect, but would require photons with ten times more energy than for insulators.
Comment: What is the difference between a conduction and a valence band?
Comment: Conduction and valence bands is what creates the difference in behavior between most insulators and conductors. If you look at the energy levels of electrons in a solid, they are gathered in what we call "bands" with an energy gap between them. In insulators, the highest energy band is completely filled, wherehas in conductors it is only partially filled : we call this band the conduction band, and the lower-energy bands the valence band. You have a full explanation on this on wikipedia : https://en.wikipedia.org/wiki/Valence_and_conduction_bands
Comment: I think this is the right answer, but there's also a practical difficulty.  As electrons leave the insulator it acquires a positive charge.  One has to figure out a way to replenish the supply of electrons.  Easy in a metal.
Answer: The outer electrons in insulators tend to have higher binding energies, so it requires more photon energy to free the electron. So, the photoelectric effect is still possible in insulators, it's just harder.
Comment: `kotozna`, could you make your answer more quantitative by comparing the energy needed for the case of metals with that in the case of insulators — to see how "harder" it is? Please also mention if experiments are performed for some insulators.

Photoelectric effect


What is the meaning of "nutrient bang for your calorie buck"?

Question: What is the meaning of "nutrient bang for your calorie buck"?
Answer: The phrase derives from the idiom "bang for your buck", which refers to getting the most value for a limited amount of money.
In this case, the "calorie buck" refers to the calories ingested (the fewer the better), and the "nutrient bang" refers to the nutritional benefits you'll receive (the more the better) from consuming whatever product the calories come from.
Answer: Thats a new one! anyways it is trying to convey the message that for the cost of that many calories, you will get the highest value of the other nutrients. People are conscious about the calorific value while buying products.Usually products rich in nutrients are more expensive than the products which give big amount of calories (and are also very available easily/cheaper). Eg: Cost of Nutrient Drinks (and another eg here) vs Soda which usually run about $4 for 12 can pack or even cheaper.


Are MCUs (Like arduinos) effected by Meltdown?

Question: I am curious if Microcontrolers are effected by the vulnerabilities of Meltdown and Spectre. These devices are not specifically CPUs, and do not run Operating Systems per-say. I'm not really sure I understand the whole speculative execution attack, but I would like to know if this can effect things like  microcontrolers, which are used to drive many IOT devices.
Comment: there are some semi-related timing attacks on MCUs, so more and more parts are using constant-time comparisons in more and more places.
Answer: In general, no, for several reasons.
First, Meltdown is an issue with how Intel chose to handle invalid memory accesses during speculative execution.  A couple of ARM designs are vulnerable to a variant of Meltdown, but other than that, it's Intel-only.
Second, Meltdown and Spectre both require that the CPU use a technique called "speculative execution".  This is a great performance boost, but it comes at the cost of a bigger, more expensive, and more power-hungry chip.  Microcontrollers are small, cheap, low-power chips -- exactly the opposite of what would use speculative execution.
Third, Meltdown and Spectre are about getting around memory protection.  The typical microcontroller only runs a single program, so applying memory protection is a waste of time and effort -- any memory a program could read using Meltdown or Spectre, the program could read just by looking at it.
Comment: Speculative execution is only a major performance boost in cases where there is a large difference between the time required to read information from the fastest and slowest parts of a computer's directly-addressable memory.  On a desktop machine, the difference would often be on the order of 100:1.  On a small micro, it would be at most 2:1.  If 10% of accesses would incur a 100:1 speed penalty, reducing that could offer a major performance boost.  If 10% or even 25% of accesses incur a 2:1 speed penalty, however, the performance boost achievable by alleviating that would be minor.


Minimum clique cover relationship with Chromatic number

Question: For a general graph, can we say that the summation of minimum clique cover and chromatic number is upper bound by n+1 where n is the total number of vertices in the graph? If not, what is the relationship. Note that I am not talking about complement graph.
Comment: Probably by chromatic number you mean the least number of colors that suffice to assign *vertices* colors so that no two adjacent vertices have the  same color, although some authors use similar terminology about *edge* colorings.  However a [minimum clique cover](https://en.wikipedia.org/wiki/Clique_cover) is a clique cover, not a number per se.  Likely you meant the clique cover number(?) so that we would take the "summation" of two integers.  I think you should try to fill in some details of your Question, esp. your thoughts on why it might be true (or difficulty you had).
Comment: Why do you say that you are not talking about complement graph? Are you saying that, in your belief, the minimum clique cover number of a graph is **NOT** the same as the chromatic number of the complementary graph?
Answer: Yes, it is true.  The sum of the chromatic number $\chi(G)$ of the graph and the minimum size of a partition of the vertex set into cliques is at most $n+1$, where $n$ is the number of vertices. Observe that partitioning the vertex set of $G$ into cliques is equivalent to partitioning the vertex set of the complement graph $\overline{G}$ into independent sets.  So the second term in the sum is the chromatic number $\chi(\overline{G})$.
To prove that $\chi(G) + \chi(\overline{G} \le n+1$, order the vertices $x_i$ so that $d(x_1) \ge d(x_2) \ge \cdots \ge d(x_n)$.  Do a greedy coloring of the vertices of $G$ in order $x_1,\ldots,x_n$, and a greedy coloring of $\overline{G}$ in order $x_n, \ldots, x_1$. It can be shown that the total number of colors used in both these colorings is at most $n+1$.
Answer: Notwithstanding your weird remark that you're "not talking about complement graph", the clique cover number of a graph $G$ (the minimum number of cliques needed to cover the vertices) is obviously equal to $\chi(\overline G),$ the chromatic number of the complementary graph $\overline G;$ see the answer to this question. So the statement you want to prove is obviously equivalent to the following:

For any graph $G$ with $n$ vertices, $\chi(G)+\chi(\overline G)\le n+1.$

Here is a straightforward proof by induction on $n.$ The statement is obviously true for $n=1.$ Consider any fixed positive integer $n$ and assume that the statement is true for all $n$-vertex graphs; I have to show that it's true for any graph on $n+1$ vertices.
Let $G$ be any (simple) graph with $n+1$ vertices; I have to show that $\chi(G)+\chi(\overline G)\le n+2.$ Choose a vertex $v$ of $G.$ By the inductive hypothesis,
$$\chi(G-v)+\chi(\overline G-v)=\chi(G-v)+\chi(\overline{G-v})\le n+1.$$
Since
$$\operatorname{deg}_G(v)+\operatorname{deg}_{\overline G}(v)=n,$$
we must have
$$\operatorname{deg}_G(v)\lt\chi(G-v)\ \text{ or }\ \operatorname{deg}_{\overline G}(v)\le\chi(\overline G-v).$$
Without loss of generality assume that
$$\operatorname{deg}_G(v)\lt\chi(G-v).$$
Then
$$\chi(G)=\chi(G-v)\ \text{ and }\ \chi(\overline G)\le\chi(\overline G-v)+1,$$
whence
$$\chi(G)+\chi(\overline G)\le\chi(G-v)+\chi(\overline G-v)+1\le n+2.$$


Cooking old beets

Question: I just harvested some beets from last year, they are red in the center and the rest is a yellow white color.  They are huge, so are they edible?
Comment: There may (or may not) be palatability/texture issues ("woody") but they should be *edible* - you just may find that you don't enjoy them much, depending just how far past their prime these are.
Comment: Hello Beet gardener! We specialize in giving clear solutions to clear-cut problems. "Are they edible" is a clear, answerable problem. "How to cook them" is not, there are tons of ways to cook beets, none better than the other. So I removed that part of your question, but the first part is fine. Welcome to the site!
Answer: I believe the beets are safe to eat.  I happen to have some beets that have been in my fridge for about 3 months.  They are also big.  What I will do in the next couple of days is what I think you should do.  Clean them, cut off a bit of the top, about a half inch and the same with the bottom, then put them in foil, put a bit of canola oil and salt on them, wrap them tight, put on a pan, in preheated 400 degree F oven for about one hour, check at 45 minutes, then cook as needed 15-45 minutes more.  I don't know how big yours are and if you are a bit iffy then bake at 350 F or 375 F, same time.  Boiling is also a way to do it but baking brings out such a sweetness in the beets it is just an unbeatable way of cooking them for me.  Good Luck.
Comment: "un-beet-able" way of cooking them. :)
Answer: Beets last ages if stored in cool conditions; if it isn't moldy and is still hard like a rock and not squishy, it should be completely safe.  If it is a gigantic beet, it's possible it might be woody, but that's a separate question from whether it is safe.
I agree with @user33210; baking beets is definitely the best way of cooking them and brings out their natural sweetness in a way that moist cooking methods cannot.  I always bake them with a little bit of salt and olive oil; just like baked potatoes.

Cooking old beets


Sat math section simple mistakes

Question: I am fixing to be a junior in high school and as such have done some extensive SAT prep, although I find that when I take practice test I can constantly score about a 770/800 on the math section. However, when I go to look at the problems that I missed they are often the result of simple and careless mistakes. I was just wondering if anyone had any suggestions on how to make less careless mistakes without spending to much time which is precious on the SAT?
Answer: I have found that making mistakes follows certain predictable patterns, so I would suggest to practice even more; after a while, you learn to recognize the kinds of situations where you know that you are liable to make some mistake, and thus force yourself to be a little more careful.
Answer: Estimate. Before solving a problem, quickly approximate what you would expect the ultimate solution to be. Before answering, ask yourself, it this value around what I initially expected? Does it make sense? For some questions, you can easily eliminate half the answer choices using this method, and prevent yourself from making a careless mistake.
Answer: For me, I know I often misread the problem.  So it makes sense for me to re-read the question, slowly, completely, to make sure I know what the question is.  A correct answer to the wrong question does not help one's SAT score.  (I guess this is covered by Bram28's answer, but maybe in an unusual way.)

Sat math section simple mistakes


Email external email when new item created

Question: I was wondering if there is a way to make a list automatically email to an external users email when a new item is created/edited. I know users can set up alerts on lists, but this only works for their outlook emails. On a separate note, when you're setting up an alert, there is a text message option. This option is greyed out for me. Is there any way to change this?
Comment: Why not use a workflow?
Comment: There are people who will not be in at work and would rather have it send to a different email.
Comment: Can you not specify an external email address as the recipient of a workflow's email task? (using a 2010 workflow if in SP 2013/SP Online)
Answer: To trigger an email, most people will tell you how to create a workflow (probably with SharePoint Designer). That's quite easy, it would work in most cases. But it's not the purpose of workflows (+ they do appear in the UI, they're slow, ...)
Another approach is to ask a developer to create a custom App (sorry we have to say "add-in" now). That app would contain a "remote event receiver", i.e. a Web service triggered when an item is added. That Web service would send the email.
About the Text Message option: it's grayed out because it's not configured (it can be configured with a texting service provider on on-prem installations). But I'm not sure it's available at all in SP online.
Comment: Using `Microsoft.SharePoint.Client.Utilities.Utility.SendEmail` won't work for every recipient. Only adresses specified as email address for valid SharePoint users will be accepted by this method. So for external email you'll need to use standard .NET API and authenticate to mail server yourself.
Answer: The easy answer, for now, is to make a SharePoint 2010 workflow in SPD and use that to send the email to an external email address since the 2010 version (unlike 2013 version) can send to external emails.
How do you send emails to external email address in sharepoint online workflows


How to arrange 1 to 15 such that the sum of any adjacent 3 numbers will be a perfect cube?

Question: The numbers 1 to 15 should be arranged in a way that any 3 adjacent numbers' sum will be a perfect cube.
Comment: The smallest sum should be 8, since it's 2's cube. And maximum sum is 27(3's cube) since the maximum possible sum can only be 42.
Answer: It is impossible.  Why?
Note that the largest possible sum of any three numbers from the set is $15+14+13=42<64=4^3$ while the smallest possible sum of any three numbers will be $1+2+3=6>1=1^3$, so the only two available cubes will be $8$ and $27$
Suppose we could arrange the numbers in such a way as to have all consecutive triples adding up to be a cube.  Consider what happens to a number larger than $8$.
Let $a_1,a_2,a_3,a_4$ be distinct members of the set $\{1,2,\dots,15\}$ with specifically $a_2>8$.
Suppose $a_1,a_2,a_3,a_4$ satisfies that $a_1+a_2+a_3$ is a cube and that $a_2+a_3+a_4$ is a cube.  Then $a_1+a_2+a_3>8$ and is a cube, so $a_1+a_2+a_3=27$ is the only possibility.  Similarly $a_2+a_3+a_4=27$ is the only possibility, but then that implies that $a_1=a_4$, a contradiction.
Comment: shouldn't you take $a_2>12$, so that you are sure that there is at least one such $a_2$ since not all three of them could be on an extreme of the arrangement?
Comment: @mau how does that contradict what I had already done?  Sure, I left out the detail that there will always be a number greater than eight at a position other than an edge, but that should have been obvious without stating.  The bound on $a_2$ could even have been relaxed to $a_2>5$, as it is only necessary to note that the sum of the triple being a cube forces what the cube is in this case.
Comment: If you had stated that there is always at least one such number, I would not have commented...


wavelet analysis/ dimension reduction

Question: How wavelet analysis works as a dimension reduction techniques?
The approximations coefficients at higher level of decomposition are the fewer retained coefficients from the originial ?Is it correct?
Answer: Well, not really. The wavelet transform (the discrete to be precise) ist just a base transformation of a signal. However, it can be used to compress signals (dimension reduction in some sense), due to the fact that the energy of most natural signals is compactified under the wavelet transform.
The continuous wavelet transform does even the opposite. It blows up your signal and introduces a lot of redundancy.
Comment: I did not get your answer very well, but, for example, to decompose the original signal(0) into level higher >0, is translated by getting few coefficients = (aj= a0/j*2), so it is similar to dimension reduction ?
Comment: If you have a special signal together with a special wavelet, it in fact does a dimension reduction, but, as I said, not in general. For instance, a pure sinusoid reduced with a shannon wavelet would be such a special case. In that case a simple fourier transform does the same, btw. But most signals are folded in a nonlinear way and can not be reduced by linear transform.

wavelet analysis/ dimension reduction


How do I tell Agent Ransack to avoid a particular path/filename pattern?

Question: I've started using Agent Ransack to search an entire directory at once looking for a particular string. The only issue I have with it so far is that if I'm searching in a Subversion working copy, it finds the string in the .svn folders, which of course is not what I want. Is there a way to tell it to ignore any directory named .svn as well as any subdirectories of a .svn directory?
Comment: It already knows, by default, to avoid `.git` directories. Just kidding :)
Answer: Apparently you can exclude files now in the free version of Agent Ransack by using the following syntax:
NOT: *.dll; *.svn-base

Comment: yep - works a treat!
Answer: It's not possible in the free version. Your best bet is to make sure you include the file extension and then it'll exclude the svn-base files, e.g.
*.cs;*.vb

Answer: As snowdude hinted FileLocator Pro (which is the paid version of "Agent Ransack"/"FileLocator Lite" same thing different name If you are willing to upgrade (or like me willing to persuade your boss it's a good investment!) then there is a feature specificity for this under Persistent Search Filter so once set it will always ignore ".svn", ".hg" and ".git"(for Yar!)
So if you have this version (or buy it) then you can click the little filter icon to the right of the "subfolders" tickbox (see this pic) and click "Exclude SVN/HG/GIT/CVS meta data folders" (even shows you the filter it uses so you can update it directly or copy it for other uses).
Hope this helps.
Answer: You can do it, if you turn on the File Name match by Regular Expression in the Tab Options. Then use a expression like .*jsp$,  this will exclude all SVN control files of the results.
Answer: You can use Agent Ransack itself for this task.
To avoid specific directories (Ex: c:\Windows), update the field "Look in": "c:\; !C:\Windows"
!C:\Windows - This will exclude Windows directory from search criteria.


Parsing source of a webpage with Objective-C

Question: Is there a way to parse a website's source on the iPhone to get the URL's of photos on that page? If so how would you do that?
Thanks
Answer: I'd say go for regular expressions - there is a one page library that wraps c regexesthat you can drop into your project.
Comment: Agree, you don't need to parse the entire doc just to get the  tags.
Answer: There is no super easy way. When I had to do it I wrote a libxml2 SAX parser. libxml2 has an html reader that works fairly well with malformed html, and libxml2 is included with the base system.
Answer: I recommend regular expressions.  There's a great open source Regex library for Cocoa called RegexKit.  For the most part, you can just drop it in your code and it'll "just work".
Getting all the urls of images wouldn't be too difficult (less than 20 lines of code) if you assume that all images are going to be in <img> tags.  You'd just grab all the image tags (something like: <img\s+[^>]+>), then iterate through those matches.  For each match, you'd pull out whatever's in the src attribute: src\s*=\s*("|')?\s*([^\s"']+)(\s|"|')
You might need to tweak that a bit, but it shouldn't be too bad.
Answer: You could try it using regular expressions, but I wouldn't recommend that.  You should have a look at NSXMLParser, assuming the webpage is coded to be XHTML compliant.  TouchXML is another good library.
Answer: take a look at Event Driven XML Parsing in the iPhone reference library
Answer: Are you OK with any approach you use not picking up on images loaded dynamically via JavaScript.
The closest thing I could see working is to parse out any JavaScript imports, load those up too, and then use a regular expression across the whole file looking for anything that ends in ".jpg/.gif/.png" and grab the full URL out from that.  The libxml approach would miss out on references to images not in img tags, but it might well be good enough.


What exactly mean minContextSlot parameter for getting program accounts?

Question: I would like to get only fresh program accounts by context slot. But when I use the getProgramAccounts method with minContextSlot parameter, nothing happens, I just get the all program accounts.
Answer: minContextSlot will return the program accounts if the effective slot for the query is at least this value.
Eg if you pass in minContextSlot = 100, but the node only has the account state up until slot = 99, it'll return an error, forcing you to retry the query.
Not sure what you mean by "fresh program accounts", but if you're trying to query accounts that were created or modified after a certain slot, it's quite difficult to do this historically unless you've indexed all the account changes yourself.
You can do this going forward in time by listening to onProgramAccountChange via WS (messages can be dropped!) or listening to an accountsDB geyser plugin (expensive).


Associativity of polar decomposition

Question: By polar decomposition, every continuous linear function $f \colon H \to K$ between Hilbert spaces can be written uniquely as $f = \widehat{f} \circ |f|$ for a positive operator $|f| \colon H \to H$ and a partial isometry $\widehat{f} \colon H \to K$ with $\ker(\widehat{f})=\ker(|f|)$. The binary operation $(f,g) \mapsto |g \circ f|$ on the set of positive operators is not associative. Is the binary operation $(f,g) \mapsto \widehat{g \circ f}$ on the set of partial isometries associative?
Answer: I have not fully checked this idea, but here goes.  I prefer the notation $P(T)$ for the partial isometry occurring in the polar decomposition of $T$.  I also got lost with three Hilbert spaces in the mix, so this answer is only for the case where $A,B,C$ all operate on the same fixed Hilbert space $H$.
In this notation, I believe the question is whether or not
$$
P(P(AB)C) = P(A P(BC))
$$
holds for all partial isometries $A,B,C$ on $H$.
I think it is easy to see that if $U$ is unitary, then for all $X$ we have $P(UX) = U P(X)$, and that if $T$ is a partial isometry, then $P(T) = T$.  From this, it seems to follow that if $A$ is assumed unitary, and $B$ and $C$ are partial isometries, we have
$P(P(AB)C) = P(AP(B)C) = P(ABC) = A P(BC)$ and $P(A P(BC)) = A P(P(BC)) = A P(BC)$ so the desired result holds.
If $A$ is not unitary, it still has a unitary dilation.  This means there is another Hilbert space $K$ and operators $X: K \to H$ and $Y: K \to K$ with the property that the operator $A'$ on $H \oplus K$ given by the block operator matrix $A' = \begin{pmatrix} A & X \cr 0 & Y \end{pmatrix}$ is unitary.  So consider the operators $B' = \begin{pmatrix} B & 0 \cr 0 & 0 \end{pmatrix}$ and $C' = \begin{pmatrix} C & 0 \cr 0 & 0 \end{pmatrix}$ on $H \oplus K$.  The operators $B'$ and $C'$ are partial isometries on $H \oplus K$, so by the work above,
$$
P(P(A'B')C') = P(A' P(B'C')).
$$
Now calculate: $A'B' = \begin{pmatrix} AB & 0 \cr  0 & 0 \end{pmatrix}$ and so a moment's thought ought to show that $P(A'B') = \begin{pmatrix} P(AB) & 0 \cr 0 & 0 \end{pmatrix}$, making the left hand side of the above equal to
$$
P(\begin{pmatrix} P(AB) C & 0 \cr 0 & 0 \end{pmatrix}) = \begin{pmatrix} P(P(AB) C) & 0 \cr 0 & 0 \end{pmatrix}.
$$
The right hand side is almost the same, but with $P(A P(BC))$ in the upper left corner, and unless I made a ridiculous error, you have what you want.  (There is a sightly more complicated unitary dilation theorem for operators between different spaces, so if there were no mistakes in this approach, maybe the same idea works in the general case too.)
Comment: The first half of your argument in fact shows the desired result to hold when A is an isometry (even when considering functions between different spaces). But unfortunately, in the second half, not every partial isometry can be dilated to an isometry. For example, $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ is a partial isometry on $\mathbb{C}^2$, and no matter what larger space we embed $\mathbb{C}^2$ in, the kernel will stay nontrivial.
Comment: Ack!  I even got the term wrong (my "dilation" is really an extension, and as you point out, things with kernels can't extend to things that don't have them).  One _can_ dilate (in the proper sense of the term) any contraction to a unitary, but the ``$A'$'' you get  will have to have some junk in the lower left corner.  Perhaps it is possible to salvage this idea by understanding what happens with the junk when you do a polar decomposition.  But maybe it is no simpler than trying another approach.  (I leave my whole answer unedited so that these comments make sense to people who read them.)
Answer: Ah, (quite) some fiddling with Mathematica gave a counterexample.
In the notation of anon's answer, take
$$
  A' = \begin{pmatrix} 1 & 1 & 1 & 1 \\\\ 1 & 1 & 1 & 0 \end{pmatrix}, \qquad
  B = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}, \qquad
  C = \begin{pmatrix} 0 & 1/\sqrt{2} \\\\ 1 & 0 \\\\ 0 &
      1/\sqrt{2} \end{pmatrix}.
$$
Then $A=P(A')$, $B$ and $C$ are partial isometries, but $P(P(CB)A) \neq P(CP(BA))$.
Notice that $C$ is a partial isometry that is not an isometry, i.e. has nontrivial kernel.
Comment: Wow, neat.  I wonder if this is a "minimal" example (or if it's not, what a minimal example looks like).  My own search was with smaller matrices than this (everything between spaces of dimension at most 3), and nothing seemed to work, so I gave up.

Associativity of polar decomposition


Is my guitar neck too thick for me?

Question: I've been playing for almost three years and now, as I'm starting to pay attention to the details in my playing and comparing it to other guitarists, I noticed my wrist position in the lower strings is really curved and it damages my wrist, I've only had only one electric guitar, an Ibanez S521, this started when I was practicing some arpeggios on the low E string between the 12th and 16th fret, when I played in the 5th string my wrist was straight in a natural way but in the 6th string when I reach from the 12th to the 16th my wrist bends completely. Could this be that my neck is not suited for my necessities? Forgot to mention, my hands are fairly big related to the average person.
Edit: I can play with the wrist fairly straight with the neck pointing at a ridiculous angle upwards
Comment: Probably more to do with the height you have the guitar on its strap.
Comment: I agree with Tim - it sounds more like a posture/playing position thing than an instrument thing. Try shortening your strap or adjusting your posture while you play.
Answer: J.P.,
There are some unknowns here.
As Tim implied, where you have the guitar strapped on (up high on the belly, or down low like a rock star) makes a huge difference in your playing comfort.
Or are you playing strictly sitting down?
An armless chair or stool is best.
The WORST scenario is when a player is attempting to phrase chords and arpeggios whilst slumped back on couch cushions, or laying back on a bed, supported by pillows. (Your arms/elbows are trapped, and your arms are not free to move and adjust comfortably, and your wrists become strained.)
If you have fairly large hands, my instinct is to tell you, "No, your guitar's neck is not too thick".
Answer: The issue with your wrist may be due to how you are sitting or standing when you play.  If you need to contort your wrist to reach the lower strings try using a strap, even when you sit, and placing the guitar body in front of your body.  Many electric guitarists sit with the guitar on their right leg which causes a contorted left hand in some cases.  Your statement "I can play with the wrist fairly straight with the neck pointing at a ridiculous angle upwards" is quite unfair, and a little prejudiced.  You have essentially solved your own problem.  As for the angle being "ridiculous" some guitarists play with the neck almost vertical, like a cello or upright bass.  As for the neck being too thick, people play the same arpeggio patterns on the classical and that type of neck is thicker than any electric guitar I've played.  Change your posture and keep the neck at whatever angle is required to relieve the stress on your wrist.
Answer: If your guitar is an Ibanez S-Series, and you have large hands, the answer to the headline question is almost certainly "no, the neck is not too thick". If anything, it might be too thin - the necks of these guitars are as slim as they come, and a thicker neck might offer a better angle for your thumb to rest on.
Could you borrow something with a heftier neck (a Les Paul or LP copy, for example) from a friend and try if you feel more comfortable with it?
Have you consulted with an experienced guitarist (your teacher, if you have one) and asked if they consider your hand position problematic?
Also, you write that this damages your wrist - are you actually feeling pain or stress in your wrist, or is this just something you're worried about because someone said you shouldn't bend your wrist?

Is my guitar neck too thick for me?


Number of not gates in multiplexer

Question: I'm reading a page on multiplexers that has the following image. I says for constructing one it used 7 not gate. But looking at the image I don't know why three of them are needed. Could't you just cut down on the number of them by making a connection such as what I've done in red bellow? Is there something else I'm missing here?
Comment: I answered a very similar question here: http://electronics.stackexchange.com/q/105096/17779
Answer: Yes, that could be removed and it would logically be correct, but there are a couple practical reasons why you wouldn't want to do that.
Those NOT gates are functioning as buffers for the signal coming in. This way, your single input signal doesn't need to drive a whole bunch of AND gate inputs. While this isn't a big deal for CMOS logic since input current is negligible (or at least very small) and doesn't affect the output voltage as much, TTL logic is more sensitive as it requires current to be drawn through the outputs. Three or four AND gates from one signal could be problematic as it will try to draw a lot of current. Even for CMOS inputs, which look close to a capacitor, the edge shapes of the incoming signals will be affected when they drive more gates (this is one reason why chips such as "clock buffers" exist and are very useful). Too many gates and suddenly your signal doesn't have enough time to finish driving every gate low before it starts to go high again (your waveform then starts to look like this and doesn't ever get all the way low or high) because it has to charge and discharge all those capacitors, making what effectively is an RC filter on your signal (R is the output impedance of your driver).
The other thing about using those as buffers is it makes the inputs all look the same electrically. Your ~E input drives 8 AND gates while the other inputs drive around 4. This means that you need to compensate for each individual input when interfacing to the circuit, rather than being able to use the same output circuit to drive the inputs. With those buffers, all of the pins, including the D pins, look the same as any other pin and make designing with this logic block easier to do.
I think it all comes down to "it makes it easier to design to use this particular circuit."
Comment: Huh. Interesting. Thanks a lot for the in depth answer! :)
Answer: In terms of the truth table, you are correct.  If you made the change you indicate, the logical function would be identical.
Most likely the reason for it is that the first NOT gate is acting as a signal buffer to provide fanout.  If you made the connection you have indicated, the then input line S1 (presumably coming from a PCB) would have to drive 9 gates in parallel.  Depending on the logic family and speed, there will be some maximum fanout ratio.  Even if the application supports a fanout of >9, it is common to run several multiplexers sharing the same control lines to mux a multi-bit signal.  By adding a second NOT gate to each input, you only have to drive a single gate and the fanout happens on chip.

Number of not gates in multiplexer


Updating opportunity with APEX code

Question: I want to add products to a specific opportunity via APEX code. Not sure how I would be able to do that. I know how to create an opportunity and insert it into the environment but wanted to make it more specific so I can insert more child objects
Answer: Have a look at the Salesforce Product and Scehdule Objects Data Model.
You will want to insert a collection of OpportunityLineItem objects that are linked to the parent Opportunity via the OpportunityId field.
The OpportunityLineItem records should also use the same Pricebook2Id as the parent Opportunity and have a link to PricebookEntry records that correspond to the Porduct2 you want to link to.
You might find it easier to setup one or two sample Opportunity Product records via the UI and then examine them via SOQL to see what needs to be done.
Comment: and, if using multi-currency, the currency of the PBE / OLI must match the currency of the Opportunity

Updating opportunity with APEX code


Normal bundle to complete intersection in $\mathbb{P}^n$

Question: Let $X\subset\mathbb{P}^n$ be a complete intersection defined by irreducible polynomials $f_1,...,f_k$ of degrees $d_1,...,d_k$. How to show that the normal bundle of $X$ is isomorphic to $\bigoplus\limits_{i=1}^k\mathcal{O}_X(d_i)$?
Answer: By definition, the normal bundle is $(\mathscr I/\mathscr I^2)^\vee$, where $\mathscr I$ is the ideal sheaf of $X$.
Since $X$ is a complete intersection, the ideal sheaf $\mathscr I$ is resolved by the Koszul complex. I.e., we have an exact sequence of the form
$$
0 \to \mathscr K \to \bigoplus_{i,j} \mathscr O(-d_i-d_j) \xrightarrow{\varphi} \bigoplus \mathscr O(-d_i) \to \mathscr I \to 0.
$$
Where $\varphi$ is multiplication by the vector $(f_1,\ldots, f_n)$. We have that $\mathscr I / \mathscr I^2 \simeq \mathscr I \otimes_{\mathscr O_{\mathbb P^n}} \mathscr O_X$. Thus, tensoring the above sequence with $\mathscr O_X$ we get
$$
\mathscr K \otimes \mathscr O_X \to \bigoplus_{i,j} \mathscr O_X(-d_i-d_j) \xrightarrow{\varphi} \bigoplus \mathscr O_X(-d_i) \to \mathscr I/\mathscr I^2 \to 0.
$$
But the middle map $\varphi$ is zero in $\mathscr O_X$. Hence
$$
\bigoplus \mathscr O_X(-d_i) \simeq \mathscr I/\mathscr I^2.
$$
Dualizing, we get the result.
Comment: @Steve Hi! This is based on the local fact that if $B=A/I$, then $I/I^2 \simeq I \otimes_B B$ as $B$-modules. To prove it, looks at $I \otimes B \to I/I^2$ given by $i \otimes b \mapsto ib$. First, check that it is well-defined.
Comment: This argument with Koszul complex seems nice. Thank you.
Comment: Dear @Fredrik Meyer, sorry for my late reply, could you kindly tell me where can I find the fact that you have mentioned $\mathscr I / \mathscr I^2 \simeq \mathscr I \otimes_{\mathscr O_{\mathbb P^n}} \mathscr O_X$? I really appreciate it.
Comment: Dear @FredrikMeyer, thanks for your help.
Answer: Since every $f_i$ is a section in $H^{0}(\mathbb{P}^n,\mathcal{O}(d_i))$, denote $F=\oplus{f_i}$ the section in $\oplus\mathcal{O}(d_i)$, obviously $Z(F)=X$. Now restric $dF$ to $\mathcal{T}_{\mathbb{P}^n}|_X$, we get an exact  sequence:
$$
0\to\mathcal{T}_X\to\mathcal{T}_{\mathbb{P}^n}|_X\to\oplus\mathcal{O}_X(d_i)\to0
$$
Compare with
$$
0\to\mathcal{T}_X\to\mathcal{T}_{\mathbb{P}^n}|_X\to N\to0
$$


How to grab the data from System.out as string?

Question: For a sort of IDE-like system I'm writing in java, I want to replicate and display a console working identically to the standard one. Therefore I need a way to read everything that is written to stdout and stderr.
Is there something similar to the Process.getInputStream() function that is applicable to the current Java process and not only to external ones?
Would like to avoid having to change all stdout commands.
Huge thanks in advance!
Answer: Use System.setOut() and System.setErr() to override the default output and error streams with your custom streams.
Answer: Something like this
  FileOutputStream f = new FileOutputStream("file.txt");

  System.setOut(new PrintStream(f));

  // this text will get redirected to youe file
  System.out.println("This is System class!!!");

Comment: Ahh so simple >< Been overcomplicating things.. Thank you!


In QGIS combine multiple polygon parts of a single feature

Question:
This is a repost of my unsolved QGIS dilema. The image shows a single selected feature that consists of two polygons and the corresponding attribute table. Some nodes of the left poly have been added and moved to illustrate the boundary/ overlap problem. Before the nodes were moved and added the two polys were snapped together and I could not get rid of the line between them. Hovering over the nodes, pop ups identify the polys as either 0 or 1. How does one create a single polygon from the two?
Comment: Have you tried to merge with the fwtools or with the toolbox ?
Answer: As I understand your issue, you currently have a multipolygon feature which has an invalid geometry since one part of it overlaps another part. There is a 2-step procedure that should help you to merge your multipolygon into one simple polygon:
First, run the v.clean method which is located in the processing toolbox, GRASS section. Just select your layer and let the default options on. This should split your multipolygon into several features.
Then, select these new features and press the Merge features button in the Advanced Editing Toolbar (which you may need to activate first). This will merge the features into one beautiful polygon.
Comment: Beautiful! It took a long time to find and start v.clean. Previously I separated the polygon parts of the single feature by "split selected multipart features to single part" but the "merge selected features" always retained the boundary between them. I ran v.clean on the selection and a new layer named "cleaned" was created with the two polygons. The "merge features" eliminated the boundary line as desired. I deleted the areas on the original poly, copied the new result from the "cleaned" layer and the problem is solved. Thank you ArMoraer.
Answer: I've just whiled away a few hours wrangling with just such a problem and couldn't solve it using any of the v.clean options using 2.18.2 for Mac.
I got lucky however, with trying the Multipart to singleparts geometry tool by choosing the attribute known to be unique resulting in a new temporary vector layer with three adjoining but separate features.
Select the adjoining features in edit mode and use the Edit>Merge Selected Features option to join the separate features into a single polygon.
Don't forget to save the temporary layer to disk using the Layers>Save As option otherwise the new layer will be lost even when saving the project.


Laplace's equation doesnt reproduce Newtonian gravitational potential

Question: Newton's law of gravitation describes the gravitational potential produced by a mass $m$ as : $G(r)=-k\frac{m}{r}$.However if you solve Laplace's equation for the gravitational potential in polar coordinates at the limit $r\rightarrow\infty$.The solution produces more powers than that of $r^{-1}$,you have odd power of the distance $r^{-(2n+1)},n = 0,1,2,...$ and despite their decreased amplitude they exist.So what is going on here?
Comment: Try computing the potential of a cubical mass. It won't be simply $1/r$.
Comment: Hint: what's happening at r = 0 when the potential is not $r^{-1}$. Is the laplace equation the good equation to consider if there is a particle somewhere (creating your field)?
Comment: @Syrocco the correct equation is the Poisson equation but the solution of the Poisson equation is the linear sum of the solution of the Laplace equation and something else so these terms appear there as well.
Comment: if you have a dirac delta mass at 0 (you need it to produce your field), then $1/r$ is indeed a solution. Because, in the distribution sense, $\Delta (1/r)=\delta(r)$. Do you think this equation is still valid if we put a different power of $r$ in the LHS, especially at 0?
Comment: As said in the answer by Ivan, with multiple sources of gravitational field (or with non natural coordinates), you will see these power appear as a serie!
Comment: You have to use the spherical symmetry of the source point-like mass.
Answer: You need to solve the Poisson equation $\Delta V =-4 \pi G \rho$ with $ \rho = q \delta(\vec{x})$.
In $\vec{x}=0$ it differs from the Laplace equation.
You can verify that the other terms in $1/r^l$ satisfy neither the Laplace equation at the origin nor the Poisson equation $\Delta V =-4 \pi G q \delta(\vec{x })$.
One can show that the solution which tends to $0$ at infinity is unique (even with a charge distribution which is not a function but a distribution). Once the uniqueness is admitted, the solution for the distribution $ \rho = q \delta(\vec{x})$ is necessarily spherically symmetric because otherwise we could generate other solutions by simple rotations.
Hope it can help and sorry for my poor english.
Answer: Each power term of the solution of Laplace's equation represents a different multipolar term (monopole, dipole, quadrupole …).
For a point mass, you are only interested in the monopole term, i.e., $r^{-1}$.
Such multipolar expansion is quite useful in the context of electromagnetism, but it is also useful in the context of gravitational. For example, gravitational wave radiations are usually associated with oscillations of the quadrupole term.


Insect/cocoon ID?

Question: Can anyone ID this? We're in the Rocky Mountains, Western US. We found it stuck to the wall of our garage, a few feet up, hidden behind a snow sled. It seems to be covered, or surrounded, in what looks like mouse hair. Is it a moth? It's hard to see scale, but altogether about the size of a thumb.
Answer: I would guess this is the cocoon of a moth in the subfamily Arctiina in the Erebidae family.
For example, see this example of an Isabella Tiger Moth cocoon in Kent Co., Maryland from The Maryland Biodiversity Project:

 Credit: Nancy Martin
You can find additional images of various species here, here, here, and here (seen without the cocoon).
According to here, Pyrrharctia isabella (or the Isabella Tiger Moth shown above) is sometimes found in the Rocky Mountain region of the US, but I am not knowledgeable enough about these pupae to definitively point toward a species.
Based on the experiences of many others, I would advise not touching this with bare hands or clothing.

Insect/cocoon ID?


2 pole or single pole breaker to feed sub panel

Question: I want to run wires about 70 feet to a panel that will feed a 30amp rv hook up and a 20amp receptacle. Should my feeder breaker be a 2-pole or single pole?
Comment: Does your RV require 120V or 240V?
Comment: The RV is all 120v. Just a single pole 30amp needed for the RV.
Comment: How much panelboard space do you have left in the panel you're running from, and what make and model is said panel?
Answer: It's up to you.  You can use a single pole breaker or double.  Single has potential to be cheaper.
If you don't plan on maxing out the outlet and the RV at once, you could safely go as cheap as a single pole 30A breaker 10-2 wire for the run and a 30 A panel if you can find one that small.
125A panels are pretty mass produced though, so you might find it cheaper ro get a 125A any way.  Just make sure its lugs can safely accept 10 AWG wire and know that you won't be able to use half of the positions in it.
Personally, I'd run 60A or 100A, 240 from a double pole breaker.  Especially if it's going to involve digging, since you can do more later with a full 240v panel and the extra amperage.  Think about EV charging in the next decade.
Comment: Thanks, you all have helped me to look at my project more closely to better formulate a plan. I'm now thinking I will go ahead and run 70amp using a 2p 70amp breaker. This will give me the ability to upgrade my RV hookup to 70 if ever I go with a larger rig.
Comment: Check my work at http://www.paigewire.com/pumpWireCalc.aspx  but I think you want awg4 for copper.  The most amps you can usually put through 8AWG is 40 or 50.  You want 70 and you want some distance.  -3 is correct
Comment: will 8-3 uf work or do I need to go to a 6-3 uf?
Comment: You're welcome!  Also, UF cable is one approach.  But while you're working on your plan, consider using THWN wires in a conduit if you're going to be digging so you don't have to dig twice.  You might even lay two conduits side by side and opt to use only one for now.  Conduit will protect your conductors better and allow for easier replacement or expansion at a later date (ethernet?, fiber?) THWN should be cheaper than UF, and if you ever damage it you might only have to replace the damaged conductor, not all of them since it's not bundled like UF. Leave caution tape buried above it too.
Comment: Do you have a conduit fill calculator? Or do you know what size I should use for 3-8awg and a #10 ground?
Comment: Using http://conduitfillcalculator.com/, you could do it with a 1/2 inch schedule 80 PVC conduit.  Using a bare ground would reduce fill capacity.  or you could use metal conduit for your ground.  I'd run two 1/2 inch conduits, or one 1 1/2 inch.  Two separates could be nice to isolate power from signal.  Larger is nice if you end up wanting to up the current.


How do I override the dates with Temporal tables?

Question: I'm trying to use temporal tables for our data lake environment to track changing data over time.  The first hurdle I hit is loading historical data to seed the tables.  Everything is loading as "now" instead of as of the point in the past.  Searching online the only solution I have found is to turn off system versioning, disable the period in the table, make changes to the table to fix the dates, then turn system versioning back on and hope that SQL Server will let me.  Is there a better way to force the datetime values to be what I need them to be?
Also, going forward, we are loading data that was updated yesterday and captured last night and being loaded into the lake early this morning.  I need to use yesterday's date as the effective date for the temporal table effective date, not the current timestamp.  Same issue, how can I force the datetime values?
Comment: If you need this type of tight control over what is essentially a type 2 slowly changing dimension, then you usually roll your own.  I don't know if temporal tables is the right fit for you.
Comment: How do you load your data ? can you provide an example of the script (T-SQL) used to get the data ?
Comment: Dominique, we are using Informatica to load the data.  For the specific situation I'm facing, the data is coming from an external source, so the timeline is: Daily market events happen, at close of market day file is generated automatically, our ETL process starts early am following day to load data.  The desire is to use the system versioned table because of the historical query functionality.  Rolling my own still presents all kinds of issues, not least of which is managing the data migration to a history table, or keeping everything in one, I can't reproduce the "as of" functionality.
Comment: Jonathan, I'm afraid you might be right.  This is a prime use-case for temporal tables, but only if I can override the default system timestamp when tracking changes.  If this were an online system, the system timestamp certainly works, but I'm in a warehouse situation, the gap between business event and data lake table insert are separated by a lot of time.
Answer: The time in a temporal table is "system time" not "business time".  That's the distinction that the feature name "System-Versioned Temporal Table" is getting at.
You inserted the data today, so it's versioned as of today.  You can use the temporal table to track the changes you make going forward, and retain change history, but it doesn't replace a slowly-changing dimension for business analytics.  As you noted, Temporal Tables belong in the "Data Lake" or "Raw" or "Staging" layer.
If you want to add additional history representing changes to the data before your initial load, you can turn system versioning off temporarily while you load data directly into the history table.


How to integrate with Paypal suggested solution: "Add payment buttons to your website"

Question: I hope to integrate Paypal using "Add payment buttons to your website" https://developer.paypal.com/webapps/developer/docs/integration/web/ since my site only needs one time purchase so this solution is perfect.
But it requires an IPN implemented as an async solution at the merchant side to receive the transaction result.   While the button redirects user to paypal.com to finish the payment process. But after that is done.  User is not automatically redirected back to merchant site.
I found on the forum that I can add an "data-return" attribute in the script element so when user is done paying there is an additional link that points to the merchant site.  Now since IPN is asynchronous, once user views the "data-return" page, IPN may not have been called yet.  My questions are:
<code>1. So what's the suggested solution of synchronizing between return page and IPN?
2. When testing, I found a lot of query strings are appended to the "data-return" URL, looks related to transaction result.  Can I use them to replace IPN so I have a SYNCHRONOUS way of verifying the transaction result?
</code>
Thanks!
Answer: Use Payment Data Transfer (PDT, for short). https://developer.paypal.com/webapps/developer/docs/classic/paypal-payments-standard/integration-guide/paymentdatatransfer/


Coefficient of determination of linear regression in Azure ML studio differs significantly from regression R square in Orange 3 for same dataset why?

Question: I tried exactly same dataset in Orange 3 and Azure ML Studio for linear regression. Coefficient of determination in Azure ML studio is .71 while Orange gives R square as .45. Why such a huge difference?
Answer: It could help if you have added screenshots of both models settings. However, what I can say is Orange and Azure ML Studio use different parameter settings. It might be possible that some are available in one of them but not in the other.
If we assume that you have exactly the same configuration in your two models, then the random parameter can also play a big role here. Even if you use the same seed, Azure ML and Orange, probably, don't use the same random generator algorithm. As a result, the outcome of your models is not replicable.
I would advise trying with cross-validation in both implementations and dig more on the evaluation metrics. If you need more help from the DS community, you might need to add more details as well.


Workflow to email one user if disapproved and another if approved, it has 6 approvers. Can this be done?

Question: I have a form library and when a form gets uploaded it sends email with packet information.  In the form it has a approve/disapproved drop-down.  What I would like to do is send email to a person based on approved/disapproved , i.e. approved = email@example.com ,  disapproved = firstname.lastname@example.com.   One thing I would like to happen is that it does not send the email to email@example.com until all approvers select "approved"
Answer: Add some custom columns to the form that are not displayed in any views in the library. Whenever an approver makes an action, update a value in one of those columns. Afterwards (each time a column is updated) check to see if all the other necessary columns have been updated.
If all the other necessary columns have not been updated then perform no action. If all the other columns are now complete then send the email. After sending the email update another column that indicates an e-mail has been sent. You will also need to check this "email sent" column each time the workflow meets the conditions to send the email. In this way only one e-mail will be sent.
I can't remember if this can all be done with variables... but it will work with columns.
Comment: Ok, You need to create 6 workflows, one for each approver. After the first task is assigned and completed you will be able to use SharePoint designer to update the value in a specified column. Workflows always pause until the task is complete. After the task is complete and the value is updated then check all of the 6 columns. If all 6 columns do not have the required values then terminate the workflow at that point (each workflow must do this). If all 6 are complete then check the "email_sent" column to see if the email was already sent. If no, send it and update the email sent column.
Comment: And SharePoint designer has the ability to pull values from a list; [check this](http://sharepoint.stackexchange.com/questions/16643/workflow-with-sharepoint-designer-2010-how-to-set-value-from-current-list-item) post for pulling values from the current item
Comment: This is the workflow updating the value for email send. This worked, except for the email send, it sends multiple emails depending how I create the workflow. I have it as such "if current item "column name" equals "value" email "User" else if current item "column name" equals "value" Update item in current item (assigned value 1) STEP 2 Stop the workflow and log "Email Sent" if current item"Email Sent column" equals "value"
Comment: You need to check or update the same column "email_sent" in every workflow when it sends an email.  If the check says "true" then terminate the workflow. If not, check all other 5 workflows to see if it needs to be sent.
Comment: Do I have to create a Timer Job for these columns.  What condition would I use to check if the column has been updated?  I appreciate your help.
Comment: Thanks Chris I am going to give a go now.  I will let you know how it goes.
Answer: In SharePoint Designer, on item change, kick off a workflow, if approval status = approved email one person, else if approval status = rejected, email someone else
Comment: Thanks I will give this a shot. Appreciate the fast response


Is it possible to host an ASPX web page, in a stand alone application, outside of IIS

Question: I have a thick GUI application (possibly running as a service, but also just as a normal application) on a desktop. I would like to expose a web interface to the application for some remote monitoring and control of the application.
I currently am hosting a WCF service that just returns HTML in the application, which works fine, but optimally I would like to use an ASP.Net application, or a silverlight application.
Is there any way to host the aspx or silverlight app from within my app?
As part of this, I would like to be able to share data between the two applications.
Answer: Yes. You don't even need Cassini, as it wraps ASPX hosting bits already present in the framework it's all inside System.Web.Hosting
MSDN has a good article on it all
Answer: The tiny webserver that is built into VS is called Cassini and you can download the source somewhere. Not sue about versions and licenses.
Cassini provides (demonstrates) the ASP Hosting stuff.
Answer: ScottHa has an article on hosting Cassini for unit tests. I've used it before, with a couple of slight modifications, for unit tests and it works great.
Not sure what differences you get by just using HTTP.SYS directly, but either should work pretty well, I think.
Answer: as i see it, the most simple solution for you is to put a browser control on your app, in the browser put a silverlight app,
if you wanna share data between the app and the silverlight you can try to expose a wcf service from the app to the silverlight if it works its also very simple.


Difference between "each" and "every"

Question: What is the difference between the following two sentences?

Each apple is red.
Every apple is red.
Answer: Strictly speaking, the two sentences mean the same thing. However, the sentence "Each apple is red" is slightly unusual, and the more natural way to express this would be "Every apple is red", or "All apples are red."
The reason is that the word each is generally used in situations where we consider the apples individually or sequentially, whereas every and all are used for generalizations. So we might say:

We spray-painted each apple red.

Here each is appropriate because every apple was painted individually. However, most people wouldn't say the following:

[?] Each apple turned red by October.

This isn't technically wrong, but it sounds unnatural. Much more usual would be to say one of the following:

Every apple turned red by October.
All the apples turned red by October.
Comment: I think that the phrase "each apple" is always used to refer to individual apples; wheras "every apple" refers to the apples as a collective whole.
Comment: And when do you say "each and every"?
Answer: Perhaps this question has already been answered, but here's my two cents:
Each is used for the individual person or item, and the focus is on the individual, not the group.
Every is also used for the individual, although the focus is shifted to the group. Consider the following quote from Dictionary.com:

Each, every are alike in having a distributive meaning. Of two or more members composing an aggregate, each directs attention to the separate members in turn: Each child  (of those considered and enumerated) received a large apple.  Every emphasizes inclusiveness or universality: Every child  (of all in existence) likes to play.

However, a couple of years ago I came across a somewhat novel usage, or perhaps archaic usage, of each: the poetic each. Simply put, this is a full synonym or every, as in the following excerpt:

The Holy Spirit is working in your heart;
  This is the day to give the Lord each part.
— "The Lord is Calling" by Greg and Heather Murray

The sense of each in the quote above is obvious. It does not mean to give the Lord one piece of your heart at a time; it means to surrender your entire heart. Now, technically, this usage is not usually considered; it may even be considered incorrect by textbook English standards. I simply want to point out its existence.
Answer: Each is used to refer to individual people. Every is used to refer to a group.
Answer: Examples
The following two examples demonstrate the difference between the words "each" and "every":
Each letter is wrapped in parentheses
<code>(A)(B)(C)(D)(E)(F)(G)
</code>
Every letter is wrapped in parentheses
<code>(ABCDEFGHIJK)
</code>
we can expound further
If the earlier examples made sense to you, then can stop reading if you like.
In general, the word "each" can replaced by the phrase "each individual"
The following two sentences are semmantically equivilant but differ in syntax:

each letter is wrapped in parentheses.
each individual letter is wrapped in parentheses.

The word "every" cannot be replaced by the phrase "every individual"
The word "every" can be used to mean either of the following:

each individual
the group taken as a whole instead of as individuals.

The word "Each" always refers to individuals.
Formal logic and mathematics tend to be less ambigous than English. In mathematical logic, we would have the following examples of sentences:

For any letter L there exists parentheses P such that L is wrapped in P

There exist parentheses P such that for any letter L we have it be the case that L is wrapped in P.

We have a set of numbers. The set itself is  big. Maybe the numbers  are big too, but we do not know about the numbers, we only know about the set.  Maybe the numbers are small, such as 0.1 but the set contains a lot of numbers.

We have a set of numbers. For each number in the set, the number is big. We do not know whether or not the set is big. Maybe the set is small.

1 implies 2.
However, 2 does not imply 1.
The word "every" can mean  3 or 4 (either the box is big or each item inside of the box is big)
If you have a box of items, the word "each" refers to individual items inside of the box but not the box itself.
The phrase "everything is wrapped in parentheses" means 1 or 2, but usually 2. That is the parentheses could be shared or separate but usually, shared.
Unlike the word "every" the word "each" is never used to say that the set is big. Instead the word "each" refers to the individual things inside of the set. You couldwrite, "each marble in the set of marbles is blue."
The phrase "every marble" is similair in meaning to the phrase "the set of marbles". Thus we could have "the set of marbles is blue" instead of "each indidual marble is blue".
As another example...
There is a suburb in which everyone owns a lawn mower. However, it is unclear whether they share the same lawn mower or have seperate ones.
For different people living in the same suburb, the lawn mowers they own could be different. There could be different lawnmowers For different folks.
Alternatvly, there could be one lawn mower such that everyone in the same suburb takes turns using that lawn mower. There could be 1 lawn mower shared across 20 different people.
If you write that,"everyone has access to a lawn mower", then the people could share the lawn mower or each person could have their own individual lawn mower. The word every is ambiguous.
If you write, "each individual person has their own lawn-mower,"then there are 20 different lawn mowers for 20 different people.
Answer: Consider the sentences:

"Every sandgrain is big"
"Each sandgrain is big"
"The set of all sand grains is big"
"Any sandgrain in the set of sandgrains is big"

1 can be equivilaant to 3 or 4.
2 is equivilant to 4 and 2 is never equivilant to 3.
If you say that "every egg is big" then that could be interpreted as:

a metaphorical basket of eggs is big, but each egg is unusually small.

the metaphorical basket of eggs is small, but each egg in the basket is large.

That is, we could have a large collection of small eggs or a small collection of individually large eggs.
"Every mouse" describes a large set.
The phrase "Every person" refers to a group of people or a set or people.
The phrase "each person" refers to individual people.

Difference between "each" and "every"


Why would dapper return correct count and null values?

Question: I wanted to play around a little bit with dapper so created two classes that represent the database and a little inner join query but for whatever reason, the results are of null value. I'm new to dapper, so from what I've read in the docs and a great answer on SO (How do I write one to many query in Dapper.Net?), it appears I have this setup correctly. Why would the result count be correct, but the values are all null?
 con.Open();
    string query = @"select Orders.OrderID as OrderID,
                     Orders.OrderNumber as Orders_OrderNumber,
                     Product.ProductNumber as Product_ProductNumber,
                     Product.ProductID as Product_ProductID,
                     Product.OrderID as Product_OrderID
                     from Orders INNER JOIN Product ON Orders.OrderID = Product.OrderID WHERE Orders.OrderNumber LIKE '%" + orderNumber + "%'";
    dynamic data = con.Query<dynamic>(query);
    Slapper.AutoMapper.Configuration.AddIdentifiers(typeof(Orders), new List<string> { "OrderID" });
    Slapper.AutoMapper.Configuration.AddIdentifiers(typeof(Product), new List<string> { "ProductID" });

    var orders = (Slapper.AutoMapper.MapDynamic<Orders>(data) as IEnumerable<Orders>).ToList();
    //var orders = Slapper.AutoMapper.Map<Orders>(data);
    return orders; // count is correct but values are null

Classes:
 public class Product
    {
        public int ProductID { get; set; }
        public int OrderID { get; set; }
        public string ProductNumber { get; set; }
    }

 public class Orders
    {
        public int OrderID { get; set; }
        public string OrderNumber { get; set; }
        public List<Product> Products { get; set; }
    }

Any help would be greatly appreciated.
Thanks!
Answer: I'd imagine your mapper isn't working as expected. Step into your mapper and make sure it's working as expected.
Answer: Seems to be
Orders.OrderNumber as Orders_OrderNumber the failure point. Try Orders.OrderNumber as OrderNumber.
Answer: Best practices with using Auto Mappers are Assert Statements,especially in case of Dynamic object mapping. See this example. Might help you find problem.
Additionally, taking a peek at Mapper Source code below :
public TDestination Map<TDestination>(object source, Action<IMappingOperationOptions> opts)
    {
        var mappedObject = default(TDestination);
        if (source != null)
        {
            var sourceType = source.GetType();
            var destinationType = typeof(TDestination);

            mappedObject = (TDestination)Map(source, sourceType, destinationType, opts);
        }
        return mappedObject;
    }

It seems it will return default value of T when an Empty Collection is returned.


automation of plugin installations

Question: My business is building and establishing websites for all kinds of businesses and companies.
So, every time I start a new site for a client of mine, I have to install the same plugins over and over again.
Is there a way to automate this? I mean, set up a "manifest" of needed plugins and run it on a site, then it will automatically install (and activate) all the plugins in the list.
There used to be a plugin called "Install Profiles", but it is very (!) old and not updated (latest WP version supported is 3.x)...
Answer: Check out  TGM-Plugin-Activation plugin.

TGM Plugin Activation is a PHP library that allows you to easily
  require or recommend plugins for your WordPress themes (and plugins).
  It allows your users to install and even automatically activate
  plugins in singular or bulk fashion using native WordPress classes,
  functions and interfaces. You can reference pre-packaged plugins,
  plugins from the WordPress Plugin Repository or even plugins hosted
  elsewhere on the internet.

automation of plugin installations


Is there a tool/shortcut to automatically align edges depending on where the camera is?

Question: I have this model. I want the selected edge to match the red line. Is it possible to do this without turning the model and using sx0?
Comment: In the Transformation Orientations panel you can choose View, if it doesn"t give the result you want you could use the knife and cut orthogonally (K to activate and A to cut orthogonally)
Answer: I am using 2.79b and there's CTRL+S > to hit for "Cursor to active" if an edge or a vertex point is selected in Edit mode.
There's also CTRL+S > "selection to cursor" for once the edge or a vertex point to be moved to cursor once they're selected/
I assume the red line is something parented to the camera if I'm not wrong.
And I believe You could boolean a line in the helmet with it, and then move the rest of the needed verts to the cut line, followed by selection of the cut line to dissolve it once the vertex points and edges has been moved.
Options for CTRL+S works with edges, as well as vertex points, so I bet that by moving verts, You could move the whole edge pretty precisely.


User click on log in when user already logged in so should throw message.

Question: I am write some code for user login. But i want when user click on log in button every ,that time checks user already logged in or not . if user logged then throw the informative message and if not throws error. How i do this.. Please help me..
Comment: If you reder link through the Drupal way using `l()` function, you dont need to do all these unnecessary things.
Comment: Please add your code to the question for context. It's difficult to understand what you're asking
Comment: Just i want to validate if user is logged in that time if user again click on log in button that time just throw information message to user like "User is already logged".. For that how i handle such situation .. Thats it......
Answer: You can check value of field uid of global variable $user:

global $user;
Contains preferences and other account information for logged-in users. When a user is not logged-in, the $user->uid property is set to 0.

http://api.drupal.org/api/drupal/developer!globals.php/global/user