How would I go about doing a PhD after an M.Eng in Biomedical Engineering?

Question: I have been accepted into Cornell for M.Eng in Biomedical Engineering. I had originally applied for an MS but since they've stopped enrolling students for MS I got into M.Eng which is an industry oriented course. I would like to get a PhD eventually in the field. Will going for an industry oriented course (M.Eng) over a research course end my hopes of getting a good PhD admit?
Comment: You don't need a master's degree to be admitted to a US PhD program.
Comment: When they told you that the MS was no longer an option, did you ask about getting admitted directly into the PhD program?
Comment: I did.... they didn't reply to that
Answer: Generally speaking, the main difference between an M.Eng and an M.S. is in research. An M.Eng is usually a course-based degree and usually does not involve any research. There is often some sort of "project" course which students will be required to take for credit, but that is still far from research.
An M.S would usually involve a research project and thesis, and it would give you exposure to the type of work a PhD student would be doing. That being said, many schools also offer course-based M.S. programs, so the name of the degree does not really mean much.
I would not recommend enrolling in an M.Eng program if you wish to study for PhD. In the U.S. in many schools having a master's degree is not a requirement for getting into a PhD program. I would advise you to inquire whether it's possible to be admitted to the PhD program at Cornell instead of the M.Eng.
For schools, the advantage of admitting students without master's degrees to PhD programs is that they will be around longer and probably have more time to do good work. In the case that the student doesn't quite meet up to their expectations, the can arrange for the student to leave with a M.S. halfway through the program. This to some extent compensates for the risk involved in admitting an undergrad (since they will likely not have much research experience).
M.S. students on the other hand have research experience. They have also written a thesis and possibly some papers. This gives the admission committee better insight into the applicants capabilities. Furthermore, a student with an M.S. degree will likely need less time to get started on their research and will not need to take as many courses, so they will probably hit the ground running once they enter the graduate program.
A student with an M.Eng degree will have a shorter PhD than the students coming in with B.S. degrees but will not have the experience the students with M.S. degrees have. So they will be at a disadvantage compared to both groups. Because of this admission to a "top 10" school would be a bit more difficult with an M.Eng degree.
Another thing to consider is that if your goal is a top 10 school, many of them give priority for PhD admission to students already pursuing M.S. degrees at their school. These are typically the schools which require master's degrees for PhD admission (such as Stanford and MIT).
Comment: Assuming I have the same research experience as an undergrad after completing the 1 year long program will I be preferred over undergraduate students applying directly for PhD?
Comment: I think this would be more of a matter of personal opinion and preference for the admission committee members. I applied directly for PhD programs as an undergrad with little research experience and I got good results in general. I ended up getting a M. Eng from a decently ranked but not top notch school and applied again and got results worse than my first time. Some of the prospective advisers I spoke to at the schools that rejected me said that my M. Eng was a disadvantage in their eyes.
Comment: That being said I still was able to get admission from a good school (quite a bit better than the place I got my M.Eng from). Not as good as Cornell though.
Answer: In the United States, the distinction between MS and M.Eng is not terribly large.  An MS is typically a somewhat more "heavy" course than an M.Eng, but the distinction is unlikely to matter to any significant degree for US Ph.D. admissions.


how does your standard security system work?

Question: Do they have lasers motion sensors etc.. more looking for the type of security system you would see in a basic house not rich / well endowed one.
Comment: This questions needs a lot more detail. What exactly are you talking about?
Comment: Need so much more information needed please. Type of asset under protection? If it is a house, what type? Rural / city area? Owner / renter of the asset?
Comment: Security system for house?
Comment: Well.. probably?
Answer: On the typical consumer-end, there are no lasers or microwave sensors. Look at what a starter security package includes, I'm sure lots of companies will be happy to explain to you what they offer if you call and ask. There's an increasing number of DIY solutions available, which run the gammut of complexity, but i think you're asking more about a typical alarm service, like Brinks or ADT.
Typically you can expect:

magnetic door open sensors on major egress points
PIR detectors in common areas, entryways, and high-value locations
a main control panel for setup/monitoring, by garage door or primary entrance
1-2 limited control panels for activation, panic, and monitoring
an outdoor siren to alert alarm activation to the public
a cellular modem to contact an alarm monitoring service
monitoring video cameras that can push to owner's phones ($$)
magnetic window open sensors and glass break sensors ($$$)


Circles Rolling Around a Loop of n Circles

Question: Let there be $n$ circles that have radius $r$. They all touch each other at a tangent, so they could be arranged in a circle. If a circle with radius $R$ were to roll around the outside of this arrangement, by how much would it be rotated? If you have trouble visualising this, think of a one pound coin rolling around a circular arrangement of two-pound coins.
Comment: Are you asking the same thing as [**this**](http://math.stackexchange.com/questions/2124414/circle-radius-1-revolving-around-two-touching-circles-radius-2)?
Comment: @ Ng Chung Tak Yes, it looks the same question, with a higher degree of difficulty, with no more personal involvement. I propose to close this kind of question without work.
Answer: When a circle of radius $R$ rolls without slipping around a circle of radius $r$, it rotates of an angle $2\pi(r/R+1)$.
In the case of $n$ circles of radius $r$ arranged around a central circle (see diagram below) the rolling (red) circle does not make a complete revolution around each blue circle, but only runs an arc corresponding to a central angle of $2(\alpha+\phi)$, where $\alpha=\pi/n$ and $\sin\phi=\cos\theta=r/(r+R)$.
In that path it then rotates of $2(\alpha+\phi)(r/R+1)$. Multiply that by the number of circles to get the rotation angle after a full revolution around all the circles: $\theta=2n(\alpha+\phi)(r/R+1)$, that is:
$$
\theta=2\left(\pi+n\arcsin{r\over r+R}\right)\left({r\over R}+1\right).
$$
Answer: A circle rolling once around a circle of equal radius makes two rotations. I didn't expect this, but try it on a couple of coins with good milling. And it makes sense, since where a circle makes one rotation rolling along a straight line equal to its circumference, it will make two rotations rolling along the circumference of an equal circle.  Working still with equal circles, one circle rolling around two tangent circles completes $2*\frac23*2 = \frac83$ rotations. This is because tangency makes one-third of each circle inaccessible to the rolling circle.  A circle rolling around three tangent circles makes $3*\frac12*2= 3= \frac93$ rotations, since half of each circle is inaccessible. Four tangent circles yield $4*\frac{5}{12}*2 = \frac{10}{3}$ rotations. Five tangent circles make $5*\frac{33}{90}*2 = \frac{11}{3}$ rotations. For $n>1$, the numerator of the fraction increases by $1$ for each increase in the number of tangent circles. The number of rotations is $\frac{n+6}{3}$. The fraction of each circle accessible to, i.e. traversable by, the rolling circle = $\frac{240-\theta_n}{360}$, where $\theta_n$ is the angle, in degrees, of a regular n-gon. Checking with a 16-sided regular polygon whose vertices are centers of sixteen tangent circles, and $\theta_{16} = 157.5$ degrees, we get $\frac{22}{3}$ rotations, as the rule predicts. This is perhaps more of an investigation than a proof, and supposes the rolling circle equal to the tangent circles. Where the rolling circle's radius is unequal to those of the tangent circles, $\frac{n+6}{3}$ must be multiplied by the inverse of the ratio $\frac{R}{r}$. The ratio is inverted since a smaller circle makes more rotations on larger circles and fewer on smaller circles.


Can any one experience the World without a language? if yes to what extent, if no, why not?

Question: By listening, speaking, seeing and reflecting on words made by the alphabets of a language - one experiences the World. How far is this correct? Can any one experience the World without a language? if yes to what extent, if no, why not?
Comment: Heidegger's lecture series, 'The Fundamental Concepts of Metaphysics', is a thorough examination of the man vs. animal relation to the world, which to me seems very relevant to the question. For a shorter (and more crude) overview, Agamben's 'The Open' is also rather interesting.
Comment: Babies experience the world without a language. Also see this [thread](http://skeptics.stackexchange.com/questions/2790/do-people-think-in-a-language) on Skeptics.
Comment: Thats correct !! that experience is not sharable unless they learn a language alphabet to express ! It will be interesting to explore more on the bias that Language introduces to the babies to share their experience on one hand - as per the codes/rules/structures learnt for a language , and on the other hand even that experience sharing would not have been possible without language tools. So the True experience of Babies' enters into a bias with the language and becomes unsharable without a language
Comment: @Dzmitry - Thanks a lot, I will greatly appreciate more inputs on what current research directs one to, in terms of USING the tools of a language (any of the known languages) evolved / developed experiencing what was before the origin of a language or what world is experienceable without the language, what is that connects the before-in-between(link/connection/that transforms)-with the tools of a language  and then sharing that experience using these known tools of a language.
Answer: Good question, sorry for bad English!
I can't answer by all means but here are two points.

It is well known that children, that grow up without a language,
get a fatal lack of human consciousness.
Buddhists assert that adult human beigns may (and should) spend a long time without
words and estimations and this even improve their minds.
I practise Vipassana more than 15 years (being high-educated in theoretical physics)
and just confirm it.
Comment: point 1 - it seems to some extent the children/ or we as human get their/our consciousness eclipsed with the language tools and same tools - we need to experience the world also. Certainly !! agree on point 2
Answer: How else would you experience the world? Is the color red itself the same as the word 'red'?
Of course we use words to describe our experience to others.
One could examine the world out there as logos by Christianity, a system of signs by semiotics in European philosohy as inspired by linguistics or information as modern physics - but this is language of an entirely different order and amongst themselves are not commensurate. In this sense no-one escapes language in their structuring of experience.
Comment: Thanks, if not with a language how would one experience the world !! Exactly ! Now let's go a step further, having learnt the language - can it help narrating, recalling the experience Patricia Kuhl(TedTalks)refers Neuroscience may term as Rocket Science or romantic poets term it as -Celestial Openness of Child's mind before formally learning a langauge. Having the tools of a Language can one experience the world that was before the language was learn, if so, how? So What that LANGUAGE would be, as I agree with you again, "no-one escapes language in ..";Much appreciated.


How to flag a contact as being deceased in the Contacts app for iOS/macOS?

Question: Death is a sad fact of life and over the years some of my old contacts have died.
I still wish to keep a record of their details, birthday, day they died, notes, spouse, kids, etc. in my Contacts app on my iPhone, Mac, and iCloud.
I am looking for a recommended way to flag those records to indicate that the person is dead. I am interested to know how other people have dealt with this.
Answer: As CDK suggests, my solution was adding a new custom date.
My problem was that I didn't want to be reminded with this person's birthday every year. Finally, I deleted the birthday and added a new custom date for "birth", so it's not recognized by iCal and there are no more notifications about it.
Answer: I've had the same "need." I guess we all do sooner or later. Here's what I did.
Card menu > Add Field > Dates. Then select the Date field drop down in the contact and select custom.
You can also change the general new contact template in the same Cart menu.
I hope this helps.
Comment: I wish Apple had [RFC 6474](https://tools.ietf.org/html/rfc6474) implemented in their Contacts app. You would then be able to add the place of birth, place of death and date of death to a vcard.
Answer: You could also add a special symbol to the contact's comments (I use «†», also known as U+2020 DAGGER), and then create a smart group, where the criterion is Note > contains > † (Alt-t).
Answer: I'm resorting to deleting birthday, and putting it in Notes. Keeps data, doesn't remind in calendar. ¯\(ツ)/¯
Answer: As others have said, you can add a custom date for deceased and that will help you keep a record. But that doesn't flag them very well for general viewing. You can't filter by the presence or absence of your custom field (at least not in 10.7).
I would also insert a keyword in the notes to use as your filter, something like #deceased. Then you can set up a filtered group called Not dead to mask all those poor folks. And conversely a Dead and gone group will help you reminisce.


TFSReg in 2010 Beta 2?

Question: does anybody know what is the equivalent of the TFSReg.exe command-line tool in 2010 Beta 2? I cannot find it anywhere, I searched the entire Program Files tree.  Was it renamed? Moved? Replaced by something else? Thanks...
Answer: It's actually been removed from TFS 2010 (since you'll be able to set a lot of those details using the TFS Administration Console UI.)  If you're needed to configure a custom service, then you can use the details that Jim Lamb provides in this MSDN Forum Post:  http://social.msdn.microsoft.com/Forums/en-IE/tfsprerelease/thread/fcec76a1-a772-4ecf-84f3-ad48d67762dc
Comment: Great question - it was a pre-release MSDN forum post so not sure where the new location will be.  I'll take a look for it.

TFSReg in 2010 Beta 2?


Why is it said that 'The longer the pipeline, higher the processor clock rate'

Question: I have been recently going through the hyper threading technology of Pentium 4. The number of pipeline stages is high in P4 and is said that it will increase the speed of clock rate. How is that possible?
My reference material
Answer: To answer this question we need to understand a few things about basic digital electronics.
Lets start by taking a look at a typical pipeline.

As we can see that each pipeline stage is followed by a storage register (the green rectangles) which holds the output of each stage. Now each stage of a pipeline consists of a combinational circuit. Combinational circuits are basically combination of various logic gates like NAND, NOR, etc. Each of these logic gates has some latency, i.e.- when you provide some input it takes some time to produce the output (few nanoseconds to picoseconds). So the maximum latency of each stage depends on the longest sequence of this logic gates present in that stage.
Now for a stage to produce a valid output on providing some input we need to provide it with sufficient clock period. So for a stage with large complex combinational circuit the latency is high and hence it will require a long clock period and hence slower clock rate. Similarly for a stage with short and simple combinational circuit the latency is low and hence it will require a shorter clock period and hence higher clock rate.
This is the reason behind, longer the pipeline, higher the processor clock rate.
As in a longer pipeline, the stages are divided into more and hence smaller sub-stages, which makes each stage in the pipeline simpler and combinational circuit shorter hence reducing the latency of each stage. This in turn makes room for higher clock rate.
Comment: but if I have the typical 5 stages which last several ps (let's say the highest of them is 300ps) and I divide this stages till the point each one last 1ps, why does the clock rate increase? Because I need more clock cycles to execute the same thing?
Answer: its little brief to write up about it ,To understand why different CPUs at different clock speeds perform in different ways, let me brief you how the CPU processes instructions.
From tech Website

A CPU processes instructions in an assembly-line manner, with
  different instructions existing in different stages of completion as
  they move down the line. For instance, each instruction on the
  original Pentium passes through the following, five-stage pipeline:
Prefetch/Fetch: Instructions are fetched from the instruction cache
  and aligned for decoding. Decode1: Instructions are decoded into the
  Pentium's internal instruction format. Branch prediction also takes
  place at this stage. Decode2: Same as above. Also, address
  computations take place at this stage. Execute: The integer hardware
  executes the instruction. Write-back: The results of the computation
  are written back to the register file. An instruction enters the
  pipeline at stage 1, and leaves it at stage 5. Since the instruction
  stream that flows into the CPU's front-end is an ordered sequence of
  instructions that are to be executed one after the other, it makes
  sense to feed them into the pipeline one after the other. When the
  pipeline is full, there is an instruction at each stage.
Each pipeline stage takes one clock cycle to complete, so the smaller
  the clock cycle, the more instructions per second the CPU can push
  through its pipeline. This is why, in general, a faster clockspeed
  means more instructions per second and therefore higher performance.
Most modern processors, however, divide their pipelines up into many
  more, smaller stages than the Pentium. The later iterations of the
  Pentium 4 had some 21 stages in their pipelines. This 21-stage
  pipeline accomplished the same basic steps (with some important
  additions for instruction reordering) as the Pentium pipeline above,
  but it sliced each stage into many small stages. Because each pipeline
  stage was smaller and took less time, the Pentium 4's clock cycles
  were much shorter and its clockspeed much higher.
In a nutshell, the Pentium 4 took many more clock cycles to do the
  same amount of work as the original Pentium, so its clockspeed was
  much higher for the equivalent amount of work. This is one core reason
  why there's little point in comparing clockspeeds across different
  processor architectures and families—the amount of work done per clock
  cycle is different for each architecture, so the relationship between
  clockspeed and performance (measured in instructions per second) is
  different.

An real world example from quantum thread:

Lets take A VERY simple processor. It is just a programmable
  calculator - instructions available are add a, b, c and subtract a, b,
  c. (a, b, c are numbers in memory. no way to load these numbers from
  constants ). One way to do it would be to do the following all in one
  clock cycle:

read the instruction and figure out what we're going to do
read memory location a
read memory location b
perform the add or subtract
write the result to location c

With this setup, the IPC is exactly 1, because one instruction takes
  one (VERY long) clock cycle. Now, let's improve this design. We're
  going to have 5 clock cycles per instruction, and each doing one of
  the 5 things above. So, on cycle 1, we decide what to do, on cycle 2,
  we read a, on cycle 3, we read b, and so on. Note that the IPC will be
  1/5th. The thing you have to remember is, ideally each of those steps
  takes 1/5th of the time, so the end result is the SAME performance.
A more advanced implementation is a pipelined processor - multicycle
  like the one described, but we do more than one thing at a time:
  1. read instruction i
  2. read a (for instruction i), and read instruction ii
  3. read b (for instruction i), a (for instruction ii), and instruction iii
  4. do the op for instruction i, read b for instruction ii, read a for instruction iii, and read instruction iv
  5. write c for instruction i, operate for ii, read b for iii, read a for iv, and read the instruction v
  6. store c for ii, operate for iii, read b for iv, read a for v, and read vi
(note that this requires the ability to do 3 or 4 memory accesses in a
  cycle, which I didn't have in the other 2, but for the sake of
  understanding the concepts this can be ignored)
A picture would really help, but I don't have one offhand. To see how
  this performs, note that a given instruction takes 5 cycles from start
  to finish, but at any time, multiple instructions are being processed.
  Also, every single cycle, one instruction is completed (well, from the
  5th cycle forward). So, the IPC is 1, even though each individual
  instruction takes a bunch of cycles, and the actual performance of the
  machine is 5 times the performance of the original, since the clock is
  5 times faster.
Now, a modern processor is MUCH more advanced than this - there are
  multiple pipelines working on multiple instructions, instructions are
  executed out of order, etc., so you can't just do a simple analysis
  like this to see how an Athlon will perform vs. a P4. In general, a
  longer pipeline lets you do less in each stage, so you can clock the
  design faster. The P4's 20 stage pipeline lets it run at up to 3ghz
  currently, whereas the shorter pipeline of the Athlon results in more
  work per clock, and therefore a slower max clock speed

If you were looking for hardware information,have a  reason over here


How do I bitmask these tiles?

Question: Using Godot. I can't figure out how to bitmask these tiles in order to autotile them.
I've examined the templates shown in the Godot Docs Tilemaps tutorial, but none seem appropriate. My own attempts to solve this through trial and error have left me confused and frustrated. The tiles are 32x32 pixels, and I'm fairly sure I need a 3x3 minimal masking layout.

My most recent attempt is almost workable:
Comment: There are a lot of tiles missing in the tileset. Where do you have it from?
Comment: The set is from here: https://elthen.itch.io/2d-pixel-art-forest-tileset
Answer: Your tileset is not ideal to use for autotiling, as many common pieces are missing, for example the L shape. I modified the tileset and used 160x160 wide tiles, this is the result:

The blue checkerboard means that the cell is ignored, so it could be either filled or empty.

How do I bitmask these tiles?


SiS 771/671 graphical card: Is there some Unix-like OS distro that works correctly with it?

Question: I have a SIS 771/671 graphical card on my laptop, and till now I've just got the conviction that it just does not works very well on Debian 7.0, Linux Mint 16, 17 and some old Ubuntu versions. From that post we see that I'm not alone and it does not works well on Lubuntu.
It's easy to deduce that Debian based system does not works good with that graphical card.
So, have someone a suggestion for some distro that may be more efficient with SIS 771/671?
Answer: I recently refurbished a Fujitsu-Siemens Esprimo Mobile v5535 laptop (specs here: https://www.notebookcheck.net/Fujitsu-Siemens-Esprimo-Mobile-V5535.9149.0.html) which sports a SiS Mirage 3+ 672MX graphics chip.
The goal was to install some lightweight Ubuntu 18.04 variant (32-bit or 64-bit, Lubuntu or Xubuntu to be exact).
After trial and error for 2 days and lots of expired/non-working URLs in various forums, I found the correct driver for SiS 671/672 graphics.
The actual working driver is located here: https://github.com/rasdark/xf86-video-sis671
And I have created a bash script to help others on quickly installing the driver here: https://gist.github.com/fevangelou/46a2885233c45011ad5c8752f18eac73
Installed successfully with Lubuntu 18.04 32-bit and Xubuntu 18.04 64-bit (kept this one in the end).
Comment: Thanks very much for this! Putting Ubuntu Mate 20.04 on an old Packard Bell EasyNote ALP-Ajax C3 laptop (mentioning the model in case it helps someone in the future) I'd had no luck at all finding a driver, but this worked straight out of the tin!
Answer: Mageia has working drivers for this card. You can even get them from there to use it a distribution which still uses X.org 1.15.
See the (Portuguese) blog post http://www.vivaolinux.com.br/dica/Configurando-SIS-67172-no-Ubuntu-1404.
Comment: I already tried that blog, and it's not Spanish, that is Portuguese.
Comment: Uh, that's embarrassing.
Answer: i have had antix 17.2 working with the sis graphics card on an esprimo mobile v5535
Answer: I  was also struggling with same issue on friends laptop Fujitsu-Siemens Esprimo Mobile v5535 .
Steps :

Install xubuntu  18.04.1:
http://cdimage.ubuntu.com/xubuntu/releases/18.04/release/xubuntu-18.04.1-desktop-amd64.iso
Download &  unzip driver from  Portuguese  page   :
https://static.vivaolinux.com.br/imagens/artigos/comunidade/sisimedia-0.9.1.zip
it comes from https://www.vivaolinux.com.br/artigo/Como-instalar-o-driver-da-placa-de-video-SiS-771671-no-Ubuntu-1804?pagina=2
( it is similar to what  @Jodka pointed out but bit newer for ubuntu 18 04,  this page had important hint for me to test  solution from  person "hmslima" +thank you "hmslima" and  google translator ! oh and also thank you to "TheSawage"  who send this page in the comment from the @fevangelou script    )
as admin run file manager    $>sudo thunar (or other file  manager)
and  copy and paste the xorg.conf file from the unziped dir to the /etc/X11/
and  copy sisimedia_drv.la, sisimedia_drv.so  to /usr/lib/xorg/modules/drivers/   and restart laptop
So Core finding  for me is  that you have to stick with 18.04.1 Ubuntu/Xubunut/Lubuntu  versions if using above ZIP.  If using  newer   > 18.04.1   the  script  made by  @fevangelou will not work ,  also Portuguese blog posts will not work. At least i could not get it working  under Ubuntus 20.04 &  18.04.4  , also i raised the  issue for  driver developer(s) that @fevangelou   uses  in the  script
https://github.com/rasdark/xf86-video-sis671/issues/6
all the best,


What are the tax implications for doing accounting on the cash vs the accrual method?

Question: What are the tax implications in the U.S. for doing ones accounting on the cash method versus the accrual method? As I understand it the main difference is in the timing (i.e. which period) of realizing income and expenses. So beyond potentially having more flexibility with an accrual basis, are there specific tax benefits/gotchas/rules for using that method?
Comment: With accrual basis you pay tax on income when its incurred, not paid, so there's less flexibility, not more. With cash-basis you only pay tax on the actual cash received, you don't pay taxes on money not yet paid to you.
Comment: In India, Cash basis is not permitted for other than non-profit. It has to be accrual system.
Comment: littleadv, I see what you're saying. So under accrual-basis wouldn't you have the converse situation of being able to incure an expense without actually ponying up the cash? Also, does "incurred" mean when you receive a check or when you deposit it or when the bank actually makes funds available (assuming a cash-basis context)?
Comment: @NatwarLath It is very interesting that India requires accrual basis for almost everybody. In the US, the default seems to be cash basis for individuals, though there presumably are some that use an accrual basis. Also, I believe that the Internal Revenue Service has to be informed when switching from one basis to the other.
Answer: In an accural basis, the date of invoice [or expected payment date, if giving a grace] is taken as the date when the payment was received. Similarly on the expense side, the date when the payment was due is taken as expense. So your profit and loss is on this basis.
On the other hand, on cash basis, if when you receive the funds. By this it means when it is available to you in the Bank account
Answer: Accrual accounting and cash accounting are two very different methods of accounting, and which one you use could have either an insignificant or major impact your annual tax return. The vast majority of individuals and some eligible small businesses use the cash method of accounting, because it is generally simpler and more intuitive than accrual basis.
In cash-based accounting, you record revenue (income) when you receive the cash, not when you earn it. For example, you would record your paycheck when the funds were in your bank account, even if the check was a month late and you had earned that money several weeks prior. In addition to recording income when received, you record expenses when you pay them, not when you owe them. For example, you would record your credit card payment when you send the funds to the credit card company, even if they sent you the bill two months prior.
On the other hand, in accrual-based accounting, you record revenue (income) when you earn it, regardless of whether or not you have actually been paid. For example, you would record your paycheck at the end of each workday, since you earned the income, even if you only got paid once or twice a month. In accrual accounting, it does not matter when you get paid; it only matters when you earn income. Likewise, you record expenses when you incur them, not when you actually pay them. For example, if you received a credit card bill on the first day of the month, you would record the bill as an expense as soon as you received it, even if you did not pay your bill until 30 days later.
Thus, because of these differences in how income and expenses are recorded, the tax implications come from the fact that in accrual, some of the income you record in one year might not actually be received until the next year, and some of the expenses you incurred might not be paid until the next year. In cash accounting, only the income you receive in a given year counts as income for that year, even if you earned more and didn't receive the funds until the next year. Likewise, if you owe a massive amount on a bill but only paid a small part during the year, you would only record the small part you paid as expense. This is why cash accounting tends to be the simpler of the two methods, but is also why it affects the annual tax return in a different way from accrual accounting.
If you have a major expense in a given year (like a costly medical bill), you could only deduct the amount you actually paid in that year from your taxes if you were on the cash system. If you were on accrual, you could deduct the entire invoice for the year you received it regardless of when you actually paid it. Likewise, on the accrual method, if you earned a massive amount of income in a given year, you would have to record all of the income you earned that year, even if you only received half of the funds that year. On the cash method, you would only need to record as income the amount you actually received.
Be aware that you while you can choose whether to use cash or accrual on your first tax return, the IRS requires you to use the same method in subsequent years. Individual taxpayers and eligible small businesses generally prefer the cash method, because it requires less tedious record-keeping and is generally easier to understand. If your first tax return was calculated using the cash-based method, you will be required by the IRS to use that method on all subsequent returns unless the IRS grants you permission to change methods.


2.80 Layer panel

Question: I opened a 2.79 project in 2.8 but some objects were in different layer panel and hidden. How can i unhide them and show on viewport?
Answer: After opening files from 2.79 the objects that were on a different layer that was not visible before saving the file have the disable in viewport restriction toggle enabled, which is not shown by default. To make the disable in viewport restriction toggle show, open the filter menu in the outliner and enable the disable in viewport in restriction toggles (the monitor icon). After that you can enable everything that was hidden back.

2.80 Layer panel


Connecting to a Sybase Database using C#

Question: How can I connect to a Sybase Database using C#.Net?
Do I need to install a specific dll or I should add something in the
<code>using
</code>
section?
Comment: Related question: http://stackoverflow.com/questions/945501/sybase-ase-ado-net-2-0-provider
Answer: You'll need to do both.  You can either get providers from Sybase (my last experience was not entirely positive... their connector consisted of an unmanaged dll and a managed wrapper that was not portable across 32/64 bit machines), or use the ones from the Mono project.  I'm sure there are several other providers like DataDirect as well.
Answer: An alternative to the answer provided so far is to download and install the ODBC drivers, then lookup connecting to ODBC in C#.


How to give preference over one network connection over another?

Question: Like most laptops, it has both ethernet and wireless networking options.  Likewise, I have a wireless router with 10/100/1000 Mbs.  90% of the time, wireless speeds suffice.  But once in a while, I have several gigs of data that needs transfering.  In those times, I would physically walk over the router and plug in directly.  The problem is that until I disable the wirelless network adapter, Windows still uses the wireless connection as opposed the ethernet connection.  Is there a way to configure Windows to give preference to one network adapter once it becomes availabe instead of disabling the other altogether?
Answer: (I know there is already accepted answer, but)...
First, from XP onwards, Windows has a feature called automatic metric. This feature should automatically prioritize traffic on adapter with highest throughput. When you enable 'better' NIC (eg. by plugging in cable) Windows should automatically route traffic via that interface. Obviously yours incorrectly thinks WiFi is faster (which seems to be reported for some of WiFi cards)
What is a metric anyway and how it's used by network software? Well,a metric is used in routing when there are multiple paths to a destination and a decision needs to be made which one is the best. The lower, the better. Imagine you're at exit gates in a stadium. There are several gates and each will eventually allow you to get out - you need to select which one is the best, eg. by looking at how many people queue at each. Windows does the same, but bases it's decision on link speed.
Your gate to 'outside' is called default route. Let's look at output from route print command, which shows you ip routing table:
> (output ommited)
Network Destination        Netmask          Gateway     Interface        Metric
>           0.0.0.0          0.0.0.0      192.168.0.1    192.168.0.12       25
>           0.0.0.0          0.0.0.0      192.168.0.1    192.168.0.22       10
(output ommited)

Those entries with 0.0.0.0 are default routes (sometimes also called quad 0 routes). Obviously I have two (with both cable and WiFi active), which one will be selected? The one with lower metric. In my case - 0.22 which happens to be my cable connection.
Now important thing - changing adapter priority via adapter setting does not change metric. This means it will not change routing decisions!
To actually change metric you need to to to open each adapter properties, then TCP/IP properties, Advanced, uncheck automatic metric and enter your own value. Adapter with lowest metric wins.
You can quickly check which interface is actually used - open Task Manager - Network, start download/upload and look at interface usage. If you need more detail - use Perfmon.
Comment: This was the piece I was missing! Thank you!
Comment: Yes - a wireless card reporting at 130 Mbit or 300 or 600 (HAH!) is going to beat a wired nic at 100 Mbit.
Answer: You have to change the priority of the network adapters.
To do this, open the Network and Sharing Center and click on Change adapter settings in the left pane. you should see a list of network adapters. Press the Alt button (if the menu bar is not already visible) and click on the Advanced menu, and then choose Advanced settings.
A new control panel will pop up, and you should see it open to the Adapters and Bindings tab. In the top list should be a list of your network adapters, with an up and down arrow button to the right. This is the priority order of what adapter it will use when connecting to the Internet. Choose your Ethernet adapter and click the up arrow button until it shows at the top. Then click OK.
That should cause it to prioritize your wired connection over WiFi whenever it is connected.
Comment: Two words: freakin' awesome!
Comment: In the mentioned menu it already shows my LAN network to have higher priority than my WLAN one. However, in the bottom right corner it still says it is connected to WLAN. Is this normal? Would it still show WLAN connectivity even if priorizing LAN?
Answer: Easier and faster, you can just delete the entire 0.0.0.0 route.
route delete 0.0.0.0

And adding back only the preferable route to the internet.
route add 0.0.0.0 mask 0.0.0.0 192.168.43.1 METRIC 1

Also ensure that legitimate traffic meant for the second route is added to the routing table so that all traffic meant for that route is not routed through the default route.
route add 10.1.0.0 mask 255.255.0.0 10.1.18.41 METRIC 1


Can two people log in at the same time in Admin?

Question: Can two or more people log in to a Magento e-commerce site at the same time? Can one person process sales while another updates products?
Comment: Possible duplicate of [Magento 2.1 Admin login issue - Someone logged into this account from another device session is terminated](https://magento.stackexchange.com/questions/126582/magento-2-1-admin-login-issue-someone-logged-into-this-account-from-another-de)
Comment: please specify magento version ?
and you want to user same user login at two place ?
Comment: Magento2 - previously we could but since upgrade to Magento 2 the first user gets bumped off when second user signs in. Same user login, two different computers.
Comment: @AlexMcGregor you should check [this answer](https://magento.stackexchange.com/a/126583/20064)
Answer: You need to enabled Admin Account Sharing option from admin to allowed logged in same user simultaneously.

Go to  Admin >> "Stores > Configuration > Advance > Admin > Security"
and change value of "Admin Account Sharing" to "Yes".

Second thing - Yes that is also possible one person do sales operations and another will do updates products with same user.
Note : But still as a part of Security Create new user for sales with limited access by using ACL functionality of magento 2 and give access to sales related operations to sales user same way for others.
Comment: @AlexMcGregor - If you found answer as helpful , then accept as solution so other user gets helped , Happy to help and keep helping :)
Answer: The above answers are correct. You can do it by using CLI also:
php bin/magento config:set admin/security/admin_account_sharing 1

Answer: To use this feature go to
Stores > Configuration > Advance > Admin > Security

and
change value of Admin Account Sharing to Yes.
Answer: Yes, multiple admins can login with the same account in admin. But you will need change settings in admin.
Stores > Configuration > Advance > Admin > Security

Change value of Admin Account Sharing to Yes.


How do you determine the size of zones?

Question: Determining the size of a zone still eludes me. How do you determine the number and size of zones when you are in a combat situation in Dresden Files?
Answer: Zones represent ease of movement and the ability to manœuver. But also, remember that…
Zones serve the scene

[t]he GM's primary tools for framing the scene are zones and scene aspects.

Zones are thus a tool to use as you see fit to frame the scene the way you want. They're decidedly not about simulating a place "right" or anything like that.
Is this a sniper battle? There are several zones between the parties, and they're long, narrow bands. This gives lots of opportunity for ducking into cover and being shot at if someone (foolishly?) tries close the distance.
Is this a fistfight? Do you need more than one zone? Consider Harry duking it out with the gang of lycanthropes by the highway. There are a few zones there, but the existence of each is strictly for scene-framing purposes:

There's the ditch by the side of the highway in which most of the fight happens. Harry doesn't move from this zone.
There's farther along the highway, where some of the lycanthropes gather and rally out of Harry's reach. So, there's a ditch, and it's split into two zones. The purpose of the split is to have "near" and "far" combatants.
There's the highway. This wouldn't even be a zone, except for the fact that vehicles stopping and people pouring out of them is a thing that happens.

So there you see how the zones are really determined by the needs of the scene.
Rules of thumb
However, there are some useful rules of thumb if the needs of the scene don't answer the question of whether something is a zone or more than one zone:

People have to be in the same zone in order to talk without screaming, or to fight in close combat. Any place where this should not be possible ought to be split two zones.
Aspects can be put on zones; therefore, a zone should cover an area that is reasonably homogeneous enough that Aspects can be used anywhere in the zone. (The entire main floor of Victor Sells' beachhouse would be one zone.)
Zone boundaries impede movement. Anywhere there is a distinct barrier that divides two unlike areas should also be a boundary between two zones.
A zone can be long and narrow, such as a hallway. It's easy to move along a hallway, so two people at different ends can easily be right in each other's faces pretty quick. Really long hallways, or hallways that are going to be used in a firefight, should probably be two long zones.
Zones determine ranges for thrown and fired weapons. You can only throw something one zone away. You can only shoot with a handgun two zones away. You start needing things like rifles to shoot three zones away. From this we can determine a few things:

Zones that are packed close together foil line of sight of firearms somehow, either through being full of stuff, being divided by doorways and walls, being on different levels, or something like that.
Large zones necessarily have terrain that makes closing with a shooter easy, and evading your target hard. A large field, if made all one zone, means that you can across it and punch someone before they shoot twice. Probably that large field should be divided by an "X" into four quarters to make running across it to get into hand-to-hand with a shooter is slightly harder.
If you feel like someone could throw a rock at someone else from a particular distance, they should be adjacent zones.

However, zones are necessarily an abstraction and one layout for a particular scene won't necessarily be the right layout for another scene in the same location. The sniper-fight layout of banded zones wouldn't make sense for a fistfight in the same street. Neither would the layout of Harry's fight with the lycanthropes by the highway make sense for a magical duel that raged across the whole highway, stopping traffic – nobody cares about half-ditches then.
So, zones are determined by how they need to serve the action that's about to unfold.
The nice thing about zones though, is that they're very forgiving. You could lay out the zones for a fight in several different ways, and all of them will work. They'll influence the scene differently of course, but all of them will be interesting. You don't have to get them perfect.
Example zone layouts
A highschool gym during a basketball game

The basketball court
The left audience stands
The right audience stands
A zone under each set of stands
The area between the end of the court and the exit doors
The same area beside the other end of the court
The hallways too the locker rooms
Of course, if you're playing out a game of basketball (maybe as some kind of supernatural game of magic and skill between the two faerie courts or something?), then you're likely to divide up the court into multiple zones for manœuvering and passing purposes, and leave the court surroundings as maybe two zones.

Victor Sells' beachhouse
Judging just from memory of how the fight and the shooting with Sells, the demon, and the scorpions went, I think these zones would suffice (though of course some other layout would also work):

The ground floor (with Aspects that reflect it being filled with crates and furniture, and later smoke, for cover)
The open part of the loft (with a zone boundary with the ground floor that has a pass value that people can fight to alter, what with all the dangling from the railing that happened)
The kitchenette
The seating area toward the back of the loft
The balcony outside
Comment: I adapted this very concept from Fate/Fudge/Dresden when I heard about it. Makes narrative combat simple. Mentally divide the scene into zones (rooms, combatant pairs/groups, &c.). Then movement is simply how many rounds it will take you to get from one zone to another. Want join the fight in an adjacent zone? One round. Want to run to the aid of a party member upstairs in the room off the hall? One zone for the stairs, one zone for the hall, one zone to enter the room; three rounds!
Comment: Of course, inhuman speed makes for getting across zones faster. :)
Comment: @Cthos That's handled by the text of the super-speed powers though. You don't need to adjust the zones much for that, except insofar as you have to lay out the zones according to the sort of use they have in the scene anyway.
Answer: Simply put, in all Fate 3rd Ed Games (of which DFRPG is one; also Spirit of the Century, Diaspora, and Starblazer Adventures), the size of the zones is story determined.
They're big enough to require shooting past, small enough to go HTH with anyone inside, and divided as needed for story purposes.
When I last played SOTC, the GM had us in several combats with very different sized zones.
Combat one: The hotel hall. Each hotel room was a zone. The halway itself was a 3 zones, each with 4 rooms off of it. Bathrooms were ignored for simplicity.
<code>+----+----+----+----+----+----+
| R1 | R2 | R3 | R4 | R5 | R6 |
+--#-+--#-+--#-+--#-+--#-+--#-+----+
|   H3    :   H2    :   H1    :    |
+--#-+--#-+--#-+--#-+--#-+--#-+ St |
| RC | RB | RA | R9 | R8 | R7 |    |
+----+----+----+----+----+----+----+
</code>
This gave us a good enough idea that he didn't actually DRAW the map, but simply described it. Numbering is mine, and the rooms aren't to scale.
The second example from the same campaign was the Queen Bee's lair - on an airfield in Norway...
Zone 1 Runway N (2,3)
Zone 2 Runway S (1,3)
Zone 3 Taxiway (1,2,4)
Zone 4 Apron (3,5,6,8)
Zone 5 Hangar front (4,5,8,9,10)
Zone 6 Hangar left side (4,5,7)
Zone 7 Hangar back (6,8,10,11)
Zone 8 Hangar Right side (4,5,7)
Zone 9 Hangar office (5, 10, 11)
Zone 10 Hangar inside main (5, 7,9,11)
Zone 11 Evil Machinery (7,9,10)
<code>     1 ----- 2
       \   /
         3        -- 6 ---------+
         |       /  /            \
         |      /  /    9 -- 11   \
         4 ------ 5 ----+  /  +--- 7
                \  \    10 ------//
                 \  \            /
                  -- 8 ---------+
</code>
Note that I used two different methods of drawing the maps, too...
You can take a map (like the hotel) and just divide off large rooms as needed, using rooms as zones premade for you...
Or, you can simply use numbers or labels, and draw connection lines. Either works fine.
And fate isn't the first game in print to use such a system. Crimson Cutlass did so back in the 80's. And all the classic Interactive Fiction works do so as well... both on computer and in gamebooks and solo-modules. It's highly flexible, and reinvented often because it's so powerful.
Comment: i don't understand your second map...
Comment: @DForck42 It shows the connections, not the edges.
Answer: The answer to his is highly situational, because like most things in the Dresden Files... there is no true answer.
There really is only a cost for moving between zones, so you don't want to make zones so big that it seems strange that your players and NPCs can easily move within such a huge area.  Take the example of having a Wizard's duel at a baseball arena.
If the whole playing field was one zone, that means that people could enter the zone from opposite ends and immediately get into close combat, or move through the zone extremely quickly.  Now that might be an appropriate zone size if all my players are Wizards with super speed or Werewolves high on the light of the moon, but is likely far too big for mortals.
From there you attempt to determine what is appropriate and dramatically acceptable.  If they are going to get into close fighting, maybe I make the area around each of the bases a zone, as well as having the benches for each team being their own zones.  Now you have a series of zones that are appropriate for the situation at hand.
If I am instead having a big chase scene where they are in cars and fighting it out while they drive through the streets, the zones of course need to be bigger.  Then it might be appropriate to have every "street" be its own zone as they maneuver through the city.
When it comes to Dresden Files, I just go with instinct.  There is no right way of doing it and there really is no wrong way either.  You will quickly know if what you made is too big or too little and can adjust accordingly.

How do you determine the size of zones?


How do we convert price per kVA to price per kW?

Question: I have been told that since kVA is a smaller "unit", price per kW is higher than price per kVA. I'm not entirely sure kVA and kW are not simply units as Power Factor is not really a conversion factor. But I could be wrong in my interpretation. Can someone please confirm this logic? Thank you.
Comment: Scalars are perfectly acceptable as conversion factors.
Comment: Depending on your contract, you might also get billed for reactive power, so conversion might get more complex
Answer: kW = kVA X Power Factor. Power factor is a number between zero and one because is is the cosine of the phase difference between voltage and current. Therefore kW is always equal to or less than kVA for a given situation.
Price/kW = Price/(kVAxpf)
Answer:
How do we convert price per kVA to price per kW?

You pay for energy by kWh (not kW) and reactive power by the kVArh (not kVA).
For an industrial supply or large business supply you may also pay for

MIC (maximum import capacity) which is your contracted supply capability.
Max demand (maximum kWh used in a 15 or 30 minute interval during the billing period). The idea here is to discourage high peak loads.
Reactive power or wattless units. Generally the utility company should be happy enough if you keep your power factor above 0.95. If it drops below this then expect to pay a surcharge for wattless units.

I have been told that since kVA is a smaller "unit",

No. Since kWh = kVArh x power factor, and power factor is always <= 1, then kVArh is larger than kWh.

price per kW is higher than price per kVA.

True.

I'm not entirely sure kVA and kW are not simply units as Power Factor is not really a conversion factor.

The kWh measures real energy used.
The kVArh is effectively energy lost by the utility company in their transmission network feeding into a reactive (inductive or capacitive) load at the customer. If they don't charge for it the problem tends not to get fixed.


Egyptian square pyramid

Question: If you have $140$ stone blocks, you can build an Egyptian square pyramid whose base has $7$ blocks on each side. At the same time, you can arrange those blocks into $35$ perfect 2 × 2 squares. Find a positive integer value of $n \gt 1000$
for which $35$ perfect $n \times n$
squares can also be arranged into an Egyptian square pyramid.
Comment: Don't you mean m x m squares?
Comment: No, I mean $n \times n$ squares.
Comment: you could have said  `n>3` it make no difference
Comment: Sooooooo this is the Square Pyramid Puzzle with an extra factor of $35$ in front? If so, the proof is *not* simple, although it is technically "elementary".
Answer:
 n=2378, the pyramid's base is 840 square

but I wrote software to find that
<code>#!/bin/bash

p=1
n=999

while true
do
 (( pp= p*(p+1)*(p*2+1) / 6 )) # formula for square pyrimadal numbers
 (( nn=35 * n * n ))           # 35 squares

 echo "p=$p n=$n pp=$pp nn=$nn" # show progress
 (( pp == nn )) && break        # stop when a solution is found
 (( n+= ( pp > nn ) ))
 (( p+= ( nn > pp ) ))

done
</code>
[edit]
We are now told that this solution can be found by manual means the next solution if there is any is up past one million, so  it's likely that this one is findable.
It seems we have a diophantine equation
$ 210n^2 = p*(p+1)*(2p+1) $
but I can't see how to proceed.
Comment: @ M Oehm: I have no problem with this solution. I just said that so that others can try manually.
Comment: @Human: If you would rather prefer a manual solution (as would I), I suggest you leave the question without an accepted answer. An question with an accepted answer will draw lesser views, and reduce the chances of seeing that elegant solution.
Comment: This can be even solved without programming.
Comment: I would also like to see how to solve this manually
Comment: not particularly well written even.
Comment: @Human: If you are looking for a way to find the answer by reasoning alone make it clear in the question and add the `no-computers` tag.
Answer: Here's a solution by hand. It's still a little brute-forcey but certainly plausible (if a tad tedious) to do without a computer. (I can confirm, as I did the whole thing on paper.) The result confirms Jasen's answer of

 $n=2378$; pyramid base size of $840$; $1187525640$ total blocks

We're searching for a number $n > 1000$ for which $35$ $n\times{}n$ squares can be arranged into an Egyptian square pyramid. Calling the total number of blocks $P$, this can be expressed as the following equation:
$P=35n^2$.
with the stipulation that $P$ is a square pyramidal number. These numbers are well known, and there is a handy formula for computing them. For any positive integer $m$, the $m$th square pyramidal number can be found using
$\frac{m(m+1)(2m+1)}{6}$.
Therefore, we're hunting for integers $n>1000$ and $m>0$ that satisfy the equation
$\frac{m(m+1)(2m+1)}{6}=35n^2$.
Rearranging to find $n$ yields:
$n=\sqrt{\frac{m(m+1)(2m+1)}{210}}$.
This tells us that the numerator is divisible by $210$, and the whole expression under the radical must be a perfect square. Since $m$ is an integer, so must the other terms in the numerator be. The simplest way to ensure the numerator is divisible by $210$ is to assume $210\vert m$. Then it's a matter of doing some fairly simple hand calculations over $m=210, 420, 630, 840$, etc., using some simple tricks to check if the result is a perfect square.

$m=210\rightarrow 1*211*421$. The last digit is $1$, so this could be a perfect square. However, the 10s digit is a $3$, so this cannot be a PS.
$m=420\rightarrow 2*421*841$. The last digit is $2$, so this is not a PS.
$m=630\rightarrow 3*631*1261$. The last digit is $3$, so this is not a PS.
$m=840\rightarrow 4*841*1681$. The last digit is $4$, so that doesn't rule this out yet. Last two digits are $84$, so this could still potentially be a PS. Alright, fine, multiplying it out yields $5654884$. If you're like me, you probably don't have all the squares up through four digits memorized. However, I do know all the ones up to $30$. Noting that $23^2=529$ and $24^2=576$, the number is between $2300$ and $2400$. Maybe someone out there knows a more efficient way to find the square root of a (fairly) large number, but for me it involves a lot of oscillating until I stumble upon it. Spoiler alert: It's $2378$. Thankfully all that work didn't go to waste, since I had no real assurance that that number was, in fact, a perfect square.

Therefore we have found $n=2378$ and $m=840$.
Comment: You could have noticed that $4$ and $841$ are both squares, so you then only need to check that $1681$ is also square. It is fairly obviously a bit more than $40^2=1600$. In fact, it is $40^2+2*40+1$ $=$ $(40+1)^2$ $=$ $41^2$, so you get $n=2*29*41$.
Comment: @JaapScherphuis: Good catch; that would make it a bit faster.
Answer: So far, the nearest thing I have found to a way of solving it by hand is the following: if you happen to remember that 841 and 1681 are both squares (which you very well might, especially if for any reason you've been looking at rational approximations to $\sqrt{2} lately) then on seeing
$210n^2=p(p+1)(2p+1)$
you might say: Hmm, perhaps we can make two of those factors on the RHS be squares and the other one deal with the 210; $p=210$ doesn't work, but aha!, $4\times210=840$ and if we take that as $p$ then both $p+1$ and $2p+1$ are squares, and we're done.
But this is entirely dependent on spotting something distinctly non-obvious; it doesn't offer an actual method that leads to the answer.
(In general, this kind of problem -- a cubic polynomial in one variable equals a quadratic in another -- is a matter of finding integer points on elliptic curves, and the usual method for this is "run SAGE or Magma on your computer and get it to solve the problem", which it will do by means of advanced mathematics and a lot of calculation. So if there's a method that solves this problem by hand without the sort of flash of insight described above, it will need to be one quite specific to the particular problem.)
Comment: Yeaahh, I kept hitting the wall of finding points on an ellipse, which seems far too complicated for a puzzler to do "by hand".

Egyptian square pyramid


Connect on-premise Virtual Machine to Azure Active Directory

Question: I want to include on-premise Virtual Machine into Azure Active Directory. I don't have much idea how to implement this. Kindly let me know the steps to perform this.
Comment: What do you want to achieve with that connection? Should AAD-Only users be able to access your OnPremises SharePoint?
Comment: Yes, I want only AAD users to acess on-premise Virtual Machine. Is there any way to achieve this?
Answer: Unfortunately you go the "wrong" direction from Microsoft's perspective. Integrating OnPremises Users to SP Online works like a charm. The way back is much more complicated.
You need to configure a TrustedIdentityTokenIssuer (TST) in SharePoint. This will make SharePoint trust another source for identities in addition to OnPremises ActiveDirectory.

Set up OnPremises ADFS and add it as an Application in Azure AD
Add Azure AD as an Identity Provider in ADFS
Create Relying Party in SharePoint and Add Pass Thru Rule
Create SPTrustedIdentityTokenIssuer in SharePoint

The whole process is described here in detail (skip step 1 in the blog as you want to enable AAD-Only users):
Using Azure AD as an Identity Provider with ADFS for SharePoint 2013
Integrating a TST into SharePoint comes with a lot of drawbacks and more complexity. Link is for FormsBasedAuthentication (FBA), but those drawbacks also apply to a TST. So test this scenario excessive and then decide if it is really the answer to your needs.


DSA parameter selection

Question: In the Digital Signature Algorithm, is there a reason why the larger prime parameter should be an $N$-bit prime where $N$ is a multiple of 64 ?
Is this strictly necessary for proper working of the algorithm?
Answer: The algorithm would work fine without the restriction, although it's debatable whether NIST owns the name DSA and you would have to call it something else, the way non-RSA-licensed implementations of RC4 were for a long time called ARCFOUR (alleged RC4) or similar. How about Schnorr Patent Evading Digital Signature SPEDS?
Why did NIST do it (originally, see next)? AFAICT 'everyone knows' they wanted to reduce the amount of testing needed to ensure that multiple, different implementations would provably meet the standard and (as a result) interoperate with each other reliably, while still allowing designers and possibly users to choose tradeoffs of cost vs security. I don't think I've seen that rationale officially documented anywhere, but they certainly have done essentially the same thing with other algorithms and standards, most notably AES (which supports only 3 key sizes and 1 block size, though the underlying algorithm Rijndael supports quite a few more).
Also note the originally permitted sizes 512 to 1024 by 64 -- which still survive in a good deal of software that hasn't been updated -- have officially been obsolete for a long time. Change Notice 1 to FIPS186-2 in Oct. 2001 restricted p to only 1024 (with q still 160); after a long delay FIPS186-3 in June 2009 added p 2048 or 3072 (with q 224 or 256 in the former case and 256 in the latter) -- while 1024 2048 3072 are multiples of 64, they are not all the multiples of 64 in that range, and do not provide nearly the 'fine tuning' that the old list did.
Although in contrast RSA and DH were defined and usually are implemented for any bit size (sufficiently large to provide the desired security) in practice people almost always use powers of two or 'halfway between' sizes: 512 768 1024 1536 2048 3072 4096 -- and FIPS186-3 specifies key generation for RSA signatures (for US govt use) that only allows n sizes 1024 2048 3072.

DSA parameter selection


Will adding Featherfalling to my protection 4 boots help?

Question: I am having a little trouble understanding how the protection enchantments work, so my question is, will it help any if I put Featherfalling IV on my Protection IV boots, or does that exceed the cap? All my armor is diamond and with only protection IV
Comment: I think it should help, given the nerf on the protection enchantment. I'm not certain offhand though.
Comment: This question is a subquestion of this question:  http://gaming.stackexchange.com/questions/102707/which-configuration-of-armor-enchantments-gives-maximum-protection
Answer: The reduction from enchantments is based on the Enchantment Protection Factor (EPF) of any relevant enchantments. The EPF of an enchantment is based on the level of the enchantment and the type- protection enchantments have a lower epf than more specific protections such as feather fall; they make up for it by working against anything.
The EPF of Prot IV is 5 and the EPF of Feather IV is 18. The cap is 25- this means a full set of Prot IV doesn't quite hit the cap.
If a single piece of equipment has both Protection and a specific protection, such as feather fall, only the best applies. This means a single Prot IV, Feather IV item will count as 18 against falling- not 23.
Adding Feather Fall IV (or Feather Fall III) to one piece would be enough to take you over the cap for falling damage.


Intuition behind conjugation in group theory

Question: I am learning group theory, and while learning automorphisms, I came across conjugation as an example in many textbooks. Though the definition itself, (and when considering the case of abelian groups), it seems pretty innocent, I have to admit that I have no intuition about whats happening, or why even such a map is important. Did anyone feel this way when you learnt it? Any examples/references that you have found useful when you learnt it?
Comment: An example of the top of my head where conjugation arises is related to base changes in linear algebra. Say that a linear map is represented by matrices $N$ and $M$ in two different bases, and that the base change matrix is $P$, then the relationship between $N$ and $M$ is $N = P^{-1}MP$.
Comment: Every group can be seen as the fundamental group of some connected topological space. Conjugation corresponds to a change of the base point of this space.
Comment: See figure 11 [here](https://terrytao.wordpress.com/2012/05/11/cayley-graphs-and-the-algebra-of-groups/), related to the pictures in the answer [here](https://math.stackexchange.com/questions/163262/visualizing-commutator-of-two-vector-fields) (because the "Lie algebra version" [i.e. the differential] of $xyx^{-1}y^{-1}$ is the Lie bracket) !
Answer: One way to think about conjugation is as a generalization of changing coordinates when rewriting matrices or, from a physical point of view, as the process of "seeing a group from a different perspective."
Let's make this precise.  Let $G$ be a group and let $X$ be a set on which $G$ acts faithfully.  For example, if $G$ is the Euclidean group of isometries of the plane, $X$ could be... well, the plane.  Let's say we have a reasonably concrete description of what a particular element $g \in G$ does to $X$.  For example, if $G, X$ are as above, then perhaps $g$ is a rotation counterclockwise by $\theta$ centered at the origin.
Given that description, what does the element $hgh^{-1}$ do?  Well, this is actually quite simple: in the description of $g$, replace all the elements $x \in X$ that occur by the corresponding elements $hx \in X$.  For example, if $h$ is a translation by a vector $v$, then $hgh^{-1}$ is a rotation counterclockwise by $\theta$ centered at $v$ instead of the origin.
In other words, what conjugation does in terms of group actions (and it is always a good idea to think of groups in terms of their actions) is it corresponds to a "change of coordinates" on the underlying set $X$.  This is a basic reason conjugation and group theory are important in understanding Newtonian mechanics (where $G$ is the Galilean group) or relativity (where $G$ is the Lorentz group) or really much of modern physics in general: in physics groups are always studied because of their actions and we always want our concepts and equations to be invariant under these actions.  (For example, mass and charge are invariant concepts.  The mass or charge of an object doesn't change when you rotate or translate it.)
This gives a very intuitive definition of a normal subgroup: it's a subgroup that "looks the same from every perspective."  For example, the subgroup of translations in the Euclidean group is always normal because the description "$g$ is a translation" is the same from every perspective (that is, it's invariant under conjugation).  It's a good exercise to look at some groups you're familiar with and see if you can identify which subgroups are normal based on this principle.  (This principle underlies the importance of normal subgroups in Galois theory as well.)
Comment: @Mariano: strongly disagree. I think it is never misguided to ask for intuition.
Comment: Heh. To be honest, I think that it simply does not mean anything to look for "intuition behind conjugation". Conjugation is conjugation, and that's that. At most, one can ask for interesting examples where conjugation is meaningful, so as to familiarize one self with the various avatars of the notion, but I tend to think that people asking for intuitions in situations like this are both misguided and in for a disappointment!
Comment: @Ben: Your and Mariano's comments aren't actually contradictory. The problem is the presumption is that there is intuition "behind" conjugation, rather than intuition about its applications, and intuition derived from working with it directly. Really, I think the idea there is intuition "behind" some abstract notion is an illusion derived from the situation where you only understand one application of the notion.
Comment: Wouldn't that give an "intuitive definition" of a *characteristic* subgroup?
Comment: @Mariano: I have to admit I do not have a good conceptualization of how to think of a group G acting on a set X while G is also being acted on by Aut(G).  So I tend to ignore outer automorphisms (except when G is abelian).
Comment: Thanks! I think Mariano is right, maybe what I was actually looking for was different ways/examples to look at conjugation afterall!
Comment: @Qiaochu +1 When I understood this about conjugation it suddenly made all of group theory much more intuitive. I think that the idea can be rescued from the problem posed by Mariano (normal vs. characteristic, inner vs. outer automorphisms) using a device like that in Per Vognsen's answer.
Comment: Mariano's comment made me wonder why inner automorphisms should be distinguished among the ways to shift between group actions. (For every $q\in\mathrm{Aut}(G)$, after all, $(g,x)\mapsto q(g)x$ is a left action.) I tried to view this in terms of natural transformations from category theory and using abstract groups as symmetry groups of themselves (with some redundancy, $Z(G)$, built in).
Answer: This is a belated follow-up to Qiaochu's nice answer and Mariano's comments on characteristic vs normal subgroups. I cannot yet comment on answers, so I thought I'd expand this to an answer.
One of the classic examples of normal but non-characteristic subgroups are the factors of a direct product. The factors are obviously normal, but they are non-characteristic because they are not invariant if you interchange the factors. This interchange defines an outer automorphism, or an external symmetry.
You can make Qiaochu's sketch of an "intuitive definition" closer to a real definition if you draw a distinction between internal and external symmetries. The question is how you can make that distinction intuitive by itself. Here's my attempt:
Imagine a space with symmetry group $\mathbb Z^2$. Let's call it lattice world. A denizen of lattice world has position but no orientation. This last part is crucial. It means that even though we can look at lattice world from our privileged outsider's perspective, tilting our heads 90 degrees and thereby interchange latitudes and longitudes (showing that the $\mathbb Z$ factors are non-characteristic), a denizen of lattice world has no orientation and therefore cannot turn, so that perspective is inaccessible to him: a change of perspective in his world involves only translations.
Comment: +1 Fully agree. An automorphism is "looking at the group from a different persective." An *inner* automorphism is when this is a perspective *the group itself can take.* If $O$ is the octahedral group, $Aut(O)$ is entirely inner because every automorphism corresponds to looking at the octahedron from a different angle, which is exactly what the group elements do. The automorphisms all happen in the language of the group itself. For the quaternion group $H$, we with our birds-eye view can permute $i,j,k$ arbitrarily; but the group itself does not have the power.
Comment: ...and all of a sudden, it makes complete sense why inner and outer automorphisms are called that. Thank you for the great examples.
Answer: One example of conjugation you probably already know from linear algebra is change of base.
The linear transformation A that takes (x,y) to (3x−y,2x) is reasonably nice and can be represented by the matrix $$A = \begin{pmatrix}3&2\\-1&0\end{pmatrix}$$
In the basis u=2x−y, v=x−y, Au = 2Ax − Ay = 2(3x−y) − 2x = 4x − 2y = 2u and Av = Ax−Ay = (3x−y) - 2x = x−y = v.  So in this basis, the linear transformation is incredibly nice and can be represented by the matrix $$\tilde A = \begin{pmatrix}2&0\\0&1\end{pmatrix}$$
If we take the matrix $$B = \begin{pmatrix} 2&1\\-1&-1\end{pmatrix}$$
then $B^{-1} A B = \tilde A$ expresses the matrix in the new basis defined by B, (u,v) = (2x-1y,1x-1y).  Note that Bx = u, and By = v.
In some sense this is what conjugation does in general: it changes the basis on which your group acts.  If you conjugate by g, it changes the way G acts by converting the old basis to the a new basis already acted on by g.
A good example to work through to see this conjugation in a new context is with permutations.  If a permutation A takes 1,2,3 to 3,2,1 and the permutation B takes 1,2,3 to 1,3,2, then the conjugated element $B^{-1} A B$ takes 1,3,2 to 2,3,1.  In other words, instead of acting on 1,2,3 it acts on 1,3,2.  In cycle notation, you can write this as A=(1,3) and B=(2,3) and then A^B = (1,2).  The "3" in (1,3) has been replaced by "2", since B replaces 3s by 2s.
Try to think of groups as acting on something.  Either let them be matrices moving vectors around, or permutations moving arbitrary things around.  Conjugation just lets you relabel the things you are moving around.
Comment: In general, if $\sigma$ is a permutation and $(x_1, x_2, \ldots, x_n)$ is another permutation, both in $S_m$ ($n\leq m$), then $(x_1, x_2, \ldots, x_n)^{\sigma}=(x_1\sigma, x_2\sigma, \ldots, x_n\sigma)$. This generalises, as an arbitrary permutation $\phi$ is a product of disjoint cycles of the form $(x_1, \ldots, x_{n_i})$ for variable $i$. This is very neat!
Comment: (Sorry, I hadn't noticed that this post was the best part of a year old - this thread had just floated to the top of the stack...)
Comment: Shouldn't the matrix rows for $A$ be 3, -1, and 2, 0?
Answer: There are several reasons:

Conjugation is a way of measuring, how commutative or otherwise your group is. Namely $aba^{-1} = b\Leftrightarrow ab=ba$. Since abelian groups are much easier to understand than arbitrary ones, this measure is important.
Conjugation is more or less the only known way of obtaining automorphisms of any given group without knowing anything further about the group (of course, these automorphisms can all turn out to be trivial if your group is abelian).
You have to bear in mind that normal subgroups and the action of conjugation were first introduced by Galois, because for a subgroup to be normal was exactly the criterion that he needed for an automorphism of a field to give an automorphism of a subfield upon restriction. This comment might be slightly over your head if you are only learning groups at the moment, but bear in mind that historically, the notion of normal subgroups arose in the context of interpreting groups as symmetries of field extensions, when Galois developed a theory to decide whether a given polynomial was solvable by using elementary operations of arithmetic and roots.
Comment: That's an amazing answer, thank you so much!
Answer: My intuition goes back to permutations.  Namely, conjugating everything by some group element can be identified with some permutation of the group.
As for the importance of conjugation, one easy answer is that it gives insight into the commutative portions of the group.  If $x$ is closed under conjugation by  $g$, that is $x = gxg^{-1}$, then we have $xg = gx$, so the elements commute.  The set of elements that are closed under conjugation by all group elements forms the center subgroup, while any subgroup that is closed under conjugation by all group elements is a normal subgroup.  These are fundamental to understanding the structure of a given group.
Comment: Hi, Hans.  I was going to say something similar.  Especially, one should understand what conjugation means in symmetric groups: in this case, it is literally just expressing things in terms of a relabelling (i.e., bijection) of the elements of the set.
Answer: I was thinking about Qiaochu's answer and Mariano's comment about characteristic subgroups, and I think I have something of an idea of why inner automorphisms should be distinguished.
Let $G$ be a group. We can consider it as a category with one object that we arbitrarily call $*$, in which the morphisms are the group elements and $1_*$ is $G$'s identity. Let $F:G\to\rm Set$ be a functor and denote $X=F*$ the target object. Then group elements are taken to maps in $\hom(X,X)$; since the elements in the group are invertibe these morphisms will be isomorphisms  so this is essentially a homomorphism $G\to\mathrm{Aut}(X)$. Effectively $F$ encodes a (left) group action on $X$. (A right action would be encoded with a contravariant functor.)
Since groups abstract the idea of symmetry through their potential actions, it's arguably natural to think of the collection of different ways (in particular the transformations between them) of using the abstract group to concretely realize symmetry (group actions) as "symmetry of symmetries."
If $F,L:G\to X$ are functors represented by the homomorphisms $\alpha,\beta:G\to\mathrm{Aut}(X)$ resp., then a natural transformation between them is an isomorphism $\eta:X\to X$ such that $\alpha\circ\eta=\eta\circ\beta$. That is, two group actions are naturally isomorphic if they are conjugate via $\mathrm{Aut}(X)$.
In particular, a given action $\alpha:G\to\mathrm{Aut}(X)$ provides us with many set automorphisms of $X$ and so conjugating by these, we have natural transformations $(g\mapsto\alpha(g))\mapsto(g\mapsto\alpha(hgh^{-1}))$ for every element $h\in G$. Ultimately then, $G\to\mathrm{Inn}(G)\hookrightarrow\mathrm{Aut}(F)$, where the aut group here tacitly comes from the functor category $\mathcal{C}^G$, allows one to use $G$ to encode its own symmetry via conjugation.
Answer: Let $ G $ be a group. With no further specifications, can we form any interesting isomorphisms from $ G $ to $ G $ ?
For each $ g \in G $, we can try to form an isomorphism $ \varphi_g $ from $ G $ to $ G $ using $ g $.
Let's try maps of the form $ t \mapsto g^k t^l g^m $ $(k,l,m \in \mathbb{Z})$. We want $ g^k (t_1 t_2)^l g^m $ $ = g^k t_1^{l} g^m g^k t_2^{l} g^m $ to hold for all $ t_1, t_2 \in G $, and this can be made to happen by setting $ \{ k = (-m), l = 1 \} $. Maps $ t \mapsto g^{-m}tg^{m} $ are bijections too, giving us a family $ \{ t \mapsto g^{-m}tg^{m} : g \in G, m \in \mathbb{Z} \} $ $ ( = \{ t \mapsto g^{-1}tg : g \in G \} ) $ of isomorphisms from $ G $ to $ G $. These isomorphisms are called conjugations.
[ We could've tried, instead of $ g^k t^l g^m $, some longer more general string of $ g $s and $ t $s, but I can't seem to get any more isomorphisms this way ]
Comment: couldn't get the intuition...
Comment: [Here](https://math.stackexchange.com/a/3929866/303300) is why we care about isomorphisms ${ G \overset{\varphi}{\to} G }$ (any constraint ${ x _1 \ldots x _n = y }$ in ${ G }$ can be rewritten as ${ \varphi (x _1) \ldots \varphi (x _n) = \varphi (y) }$). For concreteness, you can think of ${ G = S _N .}$


Authentic or Not?

Question: I browse periodically on eBay and see a lot of superhero lego minfigs. Are these real or Chinese knockoffs? It says that they're from China. I just search "lego minifigures".
http://www.ebay.com/itm/8-Sets-Super-Heroes-Building-Toys-Flash-Cyclops-Catwoman-Minifigures-Blocks-Toy-/141294442500?pt=Building_Toys_US&hash=item20e5ce1804
Comment: I don't see the word "LEGO" anywhere in the write-up...?
Comment: Possible duplicate of http://bricks.stackexchange.com/questions/3182/how-do-can-i-tell-if-minifigures-are-custom-made-or-lego-originals
Answer: They are fakes. From China. Some people do buy these as they are significantly cheaper than LEGO's version, however there are also problems. The quality varies greatly with some figures almost as good as LEGO, while others won't even fit together properly (see the close-up pictures in the auction). Also, the print rubs off of them quite easily. If you are just planning to buy them for a child who doesn't care about quality they are good alternatives. However they are not suitable for collectors. Also, consider the implications of ripping off both LEGO and Marvel/DC Comic licenses.
Comment: I agree. Too bad these knockoffs are polluting eBay.
Answer: They are knockoffs, I have ordered superhero minifigs, as well as bulk minifigs, from Ebay and they are definitely fake. That is not to say they were terrible quality(although some were) but you can tell by the way they fit together they knockoffs. The paint, ease of attachment, and sometimes the overall design is a dead giveaway.
Answer: If the brand is not mentioned and the country of origin is China, you can be 99.9% certain that it's not LEGO but Lepin, Duo Le Pin, Kazi, XingBao or one of the many other Chinese brands.
Answer: Definitely look like knock-offs to me.  The plastic just doesn't look quite right, even down to the reflections.  I'd be concerned about buying knock-off LEGOs for a child - you have no way of knowing whether the materials used in the construction is safe, and also high risk of bits snapping off.  (although that's probably true for LEGO in general).

Authentic or Not?


Stuck at loading spinner after resetting iPhone 4 (iOS 7.1.2)

Question: I'm giving my iPhone 4 (iOS 7.1.2) to someone else, and I wanted to reset it. So I went into Settings -> General -> Reset, and I picked the option to reset all content and settings. However it's been stuck at a screen with a loading spinner for about 5 hours now. How long should I expect this process to take?
Answer: It sounds like something's gone wrong with the reset process, but you should be able to recover the phone without too much trouble.
First, open iTunes and connect the phone to your computer. If the phone appears in iTunes without displaying any error messages then you should be able to restore it from the iTunes interface.
If the above doesn't work you can force the device to turn off by holding down the Power and Home buttons until the screen cuts out. At this point you should be able to turn it back on and (at the very least) restore the device via iTunes.
If that doesn't work, try putting the device into DFU mode:

Turn off the device (if it isn't off already) by holding down the Power and Home buttons until the screen goes blank.
Hold the Power button for 3 seconds
Hold Home button without letting go of the Power button, then wait 10 seconds.
Keep holding Home and let go of Power. Wait about 15 seconds just to be sure.

(Source for the above: https://discussions.apple.com/thread/5409794)
From there the device will appear non-functional but you should have no problems restoring from iTunes.


JSON Viewer WebPart

Question: Is there a JSON Viewer webpart that functions in a similar manner to XML Viewer? Should I just use a DataView Webpart?
Answer: You could use a data view web part, you'd just need to point it to your JSON data source.
You could also build it as a javascript file and use a content editor web part to import the script onto the page.

JSON Viewer WebPart


Which are the seven rivers of the Saptasindhu?

Question: The land where the bulk of the Rig Veda was composed is called Saptasindhu or Saptasaindhava. Avestan sources refer to the hapta-hindu.  Is the term used after Rig Veda and is there a unique list of the seven rivers of the saptasindhu?
Answer: The phrase seven rivers was used many times in Rig Veda.  For example:
Rig Veda VII.67.8

With one, the same, intention, ye swift movers, o'er the Seven Rivers
  hath your chariot travelled. Yoked by the Gods, your strong steeds
  never weary while speeding forward at the pole they bear you.

Rig Veda I.35.8

The earth's eight points his brightness hath illumined, three desert
  regions and the Seven Rivers. God Savitar the gold-eyed hath come
  hither, giving choice treasures unto him who worships.

However, in Rig Veda X.64.8 & 9, apart from mentioning 7 rivers, names of 3 rivers, viz., Sindhu, Sarasvatī, and Sarayu, were mentioned.

The thrice-seven wandering Rivers, yea, the mighty floods, the forest
  trees, the mountains, Agni to our aid, Kṛśānu, Tisya, archers to our
  gathering-place, and Rudra strong amid the Rudras we invoke. 9 Let the
  great Streams come hither with their mighty help, Sindhu, Sarasvatī,
  and Sarayu with waves. Ye Goddess Floods, ye Mothers, animating all,
  promise us water rich in fatness and in balm.

It is quite peculiar that Ganga, Yamuna were not mentioned, but Sarayu was mentioned.  Was that Sarayu the same one that now flowing adjacent to Ayodhya or something different river is not forthcoming.
It might be, perhaps, due to the fact that at the time of composing these mantras, the sages either not aware of Ganga flowing in the East or the Ganga, which was described to be emanated from Shiva's head, might not have existed.
Answer: 7 rivers
The term sapta Sindhu occurs in

Rigveda 4:28

त्वा युजा तव तत्सोम सख्य इन्द्रो अपो मनवे सस्रुतस्कः । अहन्नहिमरिणात्सप्त सिन्धूनपावृणोदपिहितेव खानि॥१॥

ALLIED with thee, in this thy friendship, Soma, Indra for man made waters flow together, Slew Ahi, and sent forth the Seven Rivers, and opened as it were obstructed fountains.

Rigveda 2:12:12

यः सप्तरश्मिर्वृषभस्तुविष्मानवासृजत्सर्तवे सप्त सिन्धून् । यो रौहिणमस्फुरद्वज्रबाहुर्द्यामारोहन्तं स जनास इन्द्रः॥१२॥

Who with seven guiding rivers, the Bull, the Mighty, set free the Seven great Floods to flow at pleasure; Who, thunder-armed, rent Rauhina in pieces when scaling heaven, He, O ye men, is Indra.

But Rigveda talks about several other groups of rivers.
21 rivers

Rigveda 10:64:8

त्रिः सप्त सस्रा नद्यो महीरपो वनस्पतीन्पर्वताँ अग्निमूतये । कृशानुमस्तॄन्तिष्यं सधस्थ आ रुद्रं रुद्रेषु रुद्रियं हवामहे॥८॥

The thrice-seven wandering Rivers,(21 rivers) yea, the mighty floods, the forest trees, the mountains, Agni to our aid, Krsanu, Tisya, archers to our gathering-place, and Rudra strong amid the Rudras we invoke.

8 rivers
>उत नः प्रिया प्रियासु सप्तस्वसा सुजुष्टा । सरस्वती स्तोम्या भूत्॥१०॥

Yea, she most dear amid dear stream, Seven-sistered, graciously inclined, Sarasvati hath earned our praise.

I think that seven is special no. used with rivers.


Is Donald Trump officially the Republican nominee [as of May 7th]?

Question: Googling the phrase 2016 primaries indicates that Donald Trump is the "likely Republican presidential nominee" (emphasis mine).
While Donald Trump needs 1237 delegates to secure the nomination, he currently only has 1013. However, all of his opponents (namely Cruz and Kasich) have recently dropped out of the race.
My question is, why does Google only consider him the likely nominee? Won't he, by definition, win the remaining delegates and pass the 1237 threshold? Or can something still go wrong pre-convention that takes the nomination away from Trump?
Comment: What most people are saying is that Cruz and Carson are *mathematically eliminated* from the race. As you stated, Trump already has 1013 delegates, so even if either of the other two won any of the remaining states, Trump still wins the nomination, *barring unforeseen circumstances*.
Answer: No, Trump is not the official nominee until the Republican National Convention says so.
A number of things could go wrong (or right, depending on your politics) to prevent Donald Trump from becoming the Republican party presidential candidate.
The first thing to realize is the rules for "pledged" delegates vary by state. In the Republican Party system most delegates are "bound" by their states; their vote will go to their pledged candidate in the first round. If no one gets a majority in the first round, states begin to release their delegates from their obligations for the next rounds of voting. At that point it's chaos and delegates become free to make their own choices. Formerly bound delegates will become free to vote for any eligible candidate in the second round or later rounds. Eligibility is determined by party rules and can be changed at the convention. This is a simplification of a very complicated process, more detail below.
The second thing is that selecting a presidential candidate is a party process, not a legal one. The Republican Party decides how they select their candidates, and the Democratic Party has their own. This is further split up state by state. This is part of why the process is so weird.
That said, barring serious party shenanigans, if Trump gets the magic number of pledged delegates he is in... except for below.
He could die.
A similar situation happened in the 1872 presidential election when Liberal Republican candidate Horace Greeley died after the Electorial College had been chosen but before they had voted. All but three electors switched their votes to other candidates. This didn't throw the election, Grant already had it wrapped up.
What happens to Trump's pledged delegates if he dies varies state by state.
He could not get the necessary bound delegates.
Candidates typically don't "drop out" of the race, they "suspend their campaigns". Even if Trump is the only one on the ballot it's not assured he will be awarded the delegates. The rules vary from state to state.
If Trump fails to get the necessary bound delegates in the first round, voting goes into a second round. This is known as a contested convention and it's what Ted Cruz was hoping for. At this point things begin to unravel. The rules get complicated, both state and convention rules. Once unbound delegates can vote for anyone they want, or not vote at all, though it's unclear what votes will be counted. We'll get to that.
With Cruz out this is extremely unlikely. What is more likely is attempts at the convention to take bound delegates away from Trump with various procedural shenanigans, more on that later.
He could drop out.
I was waiting for Trump's April 1st press conference to declare his candidacy the greatest April Fool's Joke of all time. Alas, it didn't happen. But if it did, things would be in turmoil and the other candidates would come racing back to the campaign trail.
Republican National Convention shenanigans.
The 2016 RNC rules are currently not known. Yes, this is a little crazy. Just like Trump is the presumptive nominee, everyone is operating under presumptive rules.
The rules of the 2016 Republican National Convention are not law. Each convention decides its own rules. It's customary to adopt the rules of the previous RNC, and the tweaks since, but this is not required! Several key rules could be changed, or their interpretations changed, to "unbind" delegates, or change how votes are counted, or to allow bound delegates to simply not vote. Rule 16 and Rule 40 are the two big ones outlining how delegates votes are counted, and who is eligible for nomination.
Under the presumptive Rule 40, only Trump and Cruz are eligible for the nomination because they're the only ones who can "demonstrate the support of a majority of the delegates from each of eight (8) or more states". This rule can change. It was added in 2012 to prevent spoiler candidates, but that's exactly what the 2016 anti-Trump people want.
The presumptive, and heavily revised since 2012, Rule 16 deals with rogue delegates. Long story short, delegates cannot act against their state's rules. If they do they are considered to have resigned and voted according to their state's rules. Like many parliamentary systems, there's some wiggle room in the rule which allows the Secretary Of The Convention to pick and choose which violations they pay attention to. And, again, Rule 16 may change.
It's complicated. Pulling the presumptive rules apart is its own answer. Even then they might not be the 2016 rules.
Such shenanigans are a part of every RNC. An overt move would shatter the Republican Party and the people's already shaky faith in their nomination system. Expect a lot of rules lawyering. This is the big unknown.
A similar situation happened at the 1912 RNC to protect President Taft's nomination as the incumbent from the popular former president Teddy Roosevelt. Denied the Republican nomination, Teddy ran in 1912 as a 3rd party candidate.
Sources

Rules Of The Republican Party As Adopted At The 2012 Republican National Convention
Everything you need to know about delegate math in the presidential primary.
Republican Delegate Allocation Rules By State
The Real Import Of Rule 40 in 2016
Throwing Out Convention Votes?
You may find more answers in this related question eventually.
Comment: "I was waiting for Trump's April 1st press conference to declare his candidacy the greatest April Fool's Joke of all time." You were not alone.
Comment: Please keep comments about the answer. If you want to talk about Trump, please [take it to chat](https://chat.stackexchange.com/rooms/6619/agora).
Comment: I would add that even though he is not yet officially the nominee, he is almost certainly going to be.
Comment: @Brytham, the wording about picking "anyone they want" is incorrect.  Only eligible candidates can be on the ballot at the convention at any point.  There is no threshold after which eligibility rules don't apply.
Comment: @BenCollins I did some more research and heavily updated the answer with more sources and the presumptive rules. No version so far has been entirely correct. There's a lot of rules to keep delegates in line, they're complicated, and they might change. To be technically correct would require its own answer. Instead I summarized the important parts as best I could and made it clear what parts are a simplification.
Comment: @Schwern I appreciate the clarification, but this still doesn't address my very narrowly focused edit: there is no set of circumstances under which a delegate can vote for "whomever they wish".  There is no such thing as a write-in at the national convention.  Delegates can only cast a vote for someone eligible to be on the ballot.  Your clarifications are helpful in that they explain that the rules have not yet been approved by the convention and are still subject to modification, but there is zero chance that the rules will be modified in such a way that "anyone" would be eligible.
Comment: @BenCollins I think we're having a terminology argument, [let's talk about that in chat](https://chat.stackexchange.com/rooms/39474/rnc-2016).
Comment: I may have missed it, but I just wanted to add that the significance of a candidate _suspending_ their campaign is that they actually still remain candidates and therefore are still eligible to become the nominee if the front runner has not amassed the required delegate votes.
Answer: Even though it is "probable" even above the 99% level, he is still not officially nominated by the Republican convention. Even after he has 1,237 delegates committed to vote for him (no matter which ballot), that would only make him the presumptive nominee. That is, we assume that he will become the nominee, but he has not yet become so.
Google "Republican presumptive nominee" and you will see the term used.
After all, a meteorite could still (G0d forbid) hit the airplane Mr. Trump is in before the convention.
Comment: @user8058 As I said, any number of things could still happen no matter how unlikely or improbable.
Comment: I think a fair follow-up (and might be what the OP was getting at with his/her last sentence) is: can the Republican establishment do anything from this point forward to *prevent* Trump from being officially nominated by the RNC?
Answer: He is not the official nominee of the Republican party until the National Convention has met and has officially elected him.
However, considering that he will likely run unopposed, this is basically just a formality. Expect the RNC to become a big PR event for him. The only interesting question which will likely be answered there is who his running mate will be.
Comment: Considering that anyone who would run with Trump would essentially be committing career suicide, it would be hilarious if he couldn't find anyone to run with him. Or worse yet, end up with Sarah Palin as VP...
Comment: He is the presumptive nominee.
Comment: @DrunkenCodeMonkey My assumption would be that he'd try to get along without a VP ... :)
Answer: A Nomination will be made at the Republican National Convention in  Cleveland, Ohio from July 18 until July 21, 2016.
If Trump is nominated at that convention and He recieves enough votes to win the nomination, he will then be refered to as the Republican Party's presidential Nominee. Until then it is presumed by most people that he will be the nominee hence he is the "presumed nominee"
Answer: Other answers have already addressed the point that the Republican National Convention will nominate a candidate, but I'd like to address an additional point.

(...) why does Google only consider him the likely nominee?
(Emphasis mine).

Because Google is just a search engine. The results they provide merely reflect what others on the Internet are writing about it.
Not only does Google not have a say in who the RNC will nominate, they aren't a media outlet either. While some journalists / bloggers / pundits may draw that conclusion, Google's job is not to draw conclusions, just to report back what can be found about a subject on the Internet.
Comment: That is NOT actually always the case, despite Google claiming so. They do actively curate search results, and tweak search algorithms (and do other, far more overt things) to influence discourse.