How did the Slap Bet count develop?

Question: In How I Met Your Mother Barney lost the Slap Bet some time ago. Being a somewhat inconsistent viewer, I can't recall the episodes that fulfilled the bet, and in my mind, at least 1 is still due. What is the count outstanding on this?
Answer: In "Slap Bet" (Season 2 Episode 9) Marshall was granted 5 slaps to be delivered any time in the future without warning.
In "Disaster Averted" (Season 7 Episode 9) Marshall was granted 3 slaps of the same kind.
5 + 3 = 8
out of which
6
slaps have been executed.
The 1st slap was executed in "Slap Bet" (Season 2 Episode 9)
The 2nd slap was executed in "Stuff" (Season 2 Episode 16)
The 3rd slap was executed in "Slapsgiving" (Season 3 Episode 9)
The 4th slap was executed in "Slapsgiving 2: Revenge of the Slap" (Season 5 Episode 9)
The 5th and 6th slap were executed in "Disaster Averted" (Season 7 Episode 9)
Hence, there are
2 slaps remaining.
On a related note, the 7th slap has a countdown at Slapbet Countdown! :D
Update: Both those slaps were delivered before the ending of the show.
Answer: According to the How I Met Your Mother wiki, the original five slaps have all been used. However, Barney offered Marshall another three as payment on a separate bet, one of which has been used. Two remain as of this date.
Comment: If I recall, when they did the extended slaps (The Ducky Tie episode), Marshall still had one of the 5 left.  But he used two back to back.  I could be wrong though.  But I do know the extended slaps came from the Ducky Tie.
Answer: All slaps were used.
In season 7 episode 9 "Disaster Averted" Barney managed to get Marshall and Lily to agree to 3 additional slaps in exchange for letting him take the Ducky Tie off. Marshall then had a total of 4 slaps which he proceeded to use 2 right away. This left Marshall with 2 slaps.
Then is season 9 episode 14 "Slapsgiving 3: Slappointment in slapmarra" Marshall told the story of how he learned the slap of a million exploding suns in an attempt to scare Barney. He slapped him under a tree next to four women and a tiger.
The last slap took place in season 9 episode 22 "The end of the aisle". Towards the end of the episode while Barney waits for Robin to walk down the aisle, Barney starts to freak out again so Marshall slaps him thus freeing him from the slap bet.

How did the Slap Bet count develop?


Resource for learning straightedge and compass constructions

Question: Does anyone know a good resource for learning about straightedge and compass constructions besides "The Elements?" I tutor geometry to middle-schoolers and high-schoolers and thought that including construction problems might give them better intuition for the subject. I know most of the constructions in book 1 of The Elements, but would appreciate it if anyone could point me in the direction of a book or other resource that has interesting construction problems, hopefully graded by difficulty.
Answer: For compass and straightedge construction (personally, I still learn best by manually doing what I'm learning), see, e.g.,:

Geometric Constructions  (mathopenref.com) gives a brief history and provides references and links to dozens of geometric constructions, each with detailed instructions.
Compass-and-Straightedge (a site created by David Eppstein with dozens of links to specific constructions.)

The following isn't a recommendation regarding pencil and paper geometric constructions, or resources that involve literal use of compass and ruler, but I've found that software like geogebra is extremely helpful when learning and/or teaching geometry, and is useful and appealing to pre-teens, all the way up to post-secondary students (and their teachers and tutors!)
Given that today's children and teens typically have developed facility with and attraction to computer applications, it is appealing to even the most resistant students!
The software helps students readily construct and manipulate geometric objects, as well as better understand, through immediate visualization, how changes in certain parameters affect resulting geometric constructions.
You'll find many applications, teaching lessons, tips, and such at geogebra.org (linked above, where you can download the software, and find links to curriculum, downloads, teaching tips, and discussion boards, etc.).
Note: Geogebra is free and well-supported software available for download to anyone and everyone.
One additional helpful package, very user friendly, (there's a minimal expense, but a lot of support) is the Geometer's SketchPad (see resources available at Dynamic Geometry).
Answer: I don't know anything really modern except DIXON which may or may not be what you want. There are two relevant sections in DORRIE.
Note that, more for yourself, there is a fair amount on such constructions in the hyperbolic plane. There is a good deal of unfamiliar stuff to learn, but the one simple statement is this: an angle is constructible in the hyperbolic plane if and only if it is constructible in the Euclidean plane. Here an angle is between two lines, or between a line and a circle that meets it, or two circles that meet. Who knew?
You might want to look at publisher's websites, places that publish school material. There is one near me (at least it was), Key Curriculum Press. Maybe they have something at the correct level and pacing.
HERE is a construction problem i solved on MSE. It could easily be in Euclid, which I have never read.


How efficient is allocating workers on auto mode?

Question: I would like to manage workers on my own, but as the city becomes larger, I found it really hard to manage them by myself. If I put workers on auto-mode, how efficient would they be? are there any limitations as to what auto-mode cannot do or will not do?
Answer: I've been pretty happy with automated workers so far.  One thing they will do is adjust things dynamically as your cities grow.  They'll swap trading posts for farms if the city needs more food and that kind of thing.
If your cities are efficient you probably will end up with trading posts on nearly all non-hill tiles, though, which is a little boring-looking.
One further note: I did notice that the auto-workers built a trading post on a marsh without actually clearing the marsh (which could ONLY help the city), so that wasn't ideal.
Comment: If you look at it defense-wise, not clearing that marsh for the 1 extra food is worth it in war time.
Answer: At the start of the game, I usually manage the first two workers for a while. I also find the auto setting of connecting cities by road is slower to eventuate than what i need to move an army or to make my autoworkers move around the empire faster. So I will usually interrupt a couple of workers to manually get the job done.
Also, when there are barbarians nearby, auto workers seek protection. If the workers are manual you have a choice as to how in danger they are so you may keep them working. Just don't forget about them or find out they have no escape. The only down side is every turn you need to give your worker the same order until threat is gone.
Also, I tend to look around the map for any useless road/railroad tiles and grab a worker whose job is to go around and remove them. +1G/+2G for removed road/rail which can add up over many turns.
Answer: I saw that automated workers like trading posts a lot. I prefer to switch between farms and mines for nearly every tile so this might be a problem.
Also I've realised that there is some kind of "best road path" between two cities so if you build a road manually to connect two cities, your automated workers might realise that there is no road on the "prefered tiles" and build a second road.
I usually micro manage my workers until there are no more tiles that can be improved


What is an acceptable on-call agreement between employee and employer?

Question: IT is full of heroics; many of us carry a paging device, like Batman waiting in his cave for the Bat Signal. What's a fair compensation for on-call schedules? Is it acceptable to handle emergency requests on weekends or vacations?
Answer: Seems like this would depend entirely on the individual and the workplace...
I work at an e-commerce company, and the website needs to be up 24x7, which means someone in the admin group is also on call 24x7.  We rotate being on call for a week between 4 people... the number of calls can vary from 0 calls for the week (rare), to getting virtually no sleep for the entire week (also rare).  Usually it's a few early AM pages of varying severity, but for that week the on call guy is basically chained to a laptop & sleeps with his cell phone.
Compensation here comes in the form of a flexible work schedule, being able to take personal days as needed, great management & an enjoyable workplace.  Personally, I just consider being on call to be a necessary evil that comes with this particular position, and I think the perks are worth the stress.  Then again, quite a few would say it's not even close to being worth it.
People's expectations would probably depend on business hours, the importance of the systems to the business, and both tangible & intangible benefits provided by the employer.
Comment: ~300 physical boxes & ~75 VMs last time I counted, but probably only a 1/4 of those are critical systems.
Comment: wow, impressive. Out of curiosity, how many servers do you support (assuming servers are the majority of your workload)?
Answer: For me, this depends 100% on expectation.  If the employer is completely up-front about this, they're going to keep (and deserve to keep) employees.  It needs to be brought up during the interview, and if the reality doesn't match up to the picture they painted, you need to document, and bring it to your review.  You agreed on a salary based on faith that they fully disclosed the company's demands on your personal time.
You're not going to be able to change business needs.  The only real use for this information is as leverage for salary bumps.
Vacation is a different story, in my book.  If I have a stand-in, and I've documented properly, there's no excuse for calling me.  It's not actually a vacation unless my phone is off.  I've helped shape this expectation by bending over backward to keep coworkers from being called on their own vacations.
The real answer is to be 100% clear on what your limits are in the interview.  It's ok with me if that costs me a job that would make me unhappy.
Answer: IMO, in a larger environment it should be a combination of off-shift staff (typcially entry-level), paid oncall for most technical staff and unpaid oncall for senior technical staff.
That sets up an environment that is healthy:

The technical staff needs to document stuff so the off-shift people can take care of routine things
The mainstream technical staff is fairly compensated and not getting paged 3 times a week
The senior technical staff has a disincentive to hold knowledge too close, as they will be woken up if they do.
Comment: Three new answers for a fairly old question. Wonder how this came up.
Comment: Not sure, I didn't look at the date. Someone must have submitted an answer that popped the topic to the top of the list.
Answer: One thing that's important to take into account is "on call" v.s. "state of readiness" (call 'em what you will) when considering just compensation.
The way I differentiate between them is basically by figuring out what I'm allowed to do.
For me, I feel an added 10-20% salary is fine if I'm expected to carry the pager and handle/redirect any incoming issues.
For the above, I can reasonably expected to be allowed to drink a beer or two, with a on-site time of an hour or two. I.e., be allowed to have a social life.
If I'm expected to be on-site in five minutes, keeping a constant watch on systems (despite having alarmsystems), I'd most likely shoot for 200-300% daily salary, with a one week rotation.
It can be difficult to get the second version approved, and so my favorite is a bit of a hybrid system.
I generally know my systems well enough that problems don't occur, so I'm happy with 10-20% pager bonus, but on the occasions an issue does occur, I want fair compensation i.e., per-ticket compensation.
How much that compensation is is up to you and your boss, obviously, but a few of the "soft" parts of the package I like to include are:
* Guaranteed time to FIX the problem v.s. WORKING AROUND IT.
* Flex time (while it's true that I'd be compensated for any issues, getting in at 8am or 9am after staying up until 4am is hardly productive).
Another great thing is reviewing cases.
Even if it just looks like a one-time issue, other people on your team may very well have useful input, and the review process itself (no, not just a incident report, but an actual honest-to-god face-to-face meeting with your team and your boss) will help you stay proactive rather than reactionary.
A quick example would be noticing that a server had a disk fill up overnight (possibly despite having been 30% free the evening before).
The reactionary thing to do would be to investigate and clean up whatever went wrong, as well as write up an incident report.
The proactive thing to do would be to investigate and clean up whatever went wrong, write up the incident report, plan a look-over on the remaining servers, order new hardware etc.
As a sidenote, that last part fits in very well with the "Guaranteed time" I mentioned earlier.
Answer: This can vary widely between employers and environments.  Where I'm currently at, we're expected to be on call, but only when it is our turn in the rotation.  Our work schedule is pretty flexible, so if you're up late the night before because of an issue, you can usually come in late the next day or even work from home.  I agree with Kara Marfia that this is something that needs to be agreed upon in advance, and that things should be spelled out pretty plainly, both what they expect and what you expect.
One thing that makes our system a bit better is that we have a Network Operations Center that is staffed with pretty bright people.  They can usually resolve issues that are well-documented without needing to page or call.
Answer: I would like to think that I am the exception, but my research has shown that there's a decent percentage (~10%) of administrators who are the sole admin in a 24x7x365 computing environment who have >50 servers. (See my research and the results of my 334-admin survey here: http://www.standalone-sysadmin.com/blog/2009/05/happy-1st-blogiversary/)
Essentially, your job as an admin is either to support a 24 hour operation or it isn't. You're hired to be on-call at times, then it's acceptable. If you weren't hired to be on-call and your employer wants to make you on-call, then that needs to be a two way street. You need reimbursed in one way or another for your extra investment in the company. Whether you take that as monetary or temporal is between you and your employer. If they don't want to reimburse you more and you disagree, chances are good that it is an at-will employment.
Ultimately, you must be the final judge of "fair", but reaching a mutually beneficial agreement should definitely be possible.
Comment: It seems like your survey was inadequately ranged. I don't have access to it, but if 50 percent of your survey sample has 200+ users, you need to broaden the scope a bit. Maybe even ask for a direct estimate, so you can calculate things like users per server. and servers per admin.
Comment: I completely agree. I even mentioned that in the blurb on the blog. You can get the raw CSV data here: http://www.noneck.net/standalone/raw_results.csv

I'd like to do the survey again this year with a little better distinction on the high end. My blog caters to small infrastructure admins. I had no idea that 50% of my readers would admin over 200 machines. Unreal.
Answer: This sounds like a basic economics question to which the answer is "what the market will bear".
Apart from actual monetary compensation I have other expectations if I am on call.
For example if I get woken up at 3am and because of that I don't get back to sleep until say 5am, then I don't think it is reasonable for my employee to expect me to be in the office at 9am.
I'f I'm up say 3-4 hours or more, don't expect to see me at all the next day unless I need to come in to clean up after the on call incident. And yeah I expect to get paid for that day I don't come into the office.
I also think that if the employer is not willing to make sure you have appropriate road warrior tools (i.e. wireless broadband and a laptop that is portable enough to carry every where) then the on call rate should go up.
If on call really means not being able to leave home, then I think it is worth more money.
Answer: Federal law has very specific rules about how much a person must be paid for being on call.  The first question is whether you are free to do what you would do ordinarily away from work, or whether you have restrictions on your activities and whereabouts.  If you cannot go and do as you please because of the conditions necessary for responding to calls, you must be paid for that time.  For example, you like to go fishing at a park 25 miles from home.  You cannot do that when you are on call, because you have to be able to get to the office within fifteen minutes.  This would arguably be time for which you had to be compensated by the hour.
Hourly computer professionals must be paid for every hour they work, but only at straight time if they are "exempt" from wage and hour laws.  An increase in salary means either 1) you are salaried exempt, which means you receive a check for your weekly (rate x 40 hours) salary even if you work 35 hours rather than 40.  It also means you receive that check if you work 55 hours -- no overtime, no comp time.  In the alternative, if you are salaried but not exempt, you must receive overtime for all work over 40 hours in a week.  Private employers may not substitute comp time for overtime pay.
Check out the Department of Labor website, subcategory Wages and Hours.  You should find a lot of useful information there.
Answer: As everyone else has already said - it depends. If it's part of the job, and stated up-front, you have nothing to quibble about.
In my current role I am the entire IT "department". I am not on call. Nor am I expected to provide 24x7 service. However, In the event my services are required out of hours I expect to have to do whatever is needed and this has been factored into my salary. Fortunately, most issues are solved with a phone call or a few minutes connected via the VPN, which is handy because it's a minimum of an hour's drive to the office. There will always be some work that must be performed out of hours (maybe two or three days a year) and for that I receive time off in compensation.
Answer: 2 person rotation 24x7 365 days. (25 weeks each)(300-500 users)  We receive a comparable pay and benefit package as salary.  For on-call, we recieve 4 hours of comp time per week.  The problem is, HR does not recognize "comp time". If we dont take it and we are laid off or let go, we lose it.  The next problem is, in the past year or so, the manager is no longer willing to allow us to use the comp time as freely.  I am woke up 3 to 4 out of 7 days.  I average a call a night.  Most issues are locked accounts or printer issues easily solved by logging in thru the VPN.  (for me, any wake up call creates an hour or 2 of up time until i can get back to sleep)  Now the manager is threatening to take away comp time all together!  Additionally, after being up all night a few weeks ago, (last email was 3:45 am), I was charged comp time when I took the next day off!  Myself and the other person are trying to find good solid information on what is "standard" and "acceptable" for on-call compensation.  If the rotation were larger, it would not be so bad.
Answer: As far as I'm concerned, if you're on call and must be available (which interferes with your "off" time), you should be paid full hourly rate for every hour you are on call.  It is so disruptive to your personal life that you should be compensated for contributing to the business and this should be considered overtime.  Employers are taking advantage of people and trying to force us all into a 24/7 society and unless the government labor laws protect us, we can only fight for our rights.


Sequence of constant rank matrices

Question: Let $A_k$ be a sequence of real, rank $r$, $n$ x $m$ matrices such that $A_k$ converges to a rank $r$ matrix $A$. Let $v_k, u_k$ be sequences of vectors such that $u_k\rightarrow u$ and $A_k v_k=u_k$. I would like to know if it is possible to show that there exist a vector $v$ such that $Av=u$. Appreciate any help.
Answer: I think it is best to settle this problem geometrically, that is if you think of matrices as linear maps from $\mathbb R^m$ to $\mathbb R^n$. The images of these maps are $r$-dimensional linear subspaces of $\mathbb R^n$. Let $X_k$ denote the image of $A_k$, then $u_k\in X_k$, and you want to prove that the limit vector $u$ belongs to the image of $A$.
Let $X$ denote the set of all possible limits of sequences such that the $k$th element of the sequence belongs to $X_k$ for every $k$ (for example, $\{u_k\}$ is one such sequence, hence $u\in X$). Clearly $X$ is a linear subspace of $\mathbb R^n$, and it contains the image of $A$. And the dimension of $X$ is no greater than $r$. Indeed, suppose that $X$ contains $r+1$ linearly independent vectors $w_1,\dots,w_{r+1}$. Each $w_i$ is a limit of a sequence of vectors $u_k^{(i)}\in X_k$. For a sufficiently large $k$, the vectors $u_k^{(1)},\dots,u_k^{(r+1)}$ are linearly independent because the set of linearly independent $(r+1)$-tuples is open. This contradicts the fact that $\dim X_k=r$.
Since $X$ is a linear subspace of dimension at most $r$ and it contains the ($r$-dimensional) image of $A$, it must coincide with that image. Since $u\in X$ by definition, it follows that $u$ belongs to the image of $A$, q.e.d.
Comment: Very      nice.
Answer: It suffices to show that you can replace the original sequence $v_k$ by a new one that is bounded.
You should start by figuring out what can go wrong, i.e, how is it possible for the sequence $v_k$ to be unbounded?
Comment: That's true. (you mean $v_k$)

Sequence of constant rank matrices


Dry hopping in secondary...

Question: What is the time frame for dry-hopping in secondary?  How soon can I bottle, and how long before tannins/off flavors leech out of the hops?
Answer: There is a wealth of information from three guys who know hops on the dry-hopping episode of Brew Strong.  Jamil, John Palmer, and Mike "Tasty" McDole (who won the Sam Adam's Longshot Competition with an awesome IPA) cover all aspects of dry hopping your beer to get the best results.
They reveal that you only need a few days of dry hopping to get great hop aroma and how long before before grassy or vegetal off-flavors begin to come out of the hops.
Answer: I try and give my dry hop 1 week 2 at the most. Hop aroma quickly diminishes so I try and get the beer out and kegged or bottles after a week. When "life" happens and you find yourself dry hopping for 3 or 4 weeks I notice less aroma but never off flavors. It must take an extended period before any negative effects begin.

Dry hopping in secondary...


Teichmuller Theory introduction

Question: What is a good introduction to Teichmuller theory, mapping class groups etc., and relation to moduli space of curves or Riemann surfaces?
Comment: http://www.math.harvard.edu/~ctm/home/text/class/harvard/275/09/html/base/rs/rs.pdf
Answer: The primer on mapping class groups, by Farb and Margalit.
Answer: John Hubbard has a recent book on Teichmuller theory which is quite good and geometric.
Comment: Its a good book, but it builds up alot of technique before it gets to defining Teichmuller spaces.  It makes it a wonderfully self-contained resource, but it can also be daunting to someone trying to read it casually.  A more mathematically mature reader should be encourage to skim/skip to the good stuff, and then refer back as needed.
Comment: This book would be on the far topologist-friendly end of the spectrum of books on the topic.
Comment: Absolutely true: it is the first of a planned series exploring four of Thurston's big theorems from the 70s.
Answer: In addition to the ones already mentioned:
J. Harer's lecture notes on the cohomology of moduli spaces (doesn't have all the proofs, but describes the main ideas related to the cell decomposition of the moduli spaces; Springer LNM something, I believe; unfortunately I'm away for the holidays and can't access Mathscinet to find a precise reference).
K. Strebel, Quadratic differentials (careful exposition of the complex analytic results used to construct the cell decomposition mentioned above; not much about moduli spaces or Teichm\"uller theory though; Springer Erbebnisse).
L. Ahlfors, Lectures on quasi-conformal mappings (construction of Teichmuller spaces).
L. Ahlfors' and L. Bers's papers in Analytic functions, Princeton, 1960.
Answer: I find "An Introduction to Teichmuller spaces" by Imayoshi and Taniguchi to be a pretty good reference. Its advantage over Hubbard is that it exists on gigapedia, but I don't know how it compares to the other books in this list.
Comment: this is the only book possibly which does calderon-zgymund to prove the existence of solutions for beltrami equations
Answer: If you're more analytically minded, I recommend
Gardiner and Lakic, Quasiconformal Teichmuller theory
and
Nag, The complex analytic theory of Teichmuller spaces
Answer: Hubbard's book is by far the most readable for the average good student -- I don't think it makes sense to begin with anything else right now. When the projected series is finished,it should be the definitive introduction to the subject.
For connections between all these subjects,there's probably no better current source then Jost's Compact Riemann Surfaces. Although the treatment of Teichmuller spaces per se is brief in the book,it contains a wealth of other important topics related to Riemann surfaces. Like everything Jost writes, it's crystal clear if compressed within an epsilson of readability. Jost makes up for the density of the text with its clarity.
Answer: McMullen's notes (http://www.math.harvard.edu/~ctm/home/text/class/harvard/275/09/html/base/rs/rs.pdf)
Answer: Looking at my bookshelf, there's a few other books that come to mind with varying levels of relevance:
Ahlfors & Sario. Riemann Surfaces.
Jost.  Compact Riemann Surfaces.
Maskit.  Kleinian Groups.
Tromba.  Teichmuller theory in Riemannian geometry.
Farkas and Kra.  Riemann Surfaces.
For my own purposes the Hubbard book is what I'd consider a natural starting point.
Answer: Ivanov has a nice review of much of the theory of mapping class groups here.  The emphasis is on mapping class groups rather than Teichmuller theory, but the latter is certainly discussed.  I find this to be a very useful reference.

Teichmuller Theory introduction


How to keep LiveCD settings?

Question: As I am now only able to run my machine from liveCD (Fedora XFCE 16 or Xubuntu 11.10), I would like to know if there is a way to keep the current session's settings (keyboard, wifi etc.) for a later session. Can the settings be stored on a USB drive for instance?
Answer: You can create a bootable flash drive with the version of Linux you choose, but by nature live cds can't preserves settings.  The memory on the disk can't be modified, and the running os in ram can't be preserved. To find out how to create a bootable flash drive, most distros provide instructions. In other words instead of saving the settings on a flash drive, keep the entire os on the usb stick. Make sure that you aren't using the drive as a "live cd" itself, but actually installing the os to the drive. Here's an example http://www.pendrivelinux.com/usb-x-ubuntu-610/ Here's a better link http://www.pendrivelinux.com/universal-usb-installer-easy-as-1-2-3/
Your other option is to find the the configuration files the are used by those individual programs and load them when you start the cd, i.e  from usb

How to keep LiveCD settings?


Formal requirements analysis

Question: I've been doing a Requirements Analysis module in my university. We've covered all kinds of formal requirements stuff, like UML, Use Case Diagrams, Sequence Diagrams and Contracts. I have to ask- are any of these things of any value whatsoever in the real world? The only Use Case Diagrams I've ever seen could be summed up in a vastly more succint and understandable way in plain English- or even code in a reasonable language of your choice- and the same goes for the other things I've mentioned.
Comment: the point of these diagrams and techniques in the long run is to internalize how to think about problem requirements, even if you never draw another use-case diagram in your life.
Answer: Requirements analysis is a process that uses tools like UML diagrams.
Diagrams facilitate communication and sometimes can be used to generate code (e.g. 1 and 2 below). They also aid in adding a visual layer to the solution that could some time help the end users and developers (e.g. 4, 5 below).
Use Case diagrams represent high level view of the interaction and don't replace the textual case description.
Some of the very important diagrams for showing requirements and design are:
1. Entity relationship diagrams
2. Class diagrams
3. Sequence diagrams
4. BPM - Business Process Models
5. Page Flow diagrams
6. Use Cases
Requirements analysis is a large subject and may use other diagrams as well.
The important thing is that you capture and communicate your knowledge about the system requirements to the developers and the end-users.
There are suite of tools that can validate some of the above diagrams (e.g. 1,2). Some tools can build a web applications automatically from (1 or 4).
Yes the diagrams are important and are used in some projects.
Comment: Plus: use case and many other diagrams are not intended solely for the developers, but for end users and stake holders that are not well versed in any other technique but pick up on these diagrams pretty immediately.
Answer: School problems are over simplified for learning. So you can't make any value judgements based on them as to the utility in the real world. Next most people who do requirements Analysis are NOT programmers and certainly could not express the requirement in code.  Third, these are most useful when doing large complex projects, so the experience of people working on smaller things may indicate that they are less than useful where the experience of people working on space shuttle type projects might be different.
The last large project I worked on had an inital requirements document of over 800 pages (which grew to be much larger by the end) sometimes having a diagram makes it easier to understand the written word.  It also helps to keep people things in terms of the process that needs to be created and helps prevent silliness such as requiring the software to have an approval from a manager fo rthe budget expenditure for a project of more than $1 million but no requirement that tells you what to do do if he rejects the project. When you use a structured approach, you are more likely to see all the pieces I think.
Are these techniques appropriate for all projects always, no of course not.  Agile projects tend to take a less complex approach to begin with and these techniques might be perceived as time wasters. From what you wrote, I would suspect you are someone who is more comforatble with the agile approach and thus may not use these techniques or see them in the projects you work on. What is critical about learning them for you may be more the analytical thought process that goes into doing them rather than the actual diagrams themselves.
Answer: Yes, these things have value in the real world.
You don't always use all of them, and you don't always polish them to perfection, but they are all useful tools to model the problem domain and be sure you're going to build the correct solution.
Each of these artifacts exists because they are better ways of documenting what's going on than plain text.  If plain English worked better, we'd use that.  (Where plain English works better, we use do use textual requirements!)
Don't be fooled by the simplicity of these things in the learning environment.  They can get quite complex in the real world.  Or, alternatively, sometimes they are not complex -- and thus become a simple way of conveying understanding.

Formal requirements analysis


Difference between "susto" and "aprensión"

Question:
Del susto y la aprensión, el rugido ha cesado.

What are the differences between "susto" and "aprensión"?
Both mean fear, no?
Or they have different degrees of fear?
Comment: I think if you look at them in a dictionary (RAE for example) the difference is evident.
Answer: When meant as "fear", aprensión is softer than susto. It certainly does not mean "fear" in a physical manner. What I mean is that you can feel susto in your skin. Your body can react to susto, but aprensión is just a conscious thought, an idea that doing something might not be in your best interest or convenience.
For example, you are on the way to see your friend John when you see his fiancée kissing another guy in a dark corner of a park. The feeling you experiment when you think about whether to tell John about this, or how to tell him, is aprensión. It is uncomfortable, awkward, and you even ask yourself "why did I have to witness that?". You are afraid of telling John what you saw, but you don't really fear the situation, you are not frightened.
On the other hand, if you happen to walk by a dark corner of a park and you see a gang of tall guys with guns walking toward you, then you will fear them, you will be asustado ("tengo susto"), but aprensión has nothing to do there. This level of fear you can feel in your sweaty skin or your growing pupils, but in the previous example, you only feel it "in your guts", it's something internal.
Answer: From RAE:

susto.

m. Impresión repentina causada por miedo, espanto o pavor.
m. Preocupación por alguna adversidad o daño que se teme.

i.e. Impression caused by sudden fear, terror or dread.

aprensión.

f. aprehensión.
f. Escrúpulo, recelo de ponerse alguien en contacto con otra persona o con algo de que le pueda venir contagio, o bien de hacer o decir algo que teme que sea perjudicial o inoportuno.
f. Opinión, figuración, idea infundada o extraña.

i.e. Qualms or misgivings. Scruple of someone get in touch with someone or something that you can come contagion, or of doing or saying something that he fears is detrimental or inappropriate.


Mehudar: a Subset of the Daled Minim?

Question: I remember seeing a teshuva about this once. Could be it was even here. If someone has daled minim and one of the minim is not "mehudar" does it then "take away" from the hidur of the other minim. That because this one is not mehudar it can no longer be called "mehudar daled minim". Thus, if someone has the option to buy one min that is mehudar even though the other ones won't be, is there an advantage in this?
Answer: I am not sure I will be able to locate the particular source you refer to, though my perusal of the Yad Moshe index would imply that it isn't in the Igros Moshe. If I find something I'll try to include it.
It would seem, however, that if one is able to obtain a single min (species) that is mehudar one should do so. Practically at my own Shul, and I imagine many others, the only difference between the standard and deluxe arbah minim is whether the Esrog is mehudar.
This seems to be fairly straight forward in the Shulchan Aruch O.C. 666:1.  The second understanding brought by the Mechaber there would seem to be the primary source (at least as far as it is codified in the Shulchan Aruch) for purchasing mehudar arbah minim. It explains (if I'm reading it correctly) that one should pay an additional third to purchase an esrog with greater hiddur. Although the Poskim (see M.B. 2 citing the M.A.) point out that this applies to other mitzvah items as well, which would in principle include sha'ar arbah minim) the advice to choose the more mehudar esrog was in no way contingent upon the hiddur of the rest of the arbah minim.

Mehudar: a Subset of the Daled Minim?


Show ${156 \choose 87} + {156 \choose 86} = {157 \choose 87}$

Question: How can we show the following equality?
$${156 \choose 87} + {156 \choose 86} = {157 \choose 87}$$
Answer: Algebraic proof:
$\binom{n}{k+1}+\binom{n}{k}=\frac{n!}{(k+1)!(n-k-1)!}+\frac{n!}{k!(n-k)!}=\frac{n!(n-k)}{(k+1)!(n-k)!}+\frac{n!(k+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)!(n-k)!}=\binom{n+1}{k+1}$
Combinatorial proof:
If we want to choose $(k+1)$ objects out of $(n+1)$ objects, we either choose the first one and then choose $k$ more from the remaining $n$ objects or not choose the first one and then choose all $(k+1)$ from the remaining $n$ objects.
Answer: In general, $$\dbinom{n}r + \dbinom{n}{r-1} = \dbinom{n+1}r$$
There are many ways to prove this, a simple way is just to manipulate the left hand side algebraically to obtain the right hand side i.e.
\begin{align}
\dbinom{n}r + \dbinom{n}{r-1} & = \dfrac{n!}{r! (n-r)!} + \dfrac{n!}{(r-1)! (n-r+1)!}\\
& = \dfrac{n!}{r! (n+1-r)!} \left(n+1-r + r \right)\\
& = \dfrac{n!}{r! (n+1-r)!} \left(n+1 \right)\\
& = \dfrac{(n+1)!}{r! (n+1-r)!}\\
& = \dbinom{n+1}r
\end{align}
Take $n=156$ and $r=87$ to get what you want.
Answer: Hint: Given a set of $157$ objects and you are choosing $87$ items out of it, you can either take the first element or not take it. How many ways are there to pick the $87$ objects in each case?
Answer: I think this is the best way to do it. Using Wolfram Alpha, we see that
$$\begin{align}
\binom{156}{87} &= 2071356239566075202805184663760798068902414000\\[12pt]
\binom{156}{86} &= 2574399897746407752057872367816991885635857400\\[12pt]
\binom{157}{87} &= 4645756137312482954863057031577789954538271400.
\end{align}
$$
From here it is easy to check that the desired equality is true.
Comment: Thank you for your input, but this was a problem I had on a test...


Rasterize a vector layer in PostGIS

Question: Related to this question asked here on gis.se and this thread on postgis-user, has anyone worked out a good solution for in-database rasterization of a vector layer using PostGIS?  It sounds like the necessary functions are available and it is possible to get a valid output, but not one that is readable by GDAL et al. because it has irregular tiles.
To be more specific, I would like to rasterize two distinct vector maps of the US, counties and NBCD mapping zones, to a particular resolution and both aligned with an existing raster.  It is possible to do this with gdal_rasterize, but with some trial and error as I recall -- mainly because the extents of the three inputs do not match, I expect.
Any thoughts?
Answer: Here is a guide on how to rasterize a vector table in PostGIS...
Comment: Hi, please do add more detail to your answer. Link only answers discouraged.
Answer: I think you could use ST_AsRaster: https://postgis.net/docs/RT_ST_AsRaster.html
Comment: Hi, please do add more detail to your answer. Link only answers are not good enough.

Rasterize a vector layer in PostGIS


What is the effective aperture of two stacked lenses for macro?

Question: Is there a way to calculate the aperture of two lenses stacked together for macro? In my specific case, I have 90mm f/2.8 and 18mm 1/3.5 lenses.
Comment: The thing is, if you have a reversed lens, the big light-gathering end is now at the back, and the small end is at the front - seems that you'd have a lot less light gathered and transmitted that way.
Comment: @MikeW Right, I didn't think of it. So probalby the asnware is there is no way of doing it.
Comment: Mike is correct, the entrance pupil diameter often changes when you reverse a lens. Many true macro lenses have a 1:1 ratio (or very nearly so) of entrance to exit pupil...so if you reverse a true macro lens, your entrance pupil through the back of the lens is about the same as through the front. The same is not necessarily true for non-macro lenses, however, and "physical aperture diameter" is the same as the "entrance pupil diameter" as seen through the entry element of the lens. When reversing a lens, the entry element is often smaller, thereby explicitly limiting pupil size.
Answer: If you want to precalculate the value, I imagine it's possible, but I don't know how.
If you have the two lenses and want to determine through experimentation what the effective f/stop is, that's doable.
First mount the one of the lenses.  Put camera in manual mode, and take a shot of something, preferably a neutral gray card.  Something that is a solid color would be best.  Set the shutter speed so that you get a good exposure.
Now add the reversed lens.  With the same fixed ISO, take shots of the same subject (card) and with a range of shutter speeds.  Because this will now be a macro shot of a small area of your original subject, that's why I suggested a solid color.  Be careful that you are not blocking your light source with the lens.
Then compare those shots and find one that matches the tones of that first shot.
The difference in shutter speeds will tell you the additional stops of light beyond f/2.8 that you've lost by adding the second lens.
So if the original shot was at 1/60, and you have a shot with the two lenses at 1/15 that roughly matches the brightness of the original, then that's two full stops (1/60 --> 1/30 --> 1/15), so your effective aperture is f/5.6  ( f/2.8 --> f/4 --> f/5.6 )
Answer: Mike is correct in his comment, the entrance pupil diameter often changes when you reverse a lens. Many true macro lenses have a 1:1 ratio (or very nearly so) of entrance to exit pupil...so if you reverse a true macro lens, your entrance pupil through the back of the lens is about the same as through the front. The same is not necessarily true for non-macro lenses, however, and "physical aperture diameter" is the same as the "entrance pupil diameter" as seen through the entry element of the lens. When reversing a lens, the entry element is often smaller, thereby explicitly limiting pupil size.
You would need to know the magnification factor of both lenses, and for the reversed lens, the magnification factor when viewing through the back of the lens. That can be difficult to learn without a detailed understanding of the lens, including explicit focal length as well as any extension length, and if a retrofocal group is present, you would also need to know the effect of that. If you can find detailed design patents for the lenses your working with, you may be able to glean enough information to discern the necessary information to calculate effective aperture of the whole system.

If you need a quick and dirty but fairly accurate way to measure the difference in aperture with the single base lens and the reversed lens, a simple approach would be to take two shots, one with the single lens, and one with the whole system, at max aperture, same shutter and ISO. Adjust the exposure of the second shot until it appears to be the same with a post processing tool (I recommend Lightroom, as it has an excellent side-by-side compare mode). When the gray card tone of the two images matches, you can compute the difference of the aperture with the second lens in stops by how many EV of exposure adjustment was necessary. Once you know the difference, you would simply need to add that to any aperture you set the base lens to to compute the necessary exposure.
Comment: Cool :) thanks for your detailed answere. On the other hand for normal photography it is just simpler to try and figure it out :D


Glade alternative for Quickly?

Question: Is there? I am not getting on with Glade one bit. Any other ones out there that are better to use? Thing I dont like about Glade is when I make a mistake and I save I cannot undo along with several other issues.
Comment: Any ideas? 5 more to go.
Answer: I am not aware of an alternative; the alternative is to construct your interfaces in code.
Answer: I don't use glade for my own projects, I find it too limiting and dislike having to switch between my text editor and a GUI program.  Also, while glade does take care of a lot of simple drudgery, you're still going to have to do the harder bits in code anyway, so why not do it all in code?
I'll generally define functions to take care of constructing frequently used widgets, like labels, buttons, and menu items.  This also helps me enforce a common style across all the widgets.
There are frameworks and toolkits out there, like Kiwi and quidgets, that aim to give you higher level abstractions, but personally I find them either way overkill or not exactly what I need, so I just use them for ideas and make my own custom widget code.
Answer: You might be able to use PyQt instead.
http://www.riverbankcomputing.com/software/pyqt/intro

Glade alternative for Quickly?


What about generic lightning cables or adapters?

Question: Are lightning cables that are not manufactured by Apple (off-brand or generic) legal?
If not, when does it become okay to sell them? Because obviously the 30 pin adapter is everywhere..
Comment: What do you mean "legal"?
Comment: I was about to ask a too-similar question - "Do the generic connectors function?" Now that they are out, both the Apple iPhone 5 and the generic cables, I hope to see more discussion on this.
Comment: Update - I just bought a sub-$5 lightening cable from an eBay vendor, and it's working fine. I'd imagine if there were legal issues, Apple would actively go after these vendors and eBay.
Answer: It's pretty much impossible to say for sure at the moment, as the cables aren't publicly available yet, nor are the licensing agreements, but we can do some informed speculation.
From a legal standpoint, if Apple is granted patents on the connector, they could go after anyone who produces a copy without a license. So far I haven't seen any evidence they've applied for a patent on it (nor can I find any about the original dock connector), but that doesn't mean it hasn't/won't happen. However it's a timely process, as a patent won't be granted right away, and litigation also takes time.
It's also worth considering that there seem to be a lot of no-name manufacturers producing dock connectors, and I suspect that while the numbers may be large in the aggregate, the individual ones are small enough that it may not be worthwhile for Apple to pursue all the small entities in court. They don't seem to have made a dent in the off-brand market for cheap 30-pin dock connectors, based on searching Amazon.
Apple can (and almost certainly will) license third parties to manufacture their own cables and connectors—they do this currently with the Made for iPod/iPhone/iPad programs. Anything that has the "MFi" logo on it is made by a company that has paid license fees to Apple:

From a technical standpoint, how difficult it is to produce depends on what's inside the connector exactly.
Apple has used various methods in the past to restrict certain dock connector features (mostly using some resistors in the connector I believe), but it's also possible that they'll use some active circuitry to prevent unlicensed copies. However I'd bet against that, as they have to make massive quantities of these, and the cost and production constraints of even a simple chip could make things problematic.
The physical connector is also unlike most other connectors out there (which come in all sorts of different sizes and shapes, but are largely made out of stamped metal with pins or an array of contact points). Everything on the Lightning connector is flush, and knowing Apple, the tolerances are probably fairly small. It may be different enough that it takes a while for other manufacturers to reproduce it consistently, but eventually I'm sure it's doable.
So my guess is that yes, we'll see cheap generic Lightning connectors, but it may take a bit longer, and the supply may ebb and flow if Apple takes legal action.
Comment: rob - I've seen 'tear-downs' on the connector that show there's some intelligence there, a bit of a processor or DRM action going on. This will be the issue if any. But, it's now December, and dozens of vendors are springing up. If the generics didn't work, I'm sure we'd have heard by now.


"When would be the best time" vs. "what would be the best time"

Question: Is it more appropriate to say

When would be the best time and date for the meeting?

or

What would be the best time and date for the meeting?

I would assume the former and not the latter, but is it just as acceptable?
Answer: Both are correct, however, In my experience "what is the best time" is more common. The questions amount to the same thing but have subtly different meanings.

When is the best time ...?

The question here is "choose the best time, now when is that time?", you are choosing the time and then giving an attribute of the time — when.

What is the best time ...?

The question here is: of all the possible times which of these do you choose? Here you are choosing the time itself, not the "when" attribute of the time. This is confusing because the time and its attribute "when" have the same name.
Consider this as a comparative case. At a school, the day is divided into eight periods, starting on the half hour with an hour off for lunch. So the first is 9–9:30, the second 9:30–10 and so on.
Now I ask:

When is the period you take Chemistry?

You'd perhaps answer "9:30am".
But if I ask:

What is the period you take Chemistry?

You'd answer "Second period."
If you didn't number the periods you might answer "the 9:30am period." Which would be similarly confusing.
Answer: I would actually argue that "what... time" is more correct than "when... time"; however, I've heard both, and would accept and understand either.
Answer: "When" is arguably more correct, but both are common, e.g. "what's a good time for you?" or "when would you like to meet?".  I suspect that "what" is in this case shorthand for "what time" or "what day", and "what day is best for the meeting?" is certainly valid.


How can I clean the glass door on my fireplace?

Question: I tried crumpled up newspaper but that was a joke. Are there any good techniques for doing this?  Mine is totally covered with creosote right now and I can barely see through it.
Answer: This is going to sound bogus but I just got this tip the other day from a friend and it worked great on my woodstove.  Take a wet paper towel and dip it into some cool ashes from inside the fireplace, just enough to get a thin coating of ashes on the wet towel.  Then scrub the glass with the wet towel.  Within a few seconds you'll see even the toughest creosote start to disappear.  I usually finish with a dry paper towel just to get it really sparkling.  I don't know how or why this works, but I suppose it's probably some combination of abrasion and a dissolving action.  It really is amazing.
Just FYI, if your glass is totally covered in creosote, it probably means your fireplace isn't burning hot enough, and/or you're burning too much sappy wood like pine, and/or you're burning wood that's not properly seasoned.  The creosote on your glass is also being deposited on the inside of your chimney, which is bad.  Pine can be burned but needs to be well seasoned and ideally should be mixed with seasoned hardwood.
Comment: Wood ashes are alkali (they're used in gardening if the soil's too acidic), so it's possible that it's like cleaning with baking soda.
Comment: Ah, I bet that's it.  I think most household cleansers also have a high pH.
Answer: Other than the ash trick, the best way to get it clean is a really hot fire.
Answer: I've had success by spraying on Windex(TM)(R), waiting 5 minutes, then rubbing hard with a paper towel.
As mentioned by @Mike, there should not be too much build up. Burn different wood or burn hotter.


Do I need to verify my domain name again for Google Apps after it expires?

Question: I have registered a domain name and created an email account for that domain using Google Apps. I want to know what will happen to my Google Apps account associated with that domain after its registration expires or if I'm not going to use it (the domain I mean).
Answer: When the domain name expires so will the MX records for that domain. As you have already verified the domain with Google theoritically you will still be able to send email as that domain (since you are using Google servers to send) however you will no longer recieve incoming mail.
Its possible that Google might detect the domain is no longer registered and prevent you from sending mail.
That being said, Google wont disable or delete your apps account when your domain expires (provided you continue to pay the subscription) - all your data will remain and the account will still function, just without email (eg. you will still be able to use your Google docs)
Answer: Under your billing information you will see exactly what you paid and for how long. If you're going to renew it after a year then you don't have to do much, but if you don't then it expires.


How to release memory in android

Question: I need to free below pointers, how to release memory  for canvas, paint and matrix?
 Canvas pcanvas = new Canvas();

 Paint mPaint = new Paint();

 Matrix matrix = new Matrix();

Thanks in advance.
Answer: Java does not have pointers and you cannot explicitly release memory.
However it is possible to leak memory in Java and Android.  If any of your objects reference unmanaged memory, you need to let them know when they can release that memory.
Because it appears that you're doing graphics related work, I'd guess that you're using Bitmaps somewhere and these use a lot of memory and the memory needs to be released.  When you're finished using a bitmap, make sure you call Bitmap.recycle().
Answer: after you are not more referencing the object you created, assign it the null value. That obviously does not release immediately the memory but int this way you are telling to the GC that the object is eligible for garbage collection.. (effective java).
Edit: as @spatulamania said, if you are managing bitmap you have to call the recycle (it call a c/c++ free I think) method in order to free memory associated with. Set to null a bitmap reference is not useful because the bitmap is native implemented and the java object is few bytes.

How to release memory in android


How do I copy smart objects from one opened file to another

Question: Lets say I have two opened images I am working on in Photo Shop.  I want to copy an entire layer or a smart object from one opened image to the other opened image.
How do i do it?
Answer: For a layer:
In your original image right click the layer in the list and select the "Duplicate Layer..." option.
You get a dialog prompting for the new layer name and (importantly in this case) a Destination drop down. It's defaulted to the current document but you can select another open document or create a new document. If you select another document you can also set it's name.
Comment: Worked perfectly for me (May 24, 2016). Thank you!
Answer: You just drag from the layers tab, the smart object and drop it to the other window with the second photo that you like to copy.
Answer: Unfortunately I've not found a consistent way to actually automate this via an action which highly increases the amount of time required for such a simple task.
For now I move the window with the master smart filter applied and manually drag the smart filter object between documents.


When do you start buying wards, and when do you stop?

Question: I just recently started playing ranked, and a lot of people keep on telling me to buy wards. What's a good time to start buying wards, and at what point of the game do you stop buying wards?
Comment: You start as soon as you can afford them, and stop when you no longer have space in your inventory.
Answer: It depends on your lane and role. If you are support you want to start with at least 2 wards and a fairy charm. As every other role. I grab a ward after my first buy because you want to secure your lane from ganks.
As for when to stop. You should never stop. I like to make sure I always have at least one ward on me in case my support might not have one. Knowing where your opponent is wins games a large portion of the time. So do not stop warding.
Answer: You should ALWAYS buy 1 or 2 wards until you are finished with your gear.

BUT it's not always possible, so here we go:
Let's split by lane:
JUNGLE

Buy 1 or 2 when you get in the base after your first walk in the
  jungle. keep trying 1 or 2 every time you get back in the base

TOP

Try buy 1 at the beginner of the match or as soon as possible, recall
  and shop 1 or 2. Keep at least the river warded.

MID

Try buy 1 at the beginner of the match or as soon as possible, recall
  and shop 1 or 2. Combine with your jungle, so he can ward one part of the river and you do the other.

BOTTOM

If you are with a support, you can focus more in your gear. But you need buy some too. So when you have a spare of money, buy a couple or it and help warding dragon and river.

Here a great link for warding map
Answer: If you play support you usually start out with wards, pots, and faerie charm. Other than that you want to buy them anytime you have alittle extra gold, ASAP.
Answer: You tend to be wards to help the team, especially useful early game, (laning phase)
wards tend to lose some of its usefulness late game.
Early on, warding for baron is useless as it didnt spawn. Later on when champions are higher level, there is a chance for people to baron thus it should be warded.
Warding is basically for vision, if you have won every lane, there isnt a point to ward
Answer: If your support: Start out by buying at least 1 wards to prevent early gank at bot.
If your Top/Mid: Wait till you go back first time and just buy a ward once you feel like it, unless you perma push I would recommend just having 1 ward up all the time and stick to that side of the lane.
If your jungler: buy wards at around 5-10 minutes and keep doing it, to prevent getting counterjungled a lot.
Mid/late game when the lane phases is gone, I would recommend just letting your support pick up wards/oracle. But of course, if you got bag space and a bit of gold left, then buy a ward just to help your team.
Answer: Wards are good at all points in the game. Normally a support player will start with 3 wards. If you are in the mid or top lane I will usually buy a ward or two the first time i go back and continue to buy wards throughout the game. The support player will buy the majority of the wards so dont go wards crazy. Warding the river or the enemy jungle will help prevent the enemy jungler from ganking and help your jungler counter jungle.
Answer: If you are the support, you buy them at level one, and you don't stop until one of the nexus' is dead. If you are another role, you buy them on your first return to base, and don't stop until you have full build. Map awareness wins games.
Answer: In my opinion you should keep buying wards to try and have a constant map control, and if the the opposing team buys a oracle, then you kill that target and start all over again.
This is just in a perfect world thou, every match is different and you have to adapt to the situation when it's needed...
Answer: It is best to begin buying them as soon as the jungler can begin ganking successfully. If it is an early ganking jungler, you might want to start with wards although usually this is only done by supports in bot lane. For mid and top you want to start picking them up after your first trip back. Never stop buying them.
Answer: Buy wards immediately when the game starts but you don't want to place them before the general level in the game is 3 or 4 because most jungler's can't do a good gank until those levels. You almost never want to stop warding in fact you actually want to buy more and extend your vision of the map. You only stop when you reach the very late game when not much is happening but team fights as long as u have baron and dragon warded.


How is this problem called?

Question: I have a set of integers $\{8, 8, 6\}$. I want to know if It is possible to get $21$ by adding a subset of them. I would like to know how is the problem called so that I can google it.
Answer: This is the Subset Sum Problem.
Answer: Your question is slightly flawed: $\{8,8,6\}$ is not a set of integers, because it contains a repetition. (Or rather, it is a set of integers, but it only contains the two elements $8$ and $6$.)
If you really did mean "set of integers", then rattle's answer is correct. If you meant "sequence of integers", as suggested by your example, then this is the Knapsack Problem. The Subset Sum Problem is a special case of the Knapsack Problem.
Answer: This "set" is an example of what is sometimes called a "multiset": it allows one member (in this case 8) to have multiple memberships in it rather than only one.
You can't get an odd number such as 21 by adding even numbers.

How is this problem called?


Does animal fat and meat contributes differently to our body?

Question:
Is animal fat harmful to our body or is it harmful if the amount eaten exceeds a certain amount?
I have heard fat is more concentrated than protein. If I am correct then we can get more energy from fat. Then why is fat not desirable?
Our ancestors used all the parts from their hunt, even the fat. How did it work for them?
Would eating fat be beneficial for an athlete? (May be for different purposes: power, endurance or building muscle)
Answer: Have you read the Wikipedia article on Fat?  It has literally all of the answers you are seeking.  However...
1: Every single thing is bad for you if you get too much of it.  Animal fat, sugar, water, you can over-do it on anything.  Fat can be bad for you but it is definitely required.  In particular, "Vitamins A, D, E, and K are fat-soluble, meaning they can only be digested, absorbed, and transported in conjunction with fats. Fats are also sources of essential fatty acids, an important dietary requirement."  Fat does a host of other things, as mentioned in the article.
2: Fat is indeed a concentrated store of energy, but there's more to life than just energy.  Protein, like other nutrients, are required for life.  Protein happens to be particularly important, providing the structure for all our cells, is required for muscle growth, and provides amino acids, which are needed to make other proteins in your body.  Proteins are used as hormones, cellular messengers, immune receptors, so on and so forth.  You couldn't live without protein.  Energy, while concentrated in fat, could come from elsewhere.
3: It worked just the same for them as for us.  I will add, however, that our ancestors did not necessarily use every part from every animal, and these days, we actually use every part of many animals, just not always for food.
4: Sure?  Energy is good, but protein is probably better.  "Athlete" is a vague term, as strength-training and endurance activities are very different from each other.  Protein is probably better for strength, carbohydrates are probably better for endurance.  But you can ask over at https://fitness.stackexchange.com/.


What's the difference between SA and SAAR?

Question: I've only recently begun working in the quantitative finance field, and I've noticed that some time series I'm given are labeled "seasonally adjusted", and some labeled with "seasonally adjusted annual rate". What's the difference between these two data transforms?
Answer: Seasonal adjustments per se are a topic in and by itself, see for example this Wikipedia page.
As for the added "AR", imagine an event with an unconditional average of '100' (to pick a base) when measure monthly.  So the annual average is 1200, and similarly, a 'SAAR' series accounts for both recurring seasonal patterns (quarterly, monthly, weekly, daily, intra-daily, ...) and also scales the measurements to an annual total.
This wasn't really a quant finance question, maybe try the Stats sister site next time around.
Comment: Thanks. I wasn't sure where to post it, so thanks for the comment.
Answer: A SAAR series is expressed as if period-over-period growth were to occur for an entire year.  Real GDP is reported this way.  A report of 3% GDP growth in a quarter does not mean GDP grew 3% in that quarter.  It means GDP would be 3% higher in one year if this quarter's growth were to occur over 4 consecutive quarters.


Homological algebra for commutative monoids?

Question: Homological algebra for abelian groups is a standard tool in many fields of mathematics. How much carries over to the setting of commutative monoids (with unit)? It seems like there is a notion of short exact sequence. Can we use this to define ext groups which classify extensions? What works and what doesn't work and why?
Comment: Of course I'd be happy if the "ext groups" weren't groups but monoids.
Comment: Monoids have an identity by definition.  Also, it's not immediately straighforward that we have a good notion of a kernel either without categorifying to some extent, as indicated in the answer.
Comment: Added the "f-1" tag since I believe this question and its answers are interesting for people working over the field with one element.
Answer: A comment on Eric's answer (which is mostly the answer I would give): what is true is that any connected commutative monoid space is (weakly) equivalent to a topological abelian group.  A non-connected topological commutative monoid (or equivalently, a non-connected simplicial commutative monoid) contains some more information than its group completion.
So there will be a nice theory of "homological algebra", via simplicial objects, which is closely related to, but not equivalent to, homological algebra of abelian groups.
I'm not aware that anyone has examined this closely, before.
Answer: I used to think about this problem in relation to a chain theory for bordism ( as mentioned by Josh Shadlen above).
The problems you have with monoids is first and foremost that the category is not balanced. That means that you can have an epimorphism that is also a monomorphism but NOT an isomorphism. eg. the inclusion of N --> Z.
Subsequently most constructions that you would like to make - notably short exact sequences and the snake lemma - fail at some level.
I made a few notes on this as part of my investigation into bordism theory / homework assignment
 here.

In this case, we have complexes of free abelian monoids whose homology takes it's values in abelian groups and yet, the long exact sequence does not come from a short exact sequence of monoid complexes.

My references include:
[Bau89] Friedrich W. Bauer. Generalised homology theories and chain complexes. Annali di Mathematica pura ed applicata, CLV:143–191, 1989.
[Bau95] Friedrich W. Bauer. Bordism theories and chain complexes. Journal of Pura and Applied Algebra, 102:251–272, 1995.
[BCF63] R.O. Burdick, P.E. Conner, and E.E. Floyd. Chain theories and their derived homology. Proceedings of the AMS, 19(5):1115–1118, Oct. 1963.
[Koc78] S. O. Kochman. A chain functor for bordism. Transactions of the American Mathematical Society, 239:167–196, 1978.
Answer: Your question can be understood as how to do Homological Algebra over the Field with one Element.
Deitmar, in http://arxiv.org/abs/math/0608179 , section 6, gives an example of what can go wrong if you try to do sheaf cohomology directly via resolutions...
You might also want to look at his http://arxiv.org/abs/math/0605429 ; in order to construct K-theory of monoids he sets up an analogue of the Q-construction. The Hom-sets in the resulting category are sort of Exts, maybe something to start with...
Durov, in http://arxiv.org/abs/0704.2030 , follows the simplicial approach for commutative monads, of which commutative monoids are a special case
Comment: I'm not sure about your first sentence.  The monoids in these papers are playing the roles of the commutative rings in ordinary algebraic geometry, not the modules.  Modules become sets.  As far as I can tell, "homological algebra over the field with one element" is a fancy phrase for homotopy theory.
Comment: True, one would have to consider "Andre-Quillen-homology" over the field with one element, but I definitely agree that this is now just fancy language.
Answer: This is an answer to one part of your question.  The paper "Extension Theories for Monoids" by Charles Wells, Semigroup Forum 16 (1978), 13-35, gives a precise answer to the specific question: How does the Beck cohomology theory for monoids classify extensions of monoids?  (It classifies Leech extensions.) The paper with corrections and a list of subsequent papers related to it may be found here. Beck's thesis is now online here.
Comment: This seems very promising. Thanks for the reference!
Comment: The entire paper "Extension Theories for Monoids" is now available here: http://www.cwru.edu/artsci/math/wells/pub/pdf/ExtThMon.pdf

Look at Grillet's paper, too.  I have not read the Novikov paper.
Comment: The reference to Grillet's and Novikov's papers is in the preamble to the Extension Theories paper on line as in the previous comment.
Answer: One interpretation of the "right" category for doing homological algebra of commutative monoids would be the category of simplicial commutative monoids.  Now up to homotopy, a space is a commutative monoid iff it is a commutative group iff it is a product of Eilenberg-MacLane spaces.  There are some details to check, but this should imply that the forgetful functor from simplicial abelian groups to simplicial commutative monoids is a Quillen equivalence for the standard model structures, with the inverse being group-completion.  This correspondence would indicate that the correct notion of "homology" of a simplicial monoid would be the homology of its group-completion as a chain complex.
However, this is presumably not what you're looking for, since you care about the monoids themselves, not just their group-completions.
Answer: Jaret Flores's PhD thesis,

Jaret Flores, Homological Algebra for Commutative Monoids, arXiv:1503.02309.

treats precisely this topic.
Answer: I agree with Charles and Eric that the natural setting for your question is the model category of simplicial commutative monoids.  However, my earlier guess that the resulting homotopy theory would admit a simple algebraic description (as in the case of simplicial abelian groups) was wrong.  For instance, consider the free commutative monoid functor from simplicial sets to simplicial commutative monoids and apply it to a space $X$ which I will assume for convenience to be connected and reduced ($X$ has only a single 0-simplex).  The resulting simplicial commutative monoid $Y$ has $\pi_0(Y) = \mathbb{N}$.  The element $1 \in \mathbb{N}$ is distinguished as the unique generator, and the connected component of $Y$ corresponding to $1 \in \pi_0(Y)$ is just $X$.  So, we can recover $X$ from $Y$.
In other words, when Charles says "A non-connected topological commutative monoid (or equivalently, a non-connected simplicial commutative monoid) contains some more information than its group completion", it is actually quite a lot more information, enough to encode an entire homotopy type!
If we apply the reduced chain complex functor to one of these free simplicial commutative monoids, we'll just get $\mathbb{N}$ in degree 0 and zero in higher degree.  The problem is that knowing the kernel of a map is 0 tells us very little (every map in the image of the free commutative monoid functor has kernel 0).
Comment:  I just pointed out this same example on my simplicial comm. monoid question. Should have checked here first...
Answer: Note that Homological algebra for abelian groups is really about homological algebra in the category of Z-modules.
The immediate neighbouring area I know of, where we take something not a module over some sort of algeba, is for group cohomology - where the standard construction is to grab the group ring, and then consider the abelian category of modules over the group ring.
So - the issue at hand, essentially, is to find yourself an abelian category that reflects properties of the monoids you want to study. Having short exact sequences is a good start, but you really want an abelian category to make the standard constructions from homological algebra work.
And if the monoids themselves don't provide all you want, you can always start looking at monoid rings instead: given M, form kM as the k-vector space spanned by a formal basis element for each element of M, and introduce an algebra structure by the monoid multiplication. After this, you can proceed in analogy to group cohomology.
Comment: As far as I can tell, just looking the monoid ring associated to a monoid doesn't cut it. For example let N be the natural numbers and consider the set E= N x N with multiplication
(a,b) (x,y) = (a + bx, by)
This is a monoid and is an extension of (N,x) by (N, +). The monoid ring of (N, x) is a polynomial ring on infinitely many generators (one for each prime). I don't see any way to extract this extension from the homological algebra of this ring.
Answer: Commutative monoids (aka N-modules) play a key role in logarithmic geometry (Ogus should have a book coming out soon), but I think the focus there is on a restricted class known as integral cones (aka fine saturated, or fs monoids).  These have decent homological properties.  I think the general setting has problems, since group completion can kill things.
Answer: I might be misunderstanding your question - are you asking about a (co)homology theory for commutative monoids, or about trying to do homological constructions in the category of commutative monoids?
In the former case, I think Grillet has some work on this, see e.g.
Grillet, Pierre-Antoine(1-TULN)
Commutative semigroup cohomology. (English summary)
Comm. Algebra 23 (1995), no. 10, 3573--3587.
The idea is basically what Mikael describes, although I only know the more primitive version which is cotriple/comonad cohomology as set up by Barr and Beck. Presumably the problem with trying to apply that set up is ensuring there are enough abelian group objects to use as coefficient modules.
Answer: category of commutative monoids  plays central roles in algebraic geometry over F_un.
In fact, for such kind of category , Durov use homotopy theory in this paper:
http://arxiv.org/abs/0704.2030
But algebraic geometry over F_un is essentially study the geometry of right exact category(which is a category with subcanonical grothendieck topology) over category with only one object, in general, the correct things we should study is right exact category over category with initial object.
So, what we need is homological algebra framework on right exact category(in particular, topos, quasi abelian category, abelian category and so on).
In fact, Alexander Rosenberg built Homological algebra on noncommutative "space" (i.e.grothendieck topos)as a noncommutative version of Grothendieck tohoku lecture. He also introduced the Higher K theory for right exact category which led the universal K theory for abelian category and exact category and long exact sequence of Higher K functor of exact categories.
Here is the link of the paper:
http://publications.ictp.it/lns/vol23.html
http://users.ictp.it/~pub_off/lectures/lns023/Rosenberg/Rosenberg.pdf
Answer: I can't give a substantive answer to this question, but differential graded abelian monoids have been sighted in the wild here:
A Chain Functor For Bordism
Author(s): Stanley O. Kochman
Source: Transactions of the American Mathematical Society, Vol. 239 (May, 1978), pp. 167-196
Published by: American Mathematical Society
Stable URL: http://www.jstor.org/stable/1997852
Answer: Presummably you could use the forgetful functor to sets and its adjoint to construct a homology theory (using the walking adjunction, simplicial sets, etc) analogously to how group homology arises from the forgetful-free adjunction to set.
The details are probably rather dense and tedious though.


Business rules, business logic, input validation

Question: This might be asked a thousand times but couldn't find the answer. I just wonder how you people handle the business logic errors? Im trying to do a nice api for my business model. Some methods have quite a lot of validation and I just wonder how should I report them if something goes wrong? Returning true or false or null sometimes won't say what was the actual cause. Sometimes it works sometimes it is not enough. I read that some people use exceptions. I've thought that I could return error messages and when everything goes fine then it would be null. Somehow error codes just don't add up for me..
Comment: This isn't really "errors".  This is "input validation".  Perhaps you should change the title.  Business Logic Errors sounds very much like the business logic has an error in it because of bad design or bad requirements analysis.
Comment: What framework are you using?  Often an application framework has a standard way of reporting these kinds of errors.  Please **update** the question to name your language, framework and other tools.
Answer: You are on the right track. Domain or business validation logic should do more than just report true or false. You want to enable the "user" of your business classes (ie the UI, or your logging system) to do something more than just report to the end-user that something is wrong.
Now, how you do that is completely up to you.
Decision to make: check everything and then report all errors/warnings/information, or stop at the first one you encounter. I would personally opt for the first one and let the UI decide how to use the info. There is a trade-off though: some validations may take longer than others and you may want to check these only if the "simple" ones succeed.
I wouldn't in any case use exceptions to do this. Some people use exceptions because most development environments have a nice way of showing any unhandled exceptions to the user. I think that is just a bad/lazy choice. And not just because it is then hard to decide where the error came from (though stack traces can help there), but also because it can(will) leave your app in an unpredictable state and may cause errors further down the line that will be even harder to debug.
Therefore, I tend to reserve exceptions for, well exceptional, circumstances. Things I didn't foresee. Validation doesn't fall into that category in my opinion - you did foresee all (current) validation concerns or you wouldn't have validated against them.
All in all, I code the validation in the domain/business classes to collect an array (or whatever I care to use) of error/warning/hint instances. They may be simple strings, they can be fully fledged classes. Usually I always return an instance to the caller and never an unassigned pointer. "Valid" is simplu indicated by a message count of zero.
Answer:
I read that some people use exceptions.

There's a reason why.   Exceptions handle exceptional situations very nicely.  If your language has exceptions, start using them.
Invalid data should raise an exception.
The application can then collect all the data validation exceptions and display them to the user.

I've thought that I could return error messages and when everything goes fine then it would be null.

A function that either returns a proper value or sometimes returns a "status" value (line null) is a bad idea.  A function should always return a proper value or throw an exception.

Somehow error codes just don't add up for me.

That's because error "codes" are a very bad idea.  Except in C programs where you have very little choice.
Comment: Could I ask why many people consider throwing exceptions is a so bad idea like the accepted answer In contrast to your opinion?
Answer: PPPHP,
Have you read: Domain Driven Design ?
This book has pretty good material on developing good domain models for isolating business rules.
When fixing bugs on projects, I tend to write down a user story for that piece of functionality to understand how it works from a users point of view. For example, yesterday, I was fixing a bug that caused
an IllegalArgumentException if a certain combo box was not populated with the required data. The user would need to reset the device and start it up again. Now, fixing this problem may seem obvious, but the user story helped clear my thinking. I ended up pacifying that particular exception and ensuring the combo box had the text "data unavailable". So when the user raises a bug or calls support, the problem is better understood and the customer has not been left with some obscure error message. Little tweaks like this, seem to make a big difference when dealing with customers.
Answer: I usually use both a success flag and a list of messages, each of which is comprised of both a code and a freeform message. If you receive a successful response with messages then you know that the messages are all warnings and that something has happened on the server side.
The reason for using both codes and messages is so that a message can be displayed to a human reader but a code can be handled by software. This way, if the freeform message changes, software which handles that error doesn't also have to change.
Obviously if you are sure that there will never be a non-human reader of your responses, you can forget the codes but that's a difficult decision to undo later.


Disable default error pages/error messages in IIS

Question: I have this web application (ASP.Net MVC 3) that on certain conditions returns a custom JSON string with a HTTP status code for an error (403, 415, 500). It is deployed on a Win 2008 R2 server with IIS 7.5
Initially I was gettting the standard HTML pages for the error instead of the JSON data. I removed the error pages for these errors in the app settings. But now my queries which should return some JSON data return a single error line. When the server gives me 403, I have the message "You do not have permission to view this directory or page." (simple line, no HTML around it).
What can I do to deactivate this and finally get what the app is returning and not what the server wants to return?
Answer: You can not disable the 403.17 (expired certificate) response in IIS.
Comment: Do you have a reference for this statement?


Email Forwarding Service to use with my domain?

Question: I just registered my domain with namecheap. One of the main purposes of me registering my domain was to use the email forwarding service. Well unfortunately i found out after i registered my domain that Namecheap only has 7 unique forwarding possibilities. That's not nearly enough. Does anyone know of a relatively cheap and reliable email forwarding service i could use with my domain?
Answer: You could use Gmail for your domain (Google Apps for Business or the regular version). It's fairly simple to set up email forwarding, or domain aliasing if you want.
Alternatively, you could just get a decent web host for less than $10/month (e.g. DreamHost) that doesn't put limits on things like the number of email forwards you're allowed to have.
Personally, I'd just switch to a better web host. Since most web hosts that limit things like email accounts or subdomains generally aren't very good anyways and will end up costing you more in time/work/money in the long run.


Central limit theorem and VaR

Question: If I have a lot of data points and number of different dependent variables, can I use central limit theorem to assume data is multivariate normal and compute my VaR? Is this the appropriate use of central limit theorem for VaR calculation?
Answer: There are several methods to compute VaR: i) historical, ii) variance-covariance, and iii) monte carlo. iv) copula techniques. I assume you are asking about approach (ii).
If the data are not multivariate normal and i.i.d. then the variance-covariance approach will not reflect true risk. For example, if there is serial correlation then risk is understated.
Your intuition around the use of the central limit theorem can be applied by using a bootstrapping approach to estimating VaR. This approach treats VaR itself as a random variable which is estimated with confidence. VaR has some highly unusual properties since it is not a coherent risk measure - so good luck! There are various posts on this site on how to do bootstrapping.
Answer: If you are not taking a mean of many values (with finite variance) then the central limit theorem does not apply.
To calculate VaR anyway you can start taking the empirical quantile or use more sophisticated estimators, as the other answer mentioned.
Sorry but it's not clear to me the role of those different dependent variables.
PS: please distinguish between VaR (Value at Risk) and VAR (vector autoregression)

Central limit theorem and VaR


Would Ubuntu be installed with the current language of Windows 7?

Question: If I install Ubuntu on Samsung Notebook that is Formatted for Spanish will Ubuntu be in English which is preferred ?
If I install Ubuntu will I lose my pictures in Google Picasa ?
I am not real computer literate but tired of Windows.
Comment: DO you want to keep Windows?
Answer: The language which Ubuntu is installed in is selected during the installation process and is not related to the language in which Windows is installed on that machine. It can also be changed after the installation process (which is not true of Windows generally).
Google does not have a Picasa app for Ubuntu, however there are many similar image management programs available in the "software centre" in Ubuntu. The pictures on your computer in Windows will be accessible from Ubuntu if you install Ubuntu alongside Windows. You will of course still be able to manage your Picasa photos online in Google+ if you do that currently.
Comment: actually there are clear steps how to run Picasa in Ubuntu
Answer: First of all, Ubuntu will allow you to choose which language you wish for your Ubuntu Install only. This selection won't affect other operating systems.
Your pictures in Google Picasa will be permanently deleted if you format (wipe) your current operating system and install Ubuntu as your only OS.
If you choose a dual boot, you will be offered to choose which operating system you wish to boot at the very moment that the computer attempts the start up.
See the answer to this question: How can I dual boot Windows and Ubuntu? in order to check out how to install a dual boot OS.
You may wish to consider a dual boot previous to the full migration in order to face the transition with ease and be able to do certain tasks in your current operating system if you are in a hurry.
If you are heading to completely remove Windows in order to replace it with Ubuntu (which is a great idea, indeed), I suggest you to first make a full backup of those things that you don't wish to lose.
Additionally you may wish to consider that Ubuntu will offer you certain alternatives (a lot of them) for your current Windows based products. Picasa's alternatives will include (but not limited to) Shotwell for the pictures management.
And my advice will be that if you migrate to Ubuntu, please try to think different and don't try to do things in the "Windows way". The tasks are easier in Ubuntu, we just need to get used to the alternatives when migrating.
Good luck!
Answer: Language
You can use whatever language and keyboard setup during Ubuntu install.
Picasa
You can install Picasa in ubuntu under Wine (Windows emulation) and it works pretty well. It is enough if you just run the setup exe file in wine or there are better howtos fixing some photo upload troubles: (How would I install Picasa 3.9?)
Backup your database, especially if you have lot of data in recognized faces, places, photo comments etc.  You can use your old database and copy it to the new installation. Doublecheck, that you upgrade to the latest version of Picasa on Windows, so database can be transferred to the very same version in Linux/Wine. But this is not a straightforward process, so you will need some experience. Better would be, if you install ubuntu aside the Widoze, so you will have dual boot and you will be able to access windows files. Then you will be able to transfer the files, photos and picasa database and check that all is working. After you can free the space used by Windoze and use it for better purpose.
Answer: You can specify your preferred language during the installation procedure.
Ubuntu will not recognize any language of other OS by their own.
Personally, I installed Ubuntu in English and then set Spanish for a specific user. You can do the same for any supported language and with all the users that you want to

Others question that will help you:

How To change the Default Language?
How can I install Ubuntu without removing Windows?


Can Sierpinski's anisotropic bicolouring of the plane, assuming the continuum hypothesis (CH), be extended to three dimensions?

Question: Sierpinski showed that, on the assumption of CH (in fact, equivalently to it), each point in the plane can be coloured (say) black or white so that every section of the plane parallel to the $x$ axis is "almost" white $-$ in the sense that all but countably many points of it are white $-$ while every section parallel to the $y$ axis is almost black. Equivalently, given CH, every line through the origin can be almost white, while every circle centred on the origin is almost black. Is there a corresponding result for three-dimensional space? For example, assuming CH, can we bicolour space so that every plane through the origin is almost white, while every sphere centred at the origin is almost black?
Answer: The answer to the last sentence in the question is no, for a silly reason, namely that a plane through the origin and a sphere about the origin have an uncountable intersection, in which only countably many points could be white and countably many black.  The obvious way to evade that silliness is to replace planes and spheres with two families of sets, say $X$ and $Y$, such that the intersection of any set in $X$ and any set in $Y$ is countable.  If you assume that, and if each of the families $X$ and $Y$ contains only $\aleph_1$ sets, then Sierpinski's construction will work.
Comment: @Andreas Blass. Thanks! Yes, of course you are right. What I was wondering was whether there was a decomposition into two families, as you described, that were geometrically recognizable. Just mapping the plane to 3D space via some arbitrary bijection would carry over the planar decomposition intact: but this wouldn't be interesting. I think that your answer points at least to $X$ and $Y$ not being of the same dimensionality.
Answer: Here is a way to realize a positive version of the
phenomenon:
Theorem. If the CH holds, then there is a coloring
of $\mathbb{R}^3$ into six colors, such that every line
parallel to one of the coordinate axes uses only three of the colors and moreover is almost
monochromatic, meaning that it uses only one color for all
but countably many points. (And each major axis direction will give rise to distinct two of the six colors.) Furthermore, the coloring on every plane parallel to the
principal axis planes uses only two colors (except on countably many lines) and exhibits the two-dimensional
Sierpinski property mentioned in the question.
The proof idea is simply to extend the two-dimensional
proof idea to three dimensions. Namely, if CH holds, then
well-order the reals in order type $\omega_1$. In the two
dimensional case, we may color a pair $(x,y)$ black or
white according to whether $x\lt y$ or not with respect to
that well-ordering. Every vertical line is almost black,
since all but countably many $y$ are above any given $x$ in
the well-order, and every horizontal line is almost white,
since only countably many $x$ are below any fixed $y$.
In three dimensions, we color a triple $(x,y,z)$ according
to the order type of the triple with respect to the
well-order, that is, according to whether $x\leq y\leq z$
or $y\leq x\leq z$ and so on, using the well-order relation here (not the usual order on $\mathbb{R}$), for each of the six
possibilities. (We may use the first occurring case when the
coordinates are not distinct, since these exceptional cases don't affect the final property.)  Any line parallel to one of the principal
axes corresponds to fixing two of the variables. The third
variable will range over all the remaining reals. The point now is that there are only three ways for a given real to sit with respect to two fixed reals, so every principal line has only three colors. Furthermore, once we fix two reals, say $x$ and $y$, then all but countably many $z$ will exceed both $x$ and $y$, meaning that only one color is used for almost all $z$ on that line. Thus, all such lines are monochromatic outside a countable
set.
Similarly, every plane parallel to a principal axis plane amounts to fixing one of the variables. If we fix $x$, say, then the coloring on that plane corresponds to the colorings of $(y,z)$, which outside the exceptional lines occuring when $y\leq x$ or $z\leq x$ in the well-order, will be determined solely by the relation of $y$ and $z$ in the well-order, putting us exactly in the two-dimensional situation.
A similar phenomenon arises in every higher dimension
$\mathbb{R}^n$, simply by using more colors: one should
simply color $(x_1,\ldots,x_n)$ by the order type of these
elements with respect to the fixed well-ordering.
Finally, let me say that one can avoid the CH assumption in
this entire discussion, if one simply re-interprets the
phrase "almost all" as meaning "fewer than continuum many"
rather than "countably many". One can achieve this by
well-ordering the reals in order type continuum rather than
order type $\omega_1$, and then applying the same argument.
The crucial fact is that every initial segment of such a
well-ordering has size less than the continuum.
Comment: @Joel: Thank you very much for this interesting answer, which has given me much to study and think about. As a first thought, from Andreas's answer, it seems there is no geometrically coherent 2-colouring, while your answer shows a geometrically nice 6-colouring. Perhaps the latter is optimal, in the sense of using the fewest colours (while being geometrically simple).
Answer: Hamkins' theorem can be pushed down to 3 colors. Give $(x_1,x_2,x_3)$ color $i$ where $x_i$ is the least coordinate according to the $\omega_1$-type ordering. (Break ties by choosing the least possible $i$.) Every plane perpendicular to the $x_i$-axis will be colored $i$, except possibly on countably many lines. Every line parallel to the $x_i$-axis will colored $i$ on at most countably many points; the rest of the line will be monochromatic in another color. If we switch to spherical coordinates, then we get an analogous theorem involving spheres, cones, planes, rays, and circles.
Comment: Hi David. Welcome to MO!
Comment: @David Milovich: Thank you! This is a very nice answer.
Comment: Yes, very nice!
Answer: Sierpinski himself proved the following striking version of his theorem for the plane:
CH is equivalent to the statement that $\mathbb R^3$ can be written as $A_x\cup A_y\cup A_z$ where each $A_u$,
$u\in\{x,y,z\}$, has a finite intersection with every line parallel to the $u$-axis.
(See http://www.math.wisc.edu/~miller/old/m873-05/setplane.ps, respectively [Simms, John C.; Sierpinski's theorem. Simon Stevin 65 (1991), no. 1-2, 69–163].
I believe this theorem is proved in Sierpisnki's 1951 paper in Fund. Math. 38, pp. 1--13, but I would have to actually walk to the library to verify this (and decipher enough of the french in the article).)
There is actually a planar version of this result of Sierpinski, due to Komjath.
A set $A\subseteq\mathbb R^2$ is a cloud if there is $p\in\mathbb R^2$ such that every line through $p$ meets $A$ in only finitely many points.
CH is equivalent to $\mathbb R^2$ being the union of 3 clouds.
In both cases the striking fact is that we go down from countable to finite, which is somehow less arbitrary, in the light of Joel's comment that the previous constructions can be done in ZFC if countable is replaced by $\lneq 2^{\aleph_0}$.
Comment: Stefan, you can access Sierpinski's paper at the following URL: http://matwbn.icm.edu.pl/spis.php?wyd=1&jez=en. The theorem of his you mention is Theorem 2 of his paper.


Is there a cd image of ubuntu under 512MB?

Question: I only have a 512 MB flash drive and was wondering if there's an ISO for Ubuntu that will fit on it. I really need to format my netbook.
Answer: There's Ubuntu minimal. The iso's size is 5-20Mb, but the actual install size is of course larger.

The Minimal CD downloads packages from online archives at installation time instead of providing them on the install CD itself.

Just to add that the final install size can be reduced down to 600MB which will mean a command line system. The same page includes instructions to install low disk space demanding CLI and GUI systems but check the used disk space before continuing from command line system to GUI system.
Answer: If you have Internet access, you could do a netinstall.  Loading either billix or netboot.me would facilitate that process, and fit on a 512MB flash drive.
Answer: One possibility is to download and install Madbox 10.10.01 which is a lightweight remix based on Ubuntu 10.10 (32-bit.)  The .iso is only 388.2 MGs
http://download.tuxfamily.org/madbox/madbox-10.10/madbox-10.10.01-i386.iso
Once installed make sure that all repositories are enabled in your /etc/apt/sources.list file (remove all # comment markers except the ones in front of the deb-src lines and you might want to change the "br" to "us" if you are us based.)
Then open a terminal and perform a sudo aptitude update
The sudo aptitude install ubuntu-netbook (which is the metapackage which will completely install the UNR / UNE version of ubuntu.)  Then shutdown and reboot and select the desired interface at logon (probably the netbook one.)
If you prefer LTS, then start with the Ubuntu 10.04 version of Madbox http://download.tuxfamily.org/madbox/madbox-10.04/madbox-10.04.02-i386.iso
This way you will not need to mess around with a command line only install (which is probably not advisable for less then intermediate to advanced users.
Hope this is helpful.
Answer: MadBox: very fast, very nice OpenBox interface, Ubuntu repositories. It's a very minimal system, but perfect if you want to build on a base but don't want to remove loads of stuff first.
I actually installed Open Office!
It should be perfect for a netbook, and recognises usb mobile broadband dongles.
Debris Linux http://debrislinux.org/ 186.2MB ISO, GNOME/OpenBox
For loads of information, this is THE place:
http://distrowatch.com/search.php?category=All&origin=All&basedon=Ubuntu&notbasedon=None&desktop=All&architecture=All&status=Active