How to create charts and tables with Inkscape?

Question: Inkscape doesn't seem to support charts or tables, so what's the best way to create a chart/table and import it into Inkscape?  Excel 2013 unfortunately doesn't seem to support svg.
Comment: What is the end result you are looking for? An SVG based chart? For print or web?
Comment: To make charts and tables for PDFs, presentations, or videos.  It seems like svg would be the best way to create graphics since I could easily adjust the size when needed...  Although I'm not a graphics guy so I could be wrong.
Comment: Well, do note that a table created as a SVG is going to lose it's "Structure" in the sense of semantics and accessibility. That may or may not be an issue for you, but if the PDF is for web distribution, for example, an SVG based PDF will be less accessible than a Table-based (HTML or Word) based PDF. However, for video, that's not really an issue at all. The biggest challenge with making a table in inkscape is that inkscape doesn't have a table tool. So everything has to be hand drawn.
Comment: This worked for me: create a rectangle, then select it and go to Path > Path effects.. and then select Grid. That's all.
Answer: @Yisela recommends Gnumeric. I would also recommend looking at LibreOffice.
What I've done in the past is:

Create my tables in LibreOffice Writer, applying formatting such as row borders, cell spacing, and so on.
Copy the table, open up LibreOffice Draw, and paste the table as a "LibreOffice Text Document" using "Paste Special".
Select just the table.
Go to File > Export and export in your preferred vector format (in the past, I've used EPS without any problems). Be sure to check the box that says "selection" otherwise Draw will export the whole page.

A similar approach can be taken with charts.

If you prefer to stick with Excel, at least for the charts part, you can install the "Save as PDF" office add-on and select your charts as PDF files which you can open just fine in Inkscape.

For both of these options, most of the actual content editing would take place in LibreOffice or Excel. Inkscape would really be for fine-tuning the output.

Finally, if you're working on a project that needs lots of tables and charts, you might want to invest some time in learning a programming language like R, with which you can generate publication quality charts and tables quite easily.
Comment: The LibreOffice Writer > Draw > SVG Export workflow works like a spell! You can also add Calc to the equation, but mind that you need to use _Paste Special_ > _Formatted Text (RTF)_ in order to copy the actual table (not a Calc object) into the Writer document.
Comment: Thanks! (been using PDF. & SVG) SVG output can work from Draw as well (better formatted with line thickness, though this was largely due to source formatting), but with both EPS and SVG output I had a issue where for each line in each cell it would add small base64 encoded images as a background (?) https://i.stack.imgur.com/FGgFs.png. Removing these (`data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAA`....) using a regular expression fixed the formatting (transparent background needed), and drastically reduced the file size (from 11 megabytes to 200kilobytes).
Comment: I think you win! I searched all over for importing excel charts to inkscape and nothing I found worked.  They actually have a save as PDF option in excel 2013 that seems to work best for me.  Thanks!
Answer: Since you mention Excel, something you could do is use Gnumeric to import your .xls files into it, and from there export them as SVG. Unlike Excel, Gnumeric has more export options that would allow you to create more complex elements without having to actually draw them on Inkscape.
There is also an extension for Inkscape called NiceCharts that is good for creating basic pie and bar charts. It hasn't been updated since 2012, but it seems to be working nicely.
Other options are Gnuplot (a command line - with several frontends - driven interactive data and function plotting tool has SVG output for graphs & charts) and Matplotlib (a python plotting library and can also output SVG). Finally, SVG charter is a perl script dedicated to generate SVG charts on web servers.
For diagrams, Graphviz can produce structural information as diagrams of abstract graphs and networks in SVG and other vector formats.
Answer: Here's simple way to create table in Inkscape....
First, create rectangle with desired dimension. Then, underMenu>Edit choose Clone,then forward to Create tiled clones. Here you can set number of columns and rows.Click create or hit Enter.Your table is created.After, simply change the size of cells,play a little....To mention, this is not as good slution like creating in excell, but for something simple fair enough....
Answer: Because I needed that, I've just created a small online tool generating a piechart in svg: github/svg piechart online. Later you can edit it with Inkscape opening the SVG file created.
Comment: Hi Remmel, welcome to the site! Awesome tool. Would you mind fleshing out your answer a bit, perhaps explaining how someone could edit this with Inkscape in case any beginners might be clueless?
Comment: That tool creates SVG file which is the same kind of file used by Inkscape. Thus you'll need using the web based tool to export into SVG the piechart and later to open it with Inkscape
Comment: The page loads, but clicking the button doesn't seem to do anything...
Answer: visit https://sourceforge.net/projects/inkscape-tables/files/inkscape-table-1.0.tar.gz/download
Download inkscape tables python script; install as extension in inkscape:
To install in debian
unpack the tar.gz file using
    foo$: tar -zvxf filename.tar.gz
open the unpacked directory.
open a terminal window in that directory and
install using //
    sudo ./install
the script will auto install and the
Tables option will be visible in inkscape Extension Menu
Answer: LatexIt, a component of MacTex distribution, is also a quick way to generate tables in image format, and then import into Inkscape. This is an alternative to using a spreadsheet software.
Answer: Since this issue is not solved (in September 2018) I would suggest my solution which was inspired by @Ananda Mahto. Her solution did not work for me although my needs were quite simple (a couple of tables filled with results).
So I followed these steps:

I used LibreOffice to create the tables and decorate them as desired.
Exported the document to pdf.
Imported that pdf to inkscape to have a base to work with.

Since inkscape support for pdf has improved over the years with the above approach I can get editable tables in my inkscape. Surely it's not integrated into the actual program but it's the best solution I came up with.


Is Lebesgue's "universal covering" problem still open?

Question: The following problem has been attributed to Lebesgue. Let "set" denote any subset of the Euclidean plane. What is the greatest lower bound of the diameter of any set which contains a subset congruent to every set of diameter 1? There are a number of interesting geometric problems of this type. Is it possible that some of them may be difficult to solve because the solution is a real irrational number which (when expressed in decimal form) is not even recursive-and so cannot be approximated in the usual way?
Comment: Are you interested in minimal diameter of a universal cover or minimal area?
Comment: The [Quanta Magazine](https://www.quantamagazine.org/amateur-mathematician-finds-smallest-universal-cover-20181115/) article about recent (2018) progress is nice.
Answer: The question is still open.  There are at least two versions.  The most popular asks for the minimal-area convex subset of the plane such that every set with diameter 1 can be translated, rotated and/or reflected to fit inside it.  Here is the best lower bound I know:

Peter Brass and Mehrbod Sharifi, A lower bound for Lebesgue's universal cover problem, Int. Jour. Comp. Geom. & Appl. 15 (2005), 537--544.

Their lower bound is 0.832, obtained through a rigorous computer-aided search for the convex set with the smallest area that contains a circle, equilateral triangle and pentagon of diameter 1.
The best upper bound I'm 100% sure of is 0.8441153, proved here:

John Baez, Karine Bagdasaryan and Philip Gibbs, The Lebesgue universal covering problem, Jour. Comp. Geom. 16 (2015), 288-299; arXiv:1502.01251
.

Our paper also reviews the history of this problem, which is rather interesting.
In 1920, Pál noted that a regular hexagon of area circumscribed around the unit circle does the job.  This has area
$$ \sqrt{3}/2 \approx 0.86602540 $$
But in the same paper, he showed you could safely cut off two corners of this hexagon, defined by fitting a dodecagon circumscribed around the unit circle inside the hexagon.  This brought the upper bound down to
$$ 2 - 2/\sqrt{3} \approx 0.84529946 $$
He guessed this solution was optimal.
In 1936, Sprague sliced tiny pieces of Pal's proposed solution and bring the upper bound down to
$$  \sim 0.84413770 $$

        (Image from Hansen's paper, added by J.O'Rourke.)
The big hexagon above is Pál's original solution.   He then inscribed a regular dodecagon inside this, and showed that you can remove two of the resulting corners, say $B_1B_2B$ and $F_1F_2F,$ and get a smaller universal covering.  But Sprague noticed that near $D$ you can also remove the part outside the circle with radius 1 centered at $B_1$, as well as the part outside the circle with radius 1 centered at $F_2.$
In 1975, Hansen showed you could slice off very tiny corners off Sprague's solution, each of which reduces the area by $6 \cdot 10^{-18}$.
In a later paper, Hansen did better:

H. Hansen, Small universal covers for sets of unit diameter, Geometriae Dedicata 42 (1992), 205--213.

He again sliced two corners off Sprague's solution, but now one reduces the area by a whopping $4 \cdot 10^{-11}$, while the other, he claimed, reduces the area by $6 \cdot 10^{-18}$.
One author, in a parody of the usual optimistic prophecies of accelerating progress, commented that

...progress on this question, which has been painfully slow in the past, may be even more painfully slow in the future.

In 1980, Duff considered nonconvex subsets of the plane with least area such that every set with diameter one can be rotated and translated to fit inside it.  He found one with area
$$  \sim 0.84413570 $$
which is smaller than the best known convex solution:

G. F. D. Duff, A smaller universal cover for sets of unit diameter, C. R. Math. Acad. Sci. 2 (1980), 37--42.

In 2015, Philip Gibbs, Karine Bagdasaryan and I wrote a paper on this topic, mentioned above.  We found a new smaller universal cover, and noticed that Hansen had made a mistake in his 1992 paper.
Hansen claimed to have removed pieces of area $4\cdot 10^{-11}$ and $6 \cdot 10^{-18}$ from Sprague's universal cover, but  the actual areas removed were $3.7507 \cdot 10^{-11}$ and $8.4460 \cdot 10^{-21}$.  So, Hansen's universal covering has area
$$ \sim 0.844137708416 $$
Our new, smaller universal covering had area
$$ \sim 0.8441153 $$
This is about $2.2 \cdot 10^{-5}$ smaller than Hansen's.
To calculate the area of our universal cover Philip used a Java program, which is available online.  Greg Egan checked our work using high-precision calculations in Mathematica, which are also available online.  See the references in our paper for these programs and also a Java applet that Gibbs created to visualize Hansen's universal covering. It's fun to look at the smallest sliver Hansen removed, because it's 30 million times longer that it is wide!
More recently Philip Gibbs wrote a paper claiming to have an even smaller universal covering, with area
$$ \sim 0.8440935944 $$

Philip Gibbs, An upper bound for Lebesgue's universal covering problem, 22 January 2018.

Gibbs is a master at this line of work, but I must admit I haven't checked all the details, so it would be good for some people to carefully check them.
I've written a slightly more detailed account of the Lebesgue universal covering problem with some pictures here:

J. Baez, Lebesgue's universal covering problem (part 1), Azimuth, 8 December 2013.

J. Baez, Lebesgue's universal covering problem (part 2), Azimuth, 3 February 2015.

J. Baez, Lebesgue's universal covering problem (part 3), Azimuth, 7 October 2018.

If anyone knows of further progress on this puzzle, please let me know!
Comment: 15 October 2018 - I've updated my answer to discuss the new improved upper bound found by Philip Gibbs.
Comment: OP asks about minimal diameter (not minimal area), which is probably not the Lebesgue's original problem, but still interesting. Do you know anything about this  question?
Comment: Philip Gibbs says this other problem is harder; I don't know anything about it.
Answer: The problem has been studied for various groups $G$ of isometries of $\mathbb R^n$.  A set $K\subset \mathbb R^n$ is called $G$-universal cover iff every set of diameter 1 is contained in $gK$ for some $g\in G$.
V. Makeev  proved that the mean width of a $T_n$-universal cover  is greater or equal to $\sqrt{2n/(2n+1)}$, where $T_n$ is the group of translations of $\mathbb R^n$. For $n=2$ the estimate is sharp; the perimeter of a $T_2$-universal cover $\geq 2\pi/\sqrt{3}$  (link).
M. Kovalev obtained a rather explicit  description of all minimal $D_2$-universal covers, where $D_2$ is the group of all isometries of $\mathbb R^2$.

Theorem. Every minimal universal $D_2$-cover $K$ is star-shaped.  There is a polar coordinate system (with the centre in $K$) such that
$$\partial K=\{(\phi,\rho(\phi)):\ 0 < \phi\leq 2\pi\},$$
where  $\rho=\rho(\phi)$ is Lipschitz and for any $\phi\in[0,2\pi]$
$$c^2\leq \rho(\phi) \leq 1 - c^2,\qquad c=1-1/\sqrt{3}.$$
Comment: It's worth noting, for nonexperts, that "minimality" here means that the set $K$ doesn't contain a subset with the desired property.  This is implied by, but does not imply, area minimization.  Indeed, "minimal" solutions to the Lebesgue universal covering problem are known, while area-minimizing ones are not.
Comment: The link to the search will only work for people with MathSciNet subscription - as [previously discussed on meta](https://meta.mathoverflow.net/q/3303).
But it seems a reasonable guess that it is supposed to be Kovalev, M. D. The smallest Lebesgue covering exists. (Russian) Mat. Zametki 40 (1986), no. 3, 401–406, 430; [MR869931](https://mathscinet.ams.org/mathscinet-getitem?mr=869931).
Answer: It's Lebesgue Minimal Problem. It's still open, though there are some bounds to the area of such set, for example there are lower bound for area $S\ge \frac{\pi}{8}+\frac{\sqrt{3}}{4}$
it's not hard to show that such set must have diameter larger or equal to $\frac{\sqrt{3}}{3}+\frac{1}{2}=1.077350...$
Our set must have a subsets congruent to the equilateral triangle with side 1(ABC) and to the circle with radius 0.5(with center O). If I - the incenter of triangle ABC, and $O \in BIC$, then consider point D which lies on the radius perpendicular to BC. Then $AD \ge AI+OD=1.077...$.
Comment: You say $O \in BIC$ but your diagrams seems to have $O \notin BIC$.  Still, it obeys $AD \ge AI + OD$.
Answer: As other answers have pointed out there are different versions and generalizations of Lebesgue's Universal Covering Problem.
His original question from 1914 in a letter to Pál has been quoted as
"What is the smallest area of a convex set in the plane that contains a congruent copy of every planar set of unit diameter?" (see "Research Problems in Discrete Geometry" Brass, Moser, Pach.)
Variations of the question ask for the minimum diameter or perimeter. Sometimes the convexity condition is relaxed
but an advantage of the convex case is that the existance of a universal cover with minimum area is then ensured by
the Blaschke selection theorem. Congruence allows reflections as well as translations and rotations but in some
statements of the problem reflections are not allowed. The universal covers found by Pál and improved by Sprague do not require reflections.
The best known convex universal cover by Hansen from 1992 does require reflections.
Hansen's area is 0.844137708435197570894066994 and Duff gave a smaller non-convex  cover of 0.84413570
(see other answers for references)
However, it is possible to improve on these upper bounds as follows: Start with a regular hexagon circumscribed round
a circle of diameter one. As Pál showed this is a universal cover. Pál considered the eight sided shape formed by
removing two corners from this hexagon using cuts made with two lines tangent to the circle in such a way that the
new edges are sides of a regular dodecagon inscribed in the hexagon. Now consider a more general case where the two cuts are still tangent to
the circle and still at an angle of 60 degrees to each other but at a small angle $\sigma$ to the edges of a dodecahedron.
The remaining shape is still a universal cover.
To see this consider the six triangles at the corners of the hexagon each outisde a line tangent to the circle
and at the same slant angle away from the edge of a regualar dodecagon. In the diagram these are labelled A,B,C,D,E and F.
Any shape of diameter one can be fitted inside the hexagon. The least distance between opposite triangles such as A and D
is one, so the shape cannot be in the interior of both triangles. This is true for each of the three pairs of opposite
triangles so the shape can only be inside the interior of at most three of these triangles. Each of the possible cases
can be checked to see that the shape can then be rotated through some multiple of sixty degrees around the centre so
that it is still inside the hexagon but not inside the triangles E and C. Therefore the hexagon with these two
triangles removed is a universal cover.

This cover can be reduced further by generalising the argument used by Sprague. First observe that any shape of diameter one
can be contained inside a curve of constant width one (such as a cirle of unit diameter, or a Reuleaux polygon) so to prove
that a set of points is a universal cover it suffices to show that it can cover a set congruent to every curve of constant
width one. When such a curve is fitted inside the hexagon it will touch each of the six sides of the hexagon at a unique
point. On the side running from D to C it must be to the left of the point P on the corner of the removed triangle C.
This means that all points near A outside an arc of radius one centred on P can be removed. Similarly the curve must touch the edge
of the hexagon from E to D somewhere right of N, the corner of the triangle E that has been removed. So all points
outside an arc of radius one centred on N can also be removed. The remaining cover then has a vertex at X where these
two arcs meet. This reduces the cover but not enough to make it smaller for any non-zero value of $\sigma$ than Sprague's universal
cover which is the case $\sigma=0$.

One more piece can be removed if we use the freedom to reflect shapes. The axis of reflection to use is the line
from the midpoint M of the side of the hexagon from E to D, to the midpoint of the opposite side. A shape fitted into
the hexagon with the corners E and C removed can be reflected about this axis provided it also does not enter the
triangles F' and D' which are the reflections of C and E about the axis. When this is the case we will choose to
reflect the shape if it touches the side from E to D at a point nearer to E than D. Remember that it touches the
opposite side at the opposite point which is therefore also reflected to be nearer the corner of the hexagon at B
than the one at A. If we draw an arc centred on M of radius one it cuts off a larger region near A and no shape that
can be reflected can be beyond this line. The point where this meets the arc centred on P is marked W.

This means that the only shapes that can have points outside that arc are ones that cannot be reflected. This means
they must enter some point inside the regions F' or D' If they have a point in D' then they cannot have a point in A'
which is the triangle whose points are at a distance of more than one from all points in D'. Draw one more arc centred
on Q at the corner of the triangle C' which is the reflection of the region F. All points in C' are at a distance of one or
more than one from points in F' so the arc will touch the region F' but not enter it. This arc will meet the arc centred on M
at a point Z and the arc centred on N at a point Y. Now consider the fate of points inside the region XYZW bounded by
the four arcs. It is a general property of curves of constant width one that if two points are inside the curve
then all points on an arc of radius one through the two points are also inside the curve. Suppose then that a shape
fitted inside the hexagon had a point in XYZW and also in F' We could then join those two points with an arc but
between the two points it would be outside the arc centered on Q and would therefore go outside the hexagon. This
is in contradiction with the premise so we conclude that no shape fitted in the hexagon can have a point in both XYZW and
F'. It can also be verified that for angles $\sigma$ less than 9 degrees the region XYZW is inside the triangle A'.
Therefore shapes with a point insode XYZW do not have points in F' or D' and can be reflected. However, we have already
determined that such shapes will not have points in this region. This proves that the region XYZW can
be removed from the universal cover.
It turns out that this is now sufficient to construct a universal cover smaller than the ones of Hansen and Duff.
Even if we restrict ourselves to the convex case and remove only the part of this region that leaves a convex shape,
the area of the universal cover for an angle $\sigma = 0.4$ degrees can be computed to be 0.8441177
There are other small pieces that can be removed from this cover to reduce it further.
This of course does not solve Lebesgue's Universal Covering Problem which is hard because of the complexity of the
problem (not because the numbers involved are irrational as asked in the question) However I can break the problem down with three conjectures
1st conjecture: The minimal convex cover for any subset of curves of width one will always fit inside a hexagon
circumscribed around a circle such that opposite sides are parallel. It can be shown that such a shape is a universal cover
but only computational evidence supports the truth of this conjecture.
2nd conjecture: For any such hexagon there is a minimal convex cover. The conjecture is that the minimum area for such a
cover is found for the case of the regular hexagon. Again computation supports this conjecture.
3rd conjecture: For a regular hexagon the minimum cover within the hexagon for any subset of curves of width one
fits inside the shape formed by removing two corners by two lines tangent to the inscribed circle and at an angle of 60 degrees
to each other (as above)
If all of these three conjectures are true then the problem of finding the minimal convex cover reduces to finding
the minimal cover inside a shape of this form. The conjectures may be hard to prove but if are true then the final problem
may be tractable using methods similar to the above proof.
Answer: Philip Gibbs wrote:

the area of the universal cover for an angle $\sigma=0.4$ degrees can be computed to be $0.8441177$.

I believe Philip later noticed an error in this area computation.  He also later noticed that there's a constraint that $\sigma$ needs to obey, which is not obeyed by this choice: the best choice of $\sigma$ is about
$$ \sigma = 1.294389444703601012^\circ $$
and the area of the universal covering this gives is about
$$ 0.844115297128419059\dots .$$
We wrote a paper on this together with Karine Bagdasaryan, and Greg Egan did some high-precision calculations that give the above numbers.   Our paper explains the details:

John C. Baez, Karine Bagdasaryan and Philip Gibbs, The Lebesgue universal covering problem.

Abstract: In 1914 Lebesgue defined a "universal covering" to be a convex subset of the plane that contains an isometric copy of any subset of diameter 1. His challenge of finding a universal covering with the least possible area has been addressed by various mathematicians: Pal, Sprague and Hansen have each created a smaller universal covering by removing regions from those known before. However, Hansen's last reduction was microsopic: he claimed to remove an area of $6 \cdot 10^{−18}$, but we show that he actually removed an area of just $8 \cdot 10^{-21}$. In the following, with the help of Greg Egan, we find a new, smaller universal covering with area less than $0.8441153$. This reduces the area of the previous best universal covering by a whopping $2.2 \cdot 10^{−5}$.


Tarski's decidability proof on real closed field and Peano arithmetic

Question: It seems very confusing that real closed field (which also can be used as the theory of real number) is decidable, while Peano arithmetic, which seems to be a subset of real closed field is undecidable.
What am I getting wrong?
Answer: As William pointed out in his answer, Peano Arithmetic is a certain first-order theory which describes properties of $\mathbb{N}$, the natural numbers (that is, $\mathbb{N}\models\text{PA}$). However, it is incomplete (as shown by Gödel), and hence it is not equal to the complete theory of the natural numbers, denoted $\text{Th}(\mathbb{N})$, which consists of all first-order statements true of $\mathbb{N}$. The situation is different with the theory of real closed fields - this is a complete theory, and $\mathbb{R}\models \text{RCF}$, so $\text{RCF} = \text{Th}(\mathbb{R})$.
Your main confusion, however, seems to be about how $\text{Th}(\mathbb{N})$ can be so much more complicated than $\text{Th}(\mathbb{R})$, despite the fact that $\mathbb{N}\subset \mathbb{R}$. If $\text{Th}(\mathbb{R})$ is decidable, why can't we use the decision procedure to decide all questions about $\mathbb{N}$?
This phenomenon is possible because there is no single first-order formula (in our language $\{0,1,+,\cdot\}$) which picks out the natural numbers as a subset of the reals. To decide whether a sentence is true about $\mathbb{N}$ by asking a question about $\mathbb{R}$, we would have to somehow ensure that all quantifiers talk only about the integers.
To illustrate the difficulty, consider the sentence $\phi:\forall x\,\exists y\, (y+y = x)$. $\phi$ is certainly false in $\mathbb{N}$. But to use the decidability of $\text{Th}(\mathbb{R})$ to see this, we would need to be able to express that for all natural numbers $x$ there is a natural number $y$ such that $y+y = x$, which we cannot do. Allowing the variables to range over $\mathbb{R}$, $\phi$ is true.
Intuitively, it is questions about divisibility like this one which make $\text{Th}(\mathbb{N})$ so much more complicated than $\text{Th}(\mathbb{R})$. The formula $\psi(x,y):\exists z (x = y\cdot z)$, expressing that $y|x$, gives access to all the complexity of the prime numbers. By contrast, the reals are extremely homogeneous: $\psi(x,y)$ is true of a pair of reals $(a,b)$ whenever $b\neq 0$.
Comment: Exactly - your definition of $\mathbb{N}$ inside $\mathbb{R}$ as the smallest subset containing $1$ and closed under $+1$ is explicitly second-order.
Comment: I am told that it is only the first-order version of complete ordered field theory (i.e. with no quantification over sets) that is complete and decidable. Oh, well.
Comment: I have constructed the set of natural numbers from the ordered field axioms, and derived the equivalent of the Peano Axioms. See: http://www.dcproof.com/DerivingPA.htm  Do I not have a first-order formula here which picks out the natural numbers?
Comment: You can't have anything like the least-upper-bound axiom in a first-order language because such an axiom quantifies over sets of numbers. As such, you cannot have any first-order theory of complete ordered fields (the real numbers).
Answer: The primary reason that the first-order theory of the reals proves to be much simpler than Diophantine (integer) analogs is that the associated geometry is much simpler. Namely, the sets that are definable by real-polynomial (in)equations - so-called semi-algebraic sets - can be decomposed into a finite number of cells (e.g. cylinders) where, on each cell, every defining polynomial  has constant sign. Therefore testing the truth of a first-order statement reduces to a finite number of tests on constant-sign cells, which is trivially decided by simply evaluating the polynomials at any point in the cell. This yields an algorithm for deciding the truth of first order statements about the reals - the so-called cylindrical algebraic decomposition (CAD) algorithm  (an effective realization of Tarski's famous method of quantifier elimination for the reals).
For example, in the one-dimensional case the sets definable in the first-order language of $\rm\:\mathbb R\:$ are simply finite unions of intervals. So, e.g. to test if $\rm\ f(x) > 0\ \Rightarrow g(x) > 0\ $ we can employ Sturm's Theorem to partition $\:\mathbb R\:$ by the finite number of roots of $\rm\:f,\:g\:,\:$ and then choose a point $\rm\:r\:$ in each interval and test if $\rm\ f(r)>0\ \Rightarrow\ g(r)>0\:.\:$  The CAD algorithm works analogously in higher dimensions, by projecting higher-dimensional cells down to lower dimensions. The key property is a result of Tarski and Seidenberg that semi-algebraic sets are preserved by projections. The essential structure has been generalized in model-theoretic study of o-minimal structures.
Contrast this to the study of (first-order) Diophantine equations. Anyone who has studied number theory quickly appreciates the immense complexity of the structure of solution sets of integer polynomial equations, e.g. Pell equations, elliptic curves, etc. Undecidability results imply that there can be no simple uniform recursive decision procedure like that for the reals. Each leap into a higher dimension may require completely novel ideas. But this shouldn't be viewed as a detriment. Rather, it yields an unending source of rich problems that will constantly challenge our ingenuity.
Answer: When you talk about decidability in this context, you mean the decidability of the theory, not any particular model.
The theory of real closed fields and peano's arithmetics are different theories. The axiom of real closed fields include all the ring axioms. Peano arithmetics does not. Peano's arithmetics include induction whereas the axiom of real closed fields do not. As theories, i.e. sets of formulas, Peano Arithmetics and Real closed fields are just different. To say one is a subset of another does not even make sense.
You may be confusing the concept of models with the theory. The real numbers may be a model of real closed fields and the natural number is a model of peano arithmetics within it. However, formulas of real closed fields are decidable. In particular in these formula, the quantifier ranges over the domain of the real closed fields. For example, real closed field may decide whether there exists a solution to some equation; however, the quantifiers range over the domain of the real closed fields. You do not know anything about whether these solution lie in the natural numbers, for instance.
Answer: As others point out, the main reason for this seemingly paradoxical situation (all natural numbers are real numbers but the theory of the first one is much more complicated than the second one) is because natural numbers are not definable in the language of RCF.
We can use the same language to talk about them, say $\{0,1,+,\cdot\}$ and natural numbers are a substructure of real numbers. However being a subsructure doesn't make things simpler. Take for example prime numbers, do you expect that their theory in the same language is simpler than the theory of natural numbers? It doesn't need to be the case. The main issue as others have point out is quantification. Quantification over natural numbers can generate very complicated sets (arithmetical hierarchy) where as quantification over reals does not, in fact if we simply add order to the language then every quantified formula is equivalent formula with no quantifiers for reals, i.e. quantifiers do not increase the complexity of sets we consider for reals and can be eliminated (and that is how it is proven that RCF is decidable). We can show that this is not possible for natural numbers, quantifiers can increase the complexity of sets. The undecidability follow essentially from the fact that we can talk about computation of machines in the theories like PA and much weaker theories about natural numbers.
It might be helpful to look at something in between: consider integers. We can define natural numbers by using the fact that every natural number is equivalent to sum of four squares. Therefore they are going to be as complex as natural numbers. Another more tricky set in the middle: consider rational numbers. Their theory is similarly undecidable, however showing that we can define natural numbers in the their theory is much more complicated (this nice result is originally due to Julia Robinson). Another example: consider complex numbers. Their theory ACF is kind of even simpler than real numbers.
To make it even more clear, if we just add a predicate to the language that tells us which real numbers are natural numbers, then theory of real numbers with this predicate becomes undecidable. The main issue is that without such predicate natural numbers are indistinguishable from other real numbers in the language.


Will artificial intelligence cause mass unemployment?

Question: Everyone is afraid of losing their job to robots. Will or does artificial intelligence cause mass unemployment?
Answer: The nuanced, boring answer is that it depends on your definition of AI. Most people wouldn't say that the rule-based systems designed in the 70's are AI. The amazing leaps in machine learning are almost taken for granted as well (think about how normal speech and facial recognition have become). This is known as the AI effect; when we become accustomed to the technology, it loses it's 'magical aspect' and is thus no longer labelled as AI.
Since AI is so diverse and difficult to define, the question becomes incredibly abstract. Did Siri cause all secretaries to become unemployed? Did TurboTax replace all accountants? Some parts of AI will affect jobs, or even make them redundant yes. On the other hand, it will give rise to new jobs as well. It is therefore impossible to generalize it as 'AI will cause massive unemployment'.
This is not a new phenomenon, however, it has been part of the human economy ever since the industrial revolution (probably even before that, but I am not a historian). The invention of the car crippled the horse-and-wagon industry, but it brought along new jobs as well.
Answer: Up to a point. Some jobs will IMHO not easily be replaced by robots, others more easily. Some could be, but I hope common sense will prevail and stop that.
Manual jobs: fruit picking, warehouse picking, and cooking are some jobs that need really subtle hand control, and precise handling of fragile items. I think those will be harder to automate than eg car factory robots.
Customer facing roles: receptionists have to do a wide variety of tasks. While some of the tasks might be aided by AI systems, a good PA or receptionist cannot easily be replaced. Also, many people would much rather interact with a fellow human than with a machine, at least in some situations.
Judgments: a lot of jobs require judging a situation, balancing risks, and making 'gut' decisions. While AI systems can do them, I think many would still require human intervention. I for one wouldn't like to be sentenced by a robo-judge, or examined by a robo-doctor. True, humans also make mistakes, but they would hopefully err in less potentially disastrous ways.
Administration: again, many tasks can be assisted, which would lead to a reduction in head count, but a general AI is still off the horizon, so you'd still need humans.
Creative arts: not likely. Would you want to read computer-generated novels, or look at computer-generated paintings?
I could go on... in general: I think AI systems will make a lot of tasks faster and easier, so you need fewer people to do them. Some jobs cannot realistically be done by machines in the near-to-mid future, so overall we need not worry too much.
Comment: Yes. Just as the horse lost it's employment as the form of transportation to the car.
Comment: When the mechanical loom was invented, people were up in arms. Steam engine, etc. Every new invention makes people think there will be mass-unemployment.
Answer: Yes , but you should be happy about it. It is like usage of mashines in industry causes "mass unemployment". Modern tendencies is unfortinatly not like that - industry has not so bit interest in full automatisation cause of huge amout of cheap work force from immigrants - it is a bad factor, not that "ai takes your job".
And further, robot what would take your job must not be realy smart - in could be just some doll without AI just making what is in programm...


Molecular Cloning- Blunt end restriction endonucleases

Question: I work in a microbiology lab where we do a lot of cloning. I have always used restriction endonucleases to cleave the DNA to have sticky ends and not blunt ends. I currently am working on a project that suggests using a blunt end restriction enzyme, such as EvoRV, and then to perform a T-tailing of the plasmid, pl4440. My question is why do I not treat my pcr'd insert with a restriction digest just like I would the plasmid as the protocol suggests? How do I know the insert has ends that are able to ligate into the plasmid if I don't design the restriction sites into the primers?
Comment: If i understand the question correctly, your are asked to make a single EcoRV cut in your backbone plasmid and make blunt ended inserts?
Comment: Well, the plasmid has a single cut in the back bone but I'm not sure how the insert will ligate correctly if I do not treat the insert to make blunt ends.
Comment: if then you can only use EcoRV and nothing else then you have to use blunt ended inserts but you will have problems with insert orientation and multiple ligations of insert!
Comment: I **think** you can turn sticky ends into blunt ends with klenow enzyme. Been too long to be sure.
Comment: @user137, yes thats correct! the OP can use DNA polymerase I, large (Klenow) fragment (from NEB) to blunt end the insert. or alternatively an EcoRV site can be designed into the primers so long as it is not present elsewhere in the insert.
Comment: Just remember that in blunt end ligation the insert can go in either direction, so you need some way to differentiate direction, such as sequencing, and that half of your clones that look right on gel will be backwards.
Answer: To know if the ends are compatible see compatible cohesive ends (isoschizomers) from NEB website.
If you are unsure then do a blunt end ligation. For blunting a sticky end (From the comments):
Use Klenow or Pfu polymerase (or any proofreading/high fidelity polymerase).
You can use Taq if you intend to to a TA cloning.


What is inside the Gate of Truth?

Question: In FMA and FMA: Brotherhood, the Gate of Truth is shown to have an eye and a bunch of arms inside of it. The arms drag people into the gate during human transformation. I know that the creature sitting in front of the Gate is Truth, but what is inside of the gate (i.e. the arms and the eye)?
Comment: [This answer](http://anime.stackexchange.com/a/3209/274) may answer your question.
Comment: I had read that. If there aren't any more specific answers given in canon, then I'll just take that answer, but I was hoping there had been a bit more description in-canon than just "God".
Comment: I'll see what else I can find. There is a bit more to it, but that's the general idea.
Answer: In the manga and Brotherhood, there are two things within the Gate of Truth: God, and all alchemical knowledge that has ever existed or will ever exist. The Fullmetal Alchemist Wiki says it best:

It is the source of alchemical knowledge and the entrance to God's domain, where no mortals are meant to tread.

As is stated in this answer, this is where mortals end up when they attempt human transmutation. They pay a toll, and are granted access to all alchemical knowledge within.

While an alchemist is within this realm (the realm of non-reality), they will inevitably learn something about alchemy, and in fact anyone in the series who recalls this information gains the ability to perform transmutation without a circle.
In the 2003 anime, there are some differences. Notably, the realm itself is the source of alchemical energy (rather than knowledge), and the Gate acts as a portal between that energy and the FMA universe. You can read more about the differences on the FMA Wiki.
Answer: When truth announces who he is he is one or all or also known as god the universe or truth behind the door is nothing but not nothing it is what you think it is a fragment of your missing mind... Presented in one of the episodes the door leads to many things presented in a cross over of Hitler and the Nazi war Edward Elric entered the portal or gate way what ever you wanna call it and entered our world when the Nazi War happened. Regarding the gate way to truth that is truth the gate way takes different forms if you ever followed the story correctly it specifically states that Gate Way is a kinda door or portal to what ever your dreams can imagine. The professor from the Germany side of the door talks among parallel worlds that are connected invisibally and can't be seen and in different dimensions there is another type of you that has it's own personality. Clearly though you can only do this by being in the Full Metal Alchemist world specifically the gate way can lead to many places. ^^' That's why there is no point to the universe because everything is infinite just read the Manga and watch the show and decode what it is trying to say... ALSO An edit even though you have different yous in different dimensions supposedly the real you is in the Full Metal Alchemist World since other dimensions logically don't exist... Who ever you are is not what you are but what you are is in the FMA world... Truthfully the show talks if you play god it back fires but also we already know everything because the answer to everything is clearly NOTHING.
Comment: Can you clarify what you're trying to say here (and if possible provide some sources)?
Answer: The Gateway of Truth is the meeting point of infinity, which was theorized when the being inside of it describes himself as God, The Truth, and yourself. Edward comes to discover that the gateway you see is actually your perception of the truth, explaining why Alphonse's and Izumi's gates look different. You can also see all of Edward's memories when he was inside the truth. However in the original Fullmetal Alchemist anime, the gateway is access to all universal knowledge, explaining why you can see John F. Kennedy and moments from World War II.

What is inside the Gate of Truth?


2005 ford sportrac

Question: Overheated and found that I had no coolant. Engine cool filled radiator and reservior to Max line with water. It has been insanely hot in California I might add. Drove home 90 miles and next day drove another 130. Found on way home slightly over heating, immediately turned off air and pulled off. Temp went down. Continued another 12 miles to home. Home found I had to add entire bottle of premixed coolant I had just bought. Reservoir was empty as well. Fan running, does not to appear have leaks. Any ideas?
Comment: Is there any coolant in your oil? Running with no coolant could cause the head gasket to destroy itself, or it might be the case the head gasket has been destroyed already (hence explaining having no coolant). The coolant could also find a way to the combustion chambers and exit via the exhaust, if there's a head gasket problem.
Answer: Assuming you are asking why it nearly overheated again, there are a few possible reasons this could be.

When you added from empty, there was some air stuck in the system.
When it worked its way thru, you ended up with a low level again.
There is a blockage in the cooling system that is causing it to not
circulate.
The coolant is leaking out.
The radiator cap is bad, which allows the system to boil off.

You have left so much detail out, we can't answer with certainty. You need to monitor the coolant level. If it is still going down and there are no external leaks, there is likely a major problem with the engine.

2005 ford sportrac


Halachic Restrictions not in the Torah, and "shall not add"

Question: How can halachic restrictions that are not in the Torah be reconciled with the Deuteronomy 4:2 passage, which says, "Ye shall not add unto the word which I command you, neither shall ye diminish from it"?
Comment: http://judaism.stackexchange.com/questions/15509/
Answer: Such halachic restrictions fall into two categories: those which we believe were given as part of the Oral Law along with the Torah, and those that were clearly put in place later by the rabbis.
The former is not adding, as it was part of the given word. E.g. Deuteronomy 12:21, to eat non-sacrificial meat, just slaughter it "in the manner I have prescribed"; the method of kosher slaughter does not actually appear in Tanach, it was transmitted orally until the Talmud codified it.
The latter make a strong point of being distinct from Biblical obligations.
Here's an excerpt from Maimonides' code, Hilchot Mamrim (laws regarding rebellion against rabbinic authority) Chapter 2:

As a rabbinic court can make decrees to prohibit that which had been [Biblically] allowed, and that prohibition stands for generation ... what is it that the Torah warned "thou shalt not add nor subtract?" Not to add nor subtract and establish something new forever and claim it as Biblical law, whether the Written Law or Oral Law.
How so? It is written in the Torah (Ex. 23:19) "thou shalt not cook a g'di in its mother's milk", and by oral tradition [of Biblical force] we learned that this verse prohibits cooking and eating the meat of any mammal, whether domesticated [such as sheep] or wild [such as deer], in milk; but chicken could be cooked and eaten with milk, as far as the Biblical law.
Should a rabbinic court go and allow wild-mammal meat in milk, they have "subtracted"; and if someone prohibits chicken in milk, claiming it was included in the Biblical prohibition on g'di, that would be "adding."
But if it was stated that: chicken [in milk] was permitted Biblically, but we [the rabbis] shall prohibit it, and inform the people that this is a decree to prevent problems; as people may say "chicken is okay because the Torah didn't literally say chicken, so it didn't say wild-animal either"; and the next person will say "any domesticated mammal is fine too except for goats", and the next person will say "goat meat is acceptable in the milk of cows or sheep, as it said its mother's, meaning that same species"; and the next person will say "I can cook a goat in goat's milk, as long as it wasn't this particular goat's mother" ... therefore, we shall prohibit all meat in milk, even chicken. This is not adding, it is creating a protective fence around the Torah. And so too all similar matters.
Answer: Excellent question.  The answer is that the Rabbis themselves are invested by the Torah with the duty to protect the basic halacha - see for example Deuteronomy 17:8-11, where it basically states that we are Scripturally bound to comply with the rulings of the Sanhedrin.  Furthermore see Pirkei Avos 1:1, which states, "... make a protective fence for the Torah."  This means that the Rabbis saw it as their duty to make enactments to protect the central Scriptural commandments, in the same way that one would put up a protective cordon several feet away from an open manhole to make sure that nobody falls in.
Your question still is pertinent - is this not "adding" to the Torah?  In order to avoid this problem, it is very important that we have to have a very clear distinction between those laws that are Scripturally mandated, and those that were enacted by the Rabbis.  You will find several practical differences in our approach to Rabbinic mitzvos as opposed to Torah mitzvos; there are a great many leniencies in practical halacha regarding Rabbinic mitzvos.
Answer: There are basically 2 types of additions which are allowed:

Those that prevent us from accidentally transgressing - as it says in the first Mishna in Pirkei Avot: Make a fence around the Torah.
Those instituted by the Sages and Prophets under the umbrella of " כְּכֹל אֲשֶׁר יוֹרֽוּךָ" - "you shall do whatever they teach you" (Devarim 17:10).

Keep in mind that not everything is written in the Torah; the Oral Law has many details missing from the Written Law. E.g.: How to make Tefillin, how to slaughter properly and how many walls a Sukkah needs.
Many Halachic Restrictions were given to Moshe at Sinai as part of the Oral Law, along with the written law.
Answer: There are  at least four answers to this question: Instead of explaining them, I will provide the relevant citations.

Maimonides which has already been cited here.
R. Judah Halevi in the Kuzari 3:41.
R. Joseph Albo, Sefer ha-Ikkarim 3:14, which is also the view of Maharal (Be'er ha-Golah, beginning).
Maharsha to Megillah 14a, as well as R. Chaim Volozhiner, Nefesh ha-Chaim 1:22.

(For an in-depth discussion of Maimonides' view, see here).
Comment: wfb, Welcome to Judaism.SE, and thanks very much for these citations!
Comment: this may be the same in depth discussion in Hebrew: http://www.scribd.com/doc/68286227/%D7%A0%D7%A6%D7%97%D7%99%D7%95%D7%AA-%D7%94%D7%AA%D7%95%D7%A8%D7%94-%D7%91%D7%9E%D7%A9%D7%A0%D7%AA-%D7%94%D7%A8%D7%9E%D7%91%D7%9D
Answer: The Verse (Devarim 4:2) states:

Ye shall not add unto the word which I command you, neither shall ye diminish from it, that ye may keep the commandments of the LORD your God which I command you.

Rashi's Commentary on the Verse says:

Verse 2: Do not add.
      For instance to place five parshiyos in the tefillin, to take five species [of fruit] for the mitzvah of lulav, to place five fringes [instead of four on one's garment]; and this also applies to (the command), "And do not subtract from them."

The various commentaries point out that all of Rashi's examples are adding or subtracting from a commandment that has a specific number associated with it. So, for example, if you have a garment with 5 corners, don't put tzitzit strings on the 5th corner. Doing so would violate this commandment. However, As the Gur Aryeh says, the Rabbinically legislated prohibitions which augment the Biblical restrictions are not considered additions, since they are clearly intended as protective measures to guard against infractions of the Biblical commandments.
The Chumash Shai LaMora brings this in the name of the "Sefer Zichron".


How to control what job offers are shown on the banners?

Question: I've set C#, JS and such as my tags of interests. Java isn't there, no Ruby-on-whatever neither. Nevertheless, most of (and some days all of) the ads are about the non-tagged technologies. How can I control what's shown to be inside a certain job banner?
Comment: Related: [Is SO monitoring its users and adapting ads based on that?](http://meta.stackoverflow.com/q/270896)
Answer: Until recently, you could not. The advertisements were only personalised towards your location, and your Stack Overflow favorite / ignored tags are not taken into account.
Recently, things have been improving however. Now advertisements are targeted at specific tags and interests, but the targeting is not customisable from the Stack Overflow user's end *.
If your Stack Overflow account is linked to your Careers account, then adding tags to your dislikes field in your Careers profile will be used to filter out job ads shown to you on Stack Overflow; this gives you at least some control over what is being shown to you.
If you wanted to see new jobs in tags you are interested in, you can set up a RSS feed or email subscription instead. Go to the Careers 2.0 homepage, set up a search, and use the icons at the top of the search jobs sidebar:

Otherwise, I'd just ignore the advertisement box. It is only in the sidebar, after all.

* Well, you can opt out from the data collection to stop all targeting. On that same page you can download the data set Stack Overflow has collected on you so you can see why you get certain job ads target at you.
Comment: I can tell you they are **not** localized to me. Unless we consider a commuter distance of 5000 kilometers, that is. One position was in New York (I'm in Stockholm) and it said "relocation" not "telecommuting". +1 for the extra info (although I'm not looking for a new job actively at the moment). However, I can point out that giving suggestion on Java positions for a C#-dude is like advertizing gay bars for a straight guy. Very unlikely to be of use. And occupies the attention that could've been devoted for something better.   :)
Comment: Hmm... I'm being approached by recruiters twice, three time a month. Stockholm is **very** hot area and nobody can find right people. So I think it's just a blunder in the algorithm. We've got great pay here and a lack of skilled professionals. On **any** platform, hehe. Maybe they're targeting worldwide and pay to SO for that exposure.
Comment: @KonradViltersten: perhaps; I tend to ignore the box in any case, so I may have missed the out-of-area posts quite easily.
Comment: @KonradViltersten: I know where you are coming from; my Careers profile lists Java as an explicit dislike, keeps the recruiters at bay. Being located in the south-east of England, I am served with London positions all the time, but perhaps there is less choice around Stockholm and extra positions from around the world are added in to make up the numbers.
Comment: Pragmatic approach. Darn effective, now that I think about it. One need not to improve every thing every time.   :)
Comment: "the targeting is not customisable from the Stack Overflow user's end" - then what is the point of these ads? Impressions? Would the CTR be about the same if the job ads *were* customizable, so the SO devs don't care about making that feature available?
Comment: @DanDascalescu: the point is to reach the people that are most likely to actually be interested in the jobs advertised. Also see [the data collected on you](http://stackoverflow.com/users/prediction-data) to see what kinds of jobs might be targeted at you.
Comment: While full control is still a ways down the road, we've just started filtering out jobs that have tags that you have explicitly disliked on your Stack Overflow Careers profile, if you've linked your SO account with your Careers account
Comment: @rossipedia: cool, I'll keep updating the answers as changes come in. :-)


How to disable dedicated GPU in Ubuntu 12.10?

Question: I want to completely disable my Nvidia GT610m card, and let Ubuntu rely only on the HD4000.
How can I achieve that?
Answer: Depending on how your notebook is wired up (obviously you can't disconnect the monitor cable from a mobile GPU), you may not be able to do this. If you can do it, it wouldn't be from within Ubuntu -- it'd be in the bios.


Why couldn't Gollum see Bilbo in the caverns if he could see Frodo at Mount Doom?

Question: If Gollum was able to see Frodo and attack him when he put on the Ring, why couldn't he see Bilbo in the caverns?
Comment: Who said Gollum was able to see Frodo?
Answer: Several reasons.  First, Gollum attacked Frodo quickly, before he had a chance to move.  Second, Frodo was focused on claiming the Ring, not on dodging the unexpected, and third, Frodo didn't really have a lot of room to maneuver, anyway.

The light sprang up again, and there on the brink of the chasm, at the very Crack of Doom, stood Frodo, black against the glare, tense, erect, but still as if he had been turned to stone.

Frodo is at a cliff-edge.  Even if he perceived the danger of Gollum, he could only, really, move away from the edge -- it would be foolhardy to run along the edge.  So Gollum could in large part anticipate when Frodo might move.

Then Frodo stirred and spoke with a clear voice, indeed with a voice clearer and more powerful than Sam had ever heard him use, and it rose above the throb and turmoil of Mount Doom, ringing in the roof and walls.
'I have come,' he said. 'But I do not choose now to do what I came to do. I will not do this deed. The Ring is mine!' And suddenly, as he set it on his finger, he vanished from Sam's sight. Sam gasped, but he had no chance to cry out, for at that moment many things happened.

Note that Frodo is speaking with a "clear" and "powerful" voice. He's not ducking, he's not watching for danger -- he's claiming enormous power and mastery in a place of great power. At this moment he feels no fear, no doubt. (But he really should have...)

Something struck Sam violently in the back, his legs were knocked from under him and he was flung aside, striking his head against the stony floor, as a dark shape sprang over him. He lay still and for a moment all went black.

And it all happened very quickly.

Sam ... saw a strange and terrible thing. Gollum on the edge of the abyss was fighting like a mad thing with an unseen foe. To and fro he swayed, now so near the brink that almost he tumbled in, now dragging back, falling to the ground, rising, and falling again. And all the while he hissed but spoke no words.

Gollum rushed Frodo while he was still visible; And though Frodo disappeared before Gollum got to him, Frodo was occupied and trapped; Once Gollum had grappled with him, invisibility was of little value -- wrestlers fight as much by feel as by sight.
Answer: While not stated explicitly in the book, the film The Return of the King shows how.
In this clip, at the 2:20 mark, Gollum clocks Sam, and then looks around for Frodo. The shot clearly shows him recognizing that he can see Frodo's footprints, and so he knows exactly where Frodo is standing.
Comment: *"While not stated explicitly in the book"* This line immediately put me off, because the film just shows what the producers made up out of possible lack of evidence, and the book is the *only* important source here. (Other maybe than some letters written by JRR later)
Comment: The book also didn't explicity showing them using the bathroom, but we can probably assume they did it regularly.
Answer: You can see this in the scene Frodo may be invisible however the dust he is kicking up is not, presumably the cavern had a lot less dust or possibly not enough light to see the dust being kicked up.


Regarding 'gui' and 'command-line' tag

Question: Given Jeff's answer, does it make sense to continue using the gui and command-line tags?
Answer: If we really are going to merge GUI and command-line answers to "how do I do X  with a [gui|command-line]" then it probably makes sense to abandon the command-line tag since we already have various tags such as shell and terminal for other types of command-line questions.  The gui tag could still be really relevant to questions that are about various graphical user interfaces.
Comment: Seems to be a different, only at the surface related question.
Answer: Jeff was merging two questions which shared a common theme. However, I think Jeff was merely trying to merge two duplicate questions, because the content was nearly identical. It appears that you purposely asked the second question and duplicated the content to demonstrate a point, which is fine. He was not attempting to merge tags, which is a far different topic.
Another way to view this question is "Should we merge the GUI and Command-line tags?" My answer there is a firm No.
The user interface is an important way to access the operating system. There are major differences between any GUI and a command-line interface, and interfacing with X/Gnome/KDE is completely different the using a shell. I would also argue that there is often very little in common between a GUI and a CLI solution.
We need a way to distinguish between these two different methods.
It's very common for an organization to manage hundreds of Un*x machines, and not a single GUI among them. Given this environment, it would be wrong to provide a bunch of GUI solutions when they clearly need a CLI solution.
Comment: Just to be clear, are you implying that the comment I left on Gilles' answer is valid, that one can go ahead and re-ask...?
Answer: Yes. Sometimes a program being GUI or command line is an important part of the requirements. If you take Jeff's reasoning to an extreme, we could merge every question into "How do I do something?"
(For what it's worth, I don't think this particular merge should have been done.)
Comment: By this you actually mean that I can go through all questions that actually required the answer to be CLI, and re-ask them, only this time to require GUI. Can you add some justification.
Comment: @Tshepang: No, I mean that for *some* questions requesting a CLI answer, it's fine to ask a separate question with similar requirements except that you want a GUI.
Comment: What are you excluding by *some*? That is, where is it not okay to re-ask?
Comment: @Tshepang: When answers to one of the questions would be mostly useless to people asking the other question and vice versa. The avi file merging is borderline: the answers are mostly disjoint (which suggests the questions should be separate), but a lot of people wouldn't mind either way (which suggests the questions should be merged).

Regarding 'gui' and 'command-line' tag


Is there an online RobotC IDE?

Question: I'm on a Linux chromebook, and am using a Vex robot, and am wondering if there is an online IDE for RobotC that will run my computer?
Answer: Unfortunately, there is no RobotC version you can run on your computer as configured - RobotC is currently (as of March 2017) only available for Windows.
Depending on your Chromebook, there may be a way to run Windows software - check out this guide:
https://www.howtogeek.com/173353/how-to-run-windows-software-on-a-chromebook/
Answer: You might be able to get it to run via wine. If that doesn't work you can definitely get it running in a virtual machine.

Is there an online RobotC IDE?


Why do my LED Christmas lights glow when plugged into a light fixture via an adapter?

Question: I have an outside light that is controlled by a light switch inside. I took the light out and put a adapter to plug in an extension cord to run LED Christmas lights. When I turn the switch off, just the first strand of lights glows very faintly so I wonder is my switch used as the ground and the lights are finding ground at the tree they are hung at.
Comment: Is the light switch inside a "smart" switch, dimmer, timer, or is it a simple snap switch?
Answer: If the switch is a dimmer, remote operated switch, or a switch with a neon backlight, it may allow a small amount of leakage power even when off. It's not enough power to light an incandescent light, but it may be enough to dimly light led lights.
Answer: Try running an extension cord over to a standard outlet and troubleshoot your lights in that configuration before you bring the adapter and switch into the picture.
You can also separately test the switch and adapter by plugging in a standard incandescent shop light.
The key is to isolate and remove as many variables as possible from the problem.
Answer: I have the same situation. When I turn my Christmas lights off outside by an outdoor multi plug switch they glow very faintly.  I noticed that some LED strands when unplugged from the wall altogether still glow faintly for a minute.  So I don't think its a leakage as much as power still in the power cords, probably 100ft or more when you add it all together and it just takes a while to dissipate.  Even if it was a leakage it's so minute it would be negligible. Maybe .01 of a watt.  If you had a multimeter you might be able to see it, maybe not.


Will the oceans swallow all of the land?

Question: I am wondering why, since the oceans could feasibly hold all of the dry land, they in fact do not.  For example, the Mariana Trench drops to -10.9km, while Mount Everest only reaches to +8.8km.  Further, the oceans cover 71% of the surface.  This causes me to suspect that the negative elevation parts of the Earth could easily hold the positive.  Due to erosion, will the oceans eventually consume the mountains?  What is the timescale involved with such a process, and what forces counteract it?
Answer: If no new mountains were built, yes. Ultimately the processes of erosion would render the continents flat, and the seas would be left shallow and filled with sediments. The reason that this doesn't occur is that tectonic processes cause 'orogenesis', the formation of new mountains. The subduction of oceanic crust at plate margins is also responsible for trenches such as the Mariana, and ultimately recycles sediments into the mantle; thereby preventing the ocean floor from filling up.
Comment: Also in general terms Continental Crust is continioulsy growing. It is an helical cycle.
Answer: Interesting question.
As long as plate tectonics persists, then, as @Arkenstein mentions, mountain building and ocean basin formation will create enough topographical variety to ensure that land exists. Moreover, as long as there is a distinction between continental and oceanic lithosphere, we will have continents.
But this will not last forever. At some point, plate tectonics will stop as Earth's internal heat production falls. At this point, assuming nothing else happened, we would gradually see the mountains erode, and the sea fill up - slightly - with sediment.  Erosion stops at sea level, give or take, but sea level itself changes over time, so you might expect a world of very shallow seas where the continents used to be.
But that isn't the only thing going on.
If plate tectonics stops, we would likely move to a Mars-style regime, with hotspots being the dominant mode of heat transport.  And as with Mars, we would see huge shield volcanoes form, basically like Iceland. So imagine an ocean planet with some massive circular islands dotted about, as the old continents eroded down.
Of course, this won't actually happen; within about a billion years, the sun will get so hot that a runaway greenhouse effect will come into operation, and the oceans will evaporate, ending pretty much all life on earth, and the possibility of a global ocean. Indeed, with no mountains to draw down CO2 by erosion, a water world may be very unstable as regards runaway greenhouse effects.
Comment: Earth's distant future is probably much more like Venus than Mars.
Comment: @David Hammen Agreed, but I was looking at tectonic styles in isolation other factors, like increasing solar radiation.
Comment: The key, I think, is life.  When most of the life dies (at 150ppm CO2) the biological systems and sinks will disappear allowing positive feedback systems to enter into the picture.  Grins.  You guys do know that the atmospheres of both plants are virtually the same?  What does that tell you about little impotent CO2?  Anyone who wants a great geology lesson or two or three should look up Nick Zentner of Central Washington University.  His lectures are watched by people all around the world and right now they are all about the PNW...can't get enough!
Answer: Note that you may have a bit of a misconception: the land heights/ocean depths, even what % of Earth is ocean, aren't too important because when land erodes, it doesn't fill in much of the depths, but instead raises the ocean bottom AND ocean top by similar amounts - the amount of water on Earth still needs somewhere to go!  It still just tends to sit on top of the land.  If you took a cup half filled with water, then dumped sand into the water, it doesn't take up all the water depth, but instead raises the water up in the cup:

The related question If the Earth were a smoothed spheroid, how deep would the ocean be? discusses how deep the oceans would be if the Earth became perfectly smoothed out and there were no change in the amount of mud (the question is, would the old sea bottom, now buried beneath the new dirt, then gradually lose some of its water?  I think the answer would be yes, but it's still true that a bit of the water may be permanently tied up into additional mud).
Erosion is indeed the process that would tend to smooth out the ground level.  However, anything that causes uneven vertical movement in the ground would work to counteract that.  Plate tectonics like mountain building and sinking plates would do so.
So I think the answer is clear that in a tectonically inactive Earth... if the Earth kept enough water during the process, we would eventually have an ocean-covered planet.
But on the other hand I'd suspect for a 4.54 billion year old planet, if it weren't for tectonics, the Earth would probably already be long since smoothed out.  So it comes down to whether plate tectonics will slow down/change in the long run, which geologists can likely better discuss (and which may be a good separate question to ask).  But either way, there will be some important uncertainties [large meteor impacts, global temperatures changes, etc]
So I'd think the best answer may probably just be: not anytime soon!
Comment: My question was not so much "where would the water go?", it was more concerning the fact that the ocean *floor* is essentially a huge pit, and dry land is a huge pile, and for the pile to fill the pit would be a lower energy state and should therefore occur spontaneously-ish.  The water is merely a catalyst.
Comment: Ah, fair enough, didn't seem that way, so didn't realize.  Certainly trend is towards entropy increase... it's just there's lots of energy (from the sun, inside the Earth), so there are resulting processes that still work against that land leveling :)


Is it important to weigh down a hop bag for dry-hopping?

Question: If so, what kind of weight should I use? Should it be enough to sink the bag to the bottom of the carboy?
Answer: It helps, but is not necessary
The more the hops are exposed to wort the better the aromas and flavors will transfer to the beer. Use something you can sanitize - smooth like marbles, coins, washers or silverware. It shouldn't lay on the bottom of the vessel; instead, suspend it in the middle of the brew. Brew Strong recorded an episode all about dry hopping.  It's a good listen with some surprising information.
Answer: I've tried weighting and not weighting and found that it makes absolutely no difference.  I hate doing unnecessary work, so I no longer weight the bags.
Answer: I know you specified carboy, but for those looking to dry-hop in the keg, it's also important not to have the bag sink to the bottom and sit over the outlet. You pull out all sorts of nasty sediment and flavors that way, particularly when using pellet hops.
I've managed to solve this by throwing both a sanitized ping pong ball and a sanitized washer in the hop bag. The washer sinks, and the ball floats, suspending the bag in the top-middle of the beer.
Comment: Ohhh that's clever.
Answer: Yes, you should weigh down the bag.
Use sanitized marbles or a spare stainless steel fitting to weigh down the bag.
Here is a link to the Brew Strong episode discussing dryhopping.
Answer: I feel the best way is to not use a hop bag. Take your hops for dry hopping out of the fridge 24 hrs before you want to add them. That will allow the oils to warm up and increase the aroma potential. I boil a small amount of water then let it cool to "warm". Add the hops dry, to a sanitized jar or measuring cup. Then add just enough water to cover the hops. Let the hops get all mushy (~3-5 min), swirl and add the hop slurry to your secondary. I have had amazing success with this method, a real noticeable difference between using a hop bag and letting them float free. I find the hops settle to the bottom and generally stay there when racking off to bottles or a keg.
Answer: Be careful of metal.  I once bought some ball bearings and used them to weight the hop bag.  When I removed them, the formerly shiny silver ball bearings were black and corroded, the beer had a powerful metallic taste and was ruined.  I now use a hop bag, and weight it with some lead sinkers that are vacuum sealed in plastic.  I once dry hopped pellets without a bag, and the filtration when racking to keg was a real PIA, so I always use a bag now
Comment: upvoted this one.  I used a bunch of steel bolts and such one time, thinking they were 'stainless', but they were in fact CHROMED, which is poison.  Pulled them out though and still drank the beer :-)
Answer: I was advised I did not have to use a bag and it was OK to just add the hops to the fermenter. The last batch was a double IPA and it turned out awsome!!!
Answer: My answer here describes my technique, which does not weigh down but rather pushes down.
Answer: Thought I would add this to the thread. Amazon has these Stainless Steel Mesh containers that look perfect for this:
https://www.amazon.com/Micron-Filter-Stainless-Cornelius-Brewing/dp/B06Y2F4BMY/
Answer: When the primary fermentation is done, the wort is already saturated with CO2, either use hop bag or directly drop the pallet will result free floating, by the time, yes the pellet will sink first but that will take longer (my experience > a week, causing a bit grassy beer), while the hop bags will stay floating with full of CO2 trapped inside, weight the bag down may have some better surface contact but not much, we think maceration will happen automatically, but in CO2 saturated hop aroma does not flow out.
I bough a lager, chilled it, then pouring it into a glass, dropped a seal bag with three hops in the bag, pushed the bag down by spoon many times, tried to get the trapped CO2 out, after for a while, I finished the beer, there was a little arome. Then now on my empty glass with hop bag left, pour a water into the glass, stir it two three times and I can smell full aroma and when drink the water, it tastes great except it is grassy
Answer: In my experience, I have found if you want that true IPA aroma you get from fruity hops when dry hopping, you definitely want that bag weighted down. It is the same as making a cup of tea. If the bag is not fully submerged, you're not wetting all of the tea leaves and as such, not making the strongest tea you possibly can. Dry hopping is same thing.
My advice as others have pointed out is to drop the hops right into the fermenting vessel. They will float at first and then you will find they will all mostly sink. When you use a hop bag, I find air gets trapped and the bag floats instead of sinking like it should. You could weigh it down, but you risk contaminating your brew (which could be a problem if you bottle condition especially).
Answer: Denny doesn't know what he is talking about. YES, they should be weighed down. Some of mine were floating on the top and the aroma was not very strong.


Why can't light penetrate solid objects?

Question: Light is combination of perpendicular electric and magnetic fields, since electric fields penetrate a conductor, why can't light travel in them?
I think my argument does sound stupid, but I can't find the flaw in it.
What do you think can be the reason?
Answer: The reasoning has to be the other way around:
Light acts on the metal and makes the electrons move. This, however, results in an energy loss, as the electrons feel a resistance and thus the radiation loses energy. This can be formulated more precisely with counteracting electric fields. That's why all good conductors are opaque.
In insulators this can not happen, as the electrons are mostly fixed at their positions so the electric field in light interacts much less with insulators. That's why some insulators are see-through (as for example glass).
Comment: @ja72 Star Trek windows?
Comment: Should *isolators* be *insulators* here? (*isolator* is the Swedish for *insulator*) (I didn't want to edit your answer too much in case I made it incorrect)
Comment: I heard that if you compress aluminum (or aluminium) enough it turns transparent.
Answer: An overview in layman's terms:
First, it is important to note that not any electric field will induce current in a conductor, because other than the fact the intensity of the field defines the speed of each charge (bigger difference of potential), the oscillation frequency of the $\mathbf{E}$ also plays a very important role, if the frequency is too high, the free electrons in the conductor will not even notice the field, so you shouldn't assume that any electric field will give rise to a current.
Second point attenuation: let's discuss what causes the EM radiation to die out when it tries to penetrate matter (solids e.g.):
EM radiation interacts with matter via a number of different phenomena, in this diagram some of the are summarized:

On left we have the initial intensity $I_0$ and photon-energy $h\nu$ of the radiation assumed here to be monochromatic, and on the right side, i.e. the part that has been transmitted (or penetrated through matter if you will), where from Beer-Lambert's law we know its intensity decays exponentially with the material's absorption coefficient and thickness. Intuitively:

The thicker the material, the more disturbance felt by the radiation, because all the phenomena (scatterings, absorption, etc) shown in the diagram become more likely as each photon meets so many atoms/electrons along its path that at one point or another it will be absorbed or scattered.
Materials with high absorption coefficients have great cross-sections for all the possible radiation-matter interactions (high probabilities.)

All this said, we come back to the frequency of the radiation, e.g. Gamma radiation , as can be produced from radioactive materials (gamma decay), does not easily interact with matter because of its highly energetic photons, which is why in nuclear reactors, there are always gamma rays that escape the facilities, no matter how many layers of thick walls have been placed. This is why the study of spectrum of EM radiation is so important, because knowing the spectrum we can predict with what types of materials it will best interact. A diagram taken from wikipedia:

Long story short, there are frequencies of EM radiation that do penetrate matter, others than don't, must have fulfilled one the interaction conditions, e.g. if they match exactly the binding energy of an electron, they're absorbed and a photoelectron is created.
Answer: There is a good explanation of this in Matter and Interactions vol II by Sherwood and Chabay. I no longer have the text; I will try to summarize its explanation as I remember it.
The electrons in a substance are analogous to charged masses on springs. The electrons in insulators are relatively tightly bound; those in conductors are loosely bound or unbound. Shining light on a substance is like driving a damped harmonic oscillator at a particular (off-resonant) frequency. The charges respond by oscillating at that frequency, but their response is out of phase with the driving frequency. The field produced by the oscillating charges  destructively interferes with the original light wave.
So if you block a beam of light with a piece of cardboard, the beam is still making oscillating electric and magnetic fields behind the cardboard, but the charges in the cardboard are shaking in precisely the right way to make fields that cancel them.
Here is a link to a college lecture where this is discussed.


How to deal with this mold on the window frame?

Question: I just discovered what I assume is mold on my window frame in the kitchen/living room (quite high humidty, but the window has been open a lot and our fan is going 24/7). Now, I've been told that dealing with black mold is dangerous, my landlord however looked at it and said it can easily be dealt with with bleach. After googling this I'm still not sure if I can or can't deal with this myself, some say I should get him to get someone to do it professionally, others recommend everything from bleach to special paint. What is the right way for me to deal with this?
Comment: There are many varieties of mold. Many of them appear black, and most are harmless outside of being cosmetically undesirable. I'd start by just wiping the stuff away with detergent and water and see if it regrows,before bothering with anything more aggressive.
Answer: That looks pretty bad. Don't use bleach, use a purpose-made mold killer product. I like Mold Armor, myself. Be careful with these products. They're super caustic and you could probably hurt yourself if you drink them or don't have good ventilation.
Comment: Aside from that stuff being rather aggressive, any other health concerns that I should be aware of? With regards to the actual mold, not just the cleaning product itself.
Comment: Why not use bleach?
Comment: Most household bleach products aren't very powerful and all the plain water in them can seep into wood, nourishing hidden mold under the surface.
Answer: I think that the landlord is responsible, but sometimes it's necessary to do things yourself... that depends on your living/renting situation.
Household (5%) bleach (dilute 1/2 cup bleach : gallon water) will absolutely work. The black mold is probably Aspergillus niger, which often produces something called aflatoxins, which can cause cancer. But the danger involved with cleaning it off of a few windows is overrated compared with chronic exposure from living with it, eating it, or cleaning it up on a daily basis (ie., people who clean or remove mold for a living). But if you are concerned, use a dust mask.
Mildew resistant paint would be a good idea, after you clean it. But more importantly, the reason for the mildew is moisture; maybe it's from humidity/condensation, but usually the problem is that the window is getting rained on, and water is collecting on the sill. Preventing mold should start with weather protection. If high humidity, is the issue then a dehumidifier should be used.
Comment: I'm mostly at concern for my health here, I don't mind cleaning it up if I'm not at risk for something. The internet seems to be full of FUD in that regard. And yeah, definitely moisture, it's an old single sided window. Although the humidity is at around 45-50% which seems to be what the UK government recommends. But yeah, since it's in the kitchen there have undoubtedly been peaks in humidity from that alone.
Comment: @justsid I see. Well, a dehumidifier should be helpful. Humidity from cooking should be vented with a hood. Often, clothes dryers are blamed for humidity, so be sure to check the vent hose for lint.
Answer: I had mold in the basement bathroom. We installed a window which can be opened, a small humidifier, and cleaned mold with bleach. Advice for bleach, I emptied window cleaning spray container and sprayed bleach rather than using a cloth or something else. Works like magic. No more mold.
Comment: In the spray bleach, did you use 100% bleach. Or did you mix with water? How many parts bleach, water? Thanks.
Comment: Used 100% bleach. No water added.


What does "securely realize" mean?

Question: I was wondering what "securely realizes" means. I see this in some cryptographic papers but I don't know what it means for a protocol to "securely realize" a function $F$.  Is it just a fancy way of saying "uses" or does it mean something else?
Comment: Generally, if the author of the paper defines the meaning of that phrase or provides a footnote with a reference, the phrase is intended to have specific technical meaning, otherwise probably not.
Comment: If F is an instance of some standard notion (e.g., a hash function) associated with a standard security definition (e.g., collision resistance), then "securely realizing F" might mean that F satisfies some security notion (e.g., it is a hash function satisfying collision resistance).

(I don't see how to "securely realize a function F," since the underlying standard notion isn't a function itself, but that might just be a wording issue in the original question.)
Answer: A standard approach in cryptography is to separate out the ideal specification (of what a scheme is supposed to achieve) from the particular instantiation (the implementation of the scheme).
To specify the security and functionality goals, sometimes cryptographers specify an "ideal functionality", which is an idealization of what we are hoping to achieve.  Then, one designs a specific instantiation -- a particular scheme -- and proves that the scheme meets the goals set out in the "ideal functionality".  We describe this by saying that the scheme securely realizes the ideal functionality.  How to prove that is another can of worms entirely... but that's what is meant by the terminology.
This kind of approach and terminology is common in, for instance, theoretical work that adopts an "ideal world/real world" formalization (including, e.g., simulation-based security; indifferentiability; universal composability; and other lines of work in cryptography).  See Wikipedia on universal composability for a high-level introduction.

What does "securely realize" mean?


Overlap with Personal Finance & Money

Question: I've seen lot of comments and close votes on question regarding anything financial as off-topic here and belonging on Personal Finance & Money. Since lot of such a question will inevitably appear, we should discuss which of questions regarding expat finances ought to be allowed and which disallowed here.
Answer: We never meant to over-train people into thinking everything should be neatly bucketed on each individual site when we introduced migrations. Overlap is good. Not sending a user that arrived to your site and actually typed something good to another site unless absolutely necessary - even better.
Migrations are for the experience of the author, don't use them for filing.
Here's a simple test and workflow.

If you remove the expat part, is it essentially the same question? Is the user not facing additional complications or twists due to being an expatriate?

Yes! -> Put it on hold with guidance, or edit it if you can.
No! -> Answer it.

I'm specifically not going to migrate any questions to any other site until after the public beta. Remember that you need to be identifying the kinds of topics in which this community can excel - and I would expect that a community full of expatriates would be happy to answer personal finance questions that I have which suffer from the complications of being an expat.
That's the gist of it, if you think about it. Stuff is hard. When you're an expat, that's amplified. If the root of the questions is about that odd twist that living abroad tends to bring to problems, answer it.
Answer: OK, here is a proposition: Let's accept those that are a legitimate questions by an expat without splitting hairs about how much applies to locals or not. Overlap is not a big problem as such.
(You can vote this up and down to express agreement or disagreement.)
Answer: Very good question. On one side some of the financial issues expat have is quite unique to their situation. A simple local answer can be painfully wrong at times.
Also it is my experience that Personal Finance & Money has very big bias towards the United States or the english speaking world in general. To best of my knowledge, a credit card rating is something that is not common world wide.
Yet I would expect quite some knowledgeable people on money.se. So even with this English bias on money.se, I would vote for migrating money question to money.se as much as possible unless it really really is obvious that it only affects expats. I would not ask the question on credit card ratings on money.se, because I expect it to be closed immediately as a stupid question. From an anglo saxon perspective everybody knows what a credit card rating is.
Also migrating these question to money.se, might dampen the anglo saxon bias on money.se and it attracts the knowlegable people.
An nice example of a good effect is a question on travel.se that got migrated to money.se. The question is spot on on travel.se, yet quite some valuable info came from money.se
Comment: I think it's called a "credit report", "credit rating" or "credit score" and goes beyond credit cards but I share your concern about the North American/Anglo bias on Personal Finance & Money. At the same time, I don't think there is any hostility towards international question there and they do make some effort to provide tags and the like. Maybe this site only needs more good quality questions and answers from other parts of the world.
Comment: Incidentally, it's true that private consumer credit reports do not exist everywhere but there is [in fact one in the Netherlands](https://nl.wikipedia.org/wiki/Bureau_Krediet_Registratie) or in Germany for example.
Comment: The Dutch example is similar to being registered at the "Banque de France" in France in the sense that it keeps records on people having problems in repayments and only in extreme conditions.
Comment: @GaelLaurans I don't feel any hostility either on money.se, only that at times it is difficult to ask a question due to a difference in context. I fully agree with you that it needs more international content, hence me being in favour for migrating questions from this site to money.se as much as possible
Comment: That's also more-or-less what a credit report is, the main difference is that it can be accessed by a broader set of people and includes more data. I think the Dutch thing is closer to a credit report than to the Banque de France files but it's another question.

Overlap with Personal Finance & Money


Stress Tensor OPE in CFT

Question: In David Tong's CFT notes, he argues that  the OPE for the holomorphic stress tensor with itself must take the form $$T(z)T(w)=\frac{c/2}{(z-w)^4}+2\frac{T(w)}{(z-w)^2}+\frac{\partial T(w)}{z-w}+finite$$ I'm a bit confused about this point. He argues that in a unitary CFT, no operators can have negative scaling dimensions, suppressing the higher order poles allowed by symmetry from appearing in the expansion. From this OPE, he derives the Virasoro algebra. Then later, when examining consequences of unitarity, he uses the Virasoro algebra to prove that no operators can have negative scaling dimension. Isn't this argument a little circular then? Is there a way to argue for the general form of the $TT$ OPE that doesn't invoke such circular reasoning?
Comment: By expanding the stress tensor in a Laurent series with the Virasoro generators as the coefficients, one may use the relationship between commutators and contour integrals of radially-ordered products to deduce the commutators of the Virasoro generators from the OPE. Are you suggesting that alternatively, one may use the Virasoro algebra to define the CFT and run it in reverse to obtain the OPE?
Comment: I'm not sure I understand what you mean by Tong "deriving the Virasoro algebra" from the OPE. The Virasoro algebra is just the (centrally extended) algebra of 2d conformal symmetry, it does not need to be "derived", it is by definition the symmetry algebra of a CFT. What do you mean by that?
Answer: What Tong is presumably doing is not "deriving the Virasoro algebra", but showing that the modes $T_n$ of the energy-momentum tensor as defined by $T(z) = T_n z^{-n-2}$ actually fulfill the commutation relations of the Virasoro algebra, that is, the non-trivial statement here is that given the OPE of $T(z)$ with itself, one can show its modes are the Virasoro generators (see also this question). Conversely, given $T(z) = L_n z^{-n-2}$ where the $L_n$ are known to be the Virasoro generators, one may derive the OPE of $T$ with itself.
However, the Virasoro algebra is present in a conformal field theory simply because it is a conformal field theory. You don't need to know anything about a stress-energy tensor to know that a quantum conformal field theory will carry a representation of the Virasoro algebra because the Virasoro algebra is the (centrally extended) algebra of infinitesimal conformal transformations. So the reasoning is not circular.
Comment: You are confused: the generators of the conformal algebra are obtained using Noether theorem. This is the relation between the Virasoro generators and the stress-tensor. It is not a coincidence, and they are not set equal because they "look alike." They are set equal because they are equals. You can, if you wish, define the conformal transformations by just looking at $\delta x^\mu$, but that is not the end of the story. The relation between $L_m$ and the conserved charges related to $T$ is a deep one, not a coincidence. You are missing the role of the Ward identities in a quantum theory
Comment: No. You are skipping the whole construction using conserved charges. That is classical mechanics.
Comment: @CGH No, the generators of the conformal algebra are obtained by looking at infinitesimal conformal transformations/conformal Killing vectors. Saying that a theory is a CFT means saying that algebra is a symmetry of the theory, and therefore the state of space carries a representation of it.  I'm not claiming it's a "coincidence" that that rep are precisely the modes of the energy-momentum tensor, but that you can know about the algebra *without* the e-m tensor, making the usage of the action of that algebra to derive something about the e-m tensor non-circular, which was OP's concern.
Answer: Let's start with the Ward Identities ((4.11) in Tong's lecture). In radial quantization for a 2D theory they read
$\delta_{\epsilon,\bar\epsilon}\Phi(w,\bar w)= \frac{1}{2\pi {\rm i}}\left( \oint_\gamma dz R\left[ T(z)\epsilon(z)\Phi(w,\bar w) \right] - \oint_\gamma d\bar z R\left[ \bar T(\bar z)\bar \epsilon(\bar z)\Phi(w,\bar w) \right] \right)$.
We know how a primary field transform under the a conformal transformation:
$\delta_{\epsilon}\Phi(w)= h (\partial\epsilon) \Phi(w) + \epsilon\partial\Phi(w) $.
(I have taken only the holomorphic part, a similar formula goes for the right moving sector. From now on, I omit the anti-holomorphic part.)
This, using Cauchy, we see that for a general conformal transformation, i.e. $\epsilon=z^n$, the OPE between $T$ and $\Phi$ has to be
$T(z)\Phi(w)= \frac{h}{(z-w)^2}\Phi(w) + \frac{1}{z-w}\partial \Phi(w)+\cdots$,
where it is understood that we are working in radial quantization, so we omit the $R$ in front of the OPE. Now, what happens if $\Phi$ is a quasi-primary? Quasi-primaries transform under the global part of the conformal group, $SO(3,1)$. This is the part that is preserved in $D>2$. The global part of the conformal group is given by $\epsilon=z^n$, $n=0,1,2$. In this case, what we obtain is
$T(z)\Phi(w)= \sum_{p\ge4} \frac{a_p(w)}{(z-w)^p}+\frac{h}{(z-w)^2}\Phi(w) + \frac{1}{z-w}\partial \Phi(w)+\cdots.$
What Tong is telling you now, is that $T$ is actually a quasi primary (actually, it is a descendant of the identity!)  When he allows a central charge extension, he is stating that, at the quantum level, you find an anomaly determined by $c\neq0.$ classically, $T$ is a primary, but not quantically. This has nothing to do with unitarity. This is just the transformation of fields. Note that $h$ hasn't been set positive.
Now, just as you defined a symmetry transformation using Noether's currents, you can define symmetry generators for the conformal algebra (Virasoro generators) using radial quantization with $T$ being the conserved current. From there, you can compute the Virasoro algebra and you find its central-extension.
Using representation theory you can see that $h\ge0$ (or from the 2 point function and cluster decomposition) by looking at the first descendant of $\Phi$, $L_{-1}\Phi$. By looking at $L_{-2}\Phi$ you find $c\ge0$.
As you see, this is a linear argument. Not a cyclic one!
Comment: @ZHC Maybe I don't understand correctly your question. First, the general form for a quasi-primary is given as the fourth equation in my answer, namely, it has an additional $\sum _{p>4} c_p (z-w)^{-p}$ terms. How to compute those terms? If you can write the theory as a Lagrangian, you can use Wick contractions, if not, you can express the term as the right descendant, for example $T = L_{-2}I$ and compute the OPE.
Comment: Right, but to prove that $h$ is nonnegative, you need to use the fact that the modes of $T$ satisfy the Virasoro algebra. So what the argument looks like to me is "If the modes satisfy the Virasoro algebra, then there are no negative weight operators in a unitary theory. If there are no negative weight operators, then the $TT$ OPE takes the appropriate form. If the OPE is of this form, then the modes satisfy the Virasoro algebra". What exactly is being argued?
Comment: I understand that as a conformally invariant quantum theory, to be as general as possible we need to centrally extend the algebra. What I'm not getting is if what we're doing here is a proof or a definition, essentially. Are we postulating that the Noether charges of $T$ generate the symmetry, just as in ordinary quantum theory, and then showing that in order for this to be true, the $TT$ OPE must look a certain way? To postulate the form of the OPE first seems a bit out of place, as the rest of the calculation seems like a long consistency check on the Noether charges generating the symmetry.
Comment: 1.- There is no "if the modes satisfy the Virasoro algebra." $T$ is the Noether conserved current. The currents define generators, the $L_m$. The generators form an algebra, the Virasoro one. 2.- The Ward Identity is valid for any symmetry in a QFT. As I showed the $TT$ OPE is the most general allowed for a QFT with conformal symmetry. 3.- There is no use of unitarity here!! 4.- We do not postulate that the charges of $T$ generate the conformal symmetry. This is how you construct the algebra!
Comment: @CGH Hello, can you explain more about how you derive OPE of quasi-primary fields? To my understanding, what you calculate is based on how quasi-primary fields transforms under a general  (instead of global) conformal transformation. Since quasi-primary fields do not obey the same transformation law as the second equation in your answer, there should be some additional terms in the variation of fields, but I don't know what these terms are and how these terms give poles of $z^{-4}$ and more.

Stress Tensor OPE in CFT