Can't find the Material Output in Cycles!
Why is the node editor showing different kinds of nodes?
Is it possible to edit the Audit fields on UPDATE? Consider this question on .  I put out a 200 bounty on the question, but did not get an answer from anyone other than myself that worked.    In fact, my answer is also the highest up-voted on it at this point.  I know that allowing OP's to retract a bounty can be a slippery slope; in fact I think that once the bounty is put up it should stay up.  But why not allow the OP to answer their own questions with a bounty and reward it to them if they receive the same up-votes as others?  Yes, I am fully aware that bounties are a risk.  People will use them for numerous reasons such as trying to draw extra attention to their issues.  However, in my specific question, the only valid answer is the one that I gave.  Had it been anyone else's question, I would have received the bounty, but since it was my own I did not. Call blocking for iPhone? How do I block unwanted callers without jailbreaking? To run programs as sudo without password, I edited /etc/sudoers with wrong content.   Then I re-edited it with Mac's default text editor(course sudo cannot be used when /etc/sudoers is wrong).  I tried to run sudo agin, and get this error:   sudo: /etc/sudoers is owned by uid 501, should be 0 sudo: no valid sudoers sources found, quitting   What should I do? I cannot remember the root password (maybe I haven't ever set root password before). Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles. Activating 'missing' standard GRASS utilities in QGIS Processing toolbox $\int f=0$ implies $f=0$ almost everywhere Let $f$ be nonnegative and measurable. Show that $\int f =0$ implies $f=0$ almost everywhere (except possibly on a set of measure zero).  My thoughts. Take a set E with $m(E) \neq 0$. Suppose $f \neq 0$ on E. That means $\int_E f \neq 0$. Since $f$ is nonnegative this implies $\int f &gt;0$. That contradicts the given information, so $f$ must be 0 almost everywhere. Is there a technical reason to use > (<) instead of != when incrementing by 1 in a 'for' loop? Add layer to a QGIS group using Python Can anyone please tell me whether the following function $d:\mathbb{R}^2 \times \mathbb{R}^2 \to  \mathbb{R}$ is really a metric or not on $\mathbb{R}^2$.  $$ d((x_1,x_2),(y_1,y_2))= \begin{cases} |x_1-y_1|, &amp;\text{ if } x_2=y_2\\ |x_1|+|y_1|+|x_2-y_2|, &amp;\text{ if } x_2 \neq y_2. \end{cases} $$  If it really is indeed a metric, please give the proof of triangle inequality.(I'm OK with the other two conditions). I often hear people saying that "I am" is the shortest sentence in the English language.  I know that there are also discussions about sentences using the imperative mood such as "Go." that would be shorter, but my question is this:  Why would we (the people saying "I am" is a full sentence) not accept "I'm" as a complete sentence?  Is there some unwritten rule about contractions that says "I'm" wouldn't be correct? Invalid field count when uploading CSV to a data extension I tried uploading non-English text to my Data Extensions but the system is throwing error named &quot;Invalid Field Count&quot;. I need a regular expression to select all the text between two outer brackets.  Example: some text(text here(possible text)text(possible text(more text)))end text  Result: (text here(possible text)text(possible text(more text))) Badge Notification Repeats Is there a way to find the number of files of a specific type without having to loop through all results inn a Directory.GetFiles() or similar method? I am looking for something like this:  int ComponentCount = MagicFindFileCount(@"c:\windows\system32", "*.dll");   I know that I can make a recursive function to call Directory.GetFiles , but it would be much cleaner if I could do this without all the iterating.  EDIT: If it is not possible to do this without recursing and iterating yourself, what would be the best way to do it? This is my first post on this forum :)  I've been trying for a while to solve this problem about symmetric difference of two sets:     Given two sets $A$ and $B$, we call symmetric difference of $A$ and $B$ the following:      $$A\mathrel{\triangle}B = (A\cup B)-(A\cap B)$$      Show that if $A\mathrel{\triangle} B = \emptyset$, then $(A\subset B) \vee (B\subset A)$.   I don’t know where to start; thank you very much for the help :) First order differential equations with 2 varriables Does anybody know how to solve the following differntial equation?  $$w^2\cdot r+\frac{dw}{dt}\cdot r+\frac{dr}{dt}\cdot w=3g$$  where w and r are a function of time so $w(t)$ and $r(t)$. $g$ is a constant  EDIT: This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.  Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.   The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration: $a_r + a_s = constant$  From circular motion we know that $a_r = \omega ^2 r$ where $r$ is the radius, and $\omega$    Further we know that $v=\omega*r$, therefore: $a_s=\frac{d}{dt}(\omega r)$  For constant acceleration $a_{total} = a_r + a_s = \omega ^2 r + \frac{d}{dt}(\omega r) =constant$  Further usefull information: Length of the track covered at a certain time $l=v*t=\omega r t$  With all this the radius will be a function of time, $r(t)$ and the angular velocity will be a function of time, $\omega (t)$  Boundary conditions: $r(0)  = 0$ $\omega (0) = 0$  Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$  Here is a diagram for clarity: How to really clear the terminal? I can issue the clear command or press Ctrl + L to clear the current Ubuntu terminal, but this just shifts previous output upwards and if you use mouse scroll or PgUP and PgDown keys it's hard to distinguish where the output of previous command ends and output of current command begins.  Is there a way to really clear the terminal so I won't see previous command results? Nautilus context menu for current directory when there's lots of files / "Open in Terminal" This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways.  Proof 1 : Euclid's. If there are finitely many primes then $p_1 p_2 ... p_n + 1$ is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes.  Proof 2 : Consider the sequence $a_n = 2^{2^n} + 1$. We have that  $$ 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), $$  so that for $m &lt; n$, $(2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1$. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them.  Proof 3 : (Note : I particularly like this one.) Define a topology on $\mathbb Z$ in the following way : a set $\mathscr N$ of integers is said to be open if for every $n \in \mathscr N$ there is an arithmetic progression $\mathscr A$ such that $n \in \mathscr A \subseteq \mathscr N$. This can easily be proven to define a topology on $\mathbb Z$. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set $$ \mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z $$ should be open and closed, but by the fundamental theorem of arithmetic, its complement in $\mathbb Z$ is the set $\{ -1, 1 \}$, which is not open, thus giving a contradiction.   Proof 4 : Let $a,b$ be coprime integers and $c &gt; 0$. There exists $x$ such that $(a+bx, c) = 1$. To see this, choose $x$ such that $a+bx \not\equiv 0 \, \mathrm{mod}$ $p_i$ for all primes $p_i$ dividing $c$. If $a \equiv 0 \, \mathrm{mod}$ $p_i$, since $a$ and $b$ are coprime, $b$ has an inverse mod $p_i$, call it $\overline{b}$. Choosing $x \equiv \overline{b} \, \mathrm{mod}$ $p_i$, you are done. If $a \not\equiv 0 \, \mathrm{mod}$ $p_i$, then choosing $x \equiv 0 \, \mathrm{mod}$ $p_i$ works fine. Find $x$ using the Chinese Remainder Theorem.   Now assuming there are finitely many primes, let $c$ be the product of all of them. Our construction generates an integer coprime to $c$, giving a contradiction to the fundamental theorem of arithmetic.  Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...)  Do you have any other nice proofs?